NCERT Solutions for Chapter 2 Fractions and Decimals Class 7 Maths
Book Solutions1
Solve:
(i)2-3/5
(ii) 4 + 7/8
(iii)3/5+2/7
(iv) 9/11−4/15
(v)7/10+2/5+3/2
(vi)2.2/3+3.1/2
(vii)8.1/2−3.5/8
Answer
(i) 2−3/5 = (10−3)/5=7/5 = 1.2/5
(ii)4+7/8 = (32+7)/8=39/8= 4.7/8
(iii)3/5+2/7 = (21+10)/35 = 31/35
(iv)9/11−4/15 = (135−44)/165 =91/165
(v) 7/10+2/5+3/2 = (7+4+15)/10 = 26/10=13/5 = 2.3/5
(vi) 2.2/3+3.1/2 = 8/3+7/2 = (16+21)/6 = 37/6 = 6.1/6
(vii)8.1/2−3.5/8 = 17/2−29/8 = (68−29)/8 = 39/8 = 4.7/8
Exercise 2.1
Page Number 31
2
Arrange the following in descending order:
(i) 2/9,2/3,8/21
(ii) 1/5,3/7,7/10
Answer
(i) 2/9,2/3,8/21⇒ 14/63,42/63,24/63 [Converting into like fractions]
⇒ 42/63>24/63>14/63 [Arranging in descending order]
Therefore, 2/3>8/21>2/9
(ii) 1/5,3/7,7/10 ⇒14/70,30/70,49/70 [Converting into like fractions]
⇒ 49/70>30/70>14/70 [Arranging in descending order]
Therefore, 7/10>3/7>1/5
Exercise 2.1
Page Number 31
3
In a “magic square”, the sum of the numbers in each row, in each column and along the diagonals is the same. Is this a magic square?
|
4/11 |
9/11 |
2/11 |
|
3/11 |
5/11 |
7/11 |
|
8/11 |
11/11 |
6/11 |
(Along the first row 4/11+9/11+2/11=15/11)(Along the first row 4/11+9/11+2/11=15/11)
Answer
Sum of first row = 4/11+9/11+2/11=15/11 [Given]
Sum of second row = 3/11+5/11+7/11=(3+5+7)/11=15/11
Sum of third row = 8/11+1/11+6/11=(8+1+6)/11=15/11
Sum of first column = 4/11+3/11+8/11=(4+3+8)/11=15/11
Sum of second column = 9/11+5/11+1/11=(9+5+1)/11=15/11
Sum of third column = 2/11+7/11+6/11=(2+7+6)/11=15/11
Sum of first diagonal (left to right) = 4/11+5/11+6/11=(4+5+6)/11=15/11
Sum of second diagonal (left to right) = 2/11+5/11+8/11=(2+5+8)/11=15/11
Since, the sum of fractions in each row, in each column and along the diagonals are same, therefore, it s a magic square.
Exercise 2.1
Page Number 31
4
A rectangular sheet of paper is 12.1/2 cm long and 10.2/3 cm wide. Find its perimeter.
Answer
The sheet of paper is in rectangular form.
Length of sheet = 12.1/2 cm and Breadth of sheet = 10.2/3 cm
Perimeter of rectangle = 2 (length + breadth)
= 2(12.1/2+10.2/3)= 2(25/2+32/3)
= 2{(25×3+32×2)/6} = 2{(75+64)/6}
= 2×139/6 = 139/3=46.1/3 cm.
Thus, the perimeter of the rectangular sheet is 46.1/3 cm.
Exercise 2.1
Page Number 31
5
Find the perimeter of (i) ΔΔABE, (ii) the rectangle BCDE in this figure. Whose perimeter is greater?
Answer
(i) In ΔABE, AB = 5/2 cm, BE = 2.3/4 cm, AE = 3.3/5 cm
The perimeter of ΔABE = AB + BE + AE
= 5/2+2.3/4+3.3/5 = 5/2+11/4+18/5
= (50+55+72)/20 = 177/20 = 8.17/20 cm
Thus, the perimeter of ΔABEΔABE is 8.17/20 cm.
(ii) In rectangle BCDE, BE = 2.3/4cm, ED = 7/6 cm
Perimeter of rectangle = 2 (length + breadth)
= 2(2.3/4+7/6) = 2(11/4+7/6)
= 2{(33+14)/12} = 47/6 = 7.5/6 cm
Thus, the perimeter of rectangle BCDE is 7.5/6 cm.
