myhelper: S.chand books class 8 maths solution chapter 15 Linear Equations exercise 15 B

EXERCISE 15 B

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Question 1

The sum of three consecutive odd natural numbers is 87. What are the three numbers ?

Sol :
Let the three consecutive odd numbers be (2x+1) , (2x+3) and(2x+5)
A.T.Q
⇒(2x+1)+(2x+3)+(2x+5)=87

⇒2x+1+2x+3+2x+5=87

⇒6x+9=87

⇒6x=87-9

⇒6x=78

⇒$x=\dfrac{78}{6}$

⇒x=13

Hence , the consecutive numbers are

⇒2x+1=2×13+1=27 ,

⇒2x+3=2×13+3=29 and

⇒2x+5=2×13+5=31

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Question 2

The sum of two numbers is 80. If the larger number exceeds four times the smaller by 5 , what is the smaller number ?

Sol :

Let the two numbers be x and y and lets assume x>y.
Also, x=4y+5..(i)
A.T.Q
x+y=80 ..(ii)
substituting value of (i) in (ii)
⇒(4y+5)+(y)=80

⇒4y+5+y=80

⇒5y+5=80

⇒5y=80-5

⇒5y=75

⇒$y=\dfrac{75}{5}$

⇒y=15

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Question 3

A girl was asked to multiply a number by $\dfrac{7}{8}$ . Instead she divided the number by $\dfrac{7}{8}$ and got the result 15 more than the correct result . What is the number ?

Sol :
Let the number be x
On dividing x by $\dfrac{7}{8}$

⇒$x \div \dfrac{7}{8}$

⇒$x \times \dfrac{8}{7}$

⇒$\dfrac{8x}{7}$

A.T.Q
⇒$x\times \dfrac{7}{8}=\dfrac{8x}{7}-15$

⇒$\dfrac{7x}{8}=\dfrac{8x}{7}-15$

⇒$15=\dfrac{8x}{7}-\dfrac{7x}{8}$

⇒$15=\dfrac{8(8x)-7(7x)}{56}$

⇒$15=\dfrac{64x-49x}{56}$

⇒$15=\dfrac{15x}{56}$

⇒$15\times 56=15x$

⇒$\dfrac{15\times 56}{15}=x$

⇒x=56

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Question 4

What is the sum of two consecutive even numbers , the difference of whose square is 84 ?

Sol :
Let the two consecutive even numbers be (2x) and (2x+2). Then ,
A.T.Q
⇒(2x+2)²-(2x)²=84

⇒{(2x)²+2²+2(2x)(2)}-4x²=84 [using(a+b)²=a²+b²+2ab]

⇒4x²+4+8x-4x²=84

⇒8x+4=84

⇒8x=84-4

⇒8x=80

⇒$x=\dfrac{80}{8}$

⇒x=10

So , the numbers are 2x=2×10=20 and 2x+2=2×10+2=22
Then, sum is 20+22=42

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Question 5

One number is 7 more than another and its square is 77 more than the square of the smaller number. What are the numbers ?

Sol :
Let the numbers are x and y.Then ,
A.T.Q

⇒x=y+7

⇒y=x-7..(i)

Also,

⇒(x)²=77+y²

⇒x²=77+(x-7)² [from (i)]

⇒x²=77+x²+49-2(x)(7) [(a-b)²=a²+b²-2ab]

⇒x²=77+x²+49-14x

⇒x²-x²+14x=77+49

⇒14x=126

⇒$x=\dfrac{126}{14}$

⇒x=9

Putting value of x in (i)

⇒y=9-7

⇒y=2

The numbers are 2 and 9

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Question 6

The numerator of a fraction is 4 less than its denominator if the numerator is decreased by 2 and the denominator is increased by 1 , then the fraction becomes $\dfrac{1}{8}$ . Find the fraction .

Sol :
Let the numerator be x and denominator be y
A.T.Q
x=y-4..(i)

Also,
⇒$\dfrac{x-2}{y+1}=\dfrac{1}{8}$

⇒8(x-2)=1(y+1)

⇒8x-16=y+1

⇒8x=y+1+16

⇒8x=y+17 ..(ii)

Substituting (i) in (ii)

⇒8(y-4)=y+17

⇒8y-32=y+17

⇒7y=17+32

⇒$y=\dfrac{49}{7}$

⇒y=7

Putting value of y in (i) , we get

⇒x=(7)-4

⇒x=3

Hence , the fraction is $\dfrac{x}{y}=\dfrac{3}{7}$

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Question 7

The digits of a two-digit number are in the ratio 2:3 and the number obtained by interchanging the digits is greater than the original number by27 . What is the original number ?

Sol:

Let the common ration be x . Then , the tens digit be 2x and once digit be 3x.

Original number be 10(tens)+(once) which can be written as 10(2x)+3x=23x

Original number 23x ..(i)

On interchanging digits 10(3x)+2x=32x

And 32x is greater than original number by 27 , which means

⇒32x=23x+27

⇒32x-23x=27

⇒9x=27

⇒$x=\dfrac{27}{9}$

⇒x=3

So , the tens digit be 2x=2(3)=6

The once digit(unit digit) be 3x=3(3)=9

Hence , the original number be 10(tens)+(once)=10(6)+9=60+9=69

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Question 8

The sum of the digits of a two- digit number is 10. The number formed by reversing the digits is 18 less than the original number. Find the original number .

