### RD Sharma Class 10 Solutions Pair Of Linear Equations In Two Variables Ex 3.10

Question 1.

Points A and B are 70 km. a part 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 meets in one hour. Find the speed of the two cars. (C.B.S.E. 2002)

Solution:

Distance between two points A and B = 70 km

Let speed of first car starting from A = x km/hr

and speed of second car starting from B =y km/hr

When these start in the some direction, they meet after 7 hours

Distance travelled by the first car = 7x km

and by second car = 7y km

When these travel in opposite direction, they meet after 1 hour then distance travelled by first car = x km

and by second car = y km

x + y = 70 ….(ii)

Adding (i) and (ii)

2x = 80

⇒ x = 40

and subtracting (i) from (ii)

2y = 60

⇒ y = 30

Speed of first car = 40 km/hr

and speed of second car = 30 km/hr

Question 2.

A sailor goes 8 km downstream in 40 minutes and returns in 1 hour. Determine the speed of the sailor in still water and the speed of the current. (C.B.S.E. 1997)

Solution:

Let the speed of sailor in still water = x km/ hr

and speed of water = y km/hr

Distance covered = 8 km

Question 3.

The boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream. Determine the speed of stream and that of the boat in still water.

Solution:

Let the speed of stream = y km/hr

and speed of boat = x km/hr

Speed of boat downstream = (x + y) km/hr

In first case, and up stream = (x – y) km/hr

Upstream distance = 30 km

and down distance = 44 km

Total time taken = 10 hrs

In second case,

upstream distance = 40 km

and downstream distance = 55 km

Total time taken = 13 hrs

Question 4.

A boat goes 24 km upstream and 28 km downstream in 6 hrs. It goes 30 km upstream and 21 km downstream in 6hrs. Find the speed of the boat in still water and also speed of the stream.

Solution:

Let the speed of boat = x km/hr

and speed of stream = y km/hr

In first case,

Distance covered upstream = 24 km

and down stream = 28 km

Total time taken = 6 hours

In second case,

Distance covered upstream = 30 km

and downstream = 21 km

Total time taken = 6= hrs.

Question 5.

A man walks a certain distance with a certain speed. If he walks 1/2 km an hour faster, he takes 1 hour less. But, if he walks 1 km an hour slower, he takes 3 more hours. Find the distance covered by the man and his original rate of walking.

Solution:

Let the distance = x km

and let certain speed = y km/hr

Question 6.

A person rowing at the rate of 5 km/h in still water, takes thrice as much time in going 40 km upstream as in going 40 km downstream. Find the speed of the stream. [NCERT Exemplar]

Solution:

Let the speed of the stream be v km/h

Given that, a person rowing in still water = 5 km/h

The speed of a person rowing in downstream = (5 + v) km/h

and the speed of a person Has rowing in upstream = (5 – v) km/h

Now, the person taken time to cover 40 km downstream,

Question 7.

Ramesh travels 760 km to his home partly by train and partly by car. He takes 8 hours if he travels 160 km. by train and the rest by car. He takes 12 minutes more if the travels 240 km by train and the rest by car. Find the speed of the train and car respectively.

Solution:

Total distance = 769 km

Let the speed of train = x km/hr

and speed of car = y km/hr

Time taken = 8 hours

In first case, distance travelled by train = 160 km

and rest distance 760 – 160 = 600 km by car

Time taken

Question 8.

A man travels 600 km partly by train and partly by car. If the covers 400 km by train and the rest by car, it takes him 6 hours and 30 minutes. But, if he travels 200 km by train and the rest by car, he takes half an hour longer. Find the speed of the train and that of the car. (C.B.S.E. 2001)

Solution:

Total journey = 600 km

Let the speed of train. = x km/hr

and speed of car = y km/hr

In first case,

Journey by train = 400 km

Journey by car = 600 – 400 = 200 km

Question 9.

Places A and B are 80 km apart from each other on a highway. A car starts from A and other from B at the same time. If they move in the same direction, they meet in 8 hours and if they move in opposite directions, they meet in 1 hour and 20 minutes. Find the speeds of the cars. [CBSE 2002]

Solution:

Distance between A and B = 80 km

One car starts from A and other from B in the same direction and they meet after 8 hours

Let the speed of the first car = x km/hr

and speed of second car = y km/hr

Distance travelled by the first car = 8x km

and by the second car = 8y km

8x – 8y = 80

⇒ x – y = 10 ….(i)

When the two cars move in opposite direction, they

Question 10.

