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.1,50,000
(b) Rs.80,000
(c) Rs.1,20,000
(d) Rs.1,00,000
The correct answers to the above question in:
Answer: (d)
Using Rule 1,
Sum borrowed = Rs.x
Simple interest after 4 years
= ${x × 4 × 5}/100$ = Rs.$x/5$
Amount lent of on compound interest = Rs.$x/2$
C.I. = P$[(1 + R/100)^T - 1]$
= $x/2[(1 + 10/100)^4 - 1]$
= $x/2[(1.1)^4 - 1]$
= $x/2$ (1.4641 - 1) = Rs.${0.4641x}/2$
${0.4641x}/2 - x/5$ = 3205
${2.3205x - 2x}/10$ = 3205
0.3205x = 32050
$x = 32050/{0.3205}$ = Rs.100000
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Read more combination of si and ci Based Quantitative Aptitude Questions and Answers
Question : 1
The simple interest on a sum of money at 4% per annum for 2 years is Rs.80. The compound interest in the same sum for the same period is
a) Rs.81.60
b) Rs.82.60
c) Rs.81.80
d) Rs.82.20
Answer »Answer: (a)
Principal = $\text"S.I. × 100"/ \text"Time × Rate"$
= ${80 × 100}/{2 × 4}$ = Rs.1000
C.I. = P$[(1 + R/100)^T - 1]$
= 1000$[(1 + 4/100)^2 - 1]$
= 1000$[(25/26)^2 - 1]$
= 1000$(676/625 - 1)$
= 1000$({676 - 625}/625)$
= ${1000 × 51}/625$= Rs.81.60
Using Rule 10,
Here, S.I. = Rs.80, R = 4%, C.I. = ?
C.I.= S.I.$(1 + R/200)$
C.I.= 80$(1 + 4/200)$
= $80(1 + 1/50)$
= 80 × $51/50$ = Rs.81.60
Question : 2
There is 40% increase in an amount in 8 years at simple interest. What will be the compound interest (in rupees) of Rs 30000 after 2 years at the same rate ?
a) 3075
b) 6150
c) 4612.5
d) 7687.5
Answer »Answer: (a)
According to the question,
If principal
= Rs.100 then interest = Rs.40.
Case I.
Rate = $\text"S.I. × 100"/ \text"Principal × Time"$
= ${40 × 100}/{100 × 8}$ = 5% per annum
Case II.
A = P$(1 + R/100)^T$
= 30000$(1 + 5/100)^2$
= 30000$(1 + 1/20)^2$
= 30000$({20 + 1}/20)^2$
= 30000$ × 21/20 × 21/20$
= Rs.33075
C. I. = Rs.(33075 - 30000) = Rs.3075
Question : 3
On a certain sum of money the compound interest for 2 years is Rs.282.15 and the simple interest for the same period of time is Rs.270. The rate of interest per annum is
a) 12.15%
b) 6.07%
c) 9%
d) 10%
Answer »Answer: (c)
Using Rule 10,
If SI on a certain sum for two years is x and CI is y, then
$y = x(r + /200)$
$282.15 = 270(1 + r/100)$
$1 + r/200 = 282.15/270$
$r/200 = 282.15/270$ - 1
$r/200 = {12.15}/270$
r = ${12.15 × 200}/270 = 9%$
Question : 4
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 : 5
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 : 6
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
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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