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.1,00,000
(b) Rs.50,000
(c) Nil
(d) Rs.40,000
The correct answers to the above question in:
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
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Read more ci with instalments Based Quantitative Aptitude Questions and Answers
Question : 1
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
Question : 2
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 : 3
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 : 4
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 : 5
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 : 6
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
compound interest Shortcuts and Techniques with Examples
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model 1 basic compound interest using formula
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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
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model 6 comparing sum in different years
Defination & Shortcuts …
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