A boat goes 12 km upstream and 40 km downstream in 8 hours. It can go 16 km upstream and 32 km downstream in the same time. Find the speed of the boat in still water and the speed of the steam. [CBSE 1999C]

Solution:

Let the speed of boat = x km/hr

and speed of stream = y km/hr

⇒ Speed of the boat upstream = (x – y) km/hr

and speed of boat down stream = x + y km/ hr

Hence speed of boat = 6 km/hr and speed streams = 2 km/hr

Question 11.

Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.

Solution:

Total distance = 300 km

Let the speed of train = x km/hr

and speed of bus = y km/hr

In first case,

Distance travelled by train = 60 km

and distance by bus = 300 – 60 = 240

and total time taken = 4 hrs

Question 12.

Ritu can row downstream 20 km in 2 hours and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.

Solution:

Let the speed of rowing in still water = x km/hr

and speed of current of water = y km/hr

Speed of downstream = (x + y) km/hr

and speed of upstream = (x – y) km/hr

Speed of rowing = 6 km/hr and speed of current = 4 km/hr

Question 13.

A motor boat can travel 30 km upstream and 28 km downstream in 7 hours. It can travel 21 km upstream and return in 5 hours. Find the speed of the boat in still water and the speed of the stream. [NCERT Exemplar]

Solution:

Let the speed of the motorboat in still water and the speed of the stream are u km/h and v km/h, respectively.

Then, a motorboat speed in downstream = (u + v) km/h

and a motorboat speed in upstream = (u – v) km/h

Motorboat has taken time to travel 30 km upstream,

Question 14.

Abdul travelled 300 km by train and 200 km by taxi, it took him 5 hours 30 minutes. But it he travels 260 km by train and 240 km by taxi he takes 6 minutes longer. Find the speed of the train and that of the taxi. (CBSE 2006C)

Solution:

Let the speed of train = x km/hr

and speed of taxi = y km/hr

In first case,

Question 15.

A train covered a certain distance at a uniform speed. If the train could have been 10 km/hr. faster, it would have taken 2 hours less than the Scheduled time. And, if the train were slower by 10 km/ hr ; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.

Solution:

Let the speed of the train = x km/hr

and time taken = y hours

Distance = speed x time = x x y = xy km

In the first case,

Speed = (x + 10) km/hr

Time = (y – 2) hours

Distance = (x + 10) (y – 2) = xy

⇒ xy – 2x+ 10y – 20 = xy

⇒ -2x + 10y – 20 = 0

⇒ x – 5y + 10 = 0 ……….(i)

In second case,

Speed of the train = (x – 10) km/hr

and time = (y + 3) hours

Distance = (x – 10) (y + 3) = xy

⇒ xy + 3x – 10y – 30 = xy

⇒ 3x – 10y – 30 = 0 ………(ii)

Multiplying (i) by 2 and (ii) by 1

Question 16.

Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of two cars ?

Solution:

Distance between A and B places = 100 km

Let the speed of first car (starting from A) = x km/hr

and speed of second car (starting from B) = y km/hr

In first case, when the cars travel in the same direction

Time when they meet = 5 hours

Distance travelled by first car in 5 hours = 5x km

and by second car = 5y = 5y km

5x – 5y = 100

⇒ x – y = 20 …(i)

In second case, when the cars travel in the opposite direction

Time when they meet = 1 hour

Distance travelled by first case = x x 1 = x km

and by second car = y x 1 = y km

x + y = 100 ……..(i)

Adding (i) and (ii)

2x = 120 ⇒ x = 60

Subtracting (i) from (ii)

2y = 80 ⇒ y = 40

Speed of first car = 60 km/hr and speed of second car = 40 km/hr

Question 17.

While covering a distance of 30 km. Ajeet takes 2 hours more than Amit. If Ajeet doubles his speed, he would take 1 hour less than Amit. Find their speeds of walking.

Solution:

Total distance = 30 km

Let speed of Ajeet = x km/hr

and speed of Amit = y km/hr

Now according to the condition,

Speed of Ajeet = 5 km/hr and speed of Amit = 7.5 km/hr

Question 18.

