Fractions

1. When Prabhu Dayal died, he left all his money for his grand-children, 3 of whom were boys and 5 were girls. In his will he insisted that each grand-child must get equal share of the total amount of ₹56,00,000.

Q1: What was the share of each child?

Solution:
Total amount = ₹56,00,000
Number of grandchildren = 3 boys + 5 girls = 8 grandchildren
Share of each child = Total amount ÷ Number of grandchildren = ₹56,00,000 ÷ 8 = ₹7,00,000

Answer: ₹7,00,000

Q2; What fraction of the money did the girls receive?

Solution:
Amount received by each girl = ₹7,00,000
Total amount received by girls = ₹7,00,000 × 5 = ₹35,00,000
Fraction of the money received by the girls = ₹35,00,000 ÷ ₹56,00,000 = 5/8

Answer: 5/8

Q3: How much did the boys receive in total?

Solution:
Amount received by each boy = ₹7,00,000
Total amount received by boys = ₹7,00,000 × 3 = ₹21,00,000

Answer: ₹21,00,000

Q4: If one of the girls did not take her share and the money is divided among the remaining grand-children, the fraction of the money received by the boys is:

Solution:
If one girl does not take her share, the number of remaining grandchildren = 8 – 1 = 7 grandchildren.
New total amount = ₹56,00,000 (still the same)
Share of each grandchild = ₹56,00,000 ÷ 7 = ₹8,00,000
Amount received by each boy = ₹8,00,000
Total amount received by boys = ₹8,00,000 × 3 = ₹24,00,000
Fraction of money received by the boys = ₹24,00,000 ÷ ₹56,00,000 = 3/7

Answer: 3/7

2. Amar is an electrician. He bought \(7\frac{1}{2}\) bundles of an electric cable where each bundle had \(202\frac{4}{5}\) m of cable.

Q1: Find the total length of the cable purchased by Amar.

Solution:
Total bundles = 7½ = 7 + ½ = 7.5 bundles
Length of each bundle = 202⅘ = 202 + 4/5 = 202.8 m
Total length of cable = 7.5 × 202.8 = 1521 m

Answer: 1521 m

Q2: If the cost of cable is ₹\(6\frac{2}{3}\) per metre, find the amount paid by Amar.

Solution:
Cost per metre = ₹6⅔ = ₹6 + 2/3 = ₹6.6667
Total amount paid = Total length × cost per metre = 1521 × 6.6667 ≈ ₹10140

Answer: ₹10140

Q3: Amar used \(2\frac{1}{2}\) bundles of cable for electric connections in the top floor of the building. What length of cable was used for the top floor?

Solution:
Bundles used for top floor = 2½ = 2 + ½ = 2.5 bundles
Length of each bundle = 202⅘ = 202.8 m
Length used for top floor = 2.5 × 202.8 = 507 m

Answer: 507 m

Q4: Amar cut a length of \(13\frac{4}{5}\) m from a bundle and divided the remaining cable of this bundle into pieces of 21 m, length each. How many pieces of 21m did he get from this bundle?

Solution:
Length cut from the bundle = 13⅘ = 13 + 4/5 = 13.8 m
Remaining length of cable = 202⅘ – 13⅘ = 202.8 – 13.8 = 189 m
Length of each piece = 21 m
Number of pieces = 189 ÷ 21 = 9 pieces

Answer: 9 pieces