5 STD MATHEMATICS TERM – II ANSWER KEY UNIT

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STD MATHEMATICS TERM – II ANSWER KEY

1

UNIT – 1 GEOMETRY

EXERCISE 1.1

1. Write the measures of the complementary angles of the following angles.

i) 45o

The sum of complementary angles is 90o A

X + 45o = 90

o

X = 90o – 45

o C

X = 45o

? 45o

0 B

ii) 30o


X + 30o = 90

o

X = 90o – 30

o C

X = 60o 30

o

?

0 B

iii) 72o


X + 72o = 90

o

X = 90o – 72

o C

X = 18o ?

72o

0 B

iv) 88o


X + 88o = 90

o

X = 90o – 88

o C

X = 2o ?

88o

0 B

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v) 38o


X + 38o = 90

o

X = 90o – 38

o C

X = 42o ?

38o

0 B

2. write the measures of the supplementary angles of the following angles.

i) 80o X

The sum of supplementary angles is 180o

X + 80o = 180

o

X = 180o – 80

o

X = 100o ? 80

o

Y 0 Z

ii) 95o

X


X + 95o = 180

o

X = 180o – 95

o

X = 85o

?

95o

Y

0

Z

iii) 110o

X


X + 110o = 180

o

X = 180o – 110

o

X = 70o

?

110o

Y 0 Z

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iv) 135o

X


X + 135o = 180

o

X = 180o – 135

o

X = 45o

? 135o

Y 0 Z

v) 150o


X

X + 150o = 180

o

X = 180o – 150

o

X = 30o

? 150o

Y 0 X

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UNIT – 2 NUMBERS

EXERCISE 2.1

I) Answer the following

1. The square number of 2 is 4

2. The square number of 5 is 25

3. The number of boxes is equal to one square number. The

number is 3

4. Which of following number is square number

a) 23 b) 54 c) 36 d) 45

5. What is next square number of 49?

a) 76 b) 95 c) 64 d) 54

EXERCISE 2.2

I ) Find the common factors for :

i) 8 and 12

The factors of 8 : 1, , , , ,8

The factors of 12 : 1, , , 3, , 6, 12

The common factors of 8, 12, are 1, 2 and 4

ii) 24 and 30

The factors of 24 : , , 4, , 8, 12, 24

The factors of 30 : , , 5, , 10, 15, 30

The common factors of 24, 30 are 1, 2, 3 and 6

iii) 20 and 30

The factors of 20 : , , 4, , , 20

The factors of 30 : , , 3, , , 15, 16, 30

The common factors of 20, 30 are 1, 2, 5 and 10

1

1

4

4

2

2

1

1

2

2

6

6

1

1

2

2

5

5

1

0

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EXERCISE 2.3

1) Choose the best answer:

1. The number divisible by 5 with no remainder

a) 14 b) 535 c) 447 d) 316

2. Pick the number which is not a multiple of 6

a) 18 b) 26 c) 72 d) 36

3. The common multiple of 4 and 8 among the given number is

a) 32 b) 84 c) 68 d) 76

4. Factors of 6

a) 1,2,3 b) 1,6 c) 1,2,3,6 d) 2,3

5. Multiple of 9 is

a) 79 b) 87 c) 29 d) 72

2) Fill in the blanks

1. Factors of 7 (1,7)

2. The only even prime number is (2)

3. L cm of 5,15 (15)

4. The numbers which divides 35without remainders are (1, 5, 7)

5. LCM of 4, 12 (12)

3) Write down the factors of the given numbers

i) 25

The factor of 25 is 1, 5, 25

ii) 36

The factors of 36 is 1, 2, 3, 4, 6, 9, 12, 36

iii) 14

The factors of 14 is 1, 2, 7, 14

iv) 16

The factors of 12 is1, 2 ,3 , 4, 6, 12

v) 12

The factors of 12 is 1, 2, 3, 4, 6, 12

4) Draw the picture of factor tree.

i)

18

9 2

3 3

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ii)

iii)

iv)

33

3

11

16

4 4

2 2

2 2

50

5 10

5 2

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5) Write down the first 5 multiples of the given numbers.

