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Unit 6 Brain Teaser | Compound Interest | DAV Solution Class 8 Secondary Mathematics
Q1A. Tick the correct option.
(i) For a given rate of interest and time, what is more profitable to the depositor?
(a) compound interest (b) simple interest (c) both are equally profitable (d) cannot be determined
(ii) A sum of Rs. 10,000 at 8% per annum for six months compounded quarterly amounts to
(a) 400 (b) 404 (c) 408 (d) 10,404
(iii) In A = p(1 + R/100)^n , A stands for
(a) time period (b) principal (c) principal+interest (d) interest
(iv) If the number of conversion periods is greater than or equal to 2, then compound interest is
(a) less than simple interest.
(b) greater than simple interest.
(c) less than or equal to simple interest.
(v) Reema wants to do a one-year deposit. She should opt for
(a) a simple interest of 10%.
(b) a compound interest of 10% compounded quarterly.
(c) a compound interest of 10% compounded annually.
(d) a compound interest of 10% compounded half-yearly.
Q1B. Answer the following questions.
(i) Find the compound interest on Rs. 1,000 at 10% per annum for two years.
(ii) If Rs. 6,000 is deposited for two years at 4% per annum compounded quarterly. then find the time period and rate to compute compound interest.
(iii) The value of a machine worth Rs. 5,00,000 depreciates at the rate of 10% every year. In how many years will its value be Rs. 3,64,500?
(iv) Find the difference between the compound interest and simple interest on Rs. 5000 for two years at 5% per annum.
(v) If Rs. 20,000 is deposited for three years at 5% compounded annually, then what will be the principal for the second year?
2. Preeti invested ₹ 50,000 at 8% per annum for 3 years and the interest is compounded annually. Calculate:
(i) The amount standing to her credit at the end of the second year.
(ii) The interest for the third year.
3. A man had ₹75,000. He invested ₹35,000 in a company which pays him 9% interest per annum and he invested rest of the money in another company which pays him 9.5% interest per annum. Find the total compound interest received by him after 2 years.
4. A certain sum of money at compound interest becomes ₹7,396 in 2 years and ₹7,950.70 in 3 years. Find the rate of interest.
5. Pooja started a business by investing ₹2,00,000. During the first three successive years, she earned a profit of 5%, 8% and 12% per annum respectively. If in each year, the profit was added on the capital at the end of the previous year, calculate her total profit after 3 years.
6. Mahesh borrowed a certain sum for 2 years at simple interest from Bhim. Mahesh lent this sum to Vishnu at the same rate for 2 years compound interest. At the end of 2 years, Mahesh received 410 as compound interest but paid ₹400 as simple interest. Find the sum and rate of interest.
7. A man invested ₹ 1,000 is invested for three years at 11% simple interest per annum and another 1,000 is invested at 10% compound interest per annum compounded annually for 3 years. Find which investment is better.
8. M/s Heera Associates let out ₹4,00,000 for 1 year at 16% p.a. compounded annually. How much they could earn if the interest is compounded (i) half-yearly (ii) quarterly?
9. Sirish borrowed a sum of ₹1,63,840 at 12.5% p.a. compounded annually. On the same day, he lent out the same amount to Sahej at the same rate of interest but compounded half-yearly. Find his gain after 2 years.
10. The annual rate of growth in population of a certain city is 8%. If its present population is ₹1,96,830, what was the population 3 years ago?
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HOTS (Higher Order Thinking Skills) Questions
HOTS 1: The difference between Compound Interest (CI) and Simple Interest (SI) on a certain sum of money for 2 years at 8% per annum is Rs 40. Find the principal sum.
Solution:
Let the principal sum be Rs P.
Rate R = 8% p.a., Time T = 2 years.
Simple Interest: SI = (P × R × T) / 100 = (P × 8 × 2) / 100 = 16P / 100.
Compound Interest Amount: A = P(1 + R/100)2 = P(1 + 8/100)2 = P(27/25)2 = 729P / 625.
CI = A – P = (729P / 625) – P = 104P / 625.
Given: CI – SI = Rs 40
(104P / 625) – (16P / 100) = 40
LCM of 625 and 100 is 2500:
[4(104P) – 25(16P)] / 2500 = 40
[416P – 400P] / 2500 = 40
16P / 2500 = 40
P = (40 × 2500) / 16 = 100,000 / 16 = Rs 6,250.
Shortcut Formula Check: Difference = P(R/100)2 → 40 = P(8/100)2 = P(64/10000) → P = 400,000 / 64 = Rs 6,250.
HOTS 2: A sum of money invested at compound interest compounded annually amounts to Rs 14,520 in 2 years and to Rs 15,972 in 3 years. Find the rate of interest per annum and the principal sum.
Solution:
In compound interest, the amount at the end of 2 years (Rs 14,520) acts as the principal for the 3rd year.
Interest for the 3rd year = Rs 15,972 – Rs 14,520 = Rs 1,452.
Rate of interest: R = (Interest × 100) / (Principal × Time)
R = (1452 × 100) / (14520 × 1) = 10% per annum.
Now find the original Principal P:
Amount after 2 years = P(1 + 10/100)2 = 14,520
P(11/10)2 = 14,520
P × (121/100) = 14,520
P = (14,520 × 100) / 121 = 120 × 100 = Rs 12,000.
HOTS 3: The population of an industrial town decreases at the rate of 5% per annum due to relocation of factories. If the present population is 41,154, what was the population 2 years ago?
Solution:
Let the population 2 years ago be P0.
Depreciation rate R = 5% per annum, n = 2 years.
Present Population = P0(1 – R/100)2
41,154 = P0(1 – 5/100)2 = P0(19/20)2
41,154 = P0 × (361 / 400)
P0 = (41,154 × 400) / 361 = 114 × 400 = 45,600.
Value Based Questions (VBQs)
Question: Mrs. Sharma deposited Rs 50,000 in a self-help microfinance bank at 8% compound interest per annum for 2 years to fund scholarships for girl students in rural schools.
(a) Calculate the total compound interest earned at the end of 2 years.
Solution:
Amount = 50,000(1 + 8/100)2 = 50,000(27/25)2 = 50,000 × 729 / 625 = 80 × 729 = Rs 58,320.
Compound Interest = Rs 58,320 – Rs 50,000 = Rs 8,320.
(b) What moral and social values are reflected in Mrs. Sharma’s decision?
Solution:
Social awareness, support for girls’ education, community empowerment, and philanthropic responsibility.
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