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.50.05
(b) Rs.50.50
(c) Rs.51.50
(d) Rs.51.25
The correct answers to the above question in:
Answer: (d)
Principal = $\text"S.I. × 100"/ \text"Time × Rate"$
= ${50 × 100}/{2 × 5}$ = Rs.500
C.I. = P$[(1 + R/100)^T - 1]$
= 500$[(1 + 5/100)^2 - 1]$
= 500$[(1 + 1/20)^2 - 1]$
= 500$[(21/20)^2 - 1]$
= 500$(441/400 - 1)$
= ${500 × 41}/400$ = Rs.51.25
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Read more combination of si and ci Based Quantitative Aptitude Questions and Answers
Question : 1
The compound interest on a certain sum of money at 5% per annum for 2 years is Rs.246. The simple interest on the same sum for 3 years at 6% per annum is
a) Rs.432
b) Rs.435
c) Rs.430
d) Rs.450
Answer »Answer: (a)
Using Rule 1,
C.I. = P$[(1 + R/100)^T - 1]$
246 = P$[(1 + 5/100)^2 - 1]$
246 = P$[(21/20)^2 - 1]$
246 = P$({441 - 400}/400)$
246 = ${41P}/400$
P = ${246 × 400}/41$ = Rs.2400
SI = $\text"Principal × Time × Rate"/100$
= ${2400 × 3 × 6}/100$ = Rs.432
Question : 2
The compound interest on a certain sum of money for 2 years at 10% per annum is Rs.420. The simple interest on the same sum at the same rate and for the same time will be
a) Rs.400
b) Rs.350
c) Rs.380
d) Rs.375
Answer »Answer: (a)
If the principal be P then
C.I. = P$[(1 + R/100)^T - 1]$
420 = P$[(1 + 10/100)^2 - 1]$
420 = P$({121 - 100}/100)$
420 = ${P × 21}/100$
P = ${420 × 100}/21$ = Rs.2000
S.I. = ${PRT}/100 = {2000 × 10 × 2}/100$ = Rs.400
Using Rule 10,
Here, C.I. = Rs.420, R = 10%, S.I. = ?
C.I. = S.I.$(1 + R/200)$
420 = S.I.$(1 + 10/200)$
420 = S.I.$(210/200)$
S.I. = ${420 × 200}/210$ = Rs.400
Question : 3
The simple interest on a sum of money for 3 years is Rs. 240 and the compound interest on the same sum, at the same rate for 2 years is Rs. 170. The rate of interest is :
a) 5$5/17$%
b) 8%
c) 12$1/2$
d) 29$1/6$%
Answer »Answer: (c)
Since, S.I. for 3 years = Rs.240
S.I. for 2 years = $240/3$× 2 = Rs.160
${PR × 2}/100$ = 160
PR = 160 × 50 = 8000...(i)
Again, C.I. - S.I. = 170 - 160 = Rs.10
${PR^2}/10000$ = 10
${8000 × R}/10000$ = 10
R = $100/8 = 25/2 = 12{1}/2$%
Question : 4
The simple interest on a certain sum of money for 2 years at 5% is Rs.1600. The compound interest at the same rate after 3 years interest compound annually, is
a) Rs.2535
b) Rs.2520
c) Rs.2555
d) Rs.2522
Answer »Answer: (d)
Principal = $\text"S.I. × 100"/\text"Time × Rate"$
= ${1600 × 100}/{5 × 2}$ = Rs.16000
C.I. = P$[(1 + R/100)^T - 1]$
= 16000$[(1 + 5/100)^3 –1]$
= 16000$[(21/20)^3 - 1]$
= $16000(9261/8000 - 1)$
= ${16000 × 1261}/8000$ = Rs.2522
Question : 5
There is 100% increase to an amount in 8 years, at simple interest. Find the compound interest of Rs.8000 after 2 years at the same rate of interest.
a) Rs.2125
b) Rs.2500
c) Rs.2250
d) Rs.2000
Answer »Answer: (a)
Using Rule 1,
Let S.I. = Rs.100, & Principal = Rs.100
Rate = $\text"S.I. × 100"/\text"Principal × Time"$
= ${100 × 100}/{100 × 8} = 25/2$%
C.I. = P$[(1 + r/100)^T - 1]$
= 8000$[(1 + 25/200)^2 - 1]$
= 8000 $(81/64 - 1)$
= ${8000 × 17}/64$ = Rs.2125
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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