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Mathematics [1 Mark Question Answer]

Compound Interest · ICSE STD 10 (ICSE - English Medium). 3 questions.

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[1 Mark Question Answer]

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Question 11 Mark
The cost of a machine depreciates by 10% every year. If its present worth is Rs.18,000; what will be its value after three years?
Answer
Applying the formula, we get;
Value after 3 years = ₹ 18,000 $\left(1-\frac{10}{100}\right)^3$
= ₹ 13,122
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Question 21 Mark
The total number of industries in a particular portion of the country is approximately 1,600. If the government has decided to increase the number of industries in the area by 20% every year; find the approximate number of industries after 2 years.
Answer
Number of industries after 2 years
= Original number of industries $\left(1+\frac{r}{100}\right)^n$
= 1,600 $\left(1+\frac{20}{100}\right)^n$
= 2,304
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Question 31 Mark
The population of a town 2 years ago was 62,500. Due to migration to cities, it decreases at the rate of 4% per annum. Find its present population.
Answer
Present population
= 62,500 x $\left(1-\frac{4}{100}\right)^2$
$=\left(62,500 \times \frac{24}{25} \times \frac{24}{25}\right)$
= 57,600
∴ Present population = 57,600
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[1 Mark Question Answer] - Mathematics STD 10 Questions - Vidyadip