Comparing Quantities Class 8 Extra Questions with Answers

These Class 8 Maths Extra Questions Chapter 8 Comparing Quantities will help students prepare well for the exams.

Class 8 Maths Chapter 8 Extra Questions Comparing Quantities

Class 8 Maths Comparing Quantities Extra Questions

Question 1.
Swathi buys a new scooter on loan. The loan is for Rs. 90,000 at rate of interest 8% per annum compounded half yearly for 2 years.
How much amount would Swathi have to pay to bank at the end of 2 years ? Show your steps.
Solution:
The loan = Rs. 90,000
Rate of interest per annum = 8%
Compounded half yearly.
So, the rate of interest = \(\frac{8}{2}\) = 4%
Time = 2 years
∴ n = 2 × 2 = 4

= 144 × 26 × 26 × \(\frac{26}{25}\) × \(\frac{26}{25}\)
= \(\frac{65804544}{625}\)
Rs 1,05,287.27

Comparing Quantities Extra Questions Class 8

Question 1.
Convert the following ratios in to percentages.
i) 1 : 2
ii) 3 : 5
Solution:

Extra Questions of Comparing Quantities Class 8

Question 1.
Keerthana purchased a dress ₹ 2200 including 10% tax. Find the price of the dress before the tax was added.
Solution:
The price includes the VAT i.e. the value added tax.
Thus, a 10% VAT means if the price with out VAT is ₹ 100 then the price including VAT is ₹ 110.
Now, when price including VAT is ₹ 110,
Original price is ₹ 100
Hence, when price including tax is ₹ 2200
Original price = ₹ \(\frac{100}{110}\) × 2200 = ₹ 2000.

Comparing Quantities Class 8 Extra Questions

Question 1.
A picnic is being planned in a school for Class VII. Girls are 60% of the total number of students and are 18 in number.
The picnic site is 55 km from the school and the transport company is charging at the rate of ₹ 12 per km. The total cost of refreshments will be ₹ 4280. Can you tell.
1. The ratio of the number of girls to the number of boys in the class ?
2. The cost per head if two teachers are also going with the class ?
3. If their first stop is at a place 22 km from the school, what per cent of the total distance of 55 km is this ₹ What per cent of the distance is left to be covered ?
Solution:
1.To find the ratio of girls to boys.
Ashima and John came up with the following answers.
They needed to know the number of boys and also the total number of students.

Ashima did this
Let the total number of students
be x. 60% of x is girls.
Therefore, 60% of x = 18
\(\frac{60}{100}\) × x = 18
or, x = \(\frac{18 \times 100}{60}\) = 30
Number of students = 30.

John used the unitary method
There are 60 girls out of 100 students.
There is one girl out of \(\frac{100}{60}\) students.
So, 18 girls are out of how many students ?
Number of students = \(\frac{100}{60}\) × 18 = 30
So, the number of boys = 30 – 18 = 12.
Hence, ratio of the number of girls to the number of boys is 18 : 12 or \(\frac{18}{12}\) = \(\frac{3}{2}\)
\(\frac{3}{2}\) is written as 3 : 2 and read as 3 is to 2.

2. To find the cost per person.

Transportation charge = Distance both ways × Rate
= ₹ (55 × 2) × 12 = ₹ 110 × 12 = ₹ 1320
Total expenses = Refreshment charge + Transportation charge
= ₹ 4280 + ₹ 1320 = ₹ 5600
Total number of persons
= 18 girls + 12 boys + 2 teachers = 32 persons
Ashima and John then used unitary method to find the cost per head.
For 32 persons, amount spent would be ₹ 5600.
The amount spent for 1 person
= ₹ \(\frac{5600}{32}\) = ₹ 175.

3. The distance of the place where first stop was made = 22 km.
To find the percentage of distance :
Ashima used this method :
\(\frac{22}{55}\) = \(\frac{22}{55}\) × \(\frac{100}{100}\) = 40%
She is mutliplying the ratio by \(\frac{100}{100}\) = 1
and converting to percentage
OR
John used the unitary method :
Out of 55 km, 22 km are travelled.
Out of 1 km, \(\frac{22}{55}\) km are travelled.
Out of 100 km, \(\frac{22}{55}\) × 100 km are travelled.
That is 40% of the total distance is travelled.