A takes 3 hours more than B to walk a distance of 30 km. But if A doubles his pace (speed) he is ahead of B by 1hours. Find their speeds.

Solution:

Let speed of A = x km/hr

and speed of B = y km/hr

Total distance in first case,

– = 3

### Pair of Linear Equations in Two Variables Class 10 Solutions Exercise 3.10

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### RD Sharma Class 10 Solutions Pair Of Linear Equations In Two Variables Ex 3.11

Question 1.

If in a rectangle, the length is increased and breadth reduced each by 2 units, the area is reduced by 28 square units. If, however the length is reduced by 1 unit and the breadth increased by 2 units, the area increased by 33 square units. Find the area of the rectangle.

Solution:

Let the length of rectangle = x units

and breadth = y units

Area = Length x breadth = x x y = xy sq. units

According to the condition given,

(x + 2) (y – 2) = xy – 28

⇒ xy – 2x + 2y – 4 = xy – 28

⇒ -2x + 2y = -28 + 4 = -24

⇒ x – y = 12 ….(i)

(Dividing by -2)

and (x – 1) (y + 2) = xy + 33

⇒ xy + 2x – y – 2 = xy + 33

⇒ 2x – y = 33 + 2

⇒ 2x – y = 35 ….(ii)

Subtracting (i), from (ii)

x = 23

Substituting the value of x in (i)

23 – y = 12

⇒ -y = 12 – 23 = -11

⇒ y = 11

Area of the rectangle = xy = 23 x 11 = 253 sq. units

Question 2.

The area of a rectangle remains the same if the length is increased by 7 metres and the breadth is decreased by 3 metres. The area remains unaffected if the length is decreased by 7 metres and breadth is increased by 5 metres. Find the dimensions of the rectangle.

Solution:

Let the length of a rectangle = x m

and breadth = y m

Area = Length x breadth = xy sq. m

According to the condition given,

(x + 7)(y – 3) = xy

⇒ xy – 3x + 7y – 21 = xy

-3x + 7y = 21 ….(i)

and (x – 7) (y + 5) = xy

⇒xy + 5x – 7y = xy

5x – 7y = 35 …(ii)

(i) from (ii)

2x = 56

⇒ x = 28

Substituting the value of x in (i)

-3 x 28 + 7y = 21

– 84 +7y = 21

⇒ 7y = 21 + 84 = 105

y = 15

Length of the rectangle = 28 m and breadth = 15 m

Question 3.

In a rectangle, if the length is increased by 3 metres and breadth is decreased by 4 metres, the area of the rectangle is reduced by 67 square metres. If length reduced by 1 metre and breath increased by 4 metres, the area is increased by is 89 sq. metres. Find the dimensions of the rectangle.

Solution:

Let length of rectangle = x m

and breadth = y m

Area = Length x breadth = xy m^{2}

According to the given condition,

(x + 3) (y – 4) = xy – 67

⇒ xy – 4x + 3y – 12 = xy – 67

⇒-4x + 3y = -67+ 12 = -55

⇒ 4x – 3y = 55 ….(i)

and (x -1) (y + 4) = xy + 89

⇒ xy + 4x – y – 4 = xy + 89

⇒ 4x – y = 89 + 4 = 93 ….(ii)

Subtracting (i) from (ii)

2y = 38

⇒ y = 19

Substituting the value of y in (i)

4x – 3 x 19 = 55

⇒ 4x – 57 = 55

⇒ 4x = 55 + 57 = 112

⇒ x = 28

Length of the rectangle = 28 m and breadth = 19 m

Question 4.

The incomes of X and Y are in the ratio of 8 : 7 and their expenditures are in the ratio 19 : 16. If each saves ₹ 1250, find their incomes.

Solution:

Incomes of X and Y are in the ratio 8 : 7

and their expenditures = 19 : 16

Let income of X = ₹ 8x

and income of Y = ₹ 7x

and let expenditures of X = 19y

and expenditure of Y = 16y

Saving in each case is save i.e. ₹ 1250

8x – 19y = 1250 ….(i)

and 7x – 16y = 1250 ….(ii)

Multiplying (i) by 7 and (ii) by 8, we get

Question 5.

A and B each has some money. If A gives ₹ 30 to B, then B will have twice the money left with A, but if B gives ₹ 10 to A, then A will have thrice as much as is left with B. How much money does each have?

