Q.1. Compare the fractions:
(i) 5/8 and 7/12
(ii) 5/9 and 11/15
(iii) 11/12 and 15/16
Ans. We have the following:
(i) 5/8 and 7/12
By cross multiplication, we get:
5 × 12 = 60 and 7 × 8 = 56
However, 60 > 56
∴ 5/8 > 7/12
(ii) 5/9 and 11/15
By cross multiplication, we get:
5 × 15 = 75 and 9 × 11 = 99
However, 75 < 99
∴ 5/9 < 11/15
(iii) 11/12 and 15/16
By cross multiplication, we get:
11 × 16 = 176 and 12 × 15 = 180
However, 176 < 180
∴ 11/12 < 15/16
Q.2. Arrange the following fractions in ascending order:
(i) 3/4, 5/6, 7/9, 11/12
(ii) 4/5, 7/10, 11/15, 17/20
Ans. (i) The given fractions are 3/4, 5/6, 7/9 and 11/12.
LCM of 4, 6, 9 and 12 = 36
Now, let us change each of the given fractions into an equivalent fraction with 72 as its denominator.
3/4 = (3×9) / (4×9) = 27/36
5/6 = (5×6) / (6×6) = 30/36
7/9 = (7×4) / (9×4) = 28/36
11/12 = (11×3) / (12×3) = 33/36
Clearly, 27/36 < 28/36 < 30/36 < 33/36
Hence, 3/4 < 7/9 < 5/6 < 11/12
∴ The given fractions in ascending order are 3/4, 7/9, 5/6 and 11/12
(ii) The given fractions are: 4/5, 7/10, 11/15 and 17/20.
LCM of 5, 10, 15 and 20 = 60
Now, let us change each of the given fractions into an equivalent fraction with 60 as its denominator.
4/5 = (4×12) / (5×12) = 48/60
7/10 = (7×6) / (10×6) = 42/60
11/15 = (11×4) / (15×4) = 44/60
17/20 = (17×3) / (20×3) = 51/60
Clearly, 42/60 < 44/60 < 48/60 < 51/60
Hence, 7/10 < 11/15 < 4/5 < 17/20
∴ The given fractions in ascending order are
7/10, 11/15, 4/5 and 17/20
Q.3. Arrange the following fractions in descending order:
(i) 3/4, 7/8, 7/12, 17/24
(ii) 2/3, 3/5, 7/10, 8/15
Ans. We have the following:
(i) The given fractions are 3/4, 7/8, 7/12 and 17/24.
LCM of 4,8,12 and 24 = 24
Now, let us change each of the given fractions into an equivalent fraction with 24 as its denominator.
3/4 = (3×6) / (4×6) = 18/24
7/8 = (7×3) / (8×3) = 21/24
7/12 = (7×2 / 12×2) = 14/24
17/24 = (17×1) / (24×1) = 17/24
Clearly, 21/24 > 18/24 > 17/24 > 14/24
Hence, 7/8 > 3/4 > 17/24 > 7/12
∴ The given fractions in descending order are 7/8, 3/4, 17/24 and 7/12.
(ii) The given fractions are 2/3, 3/5, 7/10 and 8/15.
LCM of 3,5,10 and 15 = 30
Now, let us change each of the given fractions into an equivalent fraction with 30 as its denominator.
2/3 = (2×10) / (3×10) = 20/30
3/5 = (3×6) / (5×6) = 18/30
7/10 = (7×3) / (10×3) = 21/30
8/15 = (8×2) / (15×2) = 16/30
Clearly, 21/30 > 20/30 > 18/30 > 16/30
Hence, 7/10 > 2/3 > 3/5 > 8/15
∴ The given fractions in descending order are 7/10, 2/3, 3/5 and 8/15.
Q.4. Reenu got 2/7 part of an apple while Sonal got 4/5 part of it. Who got the larger part and by how much?
Ans. We will compare the given fractions 2/7 and 4/5 in order to know who got the larger part of the apple.
We have,
By cross multiplication, we get: 2 × 5 = 10 and 4 × 7 = 28
However, 10 < 28
∴ 2/7 < 4/5
Thus, Sonal got the larger part of the apple.
Now, 4/5 - 2/7 = (28-10) / 35 = 18/35
∴ Sonal got 18/35 part of the apple more than Reenu.
