Maths Chapter-08: Comparing Quantities
EXERCISE- 8.1:
QUESTION 1:
Calculate the ratio:
(i) Rs10 to 50 paise (ii) 10kg to 500g
(iii) 60m to 20cm (iv) 20days to 50 hours
Solution;
(i) Rs.10 to 50 paise
=> 10 x 100 = 1000 paise (Since , Rs.1 = 100 paise)
Thus, the ratio = 1000/50 = 20 : 1
(ii) 10kg to 500g
10 x 1000= 10000g (Since 1kg is 1000g)
Thus, the ratio = 10000/500 = 20 : 1
(iii) 60m to 20cm
60 x 100 = 6000cm (Since 1m = 100cm)
Thus, the ratio =6000/20 = 300 : 1
(iv) 20days to 50 hours
20 x 24 = 480 hours
Thus, the ratio = 480/50 = 48 : 5
QUESTION 2:
A computer lab has 2 computers for every 4 students. For 20 students how many computers will be required?
Solution;
Given,
4 students = 2 computers
1 student requires = 2/4 computers.
Therefore, 20 students require = (2/4) x 20 = 10 computers.
QUESTION 3:
Mizoram has a population of 10 lakh and Sikkim has a population of 6 lakh. Area of Mizoram is 21,000 km2 and the area of Sikkim is 7000 km2.
(i) Find the number of people per km2 in both the states.
(ii) Which state has a lesser population?
Solution;
(i) People per km2 = Population/ Area
In Mizoram = 1000000/21000 = 47.61 ?48 people/km2
In Sikkim = 600000/7000 = 85.71 ? 86 people/km2
(ii) Sikkim has a lesser population.
EXERCISE- 8.2
QUESTION 1:
Convert the given fractional numbers to percent:
(a) 1/3 (b)6/5 (c)4/25 (d)3/7
Solution;
(a) 1/3 = (1/3) x 100% = 100/3 = 33.33%
(b) 6/5 = (6/5) x 100% = 6 x 20 = 120%
(c) 4/25 = (4/25) x 100% = 16%
(d) 2/7 = (2/7) x 100% = 28.57%
QUESTION 2:
Express the given decimal numbers in percentages.
(a) 0.31 (b)0.03 (c)19.62 (d)0.007
Solutions;
(a) 0.31 = (31/100) x 100% = 31%
(b) 0.03 = (3/100) x 100% = 3%
(c) 19.62 = (1962/100) x 100% = 1962%
(d) 0.007 = (7/1000) x 100% = 0.7%
QUESTION 3:
Find what part of the following figures is colored; also find the percentage of the colored part.
Solutions;
(i) Colored part = 1/4
Therefore, percentage of the colored portion = (1/4) x 100% = 25%
(ii) Colored portion = 3/5
Therefore, percentage of the colored portion = (3/5) x 100 =60%
(iii) Colored portion = (3/8) x 100% = 37.5%
QUESTION 4:
Calculate :
(a) 15% of 450 (b) 2% of 2 hour
(c) 10% of ?2600 (d) 50% of 5 kg
Solutions;
(a) 15% of 450 = (15/100) x 450 = 67.5
(b) 2% of 2 hours= 2% of (2×3600) secs (Since 1 hour = 60 x 60 secs)
= (2/100) x 7200 =144secs.
(c) 10% of Rs.2600 = (10/100) x 2600
=Rs.260
(d) 50% of 5kg = 50% of (5000g) (Since 1kg = 1000g)
= (50/100) x 5000
= 2500 g or 2.5kg
QUESTION 5:
Calculate the total quantity if:
(a) 5% of it is 100
(b) 10% of it is ?1060
(c) 20% of it is 800 km
(d) 90% of it is 18 minutes
(e) 9% of it is 60 liters
Solutions;
let the total quantity be Y in all above cases,
(a)5% of Y = 100
=> (5/100) x Y =100
=> Y=(100×100)/5 =2000
(b)10% of Y = 1060
=> (10/100) x Y = 1060
=> Y = (1060 x 100)/10 = Rs.10600
(c)20% of Y = 800
=> (20/100) x Y = 800
=> Y = (800 x 100)/20 = 4000km
(d) 90% of Y = 18min
=> (90/100) x Y = 18
=> Y=(18 x 100)/90 =20mins.
(e) 9% of Y = 60
=> (9/100) x Y = 60
=> Y = (60 x 100)/9 =666.66liters.
QUESTION 6:
Convert the following percentages to fractions , simplest fractions and decimal numbers:
(a)20% (b)25% (c)5% (d)120%
Solution;
| Percentage | Fractions | Simplest Fraction | Decimal Numbers. |
| 20% | 20/100 | 1/5 | 0.2 |
| 25% | 25/100 | 1/4 | 0.25 |
| 5% | 5/100 | 1/20 | 0.05 |
| 120% | 120/100 | 6/5 | 1.2 |
QUESTION 7:
12000 voters constituted a constituency, out of which 80% voted. What percentage and number of voters did not vote?
