Simple Interest MCQ Questions – Medium Level (Set 1)
Practice free Simple Interest MCQ questions with answers — medium level, covering multi-borrower problems, variable interest rates, equal instalment discharge, split investment ratios, date-based calculations, and lending profit scenarios. These questions go beyond basic formulas and test the deeper problem-solving skills needed to score well in quantitative aptitude. Attempt all questions and check your answers instantly with clear explanations.
Q1.A certain principal amount becomes 11/6 of itself in 8 years under simple interest. What is the rate of interest per annum?
View Solution & Explanation
$$
P = 6,\quad A = \frac{11}{6} \times 6 = 11,\quad T = 8 \text{ years}
$$
$$
SI = A - P = 11 - 6 = 5
$$
$$
R = \frac{SI \times 100}{P \times T}
$$
$$
R = \frac{5 \times 100}{6 \times 8} = \frac{500}{48}
$$
$$
R = 12.5\%
$$
Q2.A sum of Rs. 1,500 is borrowed at a simple interest rate of 6% per annum. What will be the total amount to be repaid after 3 years?
View Solution & Explanation
$$
P = 1500,\quad R = 6\%,\quad T = 3 \text{ years}
$$
$$
SI = \frac{P \times R \times T}{100}
$$
$$
SI = \frac{1500 \times 6 \times 3}{100} = \frac{27000}{100} = 270
$$
$$
A = P + SI
$$
$$
A = 1500 + 270 = 1770
$$
$$
\boxed{A = \text{Rs. } 1770}
$$
Q3.Ramesh lent Rs. 6,000 to Suresh for 3 years and Rs. 4,000 to Mahesh for 5 years at the same rate of simple interest. If he received a total interest of Rs. 3,800 from both, what is the rate of interest per annum?
View Solution & Explanation
$$
P_1 = 6000,\quad T_1 = 3 \text{ years}
$$
$$
P_2 = 4000,\quad T_2 = 5 \text{ years}
$$
$$
\text{Total SI} = \text{Rs. } 3800,\quad R = ?
$$
$$
SI_1 = \frac{6000 \times R \times 3}{100} = 180R
$$
$$
SI_2 = \frac{4000 \times R \times 5}{100} = 200R
$$
$$
SI_1 + SI_2 = 3800
$$
$$
180R + 200R = 3800
$$
$$
380R = 3800
$$
$$
R = \frac{3800}{380} = 10\%
$$
$$
\boxed{R = 10\%}
$$
Q4.A moneylender borrows Rs. 8,000 for 3 years at 5% p.a. simple interest and immediately lends it to someone else at 7½% p.a. for 3 years. What is his total gain in the entire transaction per year?
View Solution & Explanation
$$
P = 8000,\quad T = 3 \text{ years}
$$
$$
R_1 = 5\% \ (\text{borrowing rate}), \quad R_2 = 7.5\% \ (\text{lending rate})
$$
$$
SI_1 = \frac{P \times R_1 \times T}{100}
$$
$$
SI_1 = \frac{8000 \times 5 \times 3}{100} = 1200
$$
$$
SI_2 = \frac{P \times R_2 \times T}{100}
$$
$$
SI_2 = \frac{8000 \times 7.5 \times 3}{100} = 1800
$$
$$
\text{Total Gain} = SI_2 - SI_1 = 1800 - 1200 = 600
$$
$$
\text{Step 4: } \text{Gain per year} = \frac{\text{Total Gain}}{T} = \frac{600}{3} = 200
$$
$$
\boxed{\text{Gain per year} = \text{Rs. } 200}
$$
$$
\textbf{Quick Tip: } \text{Gain per year} = \frac{P \times (R_2 - R_1)}{100}
$$
$$
= \frac{8000 \times 2.5}{100} = 200
$$
Q5.Rs. 15,000 is divided into two parts such that the simple interest on the first part for 4 years at 10% p.a. is equal to the simple interest on the second part for 3 years at 16% p.a. Find the greater part.
