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) Rs.3050
(b) Rs.4000
(c) Rs.3000
(d) Rs.2500
The correct answers to the above question in:
Answer: (c)
P$(1 + r/100)^2$ = 4500 ...(i)
P$(1 + r/100)^4$ = 6750 .....(ii)
On dividing equation (ii) by equation (i), we get
$(1 + r/100)^2 = 6750/4500$
From equation (i),
P × $6750/4500$ = 4500
P = ${4500 × 4500}/6750$ = Rs.3,000
Using Rule 7(ii),
Here, b - a = 4 - 2 = 2
B = Rs.6750, A = Rs.4500
R% = $[(B/A)^{1/2} - 1]$ × 100%
= $[(6750/4500)^{1/2} - 1] × 100%$
= $[(3/2)^{1/2} - 1] ×$ 100%
$(3/2)^{1/2} = 1 + R/100$
$3/2 = (1 + R/100)^2$
A = P$(1 + R/100)^2$
4500 = P × $3/2$ ⇒ P = Rs.3000
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Read more comparing sum in different years Based Quantitative Aptitude Questions and Answers
Question : 1
On a certain sum of money, the simple interest for 2 years is Rs.350 at the rate of 4% per annum. It was invested at compound interest at the same rate for the same duration as before, how much more interest would be earned?
a) Rs.35
b) Rs.3.50
c) Rs.14
d) Rs.7
Answer »Answer: (d)
Principal = $\text"S.I. × 100"/ \text"Time × Rate"$
= ${350 × 100}/{2 × 4}$ = Rs.4375
C.I. = P$[(1 + R/100)^T - 1]$
= 4375$[(1 + 4/100)^2 - 1]$
= 4375$[(1 + 1/25)^T - 1]$
= 4375$[(26/25)^2 - 1]$
= 4375$(676/625 - 1)$
= ${4375 × 51}/625$ = Rs.357
Required difference
= Rs.(357 - 350) = Rs.7
Question : 2
A sum of money is invested at 20% compound interest (compounded annually). It would fetch Rs. 723 more in 2 years if interest is compounded half yearly. The sum is
a) Rs.7,500
b) Rs.15,000
c) Rs.20,000
d) Rs.30,000
Answer »Answer: (d)
Let the principal be Rs. x.
When the interest is compounded annually,
C.I. = P$[(1 + R/100)^T - 1]$
= P$[(1 + 20/100)^2 - 1]$
= P$[(6/5)^2 - 1]$
= P$(36/25 - 1)$ = Rs.${11P}/25$
When the interest is compounded half–yearly,
C.I. = P$[(1 + 10/100)^4 - 1]$
= P$[(11/10)^4 - 1]$
= P$(14641/10000 - 1)$
= Rs.${4641P}/10000$
${4641P}/10000 - {11P}/25$ = 723
${4641P - 4400P}/10000$ = 723
${241P}/10000$ = 723
P = ${723 × 10000}/241$ = Rs.30000
Question : 3
An amount of money at compound interest grows up to Rs.3,840 in 4 years and up to Rs.3,936 in 5 years. Find the rate of interest.
a) 2.05%
b) 2.5%
c) 3.5%
d) 2%
Answer »Answer: (b)
A = P$(1 + R/100)^T$
3840 = P$(1 + R/100)^4$ ....(i)
3936 = P$(1 + R/100)^5$ ...(ii)
Dividing equation (ii) by equation (i),
$3936/3840 = 1 + R/100$
$R/100 = 3936/3840$ - 1
= ${3936 - 3840}/3840 = 96/3840$
R = $96/3840 × 100$ = 2.5%
Using Rule 7(i),
Here, b - a = 5 - 4 = 1
B = Rs.3,936, A = Rs.3,840
R%= $(B/A - 1)$ × 100%
= $(3936/3840 - 1)$ × 100%
= $({3936 - 3840}/3840) × 100%$
= $96/3840 × 100% = 10/4%$ = 2.5%
Question : 4
A sum of money invested at compound interest amounts in 3 years to Rs.2,400 and in 4 years to Rs.2,520. The interest rate per annum is :
a) 12%
b) 5%
c) 10%
d) 6%
Answer »Answer: (b)
S.I. on Rs.2400 for 1 year
= Rs.(2, 520 - 2, 400) = Rs.120
Rate = $\text"S.I. × 100"/ \text"Principal × Time"$%
= ${120 × 100}/{2400 × 1}$ = 5%
Using Rule 7(i),
Here, b - a = 4 - 3 = 1
B = Rs.2520, A = Rs.2400
R% = $(B/A - 1)$ × 100%
= $[2520/2400 - 1] × 100%$
= $[{2520 - 2400}/2400] × 100%$
= $120/2400 × 100%$ = 5%
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
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model 3 combination of si & ci
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model 4 difference in ci & si
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model 5 ci with instalments
Defination & Shortcuts … -
model 6 comparing sum in different years
Defination & Shortcuts …
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