NCERT Exemplar Class 10 Maths Solutions Chapter 3 Pair Of Linear Equations In Two Variables

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NCERT Exemplar Problems Class 10 Maths Solutions Chapter 3 Pair Of Linear Equations In Two Variables

Exercise 3.1 Multiple Choice Questions (MCQs)

Question 1:
Graphically, the pair of equations
6x – 3y + 10 = 0
2x – y + 9 = 0
represents two lines which are
(a) intersecting at exactly one point
(b) intersecting exactly two points
(c) coincident
(d) parallel
Solution:
The given equations are
6x-3y+10 = 0
⇒ 2x-y+ \(\frac { 10 }{ 3 } \)= 0         [dividing by 3]… (i)
and       2x-y+9=0                                                                …(ii)
Now, table for 2x – y +  \(\frac { 10 }{ 3 } \)= 0,


Hence, the pair of equations represents two parallel lines.

Question 2:
The pair of equations x + 2y + 5 = 0 and – 3x – 6y +1 = 0 has
(a) a unique solution                                  (b) exactly two solutions
(c) infinitely many solutions                    (d) no solution
Solution:

Question 3:
If a pair of linear equations is consistent, then the lines will be
(a) parallel                                                 (b) always coincident
(c) intersecting or coincident                (d) always intersecting
Solution:

Question 4:
The pair of equations y = 0 and y = – 7 has
(a) one solution                                        (b) two solutions
(c) infinitely many solutions                  (d) no solution
Solution:
(d) The given pair of equations are y = 0 and y = – 7.

By graphically, both lines are parallel and having no solution

Question 5:
The pair of equations x = a and y = b graphically represents lines which are
(a) parallel                                                  (b) intersecting at (b, a)
(c) coincident                                            (d) intersecting at (a, b)
Solution:
(d) By graphically in every condition, if a, b>>0; a, b< 0, a>0, b< 0; a<0, b>0 but a = b≠ 0.
The pair of equations x = a and y = b graphically represents lines which are intersecting at (a, b).
If a, b > 0

Similarly, in all cases two lines intersect at (a, b).

Question 6:
For what value of k, do the equations 3x – y + 8 = 0 and 6x – ky = – 16
(a) \(\frac { 1 }{ 2 } \)                          (b) \(-\frac { 1 }{ 2 } \)                  (c) 2                         (d) -2
Solution:

Question 7:
If the lines given by 3x+ 2ky = 2 and 2x + 5y = 1 are parallel, then the value of k is
(a) \(-\frac { 5 }{ 4 } \)                       (b)  \(\frac { 2 }{ 5 } \)
(c) \(\frac { 15 }{ 4 } \)                         (d) \(\frac { 3 }{ 2 } \)
Solution:

Question 8:
The value of c for which the pair of equations cx- y = 2 and 6x – 2y = 3
will have infinitely many solutions is
(a) 3                   (b) – 3                      (c)-12                        (d) no value
Solution:
(d) Condition for infinitely many solutions

Since, c has different values.
Hence, for no value of c the pair of equations will have infinitely many solutions.

Question 9:
One equation of a pair of dependent linear equations is – 5x+ 7y – 2 = 0. The second equation can be
(a) 10x + 14y + 4=0                                    (b)-10x-14y + 4 =0
(c) -10x + 14y + 4 = 0                                 (d) 10x-14y + 4=0
Solution:
(d) Condition for dependent linear equations

Question 10:
A pair of linear equations which has a unique solution x – 2 and y = – 3 is
(a) x + y = 1 and 2x – 3y = – 5
(b) 2x+ 5y= -11 and 4x + 10y = -22
(c) 2x – y = 1 and 3x + 2y = 0
(d) x – 4y -14 = 0 and 5x – y -13 = 0
Solution:
(b) If x = 2, y = – 3 is a unique solution of any pair of equation, then these values must satisfy that pair of equations.
From option (b),        LHS = 2x + 5y = 2(2) + 5(- 3) = 4 – 15 = – 11 = RHS
and                             LHS = 4x + 10y = 4(2) + 10(- 3)= 8 – 30 = – 22 = RHS