Comparing the perimeter of triangle and that of rectangle,
8.17/20 cm >7.5/6 cm
Therefore, the perimeter of triangle ABE is greater than that of rectangle BCDE.
Exercise 2.1
Page Number 31
6
Salil wants to put a picture in a frame. The picture is 7.3/5 cm wide. To fit in the frame the picture cannot be more than 7.3/10 cm wide. How much should the picture be trimmed?
Answer
Given: The width of the picture = 7.3/5cm and the width of picture frame = 7.3/10 cm
Therefore, the picture should be trimmed = 7.3/5−7.3/10 = 38/5 = 73/10
= (76−73)/10 = 3/10 cm
Thus, the picture should be trimmed by 3/10 cm.
Exercise 2.1
Page Number 31
7
Ritu ate 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?
Answer
The part of an apple eaten by Ritu = 3/5
The part of an apple eaten by Somu = 1−3/5=(5−3)/5=2/5
Comparing the parts of apple eaten by both Ritu and Somu 3/5>2/5
Larger share will be more by 3/5−2/5=1/5 part.
Thus, Ritu’s part is 1/5 more than Somu’s part.
Exercise 2.1
Page Number 32
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?
Answer
Time taken by Michael to colour the picture = 7/12 hour
Time taken by Vaibhav to colour the picture = 3/4 hour
Converting both fractions in like fractions, 7/12 and (3×3)/(4×3)=9/12
Here, 7/12<9/12 ⇒7/12<3/4
Thus, Vaibhav worked longer time.
Vaibhav worked longer time by 3/4−7/12=(9−7)/12=2/12=1/6 hour.
Thus, Vaibhav took 1/6 hour more than Michael.
Exercise 2.1
Page Number 32
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
(a) 
(b) 
(c) 
(d) 
Answer
(i) – (d) Since 2×1/5=1/5+1/5
(ii) – (b) Since 2×1/2=1/2+1/2
(iii) – (a) Since 3×2/3=2/3+2/3+2/3
(iv) – (c) Since 3×1/4=1/4+1/4+1/4
Exercise 2.2
Page Number 36
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= 2.1/4
Answer
(i) – (c) Since 3×1/5=1/5+1/5+1/5
(ii) – (a) Since 2×1/3=1/3+1/3
(iii) – (b) Since 3×3/4=3/4+3/4+3/4
Exercise 2.2
Page Number 36
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
Answer
(i) 7×3/5 = (7×3)/5 = 21/5=4.1/5
(ii) 4×1/3 = (4×1)/3 = 4/3=1.1/3
(iii) 2×6/7 = (2×6)/7 = 12/7=1.5/7
(iv) 5×2/9 = (5×2)/9 = 10/9=1.1/9
(v) 2/3×4 = (2×4)/3 = 8/3 = 2.2/3
(vi) 5/2×6 = 5×3 = 15
(vii) 11×4/7 = (11×4)/7 = 44/7 = 6.2/7
(viii) 20×4/5 = 4 x 4 = 16
(ix) 13×1/3 = (13×1)/3 = 13/3=4.1/3
(x) 15×3/5 = 3 x 3 = 9
Exercise 2.2
Page Number 36
4
Shade:
(i) 2/3 of the triangles in box (a)
(ii) 3/5 of the squares in box (b)
(i) 1/2 of the circles in box (c)
Answer
(i) 2/3 of 9 triangles = 2/3×9 = 2 x 3 = 6 triangles
(ii) 3/5 of 15 squares = 3/5×15 = 3 x 3 = 9 squares
(iii) 1/2 of 12 circles = 1/2×12 = 6 circles
Exercise 2.2
Page Number 37
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
Answer
(a) 3×5.1/5=3×26/5=(3×26)/5=78/5=15.3/5
(b) 5×6.3/4=5×27/4=(5×27)/4=135/4=33.3/4
(c) 7×2.1/4=7×9/4=(7×9)/4=63/4=15.3/4
(d) 4×6.1/3=4×19/3=(4×19)/3=76/3=25.1/3
