Math-7 Simple Interest : 20 mixed-difficulty SI problems- each with step-by-step (digit-by-digit) solutions

Simple Interest : 20 mixed-difficulty SI problems- each with step-by-step (digit-by-digit) solutions

20 practice problems on Simple Interest tailored for competitive exams (CDS, SSC, Railways, etc.), each with a clear, step-by-step solution and a short explanation. I show every arithmetic step so there’s no confusion.

Simple Interest — 20 Practice Questions with Solutions

Problems

1. Find SI on ₹7,500 at 4% p.a. for 3 years.
2. Simple interest is ₹450 for 2 years at 5% p.a. Find the principal.
3. Find SI on ₹6,000 at 12% p.a. for 7 months.
4. A sum becomes ₹6,720 in 2 years and ₹7,056 in 3 years under simple interest. Find the principal and rate.
5. SI on ₹10,000 for 4 years equals SI on ₹15,000 for how many years at the same rate?
6. A man borrows ₹8,000 and repays ₹9,200 after 2 years. What rate was charged (simple interest)?
7. SI on a sum for 3 years at 6% is ₹180. Find the sum.
8. Two sums ₹2,000 and ₹3,000 earn simple interest. The difference of interest for 2 years is ₹50. Find the rate.
9. In how many years will simple interest on a sum at 20% p.a. be equal to the sum itself?
10. Simple interest for 3 years at R% equals simple interest for 2 years at (R + 3)%. Find R.
11. Find SI on ₹1,200 at 10% p.a. for 9 months.
12. A invests ₹2,500 for 2 years and B invests ₹4,000 for 2 years at the same rate. The difference of interests is ₹300. Find the rate.
13. At what time (years) will simple interest on a sum at 5% p.a. equal half the sum?
14. If ₹2,500 yields ₹250 as SI in 2 years, find the rate.
15. ₹5,000 is invested for 1.5 years and ₹7,000 is invested for 2 years at the same rate. If total interest is ₹1,290, find the rate.
16. A sum of ₹10,000 is split into two parts. One part is invested at 4% p.a. and the other at 6% p.a. for 3 years. If interests are equal, find the amount invested at 4%.
17. Find SI on ₹15,000 at 8% p.a. for 20 months.
18. A man lends ₹12,000; he lends ₹x at 5% and the rest at 8% for 1 year. If total interest is ₹900, find x.
19. A sum doubles under simple interest in 10 years. Find the annual rate.
20. Simple interest on ₹P for T years equals simple interest on ₹2P for 3 years at the same rate. Find T.

Solutions (step-by-step)

1 Find SI on ₹7,500 at 4% p.a. for 3 years.

SI = P x Rx T /100= 7500 x 4 x 3 /100 .
Compute numerator: 7500 x 4 =30000.

Then (30000 x 3=90000). Divide by 100 → 90000 / 100=900).
Answer: ₹900.
Explanation: Direct application of formula.

2

Use P =  100 x SI / R x T . Substitute: = 100 x 450/ 5 x 2
Denominator =10 .  Numerator =45000. (45000/ 10=4500).
Answer: ₹4,500.
Explanation: Rearranged SI formula to find principal.

3

Convert time: 7  months =7/12 years.
SI = 6000 x 12 x (7/12) /100.

Cancel 12: becomes (6000 x 7/ 100.
Compute (6000 x 7=42000). (42000/ 100=420).
Answer: ₹420.
Explanation: Use months → convert to years.

4

Difference between amounts for 3rd and 2nd year = SI for 1 year.
(A3-A2 = 7056-6720 = 336).

So P x R /100 = 336 =PR = 33600.)
Also (A2 = P + 2   PR/100} = P + 2 x 336 = P + 672 = 6720).
So (P = 6720 – 672 = 6048).
Now R = 33600/P = 33600 / 6048.

Simplify: divide by 48 → (33600 /48=700,\ 6048\div48=126). So (R = 700/126 = 50/9 \approx 5.555%).
Answer: Principal = ₹6,048, Rate = (50/9%) (≈ 5.555% p.a.).
Explanation: Use year-to-year difference to get PR/100.

5

Equate SI: 10000 x  R x 4 /100 = 15000 x R x  X/100.

 Cancel (R/100).
10000 x 4 = 15000 x X   

X = 40000/15000} = 8/3 = 2x 2/3 years.
Answer:(2 x 2/3 years (i.e. 2 years 8 months).
Explanation: SI proportional to (P\ x T).

6

Interest (=9200-8000=1200). Rate (R = 100 x SI/ (Px T) = 100 x 1200 / (8000 x 2 )
Numerator =120000  Denominator =16000.

120000/ 16000=7.5
Answer: 7.5% p.a.
Explanation: Find R from known SI.

7

P = 100 x SI / (Rx T ) } = 100 x 180 / (6 x 3). Denominator (=18). Numerator (=18000). (18000 /18=1000).
Answer: ₹1,000.
Explanation: Rearranged SI formula.

