Linear Equations in Two Variables Test

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Linear Equations in Two Variables Test - 6

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following coordinates satisfy the linear equation 7y – 5 = 8x?

A
(-5, 5)

B
(5, -5)

C
(-5, -5)

D
(5, 5)

Solution

The given linear equation is 7y - 5 = 8x.
Putting the coordinates (-5, -5) in the given equation,
7(-5) - 5 = 8(-5)
-35 - 5 = -40
-40 = -40; which is true.
Thus, (-5, -5) are the coordinates that satisfy 7y - 5 = 8x.
Question 2
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At which of the following points will the lines x - y = 0 and x + 0y = 3 intersect?

A
x = 2, y = 3

B
x = 3, y = 3

C
x = 3, y = 4

D
x = 4, y = 4

Solution

For plotting the line x - y = 0 on the graph, the table is:

x 0 1 2

y 0 1 2

For the second line, x + 0y = 3 implies x = 3.
Plotting both the lines on the graph, the graph obtained is:

Thus, the two given lines intersect at x = 3 and y = 3.
Question 3
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Find the value of m if the point (m, 8) lies on the graph which is represented by equation 8x – 5y = -8.

A
3

B
4

C
5

D
6

Solution

(m, 8) lies on the graph which is represented by the equation of line 8x – 5y = -8.
Thus, putting x = m and y = 8,
8m – 5(8) = -8
8m – 40 = -8
8m = -8 + 40
8m = 32
m = 32 ÷ 8 = 4
Question 4
1 / -0

Which of the following is the solution for both the linear equations 9x + 4y = 38 and 10x + 3y = 35 simultaneously?

A
(2, 6)

B
(2, 5)

C
(2, -5)

D
(2, 4)

Solution

The given two equations are 9x + 4y = 38 and 10x + 3y = 35.
Putting x = 2 and y = 5 in both the given equations,
9x + 4y = 38
9 × 2 + 4 × 5 = 38
18 + 20 = 38
38 = 38; which is true.
Thus, (2, 5) satisfies the equation 9x + 4y = 38.

10x + 3y = 35
10 × 2 + 3 × 5 = 35
20 + 15 = 35
35 = 35; which is true.
Thus, (2, 5) satisfies the equation 10x + 3y = 35.

Therefore, (2, 5) is the point that satisfies both the given linear equations 9x + 4y = 38 and 10x + 3y = 35 simultaneously.
Question 5
1 / -0

Which of the following lines meets the y-axis at -9?

A
3x + 18y = -18

B
3x + 2y = -18

C
2x + y = -18

D
2x + 3y = -18

Solution

The required line meets the y-axis, so the x-coordinate at that point should be equal to 0.
Putting y = -9 in 3x + 2y = -18,
3x + 2 × -9 = -18
3x - 18 = -18
3x = -18 + 18 = 0
x = 0
So, for the line 3x + 2y = -18, x = 0 at y = -9.
Thus, 3x + 2y = -18 would meet the y-axis at -9.
Question 6
1 / -0

A cook is roasting a chicken. For every kilogram the chicken weighs, he needs to cook it for 28 minutes and then an additional 30 minutes in total.

What will be the linear equation for the above information, if t (in minutes) is the total time taken to roast and k represents every kilogram?

A
t + 30k + 28 = 0

B
t – 28k = 30

C
28k - 30 = t

D
30k – 28 = t

Solution

t (in minutes) is the total time taken to roast and k represents every kilogram.
Therefore, the equation formed is:
t = 28k + 30
t – 28k = 30
Question 7
1 / -0

Which of the following statements is correct?

A
The equations y = 6 and y = - 6 represent the same line.

B
The equations y = 5 and x = 6 represent two parallel lines.

C
The equations y = 31 and y = 14 represent two parallel lines.

D
The equation 5y = 3 represents a line parallel to y-axis.

Solution

The equations y = 31 and y = 14 are both parallel to x-axis; hence, they are parallel to each other.
Question 8
1 / -0

Which of the following is the correct linear equation for the graph plotted below?

A
7x + 0y = 1

B
x - 0y = 7

C
7x + 0y = -7

D
7x - 0y = 7

Solution

A line parallel to y-axis at a distance 'c' from it is given by x = c.
Thus, x = 7 is the equation of the plotted graph.
x - 0y = 7 is the correct option.
Question 9
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On comparing the linear equation x – 2y = 3 with , the value of 3c comes out to be ...

