CBSE Class 10

RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.2

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.2

Other Exercises

• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.1
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.2
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.3
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.4
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.5
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.6
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.8
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.9
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.10
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.11
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.12
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.13
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations VSAQS
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations MCQS

Question 1.
The product of two consecutive positive integers is 306. Form the quadratic equation to find the integers, if x denotes the smaller integer. [NCERT]
Solution:
Let the first number = x
Then second number = x + 1
Their product = 306
x (x + 1) = 306
=> x² + x – 306 = 0
Required quadratic equation will be x² + x – 306 = 0

Question 2.
John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 128. Form the quadratic equation to find how many marbles they had to start with, if John had x marbles.
Solution:
No. of marbles John and Jivanti have = 45
Let number of marbles John has = x
Then number of marbles Jivanti has = 45 -x
Every one lost 5 marbles, then John’s marbles = x – 5
and Jivanti’s marbles = 45 – x – 5 = 40 – x
According to the condition,
(x – 5) (40 – x) = 128
=> 40x – x² – 200 + 5x – 128 = 0
=> -x² + 45x – 328 = 0
=> x² – 45x + 328 = 0

Question 3.
A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of articles produced in a day. On a particular day, the total cost of production was Rs. 750. If x denotes the number of toys produced that day, form the quadratic equation of find x.
Solution:
Let the number of toys in a day = x
Cost of each toy = x – 55
on a particular cost of production = Rs. 750
x (x – 55) = 750
=> x² – 55x – 750 = 0
Hence required quadratic equation will be x² – 55x – 750 = 0

Question 4.
The height of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, form the quadratic equation to find the base of the triangle.
Solution:
Let the base of a right triangle = x
Its height = x – 7
and hypotenuse = 13 cm
=> By Pythagoras Theorem
(Hypotenuse)² = (Base)² + (Height)²
(13 )² = x² + (x – 7)²
=> 169 = x² + x² – 14x + 49
=> 2x² – 14x + 49 – 169 = 0
=> 2x² – 14x – 120 = 0
=> x² – 7x – 60 = 0 (Dividing by 2)
Hence required quadratic equation will be x² – 7x – 60 = 0

Question 5.
An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore. If the average speed of the express train is 11 km/hr more than that of the passenger train, form the quadratic equation to find the average speed of express train.
Solution:
Distance between Mysore and Bangalore = 132 km
Let average speed of passenger train=x km/ hr
Then average speed of express train = (x + 11) km/hr

Question 6.
A train travels 360 km at a uniform speed. If the speed had been 5 km/hr more, it would have taken 1 hour less for the same journey. Form the quadratic equation to find the speed of the train.
Solution:
Total distance = 360 km
Let the uniform speed of the train = x km/hr
Time taken = \(\frac { 360 }{ x }\)

Hope given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.2 are helpful to complete your math homework.

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RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.1

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.1

Other Exercises

• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.1
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.2
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.3
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.4
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.5
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.6
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.8
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.9
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.10
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.11
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.12
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.13
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations VSAQS
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations MCQS

Question 1.
Which of the following are quadratic equations ?



Solution:

Question 2.
In each of the following, determine whether the given values are solutions of the given equation or not :

Solution:

Question 3.
In each of the following, find the value of k for which the given value is a solution of the given equation.

Solution:

Question 4.
Determine, if 3 is a root of the equation given below :

Solution:

Question 5.
If x = \(\frac { 2 }{ 3 }\) and x = -3 are the roots of the equation ax² + 7x + b = 0, find the values of a and b.
Solution:

Hope given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.1 are helpful to complete your math homework.

