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


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)







Hence , the consecutive numbers are

⇒2x+1=2×13+1=27 ,

⇒2x+3=2×13+3=29 and



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)
x+y=80 ..(ii)
subsituting value of (i) in (ii)








Question 3

A girl was asked to multiply a numebr by \dfrac{7}{8} . Instead she divided the numebr 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}


x\times \dfrac{7}{8}=\dfrac{8x}{7}-15






15\times 56=15x

\dfrac{15\times 56}{15}=x



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 ,

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







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


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 ,





⇒x2=77+(x-7)2 [from (i)]

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






Putting value of x in (i)



The numbers are 2 and 9


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





⇒8x=y+17 ..(ii)

Subsituting (i) in (ii)






Putting value of y in (i) , we get



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


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 by 27 . What is the original number ?


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






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


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 digi


⇒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














Putting value of y in (i) , we get




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


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


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









Hence , grapes eaten by man on first day is 8


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)








Which means 40 attempts are correct


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.


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

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


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





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


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


⇒2(x + 15) = 3x + 15

⇒2x + 30 = 3x + 15



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


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 .


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


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




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

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


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





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


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






Hence , average speed of A is 29km/hrs


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 (v+4) km/hr and time taken is 3 hours


d=(+v+4)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 (v-4) km/hr and time taken is 9 hours


d=(-v-4)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,






⇒-v=2km/hr [here , – minus sign shows direction of boat and stream]

or v = 2km/hr

Hence , the velocity of stream is 2km/hr


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 a 15% 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}\times 6x\right)=396






Hence, Total number of apples purchased is 200


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}



Money left after spending on travel must be equal to 450





Hence , total money at beginning is 1500



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