CAT — Simple & Compound Interest
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− Aman invests Rs 4000 in a bank at a certain rate of interest, compounded annually. If the ratio of the value of the investment after 3 years to the value of the investment after 5 years is 25 : 36, then the minimum number of years required for the value of the investment to exceed Rs 20000 is
Given:
Aman invests Rs. 4000 at compound interest, compounded annually.
The ratio of the amount after 3 years to the amount after 5 years is 25 : 36.
Find the minimum number of years required for the investment to exceed Rs. 20000.
Step 1: Let the annual growth factor be
1 + r = k
Then,
Amount after 3 years
= 4000k³
Amount after 5 years
= 4000k⁵
Given,
k³/k⁵ = 25/36
1/k² = 25/36
k² = 36/25
k = 6/5
Thus, the annual interest rate is
20%.
Step 2: Form the amount after n years
Amount after n years
= 4000 × (6/5)ⁿ
We need
4000 × (6/5)ⁿ > 20000
(6/5)ⁿ > 5
Step 3: Check successive powers
(6/5)⁸
= 6⁸/5⁸
= 1679616/390625
≈ 4.30
Amount after 8 years
≈ 4000 × 4.30
≈ Rs. 17200
This is less than Rs. 20000.
Now,
(6/5)⁹
= (6/5) × 4.30
≈ 5.16
Amount after 9 years
≈ 4000 × 5.16
≈ Rs. 20640
This exceeds Rs. 20000.
Final Answer
The minimum number of years required is
9
Answer: 9
Anil borrows Rs 2 lakhs at an interest rate of 8% per annum, compounded half-yearly. He repays Rs 10320 at the end of the first year and closes the loan by paying the outstanding amount at the end of the third year. Then, the total interest, in rupees, paid over the three years is nearest to
45311
51311
33130
40991
51311
Principal = Rs 2,00,000. Half-yearly rate = 4%.
Step 1: Amount at the end of the first year
= 2,00,000 × (1.04)² = 2,00,000 × 1.0816 = Rs 2,16,320
Step 2: Repayment at end of first year
Outstanding = 2,16,320 − 10,320 = Rs 2,06,000
Step 3: Outstanding at end of third year
From end of year 1 to end of year 3 = 4 half-years.
= 2,06,000 × (1.04)⁴ = 2,06,000 × 1.16985856 ≈ Rs 2,40,991
Step 4: Total amount paid
= 10,320 + 2,40,991 = Rs 2,51,311
Step 5: Total interest paid
= 2,51,311 − 2,00,000 = Rs 51,311
At a certain simple rate of interest, a given sum amounts to Rs 13920 in 3 years, and to Rs 18960 in 6 years and 6 months. If the same given sum had been invested for 2 years at the same rate as before but with interest compounded every 6 months, then the total interest earned, in rupees, would have been nearest to
3221
3180
3150
3096
3221
Interest for 3.5 years = 18960 − 13920 = ₹5040. Annual SI = 5040/3.5 = ₹1440.
Principal = 13920 − 3×1440 = ₹9600. Rate = (1440/9600)×100 = 15% p.a. → 7.5% per half-year.
CI for 2 years (4 half-years): Amount = 9600 × (1.075)⁴ ≈ 9600 × 1.33547 ≈ ₹12820.5. Interest = 12820.5 − 9600 ≈ ₹3220.5, nearest to ₹3221.
A loan of Rs 1000 is fully repaid by two installments of Rs 530 and Rs 594, paid at the end of first and second year, respectively. If the interest is compounded annually, then the rate of interest, in percentage, is
10
11
9
8
8
Step 1: Let the annual rate of interest be r%. The loan amount is Rs. 1000. At the end of the first year, the amount becomes 1000 × (1 + r/100). After paying the first installment of Rs. 530, the outstanding amount is 1000 × (1 + r/100) − 530.
Step 2: Form the equation after the second year. This outstanding amount earns interest for one more year. Hence, the amount before the second payment is [1000 × (1 + r/100) − 530] × (1 + r/100). Since the loan is fully repaid by paying Rs. 594: [1000 × (1 + r/100) − 530] × (1 + r/100) = 594.