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RD Sharma Class 10 Solutions Chapter 15 Statistics Ex 15.5

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 15 Statistics Ex 15.5

Other Exercises

• RD Sharma Class 10 Solutions Chapter 15 Statistics Ex 15.1
• RD Sharma Class 10 Solutions Chapter 15 Statistics Ex 15.2
• RD Sharma Class 10 Solutions Chapter 15 Statistics Ex 15.3
• RD Sharma Class 10 Solutions Chapter 15 Statistics Ex 15.4
• RD Sharma Class 10 Solutions Chapter 15 Statistics Ex 15.5
• RD Sharma Class 10 Solutions Chapter 15 Statistics Ex 15.6
• RD Sharma Class 10 Solutions Chapter 15 Statistics Ex VSAQS
• RD Sharma Class 10 Solutions Chapter 15 Statistics MCQS

Question 1.
Find the mode of the following data :
(i) 3, 5, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7, 4
(ii) 3, 3, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7, 4
(iii) 15, 8, 26, 25, 24, 15, 18, 20, 24, 15, 19, 15
Solution:

We see that 5 occurs in maximum times which is 5
∴ Mode = 5

we see that 3 occurs in maximum times i.e. 5
∴ Mode = 3


Here we see that 15 occurs in maximum times i.e. 54
∴ Mode = 15

Question 2.
The shirt sizes worn by a group of 200 persons, who bought the shirt from a store, are as follows :

Find the model shirt size worn by the group.
Solution:

We see that frequency of 40 is maximum which is 41
∴Mode = 40

Question 3.
Find the mode of the following distribution.

Solution:

Question 4.
Compare the modal ages of two groups of students appearing for an entrance test :

Solution:
(i) For group A

We see that class 18-20 has the maximum frequency
∴ It is a modal class

(ii)  For group B

We see that class 18-20 has the maximum frequency
∴ It is a modal class

Question 5.
The marks in science of 80 students of class X are given below: Find the mode of the marks obtained by the students in science.

Solution:

We see that class 50-60 has the maximum frequency 20
∴ It is a modal class

Question 6.
The following is the distribution of height of students of a certain class in a certain city :

Find the average height of maximum number of students.
Solution:
Writing the classes in exclusive form,

Here model class is 165.5 – 168.5
and l = 165.5, h = 3, f= 142, f₁= 118, f₂= 127

Question 7.
The following table shows the ages of the patients admitted in a hospital during a year :

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
Solution:

(i) We see that class 35-45 has the maximum frequency 23
∴ It is a modal class

= 35.375 = 35.37 years
We see that mean is less than its mode

Question 8.
The following data gives the information on the observed lifetimes (in hours) of 225 electrical components :

Determine the modal lifetimes of the components.
Solution:


We see that class 60-80 has the maximum frequency 61
∴ It is the modal class
Here l = 60, f= 6, f₁ = 52, f₂ = 38,h = 20
∴ Modal of life time (in hrs.)

Question 9.
The following table gives the daily income of 50 workers of a factory :

Find the mean, mode and median of the above data. (C.B.S.E. 2009)
Solution:

Here i = 20 and AM = 150

(ii) Max frequency  = 14
∴ Model class = 120 – 140

Question 10.
The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret, the two measures:

Solution:
Let assumed mean (A) = 32.5

(i)  We see that the class 30-35 has the maximum frequency
∴ It is the modal class
Here l = 30,f = 10, f₁ = 9, f₂ = 3, h = 5

Question 11.
Find the mean, median and mode of the following data:

Solution:
Let assumed mean (A) =175

(i) Here N = 25, \(\frac { 5 }{ 3 }\) = \(\frac { 25 }{ 2 }\) = 12.5 or 13 which lies in the class 150-200
l= 150, F= 10, f= 6, h = 50

Question 12.
A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data.

