Arithmetic 1 (Mid Size) (1 48)

Arithmetics : Solved Examples ] [ DA-1

1. NUMBER SYSTEMEx. 1. Is the number 347 prime ?Sol. Here nearest square root is 19 andprime numbers less than it are 2, 3, 5, 7,11, 13, 17. Since 347 is not divisible by anyone of these so it is a prime number.Ex. 2. Is 191 a prime number ?Sol. Since 14 is nearest square rootand prime numbers less than 14 are 2, 3,5, 7, 11, 13. Now 191 is not divisible byany one of them so it is a prime number.Ex. 3. What is the sum of first 10natural numbers ?Sol. We know that sum of first n

natural numbers = [n(n +1)]

2. The first 10

natural number are 1 to 10. Þ Sum is

=[10 (10 + 1)]

2=

(10 11)2

×

= 55

Ex. 4. Find the sum of squares of first10 natural numbers.Sol. Square of first 10 natural numbersare 1, 4, 9, 16, 25, 36, 49, 64, 81, 100.Sum of squares of first n natural numbers

= [n(n +1)(2n +1)]

63

Þ S =[10 (10 + 1)(2 10 +1)]

6×

when n = 10 Þ S =(10 11 21)

6

× × = 385.

Ex. 5. Simplify : : : : : 405 × 405Sol. 405 × 405 = (400 + 5) (400 + 5)= (400 + 5)2 = (400)2 + (5)2 + 2 × (400) × 5= 160000 + 25 + 4000 = 164025Ex. 6. Simplify : 98 × 98Sol. 98 × 98 = (100 - 2) × (100 - 2)= (100 - 2)2 = (100)2 + (2)2 - 2 × (100) × 2= 10,000 + 4 - 400 = 9604Ex. 7. Simplify : 102 × 98Sol. 102 × 98 = (100 + 2) × (100 - 2)= (100)2 - (2)2 = 10000 - 4 = 9996

||SGN||

Arithmetics : Solved ExamplesEx. 8. Simplify : 7 × [{6 (5 + 7)} ----- 2]Sol. 7 × [ { 6 (5 + 7) } - 2 ]= 7 × [(6 × 12) - 2] = 7 × [72 - 2]= 7 × 70 = 490.Ex. 9. What is the least number thatmust be added to 486 to make thisnumber exactly divisible by 7 ?Sol. We know that Dividend = (Divisor× Quotient) + Remainder. So if we divide486 by 7 we get 3 as a remainder. (486 = 69× 7 + 3) Þ The number to be added = 7 - 3= 4.Ex. 10. Find the least number of 3digits which is exactly divisible by 17.Sol. We know that the least number of3 digits is 100. If we divide 100 by 17 we get15 as a remainder. i.e. 100 = 5 × 17 + 15Þ To make it exactly divisible by 17 weshould add (17 - 15) = 2 to 100. Þ Requiredleast number is = 100 + 2 = 102Ex. 11. Find nearest number to 182which is exactly divisible by 6.Sol. If we divide 182 by 6 than we get 2as a remainder.We see that 2 < 6 Þ nearestnumber to 182 which is exactly divisible by6 will be = (182 - 2) = 180.Ex. 12. Find the value of 986 × 93 +986 × 7Sol. 986 × 93 + 986 × 7= 986 × (93 + 7) = 986 × 100 = 98600Ex. 13. Find the total of even numbersfrom 1 to 100.Sol. We know that from 1 to 100, totalnumbers will be 100.Þ There will be 50 even numbers.So sum of them is 50 × (50 + 1) = 2550[because sum of first n even numbers= n (n + 1)]Ex. 14. Find the total of odd numbersfrom 1 to 100.Sol. We know that sum of first n oddnumbers = n2. Now we have 50 odd numbersbetween 1 to 100. Þ Sum is (50)2 i.e. 2500


Ex. 7. Divide (i) 0.0081 by 9(ii) 1.48 by 0.74

Sol. (i) .0081

90081

90000= = 0.0009

(ii) 1.480.74

148074

=14874

= = 2

Ex. 8 Divide 81 by 0.09

Sol.81

0.09810009

=8100

9= = 900

Ex. 9. Find whether the followingnumbers are recurring decimals arenot :

(i) 311

(ii) 100400

(iii) 46

(iv) 47

Sol. (i) If we divide 3 by 11 we get 3

11= 0.272727....... therefore it is a recurring

number and 3

11 = 0. 27

(ii) If we divide 100 by 400 then we get 100400

= 0.25 . So it is not recurring.

(iii) 46

= 0.6666..... = 0. 6 . Therefore 46

is a recurring decimal.

(iv) 47

= 0.571428571428 .....

= 0 .571428So it is a recurring decimal.Ex. 10. Express the following intovulgar fractions :

(i) 0. 6 (ii) 0. 27

Sol. (i) 0. 6 = 0.6666.....Suppose 0.6666........ = x(where x shows a vulgar fraction)Þ 10x = 6.6666 ....Þ 10x – x = (6.6666...) - (0.6666..)

Þ 9x = 6 Þ x = 69

= 23

(ii) 0. 27 = 0.27272727..........

Suppose 0.272727....... = x

Ex. 15. Find how many numbers up to100 are divisible by 7.Sol. We see that if we divide 100 by 7We get 100 = 14 × 7 + 2. The quotient is 14therefore there are 14 numbers up to 100which are divisible by 7.

2. DECIMAL FRACTIONSEx. 1. Convert the following in tovulgar fractions. (i) 0.25 (ii) 0.371

Sol. (i) 0.25 = 025100

25100

14

= =

(ii) 0.371 = 371

1000Ex. 2. Convert

0.532.311

into vulgarfraction.

Sol.0.532.31

= 053231

53231

=

Ex. 3. Evaluate : 6.12 + 3.102 + 4.007Sol. 6.12

+ 3.102+ 4.007 13.229

Ex. 4. Evaluate : 7.1521 ----- 2.12Sol. 7.1521

– 2.1200 5.0321

Ex. 5. Evaluate :6.5371 + 2.1312 ----- 0.0067 ----- 0.808Sol. 6.5371

+ 2.1312 8.6683– 0.0067 8.6616– 0.8080 7.8536

Ex. 6. Evaluate :(i) 5.132 × 100(ii) 1.32 × 1.34

Sol. (i) 5.132 (ii) 1.32 × 100 × 1.34 513.2 1.7688(Sum of decimal places 2 + 2 = 4)


Subtracting 1 from 1 we get the remainder0. We bring down the next dividend as 46.Step III : The trial divisor is 2 × 2, i.e. , 4and the trial dividend is 04 the quotient is2. We put 2 for the next digit (towards theright) in the square root and also a annex 2to the trail divisor.The divisor becomes 22 and quotient 2.Multiplying 22 by 2 and subtracting from46 we get the remainder as 2.Step IV : We bring down the next period 41and we get the next dividend as 241. Thetrial divisor is 24 and the trial divend is 24and the quotient is 1. We put 1 for the nextdigit (towards the right) in the square rootand also a annex 1 to the trial divisor.The divisor becomes 241 and the quotient1. Multiplying 241 by 1 and subtracting from241 we get the remainder as 0.

121 .

1 14641

1

22 46

44

241 241

241

xxxThus square root of 14641 is 121.Ex. 5. Find the square root of 2116.Sol.

46 .

4 21 16

16 .86 5 16

5 16

xxxStep I : First of all we will pair the

digits from Right to Left 21 16 .Step II : The square root nearest to

that of the first period 21 is 4, the square of

Þ 100x = 27.272727......

Þ 100x - x = (27.272727....) - (0.272727......)

Þ 99x = 27

Þ x = 2799

311

=

3. SQUARE ROOTS &CUBE ROOTS

Ex. 1. Find the square root of 6561.Sol. = 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3= (3)8 = (32)4 = (34)2

= (3 × 3 × 3 × 3)2

= (81)2

Taking square root of both sides.

Þ 6561 = 81Ex. 2 Find the square root of 180625Sol. 5 × 5 × 5 × 5 × 17 × 17= (5 × 5)2 × (17)2

= (5 × 5 × 17)2 = (425)2

Þ 180625 = 425Ex. 3. Find the square root of 6084.

2 6084

2 3042

3 1521

3 507

13 169

13 13

16084 = 2 × 2 × 3 × 3 × 13 × 13= (2 × 3 × 13)2 = (78)2

Taking square root of both sides.

Þ 6084 = 78Ex. 4. Find the square root of 14641.Sol. Step I : Since 14641 have 5 digitsand it is a odd number so by pairing from

right to left we have 1 46 41 .Step II : The square root nearest to that ofthe first period 1 is 1, the square of 1 is 1.


4, the square of 4 is 16. Subtracting 16 from21 we get the remainder 5. We bring downthe next period and get the next dividend as516.

Step III : The trial divisor is 2 × 4,i.e. 8 and the trial dividend is 51 the quotientis 6. We put 6 for the next digit (towardsthe right) in the square root and also a annex6 to the trial divisor. The divisor becomes86 and the quotient 6. Multiplying 86 by 6and subtracting from 516, we get theremainder as zero. Hence the squre rootof 2116 is 46.Ex. 6. Find the square root of 3 uptotwo decimal places.Sol. 3 = 3.0000.Thus by pairing the digits we have

1.73

1 3.0000

1 .27 200 .

189 .

343 1100

1029

71

Þ 3 = 1.73Ex. 7. Find square root of 0.25.Sol. By putting zeros in right, we have

5 .

5 0. 25 00

25 .

xx .

0 25. = 0.5Ex. 8. Find square root of 0.0289

17 .

1 0. 02 89

1 .

27 189 .

189 .

xx .

0 0289. = 0.17

Ex. 9. Evaluate 7 × 63Sol. We know that a b ab× =\ 7 7× = ×63 63

= × × = × × ×7 7 9 7 7 3 3 = 7 × 3 = 21

Ex. 10. Evaluate 1728

27

Sol. =1728

27

We know thb

ab

at a =

L

NMM

O

QPP

=× × × × ×

× ×3 4

3

3 3 4 4

3 3= × × =4 4 44 4 = 4 × 2 = 8

Ex. 11. If 3 = 1.732 then find the

value of 24318

Sol.24318

3 3 3 3 318

= × × × ×

= × × =3 3 318

32

= =17322

. 0.866

4. POWERS OF NUMBERS

Ex.1. If 23

5FHG

IKJ = x then find the value

of x.

Sol. x = 23

5FHG

IKJ = =2

3

32243

5

5 \\\\\ x = 32243

Ex. 2. Solve : 27 × 29 × 23

Sol. 27 × 29 × 23 = (2)7 + 9 + 3 = (2)19

Ex. 3. Find the value of 3

3

7

5

Sol. (3)7/(3)5 = (3)7- 5 = (3)2 = 3 3 = 9


Ex. 4. Find the value of (43)2

Sol. (43)2 = (4)3 × 2 = (4)6

Ex. 5. If 32243

= x5 then find the value

of x.

Sol.32243

= x5 Þ 2 2 2 2 23 3 3 3 3

× × × ×× × × ×

= x5

Þ 2

3

5

5 = x5 or 23

5FHG

IKJ = x5 ÞÞÞÞÞ x =

23

Ex. 6. Evaluate

x

x

x

x

x

x

b

c

b c bc c

a

c a ca a

b

a b ab2 2 2 2 2 2

FHG

IKJ

×FHG

IKJ

×FHG

IKJ

+ + + + + +

Sol.

x x xb c b c bc c a c a ca a b a b ab2 2 2 2 2 2

− + + − + + − + +× ×e j e j e j

= × ×− − −x x xb c c a a b3 3 3 3 3 3

= − + − + −x b c c a a b3 3 3 3 3 3 = x0 = 1

Ex. 7. Simplify 2 34 7

2 73 2

3 5

2 3

4 2

4 2××

FHG

IKJ

× ××

FHG

IKJ

Sol. = ××

FHG

IKJ

× ××

FHG

IKJ

2 3

4 7

2 7

3 2

3 5

2 3

4 2

4 2

= × ×× × ×

+ 22 3 7

4 7 2 3

3 4 5

2 3 2 4

= ××

− −

−2 3

4 7

7 2 5 4

2 3 2

( ) ( )

( )= ×

×2 3

4 7

5

2

= × × × × ×× ×

=2 2 2 2 2 34 4 7

67

Ex. 8. 2 22 7 3 5e j e jRST

UVW÷

Sol.2

2

2 7

3 5

e j

e j = =

×

×2

2

2

2

2

3

7

5

14

15

= =−1 1

215 142

Ex. 9. Evaluate a a

a a

m n m n

n m

+ −××

Sol. a a

a a

m n m n

n m

+ −××

=+ + −

+a

a

m n m n

n m

=+

a

a

m

n m

2

= − −a m m n2 = −am n

5. HCF & LCMEx. 1. Find the HCF of 56 and 80.Sol. 56 = 2 × 2 × 2 × 780 = 2 × 2 × 2 × 2 × 5HCF = 2 × 2 × 2 = 8Ex. 2. Find the HCF of 34, 170, 85.Sol. 34 = 2 × 17 170 = 2 × 5 × 17 85 = 5 × 17 HCF = 17Ex. 3. Find HCF of 616 and 100.Sol. 110) 616 (5 550 66) 110 (1 66 44) 66 (1

44 22) 44 (2 44

xxÞÞÞÞÞ HCF = 22.Ex. 4. Find HCF of 1584, 1728, 1152.Sol.

1584) 1728 (1

1584

144) 1584 (11

144

144

144

xxx

Now 144 is HCF of 1728, 1584Hence if we divide third number by 144 wehave


ÞÞÞÞÞ HCF = 2 × 9 × 247 = 4446Ex. 8. Find HCF of 0.35, 2.45 and 14.7Sol. We will find HCF of 35, 245, 1470which is 5.Þ HCF of given numbers is = 0.05Ex. 9. Find LCM of 10, 12, 15, 18, 20.Sol.

2 10 12 15 18 20

2 5 6 15 9 10

3 5 3 15 9 5

5 5 1 5 3 5

1 1 1 3 1Þ Required LCM = 2 × 2 × 3 × 5 × 3= 180Ex. 10. Find LCM of 36, 60, 180, 90.

Sol. 36 = 2 × 2 × 3 × 3 = 22 × 32

60 = 2 × 2 × 3 × 5 = 22 × 3 × 5

180 = 2 × 2 × 3 × 3 × 5 = 22 × 32 × 5

90 = 2 × 3 × 3 × 5 = 2 × 32 × 5

Highest Powers of factors 22, 32, 5

Þ LCM = 22 × 32 × 5

= 2 × 2 × 3 × 3 × 5 = 180Ex. 11. Find LCM of 24, 63, 70.Sol.

2 24 63 70

2 12 63 35

2 6 63 35

3 3 63 35

3 1 21 35

7 1 7 35

5 1 1 5

1 1 1Þ LCM = 2 × 2 × 2 × 3 × 3 × 7 × 5= 2520

LCM of Fractions :

LCM = LCM of numerators

HCF of denominators

144) 1152 (8 1152 xxÞÞÞÞÞ 144 is HCF of 1584, 1728, 1152.Ex. 5. Find HCF of 52, 390, 650.Sol. 390) 650 (1 390 260) 390 (1 260 130) 260 (2 260 xxNow HCF of 390, 650 is 130. 52) 130 (2 104 26) 52 (2 52 xxSince 26 is divisor of 130. ÞÞÞÞÞ HCF = 26.Ex. 6. Find HCF of 1560, 1755 and1950.Sol. First we have to find HCF of anytwo numbers let us take 1560, 1950. 1560) 1950 (1 1560 390) 1560 (5 1560 xxxx\ HCF of first two numbers = 390.Now take 390, 1755 we have 390) 1755 (4

1560 195) 390 (2 390

ÞÞÞÞÞ HCF = 195. xxEx. 7. Find HCF of 84474 and 75582.Sol. 84474 = 2 × 9 × 4693 and 75582= 2 × 9 × 4199

4199) 4693 (1 4199 494) 4199 (8 3952 247) 494 (2

494 xx


=

10615265

= 10615

526

× = 5339

Ex.3. 2.1 ÷ 0.7 of 6 + 0.6 × 0.3 ----- 0.2Sol. 2.1 ÷ 4.2 + 0.6 × 0.3 - 0.2

= 214 2

.

. + 0.18 – 0.2 =

12

- 0.02

= 0.50 - 0.02 = 0.48Ex. 4. Simplify35 ----- [20 ----- {15 × 3 (8 + 6)} + 4]Sol. 35 - [20 - {45 × 14} + 4]= 35 - [20 - 630 + 4] = 35 - [- 606] = 641Ex. 5. Simplify

5 ----- [23

+ {32

----- (0.5 + 13

14

− )}]

Sol. 5 - [23

+ {32

- (0.5 + 4 312−

)}]

= 5 - [23

+ {32

- (12

- (12

+ 1

12)}]

= 5 - [23

+ {32

- 7

12}] = 5 - [

23

+ 18 7

12−

]

= 5 - (23

+ 1112

) = − LNM

OQP

= − =5 51912

4112

8 +1112

Ex. 6. Find the value of 15 27

9

2 3

2×

Sol.15 27

9

2 3

2×

= × × × ××

15 15 27 27 279 9

= 15 × 15 × 27 × 3 × 3= (15)2 × (3)2 × 27

Ex. 7. What is the value of 14

th of 15

?

Sol. 14

th of 15

= 15

× 14

= 120

Ex. 8. In a competition 14

th boys

out of the total boys took part and 17

th girls out of the total girls took part.How many participants were there inthe competition ?

