CBSE Class 10 Pair of Linear Equations in Two Variables Sure Shot Questions Set 08

Question. Two places A and B are 120 km apart from each other on a highway. One car starts from A and another from B at the same time. If they move in the same direction, they meet in 6 hours and if they move in opposite directions, they meet in 1 hour and 12 minutes. Find the speed of the cars.
Answer: Let the speed of car starting from A = \( x \) km/hr. And the speed of car starting from B = \( y \) km/h.
While moving in the same-direction let they meet at C.
Distance travelled by first car in 6 hours = AC = \( 6x \).
Distance travelled by second car in 6 hours = BC = \( 6y \).
According to the first condition.
AC = AB + BC
\( \Rightarrow 6x = 120 + 6y \)
(\( \because \) distance = Speed \( \times \) Time)
\( \Rightarrow x = 20 + y \) ....(1)
According to the second condition,
Distance travelled by first car in \( \frac{6}{5} \) hours = AD = \( \frac{6}{5}x \)
Distance travelled by second car in \( \frac{6}{5} \) hours = BD = \( \frac{6}{5}y \)
While moving in the opposite direction let they meet at D.
AD + DB = AB
\( \Rightarrow \frac{6}{5}x + \frac{6}{5}y = 120 \)
[1 hour 12 minutes = \( \frac{6}{5} \) hours]
\( \Rightarrow x + y = 120 \times \frac{5}{6} \)
\( \Rightarrow x + y = 100 \) ....(2)
Substituting \( x = 20 + y \) from equation (1) in equation (2), we get
\( (20 + y) + y = 100 \)
\( \Rightarrow 2y = 80 \)
\( \Rightarrow y = 40 \) km/hour
Putting \( y = 40 \) in equation (1), we have
\( x = 20 + 40 = 60 \) km/hour
Hence, the speed of first car = 60 km/hour. And the speed of second car = 40 km/hour.

Question. A plane left 30 minutes later than the scheduled time and in order to reach the destination 1500 km away in time, it has to increase the speed by 250 km/hr from the usual speed. Find its usual speed.
Answer: Let the usual speed of plane = \( x \) km/hr. The increased speed of the plane = \( y \) km/hr.
\( \Rightarrow y = (x + 250) \) km/hour. ....(1)
Distance = 1500 km.
According to the question,
(Scheduled time) – (time in increasing the speed) = 30 minutes.
\( \frac{1500}{x} – \frac{1500}{y} = \frac{1}{2} \) ....(2)
\( \frac{1500}{x} – \frac{1500}{x + 250} = \frac{1}{2} \) [\( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \)]
\( \Rightarrow \frac{1500x + 375000 - 1500x}{x(x + 250)} = \frac{1}{2} \)
\( \Rightarrow x(x + 250) = 750000 \)
\( \Rightarrow x^2 + 250x – 750000 = 0 \)
\( \Rightarrow x^2 + 1000x – 750x – 750000 = 0 \)
\( \Rightarrow (x – 750) (x + 1000) = 0 \)
\( \Rightarrow x = 750 \) or \( x = – 1000 \)
But speed can never be – ve
Hence, Usual speed = 750 km/hr.

