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This mock test of Compound Interest - MCQ 4 for Quant helps you for every Quant entrance exam.
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QUESTION: 1

What is the difference between the CI and SI on Rs.6500 at the rate of 4 pcpa in 2 yrs ?

Solution:

SI = 6500*4*2/100 = 520

CI = 6500*(104/100)*(104/100) = 7030.4 = 530.4

530.4 – 520 = 10.4

QUESTION: 2

The CI on a certain sum at 10% pa for 2yrs is Rs.6548. What is SI on the sum of money at 7%pa for 4 yrs(approx) ?

Solution:

CI = p [{1 + (10/100)}^{2} – 1] 6548 = p [(110/100)^{2} – 1] 6548 =p (121 – 100)/100

P = 654800/21 =31,181

SI = 31181*7*4/100 = 8731

QUESTION: 3

A sum of money amount to Rs.900 in 3 yrs and Rs.1080 in 4yrs at CI. What is the rate of CI pa ?

Solution:

(1080-900/900)*100

180*100/900 = 20%

QUESTION: 4

Find the CI on Rs.4500 at 4% pa for 2 yrs compounded half yearly(approx) ?

Solution:

CI = P [(1+(r/100) – 1] = 4500[(1+2/100)^{4} – 1

= 4500[(102*102*102*102/100*100*100*100) – 1] = 4500[1.082 – 1] = 4500*0.082 = 369

QUESTION: 5

In how many years will a sum of Rs.15625 at 8% pa compounded semi annually become Rs.17,576?

Solution:

15625*(1+(4/100))^{2n} = 17576

15625*(104/100)^{2n} = 17576

(26/25)^{2n} = 17576/15625 = (26/27)^{3}

2n = 3 N =3/2 = 1(1/2) yrs

QUESTION: 6

If 40% increase in an amount in 4 years at SI. What will be the CI of Rs.10,000 after 2 yrs at the same rate ?

Solution:

P = 100, SI = 40, T = 4

R = 100*40 / 100*4 = 10%pa

CI = 10,000(110/100)^{2} = 10000*11*11/100 = 12100

12100 – 10000 = 2100

QUESTION: 7

A sum of money is borrowed and paid in 2 annual instalments of Rs.642 each allowing 4% CI. The sum borrowed was

Solution:

P = [642/(104/100)] + [642/(104/100) ^{2})

= 642*(25/26) + 642 (25/26)^{2}

= 617.3 + 593.6 = 1210.9 = 1211

QUESTION: 8

The least number of complete years in which a sum of money put out at 10% CI will be more than doubled is

Solution:

P(110/100)^{n} > 2P

(11/10)^{n} > 2P

1.1*1.1*1.1*1.1*1.1*1.1*1.1*1.1 = 2.14 > 2

N = 8

QUESTION: 9

A sum of money doubles itself at CI in 10yrs. In how many yrs will it become 4 times ?

Solution:

P(1+(R/100))^{10} = 2P

1+(R/100)^{10} = 2

(1+(R/100))^{n} = 4 = 2^{2}

2*10 = 20yrs

QUESTION: 10

At what rate of CI pa will a sum of Rs.1000 becomes Rs.1040.4 in 2yrs ?

Solution:

1000*(1+(R/100))^{2} = 1040.4

(1+(R/100))^{2} = 1040.4/1000

(1+(R/100))^{2} = 10404/10000 = (102/100)^{2}

R = 2%

QUESTION: 11

What annual payment will discharge a debt of 1025 due in 2 years at the rate of 5% compound interest?

Solution:

QUESTION: 12

In what time will Rs. 64,000 amount to Rs.68921 at 5% per annum interest being compounded half yearly?

Solution:

R = 2.5%, A = 68921 , P = 64000 and t= 2n

A/P = [1+(R/100)]^{2n}

68921/64000 = [1+(2.5/100)]^{2n}

(41/40)^{3 }= (41/40)^{2n}

2n = 3

n=3/2 =1(1/2) years

QUESTION: 13

If the difference between the CI and SI on a sum of money at 5% per annum for 2years is Rs.16.Find the Simple Interest ?

Solution:

16 =P*(5/100)^{2}

P = 6400

SI =(6400*5*2)/100 =640

QUESTION: 14

The difference between the SI and CI on Rs.5000 at 10%per annum for 2 year is

Solution:

d = p (r/100)^{2}

= 5000 (10/100)^{2}

d = 50

QUESTION: 15

The difference SI and CI on Rs.1000 for 1 year at 10%per annum reckoned Half yearly is

Solution:

SI = (1000/100 =100

CI = [1000(1+(5/100))^{2}] – 1000 = 102.5

D = 102.5 – 100 = 2.5

QUESTION: 16

Compound interest on a certain sum of money at 20% per annum for 2 years is Rs.5984.What is the SI on the same money at 9% per annum for 6 years ?

Solution:

5984 = p [(1+(20/100)^{2}) -1] P = (5984*25)/11 = 13600

SI = (13600*9*6)/100 = 7344

QUESTION: 17

The difference between CI and SI on an amount Rs. 15000 for 2 year is Rs.96.What is the rate of interest per annum ?

Solution:

QUESTION: 18

The effective annual rate of interest corresponding to the nominal rate of 4% per annum payable half yearly is

Solution:

Let p = 100

CI = [100+(1+[2/100]^{2})] = 100 × (102/100)× (102/100)

= 104.04

Effective rate = 104.04 – 100 = 4.04%

QUESTION: 19

Rohit borrowed Rs. 1200 at 12% PA .He repaid Rs. 500 at the end of 1 year. What is the amount required to pay at the end of 2^{nd} year to discharge his loan which was calculated in CI

Solution:

CI at the end of 1^{st} year = 1200 *(1+(12 /100)) = 1344

CI = 1344-1200 = 144

500 paid then remaining amount = 1344 – 500 = 844

At the end of 2nd year

844*[(1+(12/100)] =945.28

QUESTION: 20

A sum of money invested at CI to Rs.800 in 3 years and to Rs.840 in 4 years.Find rate of interest PA ?

Solution:

CI for 3 yr = 800

CI for 4 yr = 840

CI for 1 yr = 40

R = (100 × 40)/(800×1) = 10/2 = 5%

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