Both came out with the same answer that the distance from their school of the place where they stopped at was 40% of the total distance they had to travel.
Therefore, the percent distance left to be travelled = 100% – 40% = 60%.

Question 2.
The price of a scooter was ₹ 34,000 last year. It has increased by 20% this year. What is the pride now ?
Solution:
Amita said that she would first find the increase in the price, which is 20% of ₹ 34.000. and then find the new price.
20% of ₹ 34000 = ₹ \(\frac{20}{100}\) × 34000
= ₹ 6800
New price = Old price + Increase
= ₹ 34,000 + ₹ 6,800
= ₹ 40,800
OR
Sunita used the unitary method.
20% increase means,
₹ 100 increased to ₹ 120.
So, ₹ 34,000 will increase to ?
Increased price = ₹ \(\frac{120}{100}\) × 34000
= ₹ 40,800

Similarly, a percentage decrease in price would imply finding the actual decrease followed by its subtraction the from original price.
Suppose in order to increase its sale, the price of scooter was decreased by 5%. Then let us find the price.of scooter.
Price of scooter = ₹ 34000
Reduction = 5% of ₹ 34000
= ₹ \(\frac{5}{100}\) × 34000
= ₹ 1700
New price = Old price – Reduction
= ₹ 34000 – ₹ 1700 = ₹ 32300
Discount : In other words it is called as rebate also.
It is reduction given on the Marked Price.
Amount of discount = \(\frac{\% \text { of discount }}{100}\) × Marked Price
Selling price
= Marked Price – Amount of discount

Question 3.
An item marked at ₹ 840 is sold for ₹ 714. What is the discount and discount % ?
Solution:
Discount = Marked Price – Sale Price
= ₹ 840 – ₹ 714 = ₹ 126
Since discount is on marked price, we will have to use marked price as the base.
On marked price of ₹ 840, the discount is ₹ 126.
On MP of ₹ 100, how much will the discount be ?
Discount = \(\frac{126}{840}\) × 100% = 15%
You can also find discount when discount % is given.

Question 4.
The list price of a frock is ₹ 220. A discount of 20% is announced on sales. What is the amount of discount on it and its sale price ?

Solution:
Marked price is same as the list price. 20% discount means that on ₹ 100 (MP), the discount is ₹ 20.
By unitary method, on ₹ 1 the discount will be ₹ \(\frac{20}{100}\)
On ₹ 220, discount = ₹ \(\frac{20}{100}\) × 220 = ₹ 44
The sale price = (₹ 220 – ₹ 44) or ₹ 176
Rehana found the sale price like this —
A discount of 20% means for a MP of ₹ 100, discount is ₹ 20. Hence the sale price is ₹ 80. Using unitary method, when MP is ₹ 100, sale price is ₹ 80;
when MP is ₹ 1 Sale price is ₹ \(\frac{80}{100}\)
Hence when MP is ₹ 220, sale price
= ₹ \(\frac{80}{100}\) × 220 = ₹ 176.
Even though the discount was not found, I could find the sale price directly.

Question 5.
Sohan bought a second hand refrigerator for ₹ 2,500, then spent ₹ 500 on its repairs and sold it for ₹ 3,300. Find his loss or gain per cent.
Solution:
Cost Price (CP) = ₹ 2500 + ₹ 500 (overhead expenses are added to give CP)
= ₹ 3000
Sale Price (SP) = ₹ 3300
As SP > CP, he made a profit
= ₹ 3300 – ₹ 3000 = ₹ 300
His profit on ₹ 3,000, is ₹ 300. How much would be his profit on ₹ 100?
Profit = \(\frac{300}{3000}\) × 100% = \(\frac{30}{3}\)% = 10%
P% = \(\frac{P}{C P}\) × 100

Question 6.
A shopkeeper purchased 200 bulbs for ₹ 10 each. However 5 bulbs were fused and had to be thrown away. The remaining were sold at ₹ 12 each. Find the gain or loss %.
Solution:
Cost price of 200 bulbs = ₹ 200 × 10 = ₹ 2000
5 bulbs were fused. Hence, number of bulbs left = 200 – 5 = 195
These were sold at ₹ 12 each.
The SP of 195 bulbs = ₹ 195 × 12 = ₹ 2340
He obviously made a profit (as SP > CP).
Profit = ₹ 2340 – ₹ 2000 = ₹ 340
On ₹ 2000, the profit is ₹ 340.
How much profit is made on ₹ 100 ?
Profit = \(\frac{340}{2000}\) × 100% = 17%.

