Home Quantitative Aptitude Simple and Compound Interest-Test 1

Simple and Compound Interest-Test 1

Subject :- QT

Chapter :- Simple and Compound Interest-Test 1

Questions :- 25

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Question 1 of 50

1. Question
1 points
A sum of money invested at simple interest triples itself in 8 years. How many times will it become in 20 years time?
- 8 times
- 7 times
- 6 times
- 9 times
Correct

Incorrect
Question 2 of 50

2. Question
1 points
Rs 8000 becomes Rs 10000 in 2 years at simple interest.The amount that will become Rs 6875 in 3 years at the same rate of interest is
- Rs 4850
- Rs 5000
- Rs 5500
- Rs 5275
Correct

A=P[1+rt/100]

10000=8000[1+r*2/100]

10000/8000=100+2r/100

2r=125-100

r=25/2=12.5% p.a.

let the amount which will become Rs 6875 be P.then,

6875=p[1+12.5*3/100]

6875=p[100+37.5/100]

p=6875*100/1375

p=Rs 5000

Incorrect

A=P[1+rt/100]

10000=8000[1+r*2/100]

10000/8000=100+2r/100

2r=125-100

r=25/2=12.5% p.a.

let the amount which will become Rs 6875 be P.then,

6875=p[1+12.5*3/100]

6875=p[100+37.5/100]

p=6875*100/1375

p=Rs 5000
Question 3 of 50

3. Question
1 points
The differences between the simple and compound interest on a certain sum for 3 yr at 5% p.a. is Rs 228.75.The compound interest on the sum for 2 yr at 5% p.a. is
- Rs3175
- Rs3075
- Rs3275
- Rs2975
Correct

let the sum be Rs P

S.I=PRT/100

S.I=P*5*3/100=15P/100=0.15P
$C . I = P [(1 + i)^{n} - 1] = P [(1 + 0.05)^{3} - 1] = 0.157625$
diff.between C.I and S.I=Rs 228.75

0.157625-0.15P=228.75

P=228.75/0.007625=Rs 30000
$C . I o n R s 30000 @ 5 % p . a . f o r 2 y r C . I = 30000 [(1 + 0.05)^{2} - 1] = R s 3075$

Incorrect

let the sum be Rs P

S.I=PRT/100

S.I=P*5*3/100=15P/100=0.15P
$C . I = P [(1 + i)^{n} - 1] = P [(1 + 0.05)^{3} - 1] = 0.157625$
diff.between C.I and S.I=Rs 228.75

0.157625-0.15P=228.75

P=228.75/0.007625=Rs 30000
$C . I o n R s 30000 @ 5 % p . a . f o r 2 y r C . I = 30000 [(1 + 0.05)^{2} - 1] = R s 3075$
Question 4 of 50

4. Question
1 points
Mr X invests Rs 10000 every year starting from today for next 10 yr.suppose interest rate is 8% p.a. compounded annually.calculate future value of annuity?
$[g i v e n t h a t (1 + 0.08)^{10} = 2.15892500]$
- Rs156454.88
- Rs144865.625
- Rs156554.88
- none of these
Correct

$f u t u r e v a l u e o f a n n u i t y d u e = a (1 + i) / i [(1 + i)^{n} - 1] = 10000 / 0.08 (1 + 0.08) [(1 + 0.08)^{10} - 1] = 135000 [1.15892500] = R s 156454.875 (a p p r o x)$

Incorrect

$f u t u r e v a l u e o f a n n u i t y d u e = a (1 + i) / i [(1 + i)^{n} - 1] = 10000 / 0.08 (1 + 0.08) [(1 + 0.08)^{10} - 1] = 135000 [1.15892500] = R s 156454.875 (a p p r o x)$
Question 5 of 50

5. Question
1 points
The present value of an annuity of Rs 3000 for 15 yr at 4.5% p.a. C.I is
$[g i v e n t h a t (1.045)^{15} = 1.935282]$
- Rs23809.67
- Rs32218.67
- Rs32908.67
- none of these
Correct

$P . V = a / i [(1 + i)^{n} - 1 / (1 + i)^{n}] = 3000 / 0.045 [(1 + 0.045)^{15} - 1 / (1 + 0.045)^{15}] = R s 32218.67 (a p p r o x)$

Incorrect

$P . V = a / i [(1 + i)^{n} - 1 / (1 + i)^{n}] = 3000 / 0.045 [(1 + 0.045)^{15} - 1 / (1 + 0.045)^{15}] = R s 32218.67 (a p p r o x)$
Question 6 of 50

6. Question
1 points
The rate of simple interest on a sum of money is 6% p.a. for first 3 yr , 8%p.a. for next 5 yr and 10%p.a. for more than 8 yr.if the simple interest accured by the sum for a period for 10 yr is Rs 1560.the sum is
- Rs1500
- Rs2000
- Rs3000
- Rs5000
Correct

let the sum invested be RsP.then,

S.I=Rs1560

P*6*3/100+P*8*5/100+P*10*2/100=1560

P[18/100+40/100+20/100]=1560

78P/100=1560

P=Rs 2000

Incorrect

let the sum invested be RsP.then,

S.I=Rs1560

P*6*3/100+P*8*5/100+P*10*2/100=1560

P[18/100+40/100+20/100]=1560

78P/100=1560

P=Rs 2000
Question 7 of 50

7. Question
1 points
A sum of money doubles it self in 10yr.the number of year it would treble it self is
- 25yr
- 15yr
- 20yr
- none of these
Correct