Ex. 12. Find LCM of 3

50,

620

, 140

Sol.

LCM = LCM of 3, 6, 1

HCF of 50, 20, 40=

610

=35

Ex. 13. Prove that product of twonumbers 24, 36 is equal to product oftheir LCM and HCFSol. First to find HCF we have24 = 2 × 2 × 2 × 3,36 = 2 × 2 × 3 × 3Þ HCF = 2 × 2 × 3 = 12

2 24 36

2 12 18

2 6 9

3 3 9

3 1 3

1 1Þ LCM = 2 × 2 × 2 × 3 × 3 = 72Þ HCF × LCM = 12 × 72 = 864and 24 × 36 = 864. Hence Proved.

6. SIMPLIFICATION

Ex. 1. 108 ÷ 48 of 14

+ 25

× 334

Sol. Given 108 ÷ 48 of 14

+ 25

× 334

= 108 ÷ 48 of 14

+ 25

× 154

= 108 ÷ 12 + 32

= 10812

+ 32

= 91

+ 32

= 212

= 1012

Ex. 2.

13

5 7

13

5 7

÷ +

× +

Sol. =× +

× +

13

15

71

13

51

71

=

11553

7

+

+

7


Sol. Given 14

th boys and 17

th girls

took part. i.e. 1 boy out of 4 boys took part,1 girl out of 7 girls took part.Þ 2 persons out of 11 took part.

Þ 2

11 th of total took part.

Ex. 9. Find the value of 7

11

232

++

Sol.7

11

232

++

=+

7

11

7 / 2

=+

7

127

= 79 7/

= ×7 71 9

= 499

7. AVERAGEEx. 1. Find average of first 50 naturalnumbers.

Sol. Average = Sum of 1 to 50 natural numbers50

= 1 + 2 + 3..........+50

50 =× FHG

IKJ

LNM

OQP

50512

50=

512

= 25.50Ex. 2. Find average of first ten primenumbers.Sol. First ten prime numbers are 2, 3, 5,7, 11, 13, 17, 19, 23, 29

Þ Average = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29

10

= 12910

= 12.9

Ex. 3. Average age of a class of 24students is 15 years. If age of teacheris included then average increases by1 year. What is the age of the teacher?Sol. Total age of 24 students = 24 × 15= 360. Given, average age of 25 personsincluding teacher = 16Þ Age of 16 persons = 25 × 16 = 400Þ Age of teacher = 400 - 360 = 40 years.Ex. 4. If distance between two

stations A and B is 358 km. A persontravels from A to B by speed 50 km/hrand returns with y speed of 44 km/hrThen what is average speed duringwhole journey ?Sol. If speed from A to B = x km/hr andspeed from B to A = y km/hr. Then average

= 2xyx y+ Þ According to question

Average = 2 50 44

50 44× ×

+ km/hr

= 44094

km/hr = 22047

km/hr

Ex. 5. Average of 13 numbers is 50. Ifthe average of first seven numbers is48 and average of last seven is 53 thenfind seventh number.Sol. Total of first 7 numbers = 7 × 48 =336. Total of last 7 numbers = 7 × 53 =371. Total of 13 numbers = 13 × 53 = 650Þ 7th number = 336 + 371 - 650= 707 - 650 = 57

8. PROBLEMS BASEDON AGES

Ex. 1. The age of father 5 years agowas 4 times that of the age of his son.At present father's age is three timesthat of his son. What is present age offather and son ?Sol. Suppose present age of son = xÞ 5 years ago son's age = (x - 5)Þ Father's present age = 3xÞ 5 years ago father's age = 3x - 5Þ (3x - 5) = 4 (x - 5) Þ 3x - 5 = 4x - 20Þ x = 15.Father's present age = 3 15× = 45 years.Ex. 2. The total of Anil's age and hismother's age is 34. Two years ago hismother's age was 5 times of Anil's age.What is the present age of Anil and hismother ?Sol. Suppose Anil's age two years ago = x2 years ago Anil's mother age = 5x


Loss % =20

100 × 100 = 20%

Ex. 3. The selling price of 20 books isequal to cost price of 15 books. Findloss percent.Sol. Suppose C.P. Rs. x per book Þ Total C.P. = 20 x, Total S.P. = 15 xLoss = 20x - 15x = 5x

Loss % =520

xx

× 100 = 25%

Ex. 4. A person sold a radio in Rs. 500and got 25% profit. What is the costprice of radio ?Sol. Suppose cost price = Rs. 100, thenselling price = 125. In other words, whenselling price = Rs. 125 then cost price = Rs.

100. Þ S.P. = 500 then C.P. =100125

× 500

Þ C.P. = Rs. 400Ex. 5. A retailer sells an article at aloss of 12.5%. Had he sold the articleof Rs. 51.80 more, he would haveearned a profit of 6%. Find the C.P. ofthe article.Sol. Let us assume that C.P. = Rs. 100.Lossis 12.5% i.e. Rs. 12.50. Hence S.P. = Rs.(100 – 12.50) = Rs. 87.50. To earn a profit of6%, he should sell the article for Rs. 106,i.e. for Rs. 18.50 more. In other words, ifthe difference in the two selling price is Rs.18.50, the C.P. is Rs. 100. \ for a difference

of Rs. 51.80. C.P. = 518018 50

100..

× = Rs. 280.

Ex. 6. A dealer sold a machine to ashopkeeper at 20% profit. Shopkeepersold it to a customer so as to get 25%profit for himself. The differencebetween the selling price of the dealerand that of the shopkeeper was foundto be Rs. 129. Find the initial price ofthe machine.Sol. Let us assume that he initial price ofthe machine = Rs. 100\ S.P. of the dealer = Rs. 120 so that C.P.of shopkeeper = Rs. 120

Present age of Anil = x + 2Present age of mother = 5x + 2Þ (5x + 2) + (x + 2) = 34Þ 6x = 30 Þ x = 5Þ Anils present age = 7 years,Mother's Age = 27 years.Ex. 3. Ratio of present ages ofManish and Rajesh is 4 : 5. After 5years it will be 5 : 6. What are presentages of Manish and Ravi.Sol. Suppose ages of Manish and Rajeshare 4x, 5x. After 5 years they will be

4x + 5, 5x + 5. Þ 4 55 5

xx

++

= 56

Þ 24x + 30 = 25x + 25 Þ x = 25Þ Present ages of Manish and Rajesh are20 years and 25 years.

9. PERCENTAGEEx. 1. Find 9% of Rs. 625.

Sol. 9% of Rs. 625 = 9

100 × 625 =

2254

Ex. 2. Show 58

in percentage form.

Sol.58

× 100 = 5008

= 125

2 = 62.5%

Ex. 3. How will you express 55% infraction ?

Sol. 55% = 55 × 1

100 =

1120

Ex. 4. Show 12% in decimals.

Sol. 12% = 12100

= 0.12

10. PROFIT & LOSSEx. 1. A man purchased a bicycle atRs. 600 and sold at Rs. 650, then what% did he gain or loose ?Sol. Cost price = 600. Selling price = 650Profit = 650 - 600 = 50

Gain Percentage =50600

× 100 = 813

%Ex. 2. A book is purchased for Rs. 100and sold for Rs. 80. Find loss percent.Sol. Loss = C.P. - S.P. = 100 - 80 = 20


Þ b : c = 343

: 543

× × Þ b : c = 4 : 203

Þ a : b : c = 3 : 4 : 203

Ex. 3. If a : b = 3 : 5 and sum of a andb is 56 then find a and b.Sol. Suppose a = 3x and b = 5xÞ 3x + 5y = 56 Þ x = 7 Þ a = 21, b = 35Ex. 4. If 3 : 4 :: 6 : x then find value of x.

Sol. Since 3 : 4 :: 6 : x Þ 34

6=x

Þ 3x = 24

Þ x = 8Ex. 5. Divide Rs. 1200 in the ratio 3 : 4: 5.Sol. Suppose the shares are 3x, 4x, 5x.Þ 3x + 4x + 5x = 1200 Þ x = 100Þ Shares are Rs. 300, Rs. 400, Rs. 500.Ex. 6. Find duplicate ratio of 2 : 3.Sol. We know that duplicate ratio of a : b isa2 : b2 Þ Duplicate ratio of 2 : 3 is 22 : 32

i.e. 4 : 9

12. TIME & WORKEx. 1. If A can do a work in 12 daysand B can do same work in 8 days, thenin how many days they can finish thework, working together ?Sol. A can do work in 12 days Þ A's one

day work = 1

12. B can do work in 8 days

Þ B's one day work =18

Þ (A + B)'s one day work =1

12 +

18

= 524

Þ A and B together can finish the

work in245

days.

Ex. 2. If A is twice as good as B and ifB can do a work in 6 days then in howmany days A and B will together finishthe work ?Sol. B can do work in 6 days Þ A can dowork in 3 days Þ A can do work in 3 days Þ (A + B)'s one day work

\ S.P. of shopkeeper = 125 120

100150

× = Rs. .

\ Difference between the selling prices = Rs.150 – Rs. 120 = Rs. 30.... If the difference is Rs. 30, the initial price= Rs. 100.\ If the difference is Rs. 129, the initial price

is = × =Rs Rs. . .129 100

30430 Hence, the

initial price of machine is = Rs. 430.Ex. 7. If the price of sugar falls by12.5%, a person can buy 9 kg more ofsugar for Rs. 126 than before. If theprice had fisen by 12.5%, how muchsugar could he have bought for thesum ?Sol. Let us assume that the price of sugaris Re. 1 per kg. Since the price has fallen by12.5% = (25/2)%. Sugar price becomes

Rs.175

2 100× per kg. i.e. Rs.

78

per kg.

Amount of sugar that can be purchased for

Rs. 126 is = 126 × 87

= 144 kg. i.e. Originally

he could have purchased (144 – 9) kg onlyi.e., = 135 kg. Again sugar price has

increased by 12.5% = (25/2)% i.e. sugar price

becomes Rs.225

2 100× =

98

per kg. \ Amount

of sugar that can be purchased for Rs. 126

= 126 × 89

kg = 112 kg.i.e. He can purchase (135 - 112) kg= 23 kg less than the original amount.

11. RATIO & PROPORTIONEx. 1. If a : b = 5 : 7 and b : c = 7 : 9then find a : c.Sol. a : b = 5 : 7 and b : c = 7 : 9Þ a : b : c = 5 : 7 : 9 Þ a : c = 5 : 9Ex. 2. If a : b = 3 : 4 and b : c = 3 : 5then find a : c.Sol. Since a : b = 3 : 4 and b : c = 3 : 5


13. DISCOUNTEx. 1. Find present worth of Rs. 1050due after 1 year at 5% per annum.

Sol. We know P WS DR T

. .. .

( )= ×

+ ×100

100

⇒ P W. .( )

= ×+ ×

100 1050100 5 1

= Rs. 1000

Ex. 2. Find discount on S.D. (SumDue) Rs. 2100 due after 1 year at 5%per annum.

Sol. P WS DR T

. .. .

( )= ×

+ ×100

100

P W. .( )

= ×+ ×

=100 2100100 5 1

210000105 = 2000

Þ Discount = P.W. - T.D.= 2100 - 2000 = Rs. 100Ex. 3. The present worth of a bill dueat definite time is Rs. 2200 and thetrue discount on the bill is Rs. 220.Find banker's discount and banker'sgain on it.

Sol. We know T D P W G. . . . . .= × B

Þ 220 2200= × B G. . Þ (220)2 = 2200 ×

B.G. Þ B.G. =220 220

2200×

= 22

Þ Banker's Discount = 220 + 22 = 242Ex. 4. The banker's discount on Rs.1100 is Rs. 100 due at a definite time.Find true discount and banker's gain.Sol. We know

B DAmount Rate Time

. .= × ×100

and

AmountD D

B D D= ×

−Β. . . .

. . . . T T

or AmountD D

B G= ×Β. . . .

. . T

i.e. T DB G

AmountB D

. .

. . . .= Þ

T DB G

. .

. .= =1100

100111

Þ If B.G. is Rs. 1 than T.D. = Rs. 11Þ B.D. is 12 then T.D. is = 11

=16

+13

=36

= 12

Þ A and B together will finish thework in 2 days.Ex. 3. 2 men and 3 women can do apiece of work in 10 days while 3 menand 2 women can do same work in 8days. In how many days can 2 men and1 woman complete the work ?Sol. Suppose 1 man's 1 day's work = mand 1 woman's 1 day's work = n

Þ 2m + 3n =1

10 and 3m + 2n =

18

Þ m =7

200 and n =

1100

Þ (2m + 1w)'s 1 day work =2 7200

× +

1100

=

16200

= 225

Þ 2 men and 1 woman can

complete the work in252

days.

Ex. 4. A and B together complete awork in 14 days. If A is twice as goodas B then in how many days A and Bcan separately finish the work ?Sol. Suppose B finishes the work in 2m days and A finishes the work in m days

Þ (A + B) finish the work in =23m

days.

Þ23m

= 14, m = 21 Þ A can finish in 21

days and B can finish in 42 days.Ex. 5. If A can do a work in 4 dayswhile B can do same work in 3 days,then in how many days A and B willtogether complete the work ?Sol. A can do a work in 4 days. Þ A's one

day work = 14

. B can do a work in 3 days

Þ B's one day work = 13

Þ (A + B)'s one

day work =13

+14

= 7

12Þ A + B together can finish the work

in 127

days.


Þ B.D. is 100 then T.D. is = 1112

× 100 = 91.

66 Þ B.G. = (100 - 91.66) = 8.34

14. TIME AND DISTANCEEx. 1. If a boy runs at a speed of 10meters/second then what is thedistance travelled by him in 13seconds?Sol. Speed = 10 meter/second, Time = 13seconds Þ Distance = speed × time= 10 × 13 = 130 metersEx. 2. A train covers 40 kms distancein an hour then how much distance is

travelled by it in12

hour ?

Sol. Speed = 40 km/hr, Time =12

hr

Þ Distance = 40 × 12

= 20 kms.

Ex. 3. Find time in which 100 kmsdistance can be travelled at the speedof 25 km/hr by a bicycle.Sol. Speed = 25 km/hr,Distance = 100 kms

Þ Time = distancespeed

=10025

= 4 hours

Ex. 4. Rakesh travels for 5 hours. Ifhe moves first half of his journey at aspeed of 60 km/hr and rest at a speedof 50 km/hr then find the total distancetravelled by Rakesh.Sol. Suppose total distance is x then, time

(for first half) = distancespeed

=x / 260

.....(i)

Similarly time (for second half)

= dis cespeed

tan =x / 250

.....(ii)

By (i) and (ii) ⇒ x x/ /260

250

5+ =

⇒x x

120 1005+ = ⇒

5 6600

5x x+ =

⇒ x =300011

= 2728

11 kms

15. PROBLEMS ON TRAINSEx. 1. Two trains of lengths 150 mand 100 m respactively are movingtowards each other on parallel tracksat a speed of 50 km/hr and 40 km/hrrespectively. What will be the time toclear each other from the momentthey meet ?Sol. Lengths of trains are 150 m, 100 m.Þ Total distance to be covered = 150 + 100= 250 mNow relative speed = 50 + 40 = 90 km/hr

= 90 × 5

18meter/second

= 25 meter/second

Þ Time =25025

= 10 secondsEx. 2. A train of 90 meters longcrosses a pole in 6 seconds. What isthe speed of train ?

Sol. Time = distance covered (length)speed

=906

= 15 meter/second

= 15 × 185

km/hr = 54 km/hr

Ex. 3. A train of 150 meters long isrunning with a speed of 54 km/hr. Inhow much time it will cross theplatform of length 255 meters ?Sol. Total distance to be covered= 150 + 255 = 405 m

Þ Time taken =×

405

545

18

= 40515

= 27 secondsEx. 4. Find time taken by a 120meters long train running with aspeed 36 km/hr to cross a bridge of 280meters long .Sol. Total distance to be covered= 120 + 280 = 400 meters.Speed of train = 36 km/hr

= 36 × 5

18 meter/second = 10 meter/second


time =40010

= 40 seconds

Ex. 5. Two trains of lengths 100meters and 150 meters respectively arerunning in same direction at a speedof 40 km/hr and 22 km/hr respectively.In what time will they clear eachother from the moment they meet ?Sol. The total distance to be covered= 100 + 150 = 250 m.Now the relative speed = (40 - 22)

= 18 km/hr Þ 18 ×5

18 = 5 meter/second

Þ time =2505

= 50 secondsEx. 6. Find the time taken to crass apole by a train of length 165 meterslong with speed 54 km/hr.