(e) 3.1/4×6=13/4×6=(13×3)/2=39/2=19.1/2
(f) 3.2/5×8=17/5×8=(17×8)/5=136/5=27.1/5
Exercise 2.2
Page Number 37
6
Multiply and express as a mixed fraction:
(a) 3×5.1/5
(b) 5×6.3/4
(c) 7×2.1/4
(d) 4×6.1/3
(e) 3.1/4×6
(f) 3.2/5×8
Answer
(a) 3×5.1/5=3×26/5=(3×26)/5=78/5=15.3/5
(b) 5×6.3/4=5×27/4=(5×27)/4=135/4=33.3/4
(c) 7×2.1/4=7×9/4=(7×9)/4=63/4=15.3/4
(d) 4×6.1/3=4×19/3=(4×19)/3=76/3=25.1/3
(e) 3.1/4×6=13/4×6=(13×3)/2=39/2=19.1/2
(f) 3.2/5×8=17/5×8=(17×8)/5=136/5=27.1/5
Exercise 2.2
Page Number 37
7
Find:
(a) 1/2 of (i) 2.3/4 (ii) 4.2/9
(b) 5/8 of (i) 3.5/6 (ii) 9.2/3
Answer
(a)
(i) 1/2 of 2.3/4=1/2×11/4=11/8=1.3/8
(ii) 1/2 of 4.2/9=1/2×38/9=19/9=2.1/9
(b)
(i) 5/8 of 3.5/6=5/8×23/6=115/48=2.19/48
(ii) 5/8 of 9.2/3=5/8×29/3=145/24=6.1/24
Exercise 2.2
Page Number 37
8
Vidya and Pratap went for a picnic. Their mother gave them a water bottle that contained 5 litres of 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?
Answer
Total quantity of water in bottle = 5 litres
(i) Vidya consumed = 2/5 of 5 litres = 2/5×5 = 2 litres
Thus, Vidya drank 2 litres water from the bottle.
(ii) Pratap consumed = (1−2/5) part of bottle = 5−2/5=3/5 part of bottle
Pratap consumed 3/5 of 5 litres water = 3/5×5 = 3 litres
Thus, Pratap drank 3/5 part of the total quantity of water.
Exercise 2.2
Page Number 37
1
Find:
(i) 1/4 of
(a) 1/4
(b) 3/4
(c) 4/3
(ii)1/7 of
(a) 2/9
(b) 6/5
(c) 3/10
Answer
(i) (a) 1/4 of 1/4 = 1/4×1/4=(1×1)/(4×4)=1/16
(b) 1/4 of 3/4 = 1/4×3/4=(1×3)/(4×4)=3/16
(c) 1/4 of 4/3 = 1/4×4/3=(1×4)/(4×3)=1/3
(ii) (a) 1/7 of 2/9 = 1/7×2/9=(1×2)/(7×9)=2/63
(b) 1/7 of 2/9 = 1/7×6/5=(1×6)/(7×5)=6/35
(c) 1/7 of 3/10 = 1/7×3/10=(1×3)/(7×10)=3/70
Exercise 2.3
Page Number 41
2
Multiply and reduce to lowest form (if possible):
(i) 2/3×2.2/3
(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
(vii) 4/5×12/7
Answer
(i) 2/3×2.2/3=2/3×8/3=(2×8)/(3×3)=16/9=1.7/9
(ii) 2/7×7/9=(2×7)/(7×9)=2/9
(iii) 3/8×6/4=(3×6)/(8×4)=(3×3)/(8×2)=9/16
(iv) 9/5×3/5=(9×3)/(5×5)=27/25=1.2/25
(v) 1/3×15/8=(1×15)/(3×8)=(1×5)/(1×8)=5/8
(vi) 11/2×3/10=(11×3)/(2×10)=33/20=1.3/20
(vii) 4/5×12/7=(4×12)/(5×7)=48/35=1.13/35
Exercise 2.3
Page Number 42
3
Multiply the following fractions:
(i) 2/5×5.1/4
(ii) 6.2/5×7/9
(iii) 3/2×5.1/3
(iv) 5/6×2.3/7
(v) 3.2/5×4/7
(vi) 2.3/5×3
(vii) 3.4/7×3/5
Answer
(i) 2/5×5.1/4=2/5×21/4=(2×21)/(5×4)=(1×21)/(5×2)=21/10 = 2.1/10
(ii) 6.2/5×7/9=32/5×7/9=(32×7)/(5×9)=224/45=4.44/45
(iii) 3/2×5.1/3=3/2×16/3=48/6=8
(iv) 5/6×2.3/7=5/6×17/7=85/42=2.1/42
(v) 3.2/5×4/7=17/7×4/7=68/35=1.33/35
(vi) 2.3/5×3=13/5×3/1=(13×3)/(5×1)=39/5=7.4/5
(vii) 3.4/7×3/5=25/7×3/5=(5×3)/(7×1)=15/7=2.1/7
Exercise 2.3
Page Number 42
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
Answer
(i) 2/7 of 3/4 or 3/5 of 5/8 ⇒ 2/7 x 3/4 or 3/5 x 5/8
⇒ 3/14 or 3/8 ⇒ 3/14<3/8
Thus, 3/5 of 5/8 is greater.
