Blackberry publication class 8 maths chapter comparing quantities

Exercise 9(A)   Exercise 9(B)   Exercise 9(C)   Exercise 9(D)

COMPARING QUANTITIES

EXERCISE 9(A)

QUESTION 1

Express each of the following ratios in the simplest form:
(a) 3 kg 500 g : 4 kg 250 g

Sol :

As we know that , 1 kg = 1000 g

so , 3 kg 500 g can be rewritten as

= 3 kg + 500 g   [3 kg = 3000 g ]

= 3000 g + 500 g

= 3500 g

similarly :  4 kg 250 g written as

= 4000 g + 250 g   [4 kg = 4000 g]

= 4250 g

Now , 3500 g : 4250 g can be written in ratio as

=\dfrac{3500}{4250}g

Now , to express them in simplest form , we need to find H.C.F of 3500 and 4250 , which is 250

=\dfrac{3500\div250}{4250\div250}

we get

=\dfrac{14}{17}

can be written as

= 14 : 17

 


 

(b) 8 months : 1\dfrac{1}{2} years

Sol :

8 months : 1\dfrac{1}{2} years

can be rewritten as

8 months : 1+\dfrac{1}{2} years

\dfrac{1}{2} of a year means

=12\times \dfrac{1}{2}

= 6 months

8 months : 1 years + 6 months

8 months : 12 months + 6 months  [∵ 1 year = 12 months]

8 months : 18 months

Can be written in ratio as

=\dfrac{8}{18}

=\dfrac{8\div 2}{18\div 2}

=\dfrac{4}{9}

or 4 : 9


 

(c) 36 minute : 1\dfrac{1}{2} hours

Sol :

36 minutes : 1 + \dfrac{1}{2} hours

36 minutes : 1 hour + \dfrac{1}{2} hours

1 hour = 60 minutes

\dfrac{1}{2} hours = \dfrac{60}{2} minutes

\dfrac{1}{2} hours = 30 minutes

36 minutes : 1 hour + 30 hours

36 minutes : 60 + 30 minutes

36 minutes : 90 minutes

can be written as

=\dfrac{36}{90}

=\dfrac{36\div 3}{90 \div 3}

=\dfrac{12\div 3}{30 \div 3}

=\dfrac{4\div 2}{10\div 2}

=\dfrac{2}{5}

written as

2 : 5


 

(d) 14 m : 7 m 35 cm

Sol :

Firstly we have to covert 35 cm to m

1 m = 100 cm

1 cm = \dfrac{1}{100} m

multiplying both sides by 35

35 cm = \dfrac{35}{100} m

35 cm = 0.35 m

14 m : 7 m + 0.35 m

14 : 7.35 m

written in fraction as

=\dfrac{14}{7.35}

=\dfrac{14\times 100}{7.35 \times 100}

=\dfrac{1400}{735}

H.C.F of 1400 and 735 is

=\dfrac{1400\div 35}{735\div 35}

=\dfrac{40}{21}

written as 40 : 21

 


QUESTION 2

Divide the number 952 into two parts in the ratio of 3 : 5

Sol :

Let the common ratio be x

then 3 : 5 can be written as

3x + 5x = 952

8x = 952

x = \dfrac{952}{8}

x = 119

then 3x =  3 × 119

3x = 357

and 5x = 5 × 119

5x = 595

ALTERNATE METHOD

First part =\dfrac{3}{3+5}\times 952

=\dfrac{3}{8}\times952

= 357

 

Second part =\dfrac{5}{3+5}\times 952

=\dfrac{5}{8}\times 952

= 595

 

 


QUESTION 3

A number is divided in the ratio 15 : 8 .If the second part is 56, find the number.

Sol :

Second part 8x = 56

then x=\dfrac{56}{8}

x = 7

then first part = 15x = 15 × 7 = 105

First part + Second part = Original number

= 105 +  56

= 161

 

 


QUESTION 4

Convert the following into percentage:

(a) \dfrac{7}{10}

Sol :

=\dfrac{7}{10}\times 100

=\dfrac{700}{10}

= 70%

 


(b) 1 : 4

Sol :

=\dfrac{1}{4}\times 100

=\dfrac{100}{4}

= 25 %


(c) 0.065

Sol :

= 0.065 × 100

= 6.5

 


QUESTION 5

If 11.5% of a weight is 4.6 kg, find the weight.

Sol :

11.5 % = 4.6 kg

which means 1\%= \dfrac{4.6}{11.5}

1 % = 0.4

100 % = 0.4 × 100

100% = 40 kg

 

ALTERNATE METHOD

Weight =\dfrac{4.6}{11.5}\times 100

Weight = 40 kg

 


QUESTION 6

If a number increases from 75 to 85, then find the percentage of increase.

