Let the sum be Rs. x
Then, ((x*10*7)/(100*2))-( (x*12*5)/(100*2)) = 40
7x/20)-(3x/10)=40
x = (40 * 20) = 800
Answer : Option C
Explanation :
Simple Interest for 4 years (SI) = (22400 - 14000) = Rs.8400
R = ?
T = 12 years
P = Rs. 14000
$MF#%\text{R} = \dfrac{100 \times \text{SI}}{\text{PT}} = \dfrac{100 \times 8400}{14000 \times 12} = 5\% $MF#%
The difference between the compound interest and the simple interest and the simple interest earned on a sum of money at the end of 4 years is Rs. 256.40. To find out the sum , which of the following informations given in the statements P and Q is/ are necessary ?
P : Amount of simple interest accrued after 4 years
Q : Rate of interest per annum
To find the sum , difference between C.I. and S.I. , the time and the rate of interest are needed.
`:.` Only Q is necessary
`:.` Correct answer is (B) .
principal = (P.W of Rs. 882 due 1 years hence ) + (P.W. of Rs. 882 due 2 years hence )
=`[(882)/((1 + 5/100)) + (882)/(1 + 5/100)^2]`
=` ((882 xx 20)/(21) + (882 xx 400)/(441))` = Rs. 1640.
Balance = Rs.` [{12500 xx (1 + 20/100)^3} - {2000 xx (1 + 20/100)^2 + 2000 xx (1 + 20/100) + 2000}]`
=` [(12500 xx 6/5 xx 6/5 xx 6/5) - ( 2000 xx 6/5 xx 6/5 + 2000 xx 6/5 + 2000)]`
= Rs. [21600 - (2880 + 2400 + 2000)] = Rs. 14320
Let each instalment be Rs. `x` , Then ,
`(x)/((1 + 5/100)) + (x)/((1 + 5/100))^2 = 1025`
`hArr (20x)/(21) + (400x)/(441) = 1025`
`hArr 820x = 1025 xx 441`
`hArr x = ((1025 xx 441)/(820)` = 551.25
So, value of each instalment = Rs. 551.25.
Let the value of each instalment be Rs. `x` , Then
(P.W. of Rs. `x` due 1 year hence ) + (P.W. of Rs. `x` due 2 years hence) = Rs. 2550.
`hArr (x)/((1 + 4/100)) + (x)/((1 + 4/100)^2) = 2550`
`hArr (25x)/(26) + (625x)/(676) = 2550`
`hArr 1275x = 2550 xx 676`
`hArr x = ((2550 xx 676)/(1275))`
`hArr x = 1352`
`:.` Value of each instalment = Rs. 1352.
`P(1 + 20/100)^n > 2P`
or `(6/5)^n > 2`
Now , `(6/5 xx 6/5 xx 6/5 xx 6/5) > 2`
so, n = 4 years.
`P(1 + R/100)^4 = 3P `
`rArr (1 + R/100)^4 = 3 ` .................................................(i)
Let `P(1 + R/100)^n = 27P`
`rArr (1 + R/100)^n = 27 = 3^3 = {(1 + R/100)^4 }^3` [using(i)]
`rArr (1 + R/100)^n = (1 + R/100)^12`
`rArr n= 12.`
`:.` Required time = 12 years.