(ii) 1/2 of 6/7 or 2/3 of 3/7 ⇒ 1/2 x 6/7 or 2/3 x 3/7
⇒ 3/7 or 2/7 ⇒ 3/7>2/7
Thus, 1/2 of 6/7 is greater.
Exercise 2.3
Page Number 42
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.
Answer
The distance between two adjacent saplings = 3/4 m
Saili planted 4 saplings in a row, then number of gap in saplings
Therefore, the distance between the first and the last saplings = 3×3/4 = 9/4m = 2.1/4 m
Thus the distance between the first and the last saplings is 2.1/4 m.
Exercise 2.3
Page Number 42
6
Lipika reads a book for 1.3/4 hours everyday. She reads the entire book in 6 days. How many hours in all were required by her to read the book?
Answer
Time taken by Lipika to read a book = 1.3/4 hours.
She reads entire book in 6 days.
Now, total hours taken by her to read the entire book = 1.3/4×6 = 7/4×6=21/2=10.1/2 hours
Thus, 10 hours were required by her to read the book.
Exercise 2.3
Page Number 42
7
A car runs 16 km using 1 litre of petrol. How much distance will it cover using 2.3/4 litres of petrol?
Answer
In 1 litre of pertrol, car covers the distance = 16 km
In 2.3/4 litres of petrol, car covers the distance = 2.3/4 of 16 km = 11/4×16 = 44 km
Thus, car will cover 44 km distance.
Exercise 2.3
Page Number 42
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 __________.
Answer
(a) (i) 2/3×5/10=10/30
(ii) The simplest form of 5/10 is 1/2
(b) (i) 3/5×8/15=24/75
(ii) The simplest form of 8/15 is 8/15.
Exercise 2.3
Page Number 42
1
Find:
(i) 12÷3/4
(ii) 14÷5/6
(iii) 8÷7/3
(iv) 4÷8/3
(v) 3÷2.1/3
(vi) 5÷3.4/7
Answer
(i) 12÷3/4 = 12×4/3 = 16
(ii)14÷5/6 = 14×6/5=84/5=16.4/5
(iii) 8÷7/3 = 8×3/7=24/7=3.3/7
(iv) 4÷8/3 = 4×3/8=3/2=1.1/2
(v) 3÷2.1/3 = 3÷7/3=3×3/7=9/7=1.2/7
(vi) 5÷3.4/7 = 5÷25/7=5×7/25=7/5=1.2/5
Exercise 2.4
Page Number 46
2
Find the reciprocal of each of the following fractions. Classify the reciprocals as proper fraction, 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
Answer
(i) Reciprocal of 3/7 = 7/3 → Improper fraction
(ii) Reciprocal of 5/8 = 8/5 → Improper fraction
(iii) Reciprocal of 9/7 = 7/9 → Proper fraction
(iv) Reciprocal of 6/5 = 5/6 →Proper fraction
(v)Reciprocal of 12/7 = 7/12 → Proper fraction
(vi) Reciprocal of 1/8 = 8 → Whole number
(vi) Reciprocal of 1/11 = 11 → Whole number
Exercise 2.4
Page Number 46
3
Find:
(i) 7/3÷2
(ii) 4/9÷5
(iii) 6/13÷7
(iv) 4.1/3÷3
(v) 3.1/2÷4
(vi) 4.3/7÷7
Answer
(i) 7/3÷2=7/3×1/2=(7×1)/(3×2)=7/6=1.1/6
(ii) 4/9÷5=4/9×1/5=(4×1)/(9×5)=4/45
(iii) 6/13÷7=6/13×1/7=(6×1)/(13×7)=6/91
(iv) 4.1/3÷3=13/3÷3=13/3×1/3=13/9=1.4/9
(v) 3.1/2÷4=7/2÷4=7/2×1/4=7/8
(vi) 4.3/7÷7=31/7÷7=31/7×1/7=31/49
Exercise 2.4
Page Number 46
