Aptitude_Practice Set With Solutions in PDF

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List of Aptitude Practice Question sets with Solutions in PDF

Part-I

1).The curved surface area of a right circular cylinder of radius 2 cm is 264 cm2. The

height of the cylinder, in cm, is

a) 14

b) 10.5

c) 21

d) 42

Answer: c)

Explanation: Curved surface area of cylinder =2πrh

∴2πrh =264

→2×22/7×2×h =264

→h=264×7 / 2×22×2 =21 cm

2).Base of a right pyramid is square. The length of a diagonal of the base is 12√12 cm.If

each lateral surface of the pyramid is a equilateral triangle, then its volume (in cu.cm) is

a) 208√12

b) 288√2

c) 288

d) 288√3

Answer: b)

Explanation:

Area of base =1/2 × (hypotenuse)2

=1/2 × (12√2)2 =144×2 / 2 = 144 sq.cm.

Side of triangle =12 2 × 1/√2 = 12 cm

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Slant height of pyramid =√(OB2-BF2) = √(122-62)

=√(12+6)(12-6) =√18×6

=√108 cm

∴Height of pyramid =√(OB2-FE2) = √ [(√ 108)2 -62] =√108-36

=√72 = 6√2 cm

∴Volume of pyramid =1/3 × Area of base × height

=1/3 × 144 × 6√2

=288√2 cu.cm

3).A merchant purchases a wrist watch from Rs.450 and fixes its list price in such a way

that after allowing a discount of 10%, earns a profit of 20%. Find the list price of the

wrist watch.

a) Rs.585

b) Rs.495

c) Rs.540

d) Rs.600

Answer: d)

Explanation: List price of article =Rs. x

∴x × 90/100 = 450×120 / 100

→x= 450×120 / 90 =Rs.600

4).A rides from his house at 12 noon towards B’s house at a speed of 8 km/hour. After

hour, B rides from his house towards A’s at a speed 7 km/hour. If the distance between

the two houses is 68 km the time when they meet is

a) 6 PM

b) 3 PM

c) 4 PM

d) 5 PM

Answer: d)





Explanation:

Let both meet after t hours.

∴Distance covered by A in t hours + distance covered by B in (t-1) hours = 68km

∴8t +7(t-1)hrs = 68

8t + 7t – 7 = 68

→15t = 68+7 = 75

→t=75/15 =5 hours i.e at 5 p.m.

5).Two runners cover the same distance at the rate of 15 km and 16 km per hour

respectively. Find the distance travelled when one takes 32 minutes longer than the

other.

a) 128 km

b) 64 km

c) 96 km

d) 108 km

Answer: a)

Explanation: Distance covered =x km

∴x/15 - x/16 = 32/60 =8/5

→x(16-15 / 15×16) =8/5

→x / 15×16 =8/15

→x= 16×8 =128 km.

6).On a journey across Mumbai a taxi averages 20 m.p.h. for 70% of the distance, 25

m.p.h. for 10% of the distance and 8 m.p.h. for the remainder. Then the average speed

of the whole journey is

a) 15.925 m.p.h

b) 15.25 m.p.h

c) 15 m.p.h

d) 15.625 m.p.h

Answer: d)





Explanation: Total distance =100 metre (let)

Total time =(70/20 + 10/25 + 20/8)hours

=700+80+500 / 200 =1280/200 =32/5 hours

∴Average speed =100 / 32/5 = 500/32 =125/8 =15.625 metre/hour

7).The average age of four sisters is 7 years. If the age of the mother is included, the

average is increased by 6 years. Then the age of the mother is

a) 37 years

b) 34 years

c) 32 years

d) 40 years

Answer: a

Explanation: Sum of ages of four sisters=7×4 =28 years

Sum of ages of four sisters and mother =5×13 =65 years

∴Mother’s age =65-28 =37 years.

