model 6 comparing sum in different years Practice Questions Answers Test with Solutions & More Shortcuts
compound interest PRACTICE TEST [6 - EXERCISES]
model 1 basic compound interest using formula
model 2 at ci sum becomes ‘n’ times after ‘t’ years
model 3 combination of si & ci
model 4 difference in ci & si
model 5 ci with instalments
model 6 comparing sum in different years
Question : 1 [SSC CGL Prelim 2002]
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%
Question : 2 [SSC CGL 2005]
A sum becomes Rs.4500 after two years and Rs.6750 after four years at compound interest. The sum is
a) Rs.3050
b) Rs.4000
c) Rs.3000
d) Rs.2500
Answer »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
Question : 3 [SSC CAPFs 2016]
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 : 4 [SSC CGL Tier-II 2016]
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 : 5 [SSC CGL Tier-II 2012]
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%
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model 6 comparing sum in different years Shortcuts »
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Click to Start..compound interest Shortcuts and Techniques with Examples
-
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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