RBSE Solutions for Class 8 Maths Chapter 8 Comparing Quantities Intext Questions

Rajasthan Board RBSE Solutions for Class 8 Maths Chapter 8 Comparing Quantities Intext Questions Textbook Exercise Questions and Answers.

RBSE Class 8 Maths Solutions Chapter 8 Comparing Quantities Intext Questions

(Try These - Page 119)

Comparing Quantities Class 8 Try These Question 1.
In a primary school, the parents were asked about the number of hours they spend per day in helping their children to do homework. There were 90 parents who helped for \(\frac{1}{2}\) hour to 1Page 1\(\frac{1}{2}\) hours. The distribution of parents according-to. the time for which, they said they helped is given in the adjoining figure; 20% helped for more than 1\(\frac{1}{2}\) hours per day; 30% helped for \(\frac{1}{2}\) hour to 1\(\frac{1}{2}\) hours; 50% did not help at all.
Using this, answer the following:
(i) How many parents were surveyed?
(ii) How many said that they did not help?
(iii) How many said that they helped for more than 1\(\frac{1}{2}\) hours?

Answer:
(i) Let x parents be surveyed, 30% of x helped for \(\frac{1}{2}\) hour to 1\(\frac{1}{2}\) hours.
∴ 30% of A: = 90
\(\frac{30}{100}\) × x = 90
or
x = \(\frac{90 \times 100}{30}\) = 300
Number of parents = 300

(ii) 50% parents did not help.
Number of parents did not help
= 50% of 300
= (300 × \(\frac{50}{100}\)) = 150

(iii) 20% parents helped for more than 1\(\frac{1}{2}\) hours.
∴ Number of such parents = 20% of 300
= (300 × \(\frac{20}{100}\)) = 60

(Try These - Page 121)

Class 8 Maths Comparing Quantities Try These Solutions Question 1.
A shop gives 20% discount What would the sale price of each of these be?
(a) A dress marked at ₹ 120
(b) A pair of shoes marked at ₹ 750
(c) A bag marked at ₹ 250
Answer:
(a) Given,
Marked price = ₹ 120,
Discount = 20%
∴ Discount = 20% of ₹ 120
= ₹ (\(\frac{200}{100}\) × 120) = ₹ 24
∴ S.P. = M.P. - Discount
= ₹ (120 - 24) = ₹ 96.

(b) Given,
Marked Price = ₹ 750,
Discount = 20%
∴ Discount = 20% of ₹ 750
S.P. = M.P. - Discount
= ₹ (750 - 150) = ₹ 600.

(c) Given,
Marked priee = ₹ 250,
Discount = 20%

Discount = 20% of ₹ 250
= ₹ (\(\frac{20}{100}\) × 250) = ₹ 50
∴ S.P = M.P. - Discount
= ₹ (250 - 50) = ₹ 200.

Class 8 Maths Ch 8 Try These Question 2.
A table marked at ₹ 15,000 is available for ₹ 14,400. Find the discount given and the discount percent.
Answer:
Marked price = ₹ 15000
Discount = ₹ 14400

∴ Discount = ₹ (15000 - 14400)
= ₹ 600

∴ Rate of discount = (\(\frac{600}{15000}\) × 100)%
= 4%

Class 8 Comparing Quantities Try These Question 3.
An almirah is sold at ₹ 5,225 after allowing a discount of 5%. Find its marked price.
Answer:
Let the marked price be ₹ 100 Discount = 5% of marked price = 5% of ₹ 100 = ₹ 5
∴ S.P. = M.P. - Discount
= ₹ (100 - 5) = ₹ 95
Now, when S.P. is ₹ 95, M.P. = ₹ 100
When S.P. is ₹ 1, M.P. = ₹ \(\frac{100}{95}\)
When S.P. is ₹ 5225, M.P. = ₹ (\(\frac{100}{95}\) × 5225)
= ₹ 5500
Hence the marked price of almirah is ₹ 5500.

