Home Quantitative Aptitude Simple and Compound Interest-Test 3

Simple and Compound Interest-Test 3

Subject :- QT

Chapter :- Simple and Compound Interest-Test 3

Questions :- 25

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Question 1 of 25

1. 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 2 of 25

2. 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 3 of 25

3. 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 4 of 25

4. 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 5 of 25

5. 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 6 of 25

6. 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 7 of 25

7. 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 8 of 25

8. 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 9 of 25

9. 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 10 of 25

10. 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 11 of 25

11. 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 12 of 25

12. 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 13 of 25

13. 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 14 of 25

14. 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 15 of 25

15. 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 16 of 25

16. 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 17 of 25

17. 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 18 of 25

18. 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 19 of 25

19. 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 20 of 25

20. 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 21 of 25

21. 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 22 of 25

22. 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 23 of 25

23. 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 24 of 25

24. 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 25 of 25

25. 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 3

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