View Solution & Explanation
$$
R_1 = 10\%,\quad T_1 = 4 \text{ years}, \quad R_2 = 16\%,\quad T_2 = 3 \text{ years}
$$
$$
40P_1 = 48P_2 \Rightarrow \frac{P_1}{P_2} = \frac{6}{5}
$$
$$
P_1 = \frac{6}{11} \times 15400 = 8400
$$
$$
P_2 = \frac{5}{11} \times 15400 = 7000
$$
$$
\boxed{\text{Greater Part} = P_1 = \text{Rs. } 8400}
$$
$$
\textbf{Quick Tip:}
$$
$$
\text{When } SI_1 = SI_2:
$$
$$
P_1 \times R_1 \times T_1 = P_2 \times R_2 \times T_2
$$
$$
\frac{P_1}{P_2} = \frac{R_2 \times T_2}{R_1 \times T_1}
$$
$$
\text{Just find the ratio first, then split the total! No need for simultaneous equations.}
$$
Q6.At what rate of simple interest per annum will the interest on a principal amount become equal to the principal itself in 8 years?
View Solution & Explanation
$$
SI = P,\quad T = 8 \text{ years},\quad R = ?
$$
$$
SI = \frac{P \times R \times T}{100}
$$
$$
SI = P:
$$
$$
P = \frac{P \times R \times 8}{100}
$$
$$
P \text{ on both sides:}
$$
$$
1 = \frac{R \times 8}{100}
$$
$$
\text{Step 4: Solve for } R:
$$
$$
R = \frac{100}{8} = 12.5\%
$$
$$
\boxed{R = 12.5\%}
$$
$$
\textbf{Quick Tip:}
$$
$$
\text{When } SI = P:
$$
$$
R = \frac{100}{T} \quad \text{or} \quad T = \frac{100}{R}
$$
$$
R \times T = 100
$$
Q7.Priya took a loan at a simple interest rate of 8% p.a. in the first year, which increased by 2% p.a. every subsequent year. If she paid Rs. 9,000 as total interest at the end of 3 years, what was her loan amount?
View Solution & Explanation
$$
R_1 = 8\%, \quad R_2 = 10\%, \quad R_3 = 12\%
$$
$$
T = 1 \text{ year each}, \quad \text{Total SI} = \text{Rs. } 9000, \quad P = ?
$$
$$
SI_1 = \frac{8P}{100}, \quad SI_2 = \frac{10P}{100}, \quad SI_3 = \frac{12P}{100}
$$
$$
\frac{8P}{100} + \frac{10P}{100} + \frac{12P}{100} = 9000
$$
$$
\frac{30P}{100} = 9000
$$
$$
P = \frac{9000 \times 100}{30} = 30000
$$
$$
\boxed{P = \text{Rs. } 30{,}000}
$$
$$
\textbf{Quick Tip:}
$$
$$
\text{Total SI} = \frac{P}{100} \times (R_1 + R_2 + R_3)
$$
$$
\frac{P}{100} \times (8 + 10 + 12) = 9000
$$
$$
\frac{30P}{100} = 9000 \Rightarrow P = 30000
$$
Q8.Mohan lent Rs. 5,000 to Rohan for 2 years and Rs. 3,000 to Sohan for 4 years at the same rate of simple interest. If he received a total interest of Rs. 2,200 from both, what is the rate of interest per annum?
View Solution & Explanation
$$
P_1 = 5000,\quad T_1 = 2 \text{ years}, \quad P_2 = 3000,\quad T_2 = 4 \text{ years}
$$
$$
\text{Total SI} = \text{Rs. } 2200,\quad R = ?
$$
$$
SI_1 = \frac{5000 \times R \times 2}{100} = 100R,\quad
SI_2 = \frac{3000 \times R \times 4}{100} = 120R
$$
$$
100R + 120R = 2200
$$
$$
220R = 2200
$$
$$
R = \frac{2200}{220} = 10\%
$$
$$
\boxed{R = 10\%}
$$
$$
\textbf{Quick Tip:}
$$
$$
\text{Total SI} = \frac{R}{100} \times (P_1 \times T_1 + P_2 \times T_2)
$$
$$
\frac{R}{100} \times (5000 \times 2 + 3000 \times 4) = 2200
$$
$$
\frac{R}{100} \times 22000 = 2200 \Rightarrow R = 10\%
$$
Q9.What equal annual installment will discharge a debt of Rs. 1,056 due at the end of 4 years at 4% per annum simple interest?