A=P[1+rt/100]

2P=P[1+rx10/100]

2=100+10r/100

r=10%p.a

new amount = 3P

3P=P[1+10xt/100]

3=100+10t/100

t=20yr

Incorrect

A=P[1+rt/100]

2P=P[1+rx10/100]

2=100+10r/100

r=10%p.a

new amount = 3P

3P=P[1+10xt/100]

3=100+10t/100

t=20yr
Question 8 of 50

8. Question
1 points
In what time will Rs 390625 amount to Rs 456976 at 8% p.a,when the interest is compounded semi-annually?
$[g i v e n (1.04)^{4} = 1.16986]$
- 2yr
- 4yr
- 5yr
- 7yr
Correct

P=Rs 390625

A=Rs 456976

i=8/100×1/2=0.04

time=n
$A = P (1 + i)^{n} 456976 = 390625 (1 + 0.04)^{n} 456976 / 390625 = 1 . 04^{n} 1 . 04^{4} = 1 . 04^{n} n = 4 t i m e = n / 2 = 4 / 2 = 2 y r$

Incorrect

P=Rs 390625

A=Rs 456976

i=8/100×1/2=0.04

time=n
$A = P (1 + i)^{n} 456976 = 390625 (1 + 0.04)^{n} 456976 / 390625 = 1 . 04^{n} 1 . 04^{4} = 1 . 04^{n} n = 4 t i m e = n / 2 = 4 / 2 = 2 y r$
Question 9 of 50

9. Question
1 points
A machine can be purchased for Rs 50000.machine will contribute Rs 12000 per year for next 5 yr.assume borrowing cost is 10% p.a.determine whether machine should be purchased or not
- should be purchased
- should not be purchased
- can't say about purchase
- none of these
Correct

present value of outflow machine=Rs 50000
$P . V = \frac{A}{i} [\frac{(1 + i)^{n} - 1}{(1 + i)^{n}}] = 12000 / 10 [(1 + 0.10)^{5} - 1 / (1 + 0.10)^{5}] = R s 45489.44 (a p p r o x)$
= inflow

since P.V of inflow is less as compared to outflow.so,machine should not be purchased

Incorrect

present value of outflow machine=Rs 50000
$P . V = \frac{A}{i} [\frac{(1 + i)^{n} - 1}{(1 + i)^{n}}] = 12000 / 10 [(1 + 0.10)^{5} - 1 / (1 + 0.10)^{5}] = R s 45489.44 (a p p r o x)$
= inflow

since P.V of inflow is less as compared to outflow.so,machine should not be purchased
Question 10 of 50

10. Question
1 points
How much amount is required to be invested every year so as to accumulate Rs 300000 at the end of 10 yr,if interest is compounded annually at 10%?
$[g i v e n (1.1)^{10} = 2.5937]$
- Rs18823.65
- Rs18828.65
- Rs18832.65
- Rs18882.65
Correct

F.Y= Rs 300000

let Rs a is invested annually.
$F . V = \frac{a}{i} [(1 + i)^{n} - 1] 300000 = \frac{a}{0.10} [(1 + 0.10)^{10} - 1] 30000 = 1.5937 a a = 30000 / 1.5937 = R s 18823.65 (a p p r o x)$

Incorrect

F.Y= Rs 300000

let Rs a is invested annually.
$F . V = \frac{a}{i} [(1 + i)^{n} - 1] 300000 = \frac{a}{0.10} [(1 + 0.10)^{10} - 1] 30000 = 1.5937 a a = 30000 / 1.5937 = R s 18823.65 (a p p r o x)$
Question 11 of 50

11. Question
1 points
A certain sum of money amounts to Rs 6300 in 2 yr and Rs 7875 in 3 yr 9 months at simple interest.find the rate of interest p.a.
- 20%
- 18%
- 15%
- 10%
Correct

A=P[1+rt/100]

6300=P[1+r*2/100]

6300=P[100+2r/100]…………..(1)

7875=P[1+3.75*r/100]

7875=P[100+3.75r/100]………………(2)

dividing 2 by 1

7875/6300=P[100+3.75r/100] / P[100+2r/100]

1.25=P x 100+3.75r/100+2r

3.75r-2.5r=125-100

1.25r=25

r=20%p.a.