Solution:

Let A’s money = 7 x

and B’s money = 7 y

According to the given conditions,

2(x – 30) = 7 + 30

⇒ 2x – 60 = y + 30

⇒ 2x – y = 30 + 60 = 90 ….(i)

and x + 10 = 3 (y – 10)

⇒ x + 10 = 3y – 30

⇒ x – 3y = -30 – 10 = -40 ………(ii)

Multiplying (i) by 3 and (ii) by 1,

Question 6.

ABCD is a cyclic quadrilateral such that ∠A = (4y + 20)°, ∠B = (3y – 5)°, ∠C = (+4x)° and ∠D = (7x + 5)°. Find the four angles. [NCERT]

Solution:

Question 7.

2 men and 7 boys can do a piece of work in 4 days. The same work is done in 3 days by 4 men and 4 boys. How long would it take one man and one boy to do it ?

Solution:

Let one man can do a work is = x days

Question 8.

In a ∆ABC, ∠A = x°, ∠B = (3x – 2)°, ∠C = y°. Also ∠C – ∠B = 9°. Find the three angles.

Solution:

We know that sum of three angles of a triangle = 180°

∠A + ∠B + ∠C = 180°

⇒ x° + (3x – 2)° + y° = 180°

⇒ x + 3x – 2 + y = 180

⇒ 4x + y = 180 + 2 = 182 …(i)

∠C – ∠B = 9

y – (3x – 2) = 9

⇒ y – 3x + 2 = 9

⇒ y – 3x = 9 – 2

⇒ -3x + y = 7 ….(ii)

Subtracting (ii) from (i)

7x = 175 ⇒ x = 25

Substituting the value of x in (ii)

-3 x 25 + y = 7

⇒ -75 + y = 7

⇒ y = 7 + 75 = 82

∠A = x° = 25°

∠B = 3x – 2 = 3 x 25 – 2 = 73°

∠C = 7 = 82°

Question 9.

In a cyclic quadrilateral ABCD, ∠A = (2x + 4)°, ∠B = (y + 3)°, ∠C = (2y + 10)°, ∠D = (4x – 5)°. Find the four angles.

Solution:

ABCD is a cyclic quadrilateral

and ∠A = (2x + 4)°, ∠B = (y + 3)°, ∠C = (2y + 10)°, ∠D = (4x – 5)°

The sum of opposite angles = 180°

∠A + ∠C=180° and ∠B + ∠D = 180°

⇒ 2x + 4 + 2y + 10 = 180°

⇒ 2x + 2y + 14 = 180°

⇒ 2x + 2y = 180 – 14 = 166

⇒ x + y = 83 ….(i)

(Dividing by 2)

and y + 3 + 4x – 5 = 180°

⇒ 4x + 7 – 2 = 180°

4x + 7 = 180°+ 2= 182° ….(ii)

Subtracting (i) from (ii)

3x = 99

⇒ x = 33

Substituting the value of x in (i)

33 + y = 83

⇒ y = 83 – 33 = 50

∠A = 2x + 4 = 2 x 33 + 4 = 66 + 4 = 70°

∠B = y + 3 = 50 + 3 = 53°

∠C = 2y + 10 = 2 x 50 + 10 = 100 + 10 = 110°

∠D = 4x – 5 = 4 x 33 – 5 = 132 – 5 = 127°

Question 10.

Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test ?

Solution:

Let number of right answers questions = x

and number of wrong answers questions = y

Now according to the given conditions,

3x – y = 40 ….(i)

and 4x – 2y = 50

⇒ 2x – y = 25 ….(ii)

(Dividing by 2)

Subtracting (ii) from (i), x = 15

Substituting the value of x in (i)

3 x 15 – y = 40

⇒ 45 – y = 40

⇒ y = 45 – 40 = 5

x = 15 and y = 5

Total number of questions = 15 + 5 = 20

Question 11.

In a ∆ABC, ∠A = x°, ∠B = 3x° and ∠C = 7°. If 3y – 5x = 30, prove that the triangle is right angled.