Sol. Speed = 54 km/hr = 54 ×5

18 meter/

second = 15 meter/second

Þ Time =dis ce

speedtan

= 16515

= 11 seconds

16. PARTNERSHIPEx. 1. Ashok and Sunil invest Rs.1,00,000 and Rs. 1,50,000 to start abusiness. If annual profit is Rs. 25,000then find the share of each of themSol. This is a simple partnership and ratioof their investments are 1,00,000 : 1,50,000i.e. 2 : 3 Þ Ashok's share in profit =

25000 × 25

= Rs. 10,000 and Sunil share in

profit = 25000 × 35

= Rs. 15,000Ex. 2. If profit of Rs. 90,000 is dividedin to three partners A, B and C in theratio of their investments in 2 : 3 : 4then find the share of each partnerSol. The ratio of investments is 2 : 3 : 4.Total = 2 + 3 + 4 = 9

Þ Profit of A = 90,000 × 29

= Rs. 20,000,

Profit of B = 90,000 × 39

= Rs. 30,000 and

Profit of C = 90,000 × 49

= Rs. 40,000

Ex. 3. Rajesh and Suresh start abusiness with their investments of Rs.30,000 and Rs. 40,000 respectively. After6 month Ashok joined them his theirinvestment of Rs. 40,000. If annualprofit is Rs. 9,000 then find the shareof profit for each.Sol. Ratio of their investments= (30,000 × 12), (40,000 × 12), (40,000 × 6)i.e. 3,60,000 : 4,80,000 : 2,40,000i.e. 3 : 4 : 2 and total is 3 + 4 + 2 = 9therefore share of each of them :

Rajesh's share = 90,000 × 39

= Rs. 30,000,

Suresh's share = 90,000 × 49

= Rs. 40,000

and Ashok's share = 90,000 × 29

= Rs. 20,000Ex. 4. Two partners A and B starteda business with their investments ofRs. 50,000 and Rs. 40,000. After 6month B leaves the business. Theannual profit of Rs. 7000 is to bedistributed between them. Find theshare of A.Sol. Ratio of their investments is(50,000 × 12) : (40,000 × 6)i.e. 6,00,000 : 2,40,000 i.e. 5 : 2

Þ Profit of A = 7,000 × 57

= Rs. 5000

Ex. 5. A started a business with Rs.1,00,000 and B joined him after 4months with capital of Rs. 50,000 C. Ifannual profit is Rs. 16,000 then findshare of each of them.Sol. Ration of their investments= (1,00,000 × 12) : (50,000 × 8)i.e. 12,00,000 : 4,00,000 i.e. 3 : 1.


Total = 3 + 1 = 4Now profit is Rs. 16,000

Þ Share of A = 16,000 × 34

= Rs. 12,000

Share of B = 16,000 × 14

= Rs. 4,000

Ex. 6. Nilesh started a business withRs. 30,000 and Ramesh joined himafterwards with Rs. 40,000. After oneyear if shares of profit for both of themis equal then for how many monthsdid Ramesh invest his money?Sol. Suppose Ramesh invests for xmonths Þ 30,000 × 12 = 40,000 × x.

Þ x = 36000040000

= 9

i.e. Ramesh invested for 9 months.

17. PIPES & CISTERNS.Ex. 1. If a tank can be filled by twopipes A and B in 6 hours and 8 hoursrespectively, then if both the pipes areopened simultaneously then tank willbe filled in how many hours ?Sol. Pipe A can be fill the tank in 6hours and B can be fill in 8 hours.Þ Part of tank filled in 1 hour

= 16

18

4 324

+ = + =

724

Þ tank can be filled in 247

hoursEx. 2. The pipe A can fill a tank in 4hours and another pipe B can be emptyit in 8 hours. Both the pipes are openedsimultaneously then how much timewill be taken to fill the tank?Sol. Pipe A can fill the tank in 4 hourswhile pipe B can empty it in 8 hours. Þpart of tank filleded in 1 hour =

14

18

2 18

− = − =

18

Þ tank can be fill in 8 hours.Ex. 3. A cistern is connected withthree pipes A, B and C. Pipe A and B

can fill the tank in 4 hours and 6 hoursrespectively. Pipe C can be empty itin 8 hours. If all the three pipes areopened at a time then in how muchtime tank will be full ?Sol. Pipes A and B fill it in 4 and 6 hoursrespectively, where C empties it in 8 hours.Þ Part of cistern filled in 1 hour

= 14

16

18

+ − = + − =6 4 324

724

Þ Tank will be full in 247

hours.

18. SIMPLE INTERESTEx. 1. Find simple interest on Rs.5000 at a rate of 5% p.a. for 6 months.

Sol. P = 5000, r = 5, T = 12

.

S IP r T

. .= × × = × ××100

5000 5 1100 2

= Rs. 125

Ex. 2. Find amount for a principalamount of Rs. 2000 given for two yearsat the rate 6% p.a.Sol. P = 200, r = 6, T = 2

S IP R T

. .= × × = × ×100

2000 6 2100

= Rs. 240

Amount = Principal + S.I.= 2000 + 240 = 2240Ex. 3. In what time Rs. 3000 will bedouble at the rate of 10% p.a. ?Sol. P = 3000, r = 10, T = ?, S.I. = 3000

Now TS 100.

P r3000 1003000 10

= ××

= ××

. .I

Þ T = 10 yearsEx. 4. If amount of a principal Rs.3000 is Rs. 3500 in 10 years then whatis the rate of interest ?Sol. S.I. = Amount – Principal= 3500 - 3000 = 500

Now rS 100.

P T500 1003000 10

= ××

= ××

. .I = 53

ÞÞÞÞÞ r = 53

% p.a.


Ex. 5. A sum at S.I. at 5% amounts toRs. 5500 in 2 years. What is the sum?Sol. Suppose sum is = x

Þ S IP R T x x

. .= × × = × × =100

5 2100 10

Þ xx+ =

105500 i.e.

1110

5500x = Þ Þ Þ Þ Þ x = 5000

Ex. 6. If simple interest of Rs. 3000in 2 years is Rs. 600 then what is therate of interest per annum ?Sol. S.I. = 600, P = 300, T = 2, r = ?

Þ rS 100.

P T600 1003000

= ××

= ××

. .I2

ÞÞÞÞÞ r = 10%

19. COMPOUND INTERESTEx. 1. Find compound interest on Rs.8000 at the rate of 8% per annum for 2years compounded half yearly.

Sol. Amount = P 1 +R

200

2TFHG

IKJ

(T = 2T as compounding is done half yearly)

Þ Amount = 8000 1 +8

200

4

× FHG

IKJ

= × FHG

IKJ8000 1 +

125

4

= × FHG

IKJ8000

2625

4

= × × × ×80002625

2625

2625

2625

= 9359

ÞÞÞÞÞ Interest = Rs. 1359Ex. 2. Find compound interest on Rs.400 at the rate 10% per annum for 3years compound yearly.


100

TFHG

IKJ

= 4000 1 +10100

3

× FHG

IKJ

= × × ×40001110

1110

1110

= 5324

Þ Interest = Rs. 1324.Ex. 3. Find compound interest of Rs.10,000 at 10% per annum for 1 yearcompounded quarterly


100

TFHG

IKJ

(Now T = 4T as compounding is done onquarterly basis).

= 10,000 1 +10 / 4100

4

× FHG

IKJ = × F

HGIKJ10 000, 1 +

140

4

= × × × ×10 000,4140

4140

4140

4140

= Rs. 11038 (nearly)Þ Þ Þ Þ Þ Interest = Rs. 1038Ex. 4. If compound interest on Rs.1000 at 10% p.a. is Rs. 331 compoundedyearly then what is the time period ?


100

TFHG

IKJ

Þ 1331 = 1000 1 +10100

T

× FHG

IKJ

Þ 13311000

= FHG

IKJ

1110

T

Þ 1110

1110

3 TFHG

IKJ = F

HGIKJ

ÞÞÞÞÞ T = 3 years.

20. ALLIGATION ORMIXTURE

Ex. 1. A seller mixes 8 kg wheat of Rs. 5per kg with 4 kg wheat of Rs. 4 per kg.What is cost price of mixture per kg ?Sol. Total quantity of mixture = 8 + 4 = 2 kg.Now total cost = (8 × 5) + (4 × 4) = Rs. 56

Þ Cost of mixture per kg = 5612

= 143

= Rs. 4. 66 per kgEx. 2. 4 kg rice of Rs. 10 per kg ismixed with 8 kg of Rs. 7 per kg. Whatis cost pricex of the mixture per kg.Sol. Total quantity of mixture = 4 + 8 = 12kg. Total cost = (4 × 10) + (8 × 7)

= Rs. 96 Þ Cost of mixture per kg =

= Rs. 8 per kg.Ex. 3. In what proportion rice of Rs.12 per kg be mixed with rice of Rs. 18


per kg, so that cost price of mixture isRs. 15 per kg ?Sol.C.P. (per kg) of C.P. (per kg) ofcheaper rice dearer rice Rs. 12 Rs. 18

Mean Price (m)Rs. 15

(d - m) (m - c)(18 - 15) = 3 (25 - 12) = 3Þ BY FORMULA –Quantity of cheaper rice d m

m c Quantity of dearer rice= −

−= 3

3Þ Ratio is 1 : 1Ex. 4. In what proportion mustwheat of Rs. 6 per kg be mixed withwheat of Rs. 10 per kg, so that worthof mixture is Rs. 7 per kg ?Sol.C.P. (per kg) of C.P. (per kg) ofcheaper wheat dearer wheat Rs. 6 Rs. 10

Mean Price .(m) Rs. 7 .

(d - m) (m - c) (10 - 7) = 3 (7 - 6) = 1

Quantity of cheaper wheat d mm cQuantity of dearer wheat

= −−

= 31

Þ Ratio is 3 : 1Ex. 5. In what ratio the Rs. 8 per kgrice be mixed with 12 kg rice of Rs. 16per kg so that there is a gain of 20%by selling mixture at Rs. 12 per kg ?Sol. C.P. of 1 kg of mixture

= 12 × 10020

= Rs. 10

C.P. (per kg) of C.P. (per kg) ofcheaper rice Rs. 8 dearer rice Rs. 16

Mean Price (m)Rs. 10

(d - m) = 6 (m - c) = 2

Quantity of cheaper rice d mm c Quantity of dearer rice

= −−

= =62

31

Þ Quantity of rice of Rs. 8 per kg to bemixed = 3 times of dearer rice = 36 kg.

21. VOLUME AND SURFACEEx. 1. If length, breadh and height ofa cuboid are 3 cm, 4 cm and 5 cmrespectively then find its surface area.Sol. Surface area = 2 (ab + bc + ca)= 2 (3 × 4 + 4 × 5 + 5 × 3)= 2 × 47 = 94 sq. cm.Ex. 2. If each side of a cube is 6 cmthen find its surface area and volume.Sol. Surface area = 6a2 = 6 × (6)2

= 216 sq. cm.Volume = a3 = (6)3 = 216 cub. cm.Ex. 3. If radius of base of a cone is 3cm and its height is 3 cm then whatare its curved surface area, totalsurface area and volume ?Sol. Suppose slant hight = l

Þ l = 3 42 2+ = 5.

Curved surface area = prl = p × 3 × 5= 15 p sq. cm.Total surface area = p r (r + l)= p × 3 × (3 + 5) = 24 p sq. cm and

Volume = 13

p r2 h = 13

× p × 3 × 3 × 4

= 12 p cubic cm.Ex. 4. If radis of sphere is 4 cm, thenfind its surface area and volume.Sol. Surface area = 4 p r2 = 4 × p × (4)2

= 64 p sq. cm and Volume = 43

pr3

= 43

× p × 4 × 4 × 4 = 1056

3 p cubic cm

Ex. 5. If radius of base of a cylinderis 3 cm and its height is 5 cm, then findit volumeSol. Volume = pr2h = p × 3 × 3 × 5= 45 p cubic cm.


22. RACESEx. 1. In a race of 3 kms A beats B by100 m or 20 seconds. What is A's timeover the course ?Sol. It is clear that B clears 100 m inrest 20 seconds., Þ B cover 100 m in 20 sec

Þ B cover 3 km in 20

100× 3000 sec

= 600 secÞ A covers race course in (600 - 20) seci.e. 580 sec.i.e. 9 min 40 sec.Ex. 2. A runs 1.5 times as fast as Band gives B a start of 150 m. How longshould the race course be so that A andB might reach in the same time ?

Sol.Speed of ASpeed of B

= 32

Þ In a race of 3 m A will be 1 m ahead of B.Þ 1m ahead in race of 3 m.Þ A will be ahead by 150 m in a race of150 × 3 = 450 m.Þ So the winning post is 450 m awayfrom starting point.Ex. 3. A can run 4 kms race in 6 min40 sec, while B can run in 8 min 20 sec.Then by how much distance can A beatB ?Sol. Difference of time taken= 8 min 20 sec - 6 min 40 sec. = 100 sec.Since B takes 8 min 20 sec for 4 kms i.e. 500 sec for = 400 meters

Þ 100 sec for = 4000500

× 100

= 800 metersEx. 4. In a race of 100 meters raceRavi beats Pankaj by 20 meters or 5seconds. What is Ravi's time over thecourse ?Sol. It is clear that Pankaj covers 20meters in 5 seconds.Þ Pankaj covers 100 m in 25 sec

Þ Ravi covers 100 m in (25 - 5)= 20 seconds.Ex. 5. In a race of 2 kms A can give Ba start of 35 meters and he can give astart of 65 m to C. What start B cangive to C ?Sol. When A covers 2000 m, B covers1960 m and C covers 1920 m.

i.e. When B covers 1960 m, Ccovers 1920 m.Þ When B covers 2000 m,

C covers 1920 2000

1960×

= 1959 (approx.)

i.e. 41 meters (approx.)

23. LOGARITHMSEx. 1. Find the log of 64 at base 4.Sol. log4 64 = x (suppose) Þ 4x = 64Þ Þ Þ Þ Þ x = 3Ex. 2. Find value of log10 0.01.Sol. Suppose log10 0.01 = x

i.e. 10x = 0.01 =1

102 = 10–2

ÞÞÞÞÞ x = – 2Ex. 3. Find value of log5 (25).Sol. Suppose log5 25 = xÞ log5 (5)2 = x Þ 2log5 5 = x

[... logx x = 1]

Þ x = – 2Ex. 4. Find value of log3 3 .Sol. Suppose log3 3 = xÞ log3 (3)1/3 = xÞ 1/3 (log3 3) = x

Þ x = 13

[... loga (xm) = m loga x and loga a = 1]

Ex. 5. Find value of log10 100 = ?Sol. Suppose log10 100 = x


Þ log10 (102)1/2 = xÞ log1010 = xÞ Þ Þ Þ Þ x = 1 [... loga a = 1]Ex. 6. If value of log10 2 = 0.3010 andlog10 3 = 0.4717 then find the value oflog 45.Sol. log1045 = log10 (5 × 32)

= log10 102

+ 2 log10 3

= log10 10 – log10 2 + 2 × log10 3= 1 – 0.3010 + 2 × (0.4717)= 1.6424

24. SERIESEx. 1. Find 6th term of series3, 6, 10, 15, 21, ..........Sol. 3 6 10 15 21

+ 3 + 4 + 5 + 6Þ Next term is = 21 + 7 = 28Ex. 2 31, 28, 23, 16, ?, ----- 4, ----- 17. Inabove series what is the missingnumber?Sol.31 28 23 16 ? - 4 - 17

– 3 – 5 – 7 – 9 – 11 – 13Þ Missing number = 16 - 9 = 7.Ex. 3. Find 7th term is series1, 9, 25, 49, ...........Sol. The above series in(1)2, (3)2, (5)2, (7)2

Þ nth term = (2n - 1)2

Þ 7th term = (14 - 1)= (13)2 = 169Ex. 4. In the series 1, 6, 6, 5, 11, 4, 16,3, 21, 2, 26, ...... find the next terms.Sol. This is a alternating series inwhich difference of alternate terms is asunder

+ 5 + 5 + 5 + 5 + 5

1 6 6 5 11 4 16 3 21 2 26 ?- 1 - 1 - 1 - 1 - 1

Next term is 2 + (- 1) = 0

25. CLOCKEx. 1. Find the angle between minutehand and hour hand when clock shows6.30.Sol. Angle made by the hour hand in12 hours = 360o

Þ In 6 hours 30 minutes

i.e. in 132

hours = 36012

o

× 132

= 195o

Ex.2. At what time between 6 and 7o'clock, will the two hands be at rightangle ?Sol. We see that at 6 o'clock both handare at 180o angle and space between twohands is 30 minutes but when both handsare at right angle, space between them is15 minutes i.e. minute hand is ahead of (30- 15) = 15 minutes space. But minute handgains 55 minutes in 60 minutes.Þ It will gain 15 minutes space in

= 6055

× 15 = 18011

= 164

11 minutes.

i.e. Both hands will be at right angles

at 164

11 past 6 o'clock.

(Note : Both hands will be at right angleonce again when minute hand is ahead ofhour hand).Ex. 3. At what time between 4 and 5o'clock the hands of clock will coincidetogether ?Sol. At 4 o'clock hour hand is on 4 andminute hand is at 12. i.e. minute hand is20 minute spaces behind.Þ Minute hand has to cover 20 minutes.Now minute hand gains 55 minutes in 60minutes.

Þ It gains 20 minutes in 6055

× 20 minutes.

i.e. 24011

= 219

11 minutes.

i.e. Both hands will coincide 219

11minutes past 4 o'clock.


Ex. 4. What is the angle betweenminute hand and hour hand at 4.30 ?Sol. We know that angle made by hourhand in 12 hours = 360o

Þ In 4 hours 30 minutes angle traced

= 36012

o

× 92

= 135o

26. CALENDAREx. 1. What was the day of the weekon 10th January 1958 ?Sol.1900 years = 1600 years + 300 years

= 0 odd day + 1 odd day= 1

57 years = 14 leap years + 43 ordinary years

= 28 odd days + 43 odd days= 71 odd days= 7 × 10 + 1 odd day= 1 odd day

10 days of Jan. = 7 + 3= 3 odds days.

Total odd days = 1 + 1 + 3 = 5 odd days = Friday.

Ex. 2. What was the day of week on 1January 1989 ?Sol. 1989 = 1900 years + 89 years

1900 = 1600 years + 300 years= 0 + 1 = 1 odd day

88 years = 22 leap years+ 66 ordinary years

= 44 odd days + 66 odd days= 110 days = ( 7 × 15 + 5)= 5 odd days.