8).A sells to B an iron box at 10% profit. B sells it to C for 20% profit. If C pays Rs.528,

then the cost price of A is

a) Rs.508

b) Rs.500

c) Rs.498

d) Rs.400

Answer: d)

Explanation: Cost price of box for A=Rs.x

∴x×110/100 × 120/100 = 528

9).20% of a man’s salary is paid as rent, 60% are his living expenses and 10% is paid in

LIC. If he spends the remaining Rs.3,000 on education of children, his salary is

a) Rs.30,000

b) Rs.9,000





c) Rs.90,000

d) Rs,35,000

Answer: a)

Explanation: Expenditure on (rent +living expenses +LIC) =90%

Remaining amount =10% If the total salary be Rs x, then x × 10/100 =3000

→x=3000×10 =Rs.30000

10).If (x-2)2 + (y - ½ )2 = 0 then the value of x/y is

a) 2

b) 1

c) 4

d) 1/4

Answer: c)

Explanation: (x-2)2 +(y - 1/2)2 = 0

→(x-2)2 =0 and (y - 1/2)2 =0

→x-2 =0 and y - 1/2 =0

∴x/y =2/1/2 =2×2 =4

11).A hemisphere and a cone have equal bases. If their heights are also equal, the ratio

of their curved surfaces will be

a) 1:√2

b) √2:1

c) 1:2

d) 2:1

Answer: b)

Explanation: Let the radius of the base of hemisphere be r units. Radius of the base of

cone =r units and h= r units

∴ Stand height

∴√ (r2 +r2) = √2r2 = √2r





∴Curved surface of hemisphere: curved surface of cone =2 πr 2 : πr ×√2r

= 2 : √2 = √2 : 1

12).If x=√[1 - (√5+1)/(√5-1)] then the value of 5x2 - 5x - 1 is

a) 0

b) 3

c) 4

d) 5

Answer: c)

Explanation: x= √ [(√5+1) / (√5-1) × (√5+1) / (√5+1)] =√[(√5+1)2 / (5-1)]

→ √[(√5+1)2 / 4] = (√5+1) / 2

∴5x2 - 5x - 1

→5(√5+1 / 2) 2- 5 [(√5+1) / 2] - 1

→5 [(5+1+2√5) / 4] – [(5√5+5) / 2] - 1

→5[(3+√5) / 2] – [(5√5+5) / 2] - 1

→(15 + 5√5 - 5√5 – 5 – 2) / 2

→8/2 = 4

13).When 233 is divided by 10, the remainder will be

a) 2

b) 3

c) 4

d) 8

Answer: a)

Explanation: 21=2; 22 = 4; 23 = 8; 24 = 16; 25 = 32 (i.e)

The units digits repeats itself after power 4.

Remainder after we divide 33 by 4=1

∴Unit’s digits in the product of 2=2

∴Remainder on division by 10=2





14).The value of

a) 40/11

b) 43/11

c) 46/11

d) 41/11

Answer: b)

Explanation:

15).Find the unit digit in the product (4387)245 × (621)72

a) 1

b) 2

c) 5

d) 7

Answer: d)

Explanation: 71=7; 72=49; 73=343; 74=1401; 75=16807;

i.e. The unit’s digit repeats itself after power 4.

Remainder after we divide 145 by 4=1

∴Unit’s digit in the product of (4387)245 × (621)72 = 7×1=7

16).If √2=1.4142, find the value of 2√2 + √2+ 1/(2+√2) + 1/(√2-2)

a) 1.4144



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b) 2.8284

c) 28.285

d) 2.4142

Answer: b)

Explanation: Expression =2√2 + 2 + [1 / (2+√2)] + [1 / (√2-2)]

=2√2+√2+ [(1 / (2+√2)) + (1 / (2-√2))]

=2√2+√2+ [(2-√2-2-√2) / ((2+√2)(2-√2))]

=2√2+√2+ [-2√2 / (4-2)]

=2√2+√2-√2 =2√2

=2 ×1.4142=2.8284

17).If x*y = x2 + y2 - xy then 11*13 is

a) 117

b) 147

c) 290

d) 433

Answer: b)

Explanation: x*y = x2 + y2 - xy

∴ 11*13 = 112 + 132 – 11×13

=121+169-143

=147

18).The value of [(3.2)3-0.008] / [(3.2)2+0.64+0.04] is

a) 0

b) 2.994

c) 3.208

d) 3

Answer: d)

Explanation: Expression = [(3.2)3-(0.2)3] / [(3.2)2+3.2+0.2+(0.2)2]

Let 3.2 =a and 0.2=b





∴Expression = (a3-b3) / (a2+ab+b2)

= (a-b)(22+ab+b2) / a2+ab+b2 = a-b

= 3.2-0.2 = 3

19).If,a1/3 = 11 then the value of a2 - 331a is

a) 1331331

b) 1331000

c) 1334331

d) 1330030

Answer: b)

Explanation: a1/3 =11

a=113 =1331

∴a2 -331a = a(a-331)

=1331(1331-331)

=1331 ×1000 =1331000

20).If √(1 + x/961) = 32/31, then the value of x is

a) 63

b) 61

c) 65

d) 64

Answer: a)

Explanation: √(1+x/961) = 32/31 squaring both sides.

