9th STD Maths Second Mid Term Test 2024 - Question Paper with Solutions
PART - I (6 x 1 = 6)
I. Answer all the questions.
1. Which of the following is a linear equation.
Reason: A linear equation has the highest power of the variable as 1. Option (c) is the only one that satisfies this condition.
2. If (2,3) is a solution of the linear equation \(2x + 3y = K\), then find the value of 'K'.
Solution: Substitute x = 2 and y = 3 into the equation: \(2(2) + 3(3) = 4 + 9 = 13\). So, K = 13.
3. If \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\) where \(a_1x + b_1y + c_1 = 0\) and \(a_2x + b_2y + c_2 = 0\) then the given pair of linear equations has .......... solution (s).
Reason: The condition \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\) is the standard condition for a pair of linear equations to have exactly one (unique) solution, representing two intersecting lines.
4. A chord is at a distance of 15 cm from the centre of the circle of radius 25 cm. The length of the chord is
Solution: Let the radius be \(r=25\) cm and the distance from the center be \(d=15\) cm. The radius, the distance, and half the chord length form a right-angled triangle. Let half the chord length be \(l\). By Pythagoras theorem, \(r^2 = d^2 + l^2\).
\(25^2 = 15^2 + l^2 \Rightarrow 625 = 225 + l^2 \Rightarrow l^2 = 400 \Rightarrow l = 20\) cm.
The total length of the chord is \(2 \times l = 2 \times 20 = 40\) cm.
5. The points (-5, 2) and (2, -5) lie in the .......... .
Reason: The point (-5, 2) has a negative x-coordinate and a positive y-coordinate, so it lies in the II quadrant. The point (2, -5) has a positive x-coordinate and a negative y-coordinate, so it lies in the IV quadrant.
6. The distance between the two points (2,3) and (1,4) is .......... .
Solution: Using the distance formula, \(d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\).
\(d = \sqrt{(1-2)^2 + (4-3)^2} = \sqrt{(-1)^2 + (1)^2} = \sqrt{1 + 1} = \sqrt{2}\).
PART - II (4 x 2 = 8)
II. Answer any 4 questions. 12th question is compulsory.
7. Solve by substitution method. \(x + 3y = 16\), \(2x - y = 4\)
Given equations are:
\(x + 3y = 16\) --- (1)
\(2x - y = 4\) --- (2)
From equation (1), we get \(x = 16 - 3y\). --- (3)
Substitute (3) into equation (2):
\(2(16 - 3y) - y = 4\)
\(32 - 6y - y = 4\)
\(32 - 7y = 4\)
\(-7y = 4 - 32\)
\(-7y = -28\)
\(y = 4\)
Substitute \(y=4\) into equation (3):
\(x = 16 - 3(4)\)
\(x = 16 - 12\)
\(x = 4\)
8. Solve: \(2x - y = 3\), \(3x + y = 7\)
Given equations are:
\(2x - y = 3\) --- (1)
\(3x + y = 7\) --- (2)
Adding equation (1) and (2) to eliminate y:
\((2x - y) + (3x + y) = 3 + 7\)
\(5x = 10\)
\(x = 2\)
Substitute \(x=2\) into equation (2):
\(3(2) + y = 7\)
\(6 + y = 7\)
\(y = 7 - 6\)
\(y = 1\)
9. The angles of a quadrilateral are in the ratio 2 : 4 : 5 : 7. Find all the angles.
Let the angles be \(2k, 4k, 5k,\) and \(7k\).
The sum of the angles in a quadrilateral is 360°.
\(2k + 4k + 5k + 7k = 360^\circ\)
\(18k = 360^\circ\)
\(k = \frac{360}{18}\)
\(k = 20^\circ\)
Now, we find the angles:
First angle = \(2k = 2(20^\circ) = 40^\circ\)
Second angle = \(4k = 4(20^\circ) = 80^\circ\)
Third angle = \(5k = 5(20^\circ) = 100^\circ\)
Fourth angle = \(7k = 7(20^\circ) = 140^\circ\)
10. Find the value of x°
In the given figure, triangle XYZ is inscribed in a semicircle (since one side is the diameter passing through O).
The angle subtended by a diameter at any point on the circumference is 90°.
Therefore, \(\angle XYZ = 90^\circ\).
The sum of angles in a triangle is 180°.
In \(\triangle XYZ\):
\(\angle ZXY + \angle XYZ + \angle YZX = 180^\circ\)
\(x^\circ + 90^\circ + 63^\circ = 180^\circ\)
\(x^\circ + 153^\circ = 180^\circ\)
\(x^\circ = 180^\circ - 153^\circ\)
\(x^\circ = 27^\circ\)
11. Find the distance between the points (-4, 3), (2, -3)
Let the points be A(-4, 3) and B(2, -3).