Step 3: Simplify the equation. Let 1 + r/100 = x. Then, (1000x − 530)x = 594 → 1000x² − 530x − 594 = 0. Dividing by 2: 500x² − 265x − 297 = 0.
Step 4: Solve the quadratic equation. Try r = 8%: x = 1.08. Substituting: 1000 × (1.08)² − 530 × (1.08) = 1166.4 − 572.4 = 594. The equation is satisfied. Hence, r = 8%.
Anil invests Rs 22000 for 6 years in a scheme with 4% interest per annum, compounded half-yearly. Separately, Sunil invests a certain amount in the same scheme for 5 years, and then reinvests the entire amount he receives at the end of 5 years, for one year at 10% simple interest. If the amounts received by both at the end of 6 years are equal, then the initial investment, in rupees, made by Sunil is
20860
20640
20480
20808
20808
Given:
Anil invests Rs. 22000 for 6 years at 4% per annum, compounded half-yearly.
Sunil invests a certain amount in the same scheme for 5 years and then reinvests the maturity amount for 1 year at 10% simple interest.
The final amounts received by both are equal.
Find Sunil's initial investment.
Step 1: Find Anil's maturity amount
Half-yearly interest rate
= 4%/2
= 2%
Number of half-years
= 6 × 2
= 12
Amount received by Anil
= 22000 × (1.02)¹²
Step 2: Find Sunil's maturity amount
Let Sunil's initial investment be x.
After 5 years at the same scheme,
Amount
= x × (1.02)¹⁰
This amount is reinvested for 1 year at 10% simple interest.
Final amount
= x × (1.02)¹⁰ × 1.10
Step 3: Equate the two amounts
22000 × (1.02)¹²
= x × (1.02)¹⁰ × 1.10
Dividing both sides by (1.02)¹⁰,
22000 × (1.02)²
= 1.10x
Since
(1.02)² = 1.0404,
22000 × 1.0404 = 1.10x
22888.8 = 1.10x
x = 22888.8/1.10
= 20808
Final Answer
Sunil's initial investment was
Rs. 20808
Answer:
D. 20808
Anil invests Rs. 22000 for 6 years in a certain scheme with 4% interest per annum, compounded half-yearly. Sunil invests in the same scheme for 5 years, and then reinvests the entire amount received at the end of 5 years for one year at 10% simple interest. If the amounts received by both at the end of 6 years are same, then the initial investment made by Sunil, in rupees, is
Anil invests ₹22000 for 6 years at 4% p.a., compounded half-yearly.
Rate per half-year = 4% ÷ 2 = 2%
Number of half-years = 6 × 2 = 12
Amount received by Anil = 22000 × (1.02)^12
Let Sunil's initial investment be ₹P.
He invests for 5 years in the same scheme.
Number of half-years = 5 × 2 = 10
Amount after 5 years = P × (1.02)^10
He then reinvests this amount for 1 year at 10% simple interest.
Final amount = P × (1.02)^10 × 1.10
Since both receive the same amount after 6 years,
22000 × (1.02)^12 = P × (1.02)^10 × 1.10
Cancel (1.02)^10 from both sides, 22000 × (1.02)^2 = 1.10P
Now, (1.02)^2 = 1.0404
Therefore, 22000 × 1.0404 = 1.10P → 22888.8 = 1.10P → P = 22888.8 ÷ 1.10 → P = 20808
Answer: 20808
An amount of Rs 10000 is deposited in bank A for a certain number of years at a simple interest of 5% per annum. On maturity, the total amount received is deposited in bank B for another 5 years at a simple interest of 6% per annum. If the interests received from bank A and bank B are in the ratio 10 : 13, then the investment period, in years, in bank A is
4
5
3
6
6
Step 1: Find the interest from Bank A
Interest from Bank A = (10000 × 5 × x)/100 = 500x
Maturity amount = 10000 + 500x
Step 2: Find the interest from Bank B
Interest from Bank B = [(10000 + 500x) × 6 × 5]/100 = 3000 + 150x
Step 3: Use the given ratio
500x : (3000 + 150x) = 10 : 13
13 × 500x = 10(3000 + 150x)
6500x = 30000 + 1500x
5000x = 30000 → x = 6
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