us mc 9th question paper with Mathematics solved paper

Exam Style with Complete Solutions

Time: 3 Hours
Maximum Marks: 80

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SECTION – A

MCQs — 20 × 1 = 20

Q1. Which of the following is an irrational number?

A. $\frac35$
B. $0.25$
C. $\sqrt7$
D. $-2$

Answer: C. 7\sqrt7

Solution: $\sqrt7$ cannot be expressed in the form $\frac pq$, where $p,q$ are integers and $q\neq0$. Therefore, it is irrational.

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Q2. The decimal expansion of $\frac78$ is:

A. Terminating
B. Non-terminating recurring
C. Non-terminating non-recurring
D. None

Answer: A. Terminating

Solution:$$\frac78=0.875$$

Therefore, its decimal expansion terminates.

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Q3. If $p(x)=x-5$, then the zero of $p(x)$ is:

A. 0
B. 1
C. 5
D. -5

Answer: C. 5

Solution:

For zero,$$p(x)=0$$$$x-5=0$$$$x=5$$

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Q4. Find the degree of $4x^3+2x^2-7$.

A. 1
B. 2
C. 3
D. 4

Answer: C. 3

Solution: The highest power of $x$ is $3$. Therefore, degree = 3.

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Q5. Which ordered pair satisfies $2x+y=7$?

A. (1,2)
B. (2,3)
C. (3,2)
D. (4,2)

Answer: B. (2,3)

Solution:

For $(2,3)$:$$2(2)+3=4+3=7$$

Hence, $(2,3)$ is the correct answer.

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Q6. The point $(-3,4)$ lies in:

A. First quadrant
B. Second quadrant
C. Third quadrant
D. Fourth quadrant

Answer: B. Second quadrant

Solution:
$x$ is negative and $y$ is positive. Therefore, the point lies in the second quadrant.

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Q7. Distance of $(5,-2)$ from the x-axis is:

A. 5
B. -2
C. 2
D. 7

Answer: C. 2

Solution:

Distance from x-axis = absolute value of $y$-coordinate.$$|-2|=2$$

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Q8. Two angles whose sum is $90^\circ$ are called:

A. Supplementary
B. Complementary
C. Adjacent
D. Vertically opposite

Answer: B. Complementary angles

Solution:

Angles having sum $90^\circ$ are called complementary angles.

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Q9. If two parallel lines are intersected by a transversal, corresponding angles are:

A. Unequal
B. Supplementary
C. Equal
D. Always $90^\circ$

Answer: C. Equal

Solution: Corresponding angles formed by a transversal with two parallel lines are equal.

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Q10. Sum of the angles of a triangle is:

A. $90^\circ$
B. $180^\circ$
C. $270^\circ$
D. $360^\circ$

Answer: B. 180∘180^\circ

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Q11. Each angle of an equilateral triangle is:

A. $30^\circ$
B. $45^\circ$
C. $60^\circ$
D. $90^\circ$

Answer: C. 60∘60^\circ

Solution:

Sum of angles:$$180^\circ$$

Since all three angles are equal:$$\frac{180^\circ}{3}=60^\circ$$

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Q12. A quadrilateral has:

A. 3 sides
B. 4 sides
C. 5 sides
D. 6 sides

Answer: B. 4 sides

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Q13. The diagonals of a parallelogram:

A. Are always equal
B. Bisect each other
C. Are always perpendicular
D. Never intersect

Answer: B. Bisect each other

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Q14. Area of a triangle with base 12 cm and height 5 cm:

A. 60 cm²
B. 30 cm²
C. 17 cm²
D. 24 cm²

Answer: B. 30 cm²

Solution:$$A=\frac12\times b\times h$$$$=\frac12\times12\times5$$$$=30\text{ cm}^2$$

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Q15. Volume of a cube with side 4 cm:

A. 16 cm³
B. 32 cm³
C. 64 cm³
D. 48 cm³

Answer: C. 64 cm³

Solution:$$V=a^3$$$$=4^3=64\text{ cm}^3$$

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Q16. Mean of 4, 6, 8 and 10:

A. 6
B. 7
C. 8
D. 9

Answer: B. 7

Solution:$$\text{Mean}=\frac{4+6+8+10}{4}$$$$=\frac{28}{4}=7$$

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Q17. Probability of getting a head when a fair coin is tossed once:

A. 0
B. $\frac14$
C. $\frac12$
D. 1

Answer: C. 12\frac12

Solution:

Total outcomes = 2

Favourable outcomes = 1$$P(H)=\frac12$$

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Q18. If $x=2$, find $x^2+3x+1$.

