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QUESTION: 1

Bharat borrowed Rs.180,000 on a condition that he had to pay 7.5% interest every year. He also agreed to pay the principal in equal annual instalments over 21 years. After a certain number of years, however, the rate of interest has been reduced to 7%. It is also known that at the end of the agreed period, he will have paid in all Rs.2,70,900 in interest. For how many years does he pay at the reduced interest rate?

Solution:

D. 14 years Explanation: x = interest paid at 7.5% (21-x) years interest paid at 7% ((180000*x*7.5)/100) + ((180000*7*(21-x))/100) = 270900 x = 7 21 – 7 = 14 years he paid at the reduced interest rate.

QUESTION: 2

Ankita borrows Rs.7000 at simple Interest from a lender. At the end of 3 years, she again borrows Rs.3000 and settled that amount after paying Rs.4615 as interest after 8 years from the time she made the first borrowing. what is the rate of interest?

Solution:

D. 6.5%

Explanation: SI for Rs.7000 for 8 years= (7000*r*8)/100 Again borrowed=3000 SI = (3000*r*5)/100 Total interest= [(7000*r*8)/100] + [(3000*r*5)/100] = 4615 560r + 150r = 4615 710r = 4615 r = 6.5%

QUESTION: 3

Hari borrowed some money for one year at 6% per annum simple interest and after 18 months , he again borrowed the same money at a Simple Interest of 24% per annum. In both the cases, he paid Rs.4704. Which of the following could be the amount that was borrowed by Hari in each case if interest is paid half yearly?

Solution:

D. 4200

Explanation: 12% for 6 months x = Borrowed money Take x =100% 112% of x = 4704 x = 4200

QUESTION: 4

Ravi lent out a part of Rs. 38800 is lent out at 6% per six months. The rest of the amount is lent out at 5% per annum after one year. The ratio of interest after 3 years from the time when first amount was lent is 5:4. Find the second part that was lent out at 5%.

Solution:

D. 28800

Explanation: First Part = x [x * (0.06)*6] / (388800 – x)*0.05*2 = 5/4 1.44x = 19400 – 0.5x x = 10000 Second Part = 38800 – 10000 = 28800

QUESTION: 5

Harshita lent out some money at 6% Simple Interest per annum. After one year, Rs.6800 is repaid and the rest of the amount is repaid at 5% per annum.If the second year’s interest is 11/20 of the first year’s interest, find what amount of money was lent out.

Solution:

C. 17000

Explanation: P – amount lent by Harshita I=.06*P —(i) The interest for second year as X=(P+I-6800).05 —(ii) x=(11/20)I —(iii) Put (i) in (ii) and (iii) P = 17,000

QUESTION: 6

Vikram borrows a sum of Rs.1500 at the beginning of a year. After four months Rs.2100 more is borrowed at a rate of interest double the previous one. At the end of one year, the sum of interest on both the loans is Rs.416. What is the first rate of interest per annum?

Solution:

D. 7.3%

Explanation: P = 1500

Rate of Interest = x SI = 1500x/100 = 15x P = 2100

Rate of Interest = 2x SI = 4200x/100 = 42x 57x = 416 x = 7.3%

QUESTION: 7

Rahul invested a sum of money at Simple Interest at a certain rate of interest for three years. Had it been invested at a 4% higher rate, it would have fetched Rs.480 more. Find out the Principal amount that was invested by Rahul?

Solution:

B. 4000

Explanation: x – Principal Extra amount = 4% for 3 years = 12% of x = 480 x = (480/12)*100 = 4000

QUESTION: 8

Rakesh fixes the rate of interest 6% per annum for first 3 years and for the next 4 years, 7 percent per annum and for the period beyond 7 years, 7.5 percent per annum. If Mr. Rakesh lent out Rs.1500 for 11 years, find the total interest earned by him?

Solution:

C. Rs.1140 Explanation: 6% for 3 years = 18% 7% for 4 years = 28% 7.5% for 4 years = 30% 76% of 1500 = 1140

QUESTION: 9

An equal amount of sum is invested in two schemes for four years each, both offering simple interest. When invested in scheme A at 8% per annum the sum amounts to Rs.5280. In scheme B, invested at 12% per annum it amounts to Rs.5920. What is the total sum invested?

Solution:

D. 8000

Explanation: Sum = x x + [(x*4*8)/100] = 5280 33x = (5280*25) = 4000 Total sum = 2 * 4000 = 8000

QUESTION: 10

Simple Interest on a certain sum at a certain annual rate of interest is 16% of the sum. If the numbers representing rate per cent and time in years be equal, then the rate of interest is?