Question 7.
Meenu bought two fans for ₹ 1200 each. She sold one at a loss of 5% and the other at a profit of 10%. Find the selling price of each. Also find out the total profit or loss.

Solution:
Overall CP of each fan = ₹ 1200. One is sold at a loss of 5%.
This means if CP is ₹ 100, SP = is ₹ 95.
Therefore, when CP is ₹ 1200, then
SP = ₹ \(\frac{95}{100}\) × 1200 = ₹ 1140
Also second fan is sold at a profit of 10%.
It means, if CP is ₹ 100, SP is ₹ 110.
Therefore, when CP is ₹ 1200, then
SP = \(\frac{110}{100}\) × 1200 = ₹ 1320
Was there an overall loss or gain ?
We need to find the combined CP and SP to say whether there was an overall profit or loss.
Total CP = ₹ 1200 + ₹ 1200 = ₹ 2400
Total SP = ₹ 1140 + ₹ 1320 = ₹ 2460
Since total SP > total CP, a profit of ₹ (2460 – 2400) or ₹ 60 has been made.

Question 8.
(Finding Sales Tax) The cost of a pair of roller skates at a shop was ₹ 450. The sales tax charged was 5%. Find the bill amount.

Solution:
On ₹ 100, the tax paid was ₹ 5.
On ₹ 450, the tax paid would be
= ₹ \(\frac{5}{100}\) × 450 = ₹ 22.50
Bill amount = Cost of item + Sales tax
= ₹ 450 + ₹ 22.50 = ₹ 472.50.

Question 9.
(Value Added Tax (VAT)) Waheeda bought an air cooler for ₹ 3300 including a tax of 10%. Find the price of the air cooler before VAT was added.


Solution:
The price includes the VAT, i.e., the value added tax. Thus, a 10% VAT means if the price without VAT is ₹ 100 then price including VAT is ₹ 110.
Now, when price including VAT is ₹ 110,
original price is ₹ 100.
Hence when price including tax is ₹ 3300, the original price
= ₹ \(\frac{100}{110}\) × 3300 = ₹ 3000.

Question 10.
Salim bought an article for ₹ 784 which included GST of 12%. What is the price of the article before GST was added?
Solution:
Let original price of the article be ₹ 100. GST = 12%.
Price after GST is included = ₹ (100 + 12) = ₹ 112
When the selling price is ₹ 112 then original price = ₹ 100.
When the selling price is ₹ 784, then original price = ₹ \(\frac{100}{12}\) × 784 = ₹ 700.

Question 10.
A sum of ₹ 10,000 is borrowed at a rate of interest 15% per annum for 2 years. Find the simple interest on this sum and the amount to be paid at the end of 2 years.
Solution:
On ₹ 100, interest charged for 1 year is ₹ 15.
So, on ₹ 10,000, interest charged
= \(\frac{15}{100}\) × 10000 = ₹ 1500
Interest for 2 years = ₹ 1500 × 2 = ₹ 3000
Amount to be paid at the end of 2 years = Principal + Interest
= ₹ 10000 + ₹ 3000 = ₹ 13000

Question 11.
Find CI on ₹ 12600 for 2 years at 10% per annum compounded annually.
Solution:
We have, A = P (1 + \(\frac{\mathrm{R}}{100}\))ⁿ
where Principal (P) = ₹ 12600,
Rate (R) = 10,
Number of years (n) = 2
= ₹ 12600(1 + \(\frac{10}{100}\))² = ₹ 12600(\(\frac{11}{10}\))²
= ₹ 12600 × \(\frac{11}{10}\) × \(\frac{11}{10}\) = ₹ 15246
CI = A – P
= ₹ 15246 – ₹ 12600 = ₹ 2646

Question 12.
What amount is to be repaid on a loan of ₹ 12000 for \(\frac{1}{2}\) years at 10% per annum compounded half yearly.


