model 3 combination of si & ci 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.2,250
(b) Rs.2,500
(c) Rs.2,360
(d) Rs.2,400
The correct answers to the above question in:
Answer: (d)
C.I.= P$(1 + r/100)^t$ - P
2448 = P$[(1 + r/100)^t - 1]$
or 2448 = P$[(1 + 4/100)^2 - 1]$
2448 = P$[676/625 - 1]$
2448 = P$[51/625]$
P = ${2448 × 625}/51$ = Rs.30,000
S.I. = ${30000 × 4 × 2}/100$ = Rs.2400
Using Rule 10,
Here, C.I. = Rs.2448, R = 4%, S.I. = ?
C.I.= S.I.$(1 + R/200)$
2448 = S.I.$(1 + 4/200)$
2448 = S.I.$(1 + 1/50)$
2448 = S.I.$(51/50)$
S.I. = ${2448 × 50}/51$ = Rs.2400
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Read more combination of si and ci Based Quantitative Aptitude Questions and Answers
Question : 1
A man borrowed some money from a private organisation at 5% simple interest per annum. He lended this money to another person at 10% compound interest per annum, and made a profit of Rs.26,410 in 4 years. The man borrowed
a) Rs.100000
b) Rs.200000
c) Rs.132050
d) Rs.150000
Answer »Answer: (a)
Let the principal be Rs.P. For 4 years,
S.I. = $\text"Principal × Time × Rate"/100$
= ${P × 4 × 5}/100$ = Rs.$P/5$
C.I. = P$[(1 + R/100)^T - 1]$
= P$[(1 + 10/100)^4 - 1]$
= P$[(11/10)^4 - 1]$
= P$(14641/10000 - 1) = {4641P}/10000$
According to the question,
${4641P}/10000 - P/5$ = 26410
${4641P - 2000P}/10000$ = 2641
${2641P}/10000$ = 2641
P = Rs.10000
Question : 2
The compound interest on a certain sum of money for 2 years at 5% per annum is Rs.410. The simple interest on the same sum at the same rate and for the same time is
a) Rs.405
b) Rs.400
c) Rs.350
d) Rs.300
Answer »Answer: (b)
Compound interest = P $[(1 + R/100)^T - 1]$
410 = P$[(1 + 5/100)^2 - 1]$
410 = P$[(1 + 1/20)^2 - 1]$
410 = P$[(21/20)^2 - 1]$
410 = P$(441/400 - 1)$
410 = P$(41/400)$
P = ${410 × 400}/41$ = Rs.4000
S.I. = $\text"Principal × Time × Rate"/100$
= ${4000 × 2 × 5}/100$ = Rs.400
Using Rule 10,
Here, C.I. = Rs.410, R = 5%, S.I. = ?
C.I. = S.I.$(1 + R/200)$
410 = S.I.$(1 + 5/200)$
410 = S.I.$(205/200)$
S.I. = ${410 × 200}/205$ = Rs.400
Question : 3
If the compound interest on a certain sum for two years at 12% per annum is Rs.2,544, the simple interest on it at the same rate for 2 years will be
a) Rs.2,440
b) Rs.2,400
c) Rs.2,480
d) Rs.2,500
Answer »Answer: (b)
C.I. = P$[(1 + R/100)^T - 1]$
2544 = P$[(1 + 12/100)^2 - 1]$
2544 = P$[(28/25)^2 - 1]$
2544 = P$(784/625 - 1)$
2544 = P$({784 - 625}/625)$
2544 = ${P × 159}/625$
P = ${2544 × 625}/159$ = Rs.10000
S.I. = ${P × R × T}/100$
= ${10000 × 2 × 12}/100$ = Rs.2400
Using Rule 10,
Here, C.I. = Rs.2544, R = 12%, S.I. = ?
C.I. = S.I.$(1 + R/200)$
2544 = S.I.$(1 + 12/200)$
2544 = S.I.$(212/200)$
S.I. = ${2544 × 200}/212$ = Rs.2400
Question : 4
The simple interest and compound interest (compounded annually) on a certain sum of money with a given rate for a period of 2 years are Rs.900 and Rs.954 respectively. The sum of money is
a) Rs.3750
b) Rs.3700
c) Rs.3850
d) Rs.3650
Answer »Answer: (a)
Difference of CI and SI for two years
= Rs.(954 - 900) = Rs.54
Sum= Difference in CI and SI × $(100/{Rate})^2$
Rate = ${2 × \text"Difference" × 100}/\text"Simple interest"$
= ${2 × 5400}/900 = 12%$
Sum = 54 × $(100/12)^2$
= $54 × 25/3 × 25/3$ = Rs.3750
Using Rule 10,
C.I. = Rs.954, S.I.=Rs.900, P=?
C.I.= S.I.$(1 + R/200)$
954 = 900$(1 + R/200)$
$954/900 = 1 + R/200$
$954/900 - 1 = R/200$
${954 - 900}/900 = R/200$
$54/9 = R/2$
R = 12%
Now S.I. = ${P × R × T}/100$
900 = ${P × 12 × 2}/100$
P = Rs.3750
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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