A
1

B
2

C
11

D
9

Solution

Let p(x) = x – 2y = 3

Or ax – by = c
On comparing the standard equation with the given equation, we get c = 3
So, 3c = 3 × 3 = 9
Question 10
1 / -0

Where do the points (1, 3), (2, 3), (3, 3) and (4, 3) lie on a graph?

A
On a line parallel to the y-axis

B
On a line parallel to the x-axis

C
On the y-axis

D
On the x-axis

Solution

This graph shows that the points (1, 3), (2, 3), (3, 3) and (4, 3) lie on a line parallel to the x-axis.
Question 11
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Which of the following points lies on the graph of the linear equation ax – by = c, where a, b, c ≠ 0?

A
(,)

B
(,)

C
(,)

D
(,)

Solution

The equation is ax – by = c, and the solution is (,).
(1) = c

2c - 2c = c

0 = c (False)

(2) = c

c - 2c = c

-c = c (False)

(3) = c

2c – c = c

c = c (True)

(4) = c
c - c = c

0 = c (False)
Question 12
1 / -0

Which of the following equations has (3, -4) as its solution?

A
11x + 15y = -27

B
3x + 6y = 23

C
-14x + 13y = -33

D
18x – 4y = -35

Solution

The solution (3, -4) can satisfy the equation 11x + 15y = -27.

11 × 3 + 15 × (-4) = -27
33 – 60 = -27
-27 = -27

So, the equation 11x + 15y = -27 has (3, -4) as its solution.
Question 13
1 / -0

In the linear equation ax – my + n = 0, if x = 2m, then which of the following represents the value of 'y'?

A
a +

B
2a +

C
a -

D
2a +

Solution

The given equation is ax - my + n = 0.

Putting x = 2m in the given equation, we get

a × 2m - my + n = 0

a × 2m + n = my

y =
y = 2a +
Question 14
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The sum of five times A's money and two times B's money is 10, and A's money and B's money are represented by x and y respectively. How will you represent this situation on a graph?

A

B

C

D

Solution

The linear equation for this situation is represented by 5x + 2y = 10.
When x = 0, y = 5; (0, 5) satisfies the equation.
And when y = 0, x = 2; (2, 0) satisfies the equation.
Question 15
1 / -0

Find the value of y from the given equations:

5x + y + 15 = 0, 20x – 15y = 35

A
5

B
-5

C
10

D
-10

Solution

5x + y + 15 = 0
5x + y = -15
5x = -15 – y
x = ... (1)
Now, according to the given equation 20x – 15y = 35:
Putting the value of x from equation (1),
20() – 15y = 35
4(-15 – y) – 15y = 35
-60 – 4y - 15y = 35
-19y = 35 + 60
-19y = 95
y = -5
Question 16
1 / -0

Find the value of y in the following set of equations.

+ = 5 --- (i)
+ = --- (ii)

A

B

C

D

Solution

+ = 5 --- (i)
+ = --- (ii)
Multiplying equation (i) by ,
+ = --- (iii)

So, on subtracting the equation (iii) from (ii), we can find the value of y.

y =

y = ^{y =}
Question 17
1 / -0

Some football stats are analysed and are such that:

The numbers of goals scored by four players R, M, L and X in two seasons are 50, 18, 20 and 12. The total goals scored in two seasons by these five players R, M, L, X and T are 150 and the sum of the numbers of goals scored by players B and E is equal to the number of goals scored by T. If E scored half of the goals scored by S and S scored three-fifths of the goals scored by L, then what is the number of goals scored by B in two seasons?

A
40

B
44

C
47

D
48

Solution

The number of goals scored by four players R, M, L and X in two seasons are 50, 18, 20 and 12.
Rest of the goals are scored by T and let the number be 'a'.
Total number of goals = 150
Therefore,
50 + 18 + 20 + 12 + a = 150
a + 100 = 150
a = 150 – 100 = 50
Therefore, number of goals scored by T in two seasons = 50
Sum of the numbers of goals scored by B and E is equal to the number of goals scored by T = 50
Let the number of goals scored by B in two seasons be b and that scored by E in two seasons be e.
Therefore, b + e = 50
E scored half of the number of goals scored by S.
Let the number of goals scored by S in two seasons be 's'.
Therefore, e = × s
S scored three-fifths of the number of goals scored by L.
Therefore, s = = 12
Therefore,
E scored:
e = × 12 = 6
And B scored:
b + e = 50
b = 50 – e
b = 50 – 6
b = 44
Hence, number of goals scored by B in two seasons = 44
So, option (2) is correct.
Question 18
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In a basketball match, Team A and Team B scored the number of the three pointers equal to Team C, which scored 20 three pointers. It was found that the score was miscalculated and was therefore calculated again. It was found that Team C scored 30 three pointers and Team B scored 13 three pointers. It was found that coincidentally the number of three pointers scored by Team A was same as the previous result.

