CBSE Class 7

These NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.7 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Exercise 2.7

Question 1.
Find:
(i) 0.4 ÷ 2
(ii) 0.35 ÷ 5
(iii) 2.48 ÷ 4
(iv) 65.4 ÷ 6
(v) 651.2 ÷ 4
(vi) 14.49 ÷ 7
(vii) 3.96 ÷ 4
(viii) 0.80 ÷ 5

Question 2.
Find:
(1) 4.8 ÷ 10
(ii) 52.5 ÷ 10
(iii) 0.7 ÷ 10
(iv) 33.1 ÷ 10
(V) 272.23 ÷ 10
(vi) 0.56 ÷ 10
(vii) 3.97 ÷ 10
Answer:

Question 3.
Find:
(i) 2.7 ÷ 100
(ii) 0.3 ÷ 100
(iii) 0.78 ÷ 100
(iv) 432.6 ÷ 100
(v) 23.6 ÷ 100
(vi) 98.53 ÷ 100
Answer:
(i) 2.7 ÷ 100 \(\frac{2.7}{100}\)
There are two zeros in 100
∴ The decimal point in the quotient is shifted to the left by two places
∴ \(\frac{2.7}{100}\) = 0.027

(ii) 0.3 ÷ 100
There are two zeros in 100
∴ The decimal point in the quotient is shifted to the left by two places
∴ 0.3 ÷ 100 = \(\frac{0.3}{100}\) = 0.003

(iii) 0.78 ÷ 100
There are two zeros in 100
∴ The decimal point in the quotient is shifted to the left by two places
∴ 0.78 ÷ 100 = \(\frac{0.78}{100}\) = 0.0078

(iv) 432.6 ÷ 100
There are two zeros in 100
∴ The decimal point in the quotient is shifted to the left by two places
∴ 432.6 ÷ 100 = \(\frac{432.6}{100}\) = 4.326

(v) 23.6 ÷ 100
There are two zeros in 100
The decimal point in the quotient is shifted to the left by two places
∴ 23.6 ÷ 100 = \(\frac{23.6}{100}\) = 0.236

(vi) 98.53 ÷ 100
There are two zeros in 100
∴ The decimal point in the quotient is shifted to the left by two places
∵ 98.53 ÷ 100 = \(\frac{98.53}{100}\) = 0.9853

Question 4.
(i) 7.9 ÷ 1000
(ii) 26.3 ÷ 1000
(iii) 38.53 ÷ 1000
(iv) 128.9 ÷ 1000
(v) 0.5 ÷ 1000
Answer:
(i) 7.9 ÷ 1000 = \(\frac{7.9}{1000}\) = 0.0079

(ii) 26.3 ÷ 1000 = \(\frac{26.3}{1000}\) = 0.0263

(iii) 38.53 ÷ 1000 = \(\frac{38.53}{1000}\) = 0.03853

(iv) 128.9 ÷ 1000 = \(\frac{128.9}{1000}\) = 0.1289

(v) 0.5 ÷ 1000 = \(\frac{0.5}{1000}\) = 0.0005

Question 5.
(i) 7 ÷ 3.5
(ii) 36 ÷ 0.2
(iii) 3.25 ÷ 0.5
(iv) 30.94 ÷ 0.7
(v) 0.5 ÷ 0.25
(vi) 7.75 ÷ 0.25
(vii) 76.5 ÷ 0.15
(viii) 37.8 ÷ 1.4
(ix) 2.73 ÷ 1.3
Answer:

Question 6.
A vehicle covers a distance of 43.2 km in 2.4 litres of petrol. How much distance
will it cover in one litre of petrol?
Answer:
Total distance covered 43.2 km
Quantity of petrol used = 2.4 litres
∴ Distance covered in one litre petrol
= \(\frac{\text { Total distance covered }}{\text { Total quantity of petrol }}\)
= \(\frac{43.2}{2.4}\) km
= \(\frac{432 \times 10}{24 \times 10}=\frac{432}{24}\) = l8km
Distance covered in one litre of petrol = 18km

These NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.6 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Exercise 2.6

Question 1.
Find:
(i) 0.2 × 6
(ii) 8 × 4.6
(iii) 2.71 × 5
(iv) 20.1 × 4
(v) 0.05 × 7
(vi) 211.02 × 4
(vii) 2 × 0.86
Answer:
(i) 0.2 × 6
Since 2 × 6 = 12 and there is one digit to the right of decimal point in 0.2
∴ 0.2 × 6 = 1.2