Incorrect

A=P[1+rt/100]

6300=P[1+r*2/100]

6300=P[100+2r/100]…………..(1)

7875=P[1+3.75*r/100]

7875=P[100+3.75r/100]………………(2)

dividing 2 by 1

7875/6300=P[100+3.75r/100] / P[100+2r/100]

1.25=P x 100+3.75r/100+2r

3.75r-2.5r=125-100

1.25r=25

r=20%p.a.
Question 12 of 50

12. Question
1 points
How long will Rs 12000 take to amount to Rs 14000 at 5% p.a. converted quarterly?
$[g i v e n (1.0125)^{12.4} = 1.1666]$
- 3 yr
- 3.1 yr
- 13.5 yr
- 12.4 yr
Correct

P=Rs 12000

A=Rs 14000

i=5/4%=1.25%=0.0125

no. of conversion periods=n
$A = P (1 + i)^{n} 14000 = 12000 (1 + 0.0125)^{n} 14000 / 12000 = (1.0125)^{n} (1.0125)^{12.4} = (1.0125)^{n} n = 12.4 t i m e = n / 4 = 12.4 / 4 = 3.1 y e a r s$

Incorrect

P=Rs 12000

A=Rs 14000

i=5/4%=1.25%=0.0125

no. of conversion periods=n
$A = P (1 + i)^{n} 14000 = 12000 (1 + 0.0125)^{n} 14000 / 12000 = (1.0125)^{n} (1.0125)^{12.4} = (1.0125)^{n} n = 12.4 t i m e = n / 4 = 12.4 / 4 = 3.1 y e a r s$
Question 13 of 50

13. Question
1 points
A company is considering proposal a machine either by making full payment of Rs 4000 or by leasing it for 4 yr at an annual rate of Rs 1250. which course of action is preferable , if the company can borrow money at 14% compounded annually?
- leasing is preferable
- should be purchased
- no difference
- none of these
Correct

$P . V = a / i [(1 + i)^{n} - 1 / (1 + i)^{n}] = 1250 / 0.14 [(1 + 0.14)^{4} - 1 / (1 + 0.14)^{4}] = 8928.57 x 0.4079 = R s 3642.14 (a p p r o x)$
P.V of outflows i.e.annuity paid

purchase price of machine= Rs 4000 since,purchase price of machine is more than the P.V of outflows of annuity ,so, leasing is preferable

Incorrect

$P . V = a / i [(1 + i)^{n} - 1 / (1 + i)^{n}] = 1250 / 0.14 [(1 + 0.14)^{4} - 1 / (1 + 0.14)^{4}] = 8928.57 x 0.4079 = R s 3642.14 (a p p r o x)$
P.V of outflows i.e.annuity paid

purchase price of machine= Rs 4000 since,purchase price of machine is more than the P.V of outflows of annuity ,so, leasing is preferable
Question 14 of 50

14. Question
1 points
Vipul purchases a car for Rs 550000.he gets a loan of Rs 500000 at 15% p.a. from a bank and balance of rs 50000 he pays at the time of purchase .he has to pay the whole amount of loan in 12 equal monthly installments with interest starting from the end of the first month.the money he has to pay at the end of every month is
$[g i v e n (1.0125)^{12} = 1.16075452]$
- Rs 45130.43
- Rs 45230.43
- Rs 45330.43
- none of these
Correct

Incorrect
Question 15 of 50

15. Question
1 points
If Rs 1000 be invested at interest rate of 5% and the interest rate added to the principal every 10 years,then the number of years in which it will amount to Rs 2000 is
- 50/3 yr
- 25/4 yr
- 16 yr
- 20/3 yr
Correct

Incorrect
Question 16 of 50

16. Question
1 points
The annual birth and death rates per 1000 are 39.4 and 19.4.The number of years in which the population will be doubled assuming there is no immigration or emigration is
- 35 yr
- 30 yr
- 25 yr
- none of these
Correct

Annual growth rate=annual birth rate- annual death rate=39.4-19.4=20 per 1000

rate=20/1000×100=2%

let the original population be P
$A = P (1 + i)^{n} 2 P = P (1 + 0.02)^{n} 2 = 1 . 02^{n} 1 . 02^{35} = 1 . 02^{n} n = 35 y r$

Incorrect

Annual growth rate=annual birth rate- annual death rate=39.4-19.4=20 per 1000

rate=20/1000×100=2%

let the original population be P
$A = P (1 + i)^{n} 2 P = P (1 + 0.02)^{n} 2 = 1 . 02^{n} 1 . 02^{35} = 1 . 02^{n} n = 35 y r$
Question 17 of 50

17. Question
1 points
The effective rate equivalent to nominal rate of 6% compounded monthly is
- 6.05
- 6.16
- 6.26
- 6.07
Correct

i=6%p.a=6/100×1/12=0.005

n=1 yr = 1×12 12 converted period
$E = (1 + i)^{n} - 1 = (1 + 0.005)^{12} - 1 = 0.0616 o r 6.16 %$

Incorrect

i=6%p.a=6/100×1/12=0.005

n=1 yr = 1×12 12 converted period
$E = (1 + i)^{n} - 1 = (1 + 0.005)^{12} - 1 = 0.0616 o r 6.16 %$
Question 18 of 50

18. Question
1 points
`A company establishes a sinking fund to provide for the payment of Rs 200000 debt maturing in 20 yr.contributions to the fund are to be made at the end of every yr.find the amount of each annual deposit if the rate of interest is 5% p.a
- Rs6142
- Rs6049
- Rs6052
- Rs6159
Correct

$l e t t h e a n n u a l d e p o s i t b e R s a F . Y = a / i [(1 + i)^{n} - 1] 200000 = a / 0.05 [(1 + 0.05)^{20} - 1] 10000 = a (1.6533) a = R s 6049 (a p p r o x)$