Solution:

We see that class 40-50 has the maximum frequency
∴ It is a modal class

Question 13.
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them:

Solution:
Let assumed mean (A) = 135

Here N = 68 , \(\frac { N }{ 2 }\) = \(\frac { 68 }{ 2 }\) = 34

Question 14.
100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames. Also, And the modal size of the surnames.
Solution:
Let Assumed mean (A) = 8.5

(i) Here N = 100, \(\frac { N }{ 2 }\) = \(\frac { 100 }{ 2 }\) = 50 which lies in the class 7-10
Here l = 7, F = 36, f= 40, h = 3

Question 15.
Find the mean, median and mode of the following data (C.B.S.E. 2008)

Solution:
Let assumed mean A = 70

(i) Here N = 50, then \(\frac { N }{ 2 }\) = \(\frac { 50 }{ 2 }\) = 25 which lies in the class 60-80
∴ l= 60, F= 24, f = 12 , h = 20

Question 16.
The following data gives the distribution of total monthly household expenditure of 200 families of a village. .Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:

Solution:

(i) We see that the c ass 1500-2000 has maximum frequency 40
∴ It is a modal class
Here l = 1500, f = 40 , f₁ = 24 , f₂ = 33 , h =500

Question 17.
The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Find the mode of the data.
Solution:

We see that class 4000-5000 has the maximum frequency 18
∴It is a modal class

Question 18.
The frequency distribution table of agriculture holdings in a village is given below:

Find the modal agriculture holdings of the village.
Solution:
Here the maximum class frequency is 80,
and the class corresponding to this frequency is 5-7.
So, the modal class is 5-7.
l (lower limit of modal class) = 5
f₁ (frequency of the modal class) = 80
f₀ (frequency of the class preceding the modal class) = 45
f₂(frequency of the class succeeding the modal class) = 55
h (class size) = 2

Hence, the modal agricultural holdings of the village is 6.2 hectares.

Question 19.
The monthly income of 100 families are given as below:

Solution:
In a given data, the highest frequency is 41, which lies in the interval 10000-15000.
Here, l = 10000,f₁ = 41, f₀ = 26,f₂ = 16 and h = 5000

Hope given RD Sharma Class 10 Solutions Chapter 15 Statistics Ex 15.5 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables MCQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables MCQS

Other Exercises

• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.5
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.8
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.9
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.10
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.11
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables VSAQS
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables MCQS

Mark the correct alternative in each of the following.
Question 1.
The value of k for which the system of equations. kx – y = 2, 6x – 2y = 3 has a unique solution is
(a) = 3
(b) ≠ 3
(c) ≠ 0
(d) = 0
Solution:
(b)

Question 2.
The value of k for which the system of equations 2x + 3y = 5, 4x + ky = 10 has infinitely number of solutions, is
(a) 1
(b) 3
(c) 6
(d) 0
Solution:
(c)

Question 3.
The value of k for which the system of equations x + 2y – 3 = 0 and 5x + ky + 1 = 0 has no solution, is
(a) 10
(b) 6
(c) 3
(d) 1
Solution:
(a)

Question 4.
The value of k for which the system of equations 3x + 5y = 0 and kx + 10y = 0, has a non-zero solution, is
(a) 0
(b) 2
(c) 6
(d) 8
Solution:
(c)

Question 5.
If the system of equations
2x + 3y = 7
(a + b) x + (2a – b) y = 21
has infinitely many solutions, then
(a) a = 1, b = 5
(b) a = 5, b = 1
(c) a = -1, b = 5
(d) a = 5, b = -1
Solution:
(b)

Question 6.
If the system of equations 3x + y = 1 , (2k – 1) x + (k – 1) y = 2k + 1 is inconsistant, then k =
(a) 1
(b) 0
(c) -1
(d) 2
Solution:
(d)

Question 7.
If am ≠ bl, then the system of equations
ax + by = c
lx + my = n
(a) has a unique solution
(b) has no solution
(c) has infinitely many solutions
(d) may or may not have a solution.
Solution:
(a)

The given system as a unique solution

Question 8.
If the system of equations
2x + 3y = 7
2ax + (a + b) y = 28
has infinitely many solutions, then
(a) a = 2b
(b) b = 2a
(c) a + 2b = 0
(d) 2a + b = 0
Solution:
(b)

