model 6 comparing sum in different years 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) 8 = (1.01)6y
(b) 64 = $(1.01)^{12y}$
(c) 64 = (1.04)12y
(d) $1/64$ = (1.04)12y
The correct answers to the above question in:
Answer: (b)
Rate of interest = 12% p.a.
= 1% per month
Time = 12y months
A = P$(1 + R/100)^T$
64 = 1$(1 + 1/100)^{12y}$
64 = $1(1.01)^{12y}$
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Read more comparing sum in different years Based Quantitative Aptitude Questions and Answers
Question : 1
A sum of money amounts to Rs.4,840 in 2 years and to Rs.5,324 in 3 years at compound interest compounded annually. The rate of interest per annum is :
a) 8%
b) 10%
c) 11%
d) 9%
Answer »Answer: (b)
Let the rate of interest be r% per annum,
According to the question,
4840 = P$(1 + r/100)^2$ ..... (i)
and 5324 = P$(1 + r/100)^3$....(ii)
On dividing equation (ii) by equation (i), we have,
$1 + r/100 = 5324/4840 = 1 + 484/4840$
$r/100 = 484/4840$ ⇒ r = 10%
Using Rule 7,If on compound interest, a sum becomes Rs.A in 'a' years and Rs.B in 'b' years then,(i) If b - a = 1, then, R% = $(B/A - 1)$ × 100%(ii) If b - a = 2, then, R% = $(√{B/A} - 1)$ × 100%(iii) If b - a = n then, R% = $[(B/A)^{1/n} - 1]$ × 100%where n is a whole number.
Here, b - a = 3 - 2 = 1, B = Rs.5,324, A = Rs.4,840
R% = $(B/A - 1)$ × 100%
= $(5324/4840 -1)$ × 100%
= $({5324 - 4840}/4840) × 100%$
= $484/4840 × $100% = 10%
Question : 2
The compound interest on a certain sum for two successive years are Rs.225 and Rs.238.50. The rate of interest per annum is :
a) 6%
b) 7$1/2$%
c) 10%
d) 5%
Answer »Answer: (a)
Difference = 238.50 - 225 = Rs.13.50
= S.I. on Rs.225 for 1 year
Rate = $\text"S.I. × 100"/\text"Principal × Time"$
= ${13.50 × 100}/{225 × 1}$ = 6% per annum
Using Rule 7(i),
Here, b - a = 1
B = Rs.238.50, A = Rs.225
R% = $(B/A - 1)$ × 100%
= $({238.50}/225 - 1) × 100%$
= $({238.50 - 225}/225) × 100%$
= $({13.5}/225) × 100%$ = 6%
Question : 3
The compound interest on a certain sum for 2 years at 10% per annum is Rs. 525 . The simple interest on the same sum for double the time at half the rate per cent per annum is :
a) Rs.515
b) Rs.520
c) Rs.500
d) Rs.550
Answer »Answer: (c)
C.I. = P$[(1 + R/100)^T - 1]$
525 = P$[(1 + 10/100)^2 - 1]$
525 = P$(121/100 - 1)$
525 = ${P × 21}/100$
P = ${525 × 100}/21$ = Rs.2500
Again, new rate = 5% per annum
S.I. = $\text"Principal × Time × Rate"/100$
= ${2500 × 5 × 4}/100$ = Rs.500
Question : 4
An amount of money at compound interest grows up to Rs.3,840 in 4 years and up to Rs.3,936 in 5 years. Find the rate of interest.
a) 2.05%
b) 2.5%
c) 3.5%
d) 2%
Answer »Answer: (b)
A = P$(1 + R/100)^T$
3840 = P$(1 + R/100)^4$ ....(i)
3936 = P$(1 + R/100)^5$ ...(ii)
Dividing equation (ii) by equation (i),
$3936/3840 = 1 + R/100$
$R/100 = 3936/3840$ - 1
= ${3936 - 3840}/3840 = 96/3840$
R = $96/3840 × 100$ = 2.5%
Using Rule 7(i),
Here, b - a = 5 - 4 = 1
B = Rs.3,936, A = Rs.3,840
R%= $(B/A - 1)$ × 100%
= $(3936/3840 - 1)$ × 100%
= $({3936 - 3840}/3840) × 100%$
= $96/3840 × 100% = 10/4%$ = 2.5%
Question : 5
A sum of money is invested at 20% compound interest (compounded annually). It would fetch Rs. 723 more in 2 years if interest is compounded half yearly. The sum is
a) Rs.7,500
b) Rs.15,000
c) Rs.20,000
d) Rs.30,000
Answer »Answer: (d)
Let the principal be Rs. x.
When the interest is compounded annually,
C.I. = P$[(1 + R/100)^T - 1]$
= P$[(1 + 20/100)^2 - 1]$
= P$[(6/5)^2 - 1]$
= P$(36/25 - 1)$ = Rs.${11P}/25$
When the interest is compounded half–yearly,
C.I. = P$[(1 + 10/100)^4 - 1]$
= P$[(11/10)^4 - 1]$
= P$(14641/10000 - 1)$
= Rs.${4641P}/10000$
${4641P}/10000 - {11P}/25$ = 723
${4641P - 4400P}/10000$ = 723
${241P}/10000$ = 723
P = ${723 × 10000}/241$ = Rs.30000
Question : 6
On a certain sum of money, the simple interest for 2 years is Rs.350 at the rate of 4% per annum. It was invested at compound interest at the same rate for the same duration as before, how much more interest would be earned?
a) Rs.35
b) Rs.3.50
c) Rs.14
d) Rs.7
Answer »Answer: (d)
Principal = $\text"S.I. × 100"/ \text"Time × Rate"$
= ${350 × 100}/{2 × 4}$ = Rs.4375
C.I. = P$[(1 + R/100)^T - 1]$
= 4375$[(1 + 4/100)^2 - 1]$
= 4375$[(1 + 1/25)^T - 1]$
= 4375$[(26/25)^2 - 1]$
= 4375$(676/625 - 1)$
= ${4375 × 51}/625$ = Rs.357
Required difference
= Rs.(357 - 350) = Rs.7
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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