CBSE Class 10 Mathematics Linear Equations Assignment Set 02

CBSE Class 10 Mathematics Linear Equations Assignment Set B. Students are advised to refer to the attached assignments and practise them regularly. This will help them to identify their weak areas and will help them to score better in examination. Parents should download and give the assignments to their children for practice.

Question. Find the value of k for which the given system of equations has infinitely many solutions:
i. (𝑘 − 3) 𝑥 + 3 𝑦 = 𝑘
ii. 𝑘 𝑥 + 𝑘 𝑦 = 12
Answer : K−3/𝑘 = 3/𝑘
k = 6

Question. Solve for x and y:
i. 𝑚 𝑥 – 𝑛 𝑦 = 𝑚²
+ 𝑛²;
ii. 𝑥 – 𝑦 = 2𝑛
Answer : mx – ny = m² + n²
mx – ny = 2 nm
(2) – (1)
(m – n) y = m² + n² - 2 nm = (m – n)²
Y = m – n
x – (m -n) = 2n
x = m + n

Question. Is the system of linear equations 2𝑥 + 3𝑦 – 9 = 0 𝑎𝑛𝑑 4𝑥 + 6𝑦 – 18 = 0 consistent? Justify your answer.
Answer : 2/4 = 1/2 , −9/−18 = 1/2 , 3/6 = 1/2
It has infinite number of solutions. It is consistent.

Question. Are the following pair of linear equations consistent? Justify your answer.
2𝑎𝑥 + 𝑏𝑦 = 𝑎; 4𝑎𝑥 + 2𝑏𝑦 – 2𝑎 = 0 ; 𝑎 ,𝑏 ≠0
Answer : 𝑎1/𝑎2 = 2𝑎/4𝑎 = 1/2
𝑏1/𝑏2 = 𝑏/2𝑏 = 1/2
𝑐1/𝑐2 = 𝑎/2𝑎 = 1/2
It has infinitely many solution, it is consistent

Question. For which value of a and b does the following pair of linear equations has infinite number of solutions?
i. 2𝑥 + 3𝑦 = 7
ii. 𝑎(𝑥 + 𝑦) – 𝑏 (𝑥 − 𝑦) = 3𝑎 + 𝑏 − 2
Answer : 2x + 3y = 7 𝑥(𝑎 −𝑏) + 𝑦 (𝑎 + 𝑏) = 3𝑎 + 𝑏 – 2
Since it has infinitely many solutions,
2/𝑎−𝑏 = 3/𝑎+𝑏 = 7/3𝑎+𝑏−2
After equating
a = 5b (1)
2a – 5b = 6 (2)
Solve (1) and (2)
a =5 and b= 1

Question. Find the solutions of the pair of linear equations 5𝑥 + 10𝑦 – 50 = 0 𝑎𝑛𝑑 𝑥 + 8𝑦 = 10. Hence find the value of m if 𝑦 = 𝑚𝑥 + 5.
Answer : 𝑆𝑜𝑙𝑣𝑒 𝑡ℎ𝑒 𝑔𝑖𝑣𝑒𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠
𝑥= 10, 𝑦 = 0 𝑆
𝑢𝑏 𝑥 =10 𝑎𝑛𝑑 𝑦=0 𝑖𝑛 𝑦 = 𝑚𝑥 + 5
0 = 𝑚 ×10 + 5
𝑚 = −1/2

Question. Write a pair of linear equations which has a unique solution 𝑥 = 2 𝑎𝑛𝑑 𝑦 = − 1. How many such pairs are possible?
Answer : 3x + y = 5
Infinite number of solution

Question. If 𝑥 − 4 is a factor of 𝑥³ + 𝑎𝑥² + 2𝑏𝑥 – 24 and 𝑎 – 𝑏 = 8 , find the value of 𝑎 𝑎𝑛𝑑 𝑏.
Answer : Since x - 4 is a factor of x³ + ax² + 2bx – 24
4³ + a × 4² + 2b × 4 – 24 = 0
a + 2b + 10 =0 (i)
a -b = 8 (ii)
Solve (I and (ii)
a = 2 , b = -6

