# NCERT solutions for class 7 Maths Chapter 2 Fractions And Decimals

#### Solution for Exercise 2.1

Q.1 Solve
i) $$2 - \frac{3}{5}$$
ii) $$4 + \frac{7}{8}$$
iii) $$\frac{3}{5}+\frac{2}{7}$$
iv) $$\frac{9}{11}-\frac{4}{15}$$
v) $$\frac{7}{10}+\frac{2}{5}+\frac{3}{2}$$
vi) $$2\frac{2}{3}+3\frac{1}{2}$$
vii) $$8\frac{1}{2}-3\frac{5}{8}$$

(i) $$2- \frac{3}{5} = \frac{2×5-3}{5}= \frac{7}{5} = 1\frac{2}{5}$$
ii) $$4 + \frac{7}{8} = \frac{8×4+7}{8}= \frac{39}{8}=4\frac{7}{8}$$
iii) $$\frac{3}{5}+\frac{2}{7} = \frac{7×3+5×2}{35}=\frac{21+10}{35}= \frac{31}{35}$$
iv) $$\frac{9}{11}-\frac{4}{15}= \frac{15×9-11×4}{165}= \frac{135-44}{165}= \frac{91}{165}$$
v) $$\frac{7}{10}+ \frac{2}{5}+\frac{3}{2} = \frac{7×1+2×2+3×5}{10}= \frac{7+4+15}{10} =\frac{26}{10} =\frac{26÷2}{10÷2} =\frac{13}{5} =2\frac{3}{5}$$
vi) $$2\frac{2}{3}+3\frac{1}{2} = \frac{8}{3}+\frac{7}{2}= \frac{2×8+3×7}{6}= \frac{16+21}{6}=\frac{37}{6}= 6\frac{1}{6}$$
vii) $$8\frac{1}{2}-3\frac{5}{8} = \frac{17}{2}-\frac{29}{8}= \frac{17×4-29×1}{8}= \frac{68-29}{8} = \frac{39}{8} =4\frac{7}{8}$$

Q.2 Arrange the following in descending order: (i)$$\frac{2}{9}, \frac{2}{3}, \frac{8}{21}$$
(ii) $$\frac{1}{5} ,\frac{3}{7}, \frac{7}{10}$$

(i)$$\frac{2}{9}, \frac{2}{3}, \frac{8}{21}$$
LCM of 9, 3 ,21 = 63 then
$$\frac{2}{9}=\frac{2×7}{9×7}=\frac{14}{63}$$ $$\frac{2}{3}=\frac{2×21}{3×21}=\frac{42}{63}$$ $$\frac{8}{21}=\frac{8×3}{21×3}=\frac{24}{63}$$ Hence $$\frac{42}{63} > \frac{24}{63} > \frac{14}{63}$$
$$\frac{2}{3} > \frac{8}{21} > \frac{2}{9}$$

(ii) $$\frac{1}{5} ,\frac{3}{7}, \frac{7}{10}$$
LCM of 5,7,10 = 70 then
$$\frac{1}{5}=\frac{1×14}{5×14}=\frac{14}{70}$$ $$\frac{3}{7}=\frac{3×10}{7×10}=\frac{21}{70}$$ $$\frac{7}{10}=\frac{7×7}{10×7}=\frac{49}{70}$$ Hence $$\frac{49}{70} > \frac{21}{70} > \frac{14}{70}$$
$$\frac{7}{10} > \frac{3}{7} > \frac{1}{5}$$

Q.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?

Along first row = $$\frac{4}{11}+\frac{9}{11}+\frac{2}{11} =\frac{15}{11}$$
Along second row = $$\frac{3}{11}+\frac{5}{11}+\frac{7}{11}=\frac{15}{11}$$
Along third row = $$\frac{8}{11}+\frac{1}{11}+\frac{6}{11}=\frac{15}{11}$$
Along first column= $$\frac{4}{11}+\frac{3}{11}+\frac{8}{11}=\frac{15}{11}$$
Along second column= $$\frac{9}{11}+\frac{5}{11}+\frac{1}{11}=\frac{15}{11}$$
Along third column= $$\frac{2}{11}+\frac{7}{11}+\frac{6}{11}=\frac{15}{11}$$
Along first diagonal= $$\frac{4}{11}+\frac{5}{11}+\frac{6}{11}=\frac{15}{11}$$
Along second diagonal= $$\frac{2}{11}+\frac{5}{11}+\frac{8}{11}=\frac{15}{11}$$
Since, the sum of all the fraction row wise, column wise and the diagonal wise is same i.e. $$\frac{15}{11}$$. Hence, it is a magic square.