Problems Based on Boat & Stream

Question. A boat goes 16 km upstream and 24 km downstream in 6 hours. It can go 12 km upstream and 36 km downstream in the same time. Find the speed of the boat in still water and the speed of the stream.
Answer: Let the speed of stream = \( y \) km/hr ; speed of boat in still water = \( x \) km/hr.
And the speed of boat in upstream = \( (x – y) \) km/hr.
The speed of boat in downstream = \( (x + y) \) km/hr.
According to the question,
Time taken in going 16 km upstream + time taken in going 24 km downstream = 6 hours.
\( \Rightarrow \frac{16}{x - y} + \frac{24}{x + y} = 6 \) ....(1) [\( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \)]
Again, according to the question,
Time taken in going 12 km upstream + time taken in going 36 km downstream = 6 hours.
\( \Rightarrow \frac{12}{x - y} + \frac{36}{x + y} = 6 \) ....(2)
Let \( \frac{1}{x - y} = p \), \( \frac{1}{x + y} = q \)
Equation (1) becomes \( 16p + 24q = 6 \) ....(3)
Equation (2) becomes \( 12p + 36q = 6 \) ....(4)
Multiplying equation (3) by 3 and equation (4) by 4, we get
\( 48p + 72q = 18 \) ....(5)
\( 48p + 144q = 24 \) ....(6)
Subtracting equation (5) from equation (6), we get
\( 72q = 6 \Rightarrow q = \frac{6}{72} = \frac{1}{12} \)
Putting the value of \( q \) in equation (3), we get
\( 16p + 24 (\frac{1}{12}) = 6 \)
\( \Rightarrow 16p + 2 = 6 \)
\( \Rightarrow 16p = 6 – 2 = 4 \)
\( \Rightarrow p = 1/4 \)
\( \therefore \frac{1}{x - y} = \frac{1}{4} \) and \( \frac{1}{x + y} = \frac{1}{12} \)
\( \Rightarrow x – y = 4 \) ....(7)
And, \( x + y = 12 \) ....(8)
By adding \( 2x = 16 \)
\( \Rightarrow x = 8 \)
Putting \( x = 8 \) in equation (7), we get
\( 8 – y = 4 \)
\( \Rightarrow y = 8 – 4 = 4 \)
Hence, speed of boat in still water = 8 km/hr. and speed of stream = 4 km/hr.

Question. A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km up stream and 55 km down stream. Determine the speed of the stream and that of the boat.
Answer: Let the speed of the boat in still water be \( x \) km/hr and speed of the stream be \( y \) km/hr. Then the speed of the boat downstream = \( (x + y) \) km/hr, and the speed of the boat upstream = \( (x – y) \) km/hr. Also time = distance/speed.
In the first case, when the boat goes 30 km upstream, let the time taken be \( t_1 \). Then
\( t_1 = \frac{30}{(x - y)} \)
Let \( t_2 \) be the time taken by the boat to go 44 km downstream. Then \( t_2 = \frac{44}{(x + y)} \). The total time taken, \( t_1 + t_2 \), is 10 hours Therefore, we get the equation
\( \frac{30}{(x - y)} + \frac{44}{(x + y)} = 10 \) ....(1)
In the second case in 13 hours it can go 40 km upstream and 55 km downstream. We get the equation
\( \frac{40}{x - y} + \frac{55}{x + y} = 13 \) ....(2)
Let \( \frac{1}{(x - y)} = u \) and \( \frac{1}{(x + y)} = v \) ....(3)
On substituting these values in equations (1) and (2), we get the linear pair
\( 30u + 44v = 10 \) ....(4)
\( 40u + 55v = 13 \) ....(5)
Multiplying equation (4) by 4 and equation (5) by 3, we get
\( 120u + 176v = 40 \)
\( 120u + 165v = 39 \)
On subtracting the two equations, we get
\( 11v = 1 \), i.e., \( v = \frac{1}{11} \)
Substituting the value of \( v \) in equation (4), we get
\( 30u + 4 = 10 \)
\( \Rightarrow 30u = 6 \)
\( \Rightarrow u = \frac{1}{5} \)
On putting these values of \( u \) and \( v \) in equation (3), we get
\( \frac{1}{(x - y)} = \frac{1}{5} \) and \( \frac{1}{(x + y)} = \frac{1}{11} \)
i.e., \( (x – y) = 5 \) and \( (x + y) = 11 \)
Adding these equations, we get
i.e., \( 2x = 16 \) i.e., \( x = 8 \)
Subtracting the equations, we get
\( 2y = 6 \) i.e., \( y = 3 \)
Hence, the speed of the boat in still water is 8 km/hr and the speed of the stream is 3 km/hr.

