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Simple Interest MCQ Questions – Medium Level (Set 1)

20 questions

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.

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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

Correct Answer 12.5%

P=6,A=116×6=11,T=8 yearsP = 6,\quad A = \frac{11}{6} \times 6 = 11,\quad T = 8 \text{ years}
SI=A−P=11−6=5SI = A - P = 11 - 6 = 5
R=SI×100P×TR = \frac{SI \times 100}{P \times T}
R=5×1006×8=50048R = \frac{5 \times 100}{6 \times 8} = \frac{500}{48}
R=12.5%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

Correct Answer Rs. 1,770

P=1500,R=6%,T=3 yearsP = 1500,\quad R = 6\%,\quad T = 3 \text{ years}
SI=P×R×T100SI = \frac{P \times R \times T}{100}
SI=1500×6×3100=27000100=270SI = \frac{1500 \times 6 \times 3}{100} = \frac{27000}{100} = 270
A=P+SIA = P + SI
A=1500+270=1770A = 1500 + 270 = 1770
A=Rs. 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

Correct Answer 10%

P1=6000,T1=3 yearsP_1 = 6000,\quad T_1 = 3 \text{ years}
P2=4000,T2=5 yearsP_2 = 4000,\quad T_2 = 5 \text{ years}
Total SI=Rs. 3800,R=?\text{Total SI} = \text{Rs. } 3800,\quad R = ?
SI1=6000×R×3100=180RSI_1 = \frac{6000 \times R \times 3}{100} = 180R
SI2=4000×R×5100=200RSI_2 = \frac{4000 \times R \times 5}{100} = 200R
SI1+SI2=3800SI_1 + SI_2 = 3800
180R+200R=3800180R + 200R = 3800
380R=3800380R = 3800
R=3800380=10%R = \frac{3800}{380} = 10\%
R=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

Correct Answer Rs. 200

P=8000,T=3 yearsP = 8000,\quad T = 3 \text{ years}
R1=5% (borrowing rate),R2=7.5% (lending rate)R_1 = 5\% \ (\text{borrowing rate}), \quad R_2 = 7.5\% \ (\text{lending rate})
SI1=P×R1×T100SI_1 = \frac{P \times R_1 \times T}{100}
SI1=8000×5×3100=1200SI_1 = \frac{8000 \times 5 \times 3}{100} = 1200
SI2=P×R2×T100SI_2 = \frac{P \times R_2 \times T}{100}
SI2=8000×7.5×3100=1800SI_2 = \frac{8000 \times 7.5 \times 3}{100} = 1800
Total Gain=SI2−SI1=1800−1200=600\text{Total Gain} = SI_2 - SI_1 = 1800 - 1200 = 600
Step 4: Gain per year=Total GainT=6003=200\text{Step 4: } \text{Gain per year} = \frac{\text{Total Gain}}{T} = \frac{600}{3} = 200
Gain per year=Rs. 200\boxed{\text{Gain per year} = \text{Rs. } 200}
Quick Tip: Gain per year=P×(R2−R1)100\textbf{Quick Tip: } \text{Gain per year} = \frac{P \times (R_2 - R_1)}{100}
=8000×2.5100=200= \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

Correct Answer Rs. 8,400

R1=10%,T1=4 years,R2=16%,T2=3 yearsR_1 = 10\%,\quad T_1 = 4 \text{ years}, \quad R_2 = 16\%,\quad T_2 = 3 \text{ years}
40P1=48P2⇒P1P2=6540P_1 = 48P_2 \Rightarrow \frac{P_1}{P_2} = \frac{6}{5}
P1=611×15400=8400P_1 = \frac{6}{11} \times 15400 = 8400
P2=511×15400=7000P_2 = \frac{5}{11} \times 15400 = 7000
Greater Part=P1=Rs. 8400\boxed{\text{Greater Part} = P_1 = \text{Rs. } 8400}
Quick Tip:\textbf{Quick Tip:}
When SI1=SI2:\text{When } SI_1 = SI_2:
P1×R1×T1=P2×R2×T2P_1 \times R_1 \times T_1 = P_2 \times R_2 \times T_2
P1P2=R2×T2R1×T1\frac{P_1}{P_2} = \frac{R_2 \times T_2}{R_1 \times T_1}
Just find the ratio first, then split the total! No need for simultaneous equations.\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