Incorrect

$l e t t h e a n n u a l d e p o s i t b e R s a F . Y = a / i [(1 + i)^{n} - 1] 200000 = a / 0.05 [(1 + 0.05)^{20} - 1] 10000 = a (1.6533) a = R s 6049 (a p p r o x)$
Question 19 of 50

19. Question
1 points
A person borrows Rs 5000 for 2 yr at 4% p.a. simple interest.he immediately lends to another person at 25/4% p.a for 2 yr.find his gain in the transaction per year
- Rs1125.50
- Rs125
- Rs225
- Rs167.50
Correct

S.I paid on money borrowed

=Prt/100=5000x2x4/100=Rs 400

S.I received on money lended

=Prt/100=5000x25x2/100=Rs 625

total gain in transaction for 2 yr=S.I received-S.I paid=625-400=Rs225

gain per yr=225/2=Rs112.50

Incorrect

S.I paid on money borrowed

=Prt/100=5000x2x4/100=Rs 400

S.I received on money lended

=Prt/100=5000x25x2/100=Rs 625

total gain in transaction for 2 yr=S.I received-S.I paid=625-400=Rs225

gain per yr=225/2=Rs112.50
Question 20 of 50

20. Question
1 points
A person deposited Rs 5000 in a bank.the deposit was left to accumulate at 6% compounded quarterly for the first 5 yr and at 8% compounded semi-annually for the next 8 yr.the compound amount at the end of 13 yr is
- Rs12621.50
- Rs12613.10
- Rs13613.10
- none of these
Correct

Incorrect
Question 21 of 50

21. Question
1 points
Raja aged 40 wishes his wife rani to have Rs 40 lakhs at his death.if his expectation of life is another 30 yr and he starts making equal annual investments commencing now at 3% compound interest p.a how much should he invest annually?
- Rs84077
- Rs81628
- Rs84449
- Rs84247
Correct

future value=Rs 40 lakhs

n=30 yr

i=0.03
$F . V = a / i (1 + i) [(1 + i)^{n} - 1] 4000000 = a / 0.09 (1 + 0.03) [(1 + 0.03)^{30} - 1] 4000000 = a / 0.03 x 1.03 x 1.4273 a = R s 81628 (a p p r o x)$

Incorrect

future value=Rs 40 lakhs

n=30 yr

i=0.03
$F . V = a / i (1 + i) [(1 + i)^{n} - 1] 4000000 = a / 0.09 (1 + 0.03) [(1 + 0.03)^{30} - 1] 4000000 = a / 0.03 x 1.03 x 1.4273 a = R s 81628 (a p p r o x)$
Question 22 of 50

22. Question
1 points
Two equal sum of money were lent at simple interest at 11% p.a for 7/2 yr and 9/2 yr.if the difference in the interest for two periods was Rs 412.50,then each sum is
- Rs3250
- Rs3500
- Rs3750
- Rs4350
Correct

let the sum invested be Rs P

S.I=PRt/100

[Px11x9/100×2]-[Px11x7/100×2]=412.50

22P/200=412.50

P=412.50×200/22

P=Rs 3750

Incorrect

let the sum invested be Rs P

S.I=PRt/100

[Px11x9/100×2]-[Px11x7/100×2]=412.50

22P/200=412.50

P=412.50×200/22

P=Rs 3750
Question 23 of 50

23. Question
1 points
Anshul’s father wishes to have Rs 75000 in a bank account when his first college expenses begin.how much amount his father should deposit now at 6.5% compounded annually if anshul is to start college in 8 yr hence from now?
- Rs45360
- Rs46360
- Rs55360
- Rs48360
Correct

amount = Rs 75000
$P (1 + i)^{n} = 75000 P = 75000 / 1.655 = R s 45317 (a p p r o x)$

Incorrect

amount = Rs 75000
$P (1 + i)^{n} = 75000 P = 75000 / 1.655 = R s 45317 (a p p r o x)$
Question 24 of 50

24. Question
1 points
A company may obtain a machine either by leasing it for 5 yr at an annual rent of Rs 2000 or by purchasing the machine for Rs 8100.if the company can borrow money at 18% p.a,which alternative is preferable?
- leasing
- purchasing
- can't say
- none of these
Correct

purchase cost of machine at present =Rs 8100
$p r e s e n t v a l u e o f t h e l e a s e r e n t a l = a / i [(1 + i)^{n} - 1 / (1 + i)^{n}] = 2000 / 0.18 [(1 + 0.18)^{5} - 1 / (1 + 0.18)^{5}] = 11111 x 0.5629 = R s 6254.34 (a p r o x)$

Incorrect

purchase cost of machine at present =Rs 8100
$p r e s e n t v a l u e o f t h e l e a s e r e n t a l = a / i [(1 + i)^{n} - 1 / (1 + i)^{n}] = 2000 / 0.18 [(1 + 0.18)^{5} - 1 / (1 + 0.18)^{5}] = 11111 x 0.5629 = R s 6254.34 (a p r o x)$
Question 25 of 50