Using the distance formula, \(d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\).
\(d = \sqrt{(2 - (-4))^2 + (-3 - 3)^2}\)
\(d = \sqrt{(2+4)^2 + (-6)^2}\)
\(d = \sqrt{(6)^2 + (-6)^2}\)
\(d = \sqrt{36 + 36}\)
\(d = \sqrt{72}\)
\(d = \sqrt{36 \times 2} = 6\sqrt{2}\)
12. Find the mid point of the following points (8, -2), (-8, 0)
Let the points be A(8, -2) and B(-8, 0).
Using the midpoint formula, \(M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\).
\(M = \left(\frac{8 + (-8)}{2}, \frac{-2+0}{2}\right)\)
\(M = \left(\frac{0}{2}, \frac{-2}{2}\right)\)
\(M = (0, -1)\)
PART - III (4 X 5 = 20)
III. Answer any 4 questions. 18th question is compulsory.
13. Check whether (5, -1) is a solution of the simultaneous equations \(x - 2y = 7\) and \(2x + 3y = 7\)
To check if (5, -1) is a solution, substitute \(x=5\) and \(y=-1\) into both equations.
Equation 1: \(x - 2y = 7\)
LHS = \(5 - 2(-1) = 5 + 2 = 7\).
RHS = 7.
Since LHS = RHS, the point satisfies the first equation.
Equation 2: \(2x + 3y = 7\)
LHS = \(2(5) + 3(-1) = 10 - 3 = 7\).
RHS = 7.
Since LHS = RHS, the point satisfies the second equation.
As the point (5, -1) satisfies both equations, it is a solution of the simultaneous equations.
14. Solve by elimination method: \(4a + 3b = 65\), \(a + 2b = 35\)
Given equations:
\(4a + 3b = 65\) --- (1)
\(a + 2b = 35\) --- (2)
To eliminate 'a', multiply equation (2) by 4:
\(4(a + 2b) = 4(35)\)
\(4a + 8b = 140\) --- (3)
Now, subtract equation (1) from equation (3):
\((4a + 8b) - (4a + 3b) = 140 - 65\)
\(5b = 75\)
\(b = 15\)
Substitute \(b = 15\) into equation (2):
\(a + 2(15) = 35\)
\(a + 30 = 35\)
\(a = 5\)
15. The chord of length 30 cm is drawn at a distance of 8 cm from the centre of the circle. Find the radius of the circle.
Let the length of the chord be \(AB = 30\) cm.
The distance from the center O to the chord is \(OC = 8\) cm.
The perpendicular from the center to a chord bisects the chord. So, \(AC = CB = \frac{30}{2} = 15\) cm.
Consider the right-angled triangle \(\triangle OAC\). The radius \(OA\) is the hypotenuse.
By Pythagoras theorem: \(OA^2 = OC^2 + AC^2\)
\(r^2 = 8^2 + 15^2\)
\(r^2 = 64 + 225\)
\(r^2 = 289\)
\(r = \sqrt{289}\)
\(r = 17\) cm
16. Determine whether the given points are collinear or not: (7,-2), (5, 1), (3, 4)
Let the points be A(7, -2), B(5, 1), and C(3, 4). The points are collinear if the area of the triangle formed by them is zero.
Area of \(\triangle ABC = \frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|\)
Area = \(\frac{1}{2} |7(1-4) + 5(4 - (-2)) + 3(-2 - 1)|\)
= \(\frac{1}{2} |7(-3) + 5(6) + 3(-3)|\)
= \(\frac{1}{2} |-21 + 30 - 9|\)
= \(\frac{1}{2} |30 - 30|\)
= \(\frac{1}{2} |0| = 0\)
Since the area of the triangle is 0, the points are collinear.
17. Find the points which divide the line segment joining A(-11, 4) and B(9, 8) into four equal parts.
Let the points that divide the segment AB into four equal parts be P, Q, and R.
Q is the midpoint of AB.
\(Q = \left(\frac{-11+9}{2}, \frac{4+8}{2}\right) = \left(\frac{-2}{2}, \frac{12}{2}\right) = (-1, 6)\).
P is the midpoint of AQ.
\(P = \left(\frac{-11+(-1)}{2}, \frac{4+6}{2}\right) = \left(\frac{-12}{2}, \frac{10}{2}\right) = (-6, 5)\).
