## Polynomials – Class 10th NCERT Solutions

**Some other extra questions :**

**Q1. Show that x = 3, y = 2 is not a solution of the system of linear equations 3x – 2y = 7.**

Ans. The given equations are

3x – 2y = 5

2x + y = 7.

Putting x = 3 and y =2 in (i), we get LHS

= ( 3 × 3 – 2 × 2 ) = 5 = RHS.

Thus, the values x = 3, y = 2 is not a solution of the given system of equations.

**Q2. Solve for x and y : 10x + 3y = 75 , 6x – 5y = 11.**

Ans. The given equations are :

10x + 3y = 75

6x – 5y = 11

Multiplying (i) by 5 and (ii) by 3, we get :

50x + 15y = 375

18x – 15y = 33

Adding (iii) and (iv), we get

68x = 408 ⇒ x =

⇒ x = 6

Putting x = 6 in (i), we get

(10 × 6) + 3y = 75 ⇒ 60 + 3y = 75

⇒ 3y = ( 75 – 60)

⇒ 3y = 15

⇒ y = 5

Therefore, x = 6 and y = 5.

**Q3. Solve for x and y : √2x − √3y = 0, √5x + √2y = 0.**

Ans. The given system of equation is :

√2x- √3y = 0

√5x- √2y = 0

Multiplying (i) by √2and (ii) by √3 , we get

2x – √6y = 0

√15x + √6y = 0

Adding (iii) and (iv), we get

( 2 + √15 ) x = 0 ⇒ x = 0.

Putting x = 0 in (iii), we get

( 2 × 0 ) – √6y = 0 ⇒ √6y = 0 ⇒ y = 0.

Therefore x = 0 and y = 0 is the required solution.

**Q4. The sum of two-digit number and the number formed by interchanging its digits is 110. If 10 is subtracted from the original number, the new number is 4 more than 5 times the sum of the digits of the original number. Find the original number.**

Ans. Let the tens and units digits of the required number be x and y respectively.

Then, original number = ( 10x + y).

Number formed by interchanging its digits = (10y + x).

Now, (10x + y) + (10y + x) = 110

⇒ 11(x + y) = 110

⇒ x + y = 10

Also, (10x + y) – 10 = 5(x + y) + 4 ⇒ 5x – 4y = 14

On multiplying (i) by 4 and adding (ii) to it, we get

9x = 54 ⇒ x = 6.

Putting x = 6 in (i), we get

6 + y = 10 ⇒ y = (10 – 6) = 4.

Therefore x = 6 and y = 4.

Hence, the required number is 64.

**Q5. The sum of the digits of a two-digit number is 12. The number obtained by interchanging its digits exceeds the given number by 18. Find the number.**

Ans. Let the tens digit of the required number be x and the units digit be y. Then,

x + y = 12.

Required number = (10x + y).

Number obtained on reversing the digits =

(10y + x)

Therefore (10y + x) – (10x + y) = 18

⇒ 9y – 9x = 18

⇒ y – x = 2

On adding (i) and (ii), we get

2y = 14 ⇒ y = 7.

Putting y = 7 in (i), we get

X + 7 = 12 ⇒ x = (12 – 7) = 5.

Therefore, x = 5 and y = 7.

Hence, the required number is 57.