Solution:

In a ∆ABC ,

∠A = x°, ∠B = 3x° and ∠C = 7°

But ∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)

⇒ x + 3x + y = 180

⇒4x + y = 180 ………(i)

and 3y – 5x = 30 ….(ii)

from (i) y = 180 – 4x

Substituting the value of y in (ii)

3 (180 – 4x) – 5x = 30

540 – 12x – 5x = 30

⇒ -17x = -540 + 30 = -510

⇒ 17x = 510

⇒ x = 30

y = 180 – 4x = 180 – 4 x 30 = 180 – 120 = 60

∠A = x = 30°

∠B = 3x = 3 x 30° = 90°

∠C = y = 60°

∠B of ∆ABC = 90°

∆ABC is a right angled triangle.

Question 12.

The car hire charges in a city comprise of a fixed charges together with the charge for the distance covered. For a journey of 12 km, the charge paid is ₹ 89 and for a journey of 20 km, the charge paid is ₹ 145. What will a person have to pay for travelling a distance of 30 km?

Solution:

Let the fixed charges = ₹ x

and charges per km = ₹ y

According to the given conditions

Question 13.

A part of monthly hostel charges in a college are fixed and the remaining depend on the number of days one has taken food in the mess. When a student A takes food for 20 days, he has to pay ₹ 1000 as hostel charges whereas a student B, who takes food for 26 days, pays ₹ 1180 as hostel charges. Find the fixed charged and the cost of food per day. [CBSE 2000]

Solution:

Let the fixed charge of the college hostel = ₹ x

and daily charges = ₹ y

According to the given conditions,

x + 20y = 1000 …(i)

x + 26y =1180 …(ii)

Subtracting (i) from (ii), we get

6y = 180

⇒ y = 30

Substituting the value of y in (i)

x + 20 x 30 = 1000

⇒ x + 600 = 1000

⇒ x = 1000 – 600 = 400

Fixed charges = ₹ 400 and daily charges = ₹ 30

Question 14.

Half the perimeter of a garden whose length is 4 m more them its width is 36 m. Find the dimensions of the garden.

Solution:

Let the length of the garden = x m

and width = y m

x – y = 4 …(i)

and x + y = 36 ….(ii)

Adding we get,

2x = 40

⇒ x = 20

and subtracting,

2y = 32

y = 16

Length of the garden = 20 m and width = 16 m

Question 15.

The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them

Solution:

In two supplementary angles,

Let larger angle = x

and smaller angle = y

x – y = 18°

But x + y = 180°

Adding we get,

2x = 198°

⇒ x = 9°

Subtracting the get,

2y = 162°

⇒ y = 81°

Angles are 99° and 81°

Question 16.

2 women and 5 men can together finish a piece of embroidery in 4 days while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the embroidery and that taken by 1 man alone.

Solution:

Let one woman can do the work in = x days

and one man can do the same work = y days

Question 17.

Meena went to a bank to withdraw ₹ 2000. She asked the cashier to give her ₹ 50 and ₹ 100 notes only. Meena got 25 notes in all. Find how many notes ₹ 50 and ₹ 100 she received.

Solution:

Total amount with drawn = ₹ 2000

Let number of ₹ 50 notes = x

and of ₹ 100 = y

According to the conditions

x + y = 25 …(i)

and 50x + 100y = 2000

⇒ x + 2y = 40 ….(ii)

(Dividing by 50)

Subtracting (i) from (ii), y = 15

Substituting the value of y in (i)

x + y = 25

x + 15 = 25

⇒ x = 25 – 15 = 10

Number of 50 rupees notes = 10

and number of 100 rupee notes = 15

Question 18.

There are two examination rooms A and B. If 10 candidates are sent from A to B, the number of students in each room is same. If 20 candidates are sent from B to A, the number of students in A is double the number of students in B. Find the number of students in each room.

Solution:

Let A examination room has students = x

and B room has students = y

According to the given conditions,

x – 10 = y + 10

⇒ x – y = 10 + 10

⇒ x – y = 20 ….(i)

and x + 20 = 2 (y – 20)

⇒ x + 20 = 2y – 40

⇒ x – 2y = – 40 – 20 = – 60 ….(ii)

Subtracting (ii) from (i), y = 80

Substituting the value of y in (i)

x – 80 = 20

⇒ x = 20 + 80 = 100

In examination room A, the students are = 100

and in B room = 80

Question 19.

A railway half ticket costs half the full fare and the reservation charge is the same on half ticket as on full ticket. One reserved first class ticket from Mumbai to Ahmedabad costs ₹ 216 and one full and one half reserved first class tickets cost ₹ 327. What is the basic first class full fare and what is the reservation charge?