Sol :

Let the original number be 10x+y . Where x is tens and y is unit digit

A.T.Q

⇒x+y=10 [ sum of the digit of number ]

⇒x=10-y ..(i)

On reversing the number we get 10y+x which is equal to (10x+y)-18

⇒10y+x=(10x+y)-18

⇒10y+x=10x+y-18

⇒10y-y-10x+x=-18

⇒9y-9x=-18

⇒-9(-y+x)=-18

⇒$(-y+x)=\dfrac{-18}{-9}$

⇒x-y=2

⇒(10-y)-y=2

⇒10-y-y=2

⇒-2y=2-10

⇒-2y=-8

⇒$y=\dfrac{-8}{-2}$

⇒y=4

Putting value of y in (i) , we get

⇒x=10-y

⇒x=10-4

⇒x=6

Hence , the original number be 10x+y=10(6)+4=64

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Question 9.

A man ate 100 grapes in 5 days. Each day he ate 6 more grapes than those he ate on the previous day. How many grapes did he eat the first day .

Sol :

Let the grapes eaten at first day be x . Also , each day he ate 6 more grapes

A.T.Q

1st day + 2nd day + 3rd day + 4th day + 5th day = 100 grapes

⇒(x)+(x+6)+{(x+6)+6}+{(x+6+6)+6}+{(x+6+6+6)+6}=100

⇒x+(x+6)+(x+12)+(x+18)+(x+24)=100

⇒x+x+6+x+12+x+18+x+24=100

⇒5x+60=100

⇒5x=100-60

⇒5x=40

⇒$x=\dfrac{40}{5}$

⇒x=8

Hence , grapes eaten by man on first day is 8

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Question 10

In an examination, a student scores 4 marks for every correct answer and losses 1 marks for every incorrect answer. If he attempts all 75 questions and secures 125 marks . What are the number of questions he attempted correctly ?

Sol :

+4 Marks for every correct answer is given

-1 for every incorrect answer is given

Let the correct attempt be x

then , incorrect attempt be (75-x)

A.T.Q

⇒+4(x)-1(75-x)=125

⇒4x-75+x=125

⇒5x=125+75

⇒5x=200

⇒$x=\dfrac{200}{5}$

⇒x=40

Which means 40 attempts are correct

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Question 11

Arun's age is four years more than Prashant's age 4 years ago , the ratio of their ages was 5 : 4 . Find their present ages.

Sol:

Let the Prashant's age be x (4 years ago from present)

Then Arun's age be (x+4) (4 years ago from present)

A.T.Q

$\dfrac{(x+4)}{x}=\dfrac{5}{4}$ [cross multiply]

⇒4(x+4)=5x

⇒4x+16=5x

⇒5x-4x=16

⇒x=16

Prashant's age 4 years ago be 16 And for present age (x+4) = 16+4 = 20 years

Arun's age 4 years ago be (x+4) which is 20 and now at present {(x+4)+4} ={(16+4)+4}=24 years

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Question 12

A father is three times as old as his son. 15 years hence, he will be twice as old as his son. What is the sum of their present ages ?

Sol :

Let the son age be x and father's age be 3x

In 15 years

Son age be x + 15 and Father age be 2(x+15) “OR” 3x + 15 because both equation represent fathers present age

A.T.Q

⇒2(x + 15) = 3x + 15

⇒2x + 30 = 3x + 15

⇒30-15=3x-2x

⇒15=x

Therefore, Present age of son is x=15 and

fathers age is 3x = 3(15) = 45

And sum of their present age is 45+15 = 60 years

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Question 13

4 years ago, the ratio of their ages of A and B was 2 : 3 and after 4 years it becomes 5 : 7 . Find their present ages .

Sol:

Given that, 4 years ago from present , the ratio of the ages of A and B = 2 :3

Hence we can assume that
age of A 4 years ago = 2x
age of B 4 years ago = 3x

After 4 years from present, the ratio of their ages = 5 : 7

⇒{(2x+4)+4} : {(3x+4)+4} = 5 : 7

⇒$\dfrac{2x+8}{3x+8}=\dfrac{5}{7}$

⇒7(2x+8)=5(3x+8) [cross multiplication]

⇒14x+56=15x+40

⇒56-40=15x-14x

⇒x=16

A's present age = (2x+4) = 2(16)+4 = 36

B's present age = (3x+4) = 3(16)+4 = 52

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Question 14

Length of a rectangular blackboards is 8 m more than its breadth. If its length is increased by 7 m and its breadth is decreased by 4 m , its area remains unchanged. What is the length of the blackboard ?