Question. A sailor goes 8km downstream in 40 minutes and returns in 1 hour. Determine the speed of the sailor in still water and speed of the current.
Answer: We know that 40 minutes = \( \frac{40}{60} \) hr = \( \frac{2}{3} \) hr
Let the speed of the sailor in still water be \( x \) km./hr and the speed of the current be \( y \) km/hr.
We know that speed = \( \frac{\text{distance}}{\text{time}} \)
time = \( \frac{\text{distance}}{\text{speed}} \)
speed of upstream = \( (x – y) \) km/hr and speed of downstream = \( (x + y) \) km/hr
For the first case, we get
\( \frac{2}{3} = \frac{8}{x + y} \) [\( \text{time} = \frac{\text{distance}}{\text{speed}} \)]
\( \Rightarrow 2x + 2y = 24 \)
\( \Rightarrow x + y = 12 \) ....(1)
For the second case, we get
\( 1 = \frac{8}{x - y} \) [\( \text{time} = \frac{\text{distance}}{\text{speed}} \)]
\( \Rightarrow x – y = 8 \) ....(2)
Adding equations (1) and (2), we get
\( 2x = 20 \)
\( \Rightarrow x = 10 \) km/hr
Substituting \( x = 10 \) in equation (1), we get
\( 10 + y = 12 \)
\( \Rightarrow y = 2 \) km/hr
Hence, speed of the sailor in still water and speed of the current are 10 km/hr and 2km/hr respectively.

Question. A person rows downstream 20 km in 2 hours and upstream 4 km in 2 hours. Find man’s speed of rowing in still water and the speed of the current.
Answer: Let man’s speed of rowing in still water and the speed of the current be \( x \) km/hr and \( y \) km/hr respectively.
Then, the upstream speed = \( (x – y) \) km/hr and the downstream speed = \( (x + y) \) km/hr
we know that time = \( \frac{\text{distance}}{\text{speed}} \)
First case :
\( \Rightarrow 2 = \frac{20}{x + y} \)
\( \Rightarrow 2(x + y) = 20 \)
\( \Rightarrow x + y = 10 \) ....(1)
2nd Case :
\( 2 = \frac{4}{x - y} \Rightarrow 2 (x – y) = 4 \)
\( \Rightarrow x – y = 2 \) ....(2)
Adding (1) and (2), we get
\( \Rightarrow 2x = 12 \)
\( \Rightarrow x = 6 \)
Substituting the value of \( x \) in equation (1), we get
\( 6 + y = 10 \Rightarrow y = 4 \)
Hence, man’s speed of rowing in still water and the speed of the current are 6 km/hr and 4 km/hr respectively.

Problems Based on Area

Question. If in a rectangle, the length is increased and breadth reduced each by 2 metres, the area is reduced by 28 sq. metres. If the length is reduced by 1 metre and breadth increased by 2 metres, the area increases by 33 sq. metres. Find the length and breadth of the rectangle.
Answer: Let length of the rectangle = \( x \) metres And breadth of the rectangle = \( y \) metres
Area = length \( \times \) breadth = \( xy \) sq. metres
Case 1: As per the question
Increased length = \( x + 2 \)
Reduced breadth = \( y – 2 \)
Reduced area = \( (x + 2) (y – 2) \)
Reduction in area = 28
Original Area – Reduced area = 28
\( xy – [(x + 2) (y – 2)] = 28 \)
\( \Rightarrow xy – [xy – 2x + 2y – 4] = 28 \)
\( \Rightarrow xy – xy + 2x – 2y + 4 = 28 \)
\( \Rightarrow 2x – 2y = 28 – 4 = 24 \)
\( \Rightarrow x – y = 12 \) ....(1)
Case 2 :
Reduced length = \( x – 1 \)
\( \Rightarrow \) Increased breadth = \( y + 2 \)
\( \Rightarrow \) Increased area = \( (x – 1) (y + 2) \)
Increase in area = 33
\( \therefore \) Increased area – original area = 33
\( \Rightarrow (x – 1) (y + 2) – xy = 33 \)
\( \Rightarrow xy + 2x – y – 2 – xy = 33 \)
\( \Rightarrow 2x – y = 33 + 2 = 35 \)
\( \Rightarrow 2x – y = 35 \) ....(2)
Subtracting equation (1) from equation (2), we get
\( x = 23 \)
Substituting the value of \( x \) in equation (1), we get
\( 2x – y = 12 \)
\( \Rightarrow y = 23 – 12 = 11 \)
\( \Rightarrow \) Length = 23 metres.
\( \Rightarrow \) Breadth = 11 metres.

