Simple interest worksheet featuring problems and formulas for calculating simple interest and amount.

Let me work through all the problems on this simple interest worksheet step by step.

The formulas we need:
- Simple Interest (S.I.) = (P × R × T) / 100
- Amount (A) = P + I

Problem 1: Find simple interest and amount

(a) P = $1800, R = 5%, T = 1 year
S.I. = (1800 × 5 × 1) / 100 = 9000/100 = $90
Amount = 1800 + 90 = $1890

(b) P = $2600, R = 12%, T = 3 years
S.I. = (2600 × 12 × 3) / 100 = 93600/100 = $936
Amount = 2600 + 936 = $3536

(c) P = $3125, R = 15%, T = 73 days
T in years = 73/365 = 1/5 year
S.I. = (3125 × 15 × 1/5) / 100 = (3125 × 3) / 100 = 9375/100 = $93.75
Amount = 3125 + 93.75 = $3218.75

(d) P = $5660, R = 11%, T = 9 months
T in years = 9/12 = 3/4 year
S.I. = (5660 × 11 × 3/4) / 100 = (5660 × 11 × 3) / (4 × 100) = 186780/400 = $466.95
Amount = 5660 + 466.95 = $6126.95

(e) P = $180, R = 3%, T = 1¼ year = 5/4 year
S.I. = (180 × 3 × 5/4) / 100 = (180 × 3 × 5) / (4 × 100) = 2700/400 = $6.75
Amount = 180 + 6.75 = $186.75

Problem 2: What sum would yield an interest of $36 in 3 years at 3% p.a.?
I = 36, R = 3, T = 3
36 = (P × 3 × 3) / 100
36 = 9P/100
P = 36 × 100/9 = 400
Answer: $400

Problem 3: At what rate per cent per annum will $250 amount to $330 in 4 years?
A = 330, P = 250, T = 4
I = A - P = 330 - 250 = 80
80 = (250 × R × 4) / 100
80 = 1000R/100
80 = 10R
R = 8%

Problem 4: At what rate per cent per annum will $400 yield an interest of $78 in 1½ years?
P = 400, I = 78, T = 3/2
78 = (400 × R × 3/2) / 100
78 = (400 × 3 × R) / (2 × 100)
78 = 1200R/200
78 = 6R
R = 13%

Problem 5: In what time will $400 amount to $512 if the simple interest is calculated at 14% p.a.?
P = 400, A = 512, R = 14
I = 512 - 400 = 112
112 = (400 × 14 × T) / 100
112 = 5600T/100
112 = 56T
T = 2 years

Problem 6: A sum amounts to $2400 at 15% simple interest per annum after 4 years. Find the sum.
A = 2400, R = 15, T = 4
A = P + (P × R × T)/100
2400 = P + (P × 15 × 4)/100
2400 = P + 60P/100
2400 = P + 0.6P
2400 = 1.6P
P = 2400/1.6 = 1500
Answer: $1500

Problem 7: Ken borrowed $2000 from Sam at 8% per annum. After 6 years he cleared the amount by giving $2600 cash and a watch. Find the cost of the watch.
P = 2000, R = 8, T = 6
I = (2000 × 8 × 6) / 100 = 96000/100 = $960
Total amount owed = 2000 + 960 = $2960
Cash paid = $2600
Cost of watch = 2960 - 2600 = $360

Problem 8: In how many years will $400 yield an interest of $112 at 14% simple interest?
P = 400, I = 112, R = 14
112 = (400 × 14 × T) / 100
112 = 5600T/100
112 = 56T
T = 2 years

Problem 9: In how many years will $12000 yield an interest of $13230 at 10% simple interest?
Wait, that can't be right. Let me re-read. "In how many years will $12000 yield an interest of $13230 at 10% simple interest?"

Hmm, that seems like the interest ($13230) is more than the principal ($12000). Let me check if this is correct or if it should be amount instead of interest.

Actually, looking at it again, maybe it says "$13230" as the amount, not interest. But the problem says "yield an interest of $13230". Let me solve it as stated.

I = 13230, P = 12000, R = 10
13230 = (12000 × 10 × T) / 100
13230 = 1200T
T = 13230/1200 = 11.025 years

That's about 11 years and ~9 days. This seems odd for a school problem. Let me reconsider - maybe it's meant to be the amount, not interest.

If A = 13230, then I = 13230 - 12000 = 1230
1230 = (12000 × 10 × T) / 100
1230 = 1200T
T = 1230/1200 = 1.025 years

Still odd. Let me just solve it as written with interest = $13230.

Actually wait, let me re-examine. Maybe there's a typo and it should be different numbers. But I'll solve as given.

T = 13230/1200 = 11.025 years ≈ 11 years and 9 days

Hmm, but this doesn't give a clean answer. Let me try assuming it means amount = $13230:
I = 13230 - 12000 = 1230
1230 = 1200T
T = 1.025 years

Still not clean. Let me try another interpretation - maybe the interest is $1320?
1320 = 1200T
T = 1.1 years

Or maybe $1200?
1200 = 1200T
T = 1 year

Given the context of other problems having clean answers, I suspect there might be a typo. But solving exactly as written:
T = 13230/1200 = 11.025 years

Actually, let me just provide the answer as calculated: T = 11.025 years or approximately 11 years.

Wait, I should double-check my calculation:
13230 = (12000 × 10 × T) / 100
13230 = 120000T/100
13230 = 1200T
T = 13230/1200 = 11.025

Yes, that's correct. The answer is 11.025 years.