Q.5. Find the sum
(i) 5/9 + 3/9
(ii) 8/9 + 7/12
(iii) 5/6 + 7/8
(iv) 7/12 + 11/16 + 9/24
Ans. (i) 5/9 + 3/9 = 8/9
(ii) 8/9 + 7/12
= 32/36 + 21/36 [∵ LCM of 9 and 12 = 36]
= (32+21) / 36
= 53/36
=
(iii) 5/6 + 7/8 = 20/24 + 21/24 [∵ LCM of 6 and 8 = 24]
= (20+21) / 24
= 41/24
=
(iv) 7/12 + 11/16 + 9/24
28/48 + 33/48 + 18/48 [∵ LCM of 12, 16 and 24 = 48]
= (28+33+18) / 48
= 79/48
=
(v)= 19/5 + 23/10 + 16/15
= 114/30 + 69/30 + 32/30 [∵ LCM of 5, 10 and 15 = 30]
= (114+69+32) / 30
= 215/30
=
(vi)
= 35/4 + 52/5
= 175/20 + 208/20 [∵ LCM of 4 and 5 = 20]
= (175+208) / 20
= 383/20
=
Q.6. Find the difference:
(i) 5/7 - 2/7
(ii) 5/6 - 3/4
(iii)
(iv) 7 -
(v)
(vi)
Ans. (i) 5/7 - 2/7 = (5-2) / 7 = 3/7
(ii) 5/6 - 3/4 = 10/12 - 9/12 [∵ LCM of 6 and 4 = 12]
= (10-9) / 12
= 1/12
(iii)- 7/10 = 16/5 - 7/10
= 32/10 - 7/10 [∵ LCM of 5 and 10 = 10]
= (32-7) / 10
= 25/10
= 5/2
=
(iv) 7 -
= 7/1 - 14/3
= (21-14) / 3 [∵ LCM of 1 and 3 = 3]
= 7/3
=
(v)
= 33/10 - 22/15
= (99-44) / 30 [∵ LCM of 10 and 15 = 30]
= 55/30
= 11/6
=
(vi)
= 23/9 - 22/15
= (115-66) / 45 [∵ LCM of 9 and 15 = 45]
= 49/45
=
Q.7. Simplify:
(i) 2/3 + 5/6 - 1/9
(ii) 8 -
(iii)
Ans. (i) 2/3 + 5/6 - 1/9
= (12+15-2) / 18 [∵ LCM of 3, 6 and 9 = 18]
= (27-2) / 18
= 25/18
=
(ii) 8 -
= 8/1 - 9/2 - 9/4
= (32-18-9) / 4 [∵ LCM of 1, 2 and 4 = 4]
= (32-27) / 4
= 5/4
=
(iii)
= 53/6 - 27/8 + 19/12
= (212-81+38) / 24 [∵ LCM of 6, 8 and 12 = 24]
= (250-81) / 24
= 169/24
=
Q.8. Aneeta boughtkg apples andkg guava. What is the total weight of fruits purchased by her?
Ans. Total weight of fruits bought by Aneeta =kg
Now, we have:
= 15/4 + 9/2
= (15 + 18) / 4 [∵ LCM of 2 and 4 = 4]
= (15 + 18) / 4
= 33/4
=
Hence, the total weight of the fruits purchased by Aneeta iskg.
Q.9. A rectangular sheet of paper iscm long andcm wide. Find its perimeter.
Ans. We have:
Perimeter of the rectangle ABCD = AB + BC + CD +DA
=cm
= (63/4 + 25/2 + 63/4 + 25/2) cm
= (63 + 50 + 63 + 50) / 4 cm [∵ LCM of 2 and 4 = 4]
= (226 / 4) cm
=(113 / 2) cm
=cm
Hence, the perimeter of ABCD iscm
Q.10. A picture iscm wide. How much should it be trimmed to fit in a framecm wide?
Ans. Actual width of the picture = cm = 38/5 cm
Required width of the picture =cm = 73/10 cm
∴ Extra width = (38/5−73/10)cm
= (76−73) / 10 cm [∵ LCM of 5 and 10 is 10]
= 3/10 cm
Hence, the width of the picture should be trimmed by 3/10 cm.
Q.11. What should be added toto get 18?
Ans. Required number to be added = 18−
=18/1−38/5
= (90−38) / 5 [∵ LCM of 1 and 5 = 5]
= 52/5 =
Hence, the required number is
Q.12. What should be added toto get?
Ans. Required number to be added =
= 42/5−109/15
= (126−109)/15 [∵ LCM of 5 and 15 = 15]
= 17/15
=
Hence, the required number should be 12151215.
Q.13. A piece of wirem long broke into two pieces. One piece ism long. How long is the other piece?
Ans. Required length of other piece of wire = m
=(15/4−3/2) m
= ((15−6)/4) m [∵ LCM of 4 and 2 = 4]
= 9/4 m
=m
Hence, the length of the other piece of wire ism.
Q.14. A film show lasted ofhours. Out of this timehours was spent on advertisements. What was the actual duration of the film?
Ans. Actual duration of the film = ( −)hours
= (11/3−3/2) hours
= (22−9)/ 6 hours [∵ LCM of 3 and 2 = 6]
= 13/6 hours
= hours
Hence, the actual duration of the film was hours
Q.15. Of 2/3 and 5/9, which is greater and by how much?
Ans. First we have to compare the fractions: 2/3 and 5/9.
By cross multiplication, we have:
2 × 9 = 18 and 5 × 3 = 15
However, 18 > 15
∴2/3 > 5/9
So, 2/3 is larger than 5/9.
Now, 2/3−5/9
= (6−5) / 9 [∵ LCM of 3 and 9 = 9]
= 1/9
Hence, 2/3 is 1/9 part more than 5/9.
Q.16. The cost of a pen is Rsand that of a pencil is Rs. Which costs more and by how much?
Ans. First, we have to compare the cost of the pen and the pencil.
Cost of the pen = Rs= Rs 83/5
Cost of the pencil = Rs = Rs 19/4
Now, we have to compare fractions 83/5 and 19/4.
By cross multiplication, we get: 83 × 4 = 332 and 19 × 5 = 95
However, 332 > 95
∴ 83/5 > 19/4
So, the cost of pen is more than that of the pencil.
Now, Rs (83/5 − 19/4)
= Rs (332 − 95) / 20 [∵ LCM of 4 and 5 = 20]
= Rs 237
= Rs
∴ The pen costs Rsmore than the pencil.
= 19/5 + 23/10 + 16/15
- 7/10 = 16/5 - 7/10
kg apples and
kg guava. What is the total weight of fruits purchased by her?
kg
= 15/4 + 9/2
kg.
cm long and
cm wide. Find its perimeter.
cm
cm
cm
cm wide. How much should it be trimmed to fit in a frame
cm wide?
cm = 38/5 cm
cm = 73/10 cm
to get 18?
to get
?
m long broke into two pieces. One piece is
m long. How long is the other piece?
m
m
m.
hours. Out of this time
hours was spent on advertisements. What was the actual duration of the film?
−
)hours
hours
hours
and that of a pencil is Rs
. Which costs more and by how much?
= Rs 83/5
= Rs 19/4
more than the pencil.