(Try These - Page 123)

Class 8 Maths Chapter 8 Try These Question 1.
Find selling price (SP) if a profit of 5% is made on
(a) a cycle of ₹ 700 with ₹ 50 as overhead charges.
(b) a lawn mower bought at ₹ 1150 with ₹ 50 as transportation charges.
(c) a fan bought for ₹ 560 and expenses of ₹ 40 made on its repairs.
Answer:
(a) Cost price of cycle = ₹ 700 overhead expenses = ₹ 50
Effective cost price = ₹ (700 + 50)
= ₹ 750 Profit = 5%
∴ S.P. = \(\left(\frac{100+\text { Profit } \%}{100}\right)\) × CP
= ₹ (\(\frac{100+5}{100}\) × 750)
= ₹ (\(\frac{105}{100}\) × 750)
= ₹ 787.50

(b) Cost price of a lawn mower = ₹ 1150 overhead expenses = ₹ 50
∴ Effective cost price = ₹ (1150 + 50)
= ₹ 1200
Profit = 5%
S.P. = \(\left(\frac{100+\text { Profit } \%}{100}\right)\) × CP
= ₹ (\(\frac{105}{100}\) × 12000
= ₹ (105 × 12)
= ₹ 1260

(c) Cost price of fan = ₹ 560 overhead expenses = ₹ 40
Effective cost price = ₹ (560 + 40)
= ₹ 600
Profit = 5%
S.P. = \(\left(\frac{100+\text { Profit } \%}{100}\right)\) × CP
= ₹ (105 × 6)
= ₹ 630

(Try These - Page 123)

Try These Comparing Quantities Class 8 Question 1.
A shopkeeper bought two TV sets at ₹ 10,000 each. He sold one at a profit 10% and the other at a loss of 10%. Find whether he made an overall profit or loss.
Answer:
Cost price of each TV = ₹ 10,000 one is sold at a profit of 10%
If C.P. = ₹ 100,
S.P. = ₹110 [∵ S.P. = C.P. + Profit = TOO + 10% of 100 = 110]
Therefore when C.P. is ₹ 10,000
Then, S.P = ₹ \(\frac{110}{100}\) × 10,000 = ₹ 11,000

In second case, TV at a loss of 10%
⇒ If C.P. is ₹ 100, S.P. is ₹ 90 [∵ S.P. = C.P - Loss = 100 - 10% of 100 = 90]
Therefore, when C.P. is ₹ 10,000
Then, S.P. = ₹ \(\frac{90}{100}\) × 10,000 = ₹ 9,000
Total C.P. of two sets of T. V.
= ₹ 10,000 + ₹ 10,000 = ₹ 20,000
Total S.P. of two sets of T.V.
= ₹ 11,000 + ₹ 9,000 = ₹ 20,000

∵ Total S.P. = Total C.P.
∴ He neither gains nor loss.

(Think, Discuss And Write - Page 125)

Try These Solutions Class 8 Maths Chapter 8 Question 1.
Two times a number is a 100% increase in the number. If we take half the number what would be the decrease in per-cent?
Answer:
Let the number be x. Then its half = \(\frac{x}{2}\)
Decrease % = (\(\frac{\text { Decrease }}{\text { Original Value }}\) × 100)%
= (\(\frac{x / 2}{x}\)× 100)%
= 50%

Class 8 Ch 8 Try These Question 2.
By what percent is Rs 2,000 less than ₹ 2,400? Is it the same as the percent by which ₹ 2,400 is more than ₹ 2,000?
Answer:
1st case :
Decrease % = (\(\frac{\text { Decrease }}{\text { Original Value }}\) × 100)%
= (\(\frac{400}{2400}\) × 100)% = \(\frac{50}{3}\)%
= 16\(\frac{2}{3}\)%

2nd case:
Increase % = (\(\frac{\text { Increase }}{\text { Original Value }}\) × 100)%
= (\(\frac{400}{2000}\) × 100)% = 20%
No, they are not the same.