25. Question
1 points
In how much time would the simple interest on certain sum be 0.125 times the principal at 10%p.a.?
- 5/4 yr
- 7/4 yr
- 9/4 yr
- 11/4 yr
Correct

let the principal be Rs P

S.I=PRT/100

0.125p=px10xt/100

0.125=10t/100

t=0.125×100/10

t=1.25=5/4 yr

Incorrect

let the principal be Rs P

S.I=PRT/100

0.125p=px10xt/100

0.125=10t/100

t=0.125×100/10

t=1.25=5/4 yr
Question 26 of 50

26. Question
1 points
The difference between compound interest and simple interest on a certain sum for 2 yr @10%p.a. is Rs 10.find the sum
- Rs 1010
- Rs 1095
- Rs 1000
- Rs 990
Correct

Incorrect
Question 27 of 50

27. Question
1 points
A machine worth Rs 490740 is depreciated at 15% on its opening value each year.when its value would reduce to Rs 200000:
- 5 year 6 months
- 5 year 7 months
- 5 year 5 months
- none of these
Correct

A=200000 and in case of depreciation
$A = P (1 - i)^{n} 200000 - 490740 = (1 - 0.15)^{n} 0.4075 = (0.85)^{n} 0 . 85^{5.5} = 0 . 85^{n} n = 5.5 o r 5 y r 6 m o n t h s (a p p r o x)$

Incorrect

A=200000 and in case of depreciation
$A = P (1 - i)^{n} 200000 - 490740 = (1 - 0.15)^{n} 0.4075 = (0.85)^{n} 0 . 85^{5.5} = 0 . 85^{n} n = 5.5 o r 5 y r 6 m o n t h s (a p p r o x)$
Question 28 of 50

28. Question
1 points
A sinking fund is created for redeeming debentures worth rs 5 lakhs at the end of 25 yr.hoe much provision needs to be made out of profits each year provided sinking fund investments can earn interest at 4%p.a.?
- 12006
- 12040
- 12039
- 12035
Correct

$l e t t h e a n n u a l i n v e s t m e n t b e R s a A = a / i [(1 + i)^{n} - 1] 500000 = a / 0.04 [(170.04)^{25} - 1] 20000 = a x 1.6658 a = R s 12006 (a p p r o x)$

Incorrect

$l e t t h e a n n u a l i n v e s t m e n t b e R s a A = a / i [(1 + i)^{n} - 1] 500000 = a / 0.04 [(170.04)^{25} - 1] 20000 = a x 1.6658 a = R s 12006 (a p p r o x)$
Question 29 of 50

29. Question
1 points
If the difference between simple interest and compound interest is Rs 11 at the rate of 10% for 2 yr,then find the sum
- Rs1200
- Rs1100
- Rs1000
- none of these
Correct

difference between S.I and C.I=Rs 11

now,rate=10%,i=0.1,t=2
$P [(1 + i)^{n} - 1] - P x i x t = 11 P [(1 + 0.1)^{2} - 1 - 0.1 x 2] = 11 P = R s 1100$

Incorrect

difference between S.I and C.I=Rs 11

now,rate=10%,i=0.1,t=2
$P [(1 + i)^{n} - 1] - P x i x t = 11 P [(1 + 0.1)^{2} - 1 - 0.1 x 2] = 11 P = R s 1100$
Question 30 of 50

30. Question
1 points
Find the numbers of years in which a sum doubles itself at the rate of 8%p.a.
- 23/2
- 25/2
- 19/2
- 27/2
Correct

let the principal be Rs x

2x=X[1+it]

2x=x[1+0.08t]

t=25/2 yr

Incorrect

let the principal be Rs x

2x=X[1+it]

2x=x[1+0.08t]

t=25/2 yr
Question 31 of 50

31. Question
1 points
In how many years,a sum will become double at 5% p.a. compound interest.
- 14.0 yr
- 14.1 yr
- 14.2 yr
- 14.3 yr
Correct

let the principal amount be Rs P

then,A=Rs2P and ROI=5% and i=0.05

number of years=n
$A = P (1 + i)^{n} 2 P = P (1 + 0.05)^{n} 2 = 1 . 05^{n} 1 . 05^{14.3} = 1 . 05^{n} n = 14.3 y r$

Incorrect

let the principal amount be Rs P

then,A=Rs2P and ROI=5% and i=0.05

number of years=n
$A = P (1 + i)^{n} 2 P = P (1 + 0.05)^{n} 2 = 1 . 05^{n} 1 . 05^{14.3} = 1 . 05^{n} n = 14.3 y r$
Question 32 of 50

32. Question
1 points
The time by which a sum of money is 8 times of itself if it doubles itself in 15 years.
- 42 yr
- 43 yr
- 45 yr
- 46 yr
Correct

let the principal amount be Rs P

rate of interest = r%p.a=i
$A = P (1 + i)^{n} 2 P = P (1 + i)^{15} 2 = 1 + i^{15} (1.0473)^{15} = (1 + i)^{15} 1 + i = 1.0473 i = 0.473 n o w, 8 P = P (1 + i)^{n} = 8 = (1.0473)^{n} = n = 45$

Incorrect

let the principal amount be Rs P

rate of interest = r%p.a=i
$A = P (1 + i)^{n} 2 P = P (1 + i)^{15} 2 = 1 + i^{15} (1.0473)^{15} = (1 + i)^{15} 1 + i = 1.0473 i = 0.473 n o w, 8 P = P (1 + i)^{n} = 8 = (1.0473)^{n} = n = 45$
Question 33 of 50