Question. The area of a rectangle gets reduced by 9 square units if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.
Answer: Let the length of rectangle be \( x \) units and the breadth of the rectangle be \( y \) units
Area of the rectangle = \( xy \)
Case 1 : According to the first condition,
Reduced length = \( x – 5 \)
Increased breadth = \( y + 3 \)
Reduced area = \( (x – 5) (y + 3) \)
Reduction in area = 9
Original area – Reduced area = 9
\( xy – [(x – 5) (y + 3)] = 9 \)
\( \Rightarrow xy –[xy + 3x – 5y – 15] = 9 \)
\( \Rightarrow xy – xy – 3x + 5y + 15 = 9 \)
\( \Rightarrow 3x – 5y = 6 \) ....(1)
Case 2. According to the second condition,
Increased length = \( x + 3 \)
Increased breadth = \( y + 2 \)
Increased area = \( (x + 3) (y + 2) \)
Increase in area = 67
Increased area – Original area = 67
\( \Rightarrow (x + 3) (y + 2) – xy = 67 \)
\( \Rightarrow xy + 2x + 3y + 6 – xy = 67 \)
\( 2x + 3y = 61 \) ....(2)
On solving (1) and (2), we get \( x = 17 \) units and \( y = 9 \) units
Hence, length of rectangle = 17 units, and breadth of rectangle = 9 units.

Problems Based on Geometry

Question. The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
Answer: Let the angles be \( x \) and \( y \). Then according to the question.
\( x + y = 180 \) ....(1)
and \( x = y + 18 \) ....(2)
Putting \( x = y + 18 \) from (2) in (1), we get
\( y + 18 + y = 180 \)
\( 2y = 180 – 18 \)
\( \Rightarrow 2y = 162 \)
\( \Rightarrow y = 81 \)
Putting \( y = 81 \) in (2), we get
\( x = 81 + 18 = 99 \)
Hence, angles are \( x = 99^\circ \) and \( y = 81^\circ \).

Question. In a \( \Delta ABC \), \( \angle C = 3 \angle B = 2 (\angle A + \angle B) \). Find the three angles.
Answer: \( \angle C = 2(\angle A + \angle B) \) ....(1) (given)
Adding \( 2\angle C \) on both sides of (1), we get
\( \angle C + 2\angle C = 2 (\angle A + \angle B) + 2\angle C \)
\( \Rightarrow 3\angle C = 2 (\angle A + \angle B + \angle C) \)
\( \Rightarrow \angle C = \frac{2}{3} \times 180^\circ = 120^\circ \)
Again \( \angle C = 3\angle B \) (given)
\( 120^\circ = 3\angle B \)
\( \Rightarrow \angle B = \frac{120^\circ}{3} = 40^\circ \)
But \( \angle A + \angle B + \angle C = 180^\circ \)
\( \angle A + 40^\circ + 120^\circ = 180^\circ \)
\( \Rightarrow \angle A = 180^\circ – 40^\circ – 120^\circ = 20^\circ \)
\( \angle A = 20^\circ, \angle B = 40^\circ, \angle C = 120^\circ \)