1 January = 1 odd dayTotal odd days = 1 + 5 + 1 = 7

= It was Sunday.

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ADDITIONALSOLVED EXAMPLES

AVERAGEEx. 1. Average of first 10 naturalnumbers is .............. .Sol. 1. First 10 natural numbers are 1,2, 3, 4, 5, 6, 7, 8, 9, 10.

Average =+ + + + + + + + +1 2 3 4 5 6 7 8 9 10

10

= =5510

5.5

Ex. 2. Average of even numbersbetween 31 to 50 is .............. .Sol. 2. Even numbers between 31 to 50are 32, 34, 36, 38, 40, 42, 44, 46, 48, 50

= 32+34+36+38+40+42+44+46+48+5010

= =41010

41

Ex. 3. Average of 12, 22, 32, 42, 52, 62

is .............. .Sol. 3. 12, 22, 32, 42, 52, 62

Average = + + + + +1 2 3 4 5 66

2 2 2 2 2 2= 91

6Ex. 4. Average of prime numberslying between 10 and 30 is .............. .Sol. 4. Prime numbers between 10 and 30are 11, 13, 17, 19, 23, 29.

Average = + + + + +11 13 17 19 23 296

= =1126

563

Ex. 5. Average of 5 numbers 11, 13,17, 19, 23 is .............. .

Sol. 5. Here we have to find average of11, 13, 17, 19, 23.

Average = + + + +11 13 17 19 235

= 835


Ex. 6. Average of odd numbers lyingbetween 20 and 30 is .............. .Sol. 6. Odd numbers lies between 20 to30 are 21, 23, 25, 27, 29.

Average = + + + +21 23 25 27 295

= =1255

25

Ex. 7. Average of 4 numbers is 15. Ifthree of them are 12, 18 and 20 thenfourth is .............. .Sol. 7. Average = 15.Total of 4 numbers = 4 × 15 = 60Þ 4th number = 60 – (12 + 18 + 20) = 10Ex. 8. Average of three numbers 16,20 and 2x is 18 then value of x is.............. .Sol. 8. Average = 18

Þ 16 20 2

318

+ + =x

Þ 36 + 2x = 54 Þ 2x = 18 Þ x = 9Ex. 9. In a week of April monthtemperatures were 39oF, 41oF, 43oF,44oF, 45oF, 39oF and 43oF then averageof temperatures is .............. .Sol. 9. Total of temperatures

= + + + + + +39 41 43 44 45 39 437

= =2947

42 Fo

Ex. 10. Average of two numbers is a,if one of them is a + b then other is.............. .Sol. 10. Average = a.Suppose other number = x

Þ x a b

a+ + =

2 Þ x + a + b = 2a

Þ x = a ----- bEx. 11. In a class of 35 studentsaverage age is 12 years. If a teacherjoins them then average age includingteacher is 13. What is the age ofteacher?Sol. 11. Average of 35 students = 12

Þ Total age of 35 students = 35 × 12 = 420years. Þ Average including teacher = 13Þ Total age including teacher's age= 36 × 13 = 468Þ Age of teacher = 468 – 420 = 48 yearsEx. 12. If average marks of a studentin 5 subject is 70. If he got 76 marks in6th subject then his average marks are.............. .Sol. 12. Average of 5 subjects= 5 × 70 = 350Total of six subjects = 350 + 76 = 426

Þ New average = 4266

= 71

Ex. 13. In a class of 25 students 6students obtained average 80 marksout of 100. 10 students got average 82,5 of them got average 84 while 4 of themgot average 70 marks. Then what is theaverage marks for all students ?Sol. 13. Total of 6 students' marks= 6 × 80 = 480Total of 10 students' marks = 10 × 82= 820Total of 5 students' marks = 5 × 84 = 420Total of 4 students' marks = 4 × 70 = 280Þ Total of 25 students' marks = 2000

Þ Average = 200025

= 80Ex. 14. Find average of 5 consecutivenumbers starting from 51.Sol. 14. Numbers are 51, 52, 53, 54, 55

Þ Average = + + + +51 52 53 54 555

= =2655

53

Ex. 15. Average of three numbers is16. If second is twice of first and thirdis thrice of first then what are thenumbers ?Sol. 15. Average = 16. Suppose firstnumber = x. Then second number = 2x,

Third number = 3x Þ 63x

= 16

Þ 2x = 16 Þ Þ Þ Þ Þ x = 8


Ex. 16. Average of three consecutivenumbers is 12 then first of them is.............. .Sol. 16. Suppose numbers are

x, x + 1, x + 2 Þ x x x+ + + + =( ) ( )1 2

312

Þ 3x + 3 = 36 Þ 3x = 33 Þ x = 11Ex. 17. In a group of 6 men averageage is 60 years. What is the total oftheir ages ?Sol. 17. Average = 60 years.Total = 60 × 6 = 360Ex. 18. A man spends average Rs. 200per day then how many rupees doeshe spend in a week ?Sol. 18. Average = Rs. 200.Total expendi–ture in a week = 7 × 200 = 1400Ex. 19. The average of weights of 7boys is 30 kg while average of 6 boys is31 kg then weight of seventh boy is.............. .Sol. 19. Total weight of 7 boys= 7 × 30 = 210 Kg.Total weight of 6 boys = 6 × 31 = 186 kg.Þ Weight of 7th boy = 210 – 186 = 24 kg.Ex. 20. Average of six numbers is 1.5while average of five of them is 1.6 thenwhat is the value of sixth number ?Sol. 20. Total of sixth number= 6 × 1.5 = 9Total of five numbers = 5 × 1.6 = 8Þ Sixth number = 9 – 8 = 1Ex. 21. The average weight of 4 menincreased by 2 kg when one of themweighing 95 kg was replaced byanother man. What is the weight ofnew man ?Sol. 21. Since average is increased by 2 kg.Þ Total weight increased = 2 × 4 = 8 kg.Þ Weight of New person = 95 + 8 =103 kg.Ex. 22. The average weight of 5 mendecreased by 2 kg when one of themweighing 80 kg was replaced byanother man. What is the weight ofnew man ?

Sol. 22. Since average weight decreases by2 kg.Þ Total weight decrease by 5 × 2 = 10 kg.Þ Weight of new person = 80 - 10 = 70 kg.Ex. 23. A man gets Rs. 500 for eachmonth for January to April, Rs. 600for May to August and gets Rs. 700 forrest 4 months. What is the averageamount per month he got ?Sol. 23. Total amount for January to April= 4 x 500 = Rs. 2000Total amount for May to August= 4 × 600 = Rs. 2400Total amount for Sept. to Dec.= 4 × 700 = Rs. 2800He got total Amount for 12 months= 2000 + 2400 + 2800 = 7200

Þ Average per month = 720012

= Rs. 600Ex. 24. A boy gets marks in Hindi,English and Maths 75, 76, 78respectively. If he gets 71 marks inscience then what are his averagemarks scored ?Sol. 24. Total marks in 4 subjects (75, 76,78 and 71) = 75 + 76 + 78 + 71

= 300 Þ Average marks = 3004

= 75

Ex. 25. In a school, class 9th has foursections. If number of students insection A, B and C are 32, 34 and 35respectively and average of A, B, C, Dis 34 then how many students arethere in section D ?Sol. 25. Suppose there are x student in

section Þ 32 34 35

434

+ + + =x

Þ 101 + x = 136Þ x = 136 – 101 = 35Ex. 26. A service man gets anincrement of Rs. 200 per year. He gotRs. 4000 in January 2000 then what ishis average salary per month forperiod January 2000 to December2003?


Sol. 26. Person got 4000 in January 4000Þ Annual income for year 2000= 12 × 4000 = 48000He will receive 4200 in January 2001Þ Annual income for year 2001= 12 × 4200 = 50400He will receive 4400 in January 2002Þ Annual income for year 2002= 12 × 4400 = 52800He will receive 4600 in January 2003Þ Annual income for year 2003= 12 × 4600 = 55200Þ His income for 12 × 4 = 48 months= 48000 + 50400 + 52800 + 55200

= 206400 . Average = 206440048

= 4300 per month.Ex. 27. Anil got dividends Rs. 500, Rs.600, Rs. 800 and Rs. 900 on his sharesfor consecutive four years. Then howmany average rupees per year he got?Sol. 27. Total dividend in 4 years= 500 + 600 + 800 + 900 = 2800

Þ Average = 28004

= 700 per yearEx. 28. 6 players of a cricket teamplayed in a match, two of them got 52runs each. 3 of them scored 40 runseach, while one of them scored 118runs. Then what is average score ?Sol. 28. Total players = 6Total runs = 52 + 52 + 40 + 40 + 40 + 118

= 342 Þ Average = 3426

= 57Ex. 29. Sunil, Mohan and Anil getsome money. Mohan gets twice ofSunil while Anil gets thrice of Mohan.The average amount was 60 perperson. Then how many rupees doesMohan get ?Sol. 29. Suppose Sunil has = Rs. x.Then Mohan have = Rs. 2x andAnil has = Rs. 6x

Þ Average = + +x x x2 63

= 60 (given)

Þ 9x = 180 Þ x = 20

Þ Mohan gets = 2x = 40Ex. 30. Temperature in a weak rises1o per day. If temperature on Mondaywas 35o then what is averagetemperature of week ?Sol. 30. Temperature of Monday = 35o

Þ Temperatures on rest 6 consecutive dayswill be 36o, 37o, 38o, 39o, 40o, 41o

Þ Average =+ + + + + +35 36 37 38 39 40 41

7

= 2667

= 38o

PROBLEMS BASEDON AGES

Ex. 1. The age of Rakesh's father is4 times of Rakesh's age. 5 years ago itwas 7 times that of son's age. What ispresent age of Rakesh ?Sol. 1. Suppose present ages of Rakeshand his father are x and 4x.5 years ago their ages were (x - 5) and(4x - 5) Þ 4x - 5 = 7 (x - 5)Þ 4x - 5 = 7x - 35 Þ 3x = 30 Þ x = 10Ex. 2. The age of Reena's mother isthree times that of Reena's age. After10 years it will be twice of Reena'sage. What is her mother's present age?Sol. 2. Suppose Present ages of Reena andher mother are x and 3x.After 10 years their ages : (x + 10)and (3x + 10) Þ 3x + 10 = 2 (x + 10)Þ 3x + 10 = 2x + 20Þ 3x = 30 Þ x = 10

Ex. 3. The age of a father is 52

of hisdaughter. 5 years ago it was 3 timesthat of his daughter. What isdaughter's present age ?Sol. 3. Suppose present ages of daughter

and her father are x and 52

x.

5 years ago 52

x - 5 = 3 (x - 5)

Þ 5x - 10 = 6x - 30 Þ x = 20


Ex. 4. Ratio of Ravi's and Ashok'spresent ages is 4 : 5. After 6 years itwill be 5 : 6. What is Ravi's presentage ?Sol. 4. Suppose ages are 4x and 5x.

After 6 years 4 65 6

56

xx

++

= Þ 24x + 36

= 25x + 30 Þ x = 6 Þ Ravi's present age

= 4x = 4 × 6 = 24 yearsEx. 5. Ratio of ages of two friends is2 : 3. Three years ago it was 3 : 5. Whatare their present ages ?Sol. 5. Suppose ages of friends are 2x and

3x. Three years ago 2 33 3

35

xx

−−

=

Þ 10x – 15 = 9x - 9 Þ x = 6Þ Present ages are 12 years, 18 years.Ex. 6. The age of a mother is fourtimes of her daughter. Five year agothe mother was 9 times that of herdaughter's age. What was daughter'sage five years ago?Sol. 6. Suppose present age of daughterand her mother are x and 4x.5 year ago (4x - 5) = 9 (x - 5)Þ 4x - 5 = 9x - 45 Þ 5x = 40 Þ x = 8Þ 5 years ago daughter's age= x - 5 = 8 - 5 = 3 years.Ex. 7. Sum of ages of father and hisson is 60 years while difference is 30years. What is father's age ?Sol. 7. Suppose ages of son and father arex and yÞ x + y = 60 x – y = 30Þ 2x = 90 Þ x = 45, y = 15Ex. 8. The ratio of ages of Anil andRakesh is 3 : 4. After 5 years it will be4 : 5. What is Rakesh's present age ?Sol. 8. Suppose ages of Anil and Rakeshare 3x, 4x .

After 5 years 3 54 5

45

xx

++

=

Þ 15x + 25 = 16x + 20 Þ x = 5

Þ Rakesh's present ages will be 4x = 20Ex. 9. Sum of ages of two friends is32 while difference is 2 then their agesare .............. .Sol. 9. Suppose ages are x and y.Þ x + y = 32 x – y = 2Þ 2x = 34 Þ x = 17, y = 15Ex. 10. The age of Ritu's mother is 6times that of Ritu's age. After 6 yearsit will be 3 times that of Ritu's age.What is Ritu's present age ?Sol. 10. Suppose present ages of Ritu andher mother are x and 6x.After 6 years 6x + 6 = 3 (x + 6)Þ 3x = 12 Þ x = 4Ex. 11. The Ratio of ages of twofriends is 4 : 3 while sum of their agesis 28 years. The ratio of their ages after8 years will be .............. .Sol. 11. Suppose ages are 4x and 3x. Also4x + 3x = 28 Þ x = 4Ages are 16 years and 12 years.After 8 yearsThese will be 24 years and 20 yearsÞ Ratio is 6 : 5Ex. 12. 6 years ago Sohan's age wasthree times of Sunil's age . Sohan's agewill be 5/3 times that of Sunil's ageafter 6 years. What is Sunil's agetoday?Sol. 12. SupposeSunil's age (6 year ago) = xÞ 6 years ago Anil's age = 3x

Þ 53

(x + 6 + 6) = (3x + 6 + 6)

Þ 5 (x + 12) = 3 (x + 12) Þ x = 6Sunil's age today = x + 6 = 12 years.Ex. 13. Total of present ages of A, Band C is 60 years. If ratio of their agesis 1 : 2 : 3, what is the age of C atpresent?Sol. 13. Ages of A + B + C = 60.Suppose ages are x, 2x and 3xÞ x + 2x + 3x = 60 Þ x = 10Þ Age of C = 3x = 30 years


Ex. 14. The Ratio of ages of twofriends is 3 : 5 and sum of their ages is80 years. What are their present agesin years ?Sol. 14. Suppose ages are 3x and 5x.Þ 3x + 5x = 80, x = 10.Present ages 30 and 50.Ex. 15. The average age of group of 12persons is 40 years. If one more personis included then average increases by1. What is age of new person ?Sol. 15. Total age of 12 persons= 12 × 40 = 480Total age of 13 persons = 13 × 41 = 533Þ Age of new person = 533 – 480= 53 years.Ex. 16. The ratio of present ages of Pand Q is 5 : 7 respectively. If thedifference between Q's present age andP's age after 6 years is 2, what is thetotal of present ages of P and Q?Sol. 16. Let the present ages of P and Q be5K and 7K.Þ 7K – (5K + 6) = 2 Þ 2K = 8 Þ K = 4Þ P's age = 20. Q's Age = 28 Þ P + Q = 48Ex. 17. The average age of twenty fourboys in a class and the teacher is 15.If the teacher's age is excluded, theaverage reduces to 14. What is theteacher's age ?Sol. 17. 15 × 25 – 14 × 24 = 375 – 336= 39 yearsEx. 18. The average age of Indira andMeena is 28 years. If Kala's age is addedto it, then the average of the ages ofthe three becomes 27 year, what isKala's age ?Sol. 18. (26 × 3) – (28 × 2)= 78 – 56 = 22Ex. 19. If 6 years are subtracted fromthe present age of Gulzar and theremainder is divided by 18, then we getpresent age of his grand son Anup. IfAnup is two years younger to Maheshwhose age is 5 years, than what is thepresent age of Gulzar ?

Sol. 19.G

A− =6

18. Now A = 3

Þ G = 60.Ex. 20. 3 years ago the average age ofA, B and C was 27 years and that of Band C, 5 years ago, was 20 years. A'spresent age is .............. .Sol. 20. Total age of A, B and C (today)= 3 × 27 + 3 × 3 = 90. Total age of B and C(today) = 2 × 20 + 5 × 2 = 50Þ A's age today = 40 years.Ex. 21. Father is 5 years older to themother and mother's age now is thricethe age of the daughter. The daughteris now 10 years old then what was thefather's age when the daughter wasborn ?Sol. 21. F = M + 5, M = 3D where D = 10Þ M = 30, F = 35 Þ F – 10 = 25Ex. 22. A father told his son ‘‘I was asold as you are at present, at the timeof your birth’’. If the father is 38 yearsold now, then what was the son's agefive years ago in years ?Sol. 22. F = 38, S = 19 Þ S – 5 = 14Ex. 23. Ten years ago A was half of B'sAge. If ratio of their present ages is 3 :4, then what is the total of theirpresent ages ?

Sol. 23. A – 10 = 12

(B - 10)

Þ A : B = 3 : 4 Þ B = 20, A = 15Þ A + B = 35Ex. 24. The average age of 24 boys ina class is 16. The teacher is includedin the group and one boy is excludedfrom the group, the average increasesby one. What is the age of teacher ?Sol. 24. 24 × 17 – 24 × 16 = 24Ex. 25. The ratio of ages of A and B atpresent is 2 : 3. After five years theratio of their ages will be 3 : 4. What isthe present age of A in years ?

