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

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The following question based on compound interest topic of quantitative aptitude

Questions : The time in which Rs.80,000 amounts to Rs.92,610 at 10% p.a. compound interest, interest being compounded semi annually is :

(a) 3 years

(b) 1$1/2$ years

(c) 2$1/2$ years

(d) 2 years

The correct answers to the above question in:

Answer: (b)

Using Rule 1 and 2,

Time = t half year

and R = 5% per half year

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

$92610/80000 = (1 + 5/100)^T$

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

T = 3 half years or 1$1/2$ years

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

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

Question : 1

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

Question : 2

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 : 3

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%

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