Question

11. Two equal sums were borrowed at 8% simple interest
per annum for 2 years and 3 years respectively. The
difference in the interests was Rs 56. The sums bor-
rowed were
a) Rs 690 b) Rs 700 c) Rs 740 d) Rs 780​

Answer 1

Answer:

rs 690 is the correct answer

Answer 2

Answer:

↬ GIVEN :-

• Rate (r) = 8% p.a
• First time (t1) = 2 year
• Second time (t2) = 3 year
• Difference in interest = ₹ 56

↬ TO. FIND :-

• Principal (sum) = ?

↬ SOLUTION :-

• Let the sum (Principal) is ₹ P

↬ Case(i)

When. Time is (t1)=2 year and Rate (r) = 8%

First simple interest(I1)

$\orange{ \sf \: Simple \: \: Interest = \frac{(principal) \times (rate) \times (time)}{100} }$

$\sf \: Simple \: \: Interest(I _{1}) = \frac{(p) \times (r) \times (t _{1} )}{100}$

$\sf \: Simple \: \: Interest(I _{1}) = \frac{(p) \times (8) \times (2)}{100}$

$\sf \: Simple \: \: Interest(I _{1}) = \frac{4p}{25} \: \: \: \: \: \: \: \: \: ....(i)$

Now

↬ Case(ii)

When. Time is (t2)=3 year and Rate (r) = 8%

Second Simple interest (I2)

Again using Simple interest formula :-

$\sf \: Simple \: \: Interest(I _{2}) = \frac{(p) \times (r) \times (t _{2} )}{100}$

$\sf \: Simple \: \: Interest(I _{2}) = \frac{(p) \times (8) \times (3)}{100}$

$\sf \: Simple \: \: Interest(I _{2}) = \frac{6p}{25} \: \: \: \: \: \: \: \: \: ....(ii)$

• NOW it is given that difference of interest is ₹ 56.

$\sf Second \: \: simple \: \: interest - First \: \: Simple \: \: Interest = 56$

$\sf \implies \: (I _{2}) - (I _{1}) = 56$

$\sf \implies \: \frac{6p}{25} - \frac{4p}{25} = 56$

$\sf \implies \: \frac{6p - 4p}{25} = 56$

$\sf \implies \: \frac{2p}{25} = 56$

$\sf \implies \: p = \frac{56 \times 25}{2}$

$\sf \implies \: p = 700$

• Hence Principal (Sum) is ₹ 700

Option( B) is correct