Q.4 A rectangular sheet of paper is $$12\frac{1}{2}$$ cm long and $$10\frac{2}{3}$$cm wide. Find its perimeter.

Length of sheet = $$12\frac{1}{2} = \frac{25}{2}$$
Breath of sheet =$$10\frac{2}{3} = \frac{32}{3}$$
Perimeter = 2× (length +breath)
$$= 2×[\frac{25}{2}+\frac{32}{3}]$$
= $$2× \frac{75+64}{6}= 2 ×\frac{139}{6} = \frac{139}{3}= 43\frac{1}{3}$$

Q.5 Find the perimeter of (i) triangle ABE (ii) the rectangle BCDE in this figure. Whose perimeter is greater?

(i) Perimeter of triangleABE = AB+BE+AE = $$\frac{5}{2}+2\frac{3}{4}+3\frac{3}{5} =\frac{5}{2}+\frac{11}{4}+\frac{18}{5} =\frac{50+55+72}{20}=\frac{177}{20}=8\frac{17}{20}$$
(ii) perimeter of BCDE = 2 ×( length +breath)
= $$2 × [2\frac{3}{4}+\frac{7}{6}] = 2 × [ \frac{11}{4}+ \frac{7}{6}] = 2× \frac{33+14}{12} = 2×\frac{47}{12}= \frac{47}{6} = 7\frac{5}{6}$$
Since $$8\frac{17}{20} > 7\frac{5}{6}$$
Thus perimeter of ?ABE is greater than the perimeter of the rectangle BCDE.

Q.6 Salil wants to put a picture in a frame. The picture is $$7\frac{3}{5}$$cm wide. To fit in the frame, the picture cannot be more than $$7\frac{3}{10}$$ cm wide. How much should the picture be trimmed?

From the question, it is given that,
Picture having a width of =$$7\frac{3}{5}= 38/5$$cm
Frame having a width of $$7\frac{3}{10} = 73/10$$cm
The picture should be trimmed by = [(38/5) – (73/10)]
= $$\frac{76-73}{10}$$cm = $$\frac{3}{10}$$cm
Hence, the required width to be trimmed =3/10 cm.

Q.7 Ritu ate =$$\frac{3}{5}$$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?

Let the whole apple be 1.
part of apple eaten by Ritu=3/5
Therefore part of apple eaten by her brother Somu
=1 - 3/5 = $$\frac{5-3}{5}$$ = 2/5
Since 3/5 >2/5 thus the share of Ritu was larger.
Difference =3/5-2/5 =1/5 part.

Q.8 Michael finished colouring a picture in 7/12 hour. Vaibhav finished colouring the same picture in 3/4 hour. Who worked longer? By what fraction was it longer?

Time taken by Michael = 7/12 hour
Time taken by Vaibhav = 3/4 hour
Multiply by 3 on both nominator and denominator There fore 3/4 hour =$$\frac{9}{12}$$ hour
Therefore 9/12>7/12
Hence time taken by Vaibhav was longer .
Difference = 3/4 - 7/12 = 9/12-7/12 = 2/12=1/6 hour longer

#### Solution for Exercise 2.2

Q.1 Which of the drawings (a) to (d) show:
(i) 2 × (1/5) (ii) 2 × (1/2) (iii) 3 × (2/3) (iv) 3 × (1/4)

(i) 2×(1/5) represents drawing (d)
(ii) 2×(1/2) represents drawing (b)
(iii) 3×(2/3) represents drawing (a)
(iv) 3×(1/4) represents drawing (c)

Q.2 Some pictures (a) to (c) are given below. Tell which of them show:

(i) 3 × (1/5) = (3/5)
(ii) 2 × (1/3) = (2/3)
(iii) 3 × (3/4) = $$2\frac{1}{4}$$

(i) 3×(1/5)= (3/5) represents figure (c)
(ii) 2×(1/3)=(2/3) represents figure (a)
(iii) 3×(3/4)=$$2\frac{1}{4}$$ represents figure (b)