Sol. 25. A : B = 2 : 3 Þ AB

++

=55

34

Þ A = 10, B = 15


Ex. 26. 5 years ago Ashu's mother wasthree times as old as Ashu. After 5years, she will be twice as old as Ashu.How old is Ashu today ?Sol. 26. 3 (A – 5) = (M – 5) and 2 (A + 5) =(M + 5) Þ M – 3A = – 10 and M – 2A = 5.By solving we have A = 15, M = 35Ex. 27. Age of father is 5/2 of his son.5 year ago it was three times that ofhis son. What is father's present age ?Sol. 27. Suppose 5 years ago, Age of Son =x and age of Father = 3x

Þ (3x + 5) =52

( x + 5)

Þ 6x + 10 = 5x + 25 Þ x = 15Þ Father's present age is 3x + 5 = 50 yearsEx. 28. If sum of ages of two friendsis 35 years and difference is 5 then ageof younger friend in years is ............. .Sol. 28. Suppose ages are x and yÞ x + y = 35 and x – y = 5By solving, x = 20, y = 15Ex. 30. If the ratio of ages of Mohanand Rakesh 4 : 3 and sum of their agesis 28 years then ratio of their agesafter 4 years will be .............. .Sol. 30. Suppose ages are 4x and 3x.Þ 4x + 3x = 28Þ x = 4, Ages are 16 and 12.After 4 years ratio = 20 : 16 = 5 : 4

PERCENTAGEEx. 1. How will you express 0.15 inpercentage ?

Sol. 1. 0.15 = 15100

= 15%

Ex. 2.35

= ? %

Sol. 2.35

in percentage

=35

× 100 = 3005

= 60%

Ex. 3. 8% of 250 is .............. .

Sol. 3. 8% of 250 = 8

100 × 250 = 20

Ex. 4. 17% of 400 is .............. .

Sol. 4. 17% of 400 = 17100

× 400 = 68Ex. 5. 100 ml is ............... of 2 litter.Sol. 5. 2 litres = 2000 ml

Þ 100 1

20120

100 ml

200 ml= = × = 5%

Ex. 6. 250 gram is ........... of 5 kg.Sol. 6. 250 gm, 5 kg. = 5000 gm.

Þ 250

5000120

120

120

100= = = × in % = 5%

Ex. 7. Which of the following shows

highest percentage ? 35

, 24

, 13

, 37

Sol. 7.35

in % =35

× 100% = 60%

24

in % = 24 × 100% = 50%

13

in % =13

× 100% = 33.3%

37

in % = 37

× 100% = 42.8%

Þ 35

shows highest.

Ex. 8. Which of the following shows

lowest percentage ? 27

, 18

, 35

, 45

Sol. 8.27

in % = 27

× 100% = 28.57%

18

in % =18

× 100% = 12.5%

35

in % =35

× 100% = 60%

45

in % =45

× 100% = 80%

Hence 18

is lowest %.Ex. 9. .013 = ? %

Sol. 9. .013 = 13100

in %

= 13100

× 100% = 13%


Ex. 10. 17 is ................ of 350.

Sol. 10.17340

in % = 17340

× 100% = 5%

Ex. 11. 65 is .............. of 325.

Sol. 11.65

325 in % =

65325

× 100% = 20%

Ex. 12. 33% of 500 is .............. .

Sol. 12.33500

in % = 300 × 100% = 6.6%

Ex. 13. Rs. 5 is .............. of 100 rupees.

Sol. 13.5

100 in % =

5100

× 100% = 5%

Ex. 14.35

means ............... .

Sol. 14.35

in % =35

× 100% = 60%

Ex. 15. 3 kg is ......... percent of 10 kg .

Sol. 15.3

10 in % =

310

× 100% = 30%

Ex. 16. 25% of 30 is .............. .

Sol. 16. 25% of 30 = 30 × 25

100 =

304

= 7.5Ex. 17. 12.5% of 32 is .............. .

Sol. 17. 12.5% of 32

= 12.5100

× 32 = 540 × 32 = 4

Ex. 18.4

10 means ..................... .

Sol. 18.4

10 in % =

410

× 100% = 40%

Ex. 19. If 20% of a number is 15, thennumber is .............. .Sol. 19. Suppose it is x.

Þ x × 20

100 = 15 Þ x = 75

Ex. 20. 15% of a number is 45 thennumber is .............. .Sol. 20. Suppose number is x

Þ x × 15100

= 45 Þ x = 300

Ex. 21. 45% of 56

+ ? = 20% of 4.375.

Sol. 21. Suppose it is x.

45% of 56

= 45

10056

×

20% of 4.375 = 20

1004 375× .

Þ x = 20

1004 375

45100

56

× − ×. = 0.50

Ex. 22. 22% of 250 + 70% of 350 = ?

Sol. 22.

= 55 + 245 = 300

Ex. 23. 40% of 85 + ? = 25

of 140 + 18

of 32

Sol. 23.40

100851

25

14018

32× + = × + ×x

Þ 34 + x = 56 + 4 Þ x = 26Ex. 24. 90% of 8.1 + 799 = 50% of ?Sol. 24. 90% of 8.1 + 799

= 90

100 × 8.1 + 799

= 7.29 + 799 = 8.06.29= 50% of 1612.58Ex. 25. 30% of 300 is .............. .

Sol. 25. 30% of 300 = 30

100 × 300 = 90

Ex. 26. ? ÷ 22 = 75% of 88

Sol. 26.x22

= 75 × 88

100

Þ x22

= 3 ×884

Þ x = 1452

Ex. 27. 813

% of ? = 150

Sol. 27. Suppose required number is x.

Þ 253

× 1

100 × x = 150 Þ x = 1800

Ex. 28. 5% of (25% of Rs. 1,600) is...... .Sol. 28. 5% of (25% of Rs. 16000)

= 5

1025

100× ×FHG

IKJ16000 =

5100

× 400

= Rs. 20Ex. 29. If 15% of 40 is greater than 25%of a number by 2, then number is.............. .


Sol. 29. 15% of 40 = 15100

× 40 = 6

Þ 25% of x = 25

100 × x =

x4

Þ x4

+ 2 = 6 Þ x = 1630. A candidate, who gets 20% ofthe marks, fails by 30 marks. Butanother candidate, who gets 32%marks, get 42 marks more than passingmarks. Find maximum number ofmarks.Sol. 30. Suppose marks are x.Þ 20% of x + 30 = 32% of x – 42Þ x = 600

PROFIT & LOSSEx. 1. An article is purchased for Rs.150 and sold at Rs. 195 then profitpercent is .............. .Sol. 1. C.P. = Rs. 150, S.P. = Rs. 195Profit = SP – CP = 195 – 150 = 45

Þ Profit percent =45

150 × 100 = 30%

Ex. 2. Ram purchased a book at Rs.200 and sold at Rs. 180 then his losspercent is .............. .Sol. 2. Loss = CP – SP = 200 – 180 = 20

Þ Loss percent =20200

× 100 = 10%

Ex. 3. Ravi purchased 10 mangoesfor price of 8 mangoes. The profit heearned in percentage is .............. .Sol. 3. Profit = SP – CP = 10 – 2 = 8

Þ Profit percent =28

× 100 = 25%Ex. 4. Ashok bought oranges 5 forRs. 4 and sold 4 for Rs. 3 then his losspercent is .............. .

Sol. 4. C.P. price of a mango = 45

S.P. price of a mango = 34

Loss = CP – SP =45

–34

=16 15

20−

= 120

Þ Loss percent =120 ×

54 × 100

= 254

% or 614

%

Ex. 5. I sold a table for Rs. 1000 atprofit of 25%. What was cost price forme ?Sol. 5. ... S.P. = 125 , C.P. = 100Þ S.P. = 1000, C.P. = ?

Þ C.P. =100125

× 1000 = 800

Ex. 6. I buy 6 pens for Rs. 5 and sell5 pens for Rs. 6. My gain percent is.............. .Sol. 6. C.P. of 6 pens = Rs. 5

S.P. of 6 pens = Rs.65

× 6 = Rs. 365

.

Þ Gain on 6 pens = 365

– 5 =115

Þ Gain % =11 5

5/

× 100 = 44%Ex. 7. A man buys on article for Rs.80 and marks it at Rs. 120. He thenallows a discount of 40%. What is theloss or gain percent ?Sol. 7. S.P. = 120 – 40% of 120 = Rs. 72,

Loss = Rs. 8 Þ Loss % =880

× 100 = 10%

Ex. 8. A fruit seller sells mangoes atRs. 9 per kg and thereby loses 20%. Atwhat price per kg he should have soldthem to make profit of 5% ?Sol. 8. S.P. = Rs. 9 per kg, Loss = 20%

Þ C.P. =9 100100 - 20

× =

90080

= 454

Gain expected = 5%.

Now S.P. =454

+ 5% of 454

,

S.P. =454

+4580

=94580

= 11.81Ex. 9. A man purchased a plot ofland for Rs. 44,700 and spent Rs. 5300for levelling it. At what price should

he sell the plot so as to gain 614

% ?


Sol. 9. 50,000 + 614

% of 50,000

= 50,000 +25

4 100× × 50,000

= 50,000 + 2125 = 53125Ex. 10. A man bought goods worth

Rs. 12,000. He sells14

of them at a loss

of 10%. At what price should he sellthe remaining goods so as to get aprofit of 15% on the whole transaction?Sol. 10. Total profit required = Rs. 1800,Total S.P. = Rs. 13,80014

of the goods were sold for Rs. 2700

Þ Remaining goods should be sold forRs. 11,100Ex. 11. A bought a refrigerator andpressure cooker for Rs. 8,000 and Rs.2,000 respectively. A sold them to B andB sold them to C each making a profitof 25% on refrigerator and incurring aloss of 10% on the pressure cooker. IfA were to sell them to C directly onthe same selling price at which B hadsold them to C, what percent would Ahave gained on the whole ?Sol. 11. Refrigerator : A's C.P. = Rs. 8000Gain = 25% A's S.P. = Rs. 10,000B's C.P. = Rs. 10,000 B's S.P. = 12,500C's C.P. = Rs. 12,500Pressure Cooker :A's C.P. = Rs. 2,000Loss = 10% A's S.P. = Rs. 1,800B's C.P. = Rs. 1,800 B's S.P. = 1,620C's C.P. = Rs. 1,620As per question, if A sells directly to C atthe same rates at which B was selling to Cthen total C.P. = Rs. 8000 + Rs. 2000= Rs. 10,000.Total S.P. = Rs. 12500 + Rs. 1620= Rs. 14120, Gain = Rs. 4120

Þ Gain % = 4120

10000100× = 41.2%

Ex. 12. A dealer professes to sell hisgoods at cost price but he uses a falseweight of 950 grams for a kilogram.Then gain percent of the dealer is............. .Sol. 12. C.P. of 1000 gm = Rs. 1000

(suppose) Þ S.P. of 950 gm = Rs. 1000,

S.P. of 1000 gm = 1000950

100020000

19× =

Þ Gain % =1000 19

10000100

10019

/ × = = 55

19%

Ex. 13. Somesh bought a microwaveoven and paid 10% less than itsoriginal price. He sold it with 30%profit on the price he had paid. Whatpercentage of profit did somesh earnon the original price ?Sol. 13. Let the original price be Rs. x.

Somesh bought it for (x – 10% of x) = 9x10

He sold it for Rs.910

x + 30% of

910

x =

117x100

Þ Profit earned on the original price

=117x100

– x = 17x100

Þ Profit % = × =

17x100

x 100 17 %

Ex. 14. The cost of a paint is Rs. 60per kg. A kilogram paint covers 20square feet. How much will it cost topaint the outside of a cube having 10feet each side ?Sol. 14. Surface to be painted= 6 × (10)2 = 600 sq. ft.

Paint required =60020

= 30 kg.

Þ Cost of the paint = 30 × 60 = Rs. 1800Ex. 15. A trader purchased an oldbicycle for Rs. 480. He spent 20% of thecost on it's repair. If he wants to earnRs. 144 as net profit on it, how much


percentage should he add to thepurchase price of the bicycle ?Sol. 15. C.P. = Rs. 480Cost on repair = 20% of 480 = 96 Þ NetC.P. = 480 + 96 = 576. S.P. = 576 + 144 =720 Þ Percentage to be added to the

purchase price =240480

× 100 = 50%

Ex. 16. The price of 2 sarees and 4shirts is Rs. 1,600. With the samemoney one can buy 1 saree and 6shirts. If a person wants to buy 12shirts, how much amount person willhave to pay ?Sol. 16. Let the price of saree and shirt tobe purchased be Rs. X and Rs. Y per unitrespectively.Þ 2X + 4Y = 1600 andX + 6Y = 1600, Y = 200, X = 400Þ Price of 12 shirts = Rs. 2400.Ex. 17. Jayant opened a shopinvesting Rs. 30,000. Madhu joined him2 months later, investing Rs. 45,000.They earned a profit of Rs. 54,000 aftercompletion of one year. What will beMadhu's share of profit ?Sol. 17. Jayant : Madhu = 30000 × 12 :45000 × 10 i.e. 36 : 45 = 4 : 5

Þ Madhu's Share =59

× 54000 = 30,000Ex. 18. Chamanlal purchased 20dozen note books at the rate of Rs. 48per dozen. He sold 8 dozens at 10%profit and remaining 12 dozens with20% profit. What is his profitpercentage in whole transaction ?Sol. 18. Profit = 8 × 52.80 + 12 × 57.60 –20 × 48 = 422.40 × 691.20 – 960 = 153.6

Þ Profit % =153.6960

× 100 = 16%

19. A person sold two articles atRs. 1,200 each. Based on the cost, hisprofit on one was 20% and his loss onthe other was 20%. On the whole heSol. 19. Ist Article S.P. = Rs. 1200, Gain = 20% .

C.P. = ×+

= ×FHG

IKJ =1200 100

100 20100

S.P. 100100+Gain%

QCP

IInd Article S.P. = Rs. 1200. Loss = 20%

C.P. = ×−

= ×−

FHG

IKJ =1200 100

100 201500

S.P. 100100 Loss%

QCP

Total S.P. = Rs. 2400, C.P. = Rs. 2500,Loss = Rs. 100Ex. 20. The profit made on selling atoy for Rs. 450 is Rs. 30 more than theloss incurred on selling it for Rs. 320.The cost price (in rupees) of the toy is.............. .Sol. 20. Suppose C.P. = Rs. xÞ (450 – x) = (x - 320) + 30 i.e. x = 370Ex. 21. A sells a radio to B at a profitof 20% and B sells it to C at a profit of25%. If C pays Rs. 225 for it, what didA pay for it ?Sol. 21. Suppose A paid Rs. 100 Þ B paidRs. 120 and C paid Rs. 150.If C pays Rs. 150, A pays Rs. 100Þ If C pays Rs. 225,

A pays =10015 × 225 = Rs. 15022.

Ex. 22. The C.P. of 12 articles is equalto S.P. of 9 articles. The gain percentis .............. .Sol. 22. C.P. of 12 articles= Rs. 12 (suppose) Þ S.P. of 9 articles = Rs. 12

i.e. S.P. of 12 articles =129

× 12 = Rs. 16

Þ Gain Percent =4

12 × 100 = 33

13

%

Ex. 23. By selling 45 lemons for Rs. 40,a man looses 20%. How many shouldhe sell for Rs. 24 so as to gain 20% inthe transaction ?Sol. 23. C.P. of 45 lemons should be Rs.50. To gain 20%. S.P. of 45 lemons shouldbe Rs. 60. Þ In Rs. 24, the number of

lemons that should be sold =4560

× 24 = 18

Ex. 24. By selling an article for Rs.19.50, a dealer makes a profit of 30%.


By how many rupees should heincrease his selling price so as to makeprofit of 40% ?Sol. 24. X + 30% of X = 19.50 Þ X = 15Þ X + 40% of X = 15 + 6 = 21Ex. 25. An article when sold at a gainof 5% yields Rs. 15 more than whensold at a loss of 5%. What is the costprice ?Sol. 25. X + 5% of X – 15 = X – 5% of XÞ 10% of X = 15.

Ex. 26. If S.P. of an article is43

timesof its C.P. then profit percentage is.............. .

Sol. 26. S.P. =43

× (C.P.) = 113 × (C.P.)

Ex. 27. A person purchased a radio

at 34

th of its original price and sold it

at a profit of 20% of its original price.His profit percent wasSol. 27. Suppose original price = Rs. K.

Þ C.P. = 3K4

, S.P. = K + 20% of K = 6K5

Þ Profit =6K5

– 3K4

=9K20

Þ Profit % = ×9 203 4

100KK

//

= 60

Ex. 28. I buy oranges at 4 oranges forRs. 7 and an equal number of orangesat the rate of 3 for Rs. 5. If I sell thewhole lot of oranges at the rate of 2for Rs. 4.10 then what will be my gain% or loss?Sol. 28. Suppose I buy 24 oranges(12 is LCM of 4 and 3)12 at the rate of 4 for Rs. 7 and 12 at therate of 3 for Rs. 5Hence C.P. of 24 oranges

=74

× 12 +53

× 12 = 21 + 20 = Rs. 41.

S.P. of 24 oranges at the rate of 2 for Rs.

4.10 =410.