4
Find:
(i) 2/5÷1/2
(ii) 4/9÷2/3
(iii) 3/7÷8/7
(iv) 2.1/3÷3/5
(v) 3.1/2÷8/3
(vi) 2/5÷1.1/2
(vii) 3.1/5÷1.2/3
(viii) 2.1/5÷1.1/5
Answer
(i) 2/5÷1/2=2/5×2/1=(2×2)/(5×1)=4/5
(ii) 4/9÷2/3=4/9×3/2=2/3
(iii) 3/7÷8/7=3/7×7/8=3/8
(iv) 2.1/3÷3/5=7/3÷3/5=7/3×5/3=35/9=3.8/9
(v) 3.1/2÷8/3=7/2÷8/3=7/2×3/8=(7×3)/(2×8)=21/16=1.5/16
(vi) 2/5÷1.1/2=2/5÷3/2=2/5×2/3=(2×2)/(5×3)=4/15
(vii) 3.1/5÷1.2/3=16/5÷5/3=16/5×3/5=(16×3)/(5×5)=48/25=1.23/25
(viii) 2.1/5÷1.1/5=11/5÷/65=11/5×5/6=11/6=1.5/6
Exercise 2.4
Page Number 46
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) 0.50 > 0.05
(ii) 0.7 > 0.5
(iii) 7.0 > 0.7
(iv) 1.37 < 1.49
(v) 2.03 < 2.30
(vi) 0.80 < 0.88
Exercise 2.5
Page Number 47
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
∵ 100 paise = Re. 1
∴ 1 paisa = Re. 1/100
7 paise = Re. 7/100 = Re. 0.07
7 rupees 7 paise = Rs. 7 + Re. 7/100 = Rs. 7 + Re. 0.07 = Rs. 7.07
77 rupees 77 paise = Rs. 77 + Re. 77/100 = Rs. 77 + Re. 0.77 = Rs. 77.77
50 paise = Re. 50/100 = Re. 0.50
235 paise = Re. 235/100 = Rs. 2.35
Exercise 2.5
Page Number 47
3
(i) Express 5 cm in metre and kilometer.
(ii) Express 35 mm in cm, m and km.
Answer
(i) Express 5 cm in meter and kilometer.
∵ 100 cm = 1 meter
∴ 1 cm = 1/100 meter ⇒ 5 cm = 5/100 = 0.05 meter.
Now, ∵ 1000 meters = 1 kilometers
∴ 1 meter = 1/1000 kilometer
⇒ 0.05 meter = 0.05/1000 = 0.00005 kilometer
(ii) Express 35 mm in cm, m and km.
∵ 10 mm = 1 cm
∴ 1 mm = 1/10 cm ⇒ 35 mm = 35/10 = 3.5 cm
Now, ∵ 100 cm = 1 meter
∴ 1 cm = 1/100 meter ⇒3.5 cm = 3.5/100 = 0.035 meter
Again, ∵ 1000 meters = 1 kilometers
∴ 1 meter = 1/1000 kilometer
⇒ 0.035 meter = 0.035/1000 = 0.000035 kilometer
Exercise 2.5
Page Number 47
4
Express in kg.:
(i) 200 g
(ii) 3470 g
(iii)4 kg 8 g
Answer
1000 g = 1 kg ⇒ 1 g = 1/1000 kg
200 g = (200×1/1000) kg = 0.2 kg
3470 g = (3470×1/1000) kg = 3.470 kg
4 kg 8 g = 4 kg + (8×1/1000) kg = 4 kg + 0.008 kg = 4.008 kg
Exercise 2.5
Page Number 48
5
Write the following decimal numbers in the expanded form:
(i) 20.03
(ii) 2.03
(iii) 200.03
(iv) 2.034
Answer
(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+3×1/100+4×1/1000
Exercise 2.5
Page Number 48
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) Place value of 2 in 2.56 = 2 x 1 = 2 ones
(ii) Place value of 2 in 21.37 = 2 x 10 = 2 tens
(iii) Place value of 2 in 10.25 = 2×1/10 = 2 tenths
(iv) Place value of 2 in 9.42 = 2×1/100 = 2 hundredth
(v) Place value of 2 in 63.352 = 2×1/1000 = 2 thousandth
Exercise 2.5
Page Number 48
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 travelled by Dinesh when he went from place A to place B = 7.5 km and from place B to C = 12.7 km.