8

Difference of SI for 2 years = (3000-2000) x  R x 2 / 100 = 1000 x 2R /100 = 20R.

Set 20R = 50

R = 2.5%.
Answer: 2.5% p.a.
Explanation: Interest difference depends on principal difference.

9

We want SI = P in T years: (P x R x T ) / 100 = P.

Cancel (P): Rx T/ 100=1

R x T = 100.)
Given (R=20) → 20 x T =100

T=5 years.
Answer: 5 years.
Explanation: When interest equals principal, (RT=100).

(If R not given, T = 100/R.)

10

Given 3R = 2(R+3)). Solve: 3R = 2R + 6  

R = 6%).
Answer: 6% p.a.
Explanation: Cancel P/100 and equate.

11

Time = 9 months = 9/12 = 3/4 year.
SI = 1200 x 10 x (3/4)/ 100.

Compute 1200 x 10=12000.

12000 x 3/4=9000.

9000 / 100=90).
Answer: ₹90.
Explanation: Convert months → years.

12

SI_A = 2500 x  R x 2 / 100 = 50R.
SI_B = 4000 x  R x 2/100 = 80R.

Difference = 80R – 50R = 30R = 300. So (R = 300/30 = 10%).
Answer: 10% p.a.
Explanation: Same time and rate ⇒ SI proportional to principal.

13

We need  P x R x T /100  = P/ 2.

Cancel P: R T / 100 = 1/2.

Rearr. RT = 50. For (R=5%), (T = 50/5 = 10) years.
Answer: 10 years (for (R=5%)).
Explanation: Solve for T from equation.

14

Given (P=2500), SI = 250 for (T=2 years. Use (R = 100 x SI/ (p x T)

 Numerator =25000.

 Denominator =2500 x 2=5000). (25000/ 5000=5).
Answer: 5% p.a.
Explanation: Direct rearrangement.

15

Total SI = (5000 x  R  x 1.5 / 100 ) +  (7000 x R x 2 / 100 ) .
Compute each part: (5000 x 1.5 = 7500). So first part = 7500R / 100  = 75R
Second: (7000 x 2=14000) → =  14000R/ 100 = 140R.
Total SI = 75R + 140R = 215R.

Given (215R = 1290) ⇒ (R = 1290 / 215 = 6).

(Because (215 x 6 = 1290).)
Answer: 6% p.a.
Explanation: Add interests of each investment.

16

Let amount invested at 4% be X. Then interest for 3 years at 4% =  X x4 x3/100} = 12X /100  = 0.12x).
Other part =10000- X). Interest there at 6% for 3 years =  (10000- X) x 6 x 3 / 100 = 18(10000-X) /100 = 0.18(10000- X).
Set equal: (0.12x = 0.18(10000-x)).

Multiply by 100 to avoid decimals: (12x = 18(10000-X) = 180000 – 18X).
So (12X + 18X = 180000  )

 30X = 180000

X = 6000
Answer: ₹6,000 at 4%.
Explanation: Equal interests ⇒ set their SI expressions equal.

17

Time =20 months =20/12 = 5/3 years.
SI = {15000 x 8 x (5/3) / 100}.

Compute (15000 x 8=120000). Multiply by (5/3): (120000 x 5/3 = 600000/3 = 200000).

Divide by 100 → (200000 /100 = 2000).
Answer: ₹2,000.
Explanation: Convert months → years then apply formula.

18

Let (X) be amount lent at 5%; then rest =12000-X at 8%.

For 1 year: total interest = (X x 5) /100} + (12000-X) x 8 / 100}).

Compute numerator ×100): (5X + 8(12000-X) = 5X + 96000 – 8X = 96000 – 3X.

 Divide by 100 gives total interest. Set equal to 900: ((96000 – 3X)/100 = 900).

Multiply both sides by 100: (96000 – 3X = 90000). So (-3X = -6000) ⇒ (x = 2000).
Answer: ₹2,000 lent at 5%; ₹10,000 at 8%.
Explanation: Solve linear equation from total interest.

19

If sum doubles in 10 years under SI ⇒ SI = P in 10 years.

Using ( P x R x 10/100 = P.

Cancel P: 10R / 100} =1

R = 10%.
Answer: 10% p.a.
Explanation: Doubling under SI gives RT=100 → R=100/T.

20

We want  P x Rx T /100  = {2P x Rx 3/100}.

Cancel (P x R/100): (T = 2 x 3 = 6).
Answer: T = 6 years.
Explanation: Cancel common factors quickly.

Quick Tips & Shortcuts (for exams)

• Always convert months → years by dividing by 12; days → years by 365 (or 360 if specified).
• For equal time and same rate, SI ∝ principal. For equal principal and same rate, SI ∝ time.
• If SI equals principal → (R\times T = 100).

Many problems use cancellation: factor (P/100) or (R/100) cancels — simplifies algebra