Question
On what sum will the difference between compound interest and the simple interest for $3$ years at $12\%$ be $Rs.1,123.20?$
✓
Answer
$ P=x ; t=3 \text { years; } r=12 \% \text { For S.I.: }$
$I=\frac{P \times r \times t}{100}$
$=\frac{x \times 12 \times 3}{100}$
$=\frac{9 x}{25} $
For C.I. :
$ \text { C.I. }= P \left(1+\frac{ r }{100}\right)^{ t }- P$
$= x \left(1+\frac{12}{100}\right)^3- x$
$= x \left(1+\frac{3}{25}\right)^3- x$
$=( x \times 1.12 \times 1.12 \times 1.12)- x$
$=1.404928 x - x$
$=0.404928 x $
Given C.1.- S. I. $= Rs.22.50$
$ \Rightarrow 0.404928 x-\frac{9 x}{25}=\operatorname{Rs} 1123.20$
$\Rightarrow 0.404928 x-0.36 x=\text { Rs } 1123.20$
$\Rightarrow 0.044928 x=\text { Rs } 1123.20$
$\Rightarrow x=\text { Rs } 25000 $
Hence, sum $= Rs.25,000$
$I=\frac{P \times r \times t}{100}$
$=\frac{x \times 12 \times 3}{100}$
$=\frac{9 x}{25} $
For C.I. :
$ \text { C.I. }= P \left(1+\frac{ r }{100}\right)^{ t }- P$
$= x \left(1+\frac{12}{100}\right)^3- x$
$= x \left(1+\frac{3}{25}\right)^3- x$
$=( x \times 1.12 \times 1.12 \times 1.12)- x$
$=1.404928 x - x$
$=0.404928 x $
Given C.1.- S. I. $= Rs.22.50$
$ \Rightarrow 0.404928 x-\frac{9 x}{25}=\operatorname{Rs} 1123.20$
$\Rightarrow 0.404928 x-0.36 x=\text { Rs } 1123.20$
$\Rightarrow 0.044928 x=\text { Rs } 1123.20$
$\Rightarrow x=\text { Rs } 25000 $
Hence, sum $= Rs.25,000$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
If Δ ABC , MN || BC .
If AN : AC= 5 : 8, find ar(Δ AMN) : ar(Δ ABC)
If AN : AC= 5 : 8, find ar(Δ AMN) : ar(Δ ABC)
$
\text { Find } x \text {, if } \cos (2 x-6)=\cos ^2 30^{\circ}-\cos ^2 60^{\circ}
$
→\text { Find } x \text {, if } \cos (2 x-6)=\cos ^2 30^{\circ}-\cos ^2 60^{\circ}
$
Solve the following equation: $\frac{2 x }{ x -4}+\frac{2 x -5}{ x -3}=\frac{25}{3}$
→Given $A=\left[\begin{array}{cc}2 & 0 \\ -1 & 7\end{array}\right]$ and $1=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$ and $A^2=9 A+m l$. Find $m$
→Draw a circle of radius 3.5 cm. mark a point P outside the circle at a distance of 6 cm from the centre. Construct two tangents from P to the given circle. Measure and write down the length of one tangent.
Given: ∠CAB = 75° and ∠CBA = 50°. Find the value of ∠DAB + ∠ABD.
The line $4x − 3y + 12 = 0$ meets $x-$ axis at $A.$ Write the co-ordinates of $A.$ determine the equation of the line through $A$ and perpendicular to $4x – 3y + 12 = 0$
→In the given figure, PAT is tangent to the circle with centre O at point A on its circumference and is parallel to chord BC. If CDQ is a line segment , show that:
(i) ∠BAP = ∠ADQ
(ii) ∠AOB = 2∠ADQ
(iii) ∠ADQ = ∠ADB
(i) ∠BAP = ∠ADQ
(ii) ∠AOB = 2∠ADQ
(iii) ∠ADQ = ∠ADB
Solve the equation $x^4 + 2x^3- 13x^2 + 2x + 1 = 0.$
→Find the Median of the following distribution:
→| $x$ | $3$ | $5$ | $10$ | $12$ | $8$ | $15$ |
| $f$ | $2$ | $4$ | $6$ | $10$ | $8$ | $7$ |