1+ x/961 = (32/31)2 = 1024/961

→1 + x/961 = 1024/961 - 1 = 1024-961 / 961 = 63/961

→ x=63

21).If a and b are odd numbers, then which of the following is even

a) a+b+ab

b) a+b-1





c) a+b+1

d) a+b+2ab

Answer: d)

Explanation: The sum of two odd number is even.

The same is the case with their product.

∴a+b+2ab=Even Number

22).216-1 is divisible by

a) 11

b) 13

c) 17

d) 19

Answer: c)

Explanation: 216 -1 = (28)-1

= (28+1) (28-1)

= (256+1) (256-1)

= 257 ×255.Which is exactly divisible by 17.

23).The HCF and LCM of two numbers are 12 and 924 respectively. Then the number

of such pairs is

a) 0

b) 1

c) 2

d) 4

Answer: c)

Explanation: Let the numbers be 12x and 12y where x and y are prime to each other.

∴ LCM = 12xy

∴12xy = 924





→xy = 77

∴Possible pairs = (1.77) and (7.11)

24).What is the least number which, when divided by 5,6,7,8 gives the remainder 3 but

is divisible by 9?

a) 1463

b) 1573

c) 1683

d) 1793

Answer: c)

Explanation:

LCM of 5, 6, 7, 8 = 35 × 24 = 840

∴Required Number = 840 k+3 which is exactly divisible by 9. For k=2 .it is divisible by 9.

∴Required number = 840k+3 =840×2+3 =1683

25).Three numbers are in the ratio 3:4:5. The sum of the larges and the smallest equals

the sum of the second and 52.The smallest number is

a) 20

b) 27

c) 39

d) 52

Answer: c)

Explanation: Let the number be 2x, 4x and 5x

∴5x+3x = 4x+52

→ 4x=52

→ x=13

∴The Smallest Number =3x =3 ×13 =39

26).If the radius of a circle is increased by 50% its area is increased by

a) 125%





b) 100%

c) 75%

d) 50%

Answer: c)

Explanation: Percentage increase in area = (50+50 – (50×50)/100)% =75%

27).A and B working separately can do a piece of work in 9 and 12 days respectively. If

they work for a day alternately with A beginning, the work would completed in

a) 10 2/3 days

b) 10 ½ days

c) 10 ¼ days

d) 10 1/3 days

Answer: c)

Explanation: Part of work done by A and B in first two days = 1/9 + 1/12 = 4+3 / 36

=7/36

Part of work done in first 10 days = 35/36

Remaining work = 1 - 35/36 = 1/36

Now it is the turn of A

∴Time Taken by A = 1/36 × 9 = 1/4 day

∴Total Time = 10 + 1/4 = 10 + 1/4 days

28).A and B together can do a work in 10 days. B and C together can do the same work

in 6 days. A and C together can do the work in 12 days. Then, A, B and C together can

do the work in

a) 28 days

b) 14 days

c) 5 5/7 days

d) 8 2/7 days

Answer: c)





Explanation: (A+B)’s 1 day’s work = 1/10

(B+C)’s 1day’s work = 1/6

(C+A)’s 1 day’s work = 1/12

Adding all three

2(A+B+C)’s 1day’s work = 7/40

∴All three together will complete the work in =40/7 = 5 5/7 Days

29).A does half as much work as B in three fourth of the time. If together they take 18

days to complete a work, how much time shall B take to do it alone?

a) 30 days

b) 35 days

c) 40 days

d) 45 days

Answer: a)

Explanation: Let B complete the work in x days.

∴Work done by A in 3x/4 days =1/2

∴Time taken by A in completing the work =2 × 3x/4 = 3x/2 days

(A+B)’s 1 day’s work=1/x = 2/3x = (3+2) / 3x =5/3x

∴5/3x =1/18 =3x =90

→x=30

∴Time taken by B in completing the work = 30 days

30).If a wire is bent into the shape of a square, then the area of the square so formed is

81 cm2.When the wire is rebent into a semicircular shape then the area,(in cm2) of the

semicircle with be (Take π=22/7)

a) 22

b) 44

c) 77

d) 154

Answer: c)