Problem 10: In how many years will $600 double itself at 10% simple interest?
If it doubles, A = 2P = 1200
I = 1200 - 600 = 600
600 = (600 × 10 × T) / 100
600 = 60T
T = 10 years

Problem 11: At what rate of simple interest will $5000 amount to $6050 in 3 years, 4 months?
P = 5000, A = 6050, T = 3 + 4/12 = 3 + 1/3 = 10/3 years
I = 6050 - 5000 = 1050
1050 = (5000 × R × 10/3) / 100
1050 = (50000 × R) / 300
1050 = 500R/3
3150 = 500R
R = 6.3%

Problem 12: At what rate of simple interest will the sum of money double itself in 6 years?
If it doubles in 6 years: I = P
P = (P × R × 6) / 100
100 = 6R
R = 100/6 = 16⅔% or 16.67%

Problem 13: Find the simple interest at the rate of 5% p.a. for 3 years on that principal which in 4 years, 8 months at the rate of 5% p.a. gives $1200 as simple interest.

First, find the principal:
T₁ = 4 + 8/12 = 4 + 2/3 = 14/3 years
R₁ = 5%, I₁ = 1200
1200 = (P × 5 × 14/3) / 100
1200 = (70P/3) / 100
1200 = 70P/300
1200 × 300 = 70P
360000 = 70P
P = 360000/70 = 5142.857...

Hmm, that's not a clean number. Let me recalculate:
1200 = (P × 5 × 14/3) / 100
1200 = (P × 70/3) / 100
1200 = 70P/300
1200 × 300 = 70P
360000 = 70P
P = 360000/70 = 5142.857...

This doesn't give a clean principal. Let me check if I misread the problem.

"Find the simple interest at the rate of 5% p.a. for 3 years on that principal which in 4 years, 8 months at the rate of 5% p.a. gives $1200 as simple interest."

So we need to find S.I. for 3 years at 5% on a principal P, where P gives $1200 interest in 4 years 8 months at 5%.

Actually, I can use a ratio approach:
I₁/I₂ = T₁/T₂ (since P and R are the same)
1200/I₂ = (14/3)/3
1200/I₂ = 14/9
I₂ = 1200 × 9/14 = 10800/14 = 771.428...

Still not clean. Let me try calculating P first more carefully:
1200 = P × 5 × (14/3) / 100
1200 = P × 70/300
1200 = 7P/30
P = 1200 × 30/7 = 36000/7 = 5142.857...

Then I₂ = P × 5 × 3 / 100 = (36000/7) × 15/100 = 540000/700 = 5400/7 = 771.428...

The answer is $771.43 (approximately) or exactly 5400/7.

Actually, let me express it as a fraction: 5400/7 = 771³⁄₇

Problem 14: At what rate per cent per annum will $4000 yield an interest of $410 in 2 years?
P = 4000, I = 410, T = 2
410 = (4000 × R × 2) / 100
410 = 80R
R = 410/80 = 5.125%

Problem 15: Simple interest on a certain sum is ¹⁶⁄₂₅ of the sum. Find the rate per cent and time if both are numerically equal. [Hint: (T = R), P = x, S.I. = ¹⁶₂₅ x]

S.I. = 16x/25
Also, S.I. = (P × R × T)/100 = (x × R × R)/100 = xR²/100 (since T = R)

So: 16x/25 = xR²/100
16/25 = R²/100
R² = 1600/25 = 64
R = 8

Since T = R, T = 8 years
Rate = 8% per annum, Time = 8 years

Problem 16: Simple interest on a sum of money at the end of 5 years is ⁴₅ of the sum itself. Find the rate per cent p.a.

S.I. = 4P/5
S.I. = (P × R × 5)/100
4P/5 = 5PR/100
4/5 = 5R/100
4/5 = R/20
R = 80/5 = 16%

Let me verify all my calculations one more time before presenting the final answer.

Problem 1 checks:
(a) 1800 × 5 × 1 / 100 = 90 ✓, Amount = 1890 ✓
(b) 2600 × 12 × 3 / 100 = 936 ✓, Amount = 3536 ✓
(c) 3125 × 15 × (73/365) / 100 = 3125 × 15 × 0.2 / 100 = 93.75 ✓, Amount = 3218.75 ✓
(d) 5660 × 11 × (9/12) / 100 = 5660 × 11 × 0.75 / 100 = 466.95 ✓, Amount = 6126.95 ✓
(e) 180 × 3 × 1.25 / 100 = 6.75 ✓, Amount = 186.75 ✓

Problem 2: P = 36 × 100/(3×3) = 400 ✓
Problem 3: R = 80 × 100/(250×4) = 8% ✓
Problem 4: R = 78 × 100/(400×1.5) = 13% ✓
Problem 5: T = 112 × 100/(400×14) = 2 years ✓
Problem 6: P = 2400/1.6 = 1500 ✓
Problem 7: Watch = 2960 - 2600 = 360 ✓
Problem 8: T = 112 × 100/(400×14) = 2 years ✓
Problem 9: T = 13230 × 100/(12000×10) = 11.025 years
Problem 10: T = 600 × 100/(600×10) = 10 years ✓
Problem 11: R = 1050 × 100/(5000×10/3) = 1050 × 300/50000 = 6.3% ✓
Problem 12: R = 100/6 = 16⅔% ✓
Problem 13: I = 5400/7 = 771³⁄₇ ≈ $771.43
Problem 14: R = 410 × 100/(4000×2) = 5.125% ✓
Problem 15: R = T = 8 ✓
Problem 16: R = 16% ✓

All calculations verified. Now I'll present the final answer clearly.