Question. Find a cyclic quadrilateral ABCD, \( \angle A = (2x + 4)^\circ \), \( \angle B = (y + 3)^\circ \), \( \angle C = (2y + 10)^\circ \) and \( \angle D = (4x – 5)^\circ \). Find the four angles.
Answer: \( \angle A = (2x + 4)^\circ \), and \( \angle C = (2y + 10)^\circ \);
But \( \angle A + \angle C = 180^\circ \) (Cyclic quadrilateral)
\( \Rightarrow (2x + 4)^\circ + (2y + 10)^\circ = 180^\circ \)
\( \Rightarrow 2x + 2y = 166^\circ \)
Also \( \Rightarrow \angle B = (y + 3)^\circ \), \( \angle D = (4x – 5)^\circ \)
But \( \angle B + \angle D = 180^\circ \) (Cyclic quadrilateral)
\( \Rightarrow (y + 3)^\circ + (4x – 5)^\circ = 180^\circ \)
\( \Rightarrow 4x + y = 182^\circ \)
On solving (1) and (2), we get \( x = 33^\circ, y = 50^\circ \)
\( \angle A = (2x + 4)^\circ = (66 + 4)^\circ = 70^\circ \)
\( \angle B = (y + 3)^\circ = (50 + 3)^\circ = 53^\circ \)
\( \angle C = (2y + 10)^\circ = (100 + 10)^\circ = 110^\circ \),
\( \angle D = (4x – 5)^\circ = (4 \times 33 – 5)^\circ = 127^\circ \)
\( \angle A = 70^\circ, \angle B = 53^\circ, \angle C = 110^\circ, \angle D = 127^\circ \)

Question. The area of a rectangle remains the same if the length is decreased by 7 dm and breadth is increased by 5 dm. The area remains unchanged if its length is increased by 7 dm and and breadth decreased by 3 dm. Find the dimensions of the rectangle.
Answer: Let the length and breadth of a rectangle be \( x \) and \( y \) units respectively. So, area = \( (xy) \) sq. units.
First Case : Length is decreased by 7 dm and breadth is increased by 5 dm.
According to the question,
\( xy = (x – 7) (y + 5) \)
\( \Rightarrow xy = xy + 5x – 7y – 35 \)
\( \Rightarrow 5x – 7y – 35 = 0 \) ....(1)
Second Case : Length is increased by 7 dm and breadth is decreased by 3 dm.
Here, area also remains same
so, we get
\( xy = (x + 7) (y – 3) = xy – 3x + 7y – 21 \)
\( \Rightarrow 3x – 7y + 21 = 0 \) ....(2)
So, the system of equations becomes
\( \Rightarrow 5x – 7y – 35 = 0 \) ....(3)
\( 3x – 7y + 21 = 0 \) ....(4)
Subtracting equation (4) from (3), we get
\( 2x – 56 = 0 \)
\( \Rightarrow 2x = 56 \Rightarrow x = 28 \) dm
Substituting \( x = 28 \) in equation (3), we get
\( 5 \times 28 – 7y = 35 \)
\( \Rightarrow 7y = 105 \Rightarrow y = 15 \) dm
Hence, length and breadth of the rectangle are 28 dm and 15 dm respectively.

Question. In a triangle PQR, \( \angle P = x^\circ \), \( \angle Q = (3x – 2)^\circ \), \( \angle R = y^\circ \), \( \angle R – \angle Q = 9^\circ \). Determine the three angles.
Answer: It is given that \( \angle P = x^\circ \), \( \angle Q = (3x – 2)^\circ \), \( \angle R = y^\circ \) and \( \angle R – \angle Q = 9^\circ \)
We know that the sum of three angles in a triangle is \( 180^\circ \).
So, \( \angle P + \angle Q + \angle R = x + 3x – 2 + y = 180 \)
\( \Rightarrow 4x + y = 182 \) ....(1)
It is also given that \( \angle R – \angle Q = 9^\circ \) or \( y – (3x – 2) = 9 \)
\( \Rightarrow y – 3x + 2 = 9 \)
\( \Rightarrow 3x – y = –7 \) ....(2)
Adding equation (1) with (2), we get
\( 7x = 175 \)
\( \Rightarrow x = 25 \)
Substituting \( x = 25 \) in equation (2), we get
\( 3 \times 25 – y = – 7 \)
\( \Rightarrow y = 75 + 7 = 82 \)
Thus, \( P = x^\circ = 25^\circ \)
\( Q = (3x – 2)^\circ = (3 \times 25 – 2)^\circ = 73^\circ \) and \( R = y = 82^\circ \).

