model 5 ci with instalments 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.21,20,000
(b) Rs.28,55,000
(c) Rs.28,20,000
(d) Rs.18,50,000
The correct answers to the above question in:
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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Read more ci with instalments Based Quantitative Aptitude Questions and Answers
Question : 1
A loan of Rs.12,300 at 5% per annum compound interest, is to be repaid in two equal annual instalments at the end of every year. Find the amount of each instalment.
a) Rs.6,156
b) Rs.6,651
c) Rs.6,516
d) Rs.6,615
Answer »Answer: (d)
Using Rule 9(i),
Let each instalment be x.
$x/(1 + 5/100) + x/(1 + 5/100)^2 = 12300$
${20x}/21 + (20/21)^2x = 12300$
${20x}/21(1 + 20/21)$ = 12300
${20x}/21 × 41/21 × x = 12300$
$x = {12300 × 21 × 21}/{20 × 41}$ ⇒ x = 6615
Question : 2
A man buys a scooter on making a cash down payment of Rs.16224 and promises to pay two more yearly instalments of equivalent amount in next two years. If the rate of interest is 4% per annum, compounded yearly, the cash value of the scooter, is
a) Rs.50000
b) Rs.40000
c) Rs.46000
d) Rs.46824
Answer »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$[(1 + r/100)^n - 1]$
Let principal (present worth) for first year be $P_1$ and that for two years be $P_2$.
16224 = $P_1(1 + 4/100)$
16224 = $P_1(1 + 1/25) = {26P_1}/25$
$P_1 = {16224 × 25}/26$ = Rs.15600
Again,
$16224 = P_2(1 + 4/100)^2$
16224 = $P_2(26/25)^2 = {676P_2}/625$
$P_2 = {16224 × 625}/676$ = 15000
Cash value of the scooter
= Rs.(16224 + 15600 + 15000) = Rs.46824
Question : 3
A sum of money is paid back in two annual instalments of Rs. 17, 640 each, allowing 5% compound interest compounded annually. The sum borrowed was
a) Rs.32,400
b) Rs.32,800
c) Rs.32,000
d) Rs.32,200
Answer »Answer: (b)
Using Rule 9(i),
Sum borrowed = Present worth of Rs.17640 due 1 year hence + Present worth of Rs.17640 due 2 years hence
= Rs.$(17640/{(1 + 5/100)} + 17640/{(1 + 5/100)^2})$
= Rs.$(17640 × 20/21 + 17640 × 20/21 × 20/21)$
= Rs.(16800 + 16000) = Rs.32800
Question : 4
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
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 : 6
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
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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