Correct Answer 12.5%

SI=P,T=8 years,R=?SI = P,\quad T = 8 \text{ years},\quad R = ?
SI=P×R×T100SI = \frac{P \times R \times T}{100}
SI=P:SI = P:
P=P×R×8100P = \frac{P \times R \times 8}{100}
P on both sides:P \text{ on both sides:}
1=R×81001 = \frac{R \times 8}{100}
Step 4: Solve for R:\text{Step 4: Solve for } R:
R=1008=12.5%R = \frac{100}{8} = 12.5\%
R=12.5%\boxed{R = 12.5\%}
Quick Tip:\textbf{Quick Tip:}
When SI=P:\text{When } SI = P:
R=100TorT=100RR = \frac{100}{T} \quad \text{or} \quad T = \frac{100}{R}
R×T=100R \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

Correct Answer Rs. 30,000

R1=8%,R2=10%,R3=12%R_1 = 8\%, \quad R_2 = 10\%, \quad R_3 = 12\%
T=1 year each,Total SI=Rs. 9000,P=?T = 1 \text{ year each}, \quad \text{Total SI} = \text{Rs. } 9000, \quad P = ?
SI1=8P100,SI2=10P100,SI3=12P100SI_1 = \frac{8P}{100}, \quad SI_2 = \frac{10P}{100}, \quad SI_3 = \frac{12P}{100}
8P100+10P100+12P100=9000\frac{8P}{100} + \frac{10P}{100} + \frac{12P}{100} = 9000
30P100=9000\frac{30P}{100} = 9000
P=9000×10030=30000P = \frac{9000 \times 100}{30} = 30000
P=Rs. 30,000\boxed{P = \text{Rs. } 30{,}000}
Quick Tip:\textbf{Quick Tip:}
Total SI=P100×(R1+R2+R3)\text{Total SI} = \frac{P}{100} \times (R_1 + R_2 + R_3)
P100×(8+10+12)=9000\frac{P}{100} \times (8 + 10 + 12) = 9000
30P100=9000⇒P=30000\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

Correct Answer 10

P1=5000,T1=2 years,P2=3000,T2=4 yearsP_1 = 5000,\quad T_1 = 2 \text{ years}, \quad P_2 = 3000,\quad T_2 = 4 \text{ years}
Total SI=Rs. 2200,R=?\text{Total SI} = \text{Rs. } 2200,\quad R = ?
SI1=5000×R×2100=100R,</p><p>SI2=3000×R×4100=120RSI_1 = \frac{5000 \times R \times 2}{100} = 100R,\quad</p><p>SI_2 = \frac{3000 \times R \times 4}{100} = 120R
100R+120R=2200100R + 120R = 2200
220R=2200220R = 2200
R=2200220=10%R = \frac{2200}{220} = 10\%
R=10%\boxed{R = 10\%}
Quick Tip:\textbf{Quick Tip:}
Total SI=R100×(P1×T1+P2×T2)\text{Total SI} = \frac{R}{100} \times (P_1 \times T_1 + P_2 \times T_2)
R100×(5000×2+3000×4)=2200\frac{R}{100} \times (5000 \times 2 + 3000 \times 4) = 2200
R100×22000=2200⇒R=10%\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

Correct Answer Rs. 250

Total Debt=Rs. 1056,R=4%,n=4\text{Total Debt} = \text{Rs. } 1056,\quad R = 4\%,\quad n = 4
Let each installment=x\text{Let each installment} = x
x(112100)+x(108100)+x(104100)+x(100100)=1056x\left(\frac{112}{100}\right) + x\left(\frac{108}{100}\right) + x\left(\frac{104}{100}\right) + x\left(\frac{100}{100}\right) = 1056
x100(112+108+104+100)=1056\frac{x}{100}(112 + 108 + 104 + 100) = 1056
x100×424=1056\frac{x}{100} \times 424 = 1056
x=1056×100424=250x = \frac{1056 \times 100}{424} = 250
x=Rs. 250\boxed{x = \text{Rs. } 250}
Quick Tip:\textbf{Quick Tip:}
x=Debt×100100n+R×n(n−1)2x = \frac{\text{Debt} \times 100}{100n + R \times \frac{n(n-1)}{2}}
x=1056×100100×4+4×4×32=105600424=250x = \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