i) 7

Multiples of 7 = 7, 14, 21, 28, 35

ii) 9

Multiples of 9 = 9, 18, 27, 36, 45

iii) 16

Multiples of 16 = 16, 32, 48, 64, 80

iv) 11

Multiples of 11 = 11, 22, 33, 44, 55

v) 21

Multiples of 21 = 21, 42, 63, 84, 104

6) Find the three common multiples of the given numbers

i) 24, 16

Multiples of 24 : 24, , 72, , 120,

Multiples of 16 : 16, 32, , 64, 80, , 122, 128,

The common multiples of 24, 16, are 48, 96 and 144

ii) 12, 9

Multiples of 12: 12, 24, 36, 48, 60, 72, 84, 96, 108, 120, 132

Multiples of 9 : 9, 18, 27, 36,45, 54, 63, 72,81, 90, 99,108, 117, 126, 135,144

The common multiples of 12, 9 are 36, 72, and 108

iii) 24, 36

Multiples of 24 : 24, 48, 72, 96, 120, 144,168, 192, 216….

Multiples of 36 : 36, 72,108, 144, 180, 216, 252,….

The common multiples of 24, 36 are 72, 144, and 216

7) Find Lcm of three given numbers

i) 12 and 28

L cm of 12, 28

2 12 2 28

2 6 2 14

3 3 7 7

1 1

48

48

96

96

144

144

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12 = X X 3

28 = X X 7

L cm of 12, 28 = 2 x 2 x 3x 7 =84

ii) 16 and 24

L cm of16, 24

2 16 2 24

2 8 2 12

2 4 2 6

2 2 3 3

1 1

16 = X X X 2

24 = X X X 3

L cm of 16, 24 = 2 x 2 x 2 x 2 x 3 =48

iii) 8 and 14

L cm of 8,14

2 8 2 14

2 4 7 7

2 2 1

1

8 = X 2 X 2

14 = X 7

L cm of 8, 14 = 2 x 2 x 2 x 7 =56

iv) 30 and 20

L cm of30, 20

2 30 2 20

3 15 2 10

5 5 5 5

2

2

2

2

2

2

2

2

2

2

2

2

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1 1

30 = X 3 X

20 = X 2 X

L cm of 30, 20 = 2 x 5 x 3 x 2 =60

8) Ramya visit the gym five days once and Kavitha visit the gym six days once. In

which day will they meet each other?

Sol :

Given multiples of 5 = 5, 10, 15, 25, , 35

Multiples of 6 = 6, 12, 18, 24, , 36

30th

day will they meet Ramya and kavitha.

9) Arun and shahjhan goes for walking in a circular path of a park in the same

direction. Arun takes 6 minutes to complete one round, while shahjahan takes 8

minutes to complete one round. In how many minutes will they meet each other?

Solution :

Multiples of 6 : 6, 12, 18, 24 , 30, 36, 42, 48, 54

Multiples of 8 : 8, 16, 24, 32, 40, 48, 56, 64

24 minutes will they meet Arun and shahjahan.

2

2

5

5

30

30

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UNIT – 3

PATTERNS

Exercise – 3.1

1. Find the angle of the given shapes using the equilateral triangle

i) ii)

Answer

i) ii)

2. Find the angles of a rectangle using a circle.

Given Angle of a circle is 360o.

Let us find the angle of rectangle using a circle

= 360o ÷ 4 = 90

o.

Place 4 Rectangle as shown in the above figure.

Now the angle of rectangle is 360o ÷ 4 = 90

o.

Exercise 3.2

1) Mention the time in the clock when the angle i) 180o ii) 90

o iii) 60

o

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2) Find the angle made by the bands of the clock at the given time.

11’o Clock 9’o Clock 6’o Clock

30o

90o

180o

06 : 10 06:45 06:30

120o

90o

0o

Exercise 3.3

1) Write down the collection of words by ending with ‘ENT’ and ‘IGHT’

