Page 116 - Maths Skills - 8

114 Maths

Example 6: An alloy contains 38% copper, 28% zinc and the rest is nickel. Find each quantity in grams in a
sample of 1 kg alloy.
Solution: Given: Copper = 38% Zinc = 28%
So, Nickel = [100 – (38 + 28)]% = 34%

38
Quantity of copper in 1 kg of alloy = 38% of 1 kg = × 1000 grams = 380 grams
100
28
Quantity of zinc in 1 kg of alloy = 28% of 1 kg = × 1000 grams = 280 grams
100
34
Quantity of nickel in 1 kg of alloy = 34% of 1 kg = × 1000 grams = 340 grams
100
Example 7: The price of sugar goes up by 25%. By what per cent must a housewife reduce the consumption,
so that the expenditure does not increase?
Solution: Let the initial consumption be 10 kg and its price be ` 200.

Now, the new price of 10 kg sugar is ` 250.
With ` 250, she can purchase 10 kg sugar

10
With ` 1, she can purchase kg sugar
250
10
With ` 200, she can purchase × 200 kg sugar = 8 kg sugar.
250
Reduction in sugar consumption = 10 kg – 8 kg = 2 kg
2
Hence, per cent reduction in consumption should be 10 × 100% = 20%.

1
Example 8: After incurring a loss of 33 %, a person is left with ` 23200. Find the amount he originally had.
3
Solution: Let the original amount he had be ` x.

Loss amount = ` (x – 23200)
1
According to question, we have 33 % of x = x – 23200
3
2x
⇒ 100 % of x = x – 23200 ⇒ = 23200
3 3
100 23200 3×
⇒ × x = x – 23200 ⇒ x = = ` 34800
300 2
x Thus, originally he had ` 34800.
⇒ x – = 23200
3

Example 9: The value of a machine depreciates every year by 10%. What will be its value after 2 years if its
present value is ` 60000?
Solution: Present value of the machine = ` 60000.
Value of the machine depreciates at 10% per year.

10
Decrease in value after 1 year = 10% of ` 60000 = × ` 60000 = ` 6000
100
∴ Depreciated value after first year = ` (60000 – 6000) = ` 54000