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QUESTION: 1

On a certain sum of money, after 2 years the simple interest and compound interest obtained are Rs 800 and Rs 960 respectively. What is the sum of money invested?

Solution:

Diff = 960-800 = 160

r = 2*Diff*100/SI

So r = 2*160*100/800 = 40%

Now 160 = Pr^{2} /100^{2}

QUESTION: 2

Rs 6000 becomes Rs 7200 in 3 years at a certain rate of compound interest. What will be the amount received after 9 years?

Solution:

6000[1 + r/100]^{3} = 7200

So [1 + r/100]^{3} = 6/5

So 6000[1 + r/100]^{9} = 6000*(6/5)*(6/5)*(6/5)

QUESTION: 3

A man borrows Rs 4000 at 8% compound interest for 3 years. At the end of each year he paid Rs 500. How much amount should he pay at the end of 3rd year to clear the debt?

Solution:

Amount after 1 yr = 4000[1 + 8/100] = 4320

Paid 500, so P = 4320 – 500 = 3820

Amount after 2nd yr = 3820[1 + 8/100] = 4125.6

So P= 4125.6-500 = 3625.6

Amount after 3rd yr = 3625.6[1 + 8/100] = 3915.6

QUESTION: 4

A sum of money is lent for 2 years at 20% p.a. compound interest. It yields Rs 482 more when compounded semi-annually than compounded annually. What is the sum lent?

Solution:

P[1 + (r/2)/100]^{4} – P[1 + r/100]^{2} = 482

P[1 + 10/100]^{4} – P[1 + 20/100]^{2} = 482

Solve, P = 20,000

QUESTION: 5

The compound interest obtained after 1st and 2nd year is Rs 160 and Rs 172.8 respectively on a certain sum of money invested for 2 years. What is the rate of interest?

Solution:

Difference in interest for both yrs = 172.8 – 160 = 12.8

So (r/100)*160 = 12.8

QUESTION: 6

A sum of money becomes Rs 35,280 after 2 years and Rs 37,044 after 3 years when lent on compound interest. Find the principal amount.

Solution:

P[1 + r/100]^{3} = 37,044, and P[1 + r/100]^{2} = 35,280

Divide both equations, [1 + r/100] = 37044/35280 = 21/20

So P[21/20]^{2} = 35280

QUESTION: 7

The difference between compound interest earned after 3 years at 5% p.a. and simple interest earned after 4 years at 4% p.a. is Rs 76. Find the principal amount.

Solution:

[P[1 + 5/100]^{3} – P] – P*4*4/100 = 76

P [9261/8000 – 1 – 16/100] = 76

QUESTION: 8

A sum of money is lent at simple interest and compound interest. The ratio between the difference of compound interest and simple interest of 3 years and 2 years is 35 : 11. What is the rate of interest per annum?

Solution:

Difference in 3 yrs = Pr^{2} (300+r)/100^{3}

Difference in 2 yrs = Pr^{2} /100^{2}

So Pr^{2} (300+r)/100^{3} / Pr^{2} /100^{2} = 35/11 (300+r)/100 = 35/11

QUESTION: 9

A sum of money borrowed at 5% compound interest is to paid in two annual installments of Rs 882 each. What is the sum borrowed?

Solution:

P = 882/[1 + 5/100] + 882/[1 + 5/100]^{2}

QUESTION: 10

Rs 3903 is to be divided in a way that A’s share at the end of 7 years is equal to the B’s share at the end of 9 years. If the rate of interest is 4% compounded annually, find A’s share.

Solution:

A’s share = (1 + 4/100)^{7}

B’s share = (1 + 4/100)^{9}

Divide both, B/A = (1 + 4/100)^{2} = 676/625

So A’s share = 625/(676+625) * 3903

QUESTION: 11

The difference between the total simple interest and the total compound interest compounded annually at the same rate of interest on a sum of money at the end of two years is Rs. 350. What is definitely the rate of interest per cent per annum?

Solution:

Difference = Pr^{2} /(100)^{2}

= (350×100×100)/(P×r^{2})

P is not given

QUESTION: 12

Aswin invested an amount of Rs.9000 in a fixed deposit scheme for 2 years at CI rate 6% pa. How much amount will Aswin get on maturity of the fixed amount ?

Solution:

Amount = 9000*106/100*106/100

= 9000*53/50*53/50

= 10,112

QUESTION: 13

A sum of money invested for 7years in Scheme 1 which offers SI at a rate of 8% pa. The amount received from Scheme 1 after 7 years invested for 2 years in Scheme 2 which offers CI rate of 10% pa. If the interest received from Scheme B was Rs.1638. What was the sum invested in Scheme 1 ?

Solution:

SI ⇒ Amount = x*8*7/100 + x = 56x+100x/100 = 156x/100 = 39x/25

CI⇒ 39x/25[(1+10/100)^{2} – 1] 1638 = 39x/25[121/100 – 1] = 39x/100[21/100] X

= 1638*100*25/21*39 = 5000

QUESTION: 14

Rs.5200 was partly invested in Scheme A at 10% pa CI for 2 years and Partly invested in Scheme B at 10% pa SI for 4 years. Both the schemes earn equal interests. How much was invested in Scheme A ?

Solution:

Amount invested in Scheme B = X

Amount invested in Scheme A = 5200 – x

X*10*4/100 = (5200-x)*21/100……………………[(1-10/100)^{2} -1] = 21/100

40x/100 = (5200-x)*21/100

2x/5 = (5200-x)*21/100

200x = 5200*21*5 – x*5*21

200x = 546000 – 105x

305x = 546000

X = 1790

Scheme A = 5200 – 1790 = 3410

QUESTION: 15

The CI on Rs.7000 for 3 years at 5% for first year, 7% for second year, 10% for the third year will be

Solution:

A = 7000*105/100*107/100*110/100

= 7000*1.05*1.07*1.1

= 8650.95 = 8651

CI = 8651-7000 = 1651

QUESTION: 16

Poorni and Priyanka have to clear their respective loans by paying 2 equal annual instalments of Rs.20000 each. Poorni pays at 10% pa of SI and Priyanka pays at 10% CI pa. What is the difference in their payments ?

Solution:

D =[(20,000 *110/100*110/100) – 20,000] – 20,000 *10*2/100

=[ 24200-20000]-4000

=4200 – 4000

D = 200

QUESTION: 17

A sum of Rs.3,50,500 is to be paid back in 2 equal annual instalments. How much is each instalment, if the rate of interest charged 4% per annum compound annually ?

Solution:

Total value of all the 3 instalments

[X*100/104] + [X*100*100/104] = 3,50,500

X*25/26 + x*625/676 = 3,50,500

X*25/26[1+25/26] = 3,50,500

X*25/26[51/26] = 3,50,500

X = 3,50,500*676/25*51 = 1,85,833.72 = 1,85,834

QUESTION: 18

At CI, if a certain sum of money is doubled in 4 years, then the amount will be four fold in how many years ?

Solution:

2n = 2*4 = 8 years

QUESTION: 19

X has lent some money to A at 5% pa and B at 8% pa. At the end of the year he has gain the overall interest at 7% pa.In what ratio has he lent the money to A and B ?

Solution:

5………………………….8

……………7…………….

1………………………….2

Ratio A : B = 1:2

QUESTION: 20

The difference between interest received by X and Y is Rs.315 on Rs.3500 for 3 years. What is the difference in rate of interest ?

Solution:

3500*3/100(R1-R2) = 315

R1-R2 = 315*100/10500 = 3%

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