Explanation: Side of square = √81 = 9cm

∴Total length of wire = 4×9 = 36 cm

Now perimeter of semi =circle = πr+2r

→r(π +2) =36

→r=36 / (π +2) =36 / (22/7 + 2)

=36 / (36/7) = (36/36) 7 =7cm

Area of semi –circle = πr2 / 2 = [22/(7× 2)] ×7 ×7

=77 sq.cm

31).A publisher printed 2000copies of a book at a total cost of Rs.6000.He distributed

100 copies to institutions as free samples and sold the remaining at the rate of Rs.4 per

copy. His gain or loss percent is

a) 26 1/3% Gain

b) 26 1/3%Loss

c) 26 2/3%Gain

d) 26 2/3%Loss

Answer: c)

Explanation: Remaining copies after free distribution 2000-100 =1900 Sales proceeds

of 1900 copies =1900×4=7600.00 ∴Profit =7600-6000 =1600 ∴Percentage of profit

=1600/6000×100 =26 2/3 %gain

32).7% of the selling price of an article is equal to 8% of its cost price; 9%of the selling

price exceeds 10% of its cost price by Rs.2.The amount of profit or loss is

a) Rs.85 profit

b) Rs.90 profit

c) Rs.100 loss

d) Rs.100 profit

Answer: d)

Explanation: (2)×8 70x - 80y =0





72x - 80y = 1600

2x=1600

∴x=800 Applying the value of x in the equation (1)

5600 - 8y=0

5600 = 8y; y=700

∴Profit= S.P.- C.P= 800 - 700 =100----------(1)

×10 9x - 10y = 200 ----------(2)

7x - 8y= 0 --------(1)

(9/100)x – (10/100)y = 2 ---------------------------------

(7/100)x = (8/100)y or (7/100)x – (8/100)y = 0

33).In a party the ratio of the number of ladies to that of gents is 1:3;if 3 ladies and 2

gents also join the party, the ratio is 1:2.The latest strength in the party is

a) 25

b) 19

c) 21

d) 20

Answer: c)

Explanation: Original Ratio 1:3 Let us take the actual number as x : 3x Now 3 ladies and

2 gents join The new ratio is x+3 : 3x+2 But the actual ratio is 1:2 ∴x+3 : 3x+2 = 1.2

2x+6=3x+2 4=x Now the ladies 4+3 =7 Gents 3×4+2 =14 Total =21

34).Find the value of (4.12×4.12 - 3.12×3.12) / (4.12 - 3.12)

a) 7.24

3b) 17

c) 27

d) 47

Answer: a)

Explanation:

(a2-b2 )/(a-b) = a+b ∴4.12+3.12 =7.24





35).The average weight of 7 persons decreases by 2 kg if a person with weight 73 kg is

replaced by another person. The weight of the new person is

a) 57 kg

b) 55 kg

c) 59 kg

d) 58 kg

Answer: c)

Explanation: Let total weight of 7 persons be 7x 7x-14 =7x-73+y -14+73=y 59 =y

36).The average of 200 items was 50.Later on, it was discovered that two items were

misread as 92 and 8 instead of 192 and 88.The correct average is

a) 50.7

b) 50.9

c) 50.3

d) 50.2

Answer: b)

Explanation: Total of 200 items =200×50 =10000 Now the difference in misreading

192+88-92-8 =180 The new total =10000+180 =10180 ∴The new average =10180/200

=50.9

37).A square is converted into a rectangle by increasing the length of the square by

15% and decreasing its breath by 15%.What is the ratio of the area of the rectangle to

that of the square?

a) 371/400

b) 391/400

c) 39/40

d) 351/400





Answer: b)

Explanation: Let the side of the square be 100 ∴area=10000 Now the area of the

rectangle 115×85 =9775 The ratio is 9775/10000 =391/400

38).If a person pays Rs.60 every month for 5 years, at the rate of 10% simple interest,

what is the actual amount of interest he is to get?

a) Rs.1,800

b) Rs.3,600

c) Rs.915

d) Rs.900

Answer: c)

Explanation: I= P × [n(n+1)] / (2 ×12) × r/100 =60 × (60×61)/(2 ×12) × 10/100 =915

39).A solid cylinder has total surface area of 231 cm2.Its covered surface area is two-

thirds of the total surface area. The volume of the cylinder is:

a) 266.25cm2

b) 268 cm2

c) 269.5 cm2

d) None of these

Answer: c)

Explanation: 231=2πr2+2 rl 154 =2 πrl

∴l=154/2πr ____________(1)