Correct Answer 10.5%

R1=7%,T1=6 years,SI=Rs. 1260R_1 = 7\%,\quad T_1 = 6 \text{ years},\quad SI = \text{Rs. } 1260
T2=4 years,R2=?T_2 = 4 \text{ years},\quad R_2 = ?
1260=P×7×6100=42P1001260 = \frac{P \times 7 \times 6}{100} = \frac{42P}{100}
P=1260×10042=3000P = \frac{1260 \times 100}{42} = 3000
1260=3000×R2×41001260 = \frac{3000 \times R_2 \times 4}{100}
1260=120R21260 = 120 R_2
R2=1260120=10.5%R_2 = \frac{1260}{120} = 10.5\%
R2=10.5%\boxed{R_2 = 10.5\%}
Quick Tip:\textbf{Quick Tip:}
R1×T1=R2×T2R_1 \times T_1 = R_2 \times T_2
7×6=R2×4⇒R2=424=10.5%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

Correct Answer 15 : 6 : 4

T1=4,T2=8,T3=10 yearsT_1 = 4,\quad T_2 = 8,\quad T_3 = 10 \text{ years}
R1=12%,R2=15%,R3=18%R_1 = 12\%,\quad R_2 = 15\%,\quad R_3 = 18\%
SI1=SI2=SI3SI_1 = SI_2 = SI_3
P1×R1×T1=P2×R2×T2=P3×R3×T3P_1 \times R_1 \times T_1 = P_2 \times R_2 \times T_2 = P_3 \times R_3 \times T_3
R1T1=48,R2T2=120,R3T3=180R_1T_1 = 48,\quad R_2T_2 = 120,\quad R_3T_3 = 180
P1:P2:P3=148:1120:1180P_1 : P_2 : P_3 = \frac{1}{48} : \frac{1}{120} : \frac{1}{180}
=72048:720120:720180= \frac{720}{48} : \frac{720}{120} : \frac{720}{180}
P1:P2:P3=15:6:4P_1 : P_2 : P_3 = 15 : 6 : 4
P1:P2:P3=15:6:4\boxed{P_1 : P_2 : P_3 = 15 : 6 : 4}
Quick Tip:\textbf{Quick Tip:}
P∝1R×TP \propto \frac{1}{R \times T}
P1:P2:P3=1R1T1:1R2T2:1R3T3P_1 : P_2 : P_3 = \frac{1}{R_1T_1} : \frac{1}{R_2T_2} : \frac{1}{R_3T_3}
=148:1120:1180=15:6:4= \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

Correct Answer Rs. 6,000

T1=9 months=34 years,R1=8%T_1 = 9 \text{ months} = \frac{3}{4} \text{ years},\quad R_1 = 8\%
T2=6 months=12 years,R2=10%T_2 = 6 \text{ months} = \frac{1}{2} \text{ years},\quad R_2 = 10\%
SI1−SI2=Rs. 60,P=?SI_1 - SI_2 = \text{Rs. } 60,\quad P = ?
SI1=P×8×34100=6P100,</p><pclass="mb−4leading−7text−gray−800">SI2=P×10×12100=5P100SI_1 = \frac{P \times 8 \times \frac{3}{4}}{100} = \frac{6P}{100},\quad</p><p class="mb-4 leading-7 text-gray-800">SI_2 = \frac{P \times 10 \times \frac{1}{2}}{100} = \frac{5P}{100}
6P100−5P100=60\frac{6P}{100} - \frac{5P}{100} = 60
P100=60⇒P=6000\frac{P}{100} = 60 \Rightarrow P = 6000
P=Rs. 6000\boxed{P = \text{Rs. } 6000}
SI1=6×6000100=360,</p><pclass="mb−4leading−7text−gray−800">SI2=5×6000100=300SI_1 = \frac{6 \times 6000}{100} = 360,\quad</p><p class="mb-4 leading-7 text-gray-800">SI_2 = \frac{5 \times 6000}{100} = 300
SI1−SI2=360−300=60SI_1 - SI_2 = 360 - 300 = 60
Quick Tip:\textbf{Quick Tip:}
SI1−SI2=P100(R1T1−R2T2)SI_1 - SI_2 = \frac{P}{100}(R_1T_1 - R_2T_2)
=P100(8×34−10×12)</p><p>=P100(6−5)=P100= \frac{P}{100}\left(8 \times \frac{3}{4} - 10 \times \frac{1}{2}\right)</p><p>= \frac{P}{100}(6 - 5) = \frac{P}{100}
P100=60⇒P=6000\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

