model 1 basic compound interest using formula Section-Wise Topic Notes With Detailed Explanation And Example Questions

MOST IMPORTANT quantitative aptitude - 6 EXERCISES

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The following question based on compound interest topic of quantitative aptitude

Questions : In what time will Rs.1000 amounts to Rs.1331 at 20% per annum, compounded half yearly ?

(a) 2$1/2$ years

(b) 1$1/2$ years

(c) 1 year

(d) 2 years

The correct answers to the above question in:

Answer: (b)

Using Rule 1 and 2,

Let the required time be t years. Interest is compounded half yearly.

Time = 2t half years and rate

= $20/2$ = 10%

1000$(1 + 10/100)^{2t}$ = 1331

$(11/10)^{2t} = 1331/1000$

$(11/10)^{2t} = (11/10)^3$ ⇒ 2t = 3

t = $3/2$ years or 1$1/2$ years

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Read more basic problems using formula Based Quantitative Aptitude Questions and Answers

Question : 1

The compound interest on Rs.30,000 at 7% per annum for a certain time is Rs.4,347. The time is

a) 2.5 years

b) 3 years

c) 2 years

d) 4 years

Answer: (c)

Using Rule 1,

A = P$(1 + R/100)^T$

30000 + 4347 = $30000(1 + 7/100)^T$

$34347/30000 = (107/100)^T$

$11449/10000 = (107/100)^2 = (107/100)^T$

Time = 2 years

Question : 2

A certain sum amounts to Rs.5,832 in 2 years at 8% per annum compound interest, the sum is

a) Rs.5,400

b) Rs.5,000

c) Rs.5,280

d) Rs.5,200

Answer: (b)

Using Rule 1,

5832 = P$(1 + 8/100)^2$

5832 = P$(1 + 2/25)^2$

5832 = P $× 27/25 × 27/25$

P = ${5832 × 25 × 25}/{27 × 27}$ = Rs.5000

Question : 3

The compound interest on Rs.8,000 at 15% per annum for 2 years 4 months, compounded annually is:

a) Rs.3100

b) Rs.2980

c) Rs.3109

d) Rs.3091

Answer: (c)

Using Rule 1,

Amount = P$(1 + R/100)^t$

= 8000$(1 + 15/100)^{2{1}/3}$

= 8000$(1 + 3/20)^2(1 + 3/{20 × 3})$

= $8000 × 23/20 × 23/20 × 21/20$ = Rs.11109

Compound Interest

= Rs.(11109 - 8000) = Rs.3109.

Question : 4

At what rate per cent per annum will a sum of Rs.1,000 amounts to Rs.1,102.50 in 2 years at compound interest ?

a) 6.5%

b) 5%

c) 6%

d) 5.5%

Answer: (b)

Using Rule 1,

A = P$(1 + R/100)^T$

Let rate be 'r'

${1102.50}/1000 = (1 + r/100)^2$

$11025/10000 = (1 + r/100)^2$

$(105/100)^2 = (1 + r/100)^2$

1 + $r/100 = 105/100$

$r/100 = 5/100$ = 5%

Question : 5

In how many years will a sum of Rs.800 at 10% per annum compound interest, compounded semi-annually becomes Rs.926.10 ?

a) 2$1/2$ years

b) 1$1/2$ years

c) 2$1/3$ years

d) 1$2/3$ years

Answer: (b)

Using Rule 1 and 2,

Rate = 10% per annum = 5% half yearly

A = P$(1 + R/100)^T$

926.10 = 800$(1 + 5/100)^T$

$9261/8000 = (21/20)^T$

$(21/20)^3 = (21/20)^T$

Time = 3 half years = 1$1/2$ years

Question : 6

If the rate of interest be 4% per annum for first year, 5% per annum for second year and 6% per annum for third year, then the compound interest of Rs.10,000 for 3 years will be

a) Rs.2,000

b) Rs.1,600

c) Rs.1,575.20

d) Rs.1,625.80

Answer: (c)

Using Rule 3,
If there are distinct 'rates of interest' for distinct time periods i.e.,
Rate for 1st year → $r_1$%
Rate for 2nd year → $r_2$%
Rate for 3rd year → $r_3$% and so on
Then A = P$(1 + r_1/100)(1 + r_2/100)(1 + r_3/100)$...
C.I. = A - P

Amount = P$(1 + R_1/100)(1 + R_2/100)(1 + R_3/100)$

= 10000$(1+ 4/100)(1 + 5/100)(1 + 6/100)$

= $10000 × 26/25 × 21/20 × 53/50$

A = Rs.11575.2

C.I. = Rs.(11575.2–10000) = Rs.1575.2

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