Question:
A man takes a loan of 1000 and repays it by paying 530 at the end of the first year, and 594 at the end of the second year. Find the rate of interest per annum.
A man takes a loan of 1000 and repays it by paying 530 at the end of the first year, and 594 at the end of the second year. Find the rate of interest per annum.
Show Hint
In loan installment problems with multiple-choice options, back-solving by plugging in the given interest rates is almost always faster than solving the algebraic equation.
Updated On: Jul 20, 2026
- 8%
- 10%
- 12%
- 9%
Show Solution
Verified By Collegedunia
The Correct Option is A
Approach Solution - 1
Approach: A loan repaid in instalments balances when the loan equals the present value of every instalment discounted back at the rate \(r\). Rather than crack the quadratic, plug the answer options into the present-value equation — the one that returns 1000 is the rate.
Step 1: Write the balance equation.
\[ 1000 = \frac{530}{(1+r)} + \frac{594}{(1+r)^2}. \]
Step 2: Test \(r = 8\%\) so \(1+r = 1.08\).
\(\dfrac{530}{1.08} = 490.74\).
\((1.08)^2 = 1.1664\), so \(\dfrac{594}{1.1664} = 509.26\).
Sum \(= 490.74 + 509.26 = 1000.00\). It matches exactly.
Step 3: Sanity-check \(r = 10\%\).
\(\dfrac{530}{1.1} + \dfrac{594}{1.21} = 481.82 + 490.91 = 972.73 \neq 1000\). Too low — a higher rate over-discounts. Confirms 8% is the fit.
Final answer: 8% — option (A).
Step 1: Write the balance equation.
\[ 1000 = \frac{530}{(1+r)} + \frac{594}{(1+r)^2}. \]
Step 2: Test \(r = 8\%\) so \(1+r = 1.08\).
\(\dfrac{530}{1.08} = 490.74\).
\((1.08)^2 = 1.1664\), so \(\dfrac{594}{1.1664} = 509.26\).
Sum \(= 490.74 + 509.26 = 1000.00\). It matches exactly.
Step 3: Sanity-check \(r = 10\%\).
\(\dfrac{530}{1.1} + \dfrac{594}{1.21} = 481.82 + 490.91 = 972.73 \neq 1000\). Too low — a higher rate over-discounts. Confirms 8% is the fit.
Final answer: 8% — option (A).
Was this answer helpful?
0
0
Show Solution
Verified By Collegedunia
Approach Solution -2
Approach (solve the present-value equation directly): Instead of testing options or reframing as forward growth, solve the discounting equation itself as a quadratic, using \( u = \frac{1}{1+r} \) as the unknown.
Step 1: The balance condition is
\[ 1000 = 530u + 594u^{2}, \quad \text{where } u = \frac{1}{1+r}. \]
Step 2: Rearrange into standard form:
\[ 594u^{2} + 530u - 1000 = 0. \]
Step 3: Solve using the quadratic formula. Discriminant \( = 530^{2} + 4(594)(1000) = 280900 + 2376000 = 2656900 \), and \( \sqrt{2656900} = 1630 \).
\[ u = \frac{-530 + 1630}{1188} = \frac{1100}{1188} = \frac{25}{27}. \]
Step 4: Recover \( r \): since \( u = \frac{1}{1+r} = \frac{25}{27} \), we get \( 1+r = \frac{27}{25} = 1.08 \), so \( r = 0.08 = 8\% \).
Final answer: 8%, option (A).
Step 1: The balance condition is
\[ 1000 = 530u + 594u^{2}, \quad \text{where } u = \frac{1}{1+r}. \]
Step 2: Rearrange into standard form:
\[ 594u^{2} + 530u - 1000 = 0. \]
Step 3: Solve using the quadratic formula. Discriminant \( = 530^{2} + 4(594)(1000) = 280900 + 2376000 = 2656900 \), and \( \sqrt{2656900} = 1630 \).
\[ u = \frac{-530 + 1630}{1188} = \frac{1100}{1188} = \frac{25}{27}. \]
Step 4: Recover \( r \): since \( u = \frac{1}{1+r} = \frac{25}{27} \), we get \( 1+r = \frac{27}{25} = 1.08 \), so \( r = 0.08 = 8\% \).