Question. If 2𝑥 + 𝑦 = 23 and 4𝑥 – 𝑦 = 19, find the values of 5𝑥 – 3𝑦 and 𝑦 – 2𝑥.
Answer : 𝑆𝑜𝑙𝑣𝑒 𝑡ℎ𝑒 𝑔𝑖𝑣𝑒𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠
𝑥 = 7, 𝑦 = 9
𝑆𝑜,5𝑥 – 3𝑦 = 8
𝑦 – 2𝑥 = −5

Question. There are 20 vehicles – cars and motorcycles in a parking area. If there are 56 wheel together, how many cars and motorcycles are there.
Answer : Let no of cars = x and no of motor cycles = y
According to our condition
x + y = 20 (i)
4x + 2y = 56 (ii)
Solve (i) and (ii)
x = 8 and y = 12

Long Answer type Questions

Question. The sum of the numerator and the denominator of a fraction is 3 less than twice the denominator. If the numerator and the denominator are decreased by one, the numerator becomes half the denominator. Determine the fraction.
Answer : Let the numerator be x and the denominator be y
Fraction is xy According to the first condition x + y = 2y - 3 x + y - 2y = - 3 x - y = - 3 - - - - (1) According to the second condition x - 1 = 12y - 1 2x - 1 = y - 1 2x - 2 = y - 1 2x - y = 1 - - - - - (2) Subtracting eqn (2) from eqn (1) we get - x = - 4 x = 4 Substituting x = 4 in equation 2, we get 2x4 - y = 1 8 - y = 1 y = 7
Fraction is 47

Question. Given the linear equation 7𝑥 − 5𝑦 − 4 = 0. Write another linear equation in two variables such that the geometrical representation of the pair so formed is Intersecting lines, Parallel lines, Coincident lines
Answer : (i) Any linear equation satisfying the condition a1a2 ≠ b1b2
(ii) Any linear equation satisfying the condition a1/a2 = b1/b2 ≠ c1 /c2
(ii) Any linear equation satisfying the condition a1/a2 = b1/b2 = c1/ c2

Question. A number consists of two digits. When 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 the digit in ones place be y and the digit in tens place be y.
Two digit number = 10x + y
Given 10x + yx + y = 7
⇒ 10x + y = 7 ( x + y )
∴ 10x + y - 7x - 7y = 0
3x - 6y = 0
x - 2y = 0 - - - - - - - - (1)
According to the second condition.
10x + y - 27 = 10y + x
10x + y - 10y - x = 27
9x - 9y = 27
x - y = 3 - - - - - - - (2)
Equation (1) - (2)
x - 2y - x - y = 0 - 3
x - 2y - x + y = - 3
- y = - 3
y = 3
Substituting y = 3 in equation (2), we get
x - 3 = 3
x = 6
Two - digit number = 10x + y
= 10x6 + 3
= 60 + 3 = 63
20y + y - 27 = 10y + 2y
⇒ 9y = 27
⇒y = 3
Substitute y value in eqn(1)
we get, x = 2× 3
⇒ x = 6
Hence the required number is 63.

Question. The age of the father is twice the sum of the ages of his two children. After 20 years, his age will be equal to the sum of the ages of his children. Find the age of the fathe
Answer : Let the present age of his two children be “x” years and “y” years.
Present age of father = 2(x + y) - - - - (1)
A.T.Q.
2x + y + 20 = x + 20 + y + 20
2x + 2y + 20 = x + y + 40
2x + 2y - x - y = 40 - 20
x + y = 20 - - - - (2)
Substituting eqn (2) in eqn (1), we get
Present age of father = 2 x 20
= 40 years

