ENCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals

  ENCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals

Ex 2.1 Class 7 Maths Question 1.
Solve:

Solution:

Ex 2.1 Class 7 Maths Question 2.
Arrange the following in descending order:
(i)29,23,821(ii)15,37,710
Solution:

Ex 2.1 Class 7 Maths Question 3.
In a “magic square” the sum of number in each row, in each column and along the diagonals is the same. Is this a magic square?

Solution:

Since, the sum of all the fraction row wise, column wise and the diagonal wise is same i.e. 1511.
Hence, it is a magic square.

Ex 2.1 Class 7 Maths Question 4.
A rectangular sheet of paper is 1212 cm long and 1023 cm wide. Find its perimeter.
Solution:

Ex 2.1 Class 7 Maths Question 5.
Find the perimeter of (i) ∆ABE (ii) the rectangle BCDE in this figure. Whose perimeter is greater?
Solution:
(i) Perimeter of ∆ABE


Thus perimeter of ∆ABE is greater than the perimeter of the rectangle BCDE.

Ex 2.1 Class 7 Maths Question 6.
Salil wants to put a picture in a frame. The picture is 735 cm wide. To fit in the frame, the picture cannot be more than 7310 cm wide. How much should the picture be trimmed?
Solution:
The width of the picture

Hence, the required width to be trimmed =310 cm.

Ex 2.1 Class 7 Maths Question 7.
Ritu ate =35 part of an apple and the remaining apple was eaten by her brother Somu. How much part of the apple did Somu eat? Who had the larger share? By how much?
Solution:
Let the whole part of the apple be 1.

Ex 2.1 Class 7 Maths Question 8.
Michael finished colouring a picture in 712 hour. Vaibhav finished colouring the same picture in 34 hour. Who worked longer? By what fraction was it longer?
Solution:
Time taken by Michael = 712 hour
Time taken by Vaibhav = 34 hour

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.2

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Exercise 2.2
Ex 2.2 Class 7 Maths Question 1.
Which of the drawings (a) to (d) show.
(i)2×15(iii)3×23(ii)2×12(iv)3×14

Solution:
(i) 2×15 represents drawing (d)
(ii) 2×12 represents drawing (b)
(iii) 3×23 represents drawing (a)
(ic) 3×14 represents drawing (c)

Ex 2.2 Class 7 Maths Question 2.
Some pictures (a) to (c) are given below. Tell which of them show:


Solution:
(i) 3×15=35 represents figure (c)
(ii) 2×13=23 represents figure (a)
(iii) 3×34=214 represents figure (b)

Ex 2.2 Class 7 Maths Question 3.
Multiply and reduce to lowest form and convert into a mixed fraction:

Solution:

Ex 2.2 Class 7 Maths Question 4.
Shade:
(i) 12 of the circles in box (a)
(ii) 23 of the circles in box (b)
(iii) 35 of the circles in box (c)

Solution:

Ex 2.2 Class 7 Maths Question 5.
Find:

Solution:

Ex 2.2 Class 7 Maths Question 6.
Multiply and express as a mixed fraction.

Solution:

Ex 2.2 Class 7 Maths Question 7.
Find:

Solution:

Ex 2.2 Class 7 Maths Question 8.
Vidya and Pratap went for a picnic. Their mother gave them a water bottle that contained 5 litres of water. Vidya consumed 25 of the water. Pratap consumed the remaining water.
(i) How much water did Vidya drink?
(ii) What fraction of the total quantity of water did Pratap drink?
Solution:
(i) Water consumed by Vidya = 25 of 5 litres

Water consumed by Patap
= 5 litres – 2 litres = 3 litres
(ii) Fraction of water consumed by Pratap
= 35 litres

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

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Exercise 2.3
Ex 2.3 Class 7 Maths Question 1.
Find:

Solution:

Ex 2.3 Class 7 Maths Question 2.
Multiply and reduce to lowest form (if possible):

Solution:

Ex 2.3 Class 7 Maths Question 3.
Multiply the following fractions:

Solution:

Ex 2.3 Class 7 Maths Question 4.
Which is greater:

Solution:

Ex 2.3 Class 7 Maths Question 5.
Saili plants 4 saplings, in a row, in her garden. The distance between two adjacent saplings is 34 m. Find the distance between the first and the last sapling.
Solution:
Number of saplings = 4
Distance between two adjacent saplings = 34 m
∴ Distance between the first and the last sapling

Ex 2.3 Class 7 Maths Question 6.
Lipika reads a book for 134 hours everyday. She reads the entire book in 6 days. How many hours in all were required by her to read the book?
Solution:
In 1 day Lipika needs 134 hours
In 6 days Lipika will need 6×134 hours

Hence the required hours = 1012.