Q.3 Multiply and reduce to lowest form and convert into a mixed fraction:

(i) 7×(3/5)
(ii) 4× (1/3)
(iii) 2×(6/7)
(iv) 5× (2/9)
(v) (2/3) × 4
(vi) (5/2) ×6
(vii) 11× (4/7)
(viii) 20×(4/5)
(ix) 13×(1/3)
( x) 15 ×(3 /5)

(i)7×(3/5) = 21/5 = $$4\frac{1}{5}$$
(ii) 4× (1/3) = 4/3 = $$1\frac{1}{3}$$
(iii) 2×(6/7) =12/7 =$$1\frac{5}{7}$$
(iv) 5× (2/9) = 10/9 = $$1\frac{1}{9}$$
(v) (2/3) × 4 = 8/3 =$$2\frac{2}{3}$$
(vi) (5/2) ×6 = 30/2 =$$\frac{30÷2}{2÷2} = 15/1=15$$
(vii) 11× (4/7) = 44/7 =$$6\frac{2}{7}$$
(viii) 20×(4/5) = (20/1) × (4/5)= (80/5)= 16
(ix) 13×(1/3) = 13/3 = $$4\frac{1}{3}$$
(x) 15 ×(3 /5) = 45 /5 = 15

(i) 1/2 of the circles in box (a)
(ii) 2/3 of the circles in box (b)
(iii) 3/5 of the circles in box (c)

(i) 1/2 × 12 = 12/2 =6 circles

(ii) 2/3 × 9 = 18/3 = 6 triangles
(iii) 3/5 × 15 =45/5 =9 squares

Q.6 Multiply and express as a mixed fraction.
(a) $$3× 5\frac{1}{5}$$
(b) $$5 × 6\frac{3}{4}$$
(c) $$7× 2\frac{1}{4}$$
(d) $$4× 6\frac{1}{3}$$
(e) $$3\frac{1}{4}$$× 6
(f) $$3\frac{2}{5}$$ × 8

(a) $$3× 5\frac{1}{5} = 3×26/5 =78/5 = 15\frac{3}{5}$$
(b) $$5 × 6\frac{3}{4} = 5×27/4= 135/4 = 33\frac{3}{4}$$
(c) $$7× 2\frac{1}{4} =7×9/4= 63/4 = 15\frac{3}{4}$$
(d) $$4× 6\frac{1}{3} = 4×19/3 = 76/3 = 25\frac{1}{3}$$
(e) $$3\frac{1}{4} × 6 = 13/4 ×6 = 78/4 = 39/2 = 19\frac{1}{2}$$
(f) $$3\frac{2}{5}× 8 = 17/5 ×8 = 136/5= 27\frac{1}{5}$$

Q.5 Find:
(a) 1/2 of (i) 24 (ii) 46
(b) 2/3 of (i) 18 (ii) 27
(c) 3/4 of (i) 16 (ii) 36
(d) 4/5 of (i) 20 (ii) 35

(a) (i) 24
We have, = 1/2 × 24 = 24/2 = 12
(ii) 46
We have, = 1/2× 46 =23
(b) (i) 18
We have,= 2/3 × 18 = 2 × 6 = 12
(ii) 27
We have, = 2/3 × 27= 2 × 9 =18
(c) (i) 16
We have,= 3/4 × 16= 3 × 4= 12
(ii) 36
We have= 3/4 × 36= 3 × 9= 27
(d) (i) 20
We have,= 4/5 × 20 = 4 × 4 = 16
(ii) 35
We have, = 4/5 × 35 = 4 × 7 = 28

Q.7 find :
(a) 1/2 of (i) $$2\frac{3}{4} (ii) 4\frac{2}{9}$$
(b) 5/8 of (i) $$3\frac{5}{6} (ii) 9\frac{2}{3}$$

(a)(i) $$1/2× 2\frac{3}{4} = 1/2×11/4= 11/8=1\frac{3}{8}$$
(ii) $$1/2× 4\frac{2}{9}= 1/2 ×38/9 =38/18=19/9=2\frac{1}{9}$$
(b) (i) $$5/8 × \frac{5}{6} = 5/8×23/6=115/48 =2\frac{19}{48}$$
(ii) $$\frac{5}{8}×9\frac{2}{3}= 5/8×29/3=145/24=6\frac{1}{24}$$

Q.8 Vidya and Pratap went for a picnic. Their mother gave them a water bottle that contained 5 liters water. Vidya consumed 2/5 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?

