Question

if ₹640 amount to ₹768 in 2years 6months, what will ₹850 amount to in 3years at thesame rate per cent per annum​

Answer 1

Step-by-step explanation:

ptr/100 = 168

6.4*2.5*r=168

r = 168/16

Answer 2

➤ Given :-

Principle (1) :- ₹640

Total amount (1) :- ₹768

Time (1) :- 2 years 6 months

Principle (2) :- ₹850

Time (2) :- 3 years

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➤ To Find :-

Total amount of second statement

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➤ Formula required :-

Rate of interest :-

${\tt \large \dashrightarrow \orange{\boxed{\tt \gray{\dfrac{SI \times 100}{P \times T}}}}}$

Total amount :-

${\tt \large \dashrightarrow \orange{\boxed{\tt \gray{Simple \: Interest + Principle}}}}$

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★ How to do :-

Here, we are given with principle, total amount and time in first statement. And in the second statement, we are given with principle and time. We should find the total amount of the second statement. So, first we should find the rate of interest for first statement by using first formula and then we should apply the same rate of interest for the second statement and calculate the total amount.

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➤ Solution :-

Simple interest (1) :-

${\tt \leadsto 768 - 640}$

${\tt \leadsto Rs \: \: 128}$

Rate of interest (1) :-

${\tt \leadsto \dfrac{128 \times 100}{640 \times 2.5}}$

${\tt \leadsto \dfrac{128 \times \cancel{100}}{640 \times \cancel{2.5}} = \dfrac{128 \times 40}{640 \times 1}}$

${\tt \leadsto \dfrac{\cancel{128} \times 40}{\cancel{640} \times 1} = \dfrac{1 \times 40}{5 \times 1}}$

${\tt \leadsto \cancel \dfrac{40}{5} = \boxed{ \tt 8 \bf\%}}$

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Now,

Total amount (2) :-

${\tt \leadsto \dfrac{850 \times 3 \times 8}{100} + 850}$

${\tt \leadsto \dfrac{\cancel{850} \times 3 \times 8}{ \cancel{100}} = \dfrac{17 \times 3 \times 8}{2}}$

${\tt \leadsto \dfrac{17 \times 24}{2} = \dfrac{408}{2}}$

${\tt \leadsto \cancel \dfrac{408}{2} = 204}$

${\tt \leadsto 204 + 850}$

${\tt \leadsto Rs \: \: 1054}$

$\Huge\therefore$ The total amount of the second statement is ₹1054.

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$\dashrightarrow$ Some related formulas :-

${\tt \to \boxed{\tt Simple \: Interest = \dfrac{P \times R \times T}{100}}}$

${\tt \to \boxed{\tt Principle = \dfrac{SI \times 100}{R \times T}}}$

${\tt \to \boxed{\tt Time = \dfrac{SI \times 100}{P \times R}}}$