Solution:

B. 4%

Explanation: SI = 16% of P R = T = x SI = [P*N*R]/100

4P/25 = [P*x^2]/100 x = 4%

QUESTION: 11

A person makes a fixed deposit of Rs. 20000 in Bank of India for 3 years. If the rate of interest be 13% SI per annum charged half yearly. What amount will he get after 42 months?

Solution:

A. 29100

Explanation: R=13%, T= 42 months For half year R=13/2, T= (42/12)*2= 7 half-years SI= (20000*7*6.5/100) = 9100

A= P+SI = 20000+9100=29100

QUESTION: 12

Rakesh invests Rs 12000 as fixed deposit at a bank at the rate of 10% per annum SI. But due to some pressing needs he has to withdraw the entire money after 3 years, for which the bank allowed him a lower rate of interest. If he gets Rs 3320 less than what he would have got at the end of 5 years, the rate of interest allowed by the bank is

Solution:

C. 67/9%

Explanation: P=12000, T1=10 years, T2=3 years, R1=10%, R2=? [(12000*10*5)/100 – (12000*R2*3)/100] = 3320

=> 50-3R2 = 83/3

=> R2 = 67/9%

QUESTION: 13

Ankita borrowed some money from at the rate of 6% per annum for the first 3 year, @ the rate of 9% per annum for the next 5 year and @ the rate of 13% per annum for the period beyond 8 year. If she pays a total interest of Rs.8160 at the end of 11 year. How much money did she borrow?

Solution:

B. 8000

Explanation: T1=3, T2=5, T3= Period beyond 8 year, so T3= total years-8 = 11-8 =3 Let the amount borrowed = P [(P*6*3)/100 (P*9*5)/100 (P*13*3)/100] = 8160

[102P/100] = 8160

P=8000

QUESTION: 14

A father left a will of Rs.35 lakhs between his two sons aged 8.5 and 16 such that they may get equal amounts when each of them reach the age of 21 years.

The original amount of Rs.35 lakhs has been instructed to be invested at 10% p.a. SI. How much did the elder son get at the time of the will?

Solution:

C. 21 lakh Explanation: Let x be the amount of elder son at the time of will & Younger son’s amount at the time of will (350000 -x) For elder son, T= 21-16=5 For younger son, T=21-8.5= 12.5 Amount Should be equal so, x + [(x*10*5)/100] = (3500000-x) + [((3500000-x)*10*12.5)/100] => 5x/3 = 3500000

x=2100000

QUESTION: 15

A portion of Rs.6600 is invested at a 5% per annum, while the remainder is invested at a 3% per annum. If the annual income from the portion earning a 5% per annum is twice that of the other portion, what is the total income from the two investments after one year?

Solution:

A. Rs.270 Explanation: 5x + 3y = z (Total Income) x + y = 6600 -(1) 5x= 2(3y) 5x – 6y = 0 -(2) By solving (1) and (2) we get, x = 3600 so y = 3000 (3600*5*1)/100 + (3000*3*1)/100 = 180 + 90 = 270

QUESTION: 16

Simple Interest for the sum of Rs.1230 for two year is Rs.10 more than the SI for Rs.1130 for the same duration. Find the Rate of Interest?

Solution:

A. 5%

Explanation: (1230*2*R)/100 – (1130*2*R)/100 = 10

R=5%

QUESTION: 17

Arvind borrowed Rs.800 at 6% per annum and Deepak borrowed Rs.600 @ 10% per annum. After how much time, will they both have equal debts?

Solution:

A. 50/3 yr Explanation: 800+(800*6*T)/100 = 600+(600*10*T)/100

T=50/3 yr

QUESTION: 18

The Principal on which a simple interest of Rs.55 will be obtained after 9 months at the rate of 11/3% per annum is ___________

Solution:

D. 2000

Explanation: SI =PRT/100

P = 100*55/(11/3)*(9/12)

P = 100*55*4/11

P = 2000

QUESTION: 19

The rate of Simple Interest in UBI & BOI are in the ratio of 5:4. Mr.Naveen wants to deposit his total savings in two banks in such a way that he receive equal half-yearly interest from both banks.He should deposit in both banks UBI & BOI in the ratio of

Solution:

C. 4:5

Explanation: R1 =5x R2=4x T1=T2=1/2 yr [P1*5x*(1/2)]/100 = [P2*4x*(1/2)]/100 P1:P2 = 4:5

QUESTION: 20

A sum becomes 3 times in 5 year at a certain rate of interest. Find the time in which the same amount will be eight times at the same rate of interest?

Solution:

B. 35/2 yr Explanation: T=5, SI = 3P-P =2P

2P = P*R*5/100

R = 40%

For another time SI = 8P-P =7P

7P = P*40*T/100

T=35/2

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