Question

In what time will 64,000 amount to 1,12,000
at 25% SI per annum?​

Answer 1

${\large{\boxed{\underline{\mathrm{\bf{\orange{Given:-}}}}}}}$

• Principal = 64,000
• Amount = 1,12,000
• Rate of Interest = 25%

${\large{\boxed{\underline{\mathrm{\bf{\red{To \ Find:-}}}}}}}$

• Time

${\large{\boxed{\underline{\mathrm{\bf{\pink{Formula \ Used:-}}}}}}}$

${\pink{\bigstar}} \ {\boxed{\tt{\blue{I = A-P}}}} \ {\pink{\bigstar}}$

where,

• I = Interest
• A = Amount i.e. 1,12,000
• P = Principal i.e. 64,000

${\pink{\bigstar}} \ {\boxed{\tt{\green{T = \dfrac{S.I. \times 100}{P \times R}}}}} \ {\pink{\bigstar}}$

where,

• T = Time
• S.I. = Simple Interest
• P = Principal i.e. 64,000
• R = Rate of Interest i.e. 15%

${\large{\boxed{\underline{\mathrm{\bf{\blue{Solution:-}}}}}}}$

According to the question by using the formula of Interest, we get,

$\longmapsto {\sf{I = 1,12,000 - 64,000}}$

$\longmapsto {\sf{I = 48,000}}$

Simple Interest = 48,000

According to the question by using the formula of Time, we get,

$: \ \longmapsto \ {\sf{Time = \dfrac{48,000 \times 100}{64,000 \times 25} \ years }}$

$: \ \longmapsto \ {\sf{Time = \dfrac{48,{\cancel{000}} \times {\cancel{100}}}{64,{\cancel{000}} \times {\cancel{25}}} ^{ \ 4} \ years }}$

$: \ \longmapsto \ {\sf{Time = \dfrac{48 \times 4}{64} \ years }}$

$: \ \longmapsto \ {\sf{Time = \dfrac{{\cancel{48}} ^{ \ 3} \times 4}{{\cancel{64}^{ \ 4}}} \ years }}$

$: \ \longmapsto \ {\sf{Time = \dfrac{3 \times 4}{4} \ years }}$

$: \ \longmapsto \ {\sf{Time = \dfrac{12}{4} \ years }}$

$: \ \longmapsto \ {\sf{Time = \dfrac{{\cancel{12}}}{{\cancel{4}}} ^{\ 3} \ years }}$

$: \ \Longrightarrow \ {\orange{\sf{Time = 3 \ years }}}$

${\orange{\bigstar}} \ \therefore {\boxed{\underline{\mathsf{\green{Time}{\purple{ \ is \ }{\blue{\bf{3 \ years}}}}}}}} \ {\orange{\bigstar}}$

${\huge{\green{\checkmark}}} \ {\large{\boxed{\underline{\mathrm{\bf{\orange{Verification:-}}}}}}} \ {\huge{\green{\checkmark}}}$

$: \ \longmapsto \ {\sf{S.I. = \dfrac{P \times R \times T}{100} }}$

$: \ \longmapsto \ {\sf{48,000 = \dfrac{64,{\cancel{000}} \times 25 \times 3}{{\cancel{100}}} }}$

$: \ \longmapsto \ {\sf{48,000 = 640 \times 25 \times 3 }}$

$: \ \longmapsto \ {\sf{48,000 = 640 \times 75 }}$

$: \ \longmapsto \ {\sf{48,000 = 48,000 }}$

$\Longrightarrow {\bf{\pink{LHS = RHS}}} \Longleftarrow$

${\mathsf{Hence, Verified}} \ {\large{\green{\checkmark}}}$

Answer 2

$\sf\large\underline\green{Given:-}$

$\sf{\implies Sum\:_{(principal)}=64000}$

$\sf{\implies Sum\:_{(amount)}=112000}$

$\sf{\implies Rate\:_{(interest)}=25\%}$

$\sf\large\underline\purple{To\: Find:-}$

$\sf{\implies Time\:_{(annually)}=?}$

$\sf\large\underline\purple{Solution:-}$

Given

• Amount = 1, 12, 000

• Principal = 64, 000

So,

Interest is evaluated by using the Formula,

$\bigstar \: \: { \boxed{ \green{ \tt \: Interest \: = Amount - Principal}}}$

$\rm :\implies\: \: Interest = 112000 - 64000$

$\rm :\implies\:Interest = 48000$

Now,

We have

• Principal = 64000

• Interest = 48000

• Rate of Interest = 25 % per annum

So, using formula of Simple interest,

$\bigstar \: { \boxed{ \pink{ \tt \: Interest = \dfrac{Principal \times Rate \times Time}{100} }}}$

$\rm :\implies\:48000 = \dfrac{64000 \times 25 \times Time}{100}$

$\rm :\implies\:Time = \dfrac{48000 \times 100}{64000 \times 25}$

$\large\boxed{ \green{\rm :\implies\:Time \: = \: 3 \: years}}$