Correct Answer Rs. 8,000

R1=8%,T1=4 years,R2=10%,T2=6 years,R3=12%,T3=5 yearsR_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}
Total SI=Rs. 12160,P=?\text{Total SI} = \text{Rs. } 12160,\quad P = ?
SI1=32P100,SI2=60P100,SI3=60P100SI_1 = \frac{32P}{100},\quad SI_2 = \frac{60P}{100},\quad SI_3 = \frac{60P}{100}
32P+60P+60P100=12160\frac{32P + 60P + 60P}{100} = 12160
152P100=12160\frac{152P}{100} = 12160
P=12160×100152=8000P = \frac{12160 \times 100}{152} = 8000
P=Rs. 8000\boxed{P = \text{Rs. } 8000}
SI1=2560,SI2=4800,SI3=4800SI_1 = 2560,\quad SI_2 = 4800,\quad SI_3 = 4800
2560+4800+4800=121602560 + 4800 + 4800 = 12160
Quick Tip:\textbf{Quick Tip:}
Total SI=P100(R1T1+R2T2+R3T3)\text{Total SI} = \frac{P}{100}(R_1T_1 + R_2T_2 + R_3T_3)
=P100(32+60+60)=152P100= \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

Correct Answer Rs. 1,500

T=3 years,ΔR=2%,ΔSI=Rs. 90,P=?T = 3 \text{ years},\quad \Delta R = 2\%,\quad \Delta SI = \text{Rs. } 90,\quad P = ?
ΔSI=P×ΔR×T100\Delta SI = \frac{P \times \Delta R \times T}{100}
90=P×2×3100=6P10090 = \frac{P \times 2 \times 3}{100} = \frac{6P}{100}
P=90×1006=1500P = \frac{90 \times 100}{6} = 1500
P=Rs. 1500\boxed{P = \text{Rs. } 1500}
ΔSI=1500×2×3100=90\Delta SI = \frac{1500 \times 2 \times 3}{100} = 90
Quick Tip:\textbf{Quick Tip:}
P=ΔSI×100ΔR×TP = \frac{\Delta SI \times 100}{\Delta R \times T}
=90×1002×3=1500= \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

Correct Answer Rs. 2,200

P1=Rs. 1500,R1=7%,T=3 years,R2=10%,Gain=Rs. 345,x=?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 = ?
SI1=1500×7×3100=315SI_1 = \frac{1500 \times 7 \times 3}{100} = 315
SI2=x×10×3100=30x100SI_2 = \frac{x \times 10 \times 3}{100} = \frac{30x}{100}
30x100−315=345\frac{30x}{100} - 315 = 345
30x100=660\frac{30x}{100} = 660
x=660×10030=2200x = \frac{660 \times 100}{30} = 2200
x=Rs. 2200\boxed{x = \text{Rs. } 2200}
SI2=2200×10×3100=660SI_2 = \frac{2200 \times 10 \times 3}{100} = 660
660−315=345660 - 315 = 345
Quick Tip:\textbf{Quick Tip:}
x=(SIpaid+Gain)×100R2×Tx = \frac{(SI_{\text{paid}} + \text{Gain}) \times 100}{R_2 \times T}
=(315+345)×10010×3=2200= \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