(i) Water consumed by Vidya = 2/5 of 5 litres
2/5×5=10/5=2litre
Water consumed by Patap = 5 litres – 2 litres
= 3 litres
(ii) Fraction of water consumed by Pratap = 3/5 litres

#### Solution for Exercise 2.3

Q.1 Find:
(i) 1/4 of (a) 1/4 (b) 3/5 (c) 4/3
(ii) 1/7 of (a) 2/9 (b) 6/5 (c) 3/10

(i) (a) 1/4×1/4=1/8
(b) 1/4×3/5=3/20
(c) 1/4×4/3=4/12=1/3
(ii) (a) 1/7× 2/9=2/63
(b) 1/7× 6/5 = 6/35
(c) 1/7 × 3/10 = 3/70

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

(i) 2/3×8/3=(2×8)/(3×3)=16/9= $$1\frac{7}{9}$$
(ii) 2/7×7/9=14/63=$$\frac{14÷7}{63÷7}$$=2/9
(iii) 3/8×6/4 = 18/32 =$$\frac{18÷2}{32÷2}$$=9/16
(iv) 9/5×3/5=27/25=$$1\frac{2}{25}$$
(v)1/3×15/8=15/24=$$\frac{15÷3}{24÷3}$$=5/8
(vi) 11/2×3/10=33/20=$$1\frac{13}{20}$$
(vii) 4/5×12/7 = 48/35=$$1\frac{13}{35}$$

Q.3 Multiply the following fractions:
(i) (2/5) × $$5\frac{1}{4}$$
(ii) $$6\frac{2}{5}$$ ×7/9
(iii) 3/2× $$5\frac{1}{3}$$
(iv) 5/6×$$2\frac{3}{7}$$
(v) $$3\frac{2}{5}$$ ×4/7
(vi) $$2\frac{3}{5}$$× 3
(vii)$$3\frac{4}{7}$$ × 3/5

(i) 2/5×21/4=42/20= $$\frac{42÷2}{20÷2}$$= 21/10
=$$2\frac{1}{10}$$
(ii) 32/5×7/9=224/45= $$4\frac{44}{45}$$
(iii) 3/2 ×16/3 =48/6=8
(iv) 5/6×17/7=85/42= $$2\frac{1}{42}$$
(v) 17/5×4/7=68/35= $$1\frac{33}{35}$$
(vi)13/5×3=39/5= $$7\frac{4}{5}$$
(vii) 25/7×3/5 ÷75/35=$$\frac{75÷5}{35÷5}=15/7=2\frac{1}{7}$$

Q.4 Which is greater:
(i) (2/7) of (3/4) or (3/5) of (5/8)
(ii) (1/2) of (6/7) or (2/3) of (3/7)

(i) 2/7×3/4=6/28= $$\frac{6÷2}{28÷2}$$=3/14
3/5×5/8=15/40=$$\frac{15÷5}{40÷5}$$=3/8
Numerator are same and 14 >8
Hence 3/8>3/14
(ii) 1/2×6/7=6/14=$$\frac{6÷2}{14÷2}$$=3/7
2/3×3/7=6/21=$$\frac{6÷3}{21÷3}$$=2/7
Hence 3/7>2/7

Q.5 Saili plants 4 saplings, in a row, in her garden. The distance between two adjacent saplings is 3/4 m. Find the distance between the first and the last sapling.

The distance between two adjacent saplings =3/4 m
Number of saplings planted by Saili in a row = 4
Then, number of gap in saplings = 3/4 × 4=12/4=3
Therefore The distance between the first and the last saplings = 3 × 3/4= (9/4) m = $$2\frac{1}{4}$$

Q.6 Lipika reads a book for $$1\frac{3}{4}$$hours everyday. She reads the entire book in 6 days. How many hours in all were required by her to read the book?