Question 9.
The value of k for which the system of equations
x + 2y = 5
3x + ky + 15 = 0 has no solution is
(a) 6
(b) – 6
(c) \(\frac { 3 }{ 2 }\)
(d) None of these
Solution:
(a)

k = 6

Question 10.
If 2x – 3y = 7 and (a + b) x – (a + b – 3) y = 4a + b represent coincident lines, then a and b satisfy the equation
(a) a + 5b = 0
(b) 5a + b = 0
(c) a – 56 = 0
(d) 5a – b – 0
Solution:
(c)

Question 11.
If a pair of linear equations in two variables is consistent, then the lines represented by two equations are
(a) intersecting
(b) parallel
(c) always coincident
(d) intersecting or coincident
Solution:
(d) The system of equation is coincident
The line of there equations are intersecting or coincident

Question 12.
The area of the triangle formed by the line \(\frac { x }{ a } +\frac { y }{ b } =1\) with the coordinate axes a is
(a) ab
(b) 2ab
(c) \(\frac { 1 }{ 2 }\) ab
(d) \(\frac { 1 }{ 4 }\) ab
Solution:
(c)
The triangle is formed by the line \(\frac { x }{ a } +\frac { y }{ b } =1\) with co-ordinates
It will intersect x-axis at a and y-axis at y
Area of the triangle so formed = \(\frac { 1 }{ 2 }\) x a x b = \(\frac { 1 }{ 2 }\) ab

Question 13.
The area of the triangle formed by the lines y = x, x = 6 and y = 0 is
(a) 36 sq. units
(b) 18 sq. units
(c) 9 sq. units
(d) 72 sq. units
Solution:
(b)
The triangle formed by the lines y = x, x = 6 and y = 0 will be an isosceles right triangle whose sides will be 6 units
Area = \(\frac { 1 }{ 2 }\) x 6 x 6 = 18 sq. units

Question 14.
If the system of equations 2x + 3y = 5, 4x + ky = 10 has infinitely many solutions, then k =
(a) 1
(b) \(\frac { 1 }{ 2 }\)
(c) 3
(d) 6
Solution:
(d)

Question 15.
If the system of equations kx – 5y = 2, 6x + 2y = 7 has no solution, then k =
(a) -10
(b) -5
(c) -6
(d) -15
Solution:
(d)

Question 16.
The area of the triangle formed by the lines x = 3, y = 4 and x = y is
(a) \(\frac { 1 }{ 2 }\) sq. unit
(b) 1 sq. unit
(c) 2 sq. unit
(d) None of these
Solution:
(a)
The triangle is formed by three lines x = 3, y = 4 and x = y
Its sides containing the right angle will be 1 and 1 units
Area of triangle so formed = \(\frac { 1 }{ 2 }\) x 1 x 1 sq. units = \(\frac { 1 }{ 2 }\) sq. units

Question 17.
The area of the triangle formed by the lines 2x + 3y = 12, x – y – 1 = 0 and x = 0 is
(a) 7 sq. units
(b) 7.5 sq. units
(c) 6.5 sq. units
(d) 6 sq. units

Solution:
(b) In the given graph as shown
The triangle formed by the lines
2x + 3y = 12, x – y – 1 =0 and x = 0
Its base BD = 4 + 1 = 5 units
and perpendicular from P on BD = 6 units
Area = \(\frac { 1 }{ 2 }\) x base x height
= \(\frac { 1 }{ 2 }\) x 5 x 3 = \(\frac { 15 }{ 2 }\) sq. units
= 7.5 square units

Question 18.
The sum of the digits of a two digit number is 9. if 27 is added to it, the digits of the number get reversed. The number is
(a) 25
(b) 72
(c) 63
(d) 36
Solution:
(d) Since the sum of the digits of a two-digit number is 9, therefore
x + y = 9 …(i)
It says if the digits are reversed, the new number is 27 less than the original.
Since we are looking at the number like xy, to separate them, it is actually 10x + y for x is a tens digit.
10y+ x = 10x + y + 27
Simplify it, we get 9y = 9x + 27
y = x + 3 …(ii)
Substitute (ii) into (i), we will have
x + (x + 3) = 9
=> 2x + 3 = 9
=> 2x = 6
=> x = 3
Put back into equation (i),
=> 3 + y = 9 => y = 6
The original number is 36.

