# NCERT solutions for class 10 Mathematics-Chapter-2

Hello students , In this page we provide you NCERT solutions for class 10 Science . In below  you will see class 10 science , it is available on written images content and later it will be appears on typing text content. You can scroll it and find your solution which do you want to get solutions of any questions in your chapter. Also we provide you some extra questions for your practice for regarding to your CBSE Board exam. You can share our link with your friends or classmates and to help them to understand better about their subjects and clear their doubts.

Exercise 2.1

Exercise 2.2

Exercise 2.3

#### Some other extra questions :

Que1. 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.

Que2. 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.

Que3. 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.

Que4. 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.

Que5. 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.