CBSE Class 8 Mathematics Fractions in Disguise Very Short Answer Questions with Answers

Very Short Answer Questions

Question 1. From which Latin phrase is the word ‘per cent’ derived, and what does that phrase mean?
Answer:
It comes from the Latin phrase per centum, which means “by the hundred” or “out of hundred”. So x% means x out of every 100.

Teacher's Note:
1. Cent is linked with 100 in many words.
2. This meaning explains why % always refers to a whole of 100.

Question 2. Express \( \frac{7}{20} \) as a percentage.
Answer:
\( \frac{7}{20} \) × 100 = 35, so \( \frac{7}{20} \) = 35%.
(Equivalently, \( \frac{7}{20} \) = \( \frac{35}{100} \).)

Teacher's Note:
1. Multiply a fraction by 100 to get its percentage.
2. Or make the denominator 100: 20 × 5 = 100.

Question 3. Write 84% as a fraction in its lowest terms.
Answer:
84% = \( \frac{84}{100} \) = \( \frac{21}{25} \) (dividing the numerator and the denominator by 4).

Teacher's Note:
1. Write over 100, then cancel by the HCF.
2. HCF of 84 and 100 is 4.

Question 4. Write the decimal 0.125 as a percentage.
Answer:
0.125 × 100 = 12.5, so 0.125 = 12.5%.

Teacher's Note:
1. Multiplying by 100 moves the decimal point two places right.
2. A percentage need not be a whole number.

Question 5. If 10% of a number is 24, write down 5% and 20% of the same number.
Answer:
5% is half of 10%, so 5% of the number = 24 ÷ 2 = 12.
20% is double 10%, so 20% of the number = 24 × 2 = 48.

Teacher's Note:
1. Build other percentages from 10% by halving or doubling.
2. The number itself is 240.

Question 6. The human body is, on average, about 60% water by weight. How much water does a body of mass 50 kg contain?
Answer:
60% of 50 = \( \frac{60}{100} \) × 50 = 30.
The body contains about 30 kg of water.

Teacher's Note:
1. 60% = 0.6, so multiply by 0.6.
2. Keep the unit kg in the answer.

Question 7. Find 140% of 40.
Answer:
140% of 40 = \( \frac{140}{100} \) × 40 = 1.4 × 40 = 56.

Teacher's Note:
1. A percentage above 100 gives more than the number itself.
2. 140% = 100% + 40%, so 40 + 16 = 56.

Question 8. In a class of 40 students, 18 are boys. What percentage of the class are boys?
Answer:
\( \frac{18}{40} \) × 100 = 45, so 45% of the class are boys.

Teacher's Note:
1. Part ÷ whole × 100 gives the percentage.
2. The whole here is the full class of 40.

Question 9. Write the rule used to find the percentage increase in the price of an article.
Answer:
Percentage increase = \( \frac{\text{amount of increase}}{\text{original price}} \) × 100.
The increase is always measured against the original price.

Teacher's Note:
1. The base is the old value, not the new one.
2. The same rule with ‘decrease’ gives the percentage decrease.

Question 10. A bank advertises “interest at 6% p.a.” What does ‘p.a.’ stand for, and what does it mean?
Answer:
‘p.a.’ stands for per annum, meaning “for every year”.
So 6% p.a. means interest of 6% of the principal for each year.

Teacher's Note:
1. Annum is Latin for year.
2. Rates are usually quoted per year.

Question 11. Name the three prices used in a buying-and-selling transaction in this chapter.
Answer:
The cost price (what the seller paid), the marked price (the price quoted to the customer) and the selling price (what the customer finally pays).

Teacher's Note:
1. Discount is taken on the marked price.
2. Profit or loss is taken on the cost price.

Question 12. Which ancient Indian text of the 4th century BCE describes interest rates as “per month per cent”?
Answer:
Kautilya’s Arthashastra, which fixes rates such as “a pana and a quarter per month per cent” as fair interest.

Teacher's Note:
1. The idea of ‘per hundred’ is very old in India.
2. Pana was a coin of that period.

Question 13. A badam drink mix weighs 150 g, of which 99 g is sugar. What percentage of the mix is sugar?
Answer:
\( \frac{99}{150} \) × 100 = 66, so sugar makes up 66% of the mix, about two-thirds of the packet by weight.

Teacher's Note:
1. Divide the part by the total weight.
2. Reading labels as percentages helps compare products.

Question 14. In a porridge, millet and water are mixed in the ratio 1 : 4. What percentage of the mixture is millet?
Answer:
There are 1 + 4 = 5 parts in all, so millet is \( \frac{1}{5} \) of the mixture.
\( \frac{1}{5} \) × 100 = 20%.

Teacher's Note:
1. Add the ratio terms to get the whole.
2. 1 : 4 does not mean millet is 1/4.

Question 15. A cyclist has covered 40% of a journey, which is 84 km. Find the total length of the journey.
Answer:
If 40% of the distance d is 84 km, then d = 84 × \( \frac{100}{40} \) = 210.
The journey is 210 km long.

Teacher's Note:
1. Whole = part ÷ fraction.
2. Check: 40% of 210 = 84.

Question 16. Ravi scored 36 out of 45 in Hindi and 51 out of 60 in Mathematics. In which subject did he do better?
Answer:
Hindi: \( \frac{36}{45} \) × 100 = 80%. Mathematics: \( \frac{51}{60} \) × 100 = 85%.
He did better in Mathematics.

Teacher's Note:
1. Different maximum marks: compare percentages, not raw marks.
2. Percentages put both on the base 100.