Correct Answer Rs. 3,000

R=8%,A=Rs. 3144,P=?R = 8\%,\quad A = \text{Rs. } 3144,\quad P = ?
Total days=31+28+31+30+31+30+31+7=219\text{Total days} = 31 + 28 + 31 + 30 + 31 + 30 + 31 + 7 = 219
T=219365=35 yearsT = \frac{219}{365} = \frac{3}{5} \text{ years}
A=P(1+R×T100)A = P\left(1 + \frac{R \times T}{100}\right)
3144=P(1+8×35100)3144 = P\left(1 + \frac{8 \times \frac{3}{5}}{100}\right)
3144=P(1+24500)=P×5245003144 = P\left(1 + \frac{24}{500}\right) = P \times \frac{524}{500}
P=3144×500524=3000P = \frac{3144 \times 500}{524} = 3000
P=Rs. 3000\boxed{P = \text{Rs. } 3000}
SI=3000×8×35100=3000×24500=144SI = \frac{3000 \times 8 \times \frac{3}{5}}{100} = \frac{3000 \times 24}{500} = 144
A=3000+144=3144A = 3000 + 144 = 3144
Quick Tip:\textbf{Quick Tip:}
73365=15,146365=25,219365=35,292365=45\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

Correct Answer Rs. 750

Let amount at 12%=x,amount at 10%=y\text{Let amount at } 12\% = x,\quad \text{amount at } 10\% = y
12x+10y100=195,10x+12y100=201\frac{12x + 10y}{100} = 195,\quad \frac{10x + 12y}{100} = 201
12x+10y=19500,10x+12y=2010012x + 10y = 19500,\quad 10x + 12y = 20100
22x+22y=39600⇒x+y=180022x + 22y = 39600 \Rightarrow x + y = 1800
−2x+2y=600⇒y−x=300-2x + 2y = 600 \Rightarrow y - x = 300
x=(x+y)−(y−x)2=1800−3002=750x = \frac{(x+y) - (y-x)}{2} = \frac{1800 - 300}{2} = 750
y=1050y = 1050
x=Rs. 750\boxed{x = \text{Rs. } 750}
750×12100+1050×10100=90+105=195\frac{750 \times 12}{100} + \frac{1050 \times 10}{100} = 90 + 105 = 195
1050×12100+750×10100=126+75=201\frac{1050 \times 12}{100} + \frac{750 \times 10}{100} = 126 + 75 = 201
Extra Interest=201−195=6\text{Extra Interest} = 201 - 195 = 6
Quick Tip:\textbf{Quick Tip:}
x+y=(SI1+SI2)×100R1+R2x + y = \frac{(\text{SI}_1 + \text{SI}_2) \times 100}{R_1 + R_2}
y−x=(SI2−SI1)×100R1−R2y - x = \frac{(\text{SI}_2 - \text{SI}_1) \times 100}{R_1 - R_2}
x=(x+y)−(y−x)2x = \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