R is the midpoint of QB.
\(R = \left(\frac{-1+9}{2}, \frac{6+8}{2}\right) = \left(\frac{8}{2}, \frac{14}{2}\right) = (4, 7)\).
18. Solve by cross multiplication: \(8x - 3y = 12\), \(5x = 2y + 7\)
First, write the equations in the standard form \(ax + by + c = 0\):
\(8x - 3y - 12 = 0\)
\(5x - 2y - 7 = 0\)
Using the cross-multiplication method formula:
$$ \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} $$Here, \(a_1=8, b_1=-3, c_1=-12\) and \(a_2=5, b_2=-2, c_2=-7\).
\(\frac{x}{(-3)(-7) - (-2)(-12)} = \frac{y}{(-12)(5) - (-7)(8)} = \frac{1}{(8)(-2) - (5)(-3)}\)
\(\frac{x}{21 - 24} = \frac{y}{-60 + 56} = \frac{1}{-16 + 15}\)
\(\frac{x}{-3} = \frac{y}{-4} = \frac{1}{-1}\)
Now, solve for x and y:
\(\frac{x}{-3} = -1 \Rightarrow x = 3\)
\(\frac{y}{-4} = -1 \Rightarrow y = 4\)
PART - IV (2 X 8 = 16)
IV. Answer all the question.
19. Draw the graph for \(y = 4x - 1\), (OR) \(y = \frac{2}{3}x + 5\)
Solution for \(y = 4x - 1\):
To draw the graph, we find at least three points on the line.
- If \(x=0\), \(y = 4(0) - 1 = -1\). Point: (0, -1).
- If \(x=1\), \(y = 4(1) - 1 = 3\). Point: (1, 3).
- If \(x=2\), \(y = 4(2) - 1 = 7\). Point: (2, 7).
Steps to draw the graph:
- Draw the X and Y axes on a graph sheet.
- Plot the points (0, -1), (1, 3), and (2, 7).
- Join the points with a straight line. This line is the graph of \(y=4x-1\).
Solution for \(y = \frac{2}{3}x + 5\) (OR):
To avoid fractions, choose x-values that are multiples of 3.
- If \(x=0\), \(y = \frac{2}{3}(0) + 5 = 5\). Point: (0, 5).
- If \(x=3\), \(y = \frac{2}{3}(3) + 5 = 2 + 5 = 7\). Point: (3, 7).
- If \(x=-3\), \(y = \frac{2}{3}(-3) + 5 = -2 + 5 = 3\). Point: (-3, 3).
Steps to draw the graph:
- Draw the X and Y axes on a graph sheet.
- Plot the points (0, 5), (3, 7), and (-3, 3).
- Join the points with a straight line. This line is the graph of \(y = \frac{2}{3}x + 5\).
20. In the given figure, O is the centre of the circle, If the measure of \(\angle OQR = 48^\circ\). What is the measure of \(\angle QPR\)? (OR) Show that the point (11, 2) is the centre of the circle passing through the points (1, 2), (3, -4) and (5, -6).
Solution for the first part:
Given, O is the center and \(\angle OQR = 48^\circ\).
In \(\triangle OQR\), \(OQ = OR\) (radii of the same circle).
Therefore, \(\triangle OQR\) is an isosceles triangle.
This means the angles opposite to the equal sides are equal: \(\angle ORQ = \angle OQR = 48^\circ\).
The sum of angles in \(\triangle OQR\) is 180°.
\(\angle QOR + \angle OQR + \angle ORQ = 180^\circ\)
\(\angle QOR + 48^\circ + 48^\circ = 180^\circ\)
\(\angle QOR + 96^\circ = 180^\circ\)
\(\angle QOR = 180^\circ - 96^\circ = 84^\circ\)
The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.
\(\angle QPR = \frac{1}{2} \angle QOR = \frac{1}{2} \times 84^\circ = 42^\circ\)
Solution for the (OR) part:
Let the center be C(11, 2) and the points on the circle be A(1, 2), B(3, -4), and D(5, -6).
If C is the center, then the distance from C to each point must be equal (this distance is the radius).
Distance CA:
\(CA = \sqrt{(11-1)^2 + (2-2)^2} = \sqrt{10^2 + 0^2} = \sqrt{100} = 10\) units.
Distance CB:
\(CB = \sqrt{(11-3)^2 + (2 - (-4))^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10\) units.
Distance CD:
\(CD = \sqrt{(11-5)^2 + (2 - (-6))^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\) units.
Since \(CA = CB = CD = 10\), the point C(11, 2) is equidistant from A, B, and D.