Question 11:
If x = a and y = b is the solution of the equations x- y = 2 and x + y = 4, then the values of a and b are, respectively
(a) 3 and 5                 (b) 5 and 3              (c) 3 and 1                (d) – 1 and – 3
Solution:
(c) 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)
On adding Eqs. (i) and (ii), we get
2a = 6
a = 3 and b = 1

Question 12:
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 conditions            x+y=50                                               …(i)
Also,                                      x×1+y×2=75
⇒                                              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

Question 13:
The father’s age is six times his son’s age. Four years hence, the age of the father will be four times his son’s age. The present ages (in year) of the son and the father are, respectively
(a) 4 and 24                                              (b) 5 and 30
(c) 6 and 36                                              (d) 3 and 24
Solution:
(c)
Let x yr be the present age of father and y yr be the present age of son.
Four years hence, it has relation by given condition,
(x + 4) = 4(y + 4)
⇒                                        x-4y = 12                                                               …(i)
and                                           x = 6y                                                              …(ii)
On putting the value of x from Eq. (ii) in Eq. (i), we get
6y-4y=12
⇒                                           2y = 12
⇒                                              y = 6
When y = 6, then x = 36
Hence, present age of father is 36 yr and age of son is 6 yr.

Exercise 3.2 Short Answer Type Questions

Question 1:
Do the following pair of linear equations have no solution? Justify your answer.
(i)   2x + 4y = 3 and 12y + 6x = 6
(ii)  x = 2y and y = 2x
(iii) 3x + y – 3 = 0 and 2x + – y = 2
Solution:

Question 2:
Do the following equations represent a pair of coincident lines? Justify your answer.

Solution:

Question 3:
Are the following pair of linear equations consistent? Justify your answer,

Solution:

Question 4:
For the pair of equations λx + 3y + 7 = 0 and 2x + 6y -14 = 0. To have infinitely many solutions, the value of λ should be 1. Is the statement true? Give reasons.
Solution:
No, the given pair of linear equations
λx + 3y+7 = 0 and 2x + 6y-14 = 0
Here,                         a₁= λ, b₁ = 3 c₁ =7; a₂ =2, b₂ = 6,c₂ = -14

Hence, λ = -1 does not have a unique value.                                                     ‘
So, for no value of λ the given pair of linear equations has infinitely many solutions.

Question 5:
For all real values of c, the pair of equations x – 2y = 8 and 5x – lOy = c have a unique solution. Justify whether it is true or false.
Solution:
False, the given pair of linear equations
x-2y-8=0
5x-10y=c

But if c = 40 (real value), then the ratio  \(\frac { { c }_{ 1 } }{ { c }_{ 2 } } \) becomes  \(\frac { 1 }{ 5 }\) and then the system of linear
equations has an infinitely many solutions.
Hence, ate = 40, the system of linear equations does not have a unique solution.

Question 6:
The line represented by x = 7 is parallel to the X-axis, justify whether the statement is true or not.
Solution:
Not true, by graphically, we observe that x = 7 line is parallel to y-axis and perpendicular to X-axis.

Exercise 3.3 Short Answer Type Questions

Question 1:
For which value(s) of λ, do the pair of linear equations λx + y =λ² and x + λy = 1 have
(i) no solution?                                (ii) infinitely many solutions?
(iii) a unique solution?
Solution:
The given pair of linear equations is
λx + y = λ² and x + λy = 1
a₁ = λ, b₁= 1,  c₁ = – λ²
a₂ =1,  b_2^=λ        c₂=-1

Question 2:
For which value (s) of k will the pair of equations
kx+3y = k – 3,
12 x + ky =k
has no solution?
Solution:
Given pair of linear equations is
kx + 3y = k – 3
and                                                12x + ky = k
On comparing with ax + by + c = 0, we get
a₁ = k, b₁ = 3 and c₁ = -(k-3)
a₂ = 12,b₂ = k and  c₂ = -k


Hence, required value of k for which the given pair of linear equations has no solution is -6.