In 1 day Lipika needs $$1\frac{3}{4}$$ hours
In 6 days Lipika will need 6×$$1\frac{3}{4}$$hours=6×7/4
42/4=21/2hours = $$10\frac{1}{2}$$ hours

Q.7 A car runs 16 km using 1 litre of petrol. How much distance will it cover using $$2\frac{3}{4}$$ litres of petrol.

In 1 litre of petrol, the car covers 16 km distance In $$2\frac{3}{4}$$ litres of petrol
the car will cover $$2\frac{3}{4}$$ × 16 km distance=11/4×16=44 km

Q.8 (a) (i) provide the number in the box [ ], such that (2/3) × [ ] = (10/30)
(ii) The simplest form of the number obtained in [ ] is
(b) (i) provide the number in the box [ ], such that (3/5) × [ ] = (24/75)
(ii) The simplest form of the number obtained in [ ] is

#### Solution for Exercise 2.4

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

(i) We have,
= 12 × reciprocal of 3/4
= 12 × (4/3)= 4 × 4= 16 (ii) We have,
= 14 × reciprocal of (5/6)
= 14 × (6/5)= 84/5 =$$16\frac{4}{5}$$
(iii) We have,
= 8 × reciprocal of (7/3)
= 8 × (3/7)= (24/7) = $$3\frac{3}{7}$$
(iv) We have,
= 4 × reciprocal of (8/3)
= 4 × (3/8)= 1 × (3/2)= 3/2 =$$1\frac{1}{2}$$
(v)we have,
3÷7/3= 3× reciprocal of 7/3 =3×3/7=9/7=$$1\frac{2}{7}$$
(vi)we have,
5÷25/7=5×reciprocal of 25/7 =5×7/25 = 7/5 =$$1\frac{2}{5}$$

Q.2 Find the reciprocal of each of the following fractions. Classify the reciprocals as proper fractions, improper fractions and whole numbers.
(i) 3/7 (ii) 5/8 (iii) 9/7 (iv) 6/5 (v) 12/7
(vi) 1/8 (vii) 1/11

(i) Reciprocal of 3/7 = 7/3 (ii) Reciprocal of 5/8=8/5 (iii) Reciprocal of 9/7=7/9 (iv) Reciprocal of 6/5=5/6 (v) Reciprocal of 12/7=7/12 (vi) Reciprocal of 1/8=8 (vii) Reciprocal of 1/11=11

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

(i) We have,
= (7/3) × reciprocal of 2= (7/3) × (1/2)= 7/6= $$1\frac{1}{6}$$
(ii) We have,
= (4/9) × reciprocal of 5= (4/9) × (1/5)= 4/45 (iii) we have ,
= (6/13) × reciprocal of 7= (6/13) × (1/7)= 6/91
(iv) we have,
=(13/3) × reciprocal of 3= (13/3) × (1/3)= 13/9= $$1\frac{4}{9}$$
(v) we have ,
= (7/2) × reciprocal of 4= (7/2) × (1/4)= 7/8 (vi) we have,
= (31/7) × reciprocal of 7=(31/7) ×(1/7)= 31/49

Q.5 Find:
(i) (2/5) ÷ (1/2)
(ii) (4/9) ÷ (2/3)
(iii) (3/7) ÷ (8/7)
(iv) $$2\frac{1}{3}$$÷3/5
(v) $$3\frac{1}{2}$$÷8/3
(vi) 2/5 ÷$$1\frac{1}{2}$$
(vii) $$3\frac{1}{5} ÷1\frac{2}{3}$$
(viii)$$2\frac{1}{5} ÷1\frac{1}{5}$$

(i) We have,
= (2/5) × reciprocal of 1/2= (2/5) × (2/1)= 4/5
(ii) We have,
= (4/9) × reciprocal of (2/3)= (4/9) × (3/2)=12/18= 2/3
(iii) We have,
= (3/7) × reciprocal of (8/7)= (3/7) × (7/8)=21/56= 3/8
(iv) $$2\frac{1}{3}$$=7/3
= (7/3) × reciprocal of (3/5)= (7/3) × (5/3)= 35/9
(v) $$3\frac{1}{2}$$ =7/2
= (7/2) × reciprocal of (8/3)= (7/2) × (3/8)= 21/16
(vi) 2/5 ÷$$1\frac{1}{2}$$=(2/5) × reciprocal of (3/2)
= (2/5) × (2/3)= 4/15
(vii) $$3\frac{1}{5} ÷1\frac{2}{3}$$= (16/5) × reciprocal of (5/3)
= (16/5) × (3/5)= 48/25
(viii)$$2\frac{1}{5} ÷1\frac{1}{5}$$= (11/5) × reciprocal of (6/5)
= (11/5) × (5/6)= 11/6 = $$1\frac{5}{6}$$.

