Exercise 2.3
1. Find the reciprocal of each of the following fractions and classify them as proper, improper and whole numbers
(i) 3/7
(ii) 5/8
(iii) 9/7
(iv) 6/5
(v) 12/7
(vi) 1/8
Solution:
(i) 3/7
7/3 = improper number
(ii) 5/8
8/5 = improper number
(iii) 9/7
7/9 = proper number
(iv) 6/5
5/6 = proper number
(v) 12/7
7/12 = proper number
(vi) 1/8
8 = whole number
2. Divide
(i) (3/8) by (5/9)
(ii) 3 (1/4) by (2/3)
(iii) (7/8) by 4 (1/2)
(iv) 6 (1/4) by 2 (3/5)
Solution:
(i) Given (3/8) by (5/9)
From the rule of division of fraction we know that (a/b) ÷(c/d) = (a/b) ×(d/c)
(3/8) / (5/9) = (3/8) ×(9/5)
= (3×9) / (8×5)
= (27/40)
(ii) Given 3 ( (1/4) by (2/3)
Converting 3 (1/4) to improper we get (13/4)
From the rule of division of fraction we know that (a/b) ÷ (c/d) = (a/b) ×(d/c)
(13/4) / (2/3) = (13/4) ×(3/2)
= (13×3) / (4×2)
= (39/8)
4 (2/7)
(iii) Given (7/8) by (1/2)
Converting 4 (1/2) to improper we get (9/2)
From the rule of division of fraction we know that (a/b) ÷(c/d) = (a/b) ×(d/c)
(7/8) (9/2) = (7/8) ×(2/9)
= (7×2) / (8×9)
= (14/72)
= (7/36)
(iv) Given 6 (1/4) by 2 (3/5)
Converting 6 (1/4) and 2 (3/5) to improper we get (25/4) and (13/5)
From the rule of division of fraction we know that (a/b) ÷(c/b) = (a/b) ×(d/c)
(25/4) / (13/5) = (5/13) ×(5/13)
= (25×5) / (4×13)
= (75/52)
= 2 (21/52)
3. Divide:
Solution:
4. Simplify:
(i) (3\10) ÷(10/3)
Solution:
Given (3/10) ÷ (10/3)]
= (3×3) / (10×10)
= (9/100)
(ii) 4 (3/5) ÷(4/5)
Solution:
Given 4 (3/5) ÷(4/5)
First convert the given mixed fractions into improper fraction
4 (3/5) = (23/5)]
(23/5) ÷(4/5) = (23×5) / (5×4)
= (23/4)
= 5 (3/4)
(iii) 5 (4/7) ÷1 (3/10)
Solution:
Given 5 (4/7) ÷1 (3/10)
First converting the given mixed fraction into improper fractions
(39/7) and (13/10)
(39/7) ÷(13/10) = (39×10)/ (7×13)
= (390/91)
= (30/7)
= 4 (2/7)
(iv) 4÷2(2/5)
Solution:
Given 4 ÷ 2 (2/5)
First converting the given mixed fractions into improper
2 (2/5) = (12/5)
4 ÷(12/5) = (4×5)/ 12
= (20/12)
1 (1(2/3)
5. A wire of length 12 (1/2) m is cut into 10 pieces of equal length. Find the length of each piece.
Solution:
Given total length of the wire is = 21(1/2) = (25/2) m
It is cut into 10 pieces, so length of one piece is
= (25/2)/ 10
= (25/20)
= (5/4)
= 1 (1/4) m
6. The length of a rectangular plot of area 65(1/3) m2 is 12(1/4) m. What is the width of the plot?
Solution:
Given,
The length of a rectangular plot of area 65(1/3) m2 is 12(1/4) m.
7. By what number 6(2/9) be multiplied to get 4(4/9)?
Solution:
Let x be the number which needs to be multiplied by 6 (2/9) = (56/9)
X × (56/9) = 4 (4/9)
X × (56/9) = (40/9)
X = (40/9) ×(9/56)
X = (40/56)
X = (5/7)
8. The product of two number 25 (5/6). If one of the numbers is 6 (2/3), find the other.
Solution:
Given product of two number is 25 (5/6) = (155/6)
One of the numbers i9s 6 (2/3) = (20/3)
Let the other be x
(155/6) = x× (155/3)
X = (3/20) ×(155/6)
X = (31/8)
X = 3(7/8)
9. The cost of 6 (1/4) kg of apples is Rs. 400. At what rate per kg are the alpple being sold?
Solution:
Cost of total apples = Rs 400.
10. By selling oranges at the rate of Rs 5(1/4) per orange, a fruit seller get Rs 630. How many dozens of oranges does he sell?
Solution:
By selling oranges at the rate of Rs 5(1/4) per orange, a fruit seller get Rs 630.
5(1/4) = 2(1/4)
12 apples = 1 dozen
Therefore, 120 apples = 10 dozen
11. In mid-day meal scheme 3/10 litre of milk is given to each student of a primary school. If 30 litres of milk is distributed every-day in the school, how many students are there in the school?
Solution:
Given,
3/10 litre of milk is given to each student of a primary school.
30 litres of milk is distributed everyday in the school
Number of students given 3/10 litres of milk = 1
Number of students given 1 litre of milk = 10/3
Number of students given 30 litres of milk = 10/3 × 30 = 100 Students
12. In a charity show Rs 6496 were collected by selling some tickets. If the price of each ticket was Rs 50(3/4), how many tickets were sold?