CBSE Class 10 Maths HOTs Pair Of Linear Equations In Two Variables Set 07

PRACTICE QUESTIONS

Question. A told B ,‘when I was as old as you are now ,then your age was four years less than half of my present age .If the sum of the present ages of A and B is 61 years,what is B’s present age?(in years)
(a) 9
(b) 25
(c) 43
(d) 36
Answer: (b) 25

Question. Solve for \( x \) and \( y \) using substitution method. \( 4x + \frac{y}{3} = \frac{8}{3}; \frac{x}{2} + \frac{3y}{4} = -\frac{5}{2} \)
Answer: \( x = 1, y = -4 \)

Question. Solve for \( x \) and \( y \) using substitution method. \( 0.4x + 3y = 1.2, 7x - 2y = \frac{17}{6} \)
Answer: \( x = \frac{1}{2}, y = \frac{1}{3} \)

Question. Solve for \( x \) and \( y \) using substitution method. \( 4x - y = 10; 2x - \frac{1}{2}y - 5 = 0 \)
Answer: Infinitely many solutions.

Question. Solve for \( x \) and \( y \) using substitution method. \( \frac{x}{4} + \frac{2y}{3} = 7; \frac{x}{6} + \frac{3y}{5} = 11 \)
Answer: \( x = -84, y = 42 \)

Question. Solve for \( x \) and \( y \) using substitution method. \( a^2x - b^2y = a^2 - 2b^2; b^2x + a^2y = b^2 + 2a^2 \)
Answer: \( x = 1, y = 2 \)

Question. Solve for \( x \) and \( y \) using substitution method. \( 2x - y = -10; -6x + 3y = 30 \)
Answer: Infinitely many solutions.

Question. Solve for \( x \) and \( y \) using substitution method. \( 5x + 3y = 10; 2x + \frac{6}{5}y = 4 \)
Answer: Infinitely many solutions.

Question. Solve for \( x \) and \( y \) using substitution method. \( \sqrt{7}x + \sqrt{13}y = 0; \sqrt{5}x + \sqrt{17}y = 0 \)
Answer: \( x = 0, y = 0 \)

Question. Venu Gopal has twice as many sisters as he has brothers. If Shobha, Venu’s sister has the same number of brothers as she has sisters, then find the number of brothers of Shobha.
Answer: 3 brothers.

Question. Find the different positive fractions whose denominators are 3 and 5 and whose sum is \( 2\frac{1}{15} \).
Answer: \( \frac{2}{3} \) and \( \frac{7}{5} \)

Question. When a certain two digits number is divided by the number obtained by reversing the digits, the quotient is 2 and remainder is 7. If the number is divided by the sum of its digits the quotient is 7 and the remainder is 6. Find the product of the digits of the original number.
Answer: 24 (The number is 83, \( 8 \times 3 = 24 \))

Multiple Choice Questions 

Question. If \( x = a, y = b \) is the solution of the pair of equations \( x - y = 2 \) and \( x + y = 4 \), then the respective values of \( a \) and \( b \) are
(a) 3, 5
(b) 5, 3
(c) 3, 1
(d) –1, –3
Answer: (c) 3, 1

Question. There are two positive numbers such that sum of twice the first and thrice the second is 39, while the sum of thrice the first and twice the second is 36. The larger of the two is
(a) 9
(b) 6
(c) 5
(d) 4
Answer: (a) 9

Very Short Answer Type Questions

Question. Solve for \( x \) and \( y \) by the method of elimination: \( 4x - 3y = 1; 5x - 7y = -2 \)
Answer: Given equations are
\( 4x - 3y = 1 \) ...(i)
\( 5x - 7y = -2 \) ...(ii)
For making coefficient of \( y \) equal in both the equations multiplying equation (i) with 7, we get
\( 7 \times (4x - 3y) = 7 \times 1 \)

\( \implies 28x - 21y = 7 \) ...(iii)
Multiplying equation (ii) with 3, we get
\( 3 \times (5x - 7y) = 3 \times -2 \)

\( \implies 15x - 21y = -6 \) ...(iv)
Subtracting equation (iv) from (iii), we get
\( 13x = 13 \)

\( \implies x = 1 \)
when \( x = 1 \), equation (i) becomes
\( 4 \times 1 - 3y = 1 \)

\( \implies -3y = -3 \)

\( \implies y = 1 \)
\( \therefore x = 1, y = 1 \)