(ii) 8 × 4.6
Since 8 × 46 = 368 and there is one digit to the right of decimal point in
∴ 8 × 4.6 = 36.8

(iii) 2.71 × 5
Since 271 × 5 = 1355 and there are two digits to the right of decimal point in 2.71
∴ 2.71 × 5 = 13.55

(iv) 20.1 × 4
Since 201 × 4 = 804 and there is one digit to the right of decimal point in 20.1
∴ 20.1 × 4 = 80.4

(v) 0.05 × 7
Since 5 × 7 = 35 and there are two digits to the right of decimal point
∴ 0.05 × 7 = 0.35

(vi) 211.02 × 4
Since 21102 × 4 = 84408 and there are two digits to the right of the decimal point
∴ 211.02 × 4 = 844.08

(vii) 2 × 0.86
Since 2 × 86 = 172 and there are two digits to the right of the decimal point
∴ 2 × 0.86 = 1.72

Question 2.
Find the area of rectangle whose length is 5.7 cm and breadth is 3 cm.
Answer:
Length of a rectangle = 5.7 cm
Breadth of a rectangle = 3 cm
Area of the rectangle = Length × Breadth
= 5.7 cm × 3 cm = 17.1cm²

Question 3.
Find:
(i) 1.3 × 10
(ii) 36.8 × 10
(iii) 153.7 × 10
(iv) 168.07 × 10
(v) 31.1 × 100
(vi) 156.1 × 100
(vii) 3.62 × 100
(viii) 43.07 × 100
(ix) 0.5 × 10
(x) 0.08 × 10
(xi) 0.9 × 100
(xii) 0.03 × 1000
Answer:
(i) 1.3 × 10 = 13
(ii) 36.8 × 10 = 368
(iii) 153.7 × 10 = 1537
(iv) 168.07 × 10 = 1680.7
(v) 31.1 × 100 = 3110
Note:
(1) If there is one zero, the decimal point is shifted to the right by one place.
(2) If there are two zeroes, the decimal point is shifted to the right by two places.
(vi) 156.1 × 100 = 15610
(vii) 3.62 × 100 = 362
(viii) 43.07 × 100 = 4307
(ix) 0.5 × 10 = 5
(x) 0.08 × 10 = 0.8
(xi) 0.9 × 100 = 90
(xii) 0.03 × 1000 = 30

Question 4.
A two-wheeler covers a distance of 55.3 km in one litre of petrol. How much distance will it cover in 10 litres of petrol?
Answer:
Distance covered in one litre of petrol = 55.3 km
Distance covered in 10 litres of petrol = 55.3 × 10 km = 553 km

Question 5.
Find:
(i) 2.5 × 0.3
(ii) 0.1 × 51.7
(iii) 0.2 × 316.8
(iv) 1.3 × 3.1
(v) 0.5 × 0.05
(vi) 11.2 × 0.15
(vii) 1.07 × 0.02
(viii) 10.05 × 1.05
(ix) 101.01 × 0.01
(x) 100.01 × 1.1
Answer:
(i) 25 × 3 = 75
∴ 2.5 × 0.3 = 0.75

(ii) 1 × 517 = 517
∴ 0.1 × 51.7 = 5.17

(iii) 2 × 3168 = 6336
∴ 0.2 × 316.8 = 63.36

(iv) 13 × 31 = 403
∴ 1.3 × 3.1= 4.03

(v) 5 × 5 = 25
∴ 0.5 × 0.05 = 0.025

(vi) 112 × 15 = 1680
∴ 11.2 × 0.15 = 1.680

(vii) 107 × 2 = 214
∴ 1.07 × 0.02 = 0.0214

(viii) 1005 × 105 = 105525
∴ 10.05 × 1.05 = 10.5525

(ix) 10101 × 1 = 10101
∴ 101.01 × 0.01 = 1.0101

(x) 10001 × 11 = 110011
∴ 100.01 × 1.1 = 110.011

These NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Exercise 13.1

Question 1.
Find the value of
(i) 2⁶
(ii) 9³
(iii) 11²
(iv) 5⁴
Answer:
(i) 2⁶ = 2 x 2 x 2 x 2 x 2 x 2 = 64
(ii) 9³ = 9 x 9 x 9 = 729
(iii) 11² = 11 x 11 = 121
(iv) 5⁴ = 5 x 5 x 5 x 5 = 625