#### Solution for Exercise 2.5

Q.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

(i) 0.5 or 0.05
Comparing the tenths place, we get 5 > 0
Therefore 0.5 > 0.05
(ii) 0.7 or 0.5 Comparing the tenths place, we get 7 > 5 Therefore 0.7 > 0.5
(iii) 7 or 0.7 Comparing the one’s place, we get 7 > 0 Therefore 7 > 0.7
(iv) 1.37 or 1.49
Comparing the tenths place, we get 3 < 4
Therefore 1.37 < 1.49 or 1.49 > 1.37
(v) 2.03 or 2.30
Comparing the tenths place, we get 0 < 3br> Therefore 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
Therefore .80 < 0.88 or 0.88 > 0.80

Q.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

(i)7 paise =7/100 rupees = 0.07 rupees
(ii) 7 rupees 7 paise = 7 rupees + 7/100 rupees = 7.07 rupees
(iii) 77 rupees 77 paise = 77 rupees + 77/100 rupees = 77.77 rupees
(iv) 50 paise = 50/100 rupees = 0.50 rupees
(v) 235 paise = 235/100 rupees = 2.35 rupees

Q.3 (i) Express 5 cm in metre and kilometre.
(ii) Express 35 mm in cm, m and km.

(i) 1 metre = 100 cm
1 kilometre = 1000 metre = 100 × 1000 cm = 100000 cm
Therefore 5 cm = 5/100 metre = 0.05 metre
5 cm = 5/100000 km = 0.00005 km
Hence, 5 cm = 0.05 m and 0.00005 km

(ii) 1 cm = 10 mm and 1 km = 100000 cm
Therefore 35 mm = 35/10 cm = 3.5 cm,
35 mm = 35/1000 m = 0.035 m
35 mm = 35/1000000 km = 0.000035 km
Hence, 35 mm = 3.5 cm, 0.035 m and 0.000035 km.

Q.4 Express in kg:
(i) 200 g
(ii) 3470 g
(iii) 4 kg 8 g

(i) 200g = 200/1000 kg [since 1 kg = 1000g] = 0.2 kg
(ii) 3470 g = 3470/10000 = 3.47 kg [since 1 kg = 1000 g] (iii) 4 kg 8 g = 4 kg + 8/10000kg [since 1 kg = 1000 g] = 4 kg + 0.008 kg = 4.008 kg

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

(i) 20.03 = 2 × 10 + 0 × 1 + 0 × 1/10 + 3 × 1/100
(ii) 2.03 = 2 × 1 + 0 × 1/10 + 3 × 1/100
(iii) 200.03 = 2 × 100 + 0 × 10 + 0 × 1 + 0 × 1/10 +3 × 1/100
(iv) 2.034 = 2 × 1 + 0 × 1/10 + 3 × 1/10 + 4 × 1/1000

Q.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

(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 = 2/10 = 0.2 i.e. tenths
(iv) Place value of 2 in 9.42 = 2/100 = 0.02 i.e. hundredths
(v) Place value of 2 in 63.352 = 2/1000 = 0.002 i.e. thousandths.

Q.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?

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.

Q.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?

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.

Q.9 How much less is 28 km than 42.6 km?

Since 28 km < 42.6 km
42.6 km - 28.0 km = 14.6 km
Hence, 28 km is less than 42.6 km by 14.6 km.

#### Solution for Exercise 2.6

Q.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

(i) 0.2 × 6
$$\therefore$$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= 36.8
(iii) 2.71× 5 = 13.55
(iv) 20.1 ×4 = 80.4
(v) 0.05 × 7 = 0.35
(vi) 211.02 × 4 = 844.05
(vi) 2 × 0.86 = 1.72

Q.2 Find the area of rectangle whose length is 5.7 cm and breadth is 3 cm.