Final answer: 8%, option (A).
Was this answer helpful?
0
0
Top CAT Quantitative Aptitude Questions
- In a triangle ABC,AB=AC=8cm. A circle drawn with BC as diameter passes through A. Another circle drawn with center at A passes through B and C. Then the area,in sq.cm,of the overlapping region between the two circles is
- The simple interest accrued on an amount of Rs.25,000 at the end of three years is Rs.7,500. What would be the compound interest accrued on the same amount at the same rate in the same period?
- What is the sum of all roots of the equation \(|x+4|^2–10|x+4|=24\)?
- A solid metal sphere is melted and smaller spheres of equal radii are formed. 10% of the volume of the sphere is lost in the process. The smaller spheres have a radius which is \(\frac{1}{9}\text{th}\) the larger sphere. If 10 litres of paint were needed to paint the larger sphere,how many litres are needed to paint all the smaller spheres?
- The ratio of the ages of a husband and his wife when they got married was 6:5. 4 years and 6 years after their marriage they had their 1st and 2nd children. The sum of the present ages of the husband and wife is 6.4 times the sum of the present ages of their children. The average age of the family at present is 18.5 years. Find the ratio of the ages of the husband and wife when their second child was born.
View More Questions
Top CAT SI & CI Questions
- Raj invested a certain sum of money in two schemes. The first scheme offers a simple interest of 10% per annum, and the second scheme offers a simple interest of 12% per annum. If the total interest earned from both schemes after 3 years is Rs. 1980, and the ratio of the amounts invested in the two schemes is 5:4, what is the amount invested in the second scheme?
- Vrushali invested Rs. (X + 3000) on compound interest of r% p.a. compounded annually for 2 years and Tarushi invested Rs. (X - 2000) on simple interest of (r - 2)% p.a. for 3 years. The total interest received by both of them is together Rs. 12360. If Anant invested Rs. (X + 8000) on simple interest for 3 years on 15% and received an interest of Rs. 13500, find the value of ‘r’.
- Vrushali invested Rs. (X+3000) on compound interest of r% p.a. compounded annually for 2 years and Tarushi invested Rs. (X – 2000) on simple interest of (r-2)% p.a. for 3 years. The total interest received by both of them is together Rs. 12360. If Anant invested Rs. (X + 8000) on simple interest for 3 years on 15% and received an interest of Rs. 13500, find the value of ‘r’.
- Scheme A offers 25% interest and scheme B offers 20% interest. M invested Rs.4000 in scheme A and Rs.6000 in scheme B at compound interest. After 2 years, he invested the amount received from scheme A and B in schemes B and A respectively at simple interest for 2 more years. If another person N invested Rs.6000 in scheme A and Rs.4000 in scheme B at compound interest. After 2 years, he invested the amount received from scheme A and B in schemes B and A respectively at simple interest for 2 more years, then what is the difference between total amount received by M and N after 4 years?
- John borrowed Rs. 2,10,000 from a bank at an interest rate of 10% per annum, compounded annually. The loan was repaid in two equal instalments, the first after one year and the second after another year. The first instalment was interest of one year plus part of the principal amount, while the second was the rest of the principal amount plus due interest thereon. Then each instalment, in Rs., is
View More Questions
Top CAT Questions
- Compare and contrast between Portia and Jessica characters in Merchant of Venice.
- In a triangle ABC,AB=AC=8cm. A circle drawn with BC as diameter passes through A. Another circle drawn with center at A passes through B and C. Then the area,in sq.cm,of the overlapping region between the two circles is
- The simple interest accrued on an amount of Rs.25,000 at the end of three years is Rs.7,500. What would be the compound interest accrued on the same amount at the same rate in the same period?
- What is the sum of all roots of the equation \(|x+4|^2–10|x+4|=24\)?
- A solid metal sphere is melted and smaller spheres of equal radii are formed. 10% of the volume of the sphere is lost in the process. The smaller spheres have a radius which is \(\frac{1}{9}\text{th}\) the larger sphere. If 10 litres of paint were needed to paint the larger sphere,how many litres are needed to paint all the smaller spheres?
View More Questions