33. Question
1 points
What is the rate of simple interest if a sum of money amounts to Rs 2784 in 4 yr and Rs 2688 in 3 yr?
- 1%p.a.
- 4%p.a.
- 5%p.a.
- 8%p.a.
Correct

Incorrect
Question 34 of 50

34. Question
1 points
A sum amount to Rs 1331 at a principal of Rs 1000 at 10% compounded annually.find the time.
- 3.31 yr
- 4 yr
- 3 yr
- 2 yr
Correct

P=Rs1000

A=Rs1331

i=0.10

time = n years
$A = P (1 + i)^{n} 1331 = 1000 (1 + 0.10)^{n} 1331 = (1.10)^{n} 1 . 10^{3} = 1 . 10^{n} n = 3 h e n c e, R s 1000 a m o u n t s t o R s 1331 a t 10 % p . a . C . I i n 3 y r$

Incorrect

P=Rs1000

A=Rs1331

i=0.10

time = n years
$A = P (1 + i)^{n} 1331 = 1000 (1 + 0.10)^{n} 1331 = (1.10)^{n} 1 . 10^{3} = 1 . 10^{n} n = 3 h e n c e, R s 1000 a m o u n t s t o R s 1331 a t 10 % p . a . C . I i n 3 y r$
Question 35 of 50

35. Question
1 points
Paul borrows Rs20000 on condition to repay it with compound interest at 5%p.a in annual installments of Rs2000 each.find the number of years in which the debt would be paid off.
- 10 yr
- 12 yr
- 14 yr
- 15 yr
Correct

P.v=Rs 20000

A=Rs 2000

r=5%p.a. or i=0.05

number of years =n
$P . V = \frac{A}{i} [1 - \frac{1}{(1 + i)^{n}}] 20000 = \frac{2000}{0.05} [1 - \frac{1}{1 . 05^{n}}] 0.5 = 1 - \frac{1}{1 . 05^{n}} 1 . 05^{n} = 1 / 0.5 n = 15 y r (a p p r o x)$

Incorrect

P.v=Rs 20000

A=Rs 2000

r=5%p.a. or i=0.05

number of years =n
$P . V = \frac{A}{i} [1 - \frac{1}{(1 + i)^{n}}] 20000 = \frac{2000}{0.05} [1 - \frac{1}{1 . 05^{n}}] 0.5 = 1 - \frac{1}{1 . 05^{n}} 1 . 05^{n} = 1 / 0.5 n = 15 y r (a p p r o x)$
Question 36 of 50

36. Question
1 points
in how many years,a sum of Rs 1000 compounded annually @10% will amount to Rs 1331?
- 6 yr
- 5 yr
- 4 yr
- 3 yr
Correct

P=Rs1000

A=Rs1331

i=0.10

time = n years
$A = P (1 + i)^{n} 1331 = 1000 (1 + 0.10)^{n} 1331 = (1.10)^{n} 1 . 10^{3} = 1 . 10^{n} n = 3 h e n c e, R s 1000 a m o u n t s t o R s 1331 a t 10 % p . a . C . I i n 3 y r$

Incorrect

P=Rs1000

A=Rs1331

i=0.10

time = n years
$A = P (1 + i)^{n} 1331 = 1000 (1 + 0.10)^{n} 1331 = (1.10)^{n} 1 . 10^{3} = 1 . 10^{n} n = 3 h e n c e, R s 1000 a m o u n t s t o R s 1331 a t 10 % p . a . C . I i n 3 y r$
Question 37 of 50

37. Question
1 points
The compound interest for a certain sum @ 5% p.a. for first year is Rs 25. the S.I for the same money @ 5% p.a. for two years will be
- Rs 40
- Rs 50
- Rs 60
- Rs 70
Correct

C.I = Rs 25, i = 0.05 and n = 1
$C . I = P [(1 + i)^{n} - 1] 25 = P [(1 + 0.05)^{1} - 1] 25 = P [0.05] P = 25 / 0.05 = R s 500 n o w, R = 5 % a n d T = 2 Y r S . i = P R T / 100 = 500 x 5 / 100 x 2 = s 50$

Incorrect

C.I = Rs 25, i = 0.05 and n = 1
$C . I = P [(1 + i)^{n} - 1] 25 = P [(1 + 0.05)^{1} - 1] 25 = P [0.05] P = 25 / 0.05 = R s 500 n o w, R = 5 % a n d T = 2 Y r S . i = P R T / 100 = 500 x 5 / 100 x 2 = s 50$
Question 38 of 50

38. Question
1 points
At what % rate of compound interest will a sum of money become 16 times in four Yr, if interest is being calculated compounding annually
- r=100%
- r=10%
- r=200%
- r=20%
Correct

Let the principal be P and amount = 16 P
$A = P [1 + R / 100]^{n} 16 P = P [1 + R / 100]^{4} 16 = [1 + R / 100]^{4} = 2^{4} 1 + R / 100 = 2 R / 100 = 1 R = 100 %$

Incorrect

Let the principal be P and amount = 16 P
$A = P [1 + R / 100]^{n} 16 P = P [1 + R / 100]^{4} 16 = [1 + R / 100]^{4} = 2^{4} 1 + R / 100 = 2 R / 100 = 1 R = 100 %$
Question 39 of 50