Solution:

Let the rate of fare of full ticket = ₹ x

and rate of reservation = ₹ y

According to the given conditions,

x + y = 216 …(i)

Question 20.

A wizard having powers of mystic in candations and magical medicines seeing a cock, fight going on, spoke privately to both the owners of cocks. To one he said; if your bird wins, than you give me your stake-money, but if you do not win, I shall give you two third of that. Going to the other, he promised in the same way to give three fourths. From both of them his gain would be only 12 gold coins. Find the stake of money each of the cock- owners have.

Solution:

Let the first owner of cock has state money = x gold coins

and second owner has = y gold coins

According to the given conditions,

Question 21.

The students of a class are made to stand in rows. If 3 student are extra in a row, there would be 1 row less. If 3 students are less in a row there would be 2 rows more. Find the number of students in the class.

Solution:

Let number of rows = x

and number of students in each row = y

Now according to the conditions given

Number of total students = xy

(y + 3)(x – 1) = xy

⇒ xy – y + 3x – 3 = xy

⇒3x – y = 3 …(i)

and (y – 3) (x + 2) = xy

⇒ xy + 2y – 3x – 6 = xy

⇒ 2y – 3x = 6 …(ii)

Adding (i) and (ii), y = 9

Substituting the value of y in (i)

3x – 9 = 3

⇒ 3x = 3 + 9 = 12

⇒ x = 4

Number of total students = xy = 4 x 9 = 36

Question 22.

One says, “give me hundred, friend ! I shall then become twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the amount of their respective capital ?

Solution:

Let first person has amount of money = Rs. x

and second person has = Rs. y

Now according to the conditions given,

x + 100 = 2(y – 100)

⇒ x + 100 = 2y – 200

⇒ x – 2y = – 200 – 100

⇒ x – 2y = -300 ………(i)

and 6(x – 10) = (y + 10)

⇒ 6x – 60 = y + 10

⇒ 6x – y = 10 + 60

⇒ 6x – y = 70 ……….(ii)

From (i), x = 2y – 300

Substituting the value of x in (ii)

6(2y – 300) – y = 70

⇒ 12y – 1800 – y = 70

⇒ 11y = 70 + 1800 = 1870

y = 170

x = 2y – 300 = 2 x 170 -300 = 340 – 300 = 40

First person has money = ₹ 40

and second person has = ₹ 170

Question 23.

A shopkeeper sells a saree at 8% profit and a sweater at 10% discount, thereby getting a sum of ₹ 1008. If she had sold the saree at 10% profit and the sweater at 8% discount, she would have got ₹ 1028. Find the cost price of the saree and the list price (price before discount) of the sweater. [NCERT Exemplar]

Solution:

Let the cost price of the saree and the list price of the sweater be ₹ x and ₹ y, respectively.

Question 24.

In a competitive examination, one mark is aw awarded for each correct answer while mark is deducted for every wrong answer. Jayanti answered 120 questions and got 90 marks. How many questions did she answer correctly. [NCERT Exemplar]

Solution:

Let x be the number of correct answer of the questions in a competitive examination,

then (120 – x) be the number of wrong answers of the questions.

Then, by given condition,

Hence, Jayanti answered correctly 100 questions.

Question 25.

A shopkeeper gives books pn rent for reading. She takes a fixed charge for the first two days, and an additional charge for each day thereafter. Latika paid ₹ 22 for a book kept for 6 days, while Anand paid ₹ 16 for the book kept for four days. Find the fixed charges and charge for each extra day. [NCERT Exemplar]

Solution:

Let Latika takes a fixed charge for the first two day is ₹ x and additional charge for each day thereafter is ₹ y.

Now, by first condition,

Latika paid ₹ 22 for a book kept for six days

i.e., x + 4y = 22 …(i)

and by second condition,

Anand paid ₹ 16 for a book kept for four days

i.e., x + 2y = 16 …(ii)

Now, subtracting Eq. (ii) from Eq. (i), we get

2y = 6 ⇒ y = 3

On putting the value of y in Eq. (ii), we get

x + 2 x 3 = 16

x = 16 – 6 = 10

Hence, the fixed charge = ₹ 10

and the charge for each extra day = ₹ 3

### Pair of Linear Equations in Two Variables Class 10 Solutions Exercise 3.11

pair of linear equations in two variables class 10 solutions Exercise 3.11 Q 1.

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