View Solution & Explanation
$$
\text{Total Debt} = \text{Rs. } 1056,\quad R = 4\%,\quad n = 4
$$
$$
\text{Let each installment} = x
$$
$$
x\left(\frac{112}{100}\right) + x\left(\frac{108}{100}\right) + x\left(\frac{104}{100}\right) + x\left(\frac{100}{100}\right) = 1056
$$
$$
\frac{x}{100}(112 + 108 + 104 + 100) = 1056
$$
$$
\frac{x}{100} \times 424 = 1056
$$
$$
x = \frac{1056 \times 100}{424} = 250
$$
$$
\boxed{x = \text{Rs. } 250}
$$
$$
\textbf{Quick Tip:}
$$
$$
x = \frac{\text{Debt} \times 100}{100n + R \times \frac{n(n-1)}{2}}
$$
$$
x = \frac{1056 \times 100}{100 \times 4 + 4 \times \frac{4 \times 3}{2}} = \frac{105600}{424} = 250
$$
Q10.The simple interest on a certain sum of money at the rate of 7% p.a. for 6 years is Rs. 1,260. At what rate of interest will the same amount of interest be received on the same sum in 4 years?
View Solution & Explanation
$$
R_1 = 7\%,\quad T_1 = 6 \text{ years},\quad SI = \text{Rs. } 1260
$$
$$
T_2 = 4 \text{ years},\quad R_2 = ?
$$
$$
1260 = \frac{P \times 7 \times 6}{100} = \frac{42P}{100}
$$
$$
P = \frac{1260 \times 100}{42} = 3000
$$
$$
1260 = \frac{3000 \times R_2 \times 4}{100}
$$
$$
1260 = 120 R_2
$$
$$
R_2 = \frac{1260}{120} = 10.5\%
$$
$$
\boxed{R_2 = 10.5\%}
$$
$$
\textbf{Quick Tip:}
$$
$$
R_1 \times T_1 = R_2 \times T_2
$$
$$
7 \times 6 = R_2 \times 4 \Rightarrow R_2 = \frac{42}{4} = 10.5\%
$$
Q11.A person invests money in three different schemes for 4 years, 8 years and 10 years at 12%, 15% and 18% simple interest respectively. At the completion of each scheme, he gets the same interest. What is the ratio of his investments?
View Solution & Explanation
$$
T_1 = 4,\quad T_2 = 8,\quad T_3 = 10 \text{ years}
$$
$$
R_1 = 12\%,\quad R_2 = 15\%,\quad R_3 = 18\%
$$
$$
SI_1 = SI_2 = SI_3
$$
$$
P_1 \times R_1 \times T_1 = P_2 \times R_2 \times T_2 = P_3 \times R_3 \times T_3
$$
$$
R_1T_1 = 48,\quad R_2T_2 = 120,\quad R_3T_3 = 180
$$
$$
P_1 : P_2 : P_3 = \frac{1}{48} : \frac{1}{120} : \frac{1}{180}
$$
$$
= \frac{720}{48} : \frac{720}{120} : \frac{720}{180}
$$
$$
P_1 : P_2 : P_3 = 15 : 6 : 4
$$
$$
\boxed{P_1 : P_2 : P_3 = 15 : 6 : 4}
$$
$$
\textbf{Quick Tip:}
$$
$$
P \propto \frac{1}{R \times T}
$$
$$
P_1 : P_2 : P_3 = \frac{1}{R_1T_1} : \frac{1}{R_2T_2} : \frac{1}{R_3T_3}
$$
$$
= \frac{1}{48} : \frac{1}{120} : \frac{1}{180} = 15 : 6 : 4
$$
Q12.The simple interest on a certain sum for 9 months at 8% per annum exceeds the simple interest on the same sum for 6 months at 10% per annum by Rs. 60. Find the sum.