∴77=22πr2

49/4 =r2

7/2 =r Apply this equation________(1)

We get (154×7×2)/(2×22×7) =7

Volume of a cylinder =πr2l =22/7 × 49/7 × 7 =269.5 cm2





40).How many lead shots each of 0.3 cm. in a diameter can be made from a cuboids of

dimensions 9cm ×11cm ×12cm?

a) 81,000

b) 75,000

c) 80,000

d) 84,000

Answer: d)

Explanation: (9×11×12) / (4/3)πr2

= (9 ×11× 12 ×3× 7) / (4× 22 ×.15 ×.15)

=567/.045

=.84,000

41).A man rows a km down the stream in 15 minutes and up the stream in 20 minutes.

The velocity of the stream is

a) 3 km/hour

b) 1.25km/hour

c) 1 km/hour

d) 0.75 km/hour

Answer: c)

Explanation: In 1 hour, downstream 4km

In 1 hour. Upstream 3km

∴Speed of the stream =4-3 =1 km

42). A, B, C, can complete a work in 10 days,15 days and 12 days respectively. A and B

together commence a work and they leave it after 3 days. What is the time taken by C

to complete the remaining work?

a) 4 days

b) 7 days

c) 8 days





d) 6 days

Answer: d)

Explanation: The work done by A.B.C in one day =1/10.1/15.1/12 raspy.

∴combined work done by A and B in 3 days

=(1/10 + 1/15)×3

=(3+2 / 30)×3 =5/30 × 3 =1/2

The remaining (1 - 1/2) work can be done by

C’ in ½ / 1/12 = 12/2 =6 days

43).Three persons A, B, C, earn Rs.170 per day; A and B together earn Rs.120 per day;

B and C together earn Rs.13 per day. How much does B earn per day?

a) Rs.90

b) Rs.70

c) Rs.80

d) Rs.60

Answer: c)

Explanation: A+B+C = 170

A+B = 120

B+C = 130

∴C’s earning = 17 - 120 = 50

Now B’s earning = B+50 = 130

∴B=80

44).A person sold his watch for Rs.119.The percentage of his profit was equal to the

cost price. What was its cost price?

a) Rs.70

b) Rs.80

c) Rs.75

d) Rs.90

Answer: a)





45).The value of √27+√75-√108+√48 is

a) 7√3

b) 8√3

c) 6√3

d) -6√3

Answer: c)

Explanation: √27 + √75 - √108 + √48

= 3√3 + 5√3 - 6√3 + 4√3

= 6√3

46).Find 3 consecutive numbers such that three times the middle one is greater than the

sum of the other two by 22

a) 20,21,22

b) 23,24,25

c) 21,22,23

d) 22,23,24

Answer: c)

Explanation: Let three consecutive numbers be

X; x+1; x+2

As given 3(x+1) = x + x + 2 + 22

3x+3 = 2x+24

x=21

∴Three consecutive numbers are 21.22.23

47).A’s age is ¼ of B’s and B’s age is 1/3 of C’s. If their total age is 85 years, what is B’s

age?

a) 30 years





b) 20 years

c) 25 years

d) 24 years

Answer: b)

Explanation: 1/4 × 1/3 C+ 1/3 C + C = 85

1/12 C + 1/3 C + C = 85

C+4C+12C / 12 =85

17C = 85×12

∴C= 85×12 /17 = 60

∴B’s =1/3 C = 20

48).The least number, which leaves the remainder 45, when divided by 63 or 105 or 180

is

a) 1405

b) 1125

c) 1305

d) 1205

Answer: c)

49).The area of a sector with angle 36 is 15.4 cm2. What is its perimeter. (Use, π=

22/7)

a) 17.6 cm

b) 18.2 cm

c) 18.4 cm

d) 14.8 cm

Answer: c)





Explanation: . πr 2 /10 = 15.4

πr2 = 154

r2 = 154 ×7 /22 =49

r=7

2πr =2× 22/7 × 7 = 44

36 or 1/10 th of the perimeter of the whole circle =4.4

Perimeter of the sector = 2r+4.4 =14+4.4 =18.4 cm

50).The length of the longest pole that can be placed in a room of dimensions 8 cm×6

cm×5 cm is

a) 6.5 m

b) 5 √5 m

c) 6 √5 m

d) 5.5 m

Answer: b)

Explanation: √(82+162+52) =√(64+36+25)

√125 =5√5



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