EXERCISE 7.6

1). Solve.
a)  2/3  +  1/7
Ans.
=  (2×7) / (3×7)
=  14/21 , (1×3) / (7×3)
=  3/21                                                                                                 [LCM of 3 and 7 is 21]

   14/21 + 3/21 = (14+3)/21 = 17/21
b) 3/10 + 7/15
Ans.
=  (3×3)/(10×3) = 9/30 , (7×2)/(15×2) = 14/30                        [LCM = 30]

=  9/30 + 14/30
=  (9+14)/30
=  23/30
c) 4/9 + 2/7
Ans.
=  (4×7)/(9×7) = 28/63 , (2×9)/(7×9) = 18/63                          [LCM = 63]

=  28/63 + 18/63
=  (28+18)/63
=  46/63
d) 5/7 + 1/3
Ans.
=  (5×3)/(7×3) = 15/21 , (1×7)/(3×7)
=  7/21                                                                                            [LCM = 21]

=  15/21 + 7/21
=  (15+7)/21
=  22/21
e) 2/5 + 1/6
Ans.
=  (2×6)/(5×6)
=  12/30 , (1×5)/(6×5) = 5/30                                                    [LCM = 30]

=  (12+5)/30
=  17/30
f) 4/5 + 2/3
Ans.
= (4×3)/(5×3) = 12/15 , (2×5)/(3×5) = 10/15                [LCM = 15]

= 12/15 + 10/15
= (12+10)/15
= 22/15
g)3/4 − 1/3
Ans.
= (3×3)/(4×3) = 9/12 , (1×4)/(3×4) = 4/12                    [LCM = 12]

= 9/12 − 4/12
= (9−4)/12
= 5/12
h) 5/6 – 1/3
Ans.
= (5×1)/(6×1) = 5/6 , (1×2)/(3×2) = 2/6                         [LCM = 6]

= 5/6 − 2/6
= (5−2)/6
= 3/6
= 1/6

i) 7/10 − 2/5
Ans.
= (7×1)/(10×1) = 7/10 , (2×2)/(5×2) = 4/10                   [LCM = 10]

= 7/10 − 4/10
= (7-4)/10−
= 3/10
j) 1/2 – 1/3
Ans. = (1×3)/(2×3) = ( 3)/6 , (1×2)/(3×2) = 2/6                     [LCM = 6]
= 3/6 – 2/6
= (3-2)/6
= 1/6
k) 1/2 − 1/6
Ans. = (1×3)/(2×3) = 3/6 , (1×1)/(6×1) = 1/6                           [LCM = 6]
= 3/6  1/6
= (3−1)/6
= 2/6
= 1/3
l) 6/(8 ) − 1/3
Ans.
= (6×3)/(8×3) = 18/24 , (1×8)/(3×8) = 8/24                 [ LCM = 24]

= 18/24 − 8/24
= (18-8)/24
= 10/24
= 5/12
m) 2/3 + 3/4 + 1/2
Ans.
= (2×4)/(3×4) = 8/12 , (3×3)/(4×3)
= 9/12 , (1×6)/(2×6) = 6/12                                                      [LCM  =  12]
= (8+9+6)/12
= 23/12
n) 1/2 + 1/3 + 1/6
Ans. =(1×3)/(2×3) = 3/6 , (1×2)/(3×2) =
2/6 , (1×1)/(6×1) = 1/6                                                              [LCM  = 6]
= (3+2+1)/6
= ( 6)/( 6)
= 1/1
o) 1 1/3 + 3 2/3
Ans.
1 1/3 = (1×3+1)/3 = 4/3 , 3 2/3 = (3×3+2)/3 = 11/3

= 4/3 + 11/3
= (4+11)/3
= 15/3 = 5
p) 42/3 +3 1/4
Ans.
= (4×3+2)/3 = 14/3 , (3×4+1)/4 = 13/4

= (14×4)/(3×4) = 56/12 , (13×3)/(4×3) = 39/12                   [LCM of 3 and 4 is 12]
= (56+39)/12
= 95/12
q) 16/5 – 7/5
Ans. = (16−7)/5 = 9/5
r) 4/3 – 1/2
Ans.
= (4×2)/(3×2) = 8/6 , (1×3)/(2×3) = 3/6                        [LCM of 3,2=6]

= (8−3)/6
= 5/6

s) 21/3 − 12/3
Ans.
(2×3+1)/3 = 7/3 , (1×3+2)/3 = 5/3

= (7-5)/3
= 2/3
t) 32/3 – 12/3
Ans.
= (3×3+2)/3 = 11/3 , (1×3+2)/3 = 5/3

= (11-5)/3
= 6/3
= 2

2). Sarita bought 2/5 meter of ribbon and Lalita  3/4 meter of ribbon what was the total length of the ribbon they bought?
Ans.
Length of the ribbon bought by Sarita = 2/5

Length of the ribbon bought by Lalita = 3/4
Total length of the ribbon bought by both  = 2/5 + 3/4 
Taking LCM  =  20
= (2×4)/(5×4)  +  (3×5)/(4×5)
= (8+15)/20 = 23/20
Therefore total length of ribbon bought by both Sarita and Lalita is 23/20 meters.