View Solution & Explanation
$$
T_1 = 9 \text{ months} = \frac{3}{4} \text{ years},\quad R_1 = 8\%
$$
$$
T_2 = 6 \text{ months} = \frac{1}{2} \text{ years},\quad R_2 = 10\%
$$
$$
SI_1 - SI_2 = \text{Rs. } 60,\quad P = ?
$$
$$
SI_1 = \frac{P \times 8 \times \frac{3}{4}}{100} = \frac{6P}{100},\quad
SI_2 = \frac{P \times 10 \times \frac{1}{2}}{100} = \frac{5P}{100}
$$
$$
\frac{6P}{100} - \frac{5P}{100} = 60
$$
$$
\frac{P}{100} = 60 \Rightarrow P = 6000
$$
$$
\boxed{P = \text{Rs. } 6000}
$$
$$
SI_1 = \frac{6 \times 6000}{100} = 360,\quad
SI_2 = \frac{5 \times 6000}{100} = 300
$$
$$
SI_1 - SI_2 = 360 - 300 = 60
$$
$$
\textbf{Quick Tip:}
$$
$$
SI_1 - SI_2 = \frac{P}{100}(R_1T_1 - R_2T_2)
$$
$$
= \frac{P}{100}\left(8 \times \frac{3}{4} - 10 \times \frac{1}{2}\right)
= \frac{P}{100}(6 - 5) = \frac{P}{100}
$$
$$
\frac{P}{100} = 60 \Rightarrow P = 6000
$$
Q13.Suresh borrowed a sum of money from Mahesh at 8% per annum simple interest for the first 4 years, 10% per annum for the next 6 years and 12% per annum beyond 10 years. If he pays a total interest of Rs. 12,160 at the end of 15 years, how much money did he borrow?
View Solution & Explanation
$$
R_1 = 8\%,\quad T_1 = 4 \text{ years},\quad R_2 = 10\%,\quad T_2 = 6 \text{ years},\quad R_3 = 12\%,\quad T_3 = 5 \text{ years}
$$
$$
\text{Total SI} = \text{Rs. } 12160,\quad P = ?
$$
$$
SI_1 = \frac{32P}{100},\quad SI_2 = \frac{60P}{100},\quad SI_3 = \frac{60P}{100}
$$
$$
\frac{32P + 60P + 60P}{100} = 12160
$$
$$
\frac{152P}{100} = 12160
$$
$$
P = \frac{12160 \times 100}{152} = 8000
$$
$$
\boxed{P = \text{Rs. } 8000}
$$
$$
SI_1 = 2560,\quad SI_2 = 4800,\quad SI_3 = 4800
$$
$$
2560 + 4800 + 4800 = 12160
$$
$$
\textbf{Quick Tip:}
$$
$$
\text{Total SI} = \frac{P}{100}(R_1T_1 + R_2T_2 + R_3T_3)
$$
$$
= \frac{P}{100}(32 + 60 + 60) = \frac{152P}{100}
$$
Q14.A sum of money was invested at a certain rate of simple interest for 3 years. Had it been invested at 2% higher rate, it would have fetched Rs. 90 more as interest. Find the sum of money.
View Solution & Explanation
$$
T = 3 \text{ years},\quad \Delta R = 2\%,\quad \Delta SI = \text{Rs. } 90,\quad P = ?
$$
$$
\Delta SI = \frac{P \times \Delta R \times T}{100}
$$
$$
90 = \frac{P \times 2 \times 3}{100} = \frac{6P}{100}
$$
$$
P = \frac{90 \times 100}{6} = 1500
$$
$$
\boxed{P = \text{Rs. } 1500}
$$
$$
\Delta SI = \frac{1500 \times 2 \times 3}{100} = 90
$$
$$
\textbf{Quick Tip:}
$$
$$
P = \frac{\Delta SI \times 100}{\Delta R \times T}
$$
$$
= \frac{90 \times 100}{2 \times 3} = 1500
$$
Q15.Vikram borrowed Rs. 1,500 from Anand at 7% p.a. simple interest for 3 years. He added some more money to it and lent the entire sum to Deepak for the same period at 10% p.a. simple interest. If Vikram gains Rs. 345 from the whole transaction, find the sum lent to Deepak.
