Fractions in Disguise Class 8 Extra Questions Maths Part 2 Chapter 1 - #NCSOLVE 📚

During revision, students quickly go through Class 8 Maths Extra Questions Part 2 Chapter 1 Fractions in Disguise Class 8 Extra Questions with Answers for clarity.

Class 8 Fractions in Disguise Extra Questions

Class 8 Maths Chapter 1 Fractions in Disguise Extra Questions

Fractions in Disguise Extra Questions Class 8

Question 1.
Find:
(a) 30% of 500 apples
(b) 6\(\frac {1}{4}\)% of ₹ 160
Solution:
(a) 30% of 500 apples = \(\frac {30}{100}\) × 500 = 150 apples
(b) 6\(\frac {1}{4}\)% of ₹ 160 = \(\frac{25}{4 \times 100} \times 160\) = ₹ 10

Question 2.
A football team won 12 matches out of the total number of matches they played. If their win percentage was 60, how many matches did they play in all?
Solution:
Let the total number of matches played be Q.
Here, 60% of Q = 12 matches
⇒ \(\frac {60}{100}\) × Q = 12
⇒ Q = \(\frac{12 \times 100}{60}\) = 20
Thus, the team played 20 matches in all.

Question 3.
An Android handset, when launched, was sold for ₹ 48,000. After six months, it was sold for ₹ 45,600. Determine the percentage reduction in the price of the mobile phone.
Solution:
The initial price of the handset = ₹ 48,000
The price of the handset after six months = ₹ 45,600
Reduction in the price of the mobile phone = ₹ 48,000 – ₹ 45,600 = ₹ 2400
Percentage decrease in price = \(\left(\frac{\text { Decrease in price }}{\text { Initial price }} \times 100\right) \%\)
= \(\left(\frac{2400}{48000} \times 100\right) \%\)
= 5%

Question 4.
By selling a jacket for ₹ 1600, Seema incurred a loss of 20%. At what price should she sell the jacket to gain 10%?
Solution:
SP of the jacket = ₹ 1600
Percentage of loss = 20%
Therefore, its CP = \(\left(\frac{100}{100-\mathrm{L} \%}\right) \times \mathrm{SP}\)
= \(\left(\frac{100}{100-20}\right) \times ₹ 1600\)
= \(\frac {100}{80}\) × ₹ 1600
= ₹ 2000
Now, if percentage of gain = 10%, then
SP = \(\left(\frac{100+\mathrm{P} \%}{100}\right) \times \mathrm{CP}\)
= \(\left(\frac{100+10}{100}\right) \times ₹ 2000\)
= \(\frac {110}{100}\) × ₹ 2000
= ₹ 2200
Thus, Seema should sell the jacket for ₹ 2200 to have a gain of 10%.

Question 5.
Fatima sold two coolers for ₹ 7980 each. On selling one cooler, she incurred a loss of 5%, and on selling the other, she gained 5%. Find her overall gain or loss % in the whole transaction.
Solution:
SP of the first cooler = ₹ 7980
Percentage of loss = 5%.
Therefore, its CP = \(\left(\frac{100}{100-\mathrm{L} \%}\right) \times \mathrm{SP}\)
= \(\left(\frac{100}{100-5}\right) \times ₹ 7980\)
= \(\frac {100}{95}\) × ₹ 7980
= ₹ 8400
Now, SP of the second cooler = ₹ 7980
Percentage of profit = 5%.
Therefore, its CP = \(\left(\frac{100}{100+\mathrm{P} \%}\right) \times \mathrm{SP}\)
= \(\left(\frac{100}{100+5}\right) \times ₹ 7980\)
= \(\frac {100}{105}\) × ₹ 7980
= ₹ 7600
Total SP of the two coolers = ₹ 7980 × 2 = ₹ 15,960
Total CP = ₹ 8400 + ₹ 7600 = ₹ 16,000
Since, total CP > total SP, therefore, a loss of ₹ (16,000 – 15,960) = ₹ 40 has been made.
Thus, Loss% = \(\left(\frac{\text { Loss }}{\mathrm{CP}} \times 100\right) \%\)
= \(\left(\frac{40}{16000} \times 100\right) \%\)
= \(\frac {1}{4}\)%
= 0.25%

Question 6.
The list price of a cooler is ₹ 12,508. The rate of GST is 18%. The customer requests the shopkeeper to reduce the price of the cooler (i.e., allow a discount) so that the final price, including GST, remains ₹ 12,508. Find the reduction needed in the price of the cooler.
Solution:
List price of the cooler = ₹ 12,508
Let the reduced price of the cooler be ₹ x
Selling price of the cooler = ₹ x + 18% GST on ₹ x
⇒ 12,508 = x + (\(\frac {18}{100}\) × x)
⇒ 12,508 = \(\frac {118x}{100}\)
⇒ x = \(\frac{12,508 \times 100}{118}\)
⇒ x = ₹ 10,600
So, the reduction needed in the price of the cooler = ₹ (12,508 – 10,600) = ₹ 1908

Question 7.
Find the interest (with compounding and without compounding) on a sum of ₹ 6000 borrowed for 4 years at a rate of interest of 7% per annum.
Solution:
Given, Principal (P) = ₹ 6000
Time period (t) = 4 years
Rate of interest (R) = 7% p.a.
Without Compounding
S.I = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}(t)}{100}\)
= \(\frac{6000 \times 7 \times 4}{100}\)
= ₹ 1680
With compounding, we have
A = \(\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^t\)
= \(6000\left(1+\frac{7}{100}\right)^4\)
= \(6000\left(\frac{107}{100}\right)^4\)
= \(6000 \times \frac{107}{100} \times \frac{107}{100} \times \frac{107}{100} \times \frac{107}{100}\)
= ₹ 7864.776
= ₹ 7864.78
Now, C.I. = A – P
= 7864.78 – 6000
= ₹ 1864.78
Thus, the interest is ₹ 1864.78 with compounding and ₹ 1680 without compounding.