Sol :

Firstly , we find increase 85 – 75 = 10

And increase is of 10 , in percentage we write it as

=\dfrac{\text{increase}}{\text{original number}}\times 100

=\dfrac{10}{75}\times 100

=\dfrac{1000}{75}=\dfrac{200}{15} =\dfrac{40}{3}=13\dfrac{1}{3}

or = 13.33

 

 


QUESTION 7

If a weight of 65 kg is reduced to 60 kg, then find the percentage of reduction?

Sol :

\text{Percentage of reduction} =\dfrac{\text{difference in weight}}{\text{ original weight}}\times 100

=\dfrac{65-60}{65}\times 100

=\dfrac{5}{65}\times 100

=\dfrac{500}{65}

=\dfrac{100}{13}

=7\dfrac{9}{13}

or = 7.69

 


QUESTION 8

There are 56 teachers in a schools. Out of them 42 are female teachers and the rest are male teachers.Find the ratio of :

(a) Male teachers to the female teachers.

Sol :

male teacher = total teacher – female teacher

male teacher = 56 – 42

male teacher = 14

In ratio

=\dfrac{\text{male teacher }}{\text{female teacher}}

=\dfrac{14}{42}

=\dfrac{1}{3}

or 1 : 3

(b) Female teachers to the male teachers.

Sol :

In ratio

=\dfrac{\text{female teacher}}{\text{male teacher }}

=\dfrac{42}{14}

=\dfrac{3}{1}

or 3 : 1

(c) Male teachers to the total numbers of teachers.

Sol :

In ratio

=\dfrac{\text{male teacher}}{\text{total teacher }}

=\dfrac{14}{56}

=\dfrac{7}{28}

=\dfrac{1}{4}

or 1 : 4

 


QUESTION 9

A boy secured 28% of 500 and failed in  the examination by 10 marks.Find the pass marks percentage and the pass marks.

Sol :

28\% \text{ of } 500 = \dfrac{28}{100}\times 500

28% of 500 = 140

Then passing marks will be = 140 + 10

= 150

And in percentage can be written as =\dfrac{\text{passing marks} }{\text{total marks}}\times 100

=\dfrac{150}{500}\times 100

= 30 %

Passing marks % = 30 %

 


QUESTION 10

A cake weighting 1.98 kg is cut into two parts so that the first part is heavier than the second by 20%.Find the weight of each part.

Sol :

First part + second part = total

(x + 20%) +  (x) = 1.98

x + 0.396 + x = 1.98

2x = 1.98 – 0.396

2x = 1.584

x =\dfrac{1.584}{2}

Second part = x = 0.792

AND

First part = x + 20%

= 0.792 +  0.396

= 1.188

 


QUESTION 11

The ratio of the length to the breadth of a rectangle is 5 : 3. If the perimeter of the rectangle is 144 m , what is the length of the rectangle ?

Sol :

Let the length be 5x

and Breadth be 3x

Perimeter of rectangle = 2 ( l + b)

144 = 2 ( 5x + 3x )

144 = 2 × 8x

144 = 16 x

16 x = 144

x = \dfrac{144}{16}

x = \dfrac{72}{8}

x = \dfrac{36}{4}

x = \dfrac{18}{2}

x = 9

Then length = 5x = 5 × 9 = 45

 


QUESTION 12

A fort had provisions for 150 mens for 45 days.After 10 days, 25 mens left the fort.How long will the food last at the same rate ?

Sol :

Food of 150 men for 45 days=150×45=6750 unit
after 10 days——150×10=1500 unit
and after 10 days remaining men=150-25=125
and remaining food=6750-1500=5250 unit
so,
125×d=5250
d=42
ALTERNATE METHOD

Given that fort had provision of food for 150 men for 45 days
Hence, after 10 days, the remaining food is sufficient for 150 men for 35 days

Remaining men after 10 days = 150 – 25 = 125
Assume that after 10 days,the remaining food is sufficient for 125 men for x days

More men, Less days (Indirect Proportion)

⇒Men 150 : 125 }:: x : 35

 

 


QUESTION 13

A car covers 144 km in 3 hours.If an autorickshaw covers the same distance in 4 hour, find the ratio of the speed of the car to that of the autorickshaw ?

Sol :

As we know \text{speed} = \dfrac{\text{distance}}{\text{time}}

Speed of car can be written as 144 : 3

Speed of autorickshaw can be written as 144 : 4

And together as

144 : 3 :: 144 : 4

which can be written as

\dfrac{144}{3}=\dfrac{144}{4}

\dfrac{1}{3}=\dfrac{1}{4}

\dfrac{4}{3}

can be written as

4 : 3

 


QUESTION 14

In a railway accident , 84 out of 525 passengers were injured.What % of the passengers were unhurt?

Sol :

Total number of unhurt passenger = 525 – 84

= 441

In percentage can be written as =\dfrac{\text{unhurt passenger}}{\text{total passenger}}\times 100

=\dfrac{441}{525}\times 100

= 84 %

 


 

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