Question

a sum of money was put at simple interest at a certain rate for 2 years if the sum had been put at 3% higher rate it would earned 720 more as interest find the sum​

Answer 1

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OPTION 1) Rs. 2000

2) Rs. 1600

3) Rs. 1500

4) Rs. 2200

5) Rs. 1200

6) Rs. 1800

7) Rs. 1000

8) Rs. 1400

9) Rs. 2600

10)None of these

SolutionLet the sum be Rs. x & original rate R %, Then,

{(x*(R+3)*2)/100)}-{(x*R*2)/100} = 72

Solve

6x = 72*100

x = 1200

Answer 2

Step-by-step explanation:

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✺Answer:

♦️GiveN:

Sum of money was put with simple Interest.

At a certain rate and time is 2 years.

If it would have put 3% higher rate, then earned 720 more as simple Interest.

♦️To FinD:

The sum i.e. Principal money.

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✺Explanation of Q.

The question is based on Simple Interest concept, and also we will use linear eqaution in solving this. Not much concepts not needed, just apply formula, from the equation and find the solution.

Concept to be used:

Formula for finding Simple Interest:

$\large{ \ddag \: \: { \boxed{ \rm{ \pink{(SI)= \frac{ P \times T \times R}{100}}}}}}‡$

(SI)=

100

P×T×R

♠️ Note.....

Symbols have their usual meanings.

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Now applying this formula and forming equations will solve this question.

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✺Solution:

Let the principal = P

Rate of Interest = r

Time given = 2 years

And, simple Interest = SI

By applying formula,

$\large{ \rm{ \rightarrow \: SI = \large{ \frac{P \times 2 \times r}{100}..........(1)}}}$→SI=

100

P×2×r

..........(1)

Now, It is given that

If New rate = r+3

Then New SI = SI + 720

Time taken is 2 years and principal is P

By applying Formula,

$\begin{lgathered}\large{ \rm{ \rightarrow \: new \: SI \: = \large{\frac{P\times 2 \times (r + 3)}{100}}}} \\ \\ \large{\rm{\rightarrow \: SI \:+\: 720\: = \large{\frac{P\times 2 \times (r + 3)}{100}........(2)}}}\end{lgathered}$

→newSI=

100

P×2×(r+3)

→SI+720=

100

P×2×(r+3)

........(2)

Subtracting equation (1) from (2),

$\begin{lgathered}\large{ \rm{ \rightarrow \: \frac{P \times 2 \times (r + 3)}{100} - \frac{P \times 2 \times r }{100} = \: Rs. 720 }} \\ \\ \large{ \rm{ \rightarrow \: \frac{2P(r + 3) - 2Pr }{100} = 720}}\end{lgathered}$

→

100

P×2×(r+3)

−

100

P×2×r

=Rs.720

→

100

2P(r+3)−2Pr

=720

Taking 2p common,

$\begin{lgathered}\large{ \rm{ \rightarrow \: \frac{2P( \cancel{r} + 3 - \cancel{ r})}{100} = 720}} \\ \\ \large{ \rm{ \rightarrow \: \frac{2P \times 3}{100} = 720}} \\ \\ \large{ \rm{ \rightarrow \: P = \cancel{ \frac{720 \times 100}{2 \times 3}}}} \\ \\ \large{ \rm{ \rightarrow \: \boxed{ \red{ \rm{ \: P= \: Rs. 12000}}}}}\end{lgathered}$

$→ </p><p>100</p><p>2P( </p><p>r</p><p> </p><p> +3− </p><p>r</p><p> </p><p> )</p><p> </p><p> =720</p><p>→ </p><p>100</p><p>2P×3</p><p> </p><p> =720</p><p>→P= </p><p>2×3</p><p>720×100$

→

P=Rs.12000

Thus, the Sum of initial money:

$\large{ \rm{ \therefore{ \underline{ \purple{Principal = Rs.12000}}}}}$∴

Principal=Rs.12000

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