model 6 comparing sum in different years 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) 10%
(b) 6%
(c) 9%
(d) 8%
The correct answers to the above question in:
Answer: (a)
Let the rate of interest = R% per annum.
We know that
A = P$(1 + R/100)^T$
2420 = P$(1 + R/100)^2$ ....(i)
2662 = P$(1 + R/100)^3$ ...(ii)
Dividing equation (ii) by (i),
$1 + R/100 = 2662/2420$
$R/100 = 2662/2420$ - 1
$R/100 = {2662 - 2420}/2420$
= $242/2420 = 1/10$
R = $1/10 × $100 = 10%
Using Rule 7(i),
Here, b - a = 3 - 2 = 1
B = Rs.2,662, A= Rs.2,420
R% = $(B/A - 1)$ × 100%
= $(2662/2420 -1)$ × 100%
= $[{2662 - 2420}/2420]$ × 100%
= $242/2420 × 100%$ = 10%
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Read more comparing sum in different years Based Quantitative Aptitude Questions and Answers
Question : 1
A certain amount of money at r%, compounded annually after two and three years becomes Rs.1440 and Rs.1728 respectively. r is
a) 20
b) 5
c) 15
d) 10
Answer »Answer: (a)
If the principal be Rs.P, then
A = P$(1 + R/100)^T$
1440 = P$(1 + R/100)^2$ ...(i)
and 1728 = P$(1 + R/100)^3$ ...(ii)
On dividing equation (ii) by (i),
$1728/1440 = 1 + r/100$
$r/100 = 1728/1440$ - 1
= ${1728 - 1440}/1440 = 288/1440$
r = ${288 × 100}/1440$
r = 20% per annum
Using Rule 7(i),
Here, b - a = 3 - 2 = 1
B = Rs.1728, A = Rs.1440
R% = $(B/A - 1)$ × 100%
= $(1728/1440 - 1) × 100%$
= $({1728 - 1440}/1440) × 100%$
= $[288/1440] × 100%$ = 20%
Question : 2
An amount of money appreciates to Rs.7,000 after 4 years and to Rs.10,000 after 8 years at a certain compound interest compounded annually. The initial amount of money was
a) Rs.4,300
b) Rs.4,700
c) Rs.4,100
d) Rs.4,900
Answer »Answer: (d)
A = P$(1 + R/100)^T$
7000 = P$(1 + R/100)^4$....(i)
10000 = P$(1 + R/100)^8$.......(ii)
Dividing equation (ii) by (i)
$10000/7000 = (1 + R/100)^4$
$10/7 = (1 + R/100)^4$
From equation (i),
7000 = P × $10/7$ ⇒ P = Rs.4900
Using Rule 7(iii),
Here, b - a = 8 - 4 = 4
B = Rs.10,000, A = Rs.7000
R% = $((B/A)^{1/n} - 1)$ × 100%
R% = $[(10000/7000)^{1/4} - 1]$
= $[(10/7)^{1/4} - 1]$
$1 + R/100 = (10/7)^{1/4}$
$(1 + R/100)^4 = 10/7$
7000 = $P × 10/7$
Since, Amount = P$(1 + R/100)^4$
P = Rs.4900
Question : 3
A sum of money invested at compound interest amounts to Rs.650 at the end of first year and Rs.676 at the end of second year. The sum of money is :
a) Rs.560
b) Rs.600
c) Rs.625
d) Rs.540
Answer »Answer: (c)
Interest on Rs.650 for 1 year
= 676 - 650 = Rs.26
So, r = $26/650 × 100$
r = 4% per annum
P = $A/[1 + r/100]^t = 650/[1 + 4/100]^1$
= $650/{26/25} = 650 × 25/26$ = Rs.625
Using Rule 7(i),
Here, b - a = 1
B = Rs.676, A = Rs.650
R% = $(B/A - 1)$ × 100%
= $[676/650 - 1] × 100%$
= $[{676 - 650}/650] × 100%$
= $26/650 × 100% = 100/25$ = 4%
Amount= P$(1 + R/100)^1$
650 = P$(1 + 4/100)$
P = ${650 × 100}/104$ = Rs.625
Note : A sum at a rate of interest compounded yearly becomes Rs.$A_1$, in n years and Rs. $A_2$ in (n + 1) years, then
P = $A_1(A_1/A_2)^n$
Question : 4
A sum of money at compound interest will amount to Rs.650 at the end of the first year and Rs.676 at the end of the second year. The amount of money is
a) Rs.625
b) Rs.1,300
c) Rs.1,250
d) Rs.650
Answer »Answer: (a)
Principal = Rs.P (let)
Rate = R% per annum
A = P$(1 + R/100)^T$
650 = P$(1 + R/100)$
$650/P = (1 + R/100)$ ...(i)
Again, 676 = P$(1 + R/100)^2$
676 = P$(650/P)^2$
= ${P × 650 × 650}/P^2$
P = ${650 × 650}/676$ = Rs.625
Question : 5
The compound interest on a certain sum for 2 years at 10% per annum is Rs. 525 . The simple interest on the same sum for double the time at half the rate per cent per annum is :
a) Rs.515
b) Rs.520
c) Rs.500
d) Rs.550
Answer »Answer: (c)
C.I. = P$[(1 + R/100)^T - 1]$
525 = P$[(1 + 10/100)^2 - 1]$
525 = P$(121/100 - 1)$
525 = ${P × 21}/100$
P = ${525 × 100}/21$ = Rs.2500
Again, new rate = 5% per annum
S.I. = $\text"Principal × Time × Rate"/100$
= ${2500 × 5 × 4}/100$ = Rs.500
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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