3). Naina was given 1½  piece of cake and Najma was given 1¹⁄ ³ piece of cake. Find the total amount of cake given to both of them.

Ans.
Fraction of cake Naina got = 1½]  = 3/2

Fraction of cake Najma got = 1¹⁄³  =  4/3
Total amount of cake given to both
= 3/2 + 4/3       taking    LCM  =  6

= (3×3)/(2×3)  +  (4×2)/(3×2)
= (9+8)/6   =   17/6
Therefore the total amount of cake given to both is 17/6

4). Fill in the blanks:

a)____ − 5/8 = 1/4
Ans. = 1/4 + 5/8
= (1×2+5)/8      LCM = 8
= 7/8 − 5/8 = 1/4
b) ____− 1/5 = 1/2
Ans. = 1/2 + 1/5
= (1×5+1×2)/10   LCM = 10
= (5+2)/10 = 7/10
= 7/10 – 1/5 = 1/2
c) 1/2 −____ = 1/6
Ans. = 1/2 − 1/6
= (1×3−1×1)/6 LCM = 6
= (3−1)/6 = 2/6 = 1/3
1/2 − 1/3 = 1/6

5). Complete the addition and subtraction box.


Ans.
i)   2/3 + 4/3 = 6/3 = 2

ii)  1/3 + 2/3 = 3/3 = 1
iii)  2/3 – 1/3 = 1/3
iv)  4/3 – 2/3 = 2/3
v)   2−1 = 2
Hence, the complete given box
a)

b)

Ans.
1/2 + 1/3  =  (1×3+1×2)/6   =  5/6 [LCM  =  6]

1/3 + 1/4  =  (1×4+1×3)/12  =  7/12 [LCM  =  12]
1/2 – 1/3   =  (1×3-1×2)/6  =  1/6 [ LCM  =  6 ]
5/6 – 7/12 =  (5×2-7×1)/12  =  3/12  =  1/4
Hence, the complete box is

6). A piece of wire 7/8 meter long broke into two pieces one piece was 1/4 meter long. How long is the other piece?
Ans. The total length of the wire = 7/8 meter
Length of one piece of wire = 1/4 meter
Length of the other piece = total length – length of one piece
= 7/8 − 1/4
Taking LCM = 8
= (7×1)/(8×1) − (1×2)/(4×2)
= (7−2)/8 = 5/8
Therefore, length of the other piece is 5/8 meter

7). Nandani’s house is 9/10 km from her school. She walked some distance and then took a bus for 1/2 km to reach the school. How far did she walk?
Ans. The distance of the school from house  =  9/10 km
The distance she traveled by bus  =  1/2 km
Distance walked by Nandini = total distance – distance travelled by bus
= 9/10 − 1/2
Taking LCM = 10
= (9×1)/(10×1) − (1×25)/(2×25)
= (9−5)/10 = 4/10 = 2/5
Therefore, the distance walked by Nandini is 2/5 kms.

8). Asha and Samuel have bookshelves of the same size. Asha shelf is 5/6 full of books and Samuel’s shelf is 2/5 full. Whose bookshelf is more full by what fraction?
Ans. Fraction of Asha’s bookshelf = 5/6
Fraction of Samuel’s bookshelf = 2/5
To know whose book shelf is more full we must make them into like fractions
5/6  =  (5×5)/(6×5) = 25/30     [ LCM 0f 6 and 5 is 30 ]
2/5  =  (2×6)/(5×6) = 12/30
25/(30 )  >  12/30
5/6  >  2/5
Asha’s bookshelf is more full.
Difference = 5/6 − 2/5
= (5×5)/(6×5) – (2×6)/(5×6)
= (25−12)/30 = 13/30
Therefore Asha’s bookshelf is more full by 13/30

9). Jaidev takes 2 ¹⁄ 5  minutes to walk across the school ground. Rahul takes 7/4 minutes to do the same. Who takes less time and by what fraction
Ans. Time taken by Jaidev to walk  across the school ground = 2¹⁄ 5 = 11/5
Time taken by Rahul to walk across the school ground = 7/4 minutes
Converting them into like fractions we get
11/5  =  (11×4)/(5×4) = 44/20   [LCM of  5, 4  is 20]
7/4  =  (7×5)/(4×5) = 35/20
44/20  >  35/20
11/5   >   7/4
The  time difference  =  11/5 − 7/4
= (44−35)/20
= 9/20
Therefore Rahul takes less time to walk across the ground by 9/20 minutes