Class 10 Mathematics

Linear Equations

Exercise 3.3 (NCERT) Part 2

Question 2: Form a pair of linear equations for the following problems and find their solution by substitution method.

(a) The difference between two numbers is 26 and one number is three times the other. Find them.

Solution: Let us assume the larger number is x and smaller number is y. Then we have following equations, as per question:

`x = y + 26` and `x = 3y`

Substituting the value of x from second equation in the first equation, we get;

`x = y + 26`
Or, `3y = y + 26`
Or, `2y = 26`
Or, `y = 13`


Substituting the value of y in second equation, we get;

`x = 3y`
Or, `x = 13 xx 3 = 39`
Hence, `x = 39` and `y = 13`

(b) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

Solution: Let us assume the larger angle is x and smaller angle is y. Then we have following equations;

`x = y + 18°` and `x + y = 180°`

Substituting the value of x from first equation in second equation, we get;

`x + y = 180°`
Or, `y + 18° + y = 180°`
Or, `2y = 180° - 18° = 162°`
Or, `y = 81°`

Substituting the value of y in first equation, we get;

`x = y + 18°`
Or, `x = 81° + 18° = 99°`
Hence, `x = 99°` and `y = 81°`


(c) The coach of a cricket team buys 7 bats and 6 balls for Rs, 3800. Later, she buys 3 bats and 5 balls for Rs. 1750. Find the cost of each bat and each ball.

Solution: Let us assume, price of a bat is x and that of a ball is y. Then we have following equations;

`7x + 6y = 3800` and `3x + 5y = 1750`

Let us use first equation to express one variable in terms of another variable;

`7x+6y=3800`
Or, `7x=3800-6y`
Or, `x=(3800-6y)/(7)`

Substituting the value of x in second equation, we get;

`3x+5y=1750`
Or, `3((3800-6y)/(7))+5y=1750`

Or, `(11400-18y)/(7)+5y=1750`

Or, `(11400-18y+35y)/(7)=1750`

Or, `11400+17y=1750xx7=12250`
Or, `17y=12250-11400=850`
Or, `y=850÷17=50`

Substituting the value of y in first equation, we get;

`x=(3800-6y)/(7)`

Or, `x=(3800-6xx50)/(7)`

`=(3800-300)/(7)=(3500)/(7)=500`

Hence price of a bat=Rs.500 and price of a ball=Rs.50

(d) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs. 105 and for the journey of 15 km, the charge paid is Rs. 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?

Solution: Let us assume the fixed charge is x and charge for 1 km journey is y. Then we have following equations;

`x + 10y = 105` and `x + 15y = 155`

Let us use first equation to express one variable in terms of another variable;

`x + 10y = 105`
Or, `x = 105 – 10y`

Substituting the value of x in second equation, we get;

`x + 15y = 155`
Or, `105 – 10y + 15y = 155`
Or, `105 + 5y = 155`
Or, `5y = 155 – 105 = 50`
Or, `y = 10`

Substituting the value of y in first equation, we get;

`x = 105 – 10y`
`Or, x = 105 – 10 xx 10`
Or, `x = 5`

Hence, fixed charge = Rs. 5 and per km charge = Rs. 10


(e) A fraction becomes 9/11, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction.

Solution: Let us assume, numerator is x and denominator is y. Then we have following equations;

`(x+2)/(y+2)=(9)/(11)` and `(x+3)/(y+3)=5/6`

Let us use first equation to express one variable in terms of another variable;

`(x+2)/(y+2)=(9)/(11)`

Or, `11(x+2)=9(y+2)`
Or, `11x+22=9y+18`
Or, `11x+22-18=9y`
Or, `11x+4=9y`
Or, `y=(11x+4)/(9)`

Substituting the value of y in second equation, we get;

`(x+3)/(y+3)=5/6`

Or, `6x+18=5y+15`
Or, `6x+18-15=5y`
Or, `5y=6x+3`
Or, `5((11x+4)/(9))=6x+3`

Or, `55x+20=54x+27`
Or, `55x=54x+27-20`
Or, `55x-54x=7`
Or, `x=7`

Substituting the value of x in first equation, we get;

`y=(11x+4)/(9)`

Or, `y=(11xx7+4)/(9)=(81)/(9)=9`

Hence, `x = 7` and `y = 9`

Required fraction `= 7/9`

(f) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

Solution: Let us assume Jacob’s current age is x and his son’s current age is y.

Five years hence, Jacob’s age `= x + 5` and his son’s age `= y + 5`

As per question;

`x + 5 = 3(y + 5)`
Or, `x + 5 = 3y + 15`
Or, `x = 3y + 15 – 5 = 3y + 10`

Five years ago, Jacob’s age `= x – 5` and his son’s age `= y – 5`

As per question;

`x – 5 = 7(y – 5)`
Or, `x – 5 = 7y – 35`
Or, `x = 7y – 35 + 5 = 7y – 30`

Substituting the value of x from first equation in this equation, we get;

`3y + 10 = 7y – 30`
Or, `3y + 10 + 30 = 7y`
Or, `7y – 3y = 40`
Or, `4y = 40`
Or, `y = 10`

Substituting the value of y in first equation, we get;

`x = 3y + 10`
Or, `x = 3 xx 10 + 10 = 40`

Hence, Jacob’s age = 40 years and son’s age = 10 years


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Introduction

Exercise:3.1

Exercise:3.2(Part 1)

Exercise:3.2(Part 2)

Exercise:3.3(Part 1)

Exercise:3.4 (part 1)

Exercise:3.4 (part 2)

Exercise:3.5 (part 1)

Exercise:3.5 (part 2)

Exercise:3.6 (part 1)

Exercise:3.6 (part 2)

Exercise:3.6 (part 3)