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.