Fractions, Decimals and Percentages Word Problems | PDF ... - Free Printable

Here's the complete solution to all 10 word problems, with clear explanations for each.

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Problem 1.
*A bag contains red and yellow beads. 3/5 of the beads are red. Given that there are nine red beads, calculate the total number of beads in the bag.*

✔ Solution:
We know that 3/5 of the total beads = 9 red beads.
Let total beads = \( x \).
So:
\[
\frac{3}{5}x = 9
\]
Multiply both sides by 5:
\[
3x = 45
\]
Divide by 3:
\[
x = 15
\]
✔️ Answer: 15 beads

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Problem 2.
*In a class, 2/7 of the students have a younger sibling. Given that 25 students do not have a younger sibling, calculate the total number of students in the class.*

✔ Solution:
If 2/7 have a younger sibling, then \( 1 - \frac{2}{7} = \frac{5}{7} \) do NOT have one.
We’re told 25 students = 5/7 of the class.
Let total students = \( x \).
\[
\frac{5}{7}x = 25
\]
Multiply both sides by 7:
\[
5x = 175
\]
Divide by 5:
\[
x = 35
\]
✔️ Answer: 35 students

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Problem 3.
*During a sale, all items are advertised at 10% off the retail price. A coat costs £54 in the sale. Work out the original price of the coat.*

✔ Solution:
Sale price = 90% of original price (since 10% off).
Let original price = \( x \).
\[
0.9x = 54
\]
Divide both sides by 0.9:
\[
x = \frac{54}{0.9} = 60
\]
✔️ Answer: £60

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Problem 4.
*Jonny has a length of ribbon. He cuts off 15% of the ribbon, leaving a piece measuring 68cm. What was the original length of the ribbon?*

✔ Solution:
After cutting off 15%, 85% remains → 68 cm.
Let original length = \( x \).
\[
0.85x = 68
\]
Divide both sides by 0.85:
\[
x = \frac{68}{0.85} = 80
\]
✔️ Answer: 80 cm

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Problem 5.
*In a survey, 11 out of 25 people said that they preferred summer over winter. Write this as a decimal.*

✔ Solution:
Convert fraction \( \frac{11}{25} \) to decimal.
Divide 11 by 25:
\[
11 ÷ 25 = 0.44
\]
✔️ Answer: 0.44

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Problem 6.
*The length of a rectangle increases by 10% and its height increases by 5%. Work out the total percentage increase in the area of the rectangle.*

✔ Solution:
Original area = \( L × H \)
New length = \( 1.10L \), New height = \( 1.05H \)
New area = \( 1.10L × 1.05H = 1.155 × L × H \)
So area increased by 1.155 – 1 = 0.155 = 15.5%

✔️ Answer: 15.5%

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Problem 7.
*In a biscuit tin, 15% of the biscuits are chocolate. 1/4 are plain and the rest are oat biscuits. Given that 36 biscuits are oat biscuits, work out the total number of biscuits in the tin.*

✔ Solution:
Convert all to percentages:
- Chocolate: 15%
- Plain: \( \frac{1}{4} = 25\% \)
→ Oat biscuits = 100% – 15% – 25% = 60%

We’re told 60% = 36 biscuits.
Let total = \( x \):
\[
0.60x = 36
\]
\[
x = \frac{36}{0.60} = 60
\]
✔️ Answer: 60 biscuits

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Problem 8.
*In a group of school children, only 72% remember to bring a pencil to school. Of the remaining students, 10% remembered to bring a pen. What percentage of students in the school did not remember to bring a pen or a pencil to school?*

✔ Solution:
- 72% brought pencil → 28% did NOT bring pencil.
- Of those 28%, 10% brought a pen → so 10% of 28% = \( 0.10 × 28 = 2.8\% \) brought a pen but no pencil.
→ So total who brought *something* = 72% (pencil) + 2.8% (pen only) = 74.8%
→ Therefore, those who brought *neither* = 100% – 74.8% = 25.2%

✔️ Answer: 25.2%

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Problem 9.
*A headteacher carries out a survey. He asks 800 students what their favourite subject is. 425 students say that their favourite subject is maths. What proportion of students in this school say that their favourite subject is maths? Give your answer as a decimal.*

✔ Solution:
Proportion = \( \frac{425}{800} \)
Simplify or divide:
\[
425 ÷ 800 = 0.53125
\]
✔️ Answer: 0.53125

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Problem 10.
*At a theme park, seven out of every 40 people choose to buy a fast pass. What is this as a percentage?*

✔ Solution:
Fraction = \( \frac{7}{40} \)
Convert to percentage:
\[
\frac{7}{40} × 100 = \frac{700}{40} = 17.5\%
\]
✔️ Answer: 17.5%

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✔ Final Answers Summary:

1. 15
2. 35
3. £60
4. 80 cm
5. 0.44
6. 15.5%
7. 60
8. 25.2%
9. 0.53125
10. 17.5%

Let me know if you’d like any step explained further!