Question 17. The price of 1 kg of onions rose from ₹25 to ₹32. Find the percentage increase in the price.
Answer:
Increase = 32 − 25 = ₹7.
Percentage increase = \( \frac{7}{25} \) × 100 = 28%.

Teacher's Note:
1. Divide by the original price ₹25.
2. Dividing by ₹32 would be wrong.

Question 18. The daily footfall in a library fell from 240 to 180. Find the percentage decrease.
Answer:
Decrease = 240 − 180 = 60.
Percentage decrease = \( \frac{60}{240} \) × 100 = 25%.

Teacher's Note:
1. The base is the earlier footfall 240.
2. 60/240 simplifies to 1/4.

Question 19. A trader buys a table lamp for ₹480 and sells it for ₹600. Find his profit percentage.
Answer:
Profit = 600 − 480 = ₹120.
Profit % = \( \frac{120}{480} \) × 100 = 25%.

Teacher's Note:
1. Profit % is always on the cost price.
2. 120/480 = 1/4 = 25%.

Question 20. A dealer buys a cycle for ₹4500 and sells it for ₹4050. Find his loss percentage.
Answer:
Loss = 4500 − 4050 = ₹450.
Loss % = \( \frac{450}{4500} \) × 100 = 10%.

Teacher's Note:
1. Selling below cost means a loss.
2. Loss % is also taken on the cost price.

Question 21. A shirt marked at ₹1200 is sold after a discount of 15%. Find its selling price.
Answer:
A 15% discount leaves 85% of the marked price.
Selling price = 0.85 × 1200 = ₹1020.

Teacher's Note:
1. Discount is on the marked price.
2. The discount itself is ₹180.

Question 22. The price of a mobile phone is ₹12,000 and a GST of 18% is added to it. What is the total amount payable?
Answer:
GST = 0.18 × 12000 = ₹2160.
Total payable = 12000 + 2160 = ₹14,160 (also 12000 × 1.18).

Teacher's Note:
1. GST is added on top of the price.
2. Multiplying by 1.18 gives the total in one step.

Question 23. Find the interest earned on ₹8000 kept for 1 year at 9% p.a. without compounding, and write the formula for the amount after t years.
Answer:
Interest = \( \frac{9}{100} \) × 8000 = ₹720.
Without compounding, the amount after t years is A = p(1 + rt), where r is the rate as a decimal.

Teacher's Note:
1. Without compounding, the same interest is earned every year.
2. Here r = 0.09.

Question 24. A laptop bought for ₹60,000 depreciates by 5% in a year. What is its value after one year?
Answer:
After a fall of 5% the laptop is worth 95% of ₹60,000.
Value = 0.95 × 60000 = ₹57,000.

Teacher's Note:
1. Depreciation is a loss of value with age and use.
2. A 5% fall means multiplying by 0.95.

Question 25. A shopkeeper’s sales last month were ₹90,000, the goods he sold had cost him ₹62,000, and his other expenses were ₹9,000. Find his gross profit and his net profit.
Answer:
Gross profit = 90000 − 62000 = ₹28,000.
Net profit = 28000 − 9000 = ₹19,000 after the other expenses are deducted.

Teacher's Note:
1. Gross profit ignores other expenses.
2. Net profit subtracts every other expense too.

Question 26. A number is increased by 20% and the result is then decreased by 20%. First say whether the final value is more or less than the original, and then find what percentage of the original it is.
Answer:
It is less than the original, because the 20% fall is taken on the larger, increased value.
Taking the number as 100: 100 × 1.2 = 120, then 120 × 0.8 = 96. So it is 96% of the original.

Teacher's Note:
1. Starting from 100 makes the answer read as a percentage.
2. 1.2 × 0.8 = 0.96.

Question 27. ₹20,000 is deposited for 2 years at 10% p.a. How much more money is received when the interest is compounded annually than when it is not?
Answer:
With compounding: 20000 × 1.1 × 1.1 = ₹24,200.
Without compounding: 20000 + 2 × 2000 = ₹24,000.
Compounding gives ₹200 more.

Teacher's Note:
1. The extra ₹200 is 10% interest on the first year’s ₹2000 interest.
2. Compounding means interest on interest.

Question 28. Goods are marked at 40% above their cost price and then sold at a discount of 30%. Find the profit or loss percentage.
Answer:
Let the cost price be x. Marked price = 1.4x.
Selling price = 0.7 × 1.4x = 0.98x.
This is 2% less than the cost price, so there is a loss of 2%.

Teacher's Note:
1. The discount is on the marked price, not on the cost.
2. Multiply the multipliers: 1.4 × 0.7 = 0.98.

Question 29. The length of a rectangle is decreased by 20%. By what percentage must its breadth be increased so that the area remains unchanged?
Answer:
New length = 0.8l. For the area to stay lb, the new breadth = \( \frac{l b}{0.8 l} \) = 1.25b.
So the breadth must be increased by 25%.

Teacher's Note:
1. The breadth multiplier is 1 ÷ 0.8.
2. 20% down needs 25% up to balance.

Question 30. Ariba has a number of marbles that is 125% of the number Arun has. What percentage of Ariba’s marbles is Arun’s number of marbles?
Answer:
If Arun has 100 marbles, Ariba has 125.
Arun’s number as a percentage of Ariba’s = \( \frac{100}{125} \) × 100 = 80%.

Teacher's Note:
1. Changing the base changes the percentage.
2. 125% and 80% are reciprocal: 1.25 × 0.8 = 1.