A. 9
B. 10
C. 11
D. 12

Answer: C. 11

Solution:$$x^2+3x+1$$

Putting $x=2$:$$2^2+3(2)+1$$$$=4+6+1=11$$

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Q19. Radius of a circle is 7 cm. Find circumference.

A. 22 cm
B. 44 cm
C. 49 cm
D. 154 cm

Answer: B. 44 cm

Solution:$$C=2\pi r$$$$=2\times\frac{22}{7}\times7$$$$=44\text{ cm}$$

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Q20. If three corresponding sides of two triangles are equal, the triangles are congruent by:

A. ASA
B. SAS
C. SSS
D. RHS

Answer: C. SSS

Solution: SSS stands for Side-Side-Side congruence criterion.

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SECTION – B

5 × 2 = 10 Marks

Q21. Rationalise:

$$\frac5{\sqrt3}$$

Solution:

Multiply numerator and denominator by $\sqrt3$:$$\frac5{\sqrt3}\times\frac{\sqrt3}{\sqrt3}$$$$=\frac{5\sqrt3}{3}$$

Answer:$$\boxed{\frac{5\sqrt3}{3}}$$

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Q22. Factorise:

$$x^2+7x+12$$

Solution:

We need two numbers whose product is 12 and sum is 7.

Numbers are 3 and 4.$$x^2+7x+12$$$$=x^2+3x+4x+12$$$$=x(x+3)+4(x+3)$$$$=(x+3)(x+4)$$

Answer:$$\boxed{(x+3)(x+4)}$$

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Q23. Find two solutions of:

$$2x+y=8$$

Solution:

Put $x=0$:$$2(0)+y=8$$$$y=8$$

First solution:$$(0,8)$$

Put $x=2$:$$2(2)+y=8$$$$4+y=8$$$$y=4$$

Second solution:$$(2,4)$$

Answer:$$\boxed{(0,8),(2,4)}$$

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Q24. Find the coordinates of the point 4 units left of the y-axis and 3 units above the x-axis.

Solution:

4 units left of y-axis means:$$x=-4$$

3 units above x-axis means:$$y=3$$

Therefore:$$\boxed{(-4,3)}$$

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Q25. Sides of a triangle are 6 cm, 8 cm and 10 cm. Determine whether it is right-angled.

Solution:

Largest side = 10 cm.

According to Pythagoras theorem:$$6^2+8^2=10^2$$$$36+64=100$$$$100=100$$

Therefore, the triangle is right-angled.$$\boxed{\text{Yes, it is a right-angled triangle.}}$$

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SECTION – C

6 × 3 = 18 Marks

Q26. Simplify:

$$(2+\sqrt3)(2-\sqrt3)$$

Solution:

Using:$$(a+b)(a-b)=a^2-b^2$$

Therefore,$$(2+\sqrt3)(2-\sqrt3)$$$$=2^2-(\sqrt3)^2$$$$=4-3$$$$\boxed{1}$$

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Q27. If

$$p(x)=2x^2-5x+3$$

find $p(1)$ and $p(2)$.

Solution:

For $x=1$:$$p(1)=2(1)^2-5(1)+3$$$$=2-5+3=0$$

Therefore,$$\boxed{p(1)=0}$$

For $x=2$:$$p(2)=2(2)^2-5(2)+3$$$$=8-10+3$$$$=1$$

Therefore,$$\boxed{p(2)=1}$$

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Q28. Draw the graph of:

$$x+y=5$$

Solution:

Rewrite:$$y=5-x$$

Prepare a table:

$x$	$y$
0	5
1	4
2	3
5	0

Plot the points:$$(0,5),(1,4),(2,3),(5,0)$$

and join them with a straight line.