Total distance covered by Dinesh = AB + BC
= 7.5 + 12.7 = 20.2 km
Total distance covered by Ayub = AD + DC
= 9.3 + 11.8 = 21.1 km
On comparing the total distance of Ayub and Dinesh,
21.1 km > 20.2 km
Therefore, Ayub covered more distance by 21.1 – 20.2 = 0.9 km = 900 m
Exercise 2.5
Page Number 48
8
Shyam 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
Total weight of fruits bought by Shyam = 5 kg 300 g + 3 kg 250 g = 8 kg 550 g
Total weight of fruits bought by Sarala = 4 kg 800 g + 4 kg 150 g = 8 kg 950 g
On comparing the quantity of fruits,
8 kg 550 g < 8 kg 950 g
Therefore, Sarala bought more fruits.
Exercise 2.5
Page Number 48
9
How much less is 28 km than 42.6 km?
Answer
We have to find the difference of 42.6 km and 28 km.
42.6 – 28.0 = 14.6 km
Therefore 14.6 km less is 28 km than 42.6 km
Exercise 2.5
Page Number 48
1
Find:
(i) 0.2 x 6
(ii) 8 x 4.6
(iii) 2.71 x 5
(iv) 20.1 x 4
(v) 0.05 x 7
(vi) 211.02 x 4
(vii) 2 x 0.86
Answer
(i) 0.2 x 6 = 1.2
(ii) 8 x 4.6 = 36.8
(iii) 2.71 x 5 = 13.55
(iv) 20.1 x 4 = 80.4
(v) 0.05 x 7 = 0.35
(vi) 211.02 x 4 = 844.08
(vii) 2 x 0.86 = 1.72
Exercise 2.6
Page Number 52
2
Find the area of rectangle whose length is 5.7 cm and breadth is 3 cm ?
Answer
Given: Length of rectangle = 5.7 cm and Breadth of rectangle = 3 cm
Area of rectangle = Length x Breadth
= 5.7 x 3 = 17.1cm2cm2
Thus, the area of rectangle is 17.1 cm2cm2.
Exercise 2.6
Page Number 52
3
Find:
(i) 1.3 x 10
(ii) 36.8 x 10
(iii) 153.7 x 10
(iv) 168.07 x 10
(v) 31.1 x 100
(vi) 156.1 x 100
(vii) 3.62 x 100
(viii) 3.07 x 100
(ix) 0.5 x 10
(x) 0.08 x 10
(xi) 0.9 x 100
(xii) 0.03 x 1000
Answer
(i) 1.3 x 10 = 13.0
(ii) 36.8 x 10 = 368.0
(iii) 153.7 x 10 = 1537.0
(iv) 168.07 x 10 = 1680.7
(v) 31.1 x 100 = 3110.0
(vi) 156.1 x 100 = 15610.0
(vii) 3.62 x 100 = 362.0
(viii) 43.07 x 100 = 4307.0
(ix) 0.5 x 10 = 5.0
(x) 0.08 x 10 = 0.80
(xi) 0.9 x 100 = 90.0
(xii) 0.03 x 1000 = 30.0
Exercise 2.6
Page Number 52
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
∵ In one litre, a two-wheeler covers a distance = 55.3 km
∴ In 10 litrs, a two- wheeler covers a distance = 55.3 x 10 = 553.0 km
Thus, 553 km distance will be covered by it in 10 litres of petrol.
