RS Aggarwal solution class 8 chapter 9 Percentage Exercise 9A

Exercise 9A

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Q1 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

Question 1:

Express each of the following as a fraction:
(i) 48%
(ii) 220%
(iii) 2.5%

Answer 1:

i 48%=48100=1225

ii 220%=220100=115

iii 2.5%=2.5100=251000=140


Q2 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

Question 2:

Express each of the following as a decimal:
(i) 6%
(ii) 72%
(iii) 125%

Answer 2:

i 6%=6100=0.06

ii 72%=72100=0.72

iii 125%=125100=1.25


Q3 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

Question 3:

Express each of the following as a percentage:
(i) 925
(ii) 3125
(iii) 125

Answer 3:

i 925=(925×100)%=(9×4)%=36%


ii 3125=(3125×100)%=2.4%

iii 125=(125×100)%=240%


Q4 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

Question 4:

Convert the ratio 4 : 5 to percentage.

Answer 4:

4 :5 

=45

=  45×100%

= 80%


Q5 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

Question 5:

Express 125% as a ratio.

Answer 5:

125%=125100=54 = 5:4


Q6 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

Question 6:

Which is largest in 623%,320 and 0.14?

Answer 6:

​       We have: 623%=203%             

=203×1100             

=115             

=0.06  
      Also, 320=0.15
      The third number is 0.14.Clearly, 0.15 is the largest.

Hence, 320 is the largest.


Page-122

Q7 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

Question 7:

(i) What per cent of 150 is 96?
(ii) What per cent of 5 kg is 200 g?
(iii) What per cent of 2 litres is 250 mL?

Answer 7:

i Required percentage

=96150×100%

=64%
                                    
(ii) Required percentage

=2005×1000×100%

= 4%
                                  
iii Required percentage

=2502×1000×100%

= 12.5%
                                    

Q8 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

Question 8:

Find 412% of Rs 3600.

Answer 8:

412%=92×100

 9200  of Rs 3600

=9200×3600=Rs 162


Q9 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

Question 9:

If 16% of a number is 72, find the number.

Answer 9:

Let the number be x.16% of x is 72.

16100×x=72

16x=72×100

16x=7200

x=720016=450

 The required number is 450.


Q10 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

Question 10:

A man saves 18% of his monthly income. If he saves Rs 1890 per month, what is his monthly income?

Answer 10:

Let Rs x be his monthly income.

His savings =18% of Rs x

                =Rs x×18100=Rs 9x50
Now, 9x50=1890

x=Rs 1890×509

x=Rs 10500

 His monthly income is Rs.10500.


Q11 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

Question 11:

A football team wins 7 games, which is 35% of the total games played. How many games were played in all?

Answer 11:

Let x be the total number of games played. 

Percentage of games won=35% of x
                             =x×35100=35x100
Now, 35x100=7      

x=7×10035      

x=20

 The total number games played is 20.


Q12 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

Question 12:

Amit was given an increment of 20% on his salary. If his new salary is Rs 15300, what was his salary before the increment?

Answer 12:

Let Rs x be Amit's old salary.

His salary after increment will be Rs x+20100x

According to the question, we have:

x+20100x=15300

100x+20x100=15300       LCM=100

120x100=15300

120x=15300×100

x=15300×100120

x=12750

 The old salary is Rs 12,750.


Q13 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

Question 13:

cAnswer 13:

Let x be the number of days the school was opened.

Number of days Sonal attended school=204 days

Percentage of her attendance=85% of x
                                =x×85100=85x100
Now, 85x100=204      

x=204×10085      

x=240

 The school was opened for 240 day.


Q14 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

Question 14:

A's income is 20% less than that of B. By what per cent is B's income more than A's?

Answer 14:

Let B's income be Rs 100Then, A's income=Rs 80

Therefore, B's income is more thanA's income by 

= 100-8080×100% 

= 2080×100% = 25%                                   
                                                          

=Rs125
 B's income is more than that of A's by 

125-100%, i.e., 25%.


Q15 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

Question 15:

The price of petrol goes up by 10%. By how much per cent must a motorist reduce the consumption of petrol so that the expenditure on it remains unchanged?

Answer 15:

Let the consumption of petrol originally be 1 unit and 

let its cost be Rs 100.New cost of 1 unit of petrol=Rs 110

Now, Rs 110 will yield 1 unit of petrol.i.e., 

Rs 100 will yield 1110×100, 

i.e., 1011 units of petrol.
                    