Question 3:
For which values of a and b will the following pair of linear equations has infinitely many
solutions?
x + 2y = 1
(a -b)x+(a + b)y = a + b – 2
Solution:
Given pair of linear equations are
x+2y=1
(a-b)x+(a+b)y=a+b-2
on comparing with ax+by+c=0,we get

Question 4:
Find the values of p in (i) to (iv) and p and q in (v) for the following pair of equations
(i) 3x – y – 5 = 0 and 6x – 2y – p = 0, if the lines represented by these equations are parallel.
(ii) – x + py = 1 and px – y = 1 if the pair of equations has no solution.
(iii) – 3x + 5y = 7 and 2px – 3y _{= 1,}
if the lines represented by these equations are intersecting at a unique point.
(iv) 2x + 3y – 5 = 0 and px – 6y – 8 = 0,
if the pair of equations has a unique solution.
(v) 2x + 3y = 7 and 2px + py = 28 – qy,
if the pair of equations has infinitely many solutions.
Solution:

Question 5:
Two straight paths are represented by the equations x-3y = 2 and – 2x + 6y =5. Check whether the paths cross each other or not.
Solution:
Given linear equations are

Hence, two straight paths represented by the given equations never cross each other, because they are parallel to each other.

Question 6:
Write a pair of linear equations which has the unique solution x = – 1 and y = 3. How
many such pairs can you write?
Solution:
Condition for the pair of system to have unique solution

Hence, infinitely many pairs of linear equations are possible.

Question 7:
If 2x+ y = 23 and 4x- y = 19, then find the values of 5y – 2x and  \(\frac { y }{ x } -2.\)
Solution:

Question 8:
Find the values of x and y in the following rectangle

Solution:

Question 9:

Solution:

Question 10:
Find the solution of the pair of equations  \(\frac { x }{ 10 } +\frac { y }{ 5 } -1=0\)
\(\frac { x }{ 8 } +\frac { y }{ 6 } \)= 15 and find A, if y = λx + 5.
Solution:

Question 11:
By the graphical method, find whether the following pair of equations are consistent or not. If consistent, solve them.
(i)   3x + y + 4 = 0,6x- 2y + 4 = 0
(ii)  x — 2y — 6, 3x — 6y = 0
(iii) x + y = 3, 3x + 3y =9
Solution:

Question 12:
Draw the graph of the pair of equations 2x + y= 4 and 2x – y = 4. Write the vertices of the triangle formed by these lines and the y-axis, find the area of this triangle? ’
Solution:

Question 13:
Write an equation of a Line passing through the point representing solution of the pair of Linear equations x + y = 2 and 2x – y = 1, How many such lines can we find?
Solution:
Plotting the points A (2,0) and B (0, 2), we get the straight line AB. Plotting the points C (0, – 1)andD (1/2, 0), we get the straight line CD. The lines AB and CD intersect at E(1,1), Hence, infinite lines can pass through the intersection point of linear equations x + y = 2 and 2x – y = 1 i.e., E(1,1) like as y = x, 2x + y = 3, x + 2y = 3. so on.

Question 14:
If (x + 1) is a factor of 2x³ + ax² + 2bx+l, then find the value of a and b given that 2a – 3b = 4.
Solution:
Given that, (x + 1) is a factor of f(x) = 2x^s + ax² + 2bx + 1, then f(-1) = 0.
[if (x + a) is a factor of f(x) = ax² + bx + c, then f(-) = 0]

Hence, the required values of a and b are 5 and 2, respectively.

Question 15:
If the angles of a triangle are x, y and 40° and the difference between
the two angles x and y is 30°. Then, find the value of x and y,
Solution:
Given that, x, y and 40° are the angles of a triangle.
x + y + 40° = 180°
[since, the sum of all the angles of a triangle is 180°]

Hence, the required values of x and y are 85° and 55°, respectively.

Question 16:
Two years ago, Salim was thrice as old as his daughter and six years later, he will be four year older than twice her age. How old are they now?
Solution:
Let Salim and his daughter’s age be x and y yr respectively.
Now, by first condition
Two years ago, Salim was thrice as old as his daughter.

Hence, Salim and his daughter’s age are 38 yr and 14 yr, respectively.

