Grade 10  LinearEquationsinTwoVariables
(Sample Printed worksheet)
Answer The Following  
1)  In a two digit number, the ten's digit is twice the units's digit. If 18 is added to the number, the digit interchange their places. Find the number. 
2)  Karen is three times as old as her son. After 20 years, Karen will be two times as hold as her son. Find the current age of Karen and her son. 
3)  Two numbers are in ratio 2:3. If 12 is added to both the numbers, ratio becomes 5:6. Find the numbers. 
4)  If twice the son's age in years is added to mother's age, the sum is 46. If twice the mother's age is added to the son's age, the sum is 68. Find the age of mother and son. 
5)  Three years ago Paul was four times older than his son. After three years, Paul will be 2 years more than two times the age of his son. Find the present age of Paul and his son. 
6)  7 pens and 6 pencils cost $54 and 8 pens and 9 pencils cost $66. Find the cost of one pen and one pencil separately. 
7)  In a two digit number,its unit's digit is 2 more than ten's digit. If sum of this number and the number obtained by reversing the order of its digits is 132, find the number. 
8)  The mother's age is 8 times her son's age. After 6 years, the age of the mother will be 9/2 times her son's age. The present ages of the son and the mother are, respectively 
9)  Which of the following conditions is true if the system of equations below is shown to be inconsistent a_{1}x + b_{1}y + c_{1} = 0, a_{2}x + b_{2}y + c_{2} = 0

10)  Which of the following equations has a unique solution? 3x + 2y + 3 = 0, 9x + 6y + 10 = 0 3x + 2y + 3 = 0, 12x + 8y + 10 = 0 3x + 2y + 3 = 0, 9x + 6y + 9 = 0 3x + 2y + 11 = 0, 9x + 3y + 7 = 0 
Choose correct answer(s) from given choice  
11)  For which value of a, following pair of linear equations have no solution. a x + 16 y = a 4 x + a y = a  4

12)  If a pair of linear equations is consistent, then the lines will be

13)  There are two numbers. If four times the larger of two numbers is divided by the smaller one, we get 8 as quotient and 4 as remainder. If five times the smaller of two numbers is divided by the larger one, we get 2 as quotient and 12 as remainder. Find the numbers.

14)  The pair of equations y = 8 and y = 5 has

15)  Find value of k for which equations kx  3y  3 = 0 and 3x + y + 1 = 0, will have infinitely many solutions.

16)  The pair of equations 2x + 3y + 3 = 0 and 6x  9y  9 = 0 have

17)  Find value of k such that equations 3p + kq + 9 = 0 and p  3q + 3 = 0, represents coincident lines.

18)  Barbara and Ruth have some toys. Barbara says to Ruth, "ff you give me 7 of your toys, I will have twice the number of toys left with you". Ruth replies, "if you give me 7 of your toys I will have the same number of toys left with you". Find the the number of toys with Barbara and Jason separately.

19)  9 balls and 7 bats cost $595 and 10 balls and 11 bats cost $790. Find the cost of 7 balls and 6 bats.