Correct Answer Rs. 2,760

P1+P2+P3=7930,R=5%,T1=2,T2=3,T3=4 yearsP_1 + P_2 + P_3 = 7930,\quad R = 5\%,\quad T_1 = 2,\quad T_2 = 3,\quad T_3 = 4 \text{ years}
P1(1+5×2100)</p><pclass="mb−4leading−7text−gray−800">=P2(1+5×3100)</p><pclass="mb−4leading−7text−gray−800">=P3(1+5×4100)P_1\left(1 + \frac{5 \times 2}{100}\right)</p><p class="mb-4 leading-7 text-gray-800">= P_2\left(1 + \frac{5 \times 3}{100}\right)</p><p class="mb-4 leading-7 text-gray-800">= P_3\left(1 + \frac{5 \times 4}{100}\right)
110P1100=115P2100=120P3100</p><pclass="mb−4leading−7text−gray−800">⇒110P1=115P2=120P3\frac{110P_1}{100} = \frac{115P_2}{100} = \frac{120P_3}{100}</p><p class="mb-4 leading-7 text-gray-800">\Rightarrow 110P_1 = 115P_2 = 120P_3
P1:P2:P3=1110:1115:1120P_1 : P_2 : P_3 = \frac{1}{110} : \frac{1}{115} : \frac{1}{120}
=30360110:30360115:30360120</p><pclass="mb−4leading−7text−gray−800">=276:264:253= \frac{30360}{110} : \frac{30360}{115} : \frac{30360}{120}</p><p class="mb-4 leading-7 text-gray-800">= 276 : 264 : 253
276+264+253=793276 + 264 + 253 = 793
P1=276793×7930=2760P_1 = \frac{276}{793} \times 7930 = 2760
P1=Rs. 2760\boxed{P_1 = \text{Rs. } 2760}
A1=2760×110100=3036,</p><pclass="mb−4leading−7text−gray−800">A2=2640×115100=3036,</p><pclass="mb−4leading−7text−gray−800">A3=2530×120100=3036A_1 = 2760 \times \frac{110}{100} = 3036,\quad</p><p class="mb-4 leading-7 text-gray-800">A_2 = 2640 \times \frac{115}{100} = 3036,\quad</p><p class="mb-4 leading-7 text-gray-800">A_3 = 2530 \times \frac{120}{100} = 3036
Quick Tip:\textbf{Quick Tip:}
P∝1(1+RT100)P \propto \frac{1}{\left(1 + \frac{RT}{100}\right)}
P1:P2:P3=100110:100115:100120</p><p>=276:264:253P_1 : P_2 : P_3 = \frac{100}{110} : \frac{100}{115} : \frac{100}{120}</p><p>= 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

Correct Answer Rs. 4,500

Total=Rs. 12000,Let amount to Akash=x,Amount to Bikash=12000−x\text{Total} = \text{Rs. } 12000,\quad \text{Let amount to Akash} = x,\quad \text{Amount to Bikash} = 12000 - x
RA=14%,RB=18%,T=3 years,SIA−SIB=720R_A = 14\%,\quad R_B = 18\%,\quad T = 3 \text{ years},\quad SI_A - SI_B = 720
SIA=42x100,SIB=54(12000−x)100SI_A = \frac{42x}{100},\quad SI_B = \frac{54(12000 - x)}{100}
42x100−54(12000−x)100=720\frac{42x}{100} - \frac{54(12000 - x)}{100} = 720
42x−54(12000−x)=7200042x - 54(12000 - x) = 72000
42x−648000+54x=7200042x - 648000 + 54x = 72000
96x=720000⇒x=750096x = 720000 \Rightarrow x = 7500
Amount to Bikash=12000−7500=4500\text{Amount to Bikash} = 12000 - 7500 = 4500
Amount to Bikash=Rs. 4500\boxed{\text{Amount to Bikash} = \text{Rs. } 4500}
SIA=3150,SIB=2430,SIA−SIB=720SI_A = 3150,\quad SI_B = 2430,\quad SI_A - SI_B = 720
Quick Tip:\textbf{Quick Tip:}
SI=P×R×T100,write both SIs and apply the condition directlySI = \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

Correct Answer Rs. 15,720

P=Rs. 15000,R=12%P = \text{Rs. } 15000,\quad R = 12\%
Start: 13th Jan 2019,End: 8th June 2019\text{Start: 13th Jan 2019},\quad \text{End: 8th June 2019}
Total days=18+28+31+30+31+8=146\text{Total days} = 18 + 28 + 31 + 30 + 31 + 8 = 146
T=146365=25 yearsT = \frac{146}{365} = \frac{2}{5} \text{ years}
SI=15000×12×25100</p><pclass="mb−4leading−7text−gray−800">=15000×24500</p><p>=720SI = \frac{15000 \times 12 \times \frac{2}{5}}{100}</p><p class="mb-4 leading-7 text-gray-800">= \frac{15000 \times 24}{500}</p><p>= 720
A=15000+720=15720A = 15000 + 720 = 15720
A=Rs. 15720\boxed{A = \text{Rs. } 15720}
Quick Tip:\textbf{Quick Tip:}
73=15 yr,146=25 yr,219=35 yr,292=45 yr73 = \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}

Successfully you have reached the final question

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