2 × 24 = 49.20

Þ Gain = Rs. 8.20

i.e. Gain % =8 20.41

× 100 = 20%Ex. 29. A horse and a carriagetogether Cost Rs. 8,000. If by sellingthe horse at a profit of 10% and thecarriage at a loss of 10% total profitof 2.5% is made, then what is the costprice of the horse ?Sol. 29. Suppose C.P. of Horse = Rs. xÞ C.P. of Carriage = Rs. (8000 – x)Þ 10% of x – 10% of (8000 - x) = Rs. 2.5% of8000 Þ 20% of x = 1000 Þ x = 5000Ex. 30. Pratap bought a radio and got25% discount on the original price. Hegot Rs. 40 more than the original priceby selling it at 140% of his cost price.At what price did he buy the radio ?Sol. 30. Suppose original price be

Rs. K Þ C.P. = K – 25% of K = 34K

.

S.P. = 140% of34K

= K + 40 i.e. K = 800

Þ C.P. = Rs. 600

RATIO & PROPORTIONEx. 1. If a : b = 4 : 5 and b : c = 5 : 6then a : c is .............. .Sol. 1. a : b = 4 : b Þ b : c = 5 : 6Þ a : b : c = 4 : 5 : 6 ÞÞÞÞÞ a : c = 4 : 6Ex. 2. If a : b = 3 : 5 and b : c = 4 : 6then a : b : c is .............. .Sol. 2. a : b = 3 : 5 and b : c = 4 : 6

Þ b : c = 4 × 34

: 6 × 34

= 3 : 92

Ex. 3. a : b = 3 : 5 then 2a : 3b is.............. .Sol. 3. a : b = 3 : 5Þ 2a : 3b = 2 × 3 : 3 × 5 = 6 : 15Ex. 4. If a : b = 3 : 5 then a2 : b2 is............. .Sol. 4. a : b = 3 : 5

Þ a2 : b2 = (3)2 : (5)2


Ex. 5. If a : b = 3 : 8 thena3

:b4

is

.............. .Sol. 5. a : b = 3 : 8

Þ a3

:b4

=33

84

: Þ a3

:b4

= 1 : 2

Ex. 6. If a : b :: 2 : 3 then a3 : b3 is.............. .Sol. 6. a : b = 2 : 3Þ a3 : b3 = (2)3 : (3)3

Ex. 7. If 3 : 4 :: 6 : x then value of x is.............. .Sol. 7. 3 : 4 :: 6 : x

Þ34

=6x

Þ 3x = 24 Þ x = 8

Ex. 8. If a : x :: 2a : 8 then value of xis .............. .

Sol. 8.ax

a= 28

ÞÞÞÞÞ x = 4

Ex. 9. a : 5 :: 3 : 10 then a is ............. .

Sol. 9.a5

= 3

10Þ Þ Þ Þ Þ a =

1510

= 32

Ex. 10. 9 : 10 :: x : 12 then x is .......... .

Sol. 10.9

10 =

x12

Þ 10x = 9 × 12

Þ x =9 12

10×

= 545

11. If we divide Rs. 700 into ratio3 : 4 then first share is equal to.............. .Sol. 11. Suppose shares are 3x, 4xÞ 3x + 4x = 700Þ x = 100 Þ First share = Rs. 300Ex. 12. A father has two sons, Aniland Sonil. If father wants to divide Rs.900 as following Anil : Sunil :: 4 : 5 thenshare of Sunil is .............. .Sol. 12. Suppose they got rupees 4x and5x i.e. 4x + 5x = 900 Þ x = 100Þ Sunil got = Rs. 500Ex. 13. If x : 8 :: 10 : 80 then x is equalto .............. .

Sol. 13.x8

=1080

Þ x = 1

Ex. 14.35

:45

is same as the ratio

.............. .

Sol. 14.35

:45

=35

× 5 :45

× 6 = 3 : 4

Ex. 15. The mean proportion of 6 and12 is .............. .

Sol. 15.6

12 =

12x

Þ 6x = 144 Þ x = 24

Ex. 16. The third proportion of 4 : 8is .............. .

Sol. 16.48

=8x

Þ 4x = 64 Þ Þ Þ Þ Þ x = 16

Ex. 17. The fourth proportion of 4 ::::: 6: 12 is .............. .

Sol. 17.46

=12x

Þ 4x = 72 Þ Þ Þ Þ Þ x = 18

Ex. 18. If Rs. 1400 are divided into A,B, C and D in the ratio 2 : 3 : 4 : 5 thenshare of B is .............. .Sol. 18. Suppose shares are Rs. 2x, Rs. 3x,Rs. 4x and Rs. 5x.Þ Þ Þ Þ Þ 2x + 3x + 4x + 5x = 1400 Þ Þ Þ Þ Þ x = 100Þ Þ Þ Þ Þ Share of B = 3x = 300Ex. 19. The ratio (x2 ----- y2) : (x + y)2 isequal to .............. .Sol. 19. (x2 - y2) : (x + y)2

= ( )( ) ( )x yx y x y

2 2−+ +

: (x + y)2

= (x ----- y) : (x + y)

Ex. 20. If two friends spend Rs. 5000and Rs. 3500 in the market then ratioof their expenditures is .............. .Sol. 20. 5000 : 3500 = 10 : 7Ex. 21. The third proportion of 3 : 6is .............. .

Sol. 21. 3 : 6 : x (say) Þ36

= 6x

Þ 3x = 36 Þ x = 12.Ex. 22. The fourth proportion of 2 : 4: 6 is .............. .Sol. 22. Suppose it is x. So 2 : 4 :: 6 : x


Þ 24

=6x

Þ 2x = 24 Þ x = 12.

Ex. 23. If a : b = 8 : 5 the value of ratioa + b : a ----- b is .............. .Sol. 23. a : b = 8 : 5Þ a + b : a - b = 8 + 5 : 8 - 5 = 13 : 3Ex. 24. If x + y : x ----- y :: 9 : 1 then x : y isequal to .............. .Sol. 24. x + y : x - y = 9 : 1 Þ x = 5, y = 4Ex. 25. Triplicate proportion of 2 : 3is .............. .Sol. 25. Triplicate proportion of 2 : 3 is(2)3 : (3)3 = 8 : 27Ex. 26. If x is added to both thenumbers in ratio 3 : 5 the ratiobecomes 5 : 7 then value of x is ......... .

Sol. 26.35

57

++

=xx

Þ 21 + 7x = 25 + 5x Þ 2x = 4 ÞÞÞÞÞ x = 2Ex. 27. If 3 is deducted form eachnumber of the ratio 8 : 6 : 9 then newratio is .............. .Sol. 27. 8 : 6 : 9. If 3 is deductedwe have 8 - 3, 6 - 3, 9 - 3 = 5, 3, 6.Ex. 28. Duplicate proportion of x2 : y2

is .............. .Sol. 28. Duplicate proportion is(x2)2 : (y2)2 = x4 : y4

Ex. 29. If 0.4 : 1 is equal to ratio x : 5then x is .............. .

Sol. 29.0 41 5. = x

Þ x = 5 × 0.4 = 2

Ex. 30. If x2 = 4xy then ratio x : y isequal to .............. .

Sol. 30. x2 = 4 xy Þ x = 4y Þ xy

=41

TIME & WORKEx. 1. A can do a work in 5 dayswhile B can do same work in 4 days.Then in how many days A and B willtogether finish the work ?Sol. 1. A can do a work in 5 days

Þ A's one day's work = 15

B can do a work in 4 days

Þ B's one day's work = 14

Þ (A + B)'s one day work =15

+14

= 920

Þ A + B together complete the work in209

days.Ex. 2. A is thrice as good as B. Bcompletes a work in 9 days. Then inhow many days A will complete thesame work ?Sol. 2. B completes the work in 9 days

Þ B's one days' work = 19

Þ A's one days' work =39

= 13

Þ A completes the work in 3 days.Ex. 3. A can do a work in 7 days,which B can do in 5 days. In how manydays will they finish the work, bothworking together ?Sol. 3. A can do work in 7 days

A's one day's work = 17

.

B's one day's work = 15

Þ (A + B)' One day's work =17

+15

= 1235

A and B can finish the work in3512

days.

Ex. 4. If Ram and Manoj can do awork together in 12 days, which Manojand Ramesh can do in 16 days. On aparticular work, Ram works for 5 daysthen he leaves. After that manoj workson that job for 7 days then he alsoleaves. Remaining work is completedby Ramesh in 13 days. In how manydays will Ramesh alone complete thework ?Sol. 4. Work done by Ram and Manoj in

one day = 1

12


Manoj and Ramesh in one day = 1

16Ram's 5 days' work + Manoj's 7 days' work+ Ramesh's 13 days' work = 1Also Ram's 5 days' work + Manoj (5 + 2) days'work + Ramesh (2 + 11) days' work = 1

Þ5

12 +

216

+ Ramesh's 11 day's work = 1

Þ Ramesh's 11 days' work will be

= 1 - (5

12 +

216

) = 1124

Þ Ramesh's one day work =124

Þ Ramesh can complete the work in 24days.Ex. 5. 8 men and 8 boys can finish awork in 3 days. 3 men and 12 boys canfinish the same job in 4 days. Howmany days will 8 men and 10 boysfinish the same work ?Sol. 5. 8 Men + 8 Boys can finish the work= 3 days.Þ 24 Men + 24 Boys can finish the work =1 day3 Men and 12 Boys can finish the work in 4days.Þ 12 Men + 48 Boys can finish the work =1 day.24 M + 24 B = 12 M + 48 B[In terms of work]Þ 12 M = 24 B Þ 1 M = 2 B(1 Man = 2 Boys)Now we know from given question 8 menand 8 boys can finish the work in 3 days.Þ 8 × 2 Boys + 8 Boys can finish the workin 3 days. Þ 24 Boys do the job in 3 days.Ex. 6. Rajesh can do a work in 6hours, Rajesh and Suresh can do it in4 hours and Sunil, Rajesh, and Suresh

in 223

hours. In how many hours canRajesh and Suresh, working together,do the same job ?Sol. 6. Rajesh can do a work in 6 hours.

Rajesh's one hour's work = 16

(Rajesh + Sunil)'s one hour work =14

while

Sunil, Rajesh and Suresh do work in = 223

Þ (Sunil + Rajesh + Suresh)'s- (Rajesh + Sunil)'s work in one hour

=38

-14

= 18

i.e. Suresh's one hour work = 18

Þ (Rajesh + Suresh)'s one hour work

=16

+18

=4 324+

= 724

Þ Rajesh and Sunil can do the work

in247

days.

Ex. 7. 2 men can do a piece of workin 6 days. They started the work, after2 days 2 more men joined them. In howmany days they will completeremaining work ?Sol. 7. 2 men do the work in 6 days.

2 men's one day work = 16

Þ 2 men's 2 days work =16 × 2 =

13

Þ Remaining work = 1 -13

=23

....(I)

Now one day work of 4 men

= 2 × 13

=23

....(II) . From (I) and (II)

Þ Þ Þ Þ Þ Work will be completed in 1 day.Ex. 8. A is twice as good as B andtogether they finish a work in 12 days.In how many days will A finish thework alone ?Sol. 8. Suppose B does work in 2m days.Þ A does work in m days.

Þ A's one day work =1m

Þ (A + B)'s one

day work =1

2m +

1m

= 3

2m

Þ (A + B) will do the work in23m

= 12

Þ m = 18 ÞÞÞÞÞ A will do work in 18 days.


Ex. 9. Sudhir can do12

of the work

in 4 days and Manoj can do13

work in4 days. In how many days both Sudhirand Manoj working together can do thework?

Sol. 9. Sudhir can do12

work in 4 days.

Þ Sudhir can do one work in 8 days.

Þ Sudhir' one day work = 18

, Manoj can

do13

work in 4 days.

Þ Manoj's one day work =1

3 4× =

112

Þ (Sudhir + Manoj)'s one day work

=18

+1

12 =

3 224

524

+ = .ÞÞÞÞÞ Both together

can do work in245

days.

Ex. 10. A is twice a good as B andtogether they finish a work in 18 days.In how many days will B finish thework working alone ?Sol. 10. Suppose B does work in 2m daysÞ A can do work in m day.

(A + B)'s one day work =1m

+1

2m =

32m

Þ (A + B) together can do work in =23m

days Þ 23m

= 18 Þ m = 27 ÞÞÞÞÞ B canfinish work in 2m = 2 × 27 = 54 days.Ex. 11. 16 man can reap a field in 8days. 8 men will reap the same field in.............. .Sol. 11. 16 men can reap in 8 days.Þ 1 man can reap in 8 × 16 days

ÞÞÞÞÞ 8 men can reap in8 16×

8 = 16 days.

Inverse Proportion Men No. of days

B 16 8 A 8 x

⇒ 16 : 8 :: x : 8 ⇒ 8x = 16 × 8

⇒ = ×x

16 88

= 16 days

Ex. 12. A alone can complete a workin 16 days and B alone in 12 days. Astarts the work then they work onalternate days. The total work will becompleted in .............. .

Sol. 12. In 2 days1

16 +

112

=748

of the

work is completed.

In 12 days,4248

=78

of the work in

completed. Out of the Alternately :

remaining18

of the work,1

16 work is

completed by A on 13th day.

Remaining1

16 of the work is done by

B in34

day.

Ex. 13. Rajesh and Ajay together cancomplete a piece of work in 16 days.Rajesh alone can do it in 24 days. Howlong will Ajay alone take to completethe whole work ?

Sol. 13.1

16 –

124

=148

Þ Ajay alone can complete the workin 48 days.Ex. 14. A, B and C can complete apiece of work in 24, 6 and 12 daysrespectively. All of them working sameon the work, they will complete thework in .............. .

Sol. 14.124

+16

+1

12

= + + =1 4 224

724

.

Ex. 15. Two pipes can fill A tank in15 hours and 12 hours respectively,while a third pipe can empty it in 20hours. If the tank is empty and all thethree pipes are opened, the tank willbe full in .............. .


Sol. 15. In one hour1

15 +

112

-120

=1

10 of

the tank filled.Þ Tank will be full in 10 hours .Ex. 16. If 24 men can do a piece ofwork in 27 days working 7 hours perday, then in how many days can 14men do it working at the rate of 9hours per day ?Sol. 16. 24 men can do work in 189 hours.

49 men can do the same work in189 24×

14= 324 hours = 36 days.Ex. 17. A and B can do a work in 12days, B and C in 15 days, C and A in 20days. How long will each of them taketo complete the same work ?Sol. 17.

(A + B)'s one day's work =1

12 ..... (I)

(B + C)'s one day's work =1

15 ..... (II).

(C + A)'s one day's work =120

.... (III)

Adding I, II and III Þ 2 (A + B + C)'s one

day's work =1

12 +

115

+120

=(5 + 4 + 3)

60

= 15

Þ (A + B + C)'s one day's work = 1

10

A's one day's work =1

10 -----

115

= 130

B's one day's work =1

10 -----

120

= 120

C's one day's work =1

10 -----

112

= 160

Ex. 18. A can do a piece of work in 20days and B can do same work in 30days. They work together for 4 daysand then A leaves. How long will B taketo complete the remaining work ?Sol. 18. (A + B)'s one day's work

=120

+130

=560

= 1

12

Þ (A + B)'s 4 days work =4

12 =

13

Remaining23

of the work is done by B alone

in23

× 30 = 20 days.Ex. 19. 24 men can complete a workin 16 days. 32 women can complete thesame work in 24 days. 16 men and 16women started working and workedfor 12 days. How many more men areto the added to complete the remainingwork in 2 days ?Sol. 19. 16 men can complete the work in24 days.16 women can complete the work

in(32 24)

16×

= 48 days Þ (16 m + 16 w)'s 1

day work =124

+148

= 1

16

(16 m + 16 w)'s 12 days' work =1216

= 34

.

Remaining work = 14

Now Inverse Proportion Men Days

B 24 4 A x 2Þ 24 : x :: 2 : 4 Þ 2x = 24 × 4 Þ x = 48Þ 24 more men are to be added.Ex. 20. 24 men complete a work in 12days. 36 children can complete thesame work in 18 days. In how manydays can four men and nine childrencomplete the same work ?Sol. 20. In 12 days, 24 men can complete

a work. In 18 days 24 12×

18 = 16 men can

complete the same work ..............(1)(By inverse proportion).Also in 18 days, 36 children can complete awork ...............(2)Þ 16 men = 36 children i.e. 4 men = 9children. 4 men and 9 children= 8 men or 18 childrenÞ 8 men can complete the work in24 12×

18 = 36 days.


(Alternately) 18 children can complete the

work in18 36×

18 = 36

Ex. 21. A tank can be filled throughan inlet pipe A in 30 minutes while itcan be emptied by an outlet pipe B in45 minutes. Both the pipes wereopened simultaneously and after 30minutes the pipe B was closed. Nowafter how many minutes will the tankbe full ?

Sol. 21. In one minute130

–145

=190

of

the tank will be filled. Therefore in 30

minutes13

of the tank will be filled .Now130

of the tank is filled by pipe A in one minute.

Remaining23

of the tank will be filled by

pipe A in 30 ×23

= 20 minutes.

Ex. 22. 15 men can complete a workin 10 days. 20 children can completethe same work in 15 days. In how manydays can 10 men and 10 childrentogether complete the same work ?Sol. 22. In 10 days, 15 men can completea work.

In 15 days,15 10×

15 = 10 men can complete

a work.Also in 15 days, 20 children can complete awork.Þ 10 men = 20 children i.e. 10 men + 10children = 30 childrenÞ 30 children can complete the work

in15 20×

30 = 10 days. (Inverse proportion)

Ex. 23. Suresh can complete a job in15 hours. Ashutosh alone can completethe same job in 10 hours. If Sureshworks alone for 9 hours and thenstops. How many hours will Ashutoshtake to complete the job alone ?