Question 17:
The age of the father is twice the sum of the ages of his two children. After 20 yr, his age will be equal to the sum of the ages of his children. Find the age of the father.
Solution:
Let the present age (in year) of father and his two children be x, y and z yr, respectively.
Now by given condition,                   x=2(y+z)                                                               …(i)
and after 20 yr,                          (x + 20) = (y + 20) + (z + 20)
⇒                                              y+z +40 = x +20
⇒                                                     y + z = x – 20
On putting the value of (y + z) in Eq. (i) and get the present age of father
x =2(x -20) x
= 2x – 40 = 40
Hence, the father’s age is 40 yr.

Question 18:
Two numbers are in the ratio 5 : 6. If 8 is subtracted from each of the numbers, the ratio becomes 4 : 5, then find the numbers. .
Solution:
Let the two numbers be x and y.
Then, by first Condition, ratio of these two numbers = 5:6

Hence, the required numbers are 40 and 48.

Question 19:
There are some students in the two examination halls A and B. To make the number of students equal in each hall, 10 students are sent from A to B but, if 20 students are sent from B to A, the number of students in A becomes double the number of students in B,
then find the number of students in the both halls.
Solution:
Let the number of students in halls A and 8 are x and y, respectively.
Now, by given condition,                x-10=y+10
⇒                                                    x – y = 20                                                             … (i)
and                                                    (x + 20) = 2 (y-20)
⇒                                                       x-2y=-60                                                              …(ii)
On subtracting Eq. (ii) from Eq. (i), we get
(x-y)-(x- 2 y) = 20+60 „  x-y-x + 2y~ 80 => y= 80
On putting y = 80 in Eq. (i), we get
x – 80 = 20 => x = 100 and     y = 80
Hence, 100 students are in hall A and 80 students are in hall 8.

Question 20:
A shopkeeper gives books on 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 six days, while Anand paid ₹ 16 for the book kept for four days. Find the fixed charges and the charge for each extra day.
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 + 4 y = 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

Question 21:
In a competitive examination, 1 mark is awarded for each correct answer while 1/2 mark is deducted for every wrong answer. Jayanti answered 120 questions and got 90 marks. How many questions did she answer correctly?
Solution:
Let x be the number of correct answers 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 22:
The angles of a cyclic quadrilateral ABCD are ∠A – (6x +10)°, ∠B = (5x)°, ∠C = (x+ y)° and ∠D = (3y – 10)°.Find x and y and hence the values of the four angles.
Solution:
We know that, by property of cyclic quadrilateral,
Sum of opposite angles = 180°
∠A + ∠C = (6x + 10)° + (x + y)° = 180°

Hence, the required values of x and y are 20° and 30° respectively and the values of the four angles ;.e., ZA, ZB, ZC and ZD are 130°, 100°, 50° and 80°, respectively.

Exercise 3.4 Long Answer Type Questions

Question 1:
Graphically, solve the following pair of equations 2x + y = 6 and 2x – y + 2 = 0
Find the ratio of the areas of the two triangles formed by the lines representing these equations with the X-axis and the lines with the Y-axis.
Solution:

Question 2:
Determine graphically, the vertices of the triangle formed by the lines
y = x, 3y = x and x + y = 8
Solution:


Plotting the points A(1,1) and 6(2,2), we get the straight line AB. Plotting the points C(3,1) and 0(6,2), we get the straight line CD. Plotting the points P(0, 8), Q(4, 4) and R{8, 0), we get the straight line PQR. We see that lines AB and CD intersecting the line PR on Q and D, respectively.
So, AOQD is formed by these lines. Hence, the vertices of the A00D formed by the given lines are0(0, 0),Q(4, 4)and 0(6,2).

Question 3:
Draw the graphs of the equations x = 3, x = 5 and 2x – y – 4 = 0. Also find the area of the quadrilateral formed by the lines and the X-axis.
Solution:
Given equation of lines 2x – y – 4 = 0, x = 3 and x = 5 Table for line 2x – y – 4 = 0 Also
find the area of the quadrilateral formed by the lines and the X-axis.

Hence, the required area of the quadrilateral formed by the lines and the X-axis is 8 sq units.

