Quantitative Aptitude Questions (Compound Interest) Day-41

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Practice Aptitude Questions (Compound Interest) Day-41

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1) Find the C.I accrued by Nandhini from a bank on Rs. 24500 in 2 years, when the rates of interest for successive years are 8%, and 10% respectively.

  1. Rs. 3452
  2. Rs. 2646
  3. Rs. 1960
  4. Rs. 4606
  5. None of these

2) If the S.I on a sum of money for 2 years at 10% per annum is Rs. 100, what is the C.I on the same at the same rate and for the same time?

  1. Rs. 705
  2. Rs. 605
  3. Rs. 405
  4. Rs. 105
  5. None of these
  1. The compound interest on Rs 30000 at 7% per annum for a certain time is Rs.4347. The time is?
  1. 2 yr
  2. 2.5 yr
  3. 3 yr
  4. 4 yr
  5. None of these

4) In what time will Rs.8000 become Rs.9261 at 5% per annum compounded annually?

  1. 3 years
  2. 5 years
  3. 7 years
  4. 9 years
  5. 10 years
  1. A certain sum of money amounts to Rs. 31250 in 5 years at the rate of 25% p.a compounded annually. Find the Principal.
  1. Rs. 10240
  2. Rs. 10024
  3. Rs. 10204
  4. Cannot be determined
  5. None of these

6) Find CI on Rs.15250 at 14% per annum for 2 years 6 months, compound annually.

  1. 5943.223
  2. 5956.223
  3. 5867.553
  4. 5963.553
  5. None of these

7) Split the approximate amount of Rs. 1762 between ashok and anirudh, so that the amount of ashok after 5 years is equal to the amount of Anirudh after 7 years, the interest being compound at 5% per annum.

  1. 920,824
  2. 930,812
  3. 934,830
  4. 924,838
  5. None of these

8) A sum of cash 4 times itself at compound interest in 20 years. In how many years will it become 16 times?

  1. 40
  2. 60
  3. 50
  4. 30
  5. None of these

9) Renuka invested Rs.7500 at 20% per annum for 1 year. If the interest is compounded half – yearly, then the amount received by renuka at the end of the year.

  1. 9065
  2. 9055
  3. 9075
  4. 9085
  5. 9035

10) Aruna borrows Rs.6250 from geetha at 10% CI. At the end of every year she pays Rs.1000 as part repayment. How much does she still over after 3 such installments?

  1. 5004.3
  2. 5007.32
  3. 5008.75
  4. 5017.75
  5. 5020.25

Answers:

1) Answer D

For the first year:

Principal = Rs. 24500

Rate of interest = 8% and

Time = 1 year

Therefore, interest for the first year = P×R×T/100

=24500*8*1/100

=196000/100

=RS 1960

Therefore, the amount after 1 year = Principal + Interest

= Rs. 24500+1960

=Rs. 26460

For the second year, the new principal is Rs. 26460

Rate of interest = 10% and

Time = 1 year.

Therefore, the interest for the second year

=26460*10*1/100

=264600/100

=RS 2646

Therefore, the amount after 2 year = Principal + Interest

=26460+2646

=Rs. 29,106

Therefore, the compound interest accrued = Final amount – Initial principal

=29106-24500

=Rs. 4606

2) Answer D

SI=PTR/100

P=100*SI/TR

=100*100/20

=500

Amount = (p*(1+R/100)n)

=500(1+10/100)2

= 500(110/100)2

=500(11/10)2

=500*11/10*11/10

=60500/100

=RS 605

CI=Rs (605-500)

=Rs. 105

  1. Answer A

P = 3000

R = 7%

CI=4347

Total amount = 34347

T=?

Amount = P [(1 +(R /100))n]

34347 = 30000[1 + (7 / 100)n] [34347 / 30000] = (107 / 100)n

11449 / 10000 =(107 / 100)n

(107 * 107) / (100 * 100) = (107 / 100)n

(107 / 100)2 =(107 / 100)n

:. T = 2 yr

4) Answer A

Principal = Rs.8000;
Amount = Rs.9261;
Rate = Rs.5% p.a.
Let the time be n years then,

A=P(1+R/100)n

9261=8000(1+5/100)n

9261/8000 = (105/100)n

9261/8000 = (21/20)n

(21/20)3=(21/20)n

n=3 years

  1. Answer A

31250 = P [ 1 + (25 / 100) ]5

31250 = P (125 / 100)5

31250 (4 / 5)5 = P

P = (31250 × 16 × 16 × 4) / (25 × 25 × 5) = Rs. 10240

6) Answer B

Given time = 2yrs 6 months

= 6/12 = 2 ½ yrs

Amount = P (1+R/100)2 (1+ (1/2R)/100)

=Rs [15250 (1+ (14/100))2 (1+7/100)]

=Rs [15250 (114/100)(114/100)(107/100)]

=Rs(15250 * 1.14 * 1.14 * 1.07)

=Rs.21206.223

Compound interest = Rs. (21206.223 – 15250)

=Rs.5956.223

7) Answer D

Let the two parts Rs.x and Rs.(1762 – x)

X(1+5/100)5 = (1762 – x)(1+5/100)7

x/1762 – x = (1+5/100)7 / (1+5/100)5

x / 1762 – x = (1+5/100)2

x / 1762 – x = (105/100)2

x / 1762 –x = (21/20)2

x / 1762 – x = 21*21/20*10 = 441/400

400x = 441(1762 – x)

400x = 777042 – 441x

841x = 777042

X = 923.9 = 924

So the 2 parts are Rs.924 and 838

8) Answer A

By using given condition,

P(1+R/100)20 = 4P

(1+R / 100)20 = 4

(1+R/100)20 = 22                   ————>1

Let P (1 +R/100)n = 16P

(1+R/100)n = 16 = 24        —————->2

Using (1)

(1+R / 100)n = (1+R /100)40

N=40

Thus required time = 40 years

(or)

Let us assume, the amount as 100

100 ——–> 400

(20 yr)

400 ——–> 1600

(20 yr)

16 times will be in 40 years.

9) Answer C

P= Rs.7500    R=20% per annum

R=10% per half year, T=1, Y = 2 half year

Amount = (1+R/100)n

=7500(1+10/100)2

=7500(110/100)2

=7500(11/10*11/10)

=75*11*11

=Rs.9075

10) Answer C

6250*(10/100) = 625

(6250+625) -1000 = 5875

5875*(10/100) = 587.5

(5875+587.5) -1000 = 5462.5

5462.5*(10/100) = 546.25

(5462.5+546.25)-1000 = 5008.75

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