Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.11 | Set 2

In this article, we will discuss the solutions to Exercise 3.11 Set 2 from Chapter 3 of RD Sharma's Class 10 Mathematics textbook which focuses on the "Pair of Linear Equations in the Two Variables". This chapter is vital as it introduces students to methods for solving systems of linear equations a fundamental concept in algebra. Understanding these methods prepares students for the more advanced topics in the mathematics and helps in developing problem-solving skills.

Pair of Linear Equations in Two Variables

A pair of linear equations in two variables consists of two linear equations that involve two unknowns. These equations can be solved using various methods such as substitution, elimination, or graphical representation. The goal is to find the values of the unknowns that satisfy both equations simultaneously. Solving such pairs helps in understanding how different equations intersect and provides the solutions to real-world problems involving two variables.

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 student A takes food for 20 days, he has to pay ₹ 1000 as hostel charges whereas student B, who takes food for 26 days, pays ₹ 1180 as hostel charges. Find the fixed charge and the cost of food per day. [CBSE 2000]

Solution: 

Let's assume that the fixed charge of the college hostel ₹ x

and daily charges be ₹ y

According to given condition,

x + 20y = 1000 ---------------(i)

x + 26y =1180 ------------------(ii)

Subtracting (i) from (ii) and we get,

6y = 180

= y = 30

Substituting the value of y in (i) and we get,

x + 20 × 30 = 1000

= x + 600 = 1000

= x = 1000 – 600 = 400

Hence, 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's assume that the length of the garden be x m and width y m.

According to given condition,

x – y = 4 -----------------(i)

and x + y = 36 -----------------(ii)

Adding both the eqn. we get,

2x = 40

x = 20

put the value if x in eqn. (i) and we get,

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°

put the value if x in eqn. (i) and we 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

1 women's 1 days work = 1/x

and 1 men 1 day work = 1/y

2 women and 5 men 1 day work = 2/x + 5/y

then 4 days work = 4(2/x + 5/y) = 1

similarly 3 women and 6 men's 1 day work = 3/x + 6/y

then 3 day's work = 3(3/x + 6/y) = 1

= 8/x + 20/y = 1 ------------------(i)

= 9/x + 18/y = 1 -----------------(ii)

Multiply eqn. (i) by9 and (ii) by 10 we get,

x = 18

put the value of x in eqn. (i) and we get,

y = 36

Hence one women can do the work in 36 days and one men can do the same work in 18 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's assume that the number of ₹ 50 notes = x and of ₹ 100 = y

According to given condition,

x + y = 25 ------------------(i)

and 50x + 100y = 2000

= x + 2y = 40 ------------------(ii) (Dividing by 50)

Subtracting eqn. (i) from (ii) and we get,

y = 15

Substituting the value of y in (i) and we get.

x + y = 25

x + 15 = 25

x = 25 – 15 = 10

Hence 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 given condition

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) and we get,

y = 80

Substituting the value of y in (i) and we get,

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 given condition,

x + y = 216 ------------------(i)

and 3/2 x + 2y = 327 --------------(ii)

Multiply eqn. (i) by 4 we get,

x = 210

put the value of x in (i) and we get,

y = 6

Hence rate of fare of full ticket ₹210 and reservation charges per ticket is ₹6.

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 given condition,

x - 3/4 y = 12 ---------------(i)

and y - 2/3 x = 12 ----------------(ii)

Adding eqn. (i) and (ii) we get,

x = 42

put the value of x in eqn. (i) and we get,

y = 40

Hence the first owner has 42 gold coin and second owner has 40 gold coin.

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,

According to given condition,

Number of total students = x * y

(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) and we get,

y = 9

Substituting the value of y in (i) and we get,

3x – 9 = 3

= 3x = 3 + 9 = 12

= x = 4

Number of total students = x * y = 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's first person has amount of money ₹x and second person has ₹y

According to given condition,

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) we get,

x = 2y – 300

Substituting the value of x in (ii) and we get,

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

Hence 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.

Case 1: Sells a saree at 8% profit + Sells a sweater at 10% discount = ₹1008

= (100+8)% of x + (100-10)% of y = 1008

= 1.08x + 0.9y = 1008 ----------------(i)

Case 2: Sold the saree at 10% profit + Sold the sweater at 8% discount = ₹1028

= (100+10)% of x + (100 - 8)% of y=1028

= 1.1x + 0.92y =1028 -----------------(ii)

put the value of y in eqn. (i) we get,

x = 600

put the value of x in eqn. (i) and we get,

y = 400

Hence the cost price of the saree and the list price of sweater are ₹600 and ₹400 respectively.

Question 24. In a competitive examination, one mark is aw awarded for each correct answer while 12 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.

According to given condition,

= x - (120 - x) * 1/2 = 90

= 3x / 2 = 150

= x = 100

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.

According to given 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.

Read More:

• Pair of Linear Equations in Two Variables
• Linear Equation in Two Variables