Today I am going to share practice questions for Simple Interest. Must download read SI and CI concepts here.

r×t =

S.I = Amount - Principal ⇒S.I = 920 - 800 = 120⇒ r×t =

r = 5% ⇒ Now r+3% ⇒ 5+3 = 8% ⇒ r×t =

⇒ 800+192

⇒ 992

⇒ [

⇒ 3PR + 6P - 3PR = 36000 ⇒ 6P = 36000 ⇒ P = 6000

⇒ r×t =

⇒ R =

S.I=A-P ⇒ Let P = x ⇒SI = 2x - x ⇒ S.I = x

⇒ r×t =

A = P + S.I

⇒ A = P + [ 1+

⇒ 630 = P [ 1+

Borrowed ( debt) amount = M, n = no. of installments , amount of equal installments = x ,

rate of percent = r (quarterly = r/4 % , monthly = r/12 % ).

M = nx [ 1 +

2 = 3×1 [ 1 +

⇒x =400 %

M = nx [ 1 +

Here, M = Rs. 4200, n = 5 , y = 1 , r = 10% , x = annual installment = ?

4200 = 5x [ 1 +

x = Rs. 700 per year.

Let the sum lent out at 8% = P

P

P

- On what sum of money lent out at 9% per annum simple interest for 6 years does the simple interest amount to Rs. 810?
**(RBI'82)** - A sum of Rs. 800 amounts to Rs. 920 in 3 years at simple interest. If the interest rate is increased by 3%, it would amount to how much ?
- A sum was put at simple interest at a certain rate for 3 years. Had it been put at 2% higher rate, it would have fetched Rs. 360 more. Find the sum.
- The simple interest on a sum of money is 4/9 of the principal. Find the rate percent and time, if both are numerically equal.
**(S.S.C. 2000)** - At what rate of interest per annum will a sum double itself in 8 years?
**( Bank PO'86)** - At what rate percent per annum will a sum of money double in 16 years?
**( R.R.B. 2003)** - A sum when reckoned at simple interest 2
^{1}/2% per annum amounts to Rs. 630 after 2 years. Find the sum. - A sum of Rs. 2 is lent to be paid back in 3 equal monthly installments of Re. 1 each. Find the rate per cent .
- What annual installment will discharge a debt of Rs. 1092 due in 3 years at 12% simple interest
- What annual installment will discharge a debt of Rs 4,200 due in 5 years at 10% simple interest ?
**(AAO '82)** - A sum of Rs. 3,800 is lent out in two parts in such a way that the interest on one part at 8% for 5 years is equal to that on another part at
^{1}/_{2}% for 15 years. Find the sum lent out at 8%. - A sum of Rs. 1550 is lent out into two parts, one at 8% and another one at 6%. If the total annual income is Rs. 106, find the money lent at each rate.
**(L.I.C. A.A.O 2003)** - If rs. 85 amounts to Rs. 95 in 3 years, what Rs. 102 will amount to in 5 years at the same rate percent?

### # Solution

**1. 1500**

r×t =

^{S.I}/

_{Principal}×100 ⇒9×6 =

^{810}/P×100 ⇒ 81000/54 ⇒ 1500

**2. 992**

S.I = Amount - Principal ⇒S.I = 920 - 800 = 120⇒ r×t =

^{S.I}/

_{Principal}×100 ⇒ r×3 = 120/800 ×100

r = 5% ⇒ Now r+3% ⇒ 5+3 = 8% ⇒ r×t =

^{S.I}/

_{Principal}×100 ⇒ 8×3 = S.I/800 ×100 ⇒192

⇒ 800+192

⇒ 992

**3. 6000**

⇒ [

^{P×(R+2)×3}/

_{100}] = [

^{P×(R+2)×3}/

_{100}] = 360

⇒ 3PR + 6P - 3PR = 36000 ⇒ 6P = 36000 ⇒ P = 6000

**4. R = 6**

^{2}/_{3}% T= 6 yrs 8 months⇒ r×t =

^{S.I}/

_{Principal}×100 ⇒ R×R =

^{4x/9}/

_{x}×100 ⇒ R

^{2}=

^{400x}/9

⇒ R =

^{20}/3

**5. R= 12 1/2%**

S.I=A-P ⇒ Let P = x ⇒SI = 2x - x ⇒ S.I = x

⇒ r×t =

^{S.I}/

_{Principal}×100 ⇒ r×8=

^{x}/

_{x}×100 ⇒ R = 100/8

**6.**

**R = 6**( Same As Q5.)

^{1}/_{4}%**7. P = 600**

A = P + S.I

⇒ A = P + [ 1+

^{RT}/

_{100}]

⇒ 630 = P [ 1+

^{2×5}/

_{100×2}] ⇒630 = P × 105/108 ⇒ P = 600

**8. 400%**

Borrowed ( debt) amount = M, n = no. of installments , amount of equal installments = x ,

rate of percent = r (quarterly = r/4 % , monthly = r/12 % ).

M = nx [ 1 +

^{r(n-1)}/

_{ 200}]

2 = 3×1 [ 1 +

^{r(3-1)}/

_{200×12}]

⇒x =400 %

**9. Rs.325**(Same As Q 8&10)

**10. 700 per year**

M = nx [ 1 +

^{r(n-1)}/

_{ 200}]

Here, M = Rs. 4200, n = 5 , y = 1 , r = 10% , x = annual installment = ?

4200 = 5x [ 1 +

^{10×4}/

_{200}]

x = Rs. 700 per year.

**11. P = Rs. 600**

Let the sum lent out at 8% = P

_{1}; sum lent out at

^{1}/2 % = P

_{2}[since (SI)

_{1}= (SI)

_{2}]

P

_{1}× 8×5 =P

_{2}×

^{1}/2 ×15 ⇒ P

_{1}/P

_{2}=

^{15}/

_{2*8*5}=

^{3}/16

P

_{1}= Sum lent out at 8% =

^{3}/

_{(3+16)}⇒ 3800 = Rs. 600

**12.**900

**13.**A = 122

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