Question

plss friends help me..in these Extra questions.....​

Answer 1

(i) A T.V was bought at a price of ₹19000. After one year the value of the t.v was depreciated by 5 % ?

Solution :-

☞ Principal = ₹ 19,000

☞ Reduction = 5% of 19, 000 per year

∵ By using

$➤simple \: interest \: = \frac{p \times r \times t}{100}$

In this P = Principal , r = rate and t = time

Accounting to question

$➝ \frac{19000 \times 5 \times 1}{100}$

➝ ₹ 950

✒ Value of the 1 year

⇒ ₹ 19, 000 - 950

⇒ ₹ 18.050

❒ By another method

As we know

$➣amount = p {(1 - \frac{r}{100} )}^{n}$

In which P = principal , r = rate of interest and n = time period

So ,

$→19000 {(1 - \frac{5}{100} )}^{1}$

$→19000 \times \frac{19}{20}$

Amount = ₹ 18050

_______________

(ii) What amount is to be repaid on a loan of ₹ 10000 for 1 whole 1/3 years at 10% pr annum compounded half yearly?

Solution:-

Principal for first 6 months = ₹ 10000

Time = 1 whole 1/ 3 years

Rate = 10 %

Simple Interest =

$⇒ \frac{10000 \times 10 \times \frac{1}{3} }{100}$

⇒ ₹ 333.3

A = P + I

A = 10000 + 333.3

A = ₹10333.3

This is the principal of next 1/3 years

Simple Interest

$⇒ \frac{10333.3 \times 10 \times \frac{1}{3} }{100}$

⇒₹ 3478.877

Principal of 3 period = 10333.3+ 3478.877

⇒ ₹ 13812.177

Simple Interest =

$⇒ \frac{13812.77 \times 10 \times \frac{1}{3} }{100}$

⇒₹ 460.4059

A = P + I

= 13812.77+ 460.4059

= ₹ 14272. 5829