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) 13%
(b) 9%
(c) 11%
(d) 10%
The correct answers to the above question in:
Answer: (d)
Using Rule 1,If A = Amount, P = Principal, r = Rate of Compound Interest (C.I.), n = no. of years then,A=P$(1 + r/100)^n$, C.I. = A - PC.I. = P$[(1 + r/100)^n - 1]$
P = Rs.3000, A = Rs.3993, n = 3 years
A = P$(1 + r/100)^n$
$(1 + r/100)^n = A/P$
$(1 + r/100)^3 = 3993/3000 = 1331/1000$
$(1 + r/100)^3 = (11/10)^3$
1 + $r/100 = 11/10$
$r/100 = 11/10$ - 1
$r/100 = 1/10 ⇒ r = 100/10$ = 10%
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Read more basic problems using formula Based Quantitative Aptitude Questions and Answers
Question : 1
A certain sum of money yields Rs.1261 as compound interest for 3 years at 5% per annum. The sum is
a) Rs.8000
b) Rs.9000
c) Rs.7500
d) Rs.8400
Answer »Answer: (a)
Let the principal be Rs.x. Now,
C.I. = P$[(1 + R/100)^T - 1]$
1261 = $x[(1 + 5/100)^3 - 1]$
1261 = $x(9261/8000 - 1)$
1261 = $x({9261 - 8000}/8000)$
= ${1261x}/8000$
$x = {1261 × 8000}/1261$ = Rs.8000
Question : 2
The principal, which will amount to Rs.270.40 in 2 years at the rate of 4% per annum compound interest, is
a) Rs.220
b) Rs.200
c) Rs.250
d) Rs.225
Answer »Answer: (c)
Using Rule 1,
Let the principal be Rs.P.
270.40 = P $(1 + 4/100)^2$
270.40 = P $(1 + 0.04)^2$
P = ${270.40}/{1.04 × 1.04}$ = Rs.250
Question : 3
In what time will Rs.10,000 amount to Rs.13310 at 20% per annum compounded half yearly?
a) 3 years
b) 1$1/2$ years
c) 2$1/2$ years
d) 2 years
Answer »Answer: (b)
Using Rule 1,If A = Amount, P = Principal, r = Rate of Compound Interest (C.I.), n = no. of years then,A=P$(1 + r/100)^n$, C.I. = A - PC.I. = P$[(1 + r/100)^n - 1]$
Using Rule 2,Compound interest is calculated on four basis:
Rate | Time(n) | ||
Annually | r% | t years | |
Half–yearly(Semi-annually | $r/2$% | t × 2 years | |
Quarterly | $r/4$% | t × 4 years | |
Monthly | $r/12$% | t × 12 years |
The rate of interest is compounded half yearly,
r = 10% per half year
Let time = $T/2$ years = half years
According to the question,
Amount = P$(1 + R/100)^t$
13310 = 10000$(1 + 10/100)^T$
$13310/10000 = (11/10)^T$
$(11/10)^T = 1331/1000 = (11/10)^3$
T = 3 half years =1$1/2$ years
Question : 4
At what rate per cent per annum will Rs.2304 amount to Rs.2500 in 2 years at compound interest ?
a) 4$1/3$%
b) 4$1/2$%
c) 4$1/6$%
d) 4$1/5$%
Answer »Answer: (c)
Using Rule 1,
Let the rate per cent per annum be r. Then,
2500 = 2304$(1 + r/100)^2$
$(1 + r/100)^2 = 2500/2304 = (50/48)^2$
$1 + r/100 = 50/48 = 25/24$
$r/100 = 25/24 - 1 = 1/24$
r = $100/24 = 25/6 = 4{1}/6$%
Question : 5
In how many years will Rs.2,000 amounts to Rs.2,420 at 10% per annum compound interest?
a) 1$1/2$ years
b) 3 years
c) 2 years
d) 2$1/2$ years
Answer »Answer: (c)
Using Rule 1,
According to question,
2420 = 2000$(1 + 10/100)^t$
$2420/2000 = (11/10)^t$
or $(11/10)^t = 121/100$
or, $(11/10)^t = (11/10)^2$
t = 2 years
Question : 6
A certain sum, invested at 4% per annum compound interest, compounded half yearly, amounts to Rs.7,803 at the end of one year. The sum is
a) Rs.7,700
b) Rs.7,000
c) Rs.7,500
d) Rs.7,200
Answer »Answer: (c)
Using Rule 1,
Let the sum be P.
As, the interest is compounded half-yearly,
R = 2%, T = 2 half years
A = P$(1 + R/100)^T$
7803 = P$(1 + 2/100)2$
7803 = $(1 + 1/50)^2$
7803 = P$× 51/50 × 51/50$
P = ${7803 × 50 × 50}/{51 × 51}$ = Rs.7500
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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