Solution;
Given,
Total no. of voters = 12000
Percentage who voted = 80%
Thus, percentage who did not vote = 100 -80 = 20%
Therefore, the number of people who did not vote = 20% of 12000
= (20/100) x 12000 = 2400
QUESTION 8:
Sanjay saves Rs.5000 every month, if this is 20% of his salary. What is his monthly salary?
Solution;
Let Sanjay’s salary be Y
Now, 20% of Y = 5000
(20/100) x Y = 5000
Y = Rs.25000
Therefore,Sanjay’s salary is Rs.25000
QUESTION 9:
A localdota team played 10 matches in a season. If they won 40% of their matches, in how many matches do they win?
Solution;
Given,
Number of matches played by the dota2 team =10
Percentage of matches won = 40%
Therefore, total matches they were victorious in = 40% of 10
= (40/100) x 10
=4.
EXERCISE- 8.3
QUESTION 1:
What is the profit and theloss for the following transaction.Also find what is the profit percentage and the loss percentage in the cases given:
(i) Gardening shears were bought for Rs 250 and were sold for Rs 325
(ii) A refrigerator was bought for Rs 12,000 and sold for Rs 13,500
(iii) A cupboard was bought for Rs 2,500 and sold for Rs 3,000.
(iv) A shirt was bought for Rs 250 and sold for Rs 150
Solution;
(i)the cost price of the gardening shears: Rs 250
The selling price of the gardening shears: Rs 325
Since, SP> CP therefore here it is a profit.
Therefore profit = SP – CP = Rs 325 – Rs 250
= Rs 75
Now profit % = \(\frac{profit}{C.P}\times 100\)
= \(\frac{75}{250}\times 100=\) %
Therefore, Profit = Rs 75 and Profit % = 30%
(ii) Cost prince of refrigerator = Rs 12,000
Selling price of Refrigerator = Rs 13,500
Since, S.P > C.P Therefore, it is a profit.
Therefore, Profit = SP – CP = Rs 13500 – Rs 12000 = Rs 1500
Now profit % = \(\frac{profit}{C.P}\times 100\)
= \(\frac{1500}{12000}\times 100\) = 12.5%
Therefore, Profit = Rs 1,500 and Profit % = 12.5 %
(iii) Cost price of cupboard = Rs 2500
Selling price of cupboard = Rs 3,000
Since, SP> CP, Therefore it is a profit.
Profit = SP – CP = Rs 3,000 – Rs 2,500 = Rs 500
\(\frac{1500}{12000}\times 100\)
= \(\frac{500}{2500}\times 100\) =20%
Therefore, Profit = 500 and profit % = 20%
(iv) Cost price of shirt = Rs 250
Selling price of shirt = Rs 150
Since CP > SP, therefore here it is a loss.
Therefore Loss = CP – SP = Rs 250 – Rs 150
= Rs 100
Now loss % = \(\frac{Loss}{C.P}\times 100\)
= \(\frac{100}{250}\times 100\) = 40%
Therefore , Loss = Rs 100 and Loss% = 40%
QUESTION 2:
Convert the following ratio to its percentages:
(i) 3 : 1
(ii) 2 : 3 : 5
(iii) 1 : 4
(iv) 1 : 2 : 5
Solution;
(i)3 : 1
Total part = 3 + 1 = 4
Therefore, Fractional part = \(\frac{3}{4}:\frac{1}{4}\)
=> Percentage of parts = \(\frac{3}{4}\times 100:\frac{1}{4}\times 100\)
=> Percentage of parts = 75 % : 25 %
(ii)2 : 3 : 5
Total part = 2 + 3 + 5 = 10
Therefore, Fractional part = \(\frac{2}{10}:\frac{3}{10}:\frac{5}{10}\)
=> Percentage of parts = \(\frac{2}{10}\times 100:\frac{3}{10}\times 100:\frac{5}{10}\times 100\)
=> Percentage of parts = 20 % : 30 % : 50 %
(iii)1 : 4
Total part = 1 + 4 = 5
Therefore, Fractional part = \(\frac{1}{5}:\frac{4}{5}\)
=> Percentage of parts = \(\frac{1}{5}\times 100:\frac{4}{5}\times 100\)
=> Percentage of parts = 20% : 80%
(iv)1 : 2 : 5
Total part = 1 + 2 + 5 = 8
Therefore, Fractional part = \(\frac{1}{8}:\frac{2}{8}:\frac{5}{8}\)
=> Percentage of parts = \(\frac{1}{8}\times 100:\frac{2}{8}\times 100:\frac{5}{8}\times 100\)
=> Percentage of parts = 12.5 % : 25 % : 62.5 %
QUESTION 3:
The population of a city decreased from 25,000 to 24,500. Find the decrease in percentage.
Solution;
The decrease in population of a city is from 25,000 to 24,500.
Population Decreased = 25000 – 24500 = 500
Decreased percentage = \(\frac{Population Decreased}{Original Population}\times 100\)
= \(\frac{500}{25000}\times 100\) = 2 %
Hence, the percentage of decrease in the population is 2 %.