Sol :

Case 1:

Let the breadth be x

Length of blackboard = x+8

Area of rectangular blackboard = l × b =(x) × (x+8)

= x² + 8x..(i)

Case 2:

Breadth is decreased by 4m then breadth = x-4

Length is increased by 7m then length = x+8+7 = x+15

Area of rectangular blackboard = l × b = (x-4) × (x+15)

= x(x+15) - 4(x+15)

= x²+15x-4x-60

= x²+11x-60..(ii)

Both (i) and (ii) are equal because it is given that area remains unchanged(equal)

⇒x² + 8x = x²+11x-60

⇒x²-x²+8x-11x=-60

⇒-3x=-60

⇒$x=\dfrac{-60}{-3}$

⇒x=20

Length of blackboard = x+8 = 20+8 = 28

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Question 15

Two cars A and B leave Mumbai at the same time, travelling in opposite directions . If the average speed of car A is 8 km/h more than that of B,and they are 300 km apart at the end of 6 hours. What is the average speed of car A ?

Sol :

Let x be average speed of B

Then Average speed of A = (x+8) km/hr

Distance travelled by A in 6 hrs = 6(x+8) km [Distance=speed×time]

Distance travelled by B in 6 hrs= 6(x) km

Distance travelled by A + Distance travelled by B = Total distance

⇒6(x+8)+6x=300

⇒6x+48+6x=300

⇒12x=300-48

⇒$x=\dfrac{252}{12}$

⇒x=21

Hence , average speed of A is (x+8)=21+8=29km/hrs

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Question 16

A boat takes 9 hours to travel a distance upstream and takes 3 hours to travel the same distance downstream. If the speed of the boat in still water is 4 km/h. What is the speed of the stream ?

Sol :

Let the distance covered by the boat each way be d km. And let the velocity of the stream be v km/hr.

Case 1 : (Downstream)

While going downstream the speed of the boat is (4+v) km/hr and time taken is 3 hours

[Distance=velocity×time]

d=(4+v)3..(i) [here , v is +ve because direction of stream and boat is same]

Case 2 : (Upstream)

While going upstream the speed of the boat is (4-v) km/hr and time taken is 9hours

[Distance=velocity×time]

d=(4-v)9..(ii) [here , v is -ve because direction of boat and stream is opposite]

(i)=(ii) , As the distance travelled upstream and downstream is the same,

⇒(4+v)3=(4-v)9

⇒12+3v=36-9v

⇒3v+9v=36-12

⇒12v=24

⇒$v=\dfrac{24}{12}$

⇒v=2km/hr [here , + minus sign shows direction of boat and stream is same]

Hence , the velocity of stream is 2km/hr

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Question 17

Kartik buys some apples at the rate of 10 rupees per apple . He also buys an equal number of oranges at the rate of 6 rupees per orange. He makes a15% profit on apples and 8% profit on oranges. The total profit earned by Kartik at the end of the day is 396. When all fruit is sold out , find the number of apples purchased .

Sol :

Number of apples be x and Number of oranges also be x (equal)

Cost of apples 10x and cost of oranges be 6x

⇒(15% of 10x )+(8% of 6x)=396

⇒$\left(\dfrac{15}{100}\times 10x\right)+\left(\dfrac{8}{100}\times6x\right)=396$

⇒$\dfrac{150x}{100}+\dfrac{48x}{100}=396$

⇒$\dfrac{150x+48x}{100}=396$

⇒198x=396×100

⇒$x=\dfrac{39600}{198}$

⇒x=200

Hence, Total number of apples purchased is 200

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Question 18

Bittu spends $\dfrac{1}{3}$ of the money with him on books , $\dfrac{1}{4}$of the remaining on food and $\dfrac{2}{5}$ of the remaining on travel. Now, he is left with 450 rupees . How much money did he have in the beginning?

Sol :

Total money at the beginning be x

Spending on books [1/3 of total] $=\dfrac{1}{3}\times x=\dfrac{1x}{3}$

Remaining $=x-\dfrac{1x}{3}$ $=\dfrac{3x-1x}{3}=\dfrac{2x}{3}$

Spending on foods [1/4 of remaining ]$=\dfrac{1}{4}\times\left(x-\dfrac{1x}{3}\right)$ or $=\dfrac{1}{4}\times \dfrac{2x}{3}$$=\dfrac{2x}{12}=\dfrac{x}{6}$

Remaining $=\dfrac{2x}{3}-\dfrac{x}{6}$ $=\dfrac{2(2x)-x}{6}=\dfrac{4x-x}{6}$$=\dfrac{3x}{6}=\dfrac{x}{2}$

Spending on travel , [2/5 of remaining] $=\dfrac{2}{5}\times\dfrac{x}{2}=\dfrac{x}{5}$

Remaining $=\dfrac{x}{2}-\dfrac{x}{5}$ $=\dfrac{5x-2x}{10}=\dfrac{3x}{10}$

A.T.Q

Money left after spending on travel must be equal to 450

$\dfrac{3x}{10}=450$

3x=450×10

$x=\dfrac{4500}{3}$

x=1500

Hence , total money at beginning is 1500

S.chand books class 8 maths solution chapter 15 Linear Equations exercise 15 B

EXERCISE 15 B

Question 16

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