Answer: The graph is a straight line passing through these points.

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Q29. In triangle ABC,

$$\angle A=50^\circ,\quad \angle B=65^\circ$$

Find $\angle C$.

Solution:

We know:$$\angle A+\angle B+\angle C=180^\circ$$

Therefore,$$50^\circ+65^\circ+\angle C=180^\circ$$$$115^\circ+\angle C=180^\circ$$$$\angle C=180^\circ-115^\circ$$$$\boxed{\angle C=65^\circ}$$

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Q30. Prove that the diagonals of a parallelogram bisect each other.

Given: ABCD is a parallelogram. Diagonals AC and BD intersect at O.

To prove:$$AO=OC$$

and$$BO=OD$$

Proof:

Consider triangles $AOB$ and $COD$.

Since AB ∥ CD,$$\angle ABO=\angle CDO$$

Since AB ∥ CD,$$\angle BAO=\angle DCO$$

Also, opposite sides of a parallelogram are equal:$$AB=CD$$

Therefore,$$\triangle AOB\cong\triangle COD$$

by ASA congruence.

Hence, corresponding parts are equal:$$\boxed{AO=OC}$$

and$$\boxed{BO=OD}$$

Therefore, the diagonals of a parallelogram bisect each other.

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Q31. Find mean, median and mode:

$$5,7,8,5,9,6,5,10,7$$

Solution:

Arrange in ascending order:$$5,5,5,6,7,7,8,9,10$$

Mean

$$\text{Mean}=\frac{\text{Sum of observations}}{\text{Number of observations}}$$$$=\frac{5+7+8+5+9+6+5+10+7}{9}$$$$=\frac{62}{9}$$$$\boxed{\text{Mean}=\frac{62}{9}\approx6.89}$$

Median

There are 9 observations.

Middle observation:$$\frac{9+1}{2}=5^\text{th}$$

5th observation = 7.$$\boxed{\text{Median}=7}$$

Mode

5 occurs three times, more than any other number.$$\boxed{\text{Mode}=5}$$

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SECTION – D

4 × 5 = 20 Marks

Q32. Factorise completely:

$$x^3-6x^2+11x-6$$

and find its zeroes.

Solution:

Try $x=1$:$$1-6+11-6=0$$

Therefore, $x-1$ is a factor.

Dividing:$$x^3-6x^2+11x-6$$

by $x-1$, we get:$$x^2-5x+6$$

Now factorise:$$x^2-5x+6$$$$=x^2-2x-3x+6$$$$=x(x-2)-3(x-2)$$$$=(x-2)(x-3)$$

Therefore,$$x^3-6x^2+11x-6$$$$=(x-1)(x-2)(x-3)$$

Hence the zeroes are:$$\boxed{1,2,3}$$

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OR

Simplify:$$(2x+3)^2-(2x-3)^2$$

Using:$$a^2-b^2=(a-b)(a+b)$$$$=[(2x+3)-(2x-3)][(2x+3)+(2x-3)]$$$$=(6)(4x)$$$$\boxed{24x}$$

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Q33. In triangle ABC, AB = AC. Prove that the angles opposite these equal sides are equal.

Given:$$AB=AC$$

To prove:$$\angle B=\angle C$$

Construction: Draw the angle bisector of $\angle A$, meeting BC at D.

In triangles ABD and ACD:$$AB=AC$$

Given.$$AD=AD$$

Common side.

Also,$$\angle BAD=\angle CAD$$

because AD bisects $\angle A$.

Therefore,$$\triangle ABD\cong\triangle ACD$$

by SAS congruence.

Hence, corresponding angles are equal:$$\boxed{\angle B=\angle C}$$

Therefore, angles opposite equal sides of a triangle are equal.

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OR

For a parallelogram ABCD:$$AB\parallel CD,\qquad AD\parallel BC$$

Draw diagonal AC.

In triangles ABC and CDA:$$AC=AC$$

Common side.