39. Question
1 points
Find the present value of an annuity of Rs 1000 payable at the end of each Yr for 10 Yrs. if rate of interest is 6% compounding p.a.
$[g i v e n (1.06)^{- 10} = 0.5584]$
- Rs 7360
- Rs 8360
- Rs 12000
- none of these
Correct

$V = A / r [1 - (1 + i)^{- n}] = 1000 / 0.06 [1 - (1.06)^{- 10}] = 1000 / 0.06 [1 - 0.5584] = R s 7360$

Incorrect

$V = A / r [1 - (1 + i)^{- n}] = 1000 / 0.06 [1 - (1.06)^{- 10}] = 1000 / 0.06 [1 - 0.5584] = R s 7360$
Question 40 of 50

40. Question
1 points
If the simple interest on a sum of money at 12% p.a for 2 Yrs is Rs 3600. the compound interest on the same sum for 2 Yrs at the same rate is
- Rs 3816
- Rs 3806
- Rs 3861
- Rs 3860
Correct

SI = PRT/100

3600 = Px12x2/100

P=Rs 15000
$C I = P [(1 + R / 100)^{n} - 1] = 15000 [(1 + 12 / 100)^{2} - 1] = R s 3816$

Incorrect

SI = PRT/100

3600 = Px12x2/100

P=Rs 15000
$C I = P [(1 + R / 100)^{n} - 1] = 15000 [(1 + 12 / 100)^{2} - 1] = R s 3816$
Question 41 of 50

41. Question
1 points
The future value of an annuity of Rs 5000 is made annually for 8 Yrs @ 9% compounded annually
$[g i v e n (1.09)^{8} = 1.99256]$
- Rs 55142.22
- Rs 65142.22
- Rs 65532.22
- Rs 57142.22
Correct

$F V = P [\frac{(1 + i)^{n} - 1}{i}] w h e r e P = 5000, i = 9 % a n d n = 8 F V = 5000 [\frac{(1 + 9 / 100)^{8} - 1}{0.09}] = 5000 [\frac{1.99256 - 1}{0.09}] = 55142.22$

Incorrect

$F V = P [\frac{(1 + i)^{n} - 1}{i}] w h e r e P = 5000, i = 9 % a n d n = 8 F V = 5000 [\frac{(1 + 9 / 100)^{8} - 1}{0.09}] = 5000 [\frac{1.99256 - 1}{0.09}] = 55142.22$
Question 42 of 50

42. Question
1 points
The effective annual rate of interest corresponding to nominal rate 6%p.a payable half yearly is
- 6.06%
- 6.07%
- 6.08%
- 6.09%
Correct

$n o m i n a l r a t e = 6 % p . a e f f e c t i v e r a t e = (1 + i)^{n} - 1 i = 3 % n = 2 e f f e c t i v e r a t e = (1 + 3 / 100)^{2} - 1 = 0.0609 = 6.09 %$

Incorrect

$n o m i n a l r a t e = 6 % p . a e f f e c t i v e r a t e = (1 + i)^{n} - 1 i = 3 % n = 2 e f f e c t i v e r a t e = (1 + 3 / 100)^{2} - 1 = 0.0609 = 6.09 %$
Question 43 of 50

43. Question
1 points
The cost of machinery is Rs 125000.if its useful life is estimated to be 20 yr and the rate of depreciation of its cost is 10%p.a.,then the scrap value of the machine is
$[g i v e n (0.9)^{20} = 0.1215]$
- Rs 15187
- Rs 15400
- Rs 15300
- Rs 15250
Correct

$s c r a p v a l u e = P (1 - i)^{n} P = 125000, i = 10 %, n = 20 v = 125000 [1 - 10 / 100]^{20} = 125000 x 0.1215 = 15187$

Incorrect

$s c r a p v a l u e = P (1 - i)^{n} P = 125000, i = 10 %, n = 20 v = 125000 [1 - 10 / 100]^{20} = 125000 x 0.1215 = 15187$
Question 44 of 50

44. Question
1 points
Mr.X invests P amount at simple interest rate 10% and Mr.Y invests Q amount at compound interest ate 5% compounded annually.at the end of 2 yr both get the same amount of interest , then the relationship between 2 amounts P and Q is given by
- P=41/80Q
- P=41/40Q
- P=41/100Q
- P=41/200Q
Correct

SI=Px10/100×2=P/5
$C I = Q [(1 + 5 / 100)^{2} - 1] = 0.1025 Q P / 5 = 0.1025 Q P / 5 = 1025 / 10000 Q P / 5 = 41 / 400 Q P = 41 / 80 Q$

Incorrect

SI=Px10/100×2=P/5
$C I = Q [(1 + 5 / 100)^{2} - 1] = 0.1025 Q P / 5 = 0.1025 Q P / 5 = 1025 / 10000 Q P / 5 = 41 / 400 Q P = 41 / 80 Q$
Question 45 of 50

45. Question
1 points
If the difference SI and CI is Rs 72 @ 12% for 2Yrs. calculate the amount
- Rs 8000
- Rs 6000
- Rs 5000
- Rs 4000
Correct

