NCERT Class 10 Math Chapter 3 Important Questions Answer – Pair of Linear Equations in Two Variables

Class 10
Chapter  Pair of Linear Equations in Two Variables
Subject Math
Category Important Question Answer

Class 10 Math Chapter 3 Important Question Answer

Q1. Solve the following pair of linear equations: 2x + 3y = 7 and 6x – 5y = 11.

Ans.

We have given

2x + 3y = 7    —- (i)

6x – 5y = 11   —- (ii)

Step 1 : Multiply eq (i) by 5 and eq (ii) by 3 to make the coefficients of y equal. Then we get the equations :

10x + 15y = 35

18x – 15y = 33

Step 2 : Adding both the equations, we get

28x = 68

x = \displaystyle \frac{{68}}{{28}}=\frac{{17}}{7}

Step 3 : Substituting the value of x in eq (i), we get

2 \displaystyle \left( {\frac{{17}}{7}} \right)  + 3y = 7

3y = \displaystyle 7-\frac{{34}}{7}=\frac{{49-34}}{7}=\frac{{15}}{7}

y = \displaystyle \frac{5}{7}

Therefore, the solution is x = \displaystyle \frac{{17}}{7}, y = \displaystyle \frac{5}{7}

[ Note : you can verify it by putting the value of x and y in either equations. ]

Q2. For what value of k does the following pair of linear equations have infinite number of solutions?

 (k – 1)x + (k + 1)y = 3k – 1 and 2x + 3y = 7

Ans. Here a1 = k – 1, a2 = 2, b1 = k + 1, b2 = 3, c1 = 3k – 1, c2 = 7

Now for the given pair to have a infinite number of solutions :

\displaystyle \frac{{k-1}}{2}=\frac{{k+1}}{3}=\frac{{3k-1}}{7}

Comparing (i) and (ii) we get, \displaystyle \frac{{k-1}}{2}=\frac{{k+1}}{3}

3k – 3 = 2k + 2

k = 5

Therefore, for k = 5, the given pair of equations will have infinite number of solutions.

Q3. The pair of linear equations 3x + 5y = 7 and 9x – 10y = 14 is consistent or inconsistent.

Ans. Here a1= 3, a2 = 9, b1 = 5, b2 = -10, c1 = 7, c2 = 14

From above it is clear that

Therefore, pair of linear equation is consistent.

Q4. For what values of K does the pair of linear equations 4x + Ky + 8 = 0 and 2x + 2y + 2 = 0 has unique solution?

Ans. Here a1= 4, a2 = 2, b1 = k, b2 = 2 .

Now for the given pair to have a unique solution :

\displaystyle \frac{4}{2}\ne \frac{k}{2}

k≠4

Therefore, for all values of k, except 4, the given pair of equations will have a unique solution.

Q5. For what values of K does the pair of linear equations x – ky + 4 = 0 and 2x – 6y – 5 = 0 has no solution.

Ans. Here a1= 1, a2 = 2, b1 = -k, b2 = -6, c1 = 4, c2 = -5.

Now for the given pair to have no solution :

\displaystyle \frac{1}{2}=\frac{{-k}}{{-6}}\ne \frac{4}{{(-5)}}

i.e.  \displaystyle \frac{1}{2}=\frac{k}{6}

k = 3

Therefore, for k = 3, the given pair of equations will have no solution.

Q6. Solve: x + y = 5 and 2x – 3y = 4

Ans. x + y = 5  —- (i)

2x – 3y = 4 —— (ii)

Multiplying eq (i) with 3 we get 3x + 3y = 15

Now adding,

x = \displaystyle \frac{{19}}{5}

Putting value of x in equation (i), \displaystyle \frac{{19}}{5} + y = 5

Y = 5 – \displaystyle \frac{{19}}{5} = \displaystyle \frac{6}{5}

Q7.  If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes  if we only add 1 to denominator. Find the fraction. Most Important

Ans. Let the numerator = x

Let the denominator = y

Therefore, the fraction = \displaystyle \frac{x}{y}

According to first condition,

\displaystyle \frac{{x+1}}{{y-1}}=1

x + 1 = y – 1

x – y = -2  —- (i)

According to 2nd condition,

\displaystyle \frac{x}{{y+1}}=\frac{1}{2}

2x = y + 1

2x – y = 1 —- (ii)

Subtracting eq (i) from (ii) we get,

x = 3

Putting value of x in eq (i) we get,

3 – y = -2

Y = 5

Therefore, required fraction is \displaystyle \frac{3}{5}.

Q8. 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 ?

Ans.  Let the age of Jacob be x years.

Also, let the age of his son be y years.

According to question,

(x + 5) = 3(y + 5)

x – 3y = 10 ——- (i)

(x – 5) = 7(y – 5)

x – 7y = -30 —— (ii)

Subtracting equation (ii) from (i) we get,

4y = 40

y = 10

substituting the value of y in eq (i) we get,

x = 40

Hence the present age of jacob is 40 years and that of his son is 10 years.

Q9. Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu ?.

Ans. Let the present age of Nuri be x years.

Also, let the present age of sonu be y years.

According to question,

(x – 5) = 3(y – 5)

x – 3y = -10  ——- (i)

Also,

x + 10 = 2(y + 10)

x – 2y = 10  ——- (ii)

Subtracting eq (i) from (ii) we get,

y = 20

Substituting the value of y in eq (i) we get

x – 3(20) = -10

x = -10 + 60

x = 50

Therefore, present of Nuri is 50 years and that of Sonu is 20 years.

Q10. The sum of the digits of a two-digit number is 9. Also nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

Ans. Let the unit digit of a number be x.

Also, let the tens digit of a number be y.

Then, number n = 10y + x

Number after reversing the order of digits = 10x + y

According to question,

x + y = 9  —- (i)

Also,

9(10y + x) = 2(10x + y)

88y – 11x = 0

8y – x = 0  —— (ii)

Adding eq (i) and (ii) we get

9y = 9

y = 1

Substituting the value of y in eq (i) we get

x = 8

Therefore, required number is 18.

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