Exercise 2.6
Page Number 52
5
Find:
(i) 2.5 x 0.3
(ii) 0.1 x 51.7
(iii) 0.2 x 316.8
(iv) 1.3 x 3.1
(v) 0.5 x 0.05
(vi) 11.2 x 0.15
(vii) 1.07 x 0.02
(viii) 10.05 x 1.05
(ix) 101.01 x 0.01
(x) 100.01 x 1.1
Answer
(i) 2.5 x 0.3 = 0.75
(ii) 0.1 x 51.7 = 5.17
(iii) 0.2 x 316.8 = 63.36
(iv) 1.3 x 3.1 = 4.03
(v) 0.5 x 0.05 = 0.025
(vi) 11.2 x 0.15 = 1.680
(vii) 1.07 x 0.02 = 0.0214
(viii) 10.05 x 1.05 = 10.5525
(ix) 101.01 x 0.01 = 1.0101
(x) 100.01 x 1.1 = 110.11
Exercise 2.6
Page Number 52
1
Find:
(i) 0.4 ÷ 2
(ii) 0.35 ÷ 5
(iii) 2.48 ÷ 4
(iv) 65.4 ÷ 6
(v) 651.2 ÷ 4
(v) 14.49 ÷ 7
(vii) 3.96 ÷ 4
(viii) 0.80 ÷ 5
Answer
(i) 0.4 ÷ 2 = 4/10×1/2=2/10 = 0.2
(ii) 0.35 ÷ 5 = 35/100×1/5=7/100 = 0.07
(iii)2.48 ÷ 4 = 248/100×1/4=62/100 = 0.62
(iv) 65.4 ÷ 6 = 654/10×1/6=109/10= 10.9
(v) 651.2 ÷ 4 = 6512/10×1/4=1628/10 = 162.8
(vi)14.49 ÷ 7 = 1449/100×1/7=207/100 = 2.07
(vii)3.96 ÷ 4 = 396/100×1/4=99/100 = 0.99
(viii)0.80 ÷ 5 = 80/100×1/5=16/100 = 0.16
Exercise 2.7
Page Number 55
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
Answer
(i) 4.8 ÷ 10 = 4.8/10 = 0.48
(ii) 52.5 ÷ 10 = 52.5/10 = 5.25
(iii) 0.7 ÷ 10 = 0.7/10 = 0.07
(iv) 33.1 ÷ 10 = 33.1/10 = 3.31
(v) 272.23 ÷ 10 = 272.23/10 = 27.223
(vi) 0.56 ÷ 10 = 0.56/10 = 0.056
(vii) 3.97 ÷ 10 = 3.97/10 = 0.397
Exercise 2.7
Page Number 55
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 = 27/10×1/100=27/1000 = 0.027
(ii) 0.3 ÷ 100 = 3/10×1/100=3/1000 = 0.003
(iii) 0.78 ÷ 100 = 78/100×1/100=78/10000 = 0.0078
(iv) 432.6 ÷ 100 = 4326/10×1/100=4326/1000 = 4.326
(v) 23.6 ÷ 100 = 236/10×1/100=236/1000= 0.236
(vi) 98.53 ÷ 100 = 9853/100×1/100=9853/10000 . 0.9853
Exercise 2.7
Page Number 55
4
Find:
(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 = 79/10×1/1000=79/10000 = 0.0079
(ii) 26.3 ÷ 1000 = 263/10×1/1000=263/10000 = 0.0263
(iii) 38.53 ÷ 1000 = 3853/100×1/1000=3853/100000 = 0.03853
(iv) 128.9 ÷ 1000 = 1289/10×1/1000=1289/10000= 0.1289
(v) 0.5 ÷ 1000 = 5/10×1/1000=5/10000 = 0.0005
Exercise 2.7
Page Number 55
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
Answer
(i) 7 ÷ 3.5 = 7÷35/10=7×10/35=10/5 = 2
(ii) 36 ÷ 0.2 = 36÷2/10=36×10/2 = 18 x 10 = 180
(iii) 3.25 ÷ 0.5 = 325/100÷5/10=325/100×10/5=65/10= 6.5
(iv) 30.94 ÷ 0.7 = 3094/100÷7/10=3094/100×10/7=442/10 = 44.2
(v) 0.5 ÷ 0.25 = 5/10÷25/100=5/10×100/25=10/5= 2
(vi) 7.75 ÷ 0.25 = 775/100÷25/100=775/100×100/25 = 31
(vii) 76.5 ÷ 0.15 = 765/10÷15/100=765/10×100/15 = 51 x 10 = 510
(viii) 37.8 ÷ 1.4 = 378/10÷14/10=378/10×10/14 = 27
(ix) 2.73 ÷ 1.3 = 273/100÷13/10=273/100×10/13=21/10= 2.1
Exercise 2.7
Page Number 55
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 petrol?
Answer
∵ In 2.4 litres of petrol, distance covered by the vehicle = 43.2 km
∴ In 1 litre of petrol, distance covered by the vehicle = 43.2 ÷÷ 2.4
= 432/10÷24/10 = 432/10×10/24
= 18 km
Thus, it covered 18 km distance in one litre of petrol.
Exercise 2.7
Page Number 55