Ex 2.3 Class 7 Maths Question 7.
A car runs 16 km using 1 litre of petrol. How much distance will it cover using 234 litres of petrol?
Solution:
In 1 litre of petrol, the car covers 16 km distance In 234 litres of petrol, the car will cover 234 × 16 km distance

Hence, the required distance = 44 km.

Ex 2.3 Class 7 Maths Question 8.

Solution:

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

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

Ex 2.4 Class 7 Maths Question 1.
Find:

Solution:

Ex 2.4 Class 7 Maths Question 2.
Find the reciprocal of each of the following fractions. Classify the reciprocals as proper fractions, improper fractions and whole numbers.

(i) Reciprocal of 37=73, which is improper fraction.
(ii) Reciprocal of 58=85, which is improper fraction.
(iii) Reciprocal of 97=79, which is proper fraction.
(iv) Reciprocal of 65=56, which is proper fraction.
(vi) Reciprocal of 127=712, which is proper fraction.
(vi) Reciprocal of 18=8, which is whole number.
(vii) Reciprocal of 111=11, which is whole number.

Ex 2.4 Class 7 Maths Question 3.
Find:

Solution:

Ex 2.4 Class 7 Maths Question 4.
Find:


Solution:

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

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Exercise 2.5
Ex 2.5 Class 7 Maths 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
Solution:
(i) 0.5 or 0.05
Comparing the tenths place, we get 5 > 0
∴ 0.5 > 0.05

(ii) 0.7 or 0.5
Comparing the tenths place, we get 7 > 5
∴ 0.7 > 0.5

(iii) 7 or 0.7
Comparing the one’s place, we get 7 > 0
∴ 7 > 0.7

(iv) 1.37 or 1.49
Comparing the tenths place, we get 3 < 4
∴ 1.37 < 1.49 or 1.49 > 1.3

(iv) 1.37 or 1.49

Comparing the tenths place, we get 0 < 3

∴ 2.03 < 2.30 or 2.30 > 2.03

(vi) 0.8 or 0.88 ⇒ 0.80 or 0.88
Since tenths place is same.
Comparing the hundredth place, we get 0 < 8
∴ 0.80 < 0.88 or 0.88 > 0.80

Ex 2.5 Class 7 Maths 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
Solution:
(i) Since 1 rupee = 100 paise and 1 paise = 1100 rupees
7 paise =7100 rupees = 0.07 rupees

(ii) 7 rupees 7 paise = 7 rupees + 7100 rupees
= 7.07 rupees

(iii) 77 rupees 77 paise = 77 rupees + 77100 rupees
= 77.77 rupees

(iv) 50 paise = 50100 rupees = 0.50 rupees

(v) 235 paise = 235100 rupees = 2.35 rupees

Ex 2.5 Class 7 Maths Question 3.
(i) Express 5 cm in metre and kilometre
(ii) Express 35 mm in cm, m and km.
Solution:
(i) 1 metre = 100 cm
1 kilometre = 1000 metre = 100 × 1000 cm
= 100000 cm
∴ 5 cm = 5100 metre = 0.05 metre
5 cm = 5100000 km = 0.00005 km
Hence, 5 cm = 0.05 m and 0.00005 km

(ii) 1 cm = 10 mm and 1 km = 100000 cm
∴ 35 mm = 3510 cm = 3.5 cm,
35 mm = 351000 m = 0.035 m
35 mm = 351000000 km = 0.000035 km
Hence, 35 mm = 3.5 cm, 0.035 m and 0.000035 km.

