Tech Mahindra Simple & Compound Interest Quiz 1

1120 = P [1 + \frac{(5\times \frac{12}{5})}{100}]
P= 1000

70 = \frac{(P\times 4\times \frac{7}{2})}{100}
P = 500

9 = \frac{(1\times 40\times R)}{100}
R = 22\frac{1}{2}%

\frac{(x\times 5\times 1)}{100} + \frac{[(1500 - x)\times 6\times 1]}{100} = 85
\frac{5x}{100} + 90 – \frac{6x}{100}= 85
\frac{x}{100} = 5
=> x = 500

\frac{4}{9} P = \frac{(P\times R\times R)}{100}
R = \frac{20}{3} = 6\frac{2}{3}%

Time = 2 years 4 months = 2(\frac{4}{12}) years = 2(\frac{1}{3}) years.
Amount = Rs [8000 \times (1+(\frac{15}{100}))^{2} \times (1+\frac{((\frac{1}{3})\times 15)}{100})]
=Rs. [8000 \times  (\frac{23}{20}) \times  (\frac{23}{20}) \times (\frac{21}{20})]
= Rs. 11109. .
:. C.I. = Rs. (11109 - 8000) = Rs. 3109.

Let the sum be Rs.P.then
P(1+\frac{R}{100})^{3}=6690…(i) and P(1+\frac{R}{100})^{6}=10035…(ii)
On dividing,we get (1+\frac{R}{100})^{3}=\frac{10025}{6690}=\frac{3}{2}.
Substituting this value in (i),we get:
P\times (\frac{3}{2})=6690

P=(6690\times \frac{2}{3})=4460
Hence,the sum is rs.4460.

Amount = Rs.(30000+4347) = Rs.34347
let the time be n years
Then,30000(1+\frac{7}{100})^{n} = 34347
(\frac{107}{100})^{n} = \frac{34347}{30000} = \frac{11449}{10000} = (\frac{107}{100})^{2}
n = 2 years

Clearly, Rate = 5% p.a., Time = 3 years, S.I.= Rs. 1200. . .
So principal=RS \frac{[100\times 1200]}{3\times 5}=RS 8000
Amount = Rs. 8000 \times  [1 +\frac{5}{100}]^{3} = Rs. 9261.
.. C.I. = Rs. (9261 - 8000) = Rs. 1261.

difference in C.I and S.I in 2 years =Rs.32
S.I for 1year =Rs.400
S.I for Rs.400 for one year =Rs.32
rate=\frac{(100\times 32)}{(400\times 1)}=8%
difference between in C.I and S.I for 3rd year
=S.I on Rs.832= Rs.\frac{(832\times 8\times 1)}{100}=Rs.66.56

difference 32 +66.56 = Rs 98.56