i) WENT, SENT, B ENT, RENT, TENT

ii) NIGHT, LIGHT, RIGHT, SIGHT, MIGHT


i) COAT, BOAT, GOAT

ii) RED, BED WED

UNIT – 4

MEASUREMENTS

Exercise – 4.1

1) Find in the blanks

i) 7 Kg 400g = 7400g

ii) 5g 50mg = 5050 mg

iii) 9500 mg = 9 g 500 mg

iv) 15 kg 350 g = 15350g

v) 6250g = 6 kg 250g

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2) Add the following

i) 4 kg 250 g + 3 kg 450 g ii) 75 g 430 mg + 750 g

kg g g mg

(+) 4 250 75 430

3 450 (+) 750

7 kg 700g 825g 430 mg

iii) 97 kg 45 g + 77 kg 450 g + 33 kg 250 g iv) 75 kg 400g + 30 kg 250 g

kg g kg g

(+) 97 45 75 400

77 450 (+) 30 250

33 250 105 kg 650 g

207kg 745g

3) Subtract the following

i) 40 kg 350 g - 25 kg 200 g ii) 35kg 850g – 18 kg 500g

kg g kg g

(-) 40 350 35 430

25 200 (-) 18 500

15 150 17 350

40 kg 350g – 25 kg 200g = 15kg 150 g 35 kg 850g – 18 kg 500g = 17 kg 350 g

iii) 985 kg 475 g - 275 kg 325 g iv) 700kg – 300kg 500g

kg g kg g

(-) 985 475 700 000

275 325 (-) 300 500

710kg 150g 399kg 500g

4) Multiply the following

i) 4kg 300 g x 7 ii) 17 kg 750 g x 8 iii) 25 kg 550 g x 4

kg g

4 300 17 750 25 550

x 7 x 8 x 4

30 100 142 000 102 200

4 kg 300g x 7 = 30 kg 100g 17kg 750g x 8 = 142 kg 25kg 550 x 4 = 102kg 200g

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iv) 72 kg 350 mg x 5

72 350

x 5

361 750

72g 350 mg x 5 = 361 g 750 mg

5) Divide the following

i) 99kg 990g ÷ 3 ii) 147g 630 mg ÷ 7

99 kg 990 g = 33 kg 330 g 147g 630mg = 21g 90 mg

iii) 550kg 220g ÷ 11 iv) 484g 384 mg ÷ 4

550 kg 220g – 50 kg 20g 484g 384mg = 121g 096 mg

6) What is the total weight of 7 kg 500g of cashew nut and 3 kg 350 g of pista?

kg g

Weight of cashew nut = 7 500

Weight of pista = 3 350 (+)

Total weight = 10kg 850g

7) Vimal had a sack of cotton seeds weighing 50kg 350g. He used 7kg 300g cotton seeds

to feed his cow. How much cotton seed will be remaining after feeding his cow?

kg g

Cotton seeds in the sack 50 350

Cotton seeds used for cow 7 300 (-)

Remaining cotton seeds 43 050

Remaining of cotton seeds in sack = 43 kg 50g.

8) A glass bottle can contain 25g 125mg of medicine, how much medicine can 7 such

bottles contain?

g mg

Weight of one glass bottle of medicine 25 125

Weight of 7 glass bottle of medicine 25 125 x 7

Weight of 7 glass bottle of medicine 175g 875 mg

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9) 75kg 750g of groundnut seeds is filled in five bags, how much groundnut seeds can a

bag contain?

Total Amount of groundnut seeds = 75kg 750g

Weight of 1 bags of groundnut seeds = 5 kg

Number of bags of 5 kg groundnut seeds = 75kg 750 g ÷ 5

= 15kg 150g

we can separate 75kg 750kg groundnut seeds as 5 bags groundnut seeds in 15kg 150g

bags.

Exercise - 4.2


i) Millilitre is the smallest metric measure of capacity.

ii) Kilolitre is the largest unit of volume and equals 1000 litres.

iii) 7 kl 30 l = 7030 l

iv) 5 l 400 ml = 5400 ml

v) 1300 ml = 1 l 300 ml

2) Match the following

i) 4500 ml = 6 l 500 ml (3)

ii) 3250 ml = 8 l 200 ml (4)

iii) 6500 ml = 7 l 50 ml (5)

iv) 8200 ml = 4 l 500 ml (1)

v) 7050 ml = 3 l 250 ml (2)

3) Add and write in litres

i) 400l, 50l, 500 ml ii) 3 kl, 400l, 3 ml

3kl = 3000 l

l ml l ml

400 00 3000 000

50 00 400 000

500 003

450 500 3400 l 003 ml

400l, 50l, 500ml = 450 l 500 ml 3400l 3 ml

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iv) 1400 ml : 5680ml : 280 l

