M4maths Previous Puzzles

01 Mar 2017 Interest

The difference between simple and compound interests compounded annually on a certain sum of money for 4 years at 6.5% per annum is Rs. 1191. The value of the sum (approximately) is:

OPtion

1) 40000
2) 40500
3) 41000
4) 41500
5) 42000
6) 44500
7) 43000
8) 43500
9) 42500
10)None of these

Solution

Let the sum be Rs. X

CI = [x(1+6.5/100)^4] – x = x*(106.5/100)^4 – x
SI = x*4*6.5/100
CI-SI = x*(106.5/100)^4 – x - x*26/100 = 1.2865x – x – 0.26x = 0.265x
As per question CI-SI = 1191
So 0.265x = 1191 => x = Rs. 45000

Option 10)

Correct Option: 10 Best Solution (5)

G.S.Kantharao   8 years AGO

Let P =100₹,R=6.5%,T=4years.
Compound Intrest = P(1+6.5/100)^4 -100 =28.65₹ (Approx)
Simple Intrest = 100 × 6.5 × 4/100 = 26₹
Difference of Intrest = 2. 65₹
For a difference of 2.65₹ the principle = 100₹
For a difference of 1191₹ principle is (1191*100)/2.65 = 44943₹.
If we consider some more decimals for (106.5/100)^4 we should get 45000₹.
As there are no options regarding these answers I will go with option 10.

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Harsh Poddar   8 years AGO

Amount after simple interest=P+(PRT)/100=1.26×P
Amount after compound interest=P(1+R/100)^T= 1.28646635×P
Difference=0.02646635×P= 1191
P=45000
Option 10

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DIVYANSHU SHEKHAR   8 years AGO

let sum be Rsx
now si =x*4*6.5/100 =0.26x
ci=x(1+r/100)^t-x
=x(1+6.5/100)^4 -x
=0.2864x
ci-si=1191
=>0.2864x-0.26x=1191
=>0.0264x=1191
=>x=45113.63

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Nitish kesarwani   8 years AGO

CI =A-P , SI =P*R*T/100
=let sum be p
=p(1+6.5/100)^4-P =P*6.5*4/100

so CI=p(1065/1000)-1 SI =26P/100

CI-SI=1191
0.28646635P-0.26P=1191

P=1191/0.02646635=45000
optn 10

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Naga   8 years AGO

SI = (P*4*6.5)/100
SI = (P*26)/100
CI = P*(1+6.5/100)^4 -P
CI = P*(1.065)^4-P
CI = P*1.286466-P
CI-SI = 1191
P*1.286466-P-0.26*P = 1191
P*(1.286466-1-0.26) = 1191
P* (0.026466) =1191
P = 45001.133

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