Hexaware Simple and Compound Interest Quiz 1

Given,
Amount = Rs 49, 0000
Rate of Interest = 7% pa
Time = 5 years

We know that,
A = P + SI
A = P + \frac{P*R*T}{100}[\latex]
=P(1+\frac{R*T}{100})[\latex]
Substituting the values,
49,000 = P(1+\frac{5*7}{100})[\latex]

=P(1+\frac{35}{100})[\latex]
=P\left ( \frac{135}{100}\right )[\latex]
= \frac{49000*100}{135}[\latex]

P = Rs 36296.3

From the above, given:-
P= Rs 20,0000
Time= 5 years
Amount= 30,000
R1 = ?
R2 = 7%p.a

Therefore, SI = A - P
= Rs 30,000 - Rs 20,000
= Rs 10,000

We know that,
SI = \frac{P*R*T}{100}[\latex]
R = \frac{\left ( SI * 100 \right )}{P * T}[\latex]
= \frac{\left ( 10000 * 100 \right )}{20000 * 5}[\latex]
= 10 % p.a

Now from the question,
New Principal, = Rs 20,000+ Rs 5,000
= Rs 25,000

Therefore, the new SI

= \frac{\left ( P * R * T \right )}{100}[\latex]
= \frac{\left ( 25000* 10 *5 \right )}{100}[\latex]
=Rs 12,500

New Amount = Rs (25000+ 12,500)
= Rs 37,500

From the question,
P1= Rs 32,000
P2= Rs 34,000
Combined SI = Rs 10,560
Time = 2 years

Here,
SI_{1} + SI_{2} = Combined SI

\frac{\left ( P1 * R * T \right )}{100}[\latex] + \frac{\left ( P2 * R * T \right )}{100}[\latex] = Rs 10, 560

\frac{\left ( 32000 * R * 2 \right )}{100}[\latex] + \frac{\left ( 34000 * R * 2 \right )}{100}[\latex] = 10,560

640R + 680R = 10560
1320R = 10560
R = 8 % pa

From the question
Principal = Rs 50000
Rate = 8 % pa
Interest = 16,000

SI is given by,
SI = \frac{\left ( P * R * T \right )}{100}[\latex]
T = \frac{\left ( SI * 100 \right )}{P * R}[\latex]
= \frac{\left ( 16000 * 100 \right )}{50000 * 8}[\latex]
= 4

According to the question
Amount = 2P
SI = A-P
= 2P - P
= P

Using the SI formula,
SI = \frac{\left ( P * R * T \right )}{100}[\latex]
P = \frac{\left ( P * 5 * T \right )}{100}[\latex]
Here both the P gets canceled.

T = \frac{100}{r}[\latex]
= \frac{100}{5}[\latex]
= 20 years

Given,
P = Rs 45,000
Rate = ?
CI = Rs 22,500
T = 6 years
n = 1

A = P + CI
= Rs 45,000 + Rs, 22, 500
= Rs 67, 500

Using the formula,

A = P\left ( 1+ \frac{r}{100n} \right )^{nT}[\latex]
67,500 = 45000\left ( 1+ \frac{r}{100n} \right )^{6}[\latex]

\frac{67500}{45000}[\latex] = \left ( 1+ \frac{r}{100} \right )^{6}[\latex]

^{\sqrt[6]{\left ( \frac{67500}{45500} \right )}}[\latex] = \frac{100 + r}{100}[\latex]
1.07 = \frac{100 + r}{100}[\latex]

107 = 100 + r
R = 7%

Given,
P = Rs 16,000
R = 8% pa
N = 1

A = P\left ( 1+ \frac{r}{100n} \right )^{nT}[\latex]
= 16000\left ( 1+ \frac{8}{100} \right )^{4}[\latex]
= 16000\left ( \frac{108}{100} \right )^{4}[\latex]
= 21, 760

Given,
P = 1500
A = 1900
T = 2 years

A = P\left ( 1+ \frac{r}{100n} \right )^{nT}[\latex]
1900 = 1500\left ( 1+ \frac{r}{100} \right )^{2}[\latex]

\frac{1900}{1500}[\latex] = \left ( 1 + \frac{r}{100} \right )^{2}[\latex]

\left ( \frac{100 + r}{100} \right )^{2} [\latex] = \frac{19}{15}[\latex]

\frac{100 + r}{100}[\latex] = 1.126

R = 112.6-100
= 12.6%

A = P\left ( 1+ \frac{r}{100n} \right )^{nT}[\latex]
= 45000\left ( 1+ \frac{7}{100} \right )^{4}[\latex]
= 45000\left ( \frac{107}{100} \right )^{4}[\latex]
= 45000 * 1.31
= Rs 58,950.

Given,
SI = 4000
R = 8%
T = 4 years

P = \frac{100 * SI}{R * T}[\latex]
= \frac{100 * 4000}{8 * 4}[\latex]
= 12, 500

When compounded annually,

A = P\left ( 1+ \frac{r}{100n} \right )^{nT}[\latex]
= 12,500\left ( 1+ \frac{7}{100} \right )^{4}[\latex]
= 12, 500 * 1.36
= 17, 000