Now, reduction in consumption

=1-1011=111unit
Percentage of reduction

=111×11×100%=9111%

 A motorist must reduce the consumption of petrol by 9111%.


Q16 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

Question 16:

The population of a town increases by 8% annually. If the present population is 54000, what was it a year ago?

Answer 16:

Let x be the population of the town a year ago.

Then, present population=108% of x
                             =x×108100=27x25
Now, 27x25=54000      

x=54000×2527      

x=50000

Hence, the population of the town a year ago was 50000.


Q17 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

Question 17:

The value of a machine depreciates every year by 20%. If the present value of the machine be Rs 160000, what was its value last year?

Answer 17:

Let Rs x be the value of the machine last year.

Then, present value=80% of Rs x
                       =Rsx×80100=Rs4x5
Now, 4x5=160000      

x=160000×54      

x=40000×5=200000

Hence, the value of the machine last year was Rs 2,00,000.


Q18 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

Question 18:

An alloy contains 40% copper, 32% nickel and rest zinc. Find the mass of zinc in one kg of the alloy.

Answer 18:

Mass of the alloy=1 kg

Percentage of copper=40%Percentage of nickel=32%

Percentage of zinc=100-40+32%
                     =28%                     
 Mass of zinc in 1 kg of alloy=28100×1 kg
                                     =0.28 kg=0. 28×1000 g=280 g


Q19 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

Question 19:

Balanced diet should contain 12% of proteins, 25% of fats and 63% of carbohydrates. If a child needs 2600 calories in his food daily, find in calories the amount of each of these in his daily food intake.

Answer 19:

Amount of protein=12% of 2600
                       =2600×12100=312 cal
Amount of fat=25% of 2600
                 =2600×25100=650 cal
Amount of carbohydrate=63% of 2600
                             =2600×63100=1638 cal


Q20 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

Question 20:

Gunpowder contains 75% nitre and 10% sulphur. Find the amount of gunpowder which carries 9 kg nitre. What amount of gunpowder would contain 2.5 kg sulphur?

Answer 20:

Let x be the amount of gunpowder.Amount of nitre=75%



Let x kg be the amount of gunpowder containing 9 kg of nitre.

i.e., 75% of x=9 kg

  x×75100=9

75x100=9

x=9×10075

x=12 kg
Hence, 12 kg of gunpowder contains 9 kg of nitre.

Now, amount of sulphur=10%

Let x kg be the amount of gunpowder containing 2.5 kg of sulphur

.i.e., 10% of x=2.5 kg
  x×10100=2.5

10x100=2.5

x10=2.5

x=2.5×10

x=25 kg
Hence, 25 kg of gunpowder contains 2.5 kg of sulphur.


Q21 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

Question 21:

Divide Rs 7000 among A, B and C such that A gets 50% of what B gets and B gets 50% of what C gets.

Answer 21:

Let  Rs x be the amount of money received by C.

Then, amount of money B gets =50% of Rs x

Amount of money A gets=50% of B                                         =25% of Rs x

Now, x+50% of Rs x+25% of Rs x=Rs 7000

x+x×50100+x×25100=Rs 7000

x+50x100+25x100=Rs 7000

x+50x100+25x100=Rs 7000

175x100=Rs 7000

x=Rs 7000×100175

x=Rs 4000
 C gets Rs 4000.Amount of money B gets=50% of Rs x
                                    =50% of Rs 4000=Rs 4000×50100=Rs 2000

Amount of money A gets=25% of Rs x
                                   =25% of Rs 4000=Rs 4000×25100=Rs 1000


Q22 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

Question 22:

Find the percentage of pure gold in 22-carat gold, if 24-carat gold is 100% pure.

Answer 22:

22 carat gold contains 22 parts pure gold out of 24 parts.

Also, 24 carat gold is given to be 100% pure.

 Percentage of pure gold in 22 carat gold=2224×100% 
                                                   =9123%
Hence, 22 carat gold contains 9123% of pure gold.


Q23 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage | myhelper

Question 23:

The salary of an officer is increased by 25%. By what per cent should the new salary be decreased to restore the original salary?

Answer 23:

Let the original salary be Rs 100
Then, after increment of 25% the salary becomes
= 1001+25100 

= 100125100 

= Rs 125

To restore the original salary, let the new salary be decreased by x%.
Thus, we get

1251-x100 = 100

1-x100 = 100125 = 45

x100 = 15

x = 1005 =20 %
Therefore, the new salary must be reduced by 20% to restore the original salary.

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