Question. A railway half ticket costs half the full fare, but the reservation charges are the same on a half ticket as on a full ticket. One reserved first - class ticket from the station A to B costs ₹2530. Also one reserved first class ticket and one reserved first class half ticket from A to B costs ₹3810.Find the full first class fare from station A to B and also the reservation charges for a ticket.
Answer : Let the cost of full fare be ₹ 𝑥 and the cost of half first class fare be ₹ 𝑥2, respectively and reservation charges be ₹ 𝑦 per ticket.
Case I The cost of one reserved first class ticket from the stations A to B = ₹ 2530
x + y = 2530 ... ( i )
Case II
The cost of one reserved first class ticket and one reserved first class half ticket from stations A to B = ₹ 3810
⇒ 𝑥 + 𝑦 + (𝑥/2) + 𝑦 = 3810
⇒ 𝑥 + (𝑥/2) + 𝑦 + 𝑦 = 3810
⇒3𝑥/2 + 2 𝑦 = 3810
Multiplying throughout by 2, we get
⇒ 3𝑥 + 4𝑦 = 7620 ...(𝑖𝑖)
Now, multiplying Eq. (i) by 4 and then subtracting from Eq. (ii), we get 3 𝑥 + 4 𝑦−4𝑥−4𝑦 = 7620 −10120 − 𝑥 = − 2500 ⇒ 𝑥 = 2500
On putting the value of x in Eq. (i), we get
2500 + y = 2530
⇒ 𝑦 = 30
Hence, full first - class fare from stations A to B is ₹ 2500 and the reservation for a ticket is ₹ 30.

Question. Two numbers are in the ratio 5:6. If 8 is subtracted from each of the numbers, the ratio becomes 4:5. Find the numbers.
Answer : Let the two numbers be x and y
According to the first condition
xy = 56
Cross multiplying, we get
6x = 5y
6x - 5y = 0 - - - - - - (i)
According to the second condition
x - 8y - 8 = 45
Cross multiplying, we get
5x - 8 = 4y - 8
5x - 40 = 4y - 32
5x - 4y = 40 - 32
5x - 4y = 8 - - - - - - - (ii)
Multiplying eqn (i) by 4, we get
24x - 20y = 0 - - - - (iii)
Multiplying eqn (ii) by 5, we get
25x - 20y = 40 - - - - - (iv)
Subtracting eqn (iii) - eqn(iv), we get
24x - 20y - 25x + 20y = - 40 - x = - 40
x = 40
Substituting x = 40 in eqn (i), we get
6 x 40 - 5y = 0
240 - 5y = 0
240 = 5y
y = 48
The numbers are 40 and 48.

Question. A boat takes 4 hours to go 44 km downstream and it can go 20 km upstream in the same time. Find the speed of the stream and that of the boat in still water.
Answer : Let the speed of the stream = y km/hr and
Speed of the boat in still water = x km/hr
Speed of the boat in downstream = (x + y)km/hr
Speed of the boat in upstream = (x - y)km/hr
A.T.Q
44x + y = 4
44 = 4(x + y)
x + y = 11 - - - - - - (1)
Also 20x - y = 4
20 = 4x - y
x - y = 5 - - - - - (2)
Adding eqns 1 and 2 we get, 2x = 16 x = 8
Substituting x = 8 in eqn(1), we get
8 + y = 11
y = 3
Speed of the stream = 3 km/hr and
Speed of the boat in still water = 8 km/hr

Question. There are two examination rooms A and B. If 10 candidates are sent from A to B, the number of candidates in each room is the 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.
Answer : Let the number of students in room A be x and that in room B be y. According to the first condition x - 10 = y + 10 ⟹x−y = 20−−−−−(i) According to the second condition, we get x + 20 = 2(y - 20)
x + 20 = 2y - 40
x - 2y = - 40 - 20 ⟹x−2y = −60−−−−−(ii) Eqn (i) - Eqn (ii) we get, x - y - x + 2y = 20 + 60
y = 80
Substituting y = 80 in eqn (i), we get x - 80 = 20
x = 100
Number of students in room A = 100
Number of students in room B = 80

Question. 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 one hour. What are the speeds of the two cars?
Answer : Let x km/hr be the speed of car from point A and y km/hr be the speed of car from point B.
If the car travels in the same direction,
5x – 5y = 100 x – y = 20 …………………………………(i)
If the car travels in the opposite direction,
x + y = 100………………………………(ii)
Adding equations (i) and (ii), we get
2x = 120
x = 60 km/hr
Substituting this in equation (i), we get,
60 – y = 20
y = 40 km/h
Therefore, the speed of car from point A = 60 km/hr
Speed of car from point B = 40 km/hr.

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