1400ml = 1l 400ml

5680ml = 5l 680 ml

l ml

1 400

5 680

(+) 280 000

287 080

287 l 080 ml

4) Subtracts

i) 3 kl from 15485 l ii) 15 kl from 20 kl

3 kl = 3 x 1000 = 3000 l

15485 l 20 kl

(-) 3000 l (-) 15 kl

12485 l 5 kl

3 kl from 15485 l = 12485 l 15 kl from 20 kl = 5 kl

iii) 345 ml from 5 l

5 l = 5 x 1000 = 5000 ml

(-) 345 ml

4655 ml

345 ml from 5 l = 4655 ml

= 4 l 655 ml

5) Multiply the following

i) 3 l 200 ml x 8 ii) 4 l 450 ml x 4

l ml l ml

3 200 4 450

(x) 8 (x) 4

25 600 17 800

3 l 200 ml x 8 = 25 l 600 ml 4 l 450 ml x 4 = 17 l 800 ml

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iii) 5 l 300 ml x 5 iv) 6 l 700 ml x 6

l ml l ml

5 300 6 700

(x) 5 (x) 6

26 500 40 200

5 l 300 ml x 5 = 26 l 500 ml 6 l 700 ml x 6 = 40 l 200 ml

6) Divide the following

i) 18 l 240 ml ÷ 6 ii) 20 l 600 ml ÷ 2

18 l 240 ml ÷ 6 = 3 l 40 ml 20 l 600 ml ÷ 2 = 10 l 300 ml

iii) 21 l 490 ml ÷ 7 iv) 25 l 350 ml ÷ 5

21 l 490 ml ÷ 7 = 3 l 70 ml 25 l 350 ml ÷ 5l 70 ml

7) Kalaiyarasi bought 5 l 500 ml groundnut oil and 750 ml sesame oil. How much

amount of oil did she bought in all?

l ml

Amount of oil bought groundnut oil 5 500

Amount of oil bought sesame oil (+) 0 750

6 250

Total amount of oil bought Kalaiyarasi = 6 l 250 ml

8) In a fuel station there was 70l 500ml of fuel. How much amount of fuel will be left

after selling 35 l 700ml of fuel?

l ml

Total amount of fuel 70 500

The Amount of fuel used (-) 35 700

34 800

Remaining fuel = 34 l 800 ml

9) A pot contains 9 l 500 ml of water, how much amount of water will 7 such pots

contains?

l ml

A pot contains of water = 9 500

7 pot contains of water = (x) 7

66 500

Total quantity of water = 66 l 500 ml

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10) 25 l 500 ml of milk is filled in 5 milk cans, how much amount of milk is filled in one

can?

l ml

Total Amount of milk = 25 500

Number of can = (÷) 5

Each can will get 5 l 100 ml

5 l 100 ml of milk.

UNIT – 5

INTERCONCEPT

Exercise – 5.1

1) Answer the following

S.No. Speed Time Speed x Time = Distance

i) 48 km / hr 2 hrs 48 x 2 = 96 km

ii) 35 km / hr 3 hrs 35 x 3 = 105 km

iii) 30 km / hr 5 hrs 30 x 5 = 150 km

iv) 50 km / hr 4 hrs 50 x 4 = 200 km

v) 20 km / hr 6 hrs 20 x 6 = 120 km

2)

i) Distance covered in 2 hrs at the speed of 20 miles / hr =20 x 2 = 40 miles

ii) Distance covered in 4 hrs at the speed of 65 miles / hr = 65 x 4 = 260 miles

iii) Distance covered in 5 hrs at the speed of 48 km / hr = 48 x 5 = 240 km

iv) Distance covered in 6 hrs at the speed of 80 km / hr = 80 x 6 = 480 km

v) Distance covered in 3 hours at the speed of 42 km / hr = 42 x 3 = 126 km

3) If Gopi ran 12 hrs at a speed of 14 km / hr, then how much distance did he ran?

Distance = speed x time = 14 x 12

Distance ran by Raja = 168 km

4) If Raja rides motor cycle for 4 hours at a speed of 30 km / hr then how much

distance did he go?

Distance = speed x time = 30 x 4

Distance ran by Raja = 120 km

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Exercise – 5.2


S.No. Speed Distance Distance / speed = Time

i) 35 km / hr 280 km 280 / 35 = 8 hrs

ii) 40 km / hr 360 km 360 / 40 = 9 hrs

iii) 45 km / hr 315 km 315 / 45 = 7 hrs

iv) 50 km / hr 300 km 300 / 50 = 6 hrs

v) 55 km / hr 275 km 275 / 55 = 5 hrs

2) Wilson reached 240 km at the speed of 60 km / hr how much time did he travelled?

Distance = 240 km

Speed = 60 km / hr

Time = Distance / Speed = 240 / 60 = 4 hrs

Wilson travelled 4 hrs

3) Anbarasan travelled 350 km at the speed of 70 km / hr. Find the time taken for his

travel?