View Solution & Explanation
$$
P_1 = \text{Rs. } 1500,\quad R_1 = 7\%,\quad T = 3 \text{ years},\quad R_2 = 10\%,\quad \text{Gain} = \text{Rs. } 345,\quad x = ?
$$
$$
SI_1 = \frac{1500 \times 7 \times 3}{100} = 315
$$
$$
SI_2 = \frac{x \times 10 \times 3}{100} = \frac{30x}{100}
$$
$$
\frac{30x}{100} - 315 = 345
$$
$$
\frac{30x}{100} = 660
$$
$$
x = \frac{660 \times 100}{30} = 2200
$$
$$
\boxed{x = \text{Rs. } 2200}
$$
$$
SI_2 = \frac{2200 \times 10 \times 3}{100} = 660
$$
$$
660 - 315 = 345
$$
$$
\textbf{Quick Tip:}
$$
$$
x = \frac{(SI_{\text{paid}} + \text{Gain}) \times 100}{R_2 \times T}
$$
$$
= \frac{(315 + 345) \times 100}{10 \times 3} = 2200
$$
Q16.Meena deposited a sum of money with a bank on 1st January 2015 at 8% simple interest per annum. She received Rs. 3,144 on 7th August 2015. Find the amount she deposited.
View Solution & Explanation
$$
R = 8\%,\quad A = \text{Rs. } 3144,\quad P = ?
$$
$$
\text{Total days} = 31 + 28 + 31 + 30 + 31 + 30 + 31 + 7 = 219
$$
$$
T = \frac{219}{365} = \frac{3}{5} \text{ years}
$$
$$
A = P\left(1 + \frac{R \times T}{100}\right)
$$
$$
3144 = P\left(1 + \frac{8 \times \frac{3}{5}}{100}\right)
$$
$$
3144 = P\left(1 + \frac{24}{500}\right) = P \times \frac{524}{500}
$$
$$
P = \frac{3144 \times 500}{524} = 3000
$$
$$
\boxed{P = \text{Rs. } 3000}
$$
$$
SI = \frac{3000 \times 8 \times \frac{3}{5}}{100} = \frac{3000 \times 24}{500} = 144
$$
$$
A = 3000 + 144 = 3144
$$
$$
\textbf{Quick Tip:}
$$
$$
\frac{73}{365} = \frac{1}{5}, \quad \frac{146}{365} = \frac{2}{5}, \quad \frac{219}{365} = \frac{3}{5}, \quad \frac{292}{365} = \frac{4}{5}
$$
Q17.Suresh invested some amount at 12% simple interest and another amount at 10% simple interest. He received a yearly interest of Rs. 195. Had he interchanged the amounts, he would have received Rs. 6 more as interest. How much did he invest at 12%?
View Solution & Explanation
$$
\text{Let amount at } 12\% = x,\quad \text{amount at } 10\% = y
$$
$$
\frac{12x + 10y}{100} = 195,\quad \frac{10x + 12y}{100} = 201
$$
$$
12x + 10y = 19500,\quad 10x + 12y = 20100
$$
$$
22x + 22y = 39600 \Rightarrow x + y = 1800
$$
$$
-2x + 2y = 600 \Rightarrow y - x = 300
$$
$$
x = \frac{(x+y) - (y-x)}{2} = \frac{1800 - 300}{2} = 750
$$
$$
y = 1050
$$
$$
\boxed{x = \text{Rs. } 750}
$$
$$
\frac{750 \times 12}{100} + \frac{1050 \times 10}{100} = 90 + 105 = 195
$$
$$
\frac{1050 \times 12}{100} + \frac{750 \times 10}{100} = 126 + 75 = 201
$$
$$
\text{Extra Interest} = 201 - 195 = 6
$$
$$
\textbf{Quick Tip:}
$$
$$
x + y = \frac{(\text{SI}_1 + \text{SI}_2) \times 100}{R_1 + R_2}
$$
$$
y - x = \frac{(\text{SI}_2 - \text{SI}_1) \times 100}{R_1 - R_2}
$$
$$
x = \frac{(x+y) - (y-x)}{2}
$$
Q18.A sum of Rs. 7,930 is divided into 3 parts and given as loan at 5% simple interest to Anil, Balu and Charan for 2, 3 and 4 years respectively. If the amounts received from all three are equal after their respective periods, how much loan did Anil receive?