EXERCISE 

Question. Reena has pens and pencils which together are 40 in number. If she has 5 more pencils and 5 less pens, then number of pencils would become 4 times the number of pens. Find the original number of pens and pencils.
Answer: No. of pens = 13, No. of pencils = 27

Question. 5 pens and 6 pencils together cost ₹ 9.00, and 3 pens and 2 pencils cost ₹ 5.00. Find the cost of 1 pen 1 pencil.
Answer: Cost of pen = ₹ 1.50, cost of pencil = 0.25

Question. Two numbers differ by 2 and their product is 360. Find the numbers.
Answer: 18 and 20

Question. A two-digit number is 3 more than 4 times the sum of its digits. If 18 is added to the number, the digits are reversed. Find the number.
Answer: 35

Question. A two-digit number is 4 times the sum of its digits. If 18 is added to the number, the digit are reversed. Find the number.
Answer: 24

Question. The denominator of a fraction is 4 more than twice the numerator. When both the numerator and denominator are decreased by 6, then the denominator becomes 12 times the numerator. Determine the fraction.
Answer: \( \frac{7}{18} \)

Question. The area of a rectangle gets reduced by 80 sq. units if its length is reduced by 5 units and the breadth is increased by 2 units. If we increase the length by 10 units and decrease the breadth by 5 units, the area is increased by 50 sq. units. Find the length and breadth of the rectangle.
Answer: Length = 40, Breadth = 30

Question. In \( \Delta ABC \), \( \angle A = yº \), \( \angle B = (y – 9)º \), \( \angle C = xº \). Also \( \angle B – \angle C = 48º \), find the three angles.
Answer: 82º, 73º, 25º

Question. The largest angles of the triangle is twice the sum of the other two, the smallest is one-sixth of the largest. Find the angles in degrees.
Answer: 120º, 40º, 20º

Question. The difference between two numbers is 642. When the greater is divided by the smaller, the quotient is 8 and remainder is 19. Find the numbers.
Answer: 89, 731

Question. Of the three angles of a triangle the second is one-third the first and third is 26º more than the first. How many degrees are there in each angle ?
Answer: 66º, 22º, 92º

Question. In a factory, women are 35% of all the workers, the rest of the workers being men. The number of men exceeds that of women by 252. Find the total number of workers in the factory.
Answer: 840

Question. A total of 1400 kg of potatoes were sold in three days. On the first day 100 kg less potatoes were sold than on the second day and on the third day, 3/5 of the amount sold on the first day. How many kilograms of potatoes were sold on each day ?
Answer: 500 kg, 600 kg, 300 kg

Question. The sum of a certain even number and the fourth even number after it is 68. Find the numbers.
Answer: 30, 38

Question. Fifty nine pens and forty seven pencils together cost ₹ 513, while forty seven pens and fifty nine pencils together cost ₹ 441. Find the cost of a pen and that of a pencil.
Answer: Cost of pen = ₹ 7.50, cost of pencil ₹ 1.50

Question. The sum of two numbers is 15 and sum of their reciprocals is 3/10. Find the numbers.
Answer: 5 and 10

Question. The sum of two numbers is 16 and the sum of their reciprocals is 1/3. Find the numbers.
Answer: 12 and 4

Question. Solution of the equation \( \frac{2x - 3}{5} + \frac{x + 3}{4} = \frac{4x + 1}{7} \)
(A) 7
(B) \( -\frac{41}{11} \)
(C) \( \frac{1}{11} \)
Answer: (C) \( \frac{1}{11} \)

Question. Solution of the equation \( \frac{7x - 1}{4} - \frac{1}{3} \left[ 2x - \frac{1 - x}{2} \right] = \frac{19}{3} \)
(A) 7
(B) \( -\frac{41}{11} \)
(C) \( \frac{1}{11} \)
Answer: (A) 7

Question. Solution of the equation \( \frac{4x + 5}{6} - \frac{2(2x + 7)}{3} = \frac{3}{2} \)
(A) 7
(B) \( -\frac{41}{11} \)
(C) \( \frac{1}{11} \)
Answer: (B) \( -\frac{41}{11} \)