Let the principal be Rs x

R=12% p.a, T=2 Yrs and SI for 2 Yrs

SI=PRT/100=X x 12 x 2/100=24X/100
$C I f o r 2 Y r s C I = P [(1 + R / 100)^{t} - 1] = X [(1 + 12 / 100)^{2} - 1] C I = 159 X / 625 d i f f e r e n c e = 159 X / 625 - 24 X / 100 = 72 o n s o l v i n g w e g e t X = 5000$

Incorrect

Let the principal be Rs x

R=12% p.a, T=2 Yrs and SI for 2 Yrs

SI=PRT/100=X x 12 x 2/100=24X/100
$C I f o r 2 Y r s C I = P [(1 + R / 100)^{t} - 1] = X [(1 + 12 / 100)^{2} - 1] C I = 159 X / 625 d i f f e r e n c e = 159 X / 625 - 24 X / 100 = 72 o n s o l v i n g w e g e t X = 5000$
Question 46 of 50

46. Question
1 points
If a SI on a sum of money @ 6% p.a, for 7 Yrs. is equal to twice of SI on another sum for 9 Yrs. @5% p.a, the ratio will be
- 2:15
- 7:15
- 15:7
- 1:7
Correct

Let the principals be X and Y

case I SI = X x 6 x 7/100

case II SI = Y x 5 x 9/100

case I = 2 x case II

X x 6 x 7/100 = 2 x Y x 5 x 9/100

X/Y = 15/7 = 15:7

Incorrect

Let the principals be X and Y

case I SI = X x 6 x 7/100

case II SI = Y x 5 x 9/100

case I = 2 x case II

X x 6 x 7/100 = 2 x Y x 5 x 9/100

X/Y = 15/7 = 15:7
Question 47 of 50

47. Question
1 points
By mistake a clerk, calculated the simple interest on principal for 5 months @ 6.5% p.a, instead of 6 months @ 5.5% p.a, if the error in calculation was Rs 25.40. the original sum of principal was
- Rs 60690
- Rs 60960
- Rs 90660
- Rs 90690
Correct

Let the principal be Rs X

incorrect SI = X x 6.5 x 5/100 x 12 = 32.5X / 1200

Correct SI = X x 5.5 x 6/100 x 12 = 33X/1200

error in calculation = Rs 25.40

Correct SI – incorrect SI = Rs 25.40

33X/1200-32.5X/1200 = 25.40

X = Rs 60960

Incorrect

Let the principal be Rs X

incorrect SI = X x 6.5 x 5/100 x 12 = 32.5X / 1200

Correct SI = X x 5.5 x 6/100 x 12 = 33X/1200

error in calculation = Rs 25.40

Correct SI – incorrect SI = Rs 25.40

33X/1200-32.5X/1200 = 25.40

X = Rs 60960
Question 48 of 50

48. Question
1 points
If the simple interest on Rs 1400 for 3 Yrs, is less than the simple interest on Rs 1800 for the same period by Rs 80, then the rate of interest is
- 5.67%
- 6.67%
- 7.20%
- 5.00%
Correct

CASE I P=Rs 1400, T=3Yrs, R = X%

SI = PRT/100 = 1400 x X x 3/100 = 42X

CASE II P = Rs 1800, T = 3Yrs, R = X%

SI = PRT/100 = 1800 x X x 3/100 = 54X

Given, case I – case II = 80

54X – 42X = 80

X = 80/12 = 6.67% = R

Incorrect

CASE I P=Rs 1400, T=3Yrs, R = X%

SI = PRT/100 = 1400 x X x 3/100 = 42X

CASE II P = Rs 1800, T = 3Yrs, R = X%

SI = PRT/100 = 1800 x X x 3/100 = 54X

Given, case I – case II = 80

54X – 42X = 80

X = 80/12 = 6.67% = R
Question 49 of 50

49. Question
1 points
Nominal rate of interest is 9.9% p.a, if interest is compounded monthly, what will be the effective rate of interest
$[G i v e n [\frac{4033}{4000}]^{12} = 1.1036 (a p p r o x)]$
- 10.36%
- 9.36%
- 11.36%
- 9.9%
Correct

R = 9.9% P.a.

if interest is paid compounded monthly

i=9.9/12×100=99/12x100x10=33/4000 and n=12
$e f f e c t i v e r a t e = (1 + i)^{n} - 1 = (1 + 33 / 4000)^{12} - 1 = (4033 / 4000)^{12} - 1 = 1.1036 - 1 = 0.1036 = 10.36 %$

Incorrect

R = 9.9% P.a.

if interest is paid compounded monthly

i=9.9/12×100=99/12x100x10=33/4000 and n=12
$e f f e c t i v e r a t e = (1 + i)^{n} - 1 = (1 + 33 / 4000)^{12} - 1 = (4033 / 4000)^{12} - 1 = 1.1036 - 1 = 0.1036 = 10.36 %$
Question 50 of 50

50. Question
1 points
The SI on a sum of money is 4/9 of the principal and the no. of years is equal to the rate of interest p.a.find the rate of interest p.a?
- 5%
- 20/3%
- 22/7%
- 6%
Correct

Given say principal P

SI=4/9P

T=R

SI=PRT/100

4/9P=PxRxR/100

R=20/3%

Incorrect

Given say principal P

SI=4/9P

T=R

SI=PRT/100

4/9P=PxRxR/100

R=20/3%

Simple and Compound Interest-Test 1

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