Question 2.
Express the following in exponential form:
(i) 6 x 6 x 6 x 6
(ii) t x t
(iii) b x b x b x b
(iv) 5 x 5 x 7 x 7 x 7
(v) 2 x 2 x a x a
(vi) a x a x a x c x c x c x c x d
Answer:
(i) 6 x 6 x 6 x 6 = 6⁴
(ii) t x t = t²
(iii) b x b x b x b = b⁴
(iv) 5 X 5 X 7 X 7 X 7 = 5² X 7³
(v) 2 x 2 x a x a = 2² x a²
(vi) a x a x a x c x c x c x c x d = a³ x c⁴ x d

Question 3.
Express each of the following numbers using exponential notation.
(i) 512
(ii) 343
(iii) 729
(iv) 3125
Answer:
(i) 512

512 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 2⁹

(ii) 343

343 = 7 x 7 x 7 = 7³

(iii) 729

729 = 3 x 3 x 3x 3 x 3 x 3 = 3⁶

(iv) 3125

3125 = 5 x 5 x 5 x 5 x 5 = 5⁵

Question 4.
Identify the greater number wherever
possible in each of the following.
(i) 4³ or 3⁴ (ii) 5³ or 3⁵
(iii) 2⁸ or 8²
(iv) 100² or 2¹⁰⁰
(v) 2¹⁰ or 10²
Answer:
(i) 4³ or 3⁴
4³ = 4 x 4 x 4 = 64
3⁴ = 3x3x3x3 = 81
81 > 64
3⁴ > 4³
3⁴ is greater

(ii) 5³ or 3⁵
5³ = 5 x 5 x 5 = 125
3⁵ = 3x3x3x3x3 = 243
243 > 125
3⁵ > 5³
3⁵ is greater.

(iii) 2⁸ or 8²
2⁸ = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
= 256
8² = 8 x 8 = 64
256 > 64
2⁸ > 8²
2⁸ is greater.

(iv) 100² or 2¹⁰⁰
100² = 100 x 100 = 10000
2¹⁰⁰ = (2¹⁰)¹⁰
= (2x2x2x2x2x 2 x 2 x 2 x 2 x 2)¹⁰
= (1024)¹⁰ = [(1024)²]⁵
= (1024 x 1024)⁵ = (1048576)⁵
.’.(1048576) > 10000
or (1048576)⁵ > 100²
2¹⁰⁰ is greater

(v) 2¹⁰ or 10²
2¹⁰ = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 1024
10² = 10 x 10 = 100
1024 > 100
2¹⁰ > 10²
2¹⁰ is greater.

Question 5.
Express each of the following as product of powers of their prime factors.
(i) 648
(ii) 405
(iii) 540
(iv) 3600
Ans:
(i) 648

648 = 2 x 2 x 3 x 3 x 3 x 3
= 2³ x 3⁴

(ii) 405

405 = 3 x 3 x 3 x 3 x 5 = 3⁴ x 5

(iii) 540

540 = 2 x 2 x 3 x 3 x 3 x 5
= 2² x 3³ x 5

(iv) 3600

3600 = 2 x 2 x 2 x 2 x 3 x 3 x 5 x 5
= 2⁴ x 3² x 5²

Question 6.
Simplify
(i) 2 x 10³
(ii) 7² x 2²
(iii) 2³ x 5
(iv) 3 x 4⁴
(v) 0 x 10²
(vi) 5²x 3³
(vii) 2⁴ x 3²
Answer:
(i) 2 x 10³ = 2 x 10 x 10 x 10 = 2000
(ii) 7² x 2² = 7 x 7 x 2 x 2 = 49 x 4 = 196
(iii) 2³ x 5 = 2 x 2 x 2 x 5 = 8 x 5 = 40
(iv) 3 x 4⁴ = 3 x 4 x 4 x 4 x 4 = 3 x 256 = 768
(v) 0 x 10² = 0 x 10 x 10 = 0
(vi) 5² x 3³ = 5 x 5 x 3 x 3 x 3 = 25 x 27 = 675
(vii) 2⁴ x 3² = 2 x 2 x 2 x 2 x 3 x 3 = 16 x 9 = 144
(viii) 3² x 10⁴ = 3 x 3 x 10 x 10 x 10 x 10 = 9 x 10000 = 90000

Question 7.
Simplify:
(i) (-4)³
(ii) (-3) x (-2)³
(iii) (- 3)² x (- 5)²
(iv) (-2)³ x (- 10)³
Answer:
(i) (-4)³ = (-4) x (-4) x (-4) = – 64
(ii) (-3) x (-2)³ = (- 3) x (- 2) x (- 2) x (- 2) = (- 3) x (- 8) = 24
(iii) (- 3)² x (- 5)² = (-3) x (-3) x (- 5) x (- 5) = 9 x 25 = 225
(iv) (-2)³ x (- 10)³ = (- 2) x (- 2) x (- 2) x (-10) x (-10) x (- 10) =
(- 8) x (- 1000) = 8000