Check True/False  
20)  Equations 4x + 2y = 6 and 2x + y = k will have an unique solution for all real values of k. True False 
Answers
1) 24Step 1 
Let the ten's digit be x and unit digits be y, hence number of 10x + y Now it is given that y  2x = 0 ________________________(1) and (10x + y) + 18 = 10y + x 9x  9y = 18 x  y = 2 ________________________(2) 
Step 2 
On adding two equations, y  2x + x  y = 2 x = 2 x = 2 
Step 3 
y = 2x = 2×2 = 4 Therefore number is 24 
2) 60 and 20 years
Step 1 
Let age of Karen be x years and age of be y years 
Step 2 
It is given that x = 3y => 3y  x = 0 (x + 20) = 2(y + 20) => x  2y = 20 
Step 3 
On adding above two equations (3y  x) + (x  2y) = 20 y = 20 
Step 4 
Hence their ages are x = 3y = 60 years and y = 20 years 
3) 8 and 12
Step 1 
Since numbers are in ratio 2:3, lets assume numbers are 2x and 3x 
Step 2 
When 12 is added to both numbers, numbers become (2x + 12) and (3x + 12) 
Step 3 
Since new ratio is 5:6 (2x + 12)/(3x + 12) = 5/6 ⇒ 6 (2x + 12) = 5 (3x + 12) ⇒ 12x + 72 = 15x + 60 ⇒ 12x  15x = + 60  72 ⇒ 3x = 12 ⇒ x = 12/3 ⇒ x = 4 
Step 4 
Therefore numbers are, 2x = 2 × 4 = 8 3x = 3 × 4 = 12 
4) 30 and 8 years
Step 1 
Let the age of mother be x and that of son be y. It is given that x + 2y = 46 2x + y = 68 
Step 2 
On solving above two equations we get x = 30 and y = 8 years 
5) 19 years and 7 years
Step 1 
Let x and y be the current age of Paul and his son. It is given that (x  3) = 4 (y  3) x  4y = 9 ____________________(1) 
Step 2 
It is also given that (x + 3) = 2 (y + 3) + 2 x  2y = 5 ____________________(2) 
Step 3 
On subtracting eq. 1 from 2 2y = 14 y = 7 years 
Step 4 
On substituting value of y in eq. 1 x  4(7) = 9 x = 19 years 
6) $6 and $2
Step 1 
Let the cost of a pen and pencil be $ x and $ y respectively. It is given that 7 x + 6 y = 54 8 x + 9 y = 66 
Step 2 
On multiplying first equation by 3 and second equation by 2 21 x + 18 y = 162 16 x + 18 y = 132 
Step 3 
Subtract second equation from first equation 1 => 5 x = 30 => x = (30)/(5) = $6 
Step 4 
Substitute this value in first equation => 7 × 6 + 6 y = 54 => 6 y = 54  7 × 6 = 12 => y = (12)/(6) = $2 
7) 57
Step 1 
Let the ten's digit be x and unit digits be y, hence number of 10x + y Now it is given that y  x = 2 ________________________(1) and (10x + y) + (10y + x) = 132 11x + 11y = 132 x + y = 12 ________________________(2) 
Step 2 
On adding two equations, y  x + (x + y) = 12 + 2 2y = 14 y = 7 
Step 3 
x = y  2 = 7  2 = 5 Therefore number is 57 
8) 6 years and 48 year
9)
a_{1} 
a_{2} 
b_{1} 
b_{2} 
c_{1} 
c_{2} 
10) 3x + 2y + 11 = 0, 9x + 3y + 7 = 0
11) c. 8
12) d. intersecting or coincident
13) c. 29 and 14
Step 1 
Let the larger number be x and smaller number be y 
Step 2 
We know that Dividend = (Divisor × Quotient) + Remainder Therefore we can write relationship provided as, 4x = 8y + 4 => 4x  8y  4 = 0 5y = 2x + 12 => 2x  5y + 12 = 0 
Step 3 
On solving these two equations we get x = 29 and y = 14 
14) d. no solution
15) a. 9
16) c. infinitely many solutions
17) b. 9
18) a. 49 and 35
Step 1 
Let the number of toys with Barbara and Ruth be x and y. It is given that x + 7 = 2 (y  7) x  7 = y + 7 
Step 2 
On subtracting second equation from first equation 2 × 7 = 2 (y  7)  (y + 7) 14 = y  21 y = 5 × 7 y = 35 
Step 3 
Substitute value of y in first equation, x + 7 = 2 (35  7) x = 49 
19) a. $485
Step 1 
Let the cost of a ball and bat be $ x and $ y respectively. It is given that 9 x + 7 y = 595 10 x + 11 y = 790 
Step 2 
On multiplying first equation by 11 and second equation by 7 99 x + 77 y = 6545 70 x + 77 y = 5530 
Step 3 
Subtract second equation from first equation 1 => 29 x = 1015 => x = (1015)/(29) = $35 
Step 4 
Substitute this value in first equation => 9 × 35 + 7 y = 595 => 7 y = 595  9 × 35 = 280 => y = (280)/(7) = $40 
Step 5 
Price of 7 balls and 6 bats => 7x + 6y = 7 × 35 + 6 × 40 => = $485 
20) False