Short Answer Type Questions 

Question. Solve the following pair of linear equations for \( x \) and \( y \): \( 2(ax - by) + (a + 4b) = 0; 2(bx + ay) + (b - 4a) = 0 \)
Answer: Consider equations:
\( 2(ax - by) + (a + 4b) = 0 \)

\( \implies 2ax - 2by = -a - 4b \) ...(i)
and \( 2(bx + ay) + (b - 4a) = 0 \)

\( \implies 2bx + 2ay = 4a - b \) ...(ii)
Multiply (i) by \( a \) and (ii) by \( b \) and adding, we get
\( 2(a^2 + b^2)x = (-a - 4b)a + b(4a - b) \)
\( = -a^2 - 4ab + 4ab - b^2 \)
\( = -(a^2 + b^2) \)

\( \implies x = -\frac{1}{2} \)
Substituting in (i), we get
\( -a - 2by = -a - 4b \)

\( \implies -2by = -4b \)

\( \implies y = 2 \)
\( \therefore x = -\frac{1}{2} \) and \( y = 2 \).

Question. A number consists of two digits. Where the number is divided by the sum of its digits, the quotient is 7. If 27 is subtracted from the number, the digits interchange their places, find the number.
Answer: Let digit at unit place be \( x \), and at tenth place be \( y \).
\( \therefore \) number = \( 10y + x \)
According to the question,
\( \frac{10y + x}{y + x} = 7 \)

\( \implies 10y + x = 7y + 7x \)

\( \implies 6x - 3y = 0 \)

\( \implies 2x - y = 0 \) ...(i)
Again according to the question,
\( (10y + x) - 27 = 10x + y \)
\( 9y - 9x = 27 \)

\( \implies y - x = 3 \) or \( x - y = -3 \) ...(ii)
Solving for \( x \) and \( y \), we get
\( x = 3 \) and \( y = 6 \)
\( \therefore \) Number is 63.

Long Answer Type Question

Question. Two trains each 80 m long passes each other on parallel lines. If they are going in same direction, the faster train takes one minute to pass the other completely. If they are going in opposite directions, they over take each other in three seconds. Find the speed of each train in km/hr.
Answer: Let speed of faster train = \( x \) km/h and speed of other train = \( y \) km/h
(i) When they are going in same direction, then speed of overtaking = \( (x - y) \) km/hr
distance = \( (80 + 80) = 160 \) m = 0.16 km
Time taken = \( \frac{0.16}{x - y} \) hr
A.T.Q. Time taken = 1 minute = \( \frac{1}{60} \) hr

\( \implies \frac{0.16}{x - y} = \frac{1}{60} \)

\( \implies x - y = 9.6 \) ...(i)
When trains are moving in opposite direction, then their speed of crossing = \( (x + y) \) km/hr
Time taken = \( \frac{0.16}{x + y} \) hr
A.T.Q., time taken = 3 second = \( \frac{3}{3600} = \frac{1}{1200} \) hr

\( \implies \frac{0.16}{x + y} = \frac{1}{1200} \)

\( \implies x + y = 192 \) ...(ii)
From (i) and (ii), we get
\( x - y = 9.6 \)
\( x + y = 192 \)
\( 2x = 201.6 \)

\( \implies x = 100.8 \)
When \( x = 100.8 \), eq. (ii) becomes
\( 100.8 + y = 192 \)

\( \implies y = 192 - 100.8 = 91.2 \)
\( \therefore \) Speed of faster train = 100.8 km/h
Speed of other train = 91.2 km/h

PRACTICE QUESTIONS

Question. The difference between a two digit number and the number obtained by interchanging the digits is 27. What is the difference between the two digits of the number?
(a) 9
(b) 6
(c) 12
(d) 3
Answer: (d) 3

Question. A two – digit number is formed by either subtracting 17 from nine times the sum of the digits or by adding 21 to 13 times the difference of the digits Find the difference of the digits. Find the number.
(a) 37
(b) 73
(c) 71
(d) 72
Answer: (b) 73

Question. A two –digit number is seven times the sum of its digits. The number formed by reversing the digits is 6 more than half of the original number. Find the difference of the digits of the given number.
(a) 2
(b) 3
(c) 4
(d) 5
Answer: (a) 2

Question. Solve for \( x \) and \( y \) using elimination method: \( x + 2y = 5; \frac{3x}{2} + 3y = 10 \)
Answer: No solution

Question. Solve for \( x \) and \( y \) using elimination method: \( 7x - 2y = 9; 4x + 6y = 15 \)
Answer: \( x = \frac{42}{25}, y = \frac{69}{50} \)