Question 8.
Compare the following numbers:
(i) 2.7 x 10¹²; 1.5 x 10⁸
(ii) 4 x 10¹⁴ ; 3 x 10¹⁷
Answer:
(i) 2.7 x 10¹² = \(\frac{27}{10}\) x 10¹²
= 27 x 10¹¹
(It contains 13 digits) (\(\) = a^m-n)
1.5 x 10⁸ = \(\frac{15}{10}\) x 10⁸
= 15 x 10⁷ (\(\) = a^m-n)
(It contains 9 digits)
27 x 10¹¹ > 15 x 10⁷
.’. 2.7 x 10¹² > 1.5 x 10⁸
2.7 x 10¹² is greater.

(ii) 4 x 10¹⁴ It contains 15 digits
3 x 10¹⁷ It contain 18 digits
.-. 3 x 10¹⁷ > 4 x 10¹⁴
3 x 10¹⁷ is greater.

These NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Exercise 2.5

Question 1.
Which is greater?
(i) 0.5 or 0.05
(ii) 0.7 or 0.5
(iii) 7 or 0.7
(iv) 1.37 or 1.49
(v) 2.03 or 2.30
(vi) 0.8 or 0.88
Answer:
(i) Comparing the digits at tenths place, we have
5 > 0
∴ 0.5 >0.05
So, 0.5 is greater.

(ii) Comparing the digits at tenths places, we have
7 > 5
∴ 0.7 > 0.5
So, 0.7 is greater.

(iii) Comparing the digits at ones place, we have
7 > 0
∴ 7 > 0.7
So, 7 is greater.

(iv) Since the digits at ones place are same, comparing the digits at tenths place, we have
3 < 4
∴ 1.37 <1.49
So, 1.49 is greater.

(v) Since the digits at ones place are same, comparing the digits at tenths place, we have
0 < 3
∴ 2. 03 < 2.30
So, 2.30 is greater.

(vi) 0.8 can be written as 0.80. Now, digits at tenths place are same, comparing the digits at hundredths place, we have
0 < 8
∴ 0.80 <0.88
So, 0.88 is greater.

Question 2.
Express as rupees using decimals:
(i) 7 paise
(ii) 7 rupees 7 paise
(iii) 77 rupees 77 paise
(iv) 50 paise
(v) 235 paise.
Answer:
(i) 7 paise = ₹ = 7 × \(\frac{1}{100}=\frac{7}{100}\) = ₹ 0.07

(ii) rupees 7 paise = ₹ = 7 + 7 × \(\frac{1}{100}\)
= ₹ 7 + ₹ 0.07 = ₹ 7.07

(iii) 77 rupees 77 paise = ₹ 77 + ₹ 77 × \(\frac{1}{100}\) = ₹ 77 + ₹ 0.77 = ₹ 77.77

(iv) 50 paise = ₹ 50 × \(\frac{1}{100}\) = ₹ 0.50

(v) 235 paise= 200 paise + 35 paise
= ₹ 2 + ₹ 35 × \(\frac{1}{100}\) = ₹ 2 + ₹ 0.35
= ₹ 2.35

Question 3.
(i) Express 5 cm in metre and kilometre
(ii) Express 35 mm in cm, m, and km
Answer:
We know that 100 cm = 1 m,
1000 m = 1 km
(i) 5cm = \(\frac{5}{100}\) m = 0.05 m
5 cm = \(\frac{5}{100 \times 100}\) km
= 0.00005 km.

(ii) We know that 1 cm =10 mm
35 mm = \(\frac{35}{10 \times 100}\) m = \(\frac{35}{1000}\)
35mm = \(\frac{35}{10 \times 100}\) m = \(\frac{35}{1000}\)
= 0.035 m
35 mm \(\frac{35}{10 \times 100 \times 1000}\) km
= \(\frac{35}{1000000}\) = 0.000035 km

Question 4.
Express in kg:
(i) 200 g
(ii) 3470 g
(iii) 4 kg 8 g
Answer:
We know that 1000 g = 1 kg
(i) 200 g = \(\frac{200}{1000}\) kg = \(\frac{2}{10}\) kg = 0.2 kg
(ii) 3470 g = \(\frac{3470}{1000}\) kg = \(\frac{347}{100}\) kg = 3.47 kg
(iii) 4 kg 8 g = 4 kg + \(\frac{8}{1000}\) kg
= 4 kg + 0.008 kg = 4.008 kg