Question 8.
Arvind won some prize money in a contest. He wants to donate \(\frac {2}{5}\) of the money. Express this quantity as a percentage.
Solution:
Percentage of money to be donated = \(\frac {2}{5}\) × 100% = 40%

Question 9.
Express 24% as a fraction.
Solution:
24% = \(\frac{24}{100}=\frac{6}{25}\)

Question 10.
Find
(a) 50% of 240
(b) 25% of 484
(c) 20% of 360
(d) 12\(\frac {1}{2}\)% of 800
(e) 10% of 2350
(f) 5% of 260
(g) 2% of 550
(h) 1% of 7200
(i) 13% of 2300
Solution:
(a) 50% of 240 = 240 ÷ 2 = 120
(b) 25% of 484 = 484 ÷ 4 = 121
(c) 20% of 360 = 360 ÷ 5 = 72
(d) 12\(\frac {1}{2}\)% of 800 = 800 ÷ 8 = 100
(e) 10% of 2350 = 2350 ÷ 10 = 235
(f) 5% of 260 = 260 ÷ 20 = 13
(g) 2% of 550 = 550 ÷ 50 = 11
(h) 1% of 7200 = 7200 ÷ 100 = 72
(i) 13% of 2300 = (2300 ÷ 100) × 13
= 23 × 13
= 299

Question 11.
A cyclist cycles from Meerut to Mathura (180 km) and completes 40% of the journey. How many more kilometres does he have to travel to reach Mathura?
Solution:
Distance covered = 40% of 180 km
= \(\frac {40}{100}\) × 180 km
= 72 km
Distance to be covered = (180 – 72) km = 108 km

Question 12.
A bicycle costing ₹ 1280 is sold for ₹ 1344. Find the profit percent.
Solution:
C.P = ₹ 1280, S.P = ₹ 1344
Here S.P > C.P
So, Profit = ₹ (1344 – 1280) = ₹ 64
P% = \(\frac {64}{128}\) × 100% = 5%

Question 13.
A typewriter was bought for ₹ 760 and was sold for ₹ 722. Find the loss percent.
Solution:
C.P = ₹ 760, S.P = ₹ 722
S.P < C.P
So, Loss = ₹ 760 – ₹ 722 = ₹ 38
Loss % = \(\frac {38}{760}\) × 100% = 5%

Question 14.
Find the C.P. of an item sold for ₹ 1265 at a gain of 15%.
Solution:
S.P = ₹ 1265, P% = 15%
C.P = \(\frac{100}{100+15} \times 1265\)
= \(\frac{100 \times 253}{23}\)
= ₹ 1100

Question 15.
Find the S.P. of a table costing ₹ 1200 if it was sold at a profit of 10%.
Solution:
C.P = ₹ 1200, P% = 10%
S.P = \(\frac{100+10}{100} \times 1200\) = ₹ 1320

Question 16.
(a) In a sale at item listed at ₹ 300 is sold at a discount of 20%. Find the sale price of the item.
(b) In a sale, an item listed at ₹ 500 is sold for ₹ 440. Find the discount percent.
(c) In a sale, an item was sold for ₹ 630 after a discount of 10%. Find the list price of the item.
Solution:

Question 17.
List price of a car is ₹ 5,50,000. GST is applied @28%. Find the GST to be paid and the cost of the car, including GST.
Solution:
GST = \(\frac {28}{100}\) × 5,50,000 = ₹ 1,54,000
Price of car including GST = ₹ 5,50,000 + ₹ 1,54,000 = ₹ 7,04,000

Question 18.
Find:
(a) SI if P = ₹ 500, T = 2 years and R = 15% p.a.
(b) P if SI = ₹ 288, T = 3 years and R = 8% p.a.
(c) R if P = ₹ 4,250, T = 5\(\frac {1}{2}\) years and SI = ₹ 3,740
(d) T if P = ₹ 3,250, R = 15% and SI = ₹ 975.
Solution:

Question 19.
Puja deposited ₹ 12,000 in a bank at 10% per annum for 3 years. Find the amount received by her if the interest is compounded annually.
Solution:
We have annually compounded interest.
Given, P = ₹ 12,000, r = 10% p.a, n = 3 years

Hence, Puja received ₹ 15,972 from the bank.

Question 20.
The present population of a town is 48,000. If it increases at the rate of 10% p.a., find its population after 3 years.
Solution:
We have, P = ₹ 48,000, r = 10% p.a, n = 3 years
We know the final value of growth

Hence, the population of the town after 3 years will be ₹ 63,888.

Question 21.
The value of a machinery depreciates by 15% per year. If its present value is ₹ 52,020, what was its value 2 years ago?
Solution:
We have, A = ₹ 52,020, r = 15% p.a, n = 2 years
We know, A = \(\mathrm{P}\left(1-\frac{r}{100}\right)^n\)
Putting values,

Hence, the value of the machinery 2 years ago was ₹ 72,000.

The post Fractions in Disguise Class 8 Extra Questions Maths Part 2 Chapter 1 appeared first on Learn CBSE.

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