Formulafor Amount in compound interest
A = P((1 + \(\frac{\mathrm{R}}{100}\))ⁿ)
Where
A – Amount
P – Principal
R – Rate of Interest
n – No.of cycles that interest to be calculated
i.e. If yearly ‘n’ equal to no.of years
if half-yearly n = 2(no.of years)
if quarterly n = 4(no.of years)
if monthly n = 12 (no.of years) …………
It is important
Here we have to change rate of interest accordingly. .
Which means
If half yearly calculated then (R) becomes (\(\frac{\mathrm{R}}{2}\))
If interest calculated quarterly then (R) becomes (\(\frac{\mathrm{R}}{4}\))
If interest calculated monthly then R should be taken (\(\frac{\mathrm{R}}{12}\)) as accordingly.

Question 13.
Find CI paid when a sum of ₹ 10,000 is invested for 1 year and 3 months at 8\(\frac{1}{2}\) % per annum compounded annually.
Solution:
Mayuri first converted the time in years.
1 year 3 months = 1\(\frac{3}{12}\)year = 1\(\frac{1}{4}\) years
Mayuri tried putting the values in the known formula and came up with :
A = ₹ 1000 \(\left(1+\frac{17}{200}\right)^{1 \frac{1}{4}}\)
Now she was stuck. She asked her teacher how would she find a power which is fractional ?
The teacher then gave her a hint :
Find the amount for the whole part, i.e.,
1 year in this case. Then use this as principal to get simple interest for \(\frac{1}{4}\) year more Thus
A = ₹ 10000(1 + \(\frac{17}{200}\))
= ₹ 10000 × \(\frac{217}{200}\) = ₹ 10, 850
Now this would act as principal for the next \(\frac{1}{4}\) year. We find the SI on ₹ 10,850 for \(\frac{1}{4}\) year.
SI = ₹ \(\frac{10850 \times \frac{1}{4} \times 17}{100 \times 2}\)
= ₹ \(\frac{10850 \times 1 \times 17}{800}\) = ₹ 230.56
Interest for first year
= ₹ 10850 – ₹ 10000 = ₹ 850
And, interest for the next \(\frac{1}{4}\) year = ₹ 230.56
Therefore, total compound Interest = 850 + 230.56 = ₹ 1080.56.

Applications of Compound Interest Formula
We use the concept of compound interest, in the case of

Question 14.
The population of a city was 20,000 in the year 1997. It increased at the rate of 5% p.a. Find the population at the end of the year 2000.

Solution:
There is 5% increase in population every year, so every new year has new population. Thus, we can say it is increasing in compounded form.
Population in the beginning of 1998 = 20000 (we treat this as the principal for the 1st year)
Increase at 5% = \(\frac{5}{100}\) × 20000 = 1000
Population in 1999 = 20000 + 1000
= 21000 (Treat as the Principal for the 2nd year.)
Increase at 5% = \(\frac{5}{100}\) × 21000 = 1050
Population in 2000 = 21000 + 1050 = 22050 (Treat as the Principal for the 3rd year.)
Increase at 5% = \(\frac{5}{100}\) × 22050 = 1102.5
At the end of 2000 the population = 22050 + 1102.5 = 23152.5
or, Population at the end of 2000
= 20000(1 + \(\frac{5}{100}\))³
= 20000 × \(\frac{21}{20}\) × \(\frac{21}{20}\) × \(\frac{21}{20}\)
= 23152.5
So, the estimated population = 23153.

Aruna asked what is to be done if there is a decrease. The teacher then considered the following example.

Question 15.
A TV was bought at a price of ₹ 21,000. After one year the value of the TV was depreciated by 5% (Depreciation means reduction of value due to use and age of the item). Find the value of the TV after one year.

Solution:
Principal = ₹ 21,000
Reduction = 5% of ₹ 21000 per year
= ₹ \(\frac{21000 \times 5 \times 1}{100}\) = ₹ 1050
value at the end of 1 year = ₹ 21000 – ₹ 1050 = ₹ 19,950
Alternately,
We may directly get this as follows :
value at the end of 1 year
= ₹ 21000 (1 – \(\frac{5}{100}\))
= ₹ 21000 × \(\frac{19}{20}\) = ₹ 19,950