Sol. 23. Suresh's 1 hour job = 1

15. Suresh

can finish9

15 =

35

of the job in 9 hours.

Remaining25

of the job is done by Ashutosh

alone. 1

10 of the job is done by Ashutosh in

one hour. Þ Þ Þ Þ Þ 25

of the job is done by

Ashutosh in four hours.

Ex. 24. Suresh can do13

of a work in5 days. If Ram is thrice as good asSuresh, then in how many days Ramalone will do the work ?

Sol. 24. Manish can do13

work in 5 days.

ÞÞÞÞÞ Ram can do 3 ×13

work in 5 days.

Ex. 25. A can do a job in 6 days. If Bis 50% more efficient than A, then Balone will take .............. .Sol. 25. A can do the job in 6 days. Þ A's

one day's work = 16

. Þ B's one day's work

=16

+12 ×

16

=16

+1

12 =

312

= 14

Þ B can do the job in 4 days.Ex. 26. 6 men can complete a job in 9days. After six days of their working,2 more men joined them. How manydays will all of them take to finish theremaining work ?Sol. 26. 6 men can do job in 9 days

Þ 6 men's one day's work = 19

6 men's 6 day's work =69

=23

Þ Remaining work = 1 –23

= 13

Now (6 + 2) men's one day's work

=19

+13 ×

19

=19

+ 127

427

=


Þ 8 men can do job in274

days

or 8 men can do13

job

i.e. remaining job in =13 ×

274

= 94

days.Ex. 27. A job is done by 6 men and 4women in 8 days. In how many days 3men and 2 women will do the samejob?Sol. 27. (6 men + 4 women) can do job in 8days.Þ 2 (3 men + 2 women) can do job in 8 days.or (3 men + 2 women) can do the job in =8 × 2 = 16 days.Ex. 28. The ratio of workingefficiencies of A, B and C is 2 : 3 : 4. If ajob is done by B in 9 days then A and Cwill take time in doing the jobseparately .............. .Sol. 28. Suppose A, B and C do the job in2x, 3x, 4x days respectively.Þ 3x = 9 Þ x = 3 Þ A will do it in 6 days.While B will do it in 12 days.Ex. 29. A and B working together canfinish a job in 24 days and with C theycan do the job in 18 days. How long willC take the time working alone on thesame work ?Sol. 29. (A + B) finish the job in 24 days.

(A + B)'s one day's work = 124

.

(A + B + C) finish the job in 18 days.

(A + B + C)'s one day's work =1

18Þ C's one day work

=1

18 -

124

=( )4 3

72−

=172

.

Þ C alone will complete the job in 72 days.Ex. 30. A is twice as good as B. If Acan do a piece of work in 12 days, thenB alone will be complete it in ............ .Sol. 30. A can do the work in 12 days.Þ B will do the work in 12 × 2 = 24 days.

TIME & DISTANCEEx. 1. Devesh walks at a speed of 3km/hr. Then in how much time he willcover 600 meters ?

Sol. 1. 3 km/hr = 3 ×5

18 meter/second

= 1518

56

= . Now TimeDis ce

Speed= tan

Þ Time = = ×6005 6

60065/

= 720 seconds

= 12 minutesEx. 2. Ram runs 400 meters at aspeed of 10 meter/second. In how muchtime will he reach to final end ?

Sol. 2. TimeDis ce

Speed= tan

= 40010

= 40 seconds.Ex. 3. If a person runs at speed of 10meter/second for a time period of 25seconds, then how much distance willhe cover ?Sol. 3. We know that Distance= Speed × Time = 10 × 25 = 250 metersEx. 4. A man covers 600 meters in 2minutes with a bicycle. Then in howmuch time he will cover 54 kmsdistance ?

Sol. 4. Speed = 600120

meter/second

= 5 meter/second = 5 × 185

= 18 km/hr

TimeDis ce

Speed= tan

= 5418

= 3 hours

Ex. 5. Ashok walks 360 meters in 3minutes. His speed in km/hr is .......... .

Sol. 5. Speed =×

36060 3

meter/second

= 360180

= 2 meter/second. Now 2 meter/

second = 2 × 185

km/hr =365

km/hr


Ex. 6. If a boy walks from his houseto school at speed of 4 km/hr he is lateby 15 minutes and if he walks at speedof 6 km/hr then he reaches 10 minutesearlier. The distance of school from hishouse is .............. .Sol. 6. Suppose distance of school from

house = x Þ Time t1 =x4

and t2 =x6

Since difference of time = 25 minutes

=2560

hours =5

12 hours.

Þx4

–x6

=5

12 Þ

x12

512

= ÞÞÞÞÞ x = 5 kms

Ex. 7. A car travels from A to B at aspeed of x km/hr and returns from B

to A at 45

th of his earlier speed. If the

total time taken is 27 hours, then thedistance travelled, expressed as apercentage of x, is .............. .Sol. 7. Let the distance between A and B

be k. Þ kx

kx

+ =( / )4 5

27

Þ k = 12x Þ 2k = - 24 xÞ 2k = 2400% of xEx. 8. A man makes his upwardjourney 16 km/hr and downwardjourney at 28 km/hr. What is hisaverage speed ?Sol. 8. Average speed =

28 162+

= 22 km/hrEx. 9. A man can row 6 kms in anhour in still water and he finds that ittake twice as long in rowing up as inrowing down the river. The rate of thestream is .............. .Sol. 9. Suppose rate of the stream is xkm/hr. Þ The boat will travel downstreamby (6 + x) km/hr and upstream by (6 - x)

km/hr. Þ 6 - x =12

(6 + x) ÞÞÞÞÞ x = 2.

Ex. 10. If a student goes to school at2.5 km/hr, then he reaches 6 minutes

late. If he travels at 3 km/hr then hereaches the school 10 minutes early.What is the distance between hishouse and school ?Sol. 10. Suppose x = distance of house from

school Þx x

25660 3

1060.

− = + i.e. x = 4s

Ex. 11. Gopal walks to school at 4 km/hr and reaches school 10 minutesearlier than school time. If he walks 3km/hr he reached 100 minute late. Thedistance between his house and theschool is .............. .

Sol. 11.x x4

1060 3

1060

+ = − i.e. x = 4

Ex. 12. Car A moves 175 kms in 8hours, whereas car B moves 189 kmsin 12 hours. The ratio of the speed ofcar A and car B is .............. .Sol. 12. The required ratio is

1758

18912

: i.e.175

8634

:

i.e. 175 : 126 i.e 25 : 18Ex. 13. A trains 270 meters long isrunning at a speed of 30 meters persecond. It will cross a bridge 180meters long in .............. .Sol. 13. The train will cross the bridgeafter covering a distance of (270 + 180)meters at a speed of 30 meter/second. i.e.after 15 seconds.Ex. 14. Three persons begin to walkaround a circular track. They complete

their revolutions in 1516

seconds, 1614

seconds and 1823

seconds respectively.After what time will they be togetherat the starting point again ?

Sol. 14. The LCM of 916

654

563

, ,

= =LCM of 91, 65, 56HCF of 6, 5, 3

36401 = 3640


Ex. 15. A postman walks towardsnorth a distance of 120 m to deliver aletter. He then goes towards east for adistance of 50 m for delivering anotherletter. the shortest distance betweenthe two places .............. .Sol. 15.

Ex. 16. A and B run 200 meters in 22and 25 seconds respectively. How faris B from the finishing line when Areaches to it ?Sol. 16. When A covers 200 meters, Bcovers 176 meters.Ex. 17. A car covers 27 kms distancein 6 hours. If his average speed for first12 kma is 4 km/hr, then what is hisaverage speed during remaining part?

Sol. 17. Time for first 12 kms =124

= 3

hours. Now remaining distance is = 27 - 12= 15 kms. Remaining time = 6 - 3 = 3 hoursÞ Average speed for remaining time

=153

= 5 km/hr.

Ex. 18. A man is running with a speedof 4 km/hr. He covers 400 meters in.............. .Sol. 18. Speed 4 km/hr = 4 ×

518

meter/

second =2018

meter/second

Þ 20 meters are covered in 18 secondsÞ 400 meters will be covered in 18 × 20= 360 seconds = 6 minutesEx. 19. Rakesh starts walking for hisoffice. If he walks at the speed of 4 km/hr, he reaches 20 minutes late and ifhe walks at the speed of 5 km/hr thenreaches 10 minutes early. Then how

much distance is covered by him foroffice ?Sol. 19. Suppose distance of office from hishouse = x km.

Þ x x4 5

− = 30 minutes =12

hr.

⇒ − = 5x 4x

2012

Þ 2x = 20 ÞÞÞÞÞ x = 10 kms.

Ex. 20. Two men start walking withspeed of 6 meters/second and 5 m/second respectively. They reach theend point with time gap of 50 seconds.Find the distance covered by them.............. .Sol. 20. Suppose the distance is x then

x5

− =x6

50 seconds Þ 6x

30

− =550

x

Þ x = 30 × 50 = 1500 meters = 1.5 km.Ex. 21. A girl runs with a speed of 6km/hr on a sport track. Then what ishis speed per second ?

Sol. 21. Speed 6 km/hr = 6 × 5

18 meter/

second =53

meter/second

Ex. 22. In a race Anil and Rakeshstart running with speeds of 6 km/hrand 4 km/hr respectively. If track is of1 km then what is the time gapbetween their reaching times at thefinal point ?Sol. 22. Speed of Anil = 6 km/hr Þ time

=16

hour =606

= 10 minutes

Speed of Rakesh = 4 km/hr

Þ time =14

= 604

= 15 minutes

Þ Time gap = 15 - 10 = 5 minutesEx. 23. Subhash runs on a track of400 meters at a speed of 8 meter/seconds In how much time will hefinish the running ?Sol. 23. Speed = 8 meter/second

N E

S

130 m120 m

50 m90o)


Þ Time =4008

= 50 seconds

Ex. 24. A bus covers first 20 kms at aspeed of 30 km/hr and next 30 kms at aspeed of 20 km/hr. Find the averagespeed for whole journey.

Sol. 24. Time for first 20 kms =2030

= 23

hr.

Time for next 30 kms = 3020

= 32

hr.

Þ Time for 50 kms =23

+32

=136

hrs.

Þ Average Speed =Total dis ce

Total timetan

= 5013 6/

= 30013

km/hr

Ex. 25. If Ram crosses 600 m path in6 minutes, then his speed in km/hr is.............. .

Sol. 25. Speed =Total dis ce

Total timetan

=×600

6 60 = 10

6= 5

3 meter/second

= ×53

185

km/hr = 6 km/hr

Ex. 26. 18 km/hr is equal to ............. .

Sol. 26. 18 km/hr = 185

18× meter/second

= 5 meter/second27. 3 km/hr is .............. .

Sol. 27. 3 km/hr = 35

18×

= 56

meter/secondEx. 28. 5 meter/second is equal to.............. .

Sol. 28. 5 meter/second = ×5185

km/hr =18 km/hr.Ex. 29. A car starts moving with aspeed of 60 km/hr and a taxi startsmoving at the same time with 54 km/hr. How far was taxi from car whencar had travelled 6 kms ?

Sol. 29. Time taken for 6 kms by car

=dis ce

speedtan

Þ Time = 660

110

= hr = 6

minutes Þ In this time distance covered

by taxi = speed × time = ×541

10 = 5.4 km.

Þ Distance of taxi from the car= 6 – 5.4 = 0.6 km. = 600 metersEx. 30. A car covers half of thejourney with a speed of 6 km/hr andrest half at a speed of 4 km/hr If itreaches to reaching point in 2 hours,then find the distance travelled by it.

Sol. 30. Average speed =6 4

25

+ = km/hr

Þ Distance = 5 × 2 = 10 km.

PROBLEMS ON TRAINSEx. 1. The time taken to cross a poleby a train whose length is 140 metersand speed is 72 km/hr .............. .Sol. 1.

Speed hr meter ond= = ×72 725

18 km/ / sec

= 20 meter/second Þ Time =14020

= 7 sec

Ex. 2. A non-stop train of length 150meters will cross a platform of 250meters long, with a speed of 72 km/hrin .............. .Sol. 2. Total distance = 150 + 250= 400 m Þ Speed 72 km / hr=

= ×725

18 meter / second

= 20 meter/second

Þ Time =14020

= 20 sec

Ex. 3. Two trains of lengths 100meters and 75 meters respectively arerunning towards each other onparallel tracks with speeds of 40 km/hr and 86 km/hr respectively. They willcross each other, after they meet, in.............. .


Sol. 3. Total length = 100 + 75 = 175Þ Relative speed = (86 + 40)

= 126 km/hr = ×1265

18 meter/second

= 35 meter/second Þ t = 17535

= 5 seconds.Ex. 4. A 180 meters long traincrosses a pole in 6 seconds then itsspeed in km/hr is .............. .

Sol. 4. Speed = =dis cetimetan 180

6

= 30 meter/second = ×30185

km/hr

= 108 km/hrEx. 5. If the speed of a train is 36 km/hr and length is 80 meters then it willcross a pole in .............. .

Sol. 5. Speed = 36 km/hr

= ×365


Þ Distance = 80 meters

Þ time = =8010

= 8 seconds

Ex. 6. With a speed of 54 km/hr atrain crosses a pole in 8 seconds. Thenthe length of train is .............. .Sol. 6. Speed = 54 km/hr

= ×545


Þ Distance = speed × time = 15 × 8 = 120meter Þ length of train = 120 metersEx. 7. A trains takes 3 hours to runfrom one station to another. If itreduces its speed by 12 km/hr it takes45 minutes more for the journey. Thedistance between the two stations is.............. .Sol. 7. Suppose distance between twostations = k k.m.It takes 3 hour to cover k k.m.

Þ Speed = k3

km/hr Þ 3

312

4560

=−F

HGIKJ

−kk

Þ k = 180 kms

Ex. 8. Two trains of lengths 121 mand 99 m are running in oppositedirections with speeds of 40 km/hr and32 km/hr respectively. In what timewill they completely clear each otherfrom the moment they meet ?Sol. 8. We have to find the time to covera distance of 121 + 99 = 220 meters.

Relative speed 40 + 32 = 72 km/hr Þ 7200036000

meters/second = 20 meters/second

Þ 22011

meter/second.9. A train 300 meters long isrunning at a speed of 25 meter/secondsIt will cross a bridge 200 meters longin .............. .Sol. 9. 500 m will be covered in 20seconds with a speed of 25 meter/second in50025

= 20 seconds

Ex. 10. A train running between twostations A and B arrives at itsdestination 10 minutes late when itsspeed is 50 km/hr and 50 minutes latewhen its speed is 30 km/hr. What is thedistance between the stations A andB ?Sol. 10. Suppose required distance = k

km. Þ k50

16

=k30

56

− − Þk30

k50

=23

−

Þ k = 50 kms.Ex. 11. Two trains, each 120 meters inlength, run in opposite directions withspeeds of 40 m/second and 20 m/secondrespectively. How long will it take forthe tail end of the two trains to meeteach other from the time that theirengines cross each other ?Sol. 11. We have to find the time taken intravelling (120 + 120) = 240 meters at aspeed (40 + 20) = 60 m/second (Since trainsare running in opposite directions)

Þ Time taken = 24060

= 4 seconds


Ex. 12. A train covers the distancebetween two stations in one hour. Ifthe speed of the train is increased by10 km/hr, then what will be the timetaken to cover the same distance ?Sol. 12. Let the distance be x kms. Þ Thespeed is x km/hr. If the speed is (x + 10)km/hr then the distance of x km is covered

inx

(x 10)+ hours.

Ex. 13. A train 120 m long, travellingat 90 km/hr, overtakes another traintravelling in the same direction at 72km/hr and passes it completely in 50seconds, then the length of secondtrain is .............. .Sol. 13. When two trains are running inthe same direction, then the relative motionof the faster train with respect to the slowertrain is the difference of their speeds i.e. 18km/hr.Let x be the length of the second trainin meters. Þ 50 seconds = time taken bythe faster train in moving a distance equalto the sum of the lengths of the trains. [i.e.(120 + x) meters] with the relative speed of

18 km/hr. i.e.12018000

503600

+ =x

i.e. x = 130 mEx. 14. A train 150 meters long takes20 seconds to cross a platform 450meters long. The speed of the train inmeters per seconds is

Sol. 14.60020

metersondsec

= 30 m/second

Ex. 15. A train, travelling at a speedof 45 km/hr leaves New Delhi stationat 9 a.m. Another train travelling at60 km/hr starts at 12.00 noon from NewDelhi stations in the same direction.The distance from New Delhi wherethe two trains will meet is .............. .Sol. 15. Suppose x km is the distance fromNew Delhi where the two trains meet

Þx x− =135

45 60 i.e. x = 540

Ex. 16. If a train running at 80 km/hrcrosses a tree standing by the side ofthe track in 9 seconds, then the lengthof train is .............. .Sol. 16. In 3600 seconds distance covered= 80000 meters.

Þ In 9 seconds = ×800003600

9 meters. i.e. 200

meters.Ex. 17. A train traveling at a speed at90 km/hour crosses a pole in 10seconds. What is the length of the trainin meters ?Sol. 17. In 3600 seconds distance covered= 90000 meters.