Question. The sum of the digits of a two digit number is 8 and the difference between the number and that formed by reversing the digits is 18. Find the number.
Answer: 53

Question. Students of a class are made to stand in rows. If 4 students are extra in a row, there would be two rows less. If 4 students are less in a row, there would be four more rows. Find the number of students in the class.
Answer: 96

Question. The sum of a two digit number and the number formed by reversing the order of digits is 154. If the two digits differ by 4, find the number.
Answer: 95 or 59

Question. Solve for \( x \) and \( y \): \( a^2x + b^2y = c^2 ; b^2x + a^2y = d^2 \)
Answer: \( x = \frac{a^2c^2 - b^2d^2}{a^4 - b^4}, y = \frac{a^2d^2 - b^2c^2}{a^4 - b^4} \)

Question. Solve for \( x \) and \( y \) by the method of elimination: \( 2x - 3y = 7; 5x + 2y = 10 \)
Answer: \( x = \frac{44}{19}, y = -\frac{15}{19} \)

Question. If \( p_1x + q_1y + r_1 = 0 \) and \( p_2x + q_2y + r_2 = 0 \) are two linear equations in two variables \( x \) and \( y \) such that \( p_1, q_1, r_1, p_2, q_2 \) and \( r_2 \) are consecutive positive integers in some order, then find the values of \( x \) and \( y \).
Answer: \( x = 1, y = -2 \)

Question. Vijay had some bananas and he divided them into two lots A and B. He sold the first lot at the rate of Rs. 2 for 3 bananas and the second lot at the rate of Rs. 1 per banana and got a total of Rs. 400. If he had sold the first lot at the rate of Rs. 1 per banana and the second lot at the rate of Rs. 4 for 5 bananas, his total collection would have been Rs. 460. Find his total collection if he sells all the bananas at the rate of Rs. 6 for 5 bananas.
Answer: Rs. 600

Question. Use a single graph paper and draw the graph of the following equations: \( 2y - x = 8 \); \( 5y - x = 14 \); \( y - 2x = 1 \) Obtain the vertices of the triangle so obtained.
Answer: The vertices of the triangle are \( (2, 5), (-4, 2) \) and \( (4, 6) \).

Question. Solve the following system of equations graphically: Also find the points where the lines represented by the given equations intersect the x-axis. \( 3x + 2y = 14, x - 4y = - 7 \)
Answer: The solution of the given system is \( x = 3, y = 2.5 \). The line \( 3x + 2y = 14 \) intersects the x-axis at \( (\frac{14}{3}, 0) \) and the line \( x - 4y = -7 \) intersects the x-axis at \( (-7, 0) \).

Question. A man travels 600 km partly by train and partly by car. It takes 8 hours and 40 minutes if he travels 320 km by train and the rest by car. It would take 30 minutes more if he travels 200 km by train and the rest by car. Find the speed of the train and the car separately.
Answer: Let the speed of the train be \( u \) km/h and the car be \( v \) km/h.
\( \frac{320}{u} + \frac{280}{v} = 8\frac{40}{60} = \frac{26}{3} \) ...(i)
\( \frac{200}{u} + \frac{400}{v} = 9\frac{10}{60} = \frac{55}{6} \) ...(ii)
Solving these, we get: Speed of train = 60 km/h, Speed of car = 80 km/h.

Question. The age of a father is equal to the sum of the ages of his 5 children. After 15 years, sum of the ages of the children will be twice the age of the father. Find the age of father.
Answer: Let the father's age be \( x \) and sum of children's ages be \( y \).
\( x = y \)
After 15 years, father's age = \( x + 15 \), sum of children's ages = \( y + 5(15) = y + 75 \).
\( y + 75 = 2(x + 15) \)

\( \implies x + 75 = 2x + 30 \)

\( \implies x = 45 \).
The age of the father is 45 years.

Question. Find the four angles of a cyclic quadrilateral ABCD in which \( \angle A = (2x - 1)^\circ, \angle B = (y + 5)^\circ, \angle C = (2y + 15)^\circ \) and \( \angle D = (4x - 7)^\circ \).
Answer: In a cyclic quadrilateral, opposite angles sum to \( 180^\circ \).
\( \angle A + \angle C = 180^\circ \implies 2x - 1 + 2y + 15 = 180 \implies x + y = 83 \) ...(i)
\( \angle B + \angle D = 180^\circ \implies y + 5 + 4x - 7 = 180 \implies 4x + y = 182 \) ...(ii)
Subtracting (i) from (ii): \( 3x = 99 \implies x = 33 \).
Then \( y = 83 - 33 = 50 \).
Angles are: \( \angle A = 65^\circ, \angle B = 55^\circ, \angle C = 115^\circ, \angle D = 125^\circ \).