(Try These - Page 126)

Class 8 Maths Chapter 8 Try These Solutions Question 1.
Find interest and amount to be paid on ₹ 15,000 at 5% per annum after 2 years.
Answer:
Given, P = ₹ 15,000, R = 5% p.a. and T = 2 years
Let I be the interest and A be the amount. Then,
I = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
Or
I = ₹ \(\left(\frac{15,000 \times 5 \times 2}{100}\right)\)
= ₹ 15000
and A = P + I
or A = ₹ (15,000 + 1500)
= ₹ 16,500

(Try These - Page 129)

Class 8 Maths Chapter 8 Try These Page 121 Question 1.
Find Cl on a sum of ₹ 8000 for 2 years at 5% per annum compounded annually.
Answer:
Given, P = ₹ 8000, R = 5% p.a. and n = 2 years
Amount after 2 years = P\(\left(1+\frac{\mathrm{R}}{100}\right)^{n}\)
= ₹(8000 × \(\left(1+\frac{5}{100}\right)^{2})\)
= ₹ (20 × \(\frac{21}{20} \times \frac{21}{20}\))
= ₹ (20 × 21 × 21)
= ₹ 8820
and, compound Interest = A - P
= ₹ (8820 - 8000)
= ₹ 820

(Try These - Page 130)

Ch 8 Maths Class 8 Try These Solutions Question 1.
Find the time period and rate for each.
1. A sum taken for li years at 8% per annum is compounded half yearly.
Answer:
Given,
Rate of Interest = 8% p.a.
= 4% per half-year
Time = 1\(\frac{1}{2}\) years
= 3 half-years

2. A sum taken for 2 years at 4% per annum compounded half yearly.
Answer:
Given,
Rate of Interest = 4% p.a.
= 2% per half-year
Time = 2 years = 4 half-years

(Think, Discuss And Write - Page 130)

Class 8 Maths Comparing Quantities Try These Question 1.
A sum is taken for one year at 16% p.a. If interest is compounded after every three months, how many times will interest be charged in one year?
Answer:
Given,
Rate of interest = 16% p.a. = \(\frac{16}{4}\)% per quarter
= 4% per quarter Time = 1 year = 4 quarter
Thus interest will be charged 4 times in one year at 4% per quarter.

(Try These - Page 131)

Try These Solutions Class 8 Comparing Quantities Question 1.
Find the amount to be paid
1. At the end of 2 years on ₹ 2,400 at 5% per annum compounded annually.
Answer:
Given, P = ₹ 2400, R = 5% p.a. and n = 2 years.
∴ Amount after 2 years = P\(\left(1+\frac{R}{100}\right)^{n}\)
= ₹ [2400 × \(\left(1+\frac{5}{100}\right)^{2}\)]
= ₹ [2400 × 21 × 21]
= ₹ (6 × 21 × 21)
= ₹ 2646

2. At the end of 1 year on ₹ 1,800 at 8% per annum compounded quarterly.
Answer:
Given, Principal = ₹ 1800 Rate = 8% p.a. = 2% per quarter Time = 1 year = 4 quarter
Amount = ₹ [1800 × \(\left(1+\frac{2}{100}\right)^{4}\)]
= ₹ (1800 × \(\frac{51}{50} \times \frac{51}{50} \times \frac{51}{50} \times \frac{51}{50}\))
= ₹ \(\frac{121773618}{62500}\)
= ₹ 121773618

(Try These - Page 133)

Class 8 Chapter 8 Try These Question 1.
A machinery worth ₹ 10,500 depreciated by 5%. Find its value after one year.
Answer:
Here, P = ₹ 10,500
Reduction = 5% p.a., n = 1
Reduced value after 1 year
= ₹ [10,500 × \(\left(1-\frac{5}{100}\right)^{1}\)]
= ₹ (10,500 × \(\frac{95}{100}\))
= ₹ 9975

Try These Class 8 Chapter 8 Question 2.
Find the population of a city after 2 years, which is at present 12 lakh, if the rate of increase is 4%.
Answer:
Let the population after 2 years be P₂
Then. P₂ = P × \(\left(1+\frac{4}{100}\right)^{2}\)
= 12,00,000 × \(\frac{26}{25} \times \frac{26}{25}\)
= 1920 × 676
= 12,97,920
Hence, the population after 2 years = 12,97,920