model 5 ci with instalments 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 2000]
A builder borrows Rs.2550 to be paid back with compound interest at the rate of 4% per annum by the end of 2 years in two equal yearly instalments. How much will each instalment be ?
a) Rs.1283
b) Rs.1352
c) Rs.1275
d) Rs.1377
Answer »Answer: (b)
A = Rs.2550
R = 4% per annum
n = 2 years
Let each of the two equal instalments be x
Present worth = $\text"Instalment"/(1 + r/100)^n$
$P_1 = x/(1 + 4/100)^1 = x/{1 + 1/25} = x/{26/25}$
or $P_1 = 25/26x$
Similarly,
$P_2 = (25/26)^2x = 625/676x$
$P_1 + P_2$ = A
$25/26x + 625/676x$ = 2550
${(650 + 625)x}/676 = 2550$
$1275/676x = 2550$
x = 2550 $× 676/1275$ ⇒ x = Rs.1352
Using Rule 9,If a sum 'P' is borrowed at r% annual compound interest which is to be paid in 'n' equal annual installments including interest, then(i) For n = 2, Each annual installment= $p/{(100/{100 + r}) + (100/{100 + r})^2$(ii) For n = 3, Each annual installment= $p/{(100/{100 + r}) + (100/{100 + r})^2 + (100/{100 + r})^3$
Here, P = Rs.2550, n = 2, r = 4%
Each instalment
= $p/{(100/{100 + r}) + (100/{100 + r})^2$
= $2550/{(100/{100 + 4}) + (100/{100 + 4})^2$
= $2550/{100/104 + (100/104)^2}$
= $2550/{100/104(1 + 100/104)}$
= $2550/{100/104 (204/104)}$
= ${2550 × 104 × 104}/20400 $= Rs.1352
Question : 2 [SSC CGL Prelim 2008]
Kamal took Rs.6800 as a loan which along with interest is to be repaid in two equal annual instalments. If the rate of interest is 12$1/2$%, compounded annually, then the value of each instalment is
a) Rs.4000
b) Rs.8100
c) Rs.4050
d) Rs.4150
Answer »Answer: (c)
Let the annual instalment be x
A = P$(1 + R/T)^T$
$x = P_1(1 + 25/200)$
$x = P_1 × 9/8$
$P_1 = 8/9x$
Similarly, $P_2 = 64/81x$
$P_1 + P_2$ = 6800
$8/9x + 64/81x$ = 6800
${72x + 64x}/81 = 6800$
${136x}/81 = 6800$
$x = {6800 × 81}/136$ = Rs.4050
Using Rule 9(i),
Here, P = Rs.6800, R = $25/2$%, n = 2
Each instalment
= $p/{(100/{100 + r}) + (100/{100 + r})^2$
= $6800/{(100/{100 +{25/2}}) + (100/{100 + {25/2}})^2$
= $6800/{200/225 + (200/225)^2}$
= $6800/{200/225(1 + {200/225})}$
= ${6800 × 225 × 225}/{200 × 425}$ = Rs.4050
Question : 3 [SSC CHSL 2014]
A sum of Rs.210 was taken as a loan. This is to be paid back in two equal instalments. If the rate of interest be 10% compounded annually, then the value of each instalment is
a) Rs.225
b) Rs.127
c) Rs.210
d) Rs.121
Answer »Answer: (d)
Using Rule 9(i),
Let the value of each instalment be Rs.x
Principal = Present worth of Rs.x due 1 year hence, present worth of Rs. x due 2 years hence
210 = $x/(1 + R/100) + x/(1 + R/100)^2$
210 = $x/(1 + 10/100) + x/(1 + 10/100)^2$
210 = $x/{1 + 1/10} + x/(1 + 1/10)^2$
210 = $x/{11/10} + x/(11/10)^2$
210 = ${10x}/11 + {100x}/121$
210 = ${110x + 100x}/121$
210 × 121 = 210 x
$x = {210 × 121}/210$ = Rs.121
Question : 4 [SSC CHSL 2015]
Mr. Dutta desired to deposit his retirement benefit of Rs. 3 lakhs partly to a post office and partly to a bank at 10% and 6% interests respectively. If his monthly interest income was Rs. 2000, then the difference of his deposits in the post office and in the bank was :
a) Rs.1,00,000
b) Rs.50,000
c) Nil
d) Rs.40,000
Answer »Answer: (c)
Using Rule 1,
Let the amount deposited in Post Office be Rs.x lakhs.
Amount deposited in bank = Rs.(3 - x) lakhs
According to the question,
${x × 10 × 1}/{100 × 12} + {(3 - x) × 6 × 1}/{100 × 12}$
= $2000/100000 = 1/50$
10x + 18 - 6x
= $1/50$ × 1200 = 24
4x = 24 - 18 = 6
x = $6/4$ = Rs.$3/2$ lakhs
∴ Required difference = 0
Question : 5 [SSC CHSL 2015]
The income of a company increases 20% per year. If the income is Rs. 26,64,000 in the year 2012, then its income in the year 2010 was :
a) Rs.21,20,000
b) Rs.28,55,000
c) Rs.28,20,000
d) Rs.18,50,000
Answer »Answer: (d)
Using Rule 1,
Let the income of company in 2010 be Rs.P
According to the question,
A = P$(1 + R/100)^T$
2664000 = P$(1 + 20/100)^2$
2664000 = P$(1 + 1/5)^2$
2664000 = P × $(6/5)^2$
P = ${2664000 × 5 × 5}/{6 × 6}$ = Rs.1850000
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model 5 ci with instalments Shortcuts »
Click to Read...model 5 ci with instalments Online Quiz
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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