View Solution & Explanation
$$
P_1 + P_2 + P_3 = 7930,\quad R = 5\%,\quad T_1 = 2,\quad T_2 = 3,\quad T_3 = 4 \text{ years}
$$
$$
P_1\left(1 + \frac{5 \times 2}{100}\right)
= P_2\left(1 + \frac{5 \times 3}{100}\right)
= P_3\left(1 + \frac{5 \times 4}{100}\right)
$$
$$
\frac{110P_1}{100} = \frac{115P_2}{100} = \frac{120P_3}{100}
\Rightarrow 110P_1 = 115P_2 = 120P_3
$$
$$
P_1 : P_2 : P_3 = \frac{1}{110} : \frac{1}{115} : \frac{1}{120}
$$
$$
= \frac{30360}{110} : \frac{30360}{115} : \frac{30360}{120}
= 276 : 264 : 253
$$
$$
276 + 264 + 253 = 793
$$
$$
P_1 = \frac{276}{793} \times 7930 = 2760
$$
$$
\boxed{P_1 = \text{Rs. } 2760}
$$
$$
A_1 = 2760 \times \frac{110}{100} = 3036,\quad
A_2 = 2640 \times \frac{115}{100} = 3036,\quad
A_3 = 2530 \times \frac{120}{100} = 3036
$$
$$
\textbf{Quick Tip:}
$$
$$
P \propto \frac{1}{\left(1 + \frac{RT}{100}\right)}
$$
$$
P_1 : P_2 : P_3 = \frac{100}{110} : \frac{100}{115} : \frac{100}{120}
= 276 : 264 : 253
$$
Q19.Deepak had Rs. 12,000 with him. He lent some money to Akash for 3 years at 14% simple interest and the remaining to Bikash for the same period at 18% simple interest. After 3 years, Akash paid Rs. 720 more as interest compared to Bikash. How much money did Deepak lend to Bikash?
View Solution & Explanation
$$
\text{Total} = \text{Rs. } 12000,\quad \text{Let amount to Akash} = x,\quad \text{Amount to Bikash} = 12000 - x
$$
$$
R_A = 14\%,\quad R_B = 18\%,\quad T = 3 \text{ years},\quad SI_A - SI_B = 720
$$
$$
SI_A = \frac{42x}{100},\quad SI_B = \frac{54(12000 - x)}{100}
$$
$$
\frac{42x}{100} - \frac{54(12000 - x)}{100} = 720
$$
$$
42x - 54(12000 - x) = 72000
$$
$$
42x - 648000 + 54x = 72000
$$
$$
96x = 720000 \Rightarrow x = 7500
$$
$$
\text{Amount to Bikash} = 12000 - 7500 = 4500
$$
$$
\boxed{\text{Amount to Bikash} = \text{Rs. } 4500}
$$
$$
SI_A = 3150,\quad SI_B = 2430,\quad SI_A - SI_B = 720
$$
$$
\textbf{Quick Tip:}
$$
$$
SI = \frac{P \times R \times T}{100},\quad \text{write both SIs and apply the condition directly}
$$
Q20.Kavya borrowed Rs. 15,000 at 12% per annum from a moneylender on 13th January 2019 and returned the amount on 8th June 2019 to clear her debt. What was the total amount paid by Kavya?
View Solution & Explanation
$$
P = \text{Rs. } 15000,\quad R = 12\%
$$
$$
\text{Start: 13th Jan 2019},\quad \text{End: 8th June 2019}
$$
$$
\text{Total days} = 18 + 28 + 31 + 30 + 31 + 8 = 146
$$
$$
T = \frac{146}{365} = \frac{2}{5} \text{ years}
$$
$$
SI = \frac{15000 \times 12 \times \frac{2}{5}}{100}
= \frac{15000 \times 24}{500}
= 720
$$
$$
A = 15000 + 720 = 15720
$$
$$
\boxed{A = \text{Rs. } 15720}
$$
$$
\textbf{Quick Tip:}
$$
$$
73 = \frac{1}{5}\text{ yr},\quad 146 = \frac{2}{5}\text{ yr},\quad 219 = \frac{3}{5}\text{ yr},\quad 292 = \frac{4}{5}\text{ yr}
$$
Progress Tracker
Total Questions: 20
Answered: 0
Correct: 0