Question 19.
If x = a, y = b is the solution of the systems of equations x – y = 2 and x + y = 4, then the values of a and b are, respectively
(a) 3 and 1
(b) 3 and 5
(c) 5 and 3
(d) -1 and -3
Solution:
(a) Since, x = a and y = b is the solution of the equations x – y = 2 and x + y = 4, then these values will satisfy that equations.
a – b = 2 …(i)
and a + b = 4 …(ii)
By adding (i) and (ii), we get
2a = 6
Therefore, a = 3
By putting a = 3 in (i), we get
3 – b = 2 Therefore, b = 1
Thus, a = 3 ; b = 1

Question 20.
For what value k, do the equations 3x – y + 8 = 0 and 6x – ky + 16 = 0 represent coincident lines?
(a) \(\frac { 1 }{ 2 }\)
(b) – \(\frac { 1 }{ 2 }\)
(c) 2
(d) -2
Solution:
(c)
Let 3x – y + 8 = 0 …(i)
and 6x – – ky + 16 = 0 …(ii)
Here, a₁ = 3, b₁ = -1, c₁ = 8

Question 21.
Aruna has only ₹ 1 and ₹ 2 coins with her. If the total number of coins that she has is 50 and the amount of money with her is ₹ 75, then the number of ₹ 1 and ₹ 2 coins are, respectively
(a) 35 and 15
(b) 35 and 20
(c) 15 and 35
(d) 25 and 25
Solution:
(d) Let number of ₹ 1 coins = x
and number of ₹ 2 coins = y
Now, by given condition x + y = 50 …(i)
Also, x x 1 + y x 2 = 15
=>x + 2y = 75 …(ii)
On subtracting Eq. (i) from Eq. (ii), we get
(x + 2y) – (x + y) = 75 – 50
=> y = 25
When y = 25, then x = 25

Hope given RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables VSAQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables VSAQS

Other Exercises

• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.5
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.8
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.9
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.10
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.11
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables VSAQS
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables MCQS

Answer each of the following questions either in one word or one sentence or as per requirement of the questions.
Question 1.
Write the value of k for which the system of equations x + y – 4 = 0 and 2x + ky – 3 = 0 has no solution.
Solution:

Question 2.
Write the value of k for which the system of equations 2x – y = 5 6x + ky = 15 has infinitely many solutions.
Solution:

Question 3.
Write the value of k for which the system of equations 3x – 2y = 0 and kx + 5y = 0 has infinitely many solutions.
Solution:
3x – 2y = 0
kx + 5y = 0
Here a₁ = 3, b₁ = -2, c₁ = 0

Question 4.
Write the value of k for which the system of equations x + ky = 0, 2x – y = 0 has unique solution.
Solution:

Question 5.
Write the set of values of a and b for which the following system of equations has infinitely many solutions.
2x + 3y = 7
2ax + (a + b) y = 28
Solution:

Question 6.
For what value of ft, the following pair of linear equations has infinitely many soutions.
10x + 5y – (k – 5) = 0
20x + 10y – k = 0
Solution:

Question 7.
Write the number of solutions of the following pair of linear equations :
x + 2y – 8 = 0
2x + 4y = 16 (C.B.S.E. 2009)
Solution:
x + 2y – 8 = 0 => x + 2y = 8 ….(i)

Question 8.
Write the number of solutions of the following pair of linear equations :
x + 3y – 4 = 0
2x + 6y = 7
Solution:

Hope given RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables VSAQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.