Length = 5.7 cm
Area of rectangle = length × breadth
= 5.7 × 3 = 17.1 cm²
Hence, the required area =17.1 cm²

Q.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

(i) 1.3 × 10 = 13/10 ×10 =13
(ii) 36.8 ×10 = 368/10×10= 368
(iii) 153.7 × 10 = 1537/10 ×10 = 1537
(iv) 168.07×10 = 16807/100 ×10= 16807/10 = 1680.7
(v) 31.1×100 = 311/10×100= 311×10 =3110
(vi) 156.1×100 = 1561/10×100= 1561×10 =15610
(vii) 3.62 × 100 = 362/100×100 = 362
(viii) 43.07×100 = 4307/100×100= 4307
(ix) 0.5×10 = 5/10×10= 5
(x) 0.08× 10 = 8/100 ×10= 8/10 = 0.8
(xi) 0.9 × 100 = 9/10×100= 9×10=90
(xii) 0.03 × 1000 =3/100×1000 =3/10= 0.3

Q.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?

Distance covered in 1 litre = 55.3 km
Distance covered in 10 litres = 55.3 × 10 km= 553/10×10=553
Hence, the required distance = 553 km

Q.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

(i) 0 2.5 × 0.3
$$\therefore$$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
$$\therefore$$ 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
$$\therefore$$ 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
$$\therefore$$ 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
$$\therefore$$ 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
$$\therefore$$ 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
$$\therefore$$ 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
$$\therefore$$ 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
$$\therefore$$ 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
$$\therefore$$ 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.

#### Solution for Exercise 2.7

Q.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

(i) 0.4 ÷ 2= 4/10÷2 = 4/10× 1/2 = 2/10= 0.2
(ii) 0.35 ÷ 5 = 35/100 ÷ 5 = 35/100×1/5 = 7/100 = 0.07
(iii) 2.48 ÷ 4 = 248/100÷4=248/100×1/4 =62/100=0.62
(iv) 65.4 ÷ 6 = 654/10÷6=654/10×1/6=109/10=10.9
(v) 651.2÷4 = 6512/10÷ 4 =6512/10×1/4= 1628/10=162.8
(vi) 14.49÷7= 1449/100÷7=1449/100×1/7=207/100=2.07
(vii) 3.96÷4 = 396/100÷4=396/100×1/4=99/100=0.99
(viii) 0.80÷5=80/100÷5=80/100×1/5=16/100=0.16

Q.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

(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)

Q.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

(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)

Q.4 Find:
(i) 7.9 ÷ 1000
(ii) 26.3 ÷ 1000
(iii) 38.53 ÷ 1000
(iv) 128.9 ÷ 1000
(v) 0.5 ÷ 1000

(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)

Q.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

(i) 7 ÷ 3.5 = $$\frac{7}{3.5}=\frac{70}{35}=2$$
(ii) 36÷0.2 = $$\frac{36}{0.2}=\frac{360}{2} = 180$$
(iii) 3.25÷0.5 = $$\frac{3.25}{0.5}=\frac{325×10}{5×100}=\frac{65}{10}=6.5$$
(iv) 30.94÷0.7= $$\frac{30.94}{0.7}=\frac{3094×10}{7×100}=\frac{442}{10}=44.2$$
(v) 0.5÷0.25= $$\frac{0.5}{0.25}=\frac{5×100}{25×10}=\frac{500}{250}=2$$ (vi)7.75÷0.25=$$\frac{7.75}{0.25}=\frac{775}{25}=31$$
(vii) 76.5÷0.15= $$\frac{76.5}{0.15}=\frac{765×100}{15×10}=\frac{510}{1}= 510$$
(viii) 37.8 ÷1.4= $$\frac{37.8}{1.4}=\frac{378}{14}=27$$
(ix) 2.73 ÷1.3 = $$\frac{2.73}{1.3}=\frac{273×10}{13×100}=\frac{21}{10}= 2.1$$

Q.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?

$$\frac{43.2}{2.4}=\frac{432}{24}=18 km$$