Þ In 10 seconds = ×900003600

10 meters.

i.e. 250 meters.Ex. 18. A train 270 meter long isrunning at a speed of 30 meters perseconds It will cross a bridge 180meters long in .............. .Sol. 18. The train will cross the bridgeafter covering the distance of (270 + 180)meters at a speed of 30 m/second i.e. After15 seconds.Ex. 19. A train travelling at a certainspeed crosses a stationary engine in20 seconds. To find out the speed of thetrain, which of the followinginformations (s) given in statements(s) A and B is/are necessary ?Sol. 19. (A) The length of the Engine(B) The length of the trainEx. 20. A 180 meters long traincrosses a signal post in 6 seconds.What is the speed of the train ?Sol. 20. In 6 seconds distance covered =180 meters i.e. in 1 second it will cover 30

meters. i.e. in 3600 seconds30 3600

1000×

i.e.

in 1 hour 108 kms are covered.Ex. 21. A train passes a pole onplatform in 4 seconds and platform in10 seconds. If length of the train is 80


meters then find the length ofplatform.Sol. 21. Length of train = 80 meters. Itcrosses a pole in 4 seconds

Þ Speed =804

= 20 meters/second Þ length

of platform = 10 × 20 = 200 metersEx. 22. A non-stop train covers adistance at a speed of 75 km/hr in 20minutes then what is the distancecovered by it in this time ?Sol. 22. Distance = speed × time

= 75 × 13

= 25 kms.

Ex. 23. A train 80 m long with speed54 km/hr crosses a stationary train oflength 130 meters in .............. .Sol. 23. Total length = 80 + 130

=210 meters. Speed = 54 km/hr

= ×545


Þ Time =21015

= 14 secondsEx. 24. Two trains start from pointsA and B towards each other withspeeds of 40 km/hr and 65 km/hrrespectively. If the length of A is 130m and the length of B is 80 m then inhow much time both the trains willcross each other completely from themoment their engines meet ?Sol. 24. Total distance = 80 + 130= 210 m. Relative speed = 68 + 130

= 108 km/hr. = 108 × 518

meter/second

= 30 meter/sec. Þ time = 21030

= 7 seconds

Ex. 25. A train of 90 meters longcrosses a pole with a speed of 54 km/hr in .............. .

Sol. 25. Speed = 54 km/hr = ×545

18

meter/second Þ time =9015

= 6 seconds

Ex. 26. Find the time taken by a 180m long train running with a speed of72 km/hr to cross a bridge of 120meters.Sol. 26. Speed = 72 km/hr

= ×725


Þ Total distance = 180 + 120 = 300 meters

Þ Time =30020

= 15 seconds

Ex. 27. If two train start with speedsof 40 km/hr and 94 km/hr respectivelytowards each other. If length of firstrain is 100 meters and they clear eachother in 15 seconds after they meet eachother then length of second train is.............. .Sol. 27. Relative Speed = (94 – 40)

= 54 km/hr = ×545

18 meter/second

= 15 meter/second. Suppose length of 2ndtrain is x meters.Þ Total distance = 100 + x

Þ Time taken = 15 seconds =distance

speed

Þ 15100

15= + x

Þ 225 = 100 + x Þ x = 125 metersEx. 28. If a train starts with a speedof 54 km/hr then it covers 150 metersin .............. .Sol. 28. Speed = 54 km/hr

= ×545

18 meter/second

= 15 meter/second

Þ Distance = 150 meters

Þ Time =15015

= 10 seconds.

Ex. 29. If a platform is 250 meterslong then a non- stop train runningwith speed of 36 km/hr will cross it in.............. .Sol. 29. Speed = 36 km/hr


= ×365


Þ time =25010

= 25 seconds

PIPES & CISTERNSEx. 1. If a pipe can fill a tank in 6hours then it will fill one third of thetank in .............. .Sol. 1. Pipe can fill tank in 6 hours.Þ It will fill one third part of the tank in

= 6 × 1

13 = 2 hours.

Ex. 2. A pipe can empty a tank ofcapacity 32,000 litres in 8 hours. In onehour how much quantity of water willflow away ? (in liters)Sol. 2. Pipe can empty the tank (Capi.32,000 litres) in 8 hours. Þ The part of tank

emptied in 1 hour = 18

= 18

× 32,000

= 4,000 Litres.Ex. 3. Two pipes can fill a tank in 4hours and 5 hours respectively. If boththe pipes are opened simultaneouslythen in how much time tank will befull?Sol. 3. Pipes fill the tank in 4 hours and5 hours respectively. Þ Part of the tank

filled in 1 hour = 14

+ 15

= 920

Þ Tank will

be filled in 209

hours

Ex. 4. A pipe can fill a tank in 6hours and another pipe can empty itin 8 hours then if both are openedsimultaneously then the tank will befull in .............. .Sol. 4. Part of the tank filled in 1 hour

= 16

– 18

=4 324−

=241

Þ Tank will be filled in 24 hours.Ex. 5. Pipes A and B can empty acistern in 3 hours and 4 hoursrespectively. If both the pipes opened

at a time then in how much time thecistern will be empty ?Sol. 5. Part of the tank filled in 1 hour

= 13

+ 14

=12

=12

4 3 7+

Þ Tank will be filled in127

hours

Ex. 6. A cistern is connected withthree pipes A, B and C. A and B canfill cistern in 6 hours and 8 hoursrespectively. C can empty the cisternin 12 hours. If all the three pipes areopened then in how much time cisternwill be full ?Sol. 6. A and B can fill the tank in 6 and8 hours respectively while C can empty itin 12 hours. Þ Part of the tank filled in 1

hour = 16

+ 18

– 1

12 =

24 =

524

4 3 2+ −


hours.

Ex. 7. Three pipes A, B and C canempty a tank in 2 hours, 3 hours and 4hours respectively. If all the pipes areopened simultaneously then in howmuch time tank will be empty ?Sol. 7. A, B, C can empty the tank in 2, 3and 4 hours respectivelyÞ Part of tank emptied in 1 hour

= 12

+ 13

+ 14

=12

=1312

6 4 3+ +

Þ Tank will be empty in1213

hour.Ex. 8. If three pipes can fill a tankin 4 hours, 6 hours and 8 hoursrespectively. If all the pipes are openedat a time then in how much time tankwill be full?Sol. 8. Three pipes can fill in 4, 6 and 8hours.Þ Part of tank filled in 1 hour

= 14

+ 16

+ 18

=24

=1324

6 4 3+ +


hours.


Ex. 9. A tank is connected with threepipes A, B and C. Pipe A can fill thetank in 6 hours, while pipes B and Ccan empty the tank in 4 hours and 8hours respectively. If all the threepipes are opened at a time, then thetank will be empty in .............. .

Sol. 9. Part of tank emptied in 1 hour =14

+ 18

– 16

=24

=524

6 3 4+ −

Þ Tank will be empty in245

hours.

Ex. 10. A pipe can fill a tank in 4hours and another pipe can empty itin 2 hours. If both the pipes are openedat a time then the full tank will beempty in .............. .Sol. 10. The part of tank emptied in 1 hour

= 12

– 14

= 2 1

4−

= 14

Þ Tank will be empty in 4 hours.Ex. 11. A tank can be filled by a pipein 10 hours but due to leakage inbottom the tank will be filled in 15hours. If tank is full, then in how muchtime the tank will be empty ? (inletpipe is not open.)Sol. 11. Tank can be filled without leakage

in 10 hours. Þ Tank filled in 1 hour = 1

10.

Due to leakage tank is filled in 15 hours

Þ Tank filled in 1 hour when inlet and

leakage both are open = 1

15Þ Part of tank

emptied in 1 hour (due to leakage)

= 1

10 –

115

= − =3 230

130

Þ Full tank will be empty in 30 hours whenonly leakage is open.Ex. 12. Two pipes A and B can fill atank in 30 minutes and 45 minutes

respectively. Both pipes are openedtogether and after 9 minutes, pipe Ais turned off. In how much time thetank will be full now ?Sol. 12. Pipes can fill the tank in 30minutes and 45 minutes respectively.Þ Part of tank filled in 1 minute

= 130

+ 145

= + =3 290

590

= 118

Þ In 9 minutes tank filled = 9 × 118

= 12

Þ Remaining 12

part will be filled by B in

= 452

minutes

Ex. 13. Pipes A, B and C can fill a tankin 1 hour, 2 hours and 3 hoursrespectively. If A and B are openedtogether but pipe C is opened after 10minutes, then tank will be full in thetime .............. .

Sol. 13. A and B can fill 16

11

12

× +FHG

IKJ tank

in 10 min = 16

× 32

= 14

. Now part of the

tank filled in 1 hour by A, B and C

= 1 + 12

+ 13

= + +6 3 26

= 116

Þ 6

11 part will be filled in = 1 hour

Þ 34

part will be filled in

= 1 × 116

× 34

= 118

hours

Ex. 14. If pipe A can fill a tank in 4hours and another pipe can empty itin 8 hours, then how much part of tankwill be filled after 6 hours ?

Sol. 14. Part of the tank filled in 1 hour

= 14

– 18

= 18

Þ Part of tank filled in 6 hours = 68

= 34


Ex. 15. If two pipes A and B can fill atank in 30 minutes and 60 minutesrespectively then after 10 minutes howmuch part of the tank will be filled ?Sol. 15. Part of the tank filled in 1 minute

= 130

+ 160

= 360

= 120

Þ Part of the tank filled in 10 minutes

= 10 × 120

= 12

Ex. 16. A tank can be filled by a tapat the rate 5 litres/minutes. How longwill it take to fill the tank of 20 litres?Sol. 16. 5 liters are filled in 1 minute. Þ20 liters will be filled in 4 minutes.Ex. 17. A pipe can fill a tank ofcapacity 60 litres in 1 hour. How muchpart of a tank of 45 liters (capacitywill be) filled by this pipe in 45minutes?Sol. 17. Pipe can fill 60 liters in 1 hour.

Þ It can fill 1 liter in 160

hour

Þ In 1 min (160

hours) water filled = 1 liter

Þ In 45 minutes water filled = 45 litersi.e. Full tankEx. 18. A tap can empty a tank in 8hours. If tank is full and tap is openedthen how much part of tank remainsfilled after 2 hours ?Sol. 18. Part of the tank emptied in 1 hour

= 18

Þ Part of the tank emptied in 2 hours

= 2 × 18

= 14

Þ Part of the tank which

remains filled = 1 – 14

= 34

Ex. 19. A and B two pipes can fill acistern is 4 hours and 6 hoursrespectively. The part of cistern whichremains empty after 2 hours is ......... .Sol. 19. Part of cistern filled in 1 hour =14

+ 16

= 5

12. Part of cistern filled in 2

hours = 2 × 5

12 =

56

Þ Part of the tank

which remains empty = 1 – 56

= 16

Ex. 20. Two pipes A and B can emptya cistern is 4 hours and 3 hoursrespectively. If tank is full and boththe pipes are opened then after 3 hourshow much part of cistern remainsfilled ?Sol. 20. Part of cistern emptied in 1 hr =14

+ 13

= 7

12. Part of cistern emptied in 3

hours = 32

× 7

12 =

78

Þ Part of the tank

which remains filled = 1 – 78

= 18

Ex. 21. A cistern is normally filled in4 hours. Due to some leakage 2 hoursmore are required to fill. How muchtime will be required by the leakageto empty the full tank?Sol. 21. Part of cistern filled without

leakage in 1 hour = 14

. Part of cistern filled

with leakage in 1 hour = 16

. Þ Part of

cistern emptied in 1 hour = 14

– 16

= 3 212−

= 1

12 Þ Cistern will be empty in 12 hours.

Ex. 22. Three pipes A, B and C can filla cistern in 3 hours, 4 hours and 5hours respectively. If pipes A and Bare opened together but pipe C isopened after 1 hour then how muchtime will be required to fill thecistern?Sol. 22. Part of the tank filled by A and B

in 1 hour = 13

+ 14

= 7

12

(empty part = 1 – 7

12 =

512

.

Now A, B, C together fill in 1 hour


= 13

+ 14

+ 15

= + +20 15 1260

= 5760

.Since5760

part is filled in 1 hour Þ Empty 5

12

part can be filled in = × =512

6057

2557

hours

Ex. 23. Pipe A can fill a tank in 40minutes and pipe B can empty the

tank in 50 minutes. 12

part of tank will

be filled in .............. .

Sol. 23. Part of the tank filled in 1 minute

= 140

– 150

= 1

200 Þ

1200

part filled in 1

minute Þ 12

part will be filled in 12

× 200

= 100 minutesEx. 24. A cistern is filled by a pipe Ain 4 hours while pipe B can empty itin 8 hours. If both the pipes are openedsimultaneously then what part of thetank will be filled after 3 hours ?Sol. 24. Part of cistern filled in 1 hour =14

– 18

= 18

Þ In 3 hours = 3 × 18

= 38

Ex. 25. Pipes A and B can fill a tankin 4 hours and 6 hours respectively.The ratio of parts of the tank filled in1 hour separately by A and B is ......... .Sol. 25. Part of the tank which pipe A can

fill in 1 hour = 14

. Pipe B can fill in 1 hour

= 16

Þ Ratio is 14

: 16

i.e. 6 : 4 i.e. 3 : 2

Ex. 26. Two pipes can fill a tank in 15hours and 12 hours respectively, whilea third pipe can empty it in 20 hours.If the tank is empty and all the threepipes are opened, then the tank willbe full in .............. .Sol. 26. In one hour

= 1

15 +

112

– 120

= 1

10 tank filled

Ex. 27. A tank can be filled throughan inlet pipe A in 30 minutes and canemptied by an outlet pipe in 45minutes. Both the pipes were openedsimultaneously and after 30 minutesthe pipe B was closed. Now after howmuch time (in minutes) will the tankbe full ?

Sol. 27. In 1 minute = 130

– 145

=

190

part of the tank is filled Þ In 30 minutes

= 30 × 190

= 13

of the tank is filled. Now130

is filled in 1 minute Þ Remaining 23

is

filled in = 30 ×23

= 20 minutesEx. 28. Two pipes A and B can fill atank in 20 and 24 minutes respectivelyand a third pipe C can empty 3 gallonsper minute. If all the pipes A, B and Care opened simultaneously then thetank will be full in 15 minutes. Whatis the capacity of the tank ?Sol. 28. Suppose the capacity of the tank

= x gallons. Pipes A and B can fill 120

124

+

of the tank in one minute

i.e. 11

120 of the tank i.e.

11120

x gallons.

Pipe C can empty 3 gallons per minute

Þ In one minute = 11120

x – 3 =

11 360

120

x −b g

gallons can be filled i.e. in 15 minutes

11 3608

x − Þ x =

11 3608

x − ÞÞÞÞÞ x = 120.

Ex. 29. A cistern has two taps (whichfill it in 12 min and 15 minrespectively) and an exhaust tap. Whenall the three taps are openedsimultaneously it takes 20 minutes tofill an empty cistern. When inlet pipesare closed, how long will the exhausttap take to empty full cistern ?


Sol. 29. Let tap A can fill the cistern in 12minutes Let tap B can fill the cistern in 15minutes. Let tap C can empty the cisternin x minutes

Þ 1

121

151 1

20+ − =

x Þ x = 10

Ex. 30. Three taps A, B and C can filla tank in 12, 15 and 20 hoursrespectively. If A is opened for all thetime and B and C are opened for onehour each alternately the tank will befull in.............. .

Sol. 30. In one hour tap A can fill 1

12 of

the tank. In one hour tap B can fill 1

15 of

the tank. In one hour tap C can fill 120

of

the tank. In one hour = 1

12 +

115

= 320

of

the tank is filled. In the 2nd hour = 1

12 +

120

= 2

15 of the tank is filled.

ALLIGATION OR MIXTUREEx. 1. In what ratio must tea of Rs.42 per kg be mixed with tea of Rs. 50per kg so that cost of mixture is Rs.45 per kg ?Sol. 1.Cheaper Price Dearer PriceRs. 42 (c) Rs. 50 (c)

Rs. 45 (m)Mean Price

d - m = 5 m - c = 3

Þ Quantity of cheaper Quantity of dearer t

d mm c

teaea

= −−

= 53

Þ The required ratio is 5 : 3.Ex. 2. In what ratio must rice of Rs.16 per kg be mixed with rice of Rs. 24per kg so that cost of mixture is Rs. 18per kg ?

Sol. 2.Cheaper Price Dearer PriceRs. 16 (c) Rs. 24 (d)


d – m = 6 m – c = 2

Þ Quantity of cheaper Quantity of dearer r

d mm c

riceice

= −−

= =62

31

Þ The required ratio is 3 : 1.Ex. 3. A shopkeeper mixed 10 kgwheat of Rs. 5 per kg with 15 kg wheatof Rs. 8 per kg. Then the cost price ofmixture per kg is .............. .Sol. 3. Total Quantity = 10 + 16 = 26 kg.Þ Total cost = 10 × 5 + 16 × 8 = 178Þ Cost per kg of the mixture

= 17826

= Rs. 6.84

Ex. 4. A milkman mixes 10 litres ofwater to 50 litres of milk of Rs. 16 perliter, then cost price of mixture perliter is.............. .Sol. 4. Total quantity = 10 + 50 = 60litres. Total price = 50 × 16 = Rs. 800

Þ Cost per litre of the mixture = 80060

= Rs. 13.33Ex. 5. In what ratio tea of Rs. 80 perkg be mixed with 12 kg tea of Rs. 64per kg, so that cost price of mixture isRs. 74 per kg ?Sol. 5.Cheaper Price Dearer Price Rs. 64 (c) Rs. 80 (d)


d - m = 16 m - c = 10

Þ Quantity of cheaper Quantity of dearer t

teaea

= −−

= =d mm c

610

35

Þ The required ratio is 3 : 5.

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Arithmetic 1 (Mid Size) (1 48)