Question
The difference between simple interest and compound interest on a certain sum is $Rs. 54.40$ for $2$ years at $8$ per cent per annum. Find the sum.
✓
Answer
Let principal $(P)=x$
$ \mathrm{R}=8 \%$
$ \mathrm{~T}=2$ years
$ \mathrm{SI}=\frac{\mathrm{X} \times 8 \times 2}{100}=\frac{4 X}{25}$
$ \mathrm{Cl}=\mathrm{A}-\mathrm{P}=\mathrm{X}\left(1+\frac{8}{100}\right)^2-\mathrm{X}$
$ =\mathrm{X}\left|\left[1+\frac{2}{25}\right]^2-1\right|$
$ =\mathrm{X}\left|\left[\frac{27}{25}\right]^2-1\right|$
$ =\frac{104}{625} X$
Given,
$\mathrm{Cl}=\mathrm{SI}=54.40$
$ \frac{104 X}{625}-\frac{4 X}{25}=\text { Rs. } 54.40$
$ X\left(\frac{104}{625}-\frac{4}{25} \times \frac{25}{25}\right)=54.40$
$ X\left(\frac{4}{625}\right)=54.40$
$ X=\frac{54.40 \times 625}{4}$
$X=\text { Rs. } 8500$
Thus, principal sum $= Rs. 8500$
$ \mathrm{R}=8 \%$
$ \mathrm{~T}=2$ years
$ \mathrm{SI}=\frac{\mathrm{X} \times 8 \times 2}{100}=\frac{4 X}{25}$
$ \mathrm{Cl}=\mathrm{A}-\mathrm{P}=\mathrm{X}\left(1+\frac{8}{100}\right)^2-\mathrm{X}$
$ =\mathrm{X}\left|\left[1+\frac{2}{25}\right]^2-1\right|$
$ =\mathrm{X}\left|\left[\frac{27}{25}\right]^2-1\right|$
$ =\frac{104}{625} X$
Given,
$\mathrm{Cl}=\mathrm{SI}=54.40$
$ \frac{104 X}{625}-\frac{4 X}{25}=\text { Rs. } 54.40$
$ X\left(\frac{104}{625}-\frac{4}{25} \times \frac{25}{25}\right)=54.40$
$ X\left(\frac{4}{625}\right)=54.40$
$ X=\frac{54.40 \times 625}{4}$
$X=\text { Rs. } 8500$
Thus, principal sum $= Rs. 8500$
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
Construct a triangle using the given data: $PQ = 4.8\ cm, QR = 6.3\ cm$ and $PR = 5.5\ cm$
→The radii of two circles are in the ratio $5 : 8$. If the difference between their areas is $156p\ cm^2$, find the area of the bigger circle.
→If $2160 = 2^a. 3^b. 5^c$, find $a, b$ and $c.$ Hence calculate the value of $3^a \times 2^{-b} \times 5^{-c}.$
→In the given figure, $\angle Q=90^{\circ}, P S$ is a median om $Q R$ from $P$, and $R T$ divides $P Q$ in the ratio $1: 2$. Find: $\frac{\tan \angle TSQ }{\tan \angle PRQ }$
→Solve the following systems of equations by using the method of cross multiplication:
8x - 3y = 12, 5x = 2y +7
8x - 3y = 12, 5x = 2y +7
"The points $A (3, 0), B (a, -2)$ and $C (4, -1)$ are the vertices of $\triangle ABC$ right angled at vertex $A$. Find the value of $a.$
→Draw a graph of each of the following equations:$ 3y + 2x = 11$
→Use the following figure to find$:(i) BC$, if $AB =7.2 \ cm.(ii) G E$, if $F E=4 \ cm.(iii) A E$, if $B D=4.1 \ cm(iv) D F$, if $C G=11 \ cm$.
→A room is 13 m long and 9 m wide. Find the cost of carpeting the room with a carpet 75 cm wide at ₹ 12.50 per metre.