Question 5.
Write the following decimal numbers in the expanded form:
(i) 20.03
(ii) 2.03
(iii) 200.03
(iv) 2.034
Answer:

Question 6.
Write the place value of 2 in the following decimal numbers:
(i) 2.56
(ii) 21.37
(iii) 10.25
(iv) 9.42
(v) 63.352.
Answer:
(i) In 2.56, the digit 2 is at the ones place
∴ place value of 2 is 2 × 1 = 2.

(ii) In 21.37, the digit 2 is at the tens place
∴ place value of 2 is 2 × 10 = 20.

(iii) In 10.25, the digit 2 is at the tenths place
∴ place value of 2 is 2 × \(\frac{1}{10}=\frac{2}{10}\)

(iv) In 9.42, the digit 2 is at the hundredths place
∴ The place value of 2 is 2 × \(\frac{1}{100}=\frac{2}{100}\)

(v) In 63.352, the digit 2 is at the thousandths place
∴ The place value of 2 is 2 × \(\frac{1}{1000}=\frac{2}{100}\)

Question 7.
Dinesh went from place A to place B and from there to place C. A is 7.5 km from B and B is 12.7 km from C. Ayub went from place A to place D and from there to place C. D is 9.3 km from A and C is 11.8 km from D. Who travelled more and by how much?
Answer:
Distance from A to B = 7.5 km
Distance from B to C = 12.7 km
Distance from A to C through
B = (7.5 + 12.7) km
= 20.2 km
∴ Distance travelled by Dinesh = 20.2 km

Again
Distance from A to D = 9.3 km
Distance from D to C = 11.8 km
Distance from A to C through D
= (9.3 + 11.8) km
= 21.1 km
∴ Distance travelled by Ayub = 21.1 km 21. 1 > 20.2
∴ Ayub travelled more distance.
Difference = 21.1 km – 20.2 km
= 0.90 km (or) 900 m
∴ Ayub travelled more distance by 900 m

Question 8.
Shyama bought 5 kg 300 g apples and 3 kg 250 g mangoes. Sarala bought 4 kg 800 g oranges and 4 kg 150 g bananas. Who bought more fruits?
Answer:
Since, Shyama bought 5 kg 300 g apples and 3 kg 250 g mangoes
Total fruits bought by Shyama = 5 kg
300 g + 3 kg 250 g = 8 kg 550 g
Since, Sarala bought 4 kg 800 g oranges and 4 kg 150 g bananas
Total fruits bought by Sarala
= 4 kg 800 g + 4 kg 150 g = 8 kg 950 g
8.950 > 8.550
So, Sarala bought more fruits
Difference in weight = 8 kg 950 g – 8 kg 550 g = 0 kg 400 g
= 400 g or \(\frac{400}{1000}\)kg = 0.4 kg

Question 9.
How much less is 28 km than 42.6 km?
Answer:
Difference = 42.6 km – 28 km = 14.6 km 28 km is less than 42.6 km by 14.6 km

These NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Exercise 2.4

Question 1.
Find:
(i) 12 ÷ \(\frac { 3 }{ 4 }\)
(ii) 14 ÷ \(\frac { 5 }{ 6 }\)
(iii) 8 ÷ \(\frac { 7 }{ 3 }\)
(iv) 4 ÷ \(\frac { 8 }{ 3 }\)
(v) 3 ÷ 2\(\frac { 1 }{ 3 }\)
(vi) 5 ÷ 3\(\frac { 4 }{ 7 }\)
Answer:

Question 2.
Find the reciprocal of each of the following fractions. Classify the reciprocals as proper fractions, improper fractions and whole numbers.
(i) \(\frac { 3 }{ 7 }\)
(ii) \(\frac { 5 }{ 8 }\)
(iii) \(\frac { 9 }{ 7 }\)
(iv) \(\frac { 6 }{ 5 }\)
(v) \(\frac { 12 }{ 7 }\)
(vi) \(\frac { 1 }{ 8 }\)
(vii) \(\frac { 1 }{ 11 }\)
Answer:
(i) The reciprocal of \(\frac { 3 }{ 7 }\) is \(\frac { 7 }{ 3 }\) .
It is an improper fraction.