QUESTION 4:
Reah and Reuben bought a caravanfor Rs. 3,50,000. The next year, the price raised to Rs 3,70,000. What is the percentage of price increased?
Solution;
Increased in price of the caravan from Rs 3, 50,000 to 3, 70,000
Amount change = Rs 3, 70,000 – Rs 3, 50,000 = Rs 20,000
Therefore, Increased percentage = \(\frac{Amount of change}{Original Amount}\times 100\)
\(\frac{20000}{350000}\times 100 = 5\frac{5}{7}\) %
Hence, The percentage of price increased is \(5\frac{5}{7}\) %
QUESTION 5:
Harismitha bought a T.V for Rs 10,000 and sold it at a profit of 20%. How much money did she get for it?
Solution;
The cost price of T.V = Rs 10000
Profit percent = 20%
Now, Profit = Profit % of CP
= \(\frac{20}{100}\times 10000\)
= Rs 2000
Selling price = CP + Profit = Rs 10000 + Rs 2000 = Rs 12000
She gets Rs 12000 on selling the T.V
QUESTION 6:
Nancy sold a washing machine for Rs 13500. She Lost 20% in the bargain. What is the price that she bought the washing machine for?
Solution;
Selling price of washing machine= Rs 13,500
Loss percent = 20%
Let the cost price of washing machine be Rs x
Since, Loss = Loss % of CP
=> Loss = 20 % of Rs x = \(\frac{20}{100}\times x=\frac{x}{5}\)
Therefore, SP = CP – Loss
\(\\=>13500=x-\frac{x}{5}\\=>13500=\frac{4x}{5}\\=>x=\frac{13500\times 5}{4}=Rs16,875\)
Hence, The cost price of washing machine is Rs 16,875.
QUESTION 7:
(a) Chalk contains Calcium, Carbon and Oxygen in the ratio 10: 3 : 12. Find the percentage of carbon in chalk.
(b) If in a stick of chalk, Carbon is 3g, what is the weight of the chalk stick?
Solution;
(a) Given ratio = 10 : 3 : 12
Total part = 10 + 3 + 12 = 25
Part of carbon = \(\frac{3}{25}\)
Percentage of Carbon part in Chalk = \(\frac{3}{25}\) x 100 = 12 %
(b) Quality of Carbon in chalk stick = 3g
Let the weight of chalk be x g.
Then, 12 % of x = 3
=> \(\frac{12}{100}\) x X = 3
=> x = \(\frac{3\times 100}{12}\) = 25g
Hence, the weight of chalk stick is 25g.
QUESTION 8:
Amulya buys a book for Rs 275 and sells it at a loss of 15% . How much does she sell it for?
Solution;
The cost of a book = Rs 275
Loss percentage = 15 %
Loss = Loss % of CP = 15 % of 275
= \(\frac{15}{100}\times 275\) = Rs 41.25
Therefore , SP = CP – Loss = 275 – 41.25 = Rs 233.75
Hence, Amulya sold the book for Rs 233.75
QUESTION 9:
Find the amount to be paid at the end of 3 years in the following cases:
(i ) Principal = Rs 1200 at 12 % p. a
(ii) Principal = Rs 7500 at 5 % p. a
Solution;
(i) Here, Principal (p) = 1200
Rate ( r ) = 12 % p a
Time ( T ) = 3 years
\(Simple interest = \frac{P \times R\times T}{100} = \frac{1200\times 12\times 3}{100}\)
=432
Now, Amount = Principal + Simple Interest
= Rs 1200 + Rs 432
= Rs 1632
(ii) Here, Principal (p) = 7500
Rate ( r ) = 5 % p a
Time ( T ) = 3 years
\(Simple interest = \frac{P \times R\times T}{100} = \frac{7500\times 5\times 3}{100}\)
= 1125
Now, Amount = Principal + Simple Interest
= Rs 7500 + Rs 1125
= Rs 8625
QUESTION 10:
What rate gives Rs 280 as interest on a sum onRs 56,000 in 2 years?
Solution;
Here, Principal (P ) = Rs 56000
Simple Interest (SI ) = Rs 280
Time ( T) = 2 yrs
Simple interest = \(\frac{P \times R\times T}{100}\)
=> 280 = \(\frac{56000\times R\times 2}{100}\)
=> R = \(\frac{280\times 100}{56000\times 2}\)
=> Hence, the rate of interest on sum is 0.25%
QUESTION 11:
If Deepak gives an interest of Rs 45 for one year at 9% rate per annum. What is the sum he has borrowed from Nancy?
Solution;
Simple Interest = Rs 45
Rate (R ) = 9%
Time ( T) = 1 year
Simple interest = \(\frac{P \times R\times T}{100}\)
=> 45 = \(\frac{P\times 9\times 1}{100}\)
=> P = \(\frac{45\times 100}{9}\)
=> P = Rs 500
Hence, she had borrowed Rs 500.