Question. A man starts his job with a certain monthly salary and earns a fixed increment every year. If his salary was Rs. 1500 after 4 years of service and Rs. 1800 after 10 years of service, what was his starting salary and what is the annual increment?
Answer: Let starting salary be \( x \) and annual increment be \( y \).
\( x + 4y = 1500 \) ...(i)
\( x + 10y = 1800 \) ...(ii)
Subtracting (i) from (ii): \( 6y = 300 \implies y = 50 \).
\( x + 4(50) = 1500 \implies x = 1300 \).
Starting salary = Rs. 1300, Annual increment = Rs. 50.

Question. The incomes of X and Y are in the ratio of 8 : 7 and their expenditures are in the ratio of 19 : 16. If each saves Rs. 1250, find their income.
Answer: Let incomes be \( 8x, 7x \) and expenditures be \( 19y, 16y \).
\( 8x - 19y = 1250 \) ...(i)
\( 7x - 16y = 1250 \) ...(ii)
Solving for \( x \): \( x = 750 \).
Income of X = \( 8 \times 750 = Rs. 6000 \), Income of Y = \( 7 \times 750 = Rs. 5250 \).

Question. A bag contains 94 coins of 50 paise and 25 paise denominations. If the total worth of these coins be Rs. 29.75, find the number of coins of each kind.
Answer: Let 50p coins be \( x \) and 25p coins be \( y \).
\( x + y = 94 \) ...(i)
\( 0.50x + 0.25y = 29.75 \) \( \implies 2x + y = 119 \) ...(ii)
Subtracting (i) from (ii): \( x = 25 \).
\( y = 94 - 25 = 69 \).
50 paise coins = 25, 25 paise coins = 69.

Question. A man travels 600 km partly by train and partly by car. If he covers 400 km by train and rest by car, it takes him 6 hours and 30 minutes. But if he travels 200 km by train and rest by car. He takes half an hour longer. Find the speed of the train and that of car.
Answer: Let speed of train be \( u \), speed of car be \( v \).
\( \frac{400}{u} + \frac{200}{v} = 6.5 \)
\( \frac{200}{u} + \frac{400}{v} = 7 \)
Solving these: Speed of train = 100 km/h, Speed of car = 80 km/h.

Question. Solve for x and y: \( 3x - \frac{y + 7}{11} + 2 = 10 \), \( 2y + \frac{x + 11}{7} = 10 \)
Answer: \( x = 3, y = 4 \).

Question. Solve for x and y: \( 631x + 279y = 910 \), \( 279x + 631y = 910 \)
Answer: \( x = 1, y = 1 \).

Question. Solve for x and y: \( 254x + 309y = - 55 \), \( 309x + 254y = 55 \)
Answer: \( x = 1, y = -1 \).

Question. Solve for x and y: \( 3x + 2y = 2x + y + 3 = 4x + 3y - 3 \)
Answer: Equating first two: \( x + y = 3 \). Equating last two: \( 2x + 2y = 6 \implies x + y = 3 \).
The system has infinitely many solutions satisfying \( x + y = 3 \). One such solution is \( (3, 0) \).

Question. Solve for x and y: \( 6x + 3y = 8x + 9y - 5 = 10x + 12y - 8 \)
Answer: \( x = 1, y = \frac{1}{2} \).

Question. Ratio between the girls and boys in a class of 40 students is 2 : 3. Five new students joined the class. How many of them must be boys so that the ratio between girls and boys becomes 4 : 5?
Answer: Initial: Girls = 16, Boys = 24.
Let \( n \) boys join and \( (5 - n) \) girls join.
\( \frac{16 + 5 - n}{24 + n} = \frac{4}{5} \)

\( \implies 5(21 - n) = 4(24 + n) \)

\( \implies 105 - 5n = 96 + 4n \)

\( \implies 9n = 9 \implies n = 1 \).
Wait, let's re-calculate. Total students = 45. Ratio 4:5 means 20 girls and 25 boys.
New girls = 20, original = 16. So 4 girls join.
New boys = 25, original = 24. So 1 boy joins.
Answer: 1 boy.