(ii) The reciprocal of \(\frac { 5 }{ 8 }\) is \(\frac { 8 }{ 5 }\)
It is an improper fraction.

(iii) The reciprocal of \(\frac { 9 }{ 7 }\) is \(\frac { 7 }{ 9 }\) .
It is a proper fraction.

(iv) The reciprocal of \(\frac { 6 }{ 5 }\) is \(\frac { 5 }{ 6 }\) .
It is a proper fraction.

(v) The reciprocal of \(\frac { 12 }{ 7 }\) is \(\frac { 7 }{ 12 }\) .
It is a proper fraction.

(vi) The reciprocal of \(\frac { 1 }{ 8 }\) of is 8.
It is a whole number.

(vii) The reciprocal of \(\frac { 1 }{ 11 }\) is 11.
It is a whole number.

Question 3.
Find:
(i) \(\frac{7}{3}\) ÷ 2
(ii) \(\frac{4}{9}\) ÷ 5
(iii) \(\frac{6}{13}\) ÷ 7
(iv) 4\(\frac{1}{3}\) ÷ 3
(v) 3\(\frac{1}{2}\) ÷ 4
(vi) 4\(\frac{3}{7}\) ÷ 7
Answer:

Question 4.
Find:

Answer:

These NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Exercise 2.3

Question 1.
Find:
(i) \(\frac{1}{4}\) of (a) \(\frac{1}{4}\), (b) \(\frac{3}{5}\), (c) \(\frac{4}{3}\)
(ii) \(\frac{1}{7}\) of (a) \(\frac{2}{9}\), (b) \(\frac{6}{5}\), (c) \(\frac{3}{10}\)
Answer:

Question 2.
Multiply and reduce to lowest form (if possible)
(i) \(\frac { 2 }{ 3 }\) × 2 \(\frac { 2 }{ 3 }\)
(ii) \(\frac{2}{7} \times \frac{7}{9}\)
(iii) \(\frac{3}{8} \times \frac{6}{4}\)
(iv) \(\frac{9}{5} \times \frac{3}{5}\)
(v) \(\frac{1}{3} \times \frac{15}{8}\)
(vi) \(\frac{11}{2} \times \frac{3}{10}\)
(vii) \(\frac{4}{5} \times \frac{12}{7}\)
Answer:

Question 3.
Multiply the following fractions:

Answer:

= \(\frac{5 \times 3}{7 \times 1}=\frac{15}{7}\) = 2\(\frac{1}{7}\)

Question 4.
Which is greater?
(i) \(\frac{2}{7}\) of \(\frac{3}{4}\)or \(\frac{3}{5}\) of \(\frac{5}{8}\)
(ii) \(\frac{1}{2}\) of \(\frac{6}{7}\) or \(\frac{2}{3}\) of \(\frac{3}{7}\)
Answer:

Question 5.
Saili plants 4 saplings in a row in his garden. The distance between two \(\frac{3}{4}\) adjacent saplings is m. Find the distance between the first and the last sapling.
Answer:

Let A, B, C and D be the four saplings planted in a row.
Distance between two adjacent saplings \(\frac{3}{4}\)m
∴ Distance between the first and the last saplmg = AD = 3 × AB
3 \(\frac{3}{4}=\frac{3 \times 3}{4}=\frac{9}{4}\)m = 2\(\frac{3}{4}\)m
Distance = 2 \(\frac{1}{4}\) m

Question 6.
Lipika reads a book for 1\(\frac{3}{4}\) hours every day. She reads the entire book in 6 days.
How many hours in all were required by her to read the book?
Answer:
Total Number of days = 6
Reading time for one day = 1\(\frac{3}{4}\) hours
Total number of reading hours
= 6 × 1\(\frac{3}{4}\) hours
6 × 7 \(\frac{7}{4}=\frac{6 \times 7}{4}=\frac{3 \times 7}{2}=\frac{21}{2}\)
= 10 \(\frac { 1 }{ 2 }\) hours
Total time taken to read the book
= 10 \(\frac { 1 }{ 2 }\) hours

Question 7.
A car runs 16 km using one litre of petrol. How much distance will it cover using 2\(\frac { 3 }{ 4 }\) litres of petrol.
Answer:
Distance covered in 1 litre of petrol = 16 km
Distance covered in 2 \(\frac { 3 }{ 4 }\) litres of petrol = 16 × 2\(\frac { 3 }{ 4 }\)
16 × \(\frac { 11 }{ 4 }\) km
4 × 11 = 44 km
∴ Distance covered = 44 km.

Question 8.

Answer: