<\/span><\/h2>\nMark the correct alternative in each of the following:
<\/strong>Question 1.
<\/strong>If (4, 19) is a solution of the equation y = ax + 3, then a =<\/strong>
(a) 3 <\/strong>
(b) 4<\/strong>
(c) 5 <\/strong>
(d) 6
<\/strong>Solution:
<\/strong>\u2235 (4, 19) is a solution to the equation
y = ax + 3
\u2234 x = 4, y= 19 will satisfy the equation
\u2234 19 = a x 4 + 3 = 4a + 3
4a = 19-3 = 16 \u21d2 a= 16<\/span>4<\/span><\/span><\/span><\/span><\/span> = 4
\u2234 a = 4 (b)<\/strong><\/p>\nQuestion 2.
<\/strong>If (a, 4) lies on the graph of 3x + y = 10, then the value of a is<\/strong>
(a) 3 <\/strong>
(b) 1<\/strong>
(c) 2 <\/strong>
(d) 4
<\/strong>Solution:
<\/strong>\u2235 (a, 4) is the solution of equation 3x + y = 10
\u2234 x = a, y = 4 will satisfy the equation
\u2234 Substituting the value of x and y in the equation
3 xa + 4= 10 \u21d2 3a =10- 4 = 6
\u21d2 a = 6<\/span>3<\/span><\/span><\/span><\/span><\/span> = 2
\u2234 a = 2 (c)<\/strong><\/p>\n <\/div>\n
Question 3.
<\/strong>The graph of the linear equation 2x \u2013 y= 4 cuts x-axis at<\/strong>
(a) (2, 0) <\/strong>
(b) (-2, 0)<\/strong>
(c) (0, -4) <\/strong>
(d) (0, 4)
<\/strong>Solution:
<\/strong>\u2235 graph of the equation,
2x \u2013 y = 4 cuts x-axis
\u2234 y = 0
\u2234 2x \u2013 0 = 4 \u21d2 2x = 4
\u21d2 x = 4<\/span>2<\/span><\/span><\/span><\/span><\/span> = 2
\u2234 The line cuts x-axis at (2, 0) (a)<\/strong><\/p>\nQuestion 4.
<\/strong>How many linear equations are satisfied by x = 2 and y = -3 ?<\/strong>
(a) Only one <\/strong>
(b) Two<\/strong>
(c) Three <\/strong>
(d) Infinitely many
<\/strong>Solution:
<\/strong>\u2235 From a point, an infinite number of lines can pass.
\u2234 The solution x = 2, y = -3 is the solution of infinitely many linear equations. (d)<\/strong><\/p>\nQuestion 5.
<\/strong>The equation x \u2013 2 = 0 on the number line is represented by<\/strong>
(a) aline <\/strong>
(b) a point<\/strong>
(c) infinitely many lines<\/strong>
(d) two lines<\/strong>
Solution:
<\/strong>The equation x \u2013 2 = 0
\u21d2 x = 2
\u2234 It is represented by a point on a number line. (b)<\/strong><\/p>\nQuestion 6.
<\/strong>x = 2, y = -1 is a solution of the linear equation<\/strong>
(a) x + 2y = 0 <\/strong>
(b) x + 2y = 4<\/strong>
(c) 2x + y = 0 <\/strong>
(d) 2x + y = 5
<\/strong>Solution:
<\/strong>x = 2, y = -1
Substituting the values of x and y in the equations one by one, we get (a) x + 2y = 0
\u21d2 2 + 2(-1) = 0
\u21d2 2 \u2013 2 = 0
\u21d2 0 = 0 which is true (a)<\/strong><\/p>\nQuestion 7.
<\/strong>If (2k \u2013 1, k) is a solution of the equation 10x \u2013 9y = 12, then k =<\/strong>
(a) 1 <\/strong>
(b) 2<\/strong>
(c) 3 <\/strong>
(d) 4
<\/strong>Solution:
<\/strong>\u2235 (2k \u2013 1, k) is a solution of the equation 10x \u2013 9y = 12
Substituting the value of x and y in the equation
10(2k \u2013 1) \u2013 9k = 12
\u21d2 20k \u2013 10-9k= 12
\u21d2 20k \u2013 9k = 12 + 10
\u21d2 11k = 22
\u21d2 k =22<\/span>11<\/span><\/span><\/span><\/span><\/span> = 2
<\/sup>\u2234 k = 2 (b)<\/strong><\/p>\nQuestion 8.
<\/strong>The distance between the graph of equation x = \u2013 3 and x = 2 is<\/strong>
(a) 1 <\/strong>
(b) 2<\/strong>
(c) 3 <\/strong>
(d) 5<\/strong>
Solution:
<\/strong>The distance between the graphs of the equation
x = -3 and x = 2 will be
2(-3) = 2+ 3 = 5 (b) <\/strong><\/p>\nQuestion 9.
<\/strong>The distance between the graphs of the equations y = -1 and y = 3 is<\/strong>
(a) 2 <\/strong>
(b) 4<\/strong>
(c) 3 <\/strong>
(d) 1
<\/strong>Solution:
<\/strong>The distance between the graphs of the equation
y = -1 and y = 3
is 3 \u2013 (-1) = 3 + 1 = 4 (b)<\/strong><\/p>\nQuestion 10.
<\/strong>If the graph of equation 4x + 3y = 12 cuts the co-ordinate axes at A and B, then the hypotenuse of right triangle AOB is of length<\/strong>
(a) 4 units<\/strong>
(b) 3 units<\/strong>
(c) 5 units <\/strong>
(d) none of these
<\/strong>Solution:
<\/strong>Equation is 4x + 3y = 12
If it cuts the x-axis, then y = 0
\u2234 4x x 3 x 0 = 12
\u21d2 4x = 12 \u21d2 x = 12<\/span>4<\/span><\/span><\/span><\/span><\/span> = 3
OA = 3 units
\u2234 The point of intersection of the x-axis is (3, 0)
Again if it cuts the y-axis, then x = 0, Y= 0
\u2234 4x x 3 x 0 = 12
\u21d2 4x = 12 \u21d2 x = 12<\/span>3<\/span><\/span><\/span><\/span><\/span> = 4
\u21d2 OB = 4 units
\u2234 The point of intersection is (0, 4)
\u2234 In right \u0394AOB,
AB2<\/sup> = AO2<\/sup> + OB2
<\/sup>= (3)2<\/sup> + (4)2<\/sup>
= 9 + 16 = 25
= (5)2
<\/sup>\u2234 AB = 5 units (c)<\/strong><\/p>\nThis is the complete blog on RD Sharma Class 9 Solutions Chapter 7 MCQsTo know more about the CBSE<\/a> Class 9 Maths exam, ask in the comments.<\/p>\n
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