Ex 2.5 Class 7 Maths Question 4.
Express in kg:
(i) 200 g
(ii) 3470 g
(iii) 4 kg 8 g
Solution:
(i) 200g = 2001000 kg [∵ 1 kg = 1000g]
= 0.2 kg
(ii) 3470 g = 347010000 = 3.47 kg [∵ 1 kg = 1000 g]
(iii) 4 kg 8 g = 4 kg + 810000kg [∵ 1 kg = 1000 g]
= 4 kg + 0.008 kg = 4.008 kg

Ex 2.5 Class 7 Maths Question 5.
Write the following decimal numbers in the expanded form:
(i) 20.03
(ii) 2.03
(iii) 200.03
(iv) 2.034
Solution:
(i) 20.03 = 2 × 10 + 0 × 1 + 0 × 110 + 3 × 1100
(ii) 2.03 = 2 × 1 + 0 × 110 + 3 × 1100
(iii) 200.03 = 2 × 100 + 0 × 10 + 0 × 1 + 0 × 110 +3 × 1100
(iv) 2.034 = 2 × 1 + 0 × 110 + 3 × 110 + 4 × 11000

Ex 2.5 Class 7 Maths 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
Solution:
(i) Place value of 2 in 2.56 = 2 × 1 = 2 i.e. ones
(ii) Place value of 2 in 21.37 = 2 × 10 = 20 i.e. tens
(iii) Place value of 2 in 10.25 = 210 = 0.2 i.e. tenths
(iv) Place value of 2 in 9.42 = 2100 = 0.02 i.e. hundredths
(v) Place value of 2 in 63.352 = 21000 = 0.002 i.e. thousandths.

Ex 2.5 Class 7 Maths 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?
Solution:
Distance travelled by Dinesh from A to C
= AB + BC
= 7.5 km +- 12.7 km
= 20.2 km

Distance travelled by Ayub from A to C
= AD + DC
= 9.3 km + 11.8 km = 21.1 km
Since 21.1 km > 20.2 km.
Hence, Ayub travelled more distance.

Ex 2.5 Class 7 Maths 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?
Solution:
Fruits bought by Shyama
= 5 kg 300 g apples + 3 kg 250 g mangoes
= 5.300 kg apples + 3.250 kg mangoes
= 8.550 kg of fruits
Fruits bought by Sarala
= 4 kg 800 g oranges +- 4 kg 150 g bananas
= 4.800 kg oranges + 4.150 kg bananas
= 8.950 kg of fruits
Since 8.950 kg > 8.550 kg
Hence, Sarala bought more fruits.

Ex 2.5 Class 7 Maths Question 9.
How much less is 28 km than 42.6 km?
Solution:
Since 28 km < 42.6 km

Hence, 28 km is less than 42.6 km by 14.6 km.

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

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Exercise 2.6
Ex 2.6 Class 7 Maths 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
Solution:
(i) 0.2 × 6
∵ 2 × 6 = 12 and we have 1 digit right to the decimal point in 0.2.
Thus 0.2 × 6 = 1.2

(ii) 8 × 4.6
∵ 8 × 46 = 368 and there is one digit right to the decimal point in 4.6.
Thus 8 × 4.6 = 36.8

(iii) 2.71 × 5
∵ 271 × 5 = 1355 and there are two digits right to the decimal point in 2.71.
Thus 2.71 × 5 = 13.55

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

(v) 0.05 × 7
∵ 5 × 7 = 35 and there are 2 digits right to the decimal point in 0.05.
Thus 0.05 × 7 = 0.35

(vi) 211.02 × 4
∵ 21102 × 4 = 84408 and there are 2 digits right to the decimal point in 211.02.
Thus 211.02 × 4 = 844.08

(vii) 2 × 0.86
∵ 2 × 86 = 172 and there are 2 digits right to the decimal point in 0.86.
Thus 2 × 0.86 = 1.72

Ex 2.6 Class 7 Maths Question 2.
Find the area of rectangle whose length is 5.7 cm and breadth is 3 cm.
Solution:
Length = 5.7 cm
Breadth = 3 cm
Area of rectangle = length × breadth
= 5.7 × 3 = 17.1 cm2
Hence, the required area =17.1 cm2

Ex 2.6 Class 7 Maths 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
Solution:

Ex 2.6 Class 7 Maths Question 4.
A two-wheeler covers a distance of 55.3 km in one litre of petrol. How much distance will it cover is 10 litres of petrol?
Solution:
Distance covered in 1 litre = 55.3 km
Distance covered in 10 litres = 55.3 × 10 km

Hence, the required distance = 553 km

Ex 2.6 Class 7 Maths 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) 100.01 × 1.1
(x) 100.01 × 1.1
Solution:
(i) 0 2.5 × 0.3
∵ 25 × 3 = 75 and there are 2 digits (1 + 1) right to the decimal points in 2.5 and 0.3.
Thus 2.5 × 0.3 = 0.75

(ii) 0.1 × 51.7
∵ 1 × 517 = 517 and there are two digits (1 + 1) right to the decimal places in 0.1 and 51.7.