What was the number of three pointers scored by Team D, if the numbers of three pointers of Team B (actual), Team A (actual) and Team B (miscalculated) when add up make the same score as Team D?

A
37

B
3

C
33

D
30

Solution

Given,
Team C scored 20 three pointers when the result was miscalculated.
Let the number of three pointers scored by Team A when the core was miscalculated be a.
And the number of three pointers scored by Team B when the score was miscalculated be b.
Therefore,
a + b = 20 ... (1)
Now, when the score was calculated again:
Team B scored 13 three pointers and Team C scored 30 three pointers.
Therefore,
Team A scored 30 – 13 = 17.
Therefore, Team A scored 17 three pointers in actual and it is the same number of three pointers when the result was miscalculated.
Now, equation (1) becomes:
17 + b = 20
b = 20 – 17
b = 3
Miscalculated result for Team B = 3 three pointers
Now,
The numbers of three pointers of Team B (actual), Team A (actual) and Team B (miscalculated) when add up makes the same score as Team D.
Number of three pointers scored by Team D = 17 + 13 + 3 = 33
Hence, 33 is the answer.
Question 19
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Which of the given options represents the same condition as discussed below?

The price of a chocolate ice-cream and a vanilla ice-cream when added together is Rs. 20. The price of a vanilla ice-cream is quarter of the price of an ice-cream brick. The price of an orange ice-cream when doubled is Rs. 4 more than the price of a chocolate ice-cream.
(Assume the price of a chocolate ice-cream be Rs. a, the price of a vanilla ice-cream be Rs. b, Price of an ice-cream brick = Rs. c,Price of an orange ice-cream = Rs. d)

A
8c – d = 96

B
c – 8d = 64

C
8c + d = 64

D
8d + c = 96

Solution

Given,
Total price of a chocolate ice-cream and a vanilla ice-cream = Rs. 20
Let the price of a chocolate ice-cream be Rs. a.
And the price of a vanilla ice-cream be Rs. b.
Price of an ice-cream brick = Rs. c
Price of an orange ice-cream = Rs. d
Therefore, a + b = 20 ... (1)
Now,
The price of a vanilla ice-cream is quarter of the price of an ice-cream brick.
b =
Put this in equation (1).
a + = 20
= 20
4a + c = 4 × 20
4a + c = 80 ... (2)
Now,
The price of an orange ice-cream when doubled is Rs. 4 more than the price of a chocolate ice-cream.
2d = a + 4
a = 2d – 4
Put this in equation (2).
4(2d – 4) + c = 80
8d – 16 + c = 80
8d + c = 96
Hence, option 4 is the answer.
Question 20
1 / -0

Ram is paid Rs. 50 for delivering letters at the end of the week. On delivering each letter, he is paid Rs. 2. Shyam is paid Rs. 20 for delivering birthday cards at the end of each week. On delivering each card, he is paid Rs. 5. The money that Ram gets after a week is exactly the same as the money Shyam gets after two weeks.
(Assume that the number of letters delivered by Ram is x and the number of birthday cards delivered by Shyam is y.)

Which of the given options satisfies the above situation?

A
10y - x = 5

B
5y - x = 5

C
10y - x = 10

D
5y - 2x = 10

Solution

Given: The number of letters delivered by Ram is 'x' and the number of birthday cards delivered by Shyam is 'y'.

Ram is paid Rs. 50 for delivering the letters at the end of the week. On delivering each letter he is paid Rs. 2.
Therefore, 50 + 2x is the total amount of money that Ram gets after a week.
Shyam is paid Rs. 20 for delivering birthday cards at the end of each week. On delivering each card he is paid Rs. 5.
Therefore, 20 + 5y is the total amount of money that Shyam gets after a week.
Therefore, 2(20 + 5y) will be the money that Shyam gets after two weeks.
It is given that Shyam gets the same amount of money in two week as Ram gets in one week.
Therefore,
50 + 2x = 2(20 + 5y)
50 + 2x = 40 + 10y
50 - 40 = 10y - 2x
10 = 10y - 2x
2(5y - x) = 10
5y - x = 5
Hence, option 2 is the answer.
Question 21
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Fill in the blanks:

1. The diameter of a circle is 14 cm and its centre is at (0, 1). It intersects the Y-axis at Y = A .
2. If X = 2 and Y = 3 are a solution for the equation 4X + kY = 17, then the value of k is B .
3. A line with equation Y - 7 = 0 cuts the X-axis at C point(s).
4. In equation= -10, the solution is ( D , 20)

A
A = 8, B = 3, C = one, D = 12

B
A = 8, B = 3, C = infinite, D = 14

C
A = 8, B = 3, C = 0, D = 14

D
A = 8, B = 4, C = one, D = 14

Solution

1. The diameter of the circle is 14 cm; so the radius is 7 cm. If the centre is at Y = 1, then the circle will intersect the Y-axis at Y = 1 + 7 = 8.

