Question

22. A man donated 580,000 to provide money for scholarship. The sum was invested at 6 % p.a. How many scholarships can be paid out of the interest, if the value of each scholarship is 3200 per month?​

Answer 1

Answer:

10

Step-by-step explanation:

Amount donated by man = Rs. 5,80,000

Interest earned at the rate of 6%

p.a = 6% of 5,80,000

= 6/100× 5,80,000

= Rs. 34,800

∴ value of each scholarship = Rs 3200

∴ Number of scholarship = 34800/3200

= 10 (approx)

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Answer 2

I lent 3,000 to Anindita on 1st March, 2001 at 10 %. She cleared the debt on 18th December, 2001 by giving me 3,200 and a pen. Find the cost of the pe.

I am answering this question ok

Solution

$P \: = \: ₹3000 \\ Rate \: of \: intrest \: = \: 10 \: \% \\ T = \frac{293}{365} = 0.80 \: years \\ \color{aqua}(293 \: because \: she \: retured \: the \: money \: after \: 293 \: days)$

$I = \frac{P \times R \times T}{100} \\ \\ I = \frac{3 \cancel0 \cancel0 \cancel0 \times 1\cancel0 \times 0 \cancel.80}{1\cancel0\cancel0\cancel0\cancel0}$

$I = 3 \times 80 = ₹240$

$\\ Amount \: = \: 3000 + 240 \\ = ₹3240$

$She \: returned \: ₹3200 \: and \: a \: pen \: \\ \therefore \: cost \: of \: pen \: = \: 3240 - 3200 = ₹40$

$\color{aqua}The \: pen \: costs \: ₹ \: 4 0$

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$\large \color{lime} \: \bigstar \: More \: Formulas \: \\ \\ \color{aqua}Si \: = \: \frac{P \times R \times T}{100} \\ \\ \\ \color{aqua}Amount \: = \: Principal\:+\:Intrest \\ \\ \\ \color{aqua}Formula \: for \: Ci \: = \: A\:=\:P( {\frac{1 + r}{n} )}^{n \times t } \\ \color{aqua}where \: \: A =\:final\: amount \\\color{aqua} P\:=\: Principal \\ \color{aqua}r \: = \:rate \: of \: intrest \\ \color{aqua} n \: = \: number \: of \: times \: interest \: applied \: per \: time \: period \\ \color{aqua}t \: = \: number \: of \: time \: periods \: elapsed$