(iii) 0.2 × 316.8
∵ 2 × 3168 = 6336 and there are 2 digits (1 + 1) right to the decimal points in 0.2 and 316.8.
Thus 0.2 × 316.8 = 63.36.

(iv) 1.3 × 3.1
∵ 13 × 31 = 403 and there are 2 digits (1 + 1) right to the decimal points in 1.3 and 3.1.
Thus 1.3 × 3.1 – 4.03

(v) 0.5 × 0.05
∵ 5 × 5 = 25 and there are 3 digits (1 + 2) right to the decimal points in 0.5 and 0.05.
Thus 0.5 × 0.05 = 0.025

(vi) 11.2 × 0.15
∵ 112 × 15 = 1680 and there are 3 digits (1 + 2) right to the decimal points in 11.2 and 0.15.
Thus 11.2 × 0.15 = 1.680

(vii) 1.07 × 0.02
∵ 107 × 2 = 214 and there are 4-digits (2 + 2) right to the decimal places is 1.07 × 0.02.
Thus 1.07 × 0.02 = 0.0214

(viii) 10.05 × 1.05
∵ 1005 × 105 = 105525 and there are 4 digits (2 + 2) right to the decimal places in 10.05 × 1.05.
Thus 10.05 × 1.05 = 10.5525

(ix) 101.01 × 0.01
∵ 10101 × 1 = 10101 and there are 4 digits (2 + 2) right to the decimal places in 101.01 and 0.01.
Thus 101.01 × 0.01 = 1.0101

(x) 100.01 × 1.1
∵ 10001 × 11 = 110011 and there are 3 digits (2 + 1) right to the decimal points in 100.01 and 1.1.
Thus 100.01 × 1.1 = 110.011.

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

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

Ex 2.7 Class 7 Maths Question 2.
Find:
(i) 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
Solution:
(i) 4.8 ÷ 10 = 0.48 (Shifting the decimal point to the left by 1 place)
(ii) 52.5 ÷ 10 = 5.25 (Shifting the decimal point to the left by 1 place)
(iii) 0.7 ÷ 10 = 0.07 (Shifting the decimal point to the left by 1 place)
(iv) 33.1 ÷ 10 = 3.31 (Shifting the decimal point to the left by 1 place)
(v) 272.23 ÷ 10 = 27.223 (Shifting the decimal point to the left by 1 place)
(vi) 0.56 4- 10 = 0.056(Shifting the decimal point to the left by 1 place)
(vii) 3.97 ÷ 10 = 0.397(Shifting the decimal point to the left by 1 place)

Ex 2.7 Class 7 Maths 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
Solution:
Solution:
(i) 2.7 ÷ 100 = 0.027(Shifting the decimal point to the left by 2 places)
(ii) 0.3 ÷ 100 = 0.003(Shifting the decimal point to the left by 2 places)

(iii) 0.78 ÷ 100 = 0.0078 (Shifting the decimal point to the left by 2 places)
(iv) 432.6 ÷ 100 = 4.326 (Shifting the decimal point to the left by 2 places)
(v) 23.6 ÷ 100 = 0.236 (Shifting the decimal point to the left by 2 places)
(vi) 98.53 ÷ 100 = 0.9853 (Shifting the decimal point to the left by 2 places)

Ex 2.7 Class 7 Maths Question 4.
Find:
(i) 7.9 ÷ 1000
(ii) 26.3 ÷ 1000
(iii) 38.53 ÷ 1000
(iv) 128.9 ÷ 1000
(v) 0.5 ÷ 1000
Solution:
(i) 7.9 ÷ 1000 = 0.0079 (Shifting the decimal point to the left by 3 places)
(ii) 26.3 ÷ 1000 = 0.0263 (Shifting the decimal point to the left by 3 places)
(iii) 38.53 ÷ 1000 = 0.03853 (Shifting the decimal point to the left by 3 places)
(iv) 128.9 ÷ 1000 = 0.1289 (Shifting the decimal point to the left by 3 places)
(v) 0.5 ÷ 1000 = 0.0005 (Shifting the decimal point to the left by 3 places)

Ex 2.7 Class 7 Maths Question 5.
Find:
(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
Solution:

Ex 2.7 Class 7 Maths 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?
Solution:
2.4 litres of petrol is required to cover 43.2 km distance
∴ 1 litre of petrol will be required to cover 43.2

Hence, the required distance = 18 km

please also read =class 7 chapter 1