Distance = 350 km

Speed = 70 km / hr


Anbarasan travelled 5 hrs

4) Nazar travelled 360 km at the speed of 90km / hour find the time taken for his

travel?

Distance = 360 km

Speed = 90 km / hr


Nazar travelled 4 hrs

5) Fathima reached 480 km at the speed of 120 km/hr how much time did she

travelled?

Distance = 480 km

Speed = 120 km / hr

Time = Distance / Speed = 480 /120 = 4 hrs

Fathima travelled 4 hrs

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Exercise – 5.3


S.No. Distance Money Total Cost

i) 180 km Rs.5 per km 180 x 5 = Rs.900

ii) 220 km Rs.8 per km 220 x 8 = Rs.1760

iii) 315 km Rs.4 per km 315 x 4 = Rs.1260

iv) 420 km Rs.6 per km 420 x 6 = Rs.2520

v) 580 km Rs.3 per km 580 x 3 = Rs.1740

2) Sneha spend Rs.7 per km for a trip and she travelled 850 km. How much is the total

cost of the trip?

Cost per kilometer is Rs.7

Cost of 850 kilometer = 850 x 7

= Rs.5950

The amount spent by Sneha = Rs.5950

3) Prabhu spent Rs.9 per / km for a trip and he travelled 580 km how much is the total

cost of the trip?

Cost per kilometer is Rs.9

Cost of 580 kilometer = 580 x 9

= Rs.5220

The amount spent by Prabhu = Rs.5220

Exercise – 5.4


i) 3 km 500 m = 3½ km ii) 25 km 250 m = 25¼ km

ii) 17 km 750 m = 17¾ km iv) 35 km 250m = 35 ¼ km

v) 45 km 750 m = 45¾ km

2) Convert into hours : (In fraction)

i) 10 minutes = 10 x 1/60 = 10/60 = 1/6 hrs

ii) 25 minutes = 25 x 1/60 = 25/60 = 5/12 hrs

iii) 36 minutes = 36 x 1/60 = 36/60 = 3/5 hrs

iv) 48 minutes = 48 x 1/60 = 48/60 = 4/5 hrs

v) 50 minutes = 50 x 1/60 = 50/60 = 5/6 hrs

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3) Convert into minutes

i) 5/6 hrs = 5/6 x 60 = 50 min

ii) 8/10 hrs = 8/10 x 60 = 48 min

iii) 4/6 hrs = 4/6 x 60 = 40 min

iv) 5/10 hrs = 5/10 x 60 = 30 min

v) 6/10 hrs = 6/10 x 60 = 36 min

4) Math the following

i) ½ part of Rs.1 - Rs.100 (v)

ii) ¼ part of Rs.4 - 50 paise (i)

iii) ½ part of Rs.10 - Rs.75 (iv)

iv) ¾ part of Rs.100 - Rs.1 (ii)

v) ½ part of 200 - Rs.5 (iii)

5) Write the ¼, ½ and ¾ parts of the following

i) Rs.200

¼ part of Rs.200 = Rs.50

½ part of Rs.200 = Rs.100

¾ part of Rs.200 = Rs.150

ii) Rs.10,000

¼ part of Rs.10,000 = Rs.2500

½ part of Rs.10,000 = Rs.5000

¾ part of Rs.10,000 = Rs.7500

iii) Rs.78,000

¼ part of Rs.8000 = Rs.2000

½ part of Rs.8000 = Rs.4000

¾ part of Rs.8000 = Rs.6000

iv) Rs.24,000

¼ part of Rs.24,000 = Rs.76000

½ part of Rs.24,000 = Rs.12000

¾ part of Rs.24,000 = Rs.18000

v) Rs.50,000

¼ part of Rs.50,000 = Rs.12500

½ part of Rs.50,000 = Rs.25000

¾ part of Rs.50,000 = Rs.37500

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UNIT – 6

INFORMATION PROCESSING

Exercise – 6.1

1) Choose the correct answer

i) Which shape have 6 triangles?

Answer = B

ii) Find which garland has made up of 12 beads

Answer = C


4, 4, 3, 5, 4, 4, 3 , 5 , 4 , 4 , 3 , 5 ,

ii) 1, 1, 2, 3, 5, 8, 13 , 21 , 34 , 55 ,

3)

3 + 4 + 5 + 3 – 4 = 11

44 – 33 = 11

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5 STD MATHEMATICS TERM – II ANSWER KEY UNIT