Question

Suppose Sudheer has invested Rs. 15,000 at 8% simple interest per year. With the return from investment, he wants to buy a washing machine that costs Rs. 19,000. For what period should he invest Rs. 15,000 so that he has enough money to buy a washing machine?​

Answer 1

Answer:

3 years and 4 months

I=Pnr/100

p= prinipal

n-number of years

r%= rate of interest

I-interest earned

here,

prinipal 1500

the interest to be earned (19000-15000)

=4000

2

5.0

I-15000xnx8/100

I=1200n

4000-1200n

n-4000/1200n

n=10/3

Step-by-step explanation:

Answer 2

$\large\underline\mathrm{Given:-}$

• Principal = 15000
• Rate = 8 %

$\large\underline\mathrm{To \: find:-}$

• Find the number of years .?

$\large\underline\mathrm{Solution:-}$

We know,

The principal and the rate of interest. The interest is the amount sudhir needs in addition to 1500 to buy the washing maching.

$\: The \: \: formula \: \: simple \: \: interest \: \: is \: \: {I} \: \: = \: \: \frac{p \: \times \: n \: \times \: r}{100}$

• P = Principal
• n = Number of years
• R = rate of interest
• I = Interest earned

Here, The principal = Rs. 15000.

The money required by Sudhir for buying a washing machine = Rs. 19000

So, the interest to be earned = Rs (19000 - 15000)

= Rs 4000

The number of year for which Rs 15000 is deposited = n

The interest on Rs 15000 for n year at the rate of 8% = I

Then,

$\: \: \: \: \: \: \leadsto \: \: {I} \: \: = \: \: \frac{p \: \times \: n \: \times \: r}{100}$

$\: \: \: \: \: \: \leadsto \: \: {I} \: \: = \: \: \frac{{\cancel{15000}} \: \times \: n \: \times \: {8}}{\cancel{100}}$

$\: \: \: \: \: \: \leadsto \: \: {I} \: \: = \: \: {{150} \: \times \: n \: \times \: {8}}$

$\: \: \: \: \: \: \leadsto \: \: {I} \: \: = \: \: {{1200} \: n}$

So, the relationship between the number of years and interest if Rs 15000 is inversted at an annual interest rate of 8% .

We have,

The period in which the interest earned is Rs. 4000

$\: \: \: \: \: \: \leadsto \: \: {4000} \: \: = \: \: {{1200} \: n}$

$\: \: \: \: \: \: \leadsto \: \: {n} \: \: = \: \: \frac{4000}{1200}$

$\: \: \: \: \: \: \leadsto \: \: {n} \: \: = \: \: \frac{{40}{\cancel{00}}}{{12}{{\cancel{00}}}}$

$\: \: \: \: \: \: \leadsto \: \: {n} \: \: = \: \: \frac{40}{12}$

$\: \: \: \: \: \: \leadsto \: \: {n} \: \: = \: \: {3}\frac{1}{3}$

$\: Since \: \: {n} \: \: = \: \: {3}\frac{1}{3} \: \: and \: \: one \: \: third \: \: of \: \: a \: \: years \: \: is \: \: {4}$

Hence, Sudhir can buy washing machine after 3 years and 4 months.