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
(a) 2.5 years
(b) 3 years
(c) 2 years
(d) 4 years
The correct answers to the above question in:
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
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Read more basic problems using formula Based Quantitative Aptitude Questions and Answers
Question : 1
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 »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 : 2
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 »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 : 3
The compound interest on Rs.6,000 at 10% per annum for 1$1/2$ years, when the interest being compounded annually, is
a) Rs.900
b) Rs.910
c) Rs.930
d) Rs.870
Answer »Answer: (c)
Amount = 6000$(1 + 10/100) × (1 + {{1/2} × 10}/100)$
= $6000 × 11/10 × 21/20$ = Rs.6930
Using Rule 4,If the time is in fractional form i.e.,t = nF, thenA = P$(1 + r/100)^n(1 + {rF}/100)$e.g. t =3$5/7$ yrs, thenA = P$(1 + r/100)^3(1 + r/100 × 5/7)$
Here, t = nF
A = P$(1 + r/100)^n(1 + {rF}/100)$
CI = Rs.(6930 - 6000) = Rs.930
Question : 4
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
Answer »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
Question : 5
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 »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 : 6
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 »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
compound interest Shortcuts and Techniques with Examples
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model 1 basic compound interest using formula
Defination & Shortcuts … -
model 2 at ci sum becomes ‘n’ times after ‘t’ years
Defination & Shortcuts … -
model 3 combination of si & ci
Defination & Shortcuts … -
model 4 difference in ci & si
Defination & Shortcuts … -
model 5 ci with instalments
Defination & Shortcuts … -
model 6 comparing sum in different years
Defination & Shortcuts …
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