Question. Solution of the equation \( \frac{2y - 3}{5} + \frac{y - 3}{4} = \frac{4y + 1}{7} \)
(A) \( \frac{8}{5} \)
(B) \( \frac{209}{11} \)
(C) 1
Answer: (B) \( \frac{209}{11} \)

Question. Solution of the equation \( \frac{3}{x - 1} + \frac{4}{x - 2} = \frac{7}{x - 3} \), \( x \neq 1, 2, 3 \)
(A) \( \frac{8}{5} \)
(B) \( \frac{209}{11} \)
(C) 1
Answer: (A) \( \frac{8}{5} \)

Question. Solution of the equation \( (x + 1) (2x + 1) = (x + 3) (2x + 3) – 14 \)
(A) \( \frac{8}{5} \)
(B) \( \frac{209}{11} \)
(C) 1
Answer: (C) 1

Question. The age of a father is twice that of the elder son. Ten years hence the age of the father will be three times that of the younger son. If the difference of ages of the two sons is 15 years, then find the age of the father.
Answer: 50 years

Question. If \( 2^x – 2^{x–1} = 4 \), then find \( x^x \).
Answer: 27

Question. If \( 2x^2 + xy – 3y^2 + x + ay – 10 = (2x + 3y + b) (x – y – 2) \), then find the value of a and b.
Answer: –11 and 5

Question. A & B together have 45 coins. Both of them lost 5 coins each, and the product of the number of coins they now have is 124. Form the quadratic equation to find how many coins they had to start with, if A had x coins ?
Answer: \( x^2 – 45x + 324 = 0 \)

Question. If we divide 64 into two parts such that one part is three times the other, then find two parts.
Answer: 48

Question. For what values of k will the following pair of linear equations have infinitely many solutions:
\( 2x – 3y = 7, (k + 1)x + (1 – 2k)y = (5k – 4) \).
Answer: k = 5

Question. Find the values of k for which the system of equations \( kx – y = 2, 6x – 2y = 3 \) has (i) a unique solution, (ii) no solution, (iii) is there a value of k for which the given system has infinitely many solutions?
Answer: (i) \( k \neq 3 \), (ii) k = 3, (iii) no

Question. The students of a class are made to stand in rows. If 4 students are extra in each row, there would be 2 rows less. If 4 students are less in each row, there would be 4 rows more. Find the number of students in the class.
Answer: 96

Question. The monthly incomes of A and B are in the ratio 8 : 7 and their expenditures are in the ratio 19 : 16. If each saves ₹ 5000 per month, find the monthly income of each.
Answer: A = ₹ 24000, B = ₹ 21000

Question. The sum of two numbers is 1000 and the difference between their squares is 256000. Find the numbers.
Answer: 628, 372

Question. Places A and B are 160 km apart on a highway. One car starts from A and another from B at the same time. If they travel in the same direction, they meet in 8 hours. But, if they travel towards each other, they meet in 2 hours. Find the speed of each car.
Answer: A = 50 km/hr., B = 30 km/hr.

Question. The area of a rectangle gets reduced by 8 \( m^2 \), when its length is reduced by 5 m and its breadth is increased by 3 m. If we increase the length by 3 m and breadth by 2 m, the area is increased by 74 \( m^2 \). Find the length and the breadth of the rectangle.
Answer: 19m, 10m

Question. Taxi charges in a city consist of fixed charges per day and the remaining depending upon the distance travelled in kilometers. If a person travels 110 km, he pays ₹ 1130, and for travelling 200 km, he pays ₹ 1850. Find the fixed charges per day and the rate per km.
Answer: ₹ 250, ₹ 8 per km

Question. Points A and B are 70 km apart on a highway. A car starts from A and another car starts from B simultaneously. If they travel in the same direction, they meet in 7 hours. But, if they travel towards each other, they meet in 1 hour. Find the speed of each car.
Answer: A = 40 kmph, B = 30 kmph

Question. If twice the son's age in years is added to the mother's age, the sum is 70 years. But, if twice the mother's age is added to the son's age, the sum is 95 years. Find the age of the mother and that of the son.
Answer: 40 years, 15 years.