2. X = 2 and Y = 3 are a solution for the equation 4X + kY = 17
Putting X = 2 and Y = 3 in the equation, we get
4(2) + k(3) = 17
3k = 17 – 8 = 9
k = 3

3. A line with equation Y – 7 = 0 will not touch X-axis as the slope of this line is undefined. This is a horizontal line. So, there are zero points of intersection on X-axis.

4. = -10
Put y = 20 from the given solution of the line.
= -10

- 16 = -10
= 6
x == 14

Question 22

1 / -0

In a rectangle, the width is 12 cm more than one quarter of the length. Length is denoted by X cm and perimeter iscm.
(i). If the value of Y is 868, then what is the width of the rectangle?
(ii). If the width of the rectangle is 27 cm, then what is the value of Y?

Select one alternative among A, B, C, D from below.

	(i)	(ii)
A	22	1218
B	40	1218
C	60	2436
D	22	609

A

B

C

D

Solution

Length of the rectangle, L = X cm
Width of the rectangle, W =cm
Semi-perimeter of the rectangle == L + W
X +=
4X + X + 48 =
5X + 48 =
I: If Y = 868, then
5X + 48 =
5X = 248 – 48
X = 200 ÷ 5 = 40

Width of the rectangle ==+ 12 = 10 + 12 = 22 cm
II: If the width is 27 cm,
Length of the rectangle, L = (27 – 12) × 4 = 15 × 4 = 60 cm = X

5X + 48 =
5(60) + 48 =
348 =
Y == 1218

Hence, the correct option is 1.

Question 23
1 / -0

Which of the following sets of equations represents the triangle given in the graph?

A
AB is X + 3Y = 11.
BC is X – Y = 2.
AC is 3X – 2Y = -2.

B
AB is X + Y = 11.
BC is X – 2Y = 2.
AC is 3X – Y = -2.

C
AB is 2X + Y = 11.
BC is 9Y - 5X = 12.AC is 5Y – 7X = 9

D
AB is X + 2Y = 11.
BC is 9Y - 5X = 2.
AC is 5Y – 7X = 18

Solution

AB passes through (1, 5) and (5, 3).
X + 2Y = 11 satisfies it.

Putting X = 1, we get
1 + 2Y = 11
2Y = 10
Y = 5

Putting X = 5, we get
5 + 2Y = 11
2Y = 6
Y = 3

BC passes through (5, 3) and (-4, -2).
9Y - 5X = 2 satisfies it.

Putting X = 5, we get
9Y - 25 = 2
Y = 3

Putting X = -4, we get
9Y + 20 = 2
Y = -2

AC passes through (1, 5) and (-4, -2).
5Y - 7X = 18 satisfies it.

Putting X = 1, we get
5Y - 7 = 18
Y = 5

Putting X = -4, we get
5Y - 7X = 18
5Y + 28 = 18
Y = -2
Hence, option 4 is correct.
Question 24
1 / -0

Fill in the blanks.

(I) A hotel charges Rs. 4450 as fixed rent and Rs. 500 per day. The total amount in terms of 'y' if Amit stayed for 'x' days is _____P____.
(II) Mani has Rs. 1400 more than thrice the amount of money Sanpreet has. If 'y' represents the amount of money Mani has and 'x' represents the amount of money Sanpreet has, then the equation ____Q___ represents this situation.
(III) The length of a rectangle is 6 more than its width. If the perimeter of the rectangle is 104 cm, then the length of the rectangle is ___R____ cm.
(IV) The equation P = 4s describes the relationship between perimeter P of a square and length of each side s. The coordinates that satisfy the equation are ____S_____.