Since $AB\parallel CD$,$$\angle BAC=\angle DCA$$

Since $AD\parallel BC$,$$\angle BCA=\angle DAC$$

Therefore,$$\triangle ABC\cong\triangle CDA$$

by ASA.

Hence,$$\boxed{AB=CD}$$

and$$\boxed{BC=AD}$$

Thus, opposite sides of a parallelogram are equal.

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Q34. A cylindrical tank has radius 7 m and height 10 m. Find CSA, TSA and volume.

Given:$$r=7\text{ m}$$$$h=10\text{ m}$$$$\pi=\frac{22}{7}$$

1. Curved Surface Area

Formula:$$CSA=2\pi rh$$$$=2\times\frac{22}{7}\times7\times10$$$$=440$$$$\boxed{CSA=440\text{ m}^2}$$

2. Total Surface Area

Formula:$$TSA=2\pi r(r+h)$$$$=2\times\frac{22}{7}\times7(7+10)$$$$=44\times17$$$$\boxed{TSA=748\text{ m}^2}$$

3. Volume

$$V=\pi r^2h$$$$=\frac{22}{7}\times7^2\times10$$$$=22\times7\times10$$$$\boxed{V=1540\text{ m}^3}$$

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Q35. Find mean, median, mode and range:

$$12,15,18,10,15,20,17,15,13,15$$

Step 1: Arrange the data

$$10,12,13,15,15,15,15,17,18,20$$

Step 2: Mean

Sum:$$12+15+18+10+15+20+17+15+13+15=150$$

Number of observations:$$n=10$$

Therefore:$$\text{Mean}=\frac{150}{10}$$$$\boxed{\text{Mean}=15}$$

Step 3: Median

There are 10 observations.

Median:$$=\frac{5^\text{th}+6^\text{th}}2$$

Both are 15.$$\text{Median}=\frac{15+15}{2}$$$$\boxed{\text{Median}=15}$$

Step 4: Mode

15 occurs four times.

Therefore:$$\boxed{\text{Mode}=15}$$

Step 5: Range

$$\text{Range}=\text{Largest}-\text{Smallest}$$$$=20-10$$$$\boxed{\text{Range}=10}$$

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SECTION – E

Case-Based Questions — 3 × 4 = 12

Q36. Coordinate Geometry

Given:$$A(2,3),\quad B(8,3),\quad C(8,7),\quad D(2,7)$$

(a) What is the abscissa of A?

For $A(2,3)$, the x-coordinate is 2.$$\boxed{2}$$

(b) What is the ordinate of C?

For $C(8,7)$, the y-coordinate is 7.$$\boxed{7}$$

(c) Find AB and BC.

For AB:$$AB=8-2=6$$$$\boxed{AB=6\text{ units}}$$

For BC:$$BC=7-3=4$$$$\boxed{BC=4\text{ units}}$$

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Q37. Circular Flower Bed

Radius:$$r=7\text{ m}$$

(a) Formula for circumference

$$\boxed{C=2\pi r}$$

(b) Find circumference

$$C=2\times\frac{22}{7}\times7$$$$\boxed{C=44\text{ m}}$$

(c) Find area

Formula:$$A=\pi r^2$$$$=\frac{22}{7}\times7\times7$$$$=154$$$$\boxed{A=154\text{ m}^2}$$

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Q38. Statistics

Data:$$2,4,3,5,4,6$$

(a) Number of students

There are 6 observations.$$\boxed{6}$$

(b) Sum of observations

$$2+4+3+5+4+6=24$$$$\boxed{24}$$

(c) Mean and Median

Mean

$$\text{Mean}=\frac{24}{6}=4$$$$\boxed{\text{Mean}=4}$$

Arrange the data:$$2,3,4,4,5,6$$

There are 6 observations.$$\text{Median}=\frac{3^\text{rd}+4^\text{th}}2$$$$=\frac{4+4}{2}$$$$\boxed{\text{Median}=4}$$

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Final Answer Summary

Section	Questions	Marks
A – MCQ	20	20
B – Very Short	5	10
C – Short Answer	6	18
D – Long Answer	4	20
E – Case Study	3	12
Total	38	80

Total = 80 Marks.