Question 4:
The cost of 4 pens and 4 pencils boxes is 1100. Three times the cost of a pen is ₹ 15 more than the cost of a pencil box. Form the pair of linear equations for the above situation. Find the cost of a pen and a pencil box.
Solution:
Let the cost of a pen be ₹ x and the cost of a pencil box be ₹ y.
Then, by given condition,


Hence, the cost of a pen and a pencil box are ₹ 10 and ₹ 15, respectively.

Question 5:
Determine, algebraically, the vertices of the triangle formed by the lines
3 x – y = 3
2x – 3y = 2
and x + 2y = 8
Solution:
Given equation of lines are
3 x – y = 3                                   …………..(i)
2x – 3y = 2                                   ………….(ii)
and x + 2y = 8                                  ………….(iii)
Let lines (i), (ii) and (iii) represent the sides of a ΔABC i.e., AB, BC and CA, respectively. On solving lines (i) and (ii), we will get the intersecting point B.
On multiplying Eq. (i) by 3 in Eq. (i) and then subtracting, we get

So, the coordinate of point or vertex A is (2, 3).
Hence, the vertices of the ΔABC formed by the given lines are A (2, 3), B (1, 0) and C(4,2).

Question 6:
Ankita travels 14 km to her home partly by rickshaw and partly by bus. She takes half an hour, if she travels 2 km by rickshaw and the remaining distance by bus. On the other hand, if she travels 4 km by rickshaw and the remaining distance by bus, she takes 9 min longer. Find the speed of the rickshaw and of the bus.
Solution:

Question 7:
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.
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,

Hence, the speed of the stream is 2.5 km/h,

Question 8:
A motorboat can travel 30 km upstream and 28 km downstream in 7 h. It can travel 21 km upstream and return in 5 h. Find the speed of the boat in still water and the speed of the stream.
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,


Hence, the speed of the motorboat in still water is 10 km/h and the speed of the stream 4 km/h.

Question 9:
A two-digit number is obtained by either multiplying the sum of the digits by 8 and then subtracting 5 or by multiplying the difference of the digits by 16 and then adding 3. Find the number.
Solution:

Question 10:
A railway half ticket cost 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 stations A to B costs ₹ 2530. Also, one reserved first class ticket and one reserved first class half ticket from stations A to B costs ₹ 3810. Find the full first class fare from stations A to B and also the reservation charges for a ticket.
Solution:
Let the cost of full and half first class fare be ₹  \(\frac { X }{ 2 } \) and ₹ respectively and reservation charges be  ₹ y per ticket.

Hence, full first class fare from stations A to 6 is ₹ 2500 and the reservation for a ticket is ₹30.

Question 11:
A shopkeeper sells a saree at 8% profit and a sweater at 10% discount, thereby, getting a sum ₹ 1008. If she had sold the saree at 10% profit and the sweater at 8% discount, she would have got ₹ 1028 then find the cost of the saree and the list price (price before discount) Of the sweater.
Solution:
Let the cost price of the saree and the list price of the sweater be ₹ x and ₹ y, respectively.


Hence, the cost price of the saree and the list price (price before discount) of the sweater are ₹ 600 and  ₹ 400, respectively.

Question 12:
Susan invested certain amount of money in two schemes A and B, which offer interest at the rate of 8% per annum and 9% per annum, respectively. She received ₹ 1860 as annual interest. However, had she interchanged the amount of investments in the two schemes, she would have received ₹ 20 more as annual interest. How much money did she invest in each scheme?
Solution:
Let the amount of investments in schemes A and 6 be ₹ x and ₹ y, respectively.
Case I Interest at the rate of 8% per annum on scheme A+ Interest at the rate of 9% per
annum on scheme 6 = Total amount received

Question 13:
Vijay had some bananas and he divided them into two lots A and B. He sold the first lot at the rate of ₹ 2 for 3 bananas and the second lot at the rate of ₹ 1 per banana and got a total of ₹ 400 If he had sold the first lot at the rate of ₹ 1 per banana and the second lot at the rate of ₹ 4 for 5 bananas, his total collection would have been ₹ 460. Find the total nmber of bananas he had.
Solution:
Let the number of bananas in lots A and B be x and y, respectively

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