A

(P) (Q) (R) (S)

Rs. (500 + 4450x) y = 3 + x + 1400 27 (0, 1)

B

(P) (Q) (R) (S)

Rs. (4450 + 500x) y = 3x + 1400 29 (3, 12)

C

(P) (Q) (R) (S)

Rs. (4450 × 500x) y = 3x × 1400 23 (-1, 5)

D

(P) (Q) (R) (S)

Rs.(4950x) y = 3x + 4200 30 (3, 9)

Solution

(I) y = Total charges
So, Total charges = Fixed charges + (Charges per day × Number of days)
y = 4450 + 500x

(II) Mani has Rs. y and Sanpreet has Rs. x.
According to the question,
Mani has Rs. 1400 more than thrice the amount of money Sanpreet has.
So, y = 1400 + 3x
Or, y = 3x + 1400

(III) Width = w
Length, l = w + 6
Perimeter = 2(l + w)
104 = 2(6 + w + w)
104 = 2(6 + 2w)
104 = 12 + 4w
104 – 12 = 4w
92 = 4w
w = 23 cm
l = 23 + 6 = 29 cm

(IV) Take a point (3, 12).
Put s = 3 in P = 4s.
P = 4(3) = 12
Hence, point (3, 12) satisfies the given equation.
Question 25
1 / -0

Which of the following equations are represented by the following graph?

A
8x – 7y = 16, 22x + 19y = -2

B
7y – 2x = -9, 3x + 5y = 0

C
8x + 7y = 18 , 22x – 19 y = 2

D
9x – 7y = 6, 3x + 7y = 6

Solution

The following two equations will satisfy the given graph:

8x – 7y = 16 ... (i)

Putting (x, y) = (2, 0), we get

16 - 0 = 16
16 = 16

and for (x, y) = (1.125, -1), we get

9 + 7 = 16
16 = 16

22x + 19y = -2 ... (ii)

When (x, y) = (2.5, -3),

55 - 57 = -2
-2 = -2

and (x, y) = (-7, 8), we get

-154 + 152 = -2
-2 = -2

Hence option (1) is correct.

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(P)	(Q)	(R)	(S)
Rs. (500 + 4450x)	y = 3 + x + 1400	27	(0, 1)

(P)	(Q)	(R)	(S)
Rs. (4450 + 500x)	y = 3x + 1400	29	(3, 12)

(P)	(Q)	(R)	(S)
Rs. (4450 × 500x)	y = 3x × 1400	23	(-1, 5)

(P)	(Q)	(R)	(S)
Rs.(4950x)	y = 3x + 4200	30	(3, 9)

Linear Equations in Two Variables Test - 6

Result

Linear Equations in Two Variables Test - 6

Score

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Rank

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SHARING IS CARING

Question 1

(-5, 5)

(5, -5)

(-5, -5)

(5, 5)

Solution

Question 2

x = 2, y = 3

x = 3, y = 3

x = 3, y = 4

x = 4, y = 4

Solution

Question 3

3

4

5

6

Solution

Question 4

(2, 6)

(2, 5)

(2, -5)

(2, 4)

Solution

Question 5

3x + 18y = -18

3x + 2y = -18

2x + y = -18

2x + 3y = -18

Solution

Question 6

t + 30k + 28 = 0

t – 28k = 30

28k - 30 = t

30k – 28 = t

Solution

Question 7

The equations y = 6 and y = - 6 represent the same line.

The equations y = 5 and x = 6 represent two parallel lines.

The equations y = 31 and y = 14 represent two parallel lines.

The equation 5y = 3 represents a line parallel to y-axis.

Solution

Question 8

7x + 0y = 1

x - 0y = 7

7x + 0y = -7

7x - 0y = 7

Solution

Question 9

1

2

11

9

Solution

Question 10

On a line parallel to the y-axis

On a line parallel to the x-axis

On the y-axis

On the x-axis

Solution

Question 11

(,)

(,)

(,)

(,)

Solution

Question 12

11x + 15y = -27

3x + 6y = 23

-14x + 13y = -33

18x – 4y = -35

Solution

AB is X + 3Y = 11.
BC is X – Y = 2.
AC is 3X – 2Y = -2.

AB is X + Y = 11.
BC is X – 2Y = 2.
AC is 3X – Y = -2.

AB is 2X + Y = 11.
BC is 9Y - 5X = 12.AC is 5Y – 7X = 9

AB is X + 2Y = 11.
BC is 9Y - 5X = 2.
AC is 5Y – 7X = 18

(P) (Q) (R) (S)

Rs. (500 + 4450x) y = 3 + x + 1400 27 (0, 1)

(P) (Q) (R) (S)

Rs. (4450 + 500x) y = 3x + 1400 29 (3, 12)

(P) (Q) (R) (S)

Rs. (4450 × 500x) y = 3x × 1400 23 (-1, 5)