us mc 9th question paper

Time: 3 Hours
Maximum Marks: 80

General Instructions

1. All questions are compulsory.
2. The question paper consists of Sections A, B, C, D and E.
3. Show all necessary steps in the solutions.
4. Use $\pi=\frac{22}{7}$, wherever required.
5. Draw neat diagrams wherever necessary.

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SECTION – A

Multiple Choice Questions

20 × 1 = 20 Marks

Q1. Which of the following is a rational number?

A. $\sqrt{2}$
B. $\sqrt{3}$
C. $\frac{7}{9}$
D. $\pi$

Answer: C. 79\frac{7}{9}

Solution:

A rational number can be written in the form$$\frac{p}{q}$$

where $p,q$ are integers and $q\neq0$.

Therefore,$$\boxed{\frac{7}{9}}$$

is rational.

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Q2. Simplify:

$$\sqrt{144}$$

A. 10
B. 11
C. 12
D. 14

Answer: C. 12

Solution:$$12\times12=144$$

Therefore,$$\boxed{\sqrt{144}=12}$$

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Q3. The degree of the polynomial

$$7x^3-4x+9$$

is:

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

Answer: C. 3

Solution:

The highest power of $x$ is 3.$$\boxed{\text{Degree}=3}$$

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Q4. If

$$p(x)=2x-6$$

then the zero of $p(x)$ is:

A. 2
B. 3
C. 4
D. 6

Answer: B. 3

Solution:$$p(x)=0$$$$2x-6=0$$$$2x=6$$$$\boxed{x=3}$$

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Q5. Which point lies on the y-axis?

A. $(3,4)$
B. $(5,0)$
C. $(0,6)$
D. $(2,7)$

Answer: C. (0,6)(0,6)

Solution:

Every point on the y-axis has$$x=0$$

Therefore,$$\boxed{(0,6)}$$

lies on the y-axis.

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

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

Answer: C. Third quadrant

Solution:

Here,$$x<0,\qquad y<0$$

Therefore,$$\boxed{\text{Third quadrant}}$$

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Q7. The complement of $28^\circ$ is:

A. $52^\circ$
B. $62^\circ$
C. $72^\circ$
D. $152^\circ$

Answer: A. 62∘62^\circ

Solution:$$90^\circ-28^\circ$$$$\boxed{62^\circ}$$

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Q8. Vertically opposite angles are:

A. Supplementary
B. Equal
C. Complementary
D. Unequal

Answer: B. Equal

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Q9. If two parallel lines are cut by a transversal, alternate interior angles are:

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

Answer: A. Equal

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Q10. The sum of the interior 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. A triangle having all three sides equal is called:

A. Scalene triangle
B. Isosceles triangle
C. Equilateral triangle
D. Right triangle

Answer: C. Equilateral triangle

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Q12. The sum of the angles of a quadrilateral is:

A. $180^\circ$
B. $270^\circ$
C. $360^\circ$
D. $540^\circ$

Answer: C. 360∘360^\circ

Solution:

A quadrilateral can be divided into two triangles.$$2\times180^\circ$$$$\boxed{360^\circ}$$

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Q13. In a parallelogram, opposite sides are:

A. Equal and parallel
B. Only equal
C. Only perpendicular
D. Unequal

Answer: A. Equal and parallel

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Q14. Find the area of a parallelogram with base 15 cm and height 6 cm.

A. 21 cm²
B. 45 cm²
C. 90 cm²
D. 180 cm²

Answer: C. 90 cm²

Solution:$$A=b\times h$$$$=15\times6$$$$\boxed{A=90\text{ cm}^2}$$

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Q15. Find the area of a circle whose radius is 14 cm.

A. 308 cm²
B. 616 cm²
C. 154 cm²
D. 88 cm²

Answer: B. 616 cm²

Solution:$$A=\pi r^2$$$$=\frac{22}{7}\times14\times14$$$$=22\times2\times14$$$$\boxed{A=616\text{ cm}^2}$$

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Q16. Find the volume of a cube with side 5 cm.

A. 25 cm³
B. 75 cm³
C. 100 cm³
D. 125 cm³

Answer: D. 125 cm³

Solution:$$V=a^3$$$$=5^3$$$$=5\times5\times5$$$$\boxed{V=125\text{ cm}^3}$$

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Q17. Find the mean of 5, 10, 15 and 20.

A. 10
B. 12.5
C. 15
D. 20

Answer: B. 12.5

Solution:$$\text{Mean}=\frac{5+10+15+20}{4}$$$$=\frac{50}{4}$$$$\boxed{12.5}$$

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Q18. A die is thrown once. The probability of getting an even number is:

A. $\frac{1}{6}$
B. $\frac{1}{3}$
C. $\frac{1}{2}$
D. $\frac{2}{3}$

Answer: C. 12\frac{1}{2}

Solution:

Even numbers are:$$2,4,6$$

Favourable outcomes:$$3$$

Total outcomes:$$6$$

Therefore,$$P(E)=\frac{3}{6}$$$$\boxed{P(E)=\frac{1}{2}}$$

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Q19. If $x=3$, find:

$$x^2+2x+4$$

A. 15
B. 17
C. 19
D. 21

Answer: B. 19

Solution:$$x^2+2x+4$$$$=3^2+2(3)+4$$$$=9+6+4$$$$\boxed{19}$$

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Q20. Two triangles are congruent by SAS when:

A. Three sides are equal
B. Two sides and included angle are equal
C. Three angles are equal
D. One side is equal

Answer: B. Two sides and included angle are equal

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SECTION – B

Short Answer Questions

5 × 2 = 10 Marks

Q21. Simplify:

$$\sqrt{72}+\sqrt{32}$$

Solution:$$\sqrt{72}=\sqrt{36\times2}$$$$=6\sqrt{2}$$

Similarly,$$\sqrt{32}=\sqrt{16\times2}$$$$=4\sqrt{2}$$

Therefore,$$6\sqrt{2}+4\sqrt{2}$$$$\boxed{=10\sqrt{2}}$$

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Q22. Factorise:

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

Solution:

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

The numbers are $-3$ and $-4$.$$x^2-7x+12$$$$=x^2-3x-4x+12$$$$=x(x-3)-4(x-3)$$$$\boxed{(x-3)(x-4)}$$

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Q23. Find two solutions of:

$$x+y=9$$

Solution:

Putting $x=0$,$$0+y=9$$$$y=9$$

Therefore,$$\boxed{(0,9)}$$

Putting $x=3$,$$3+y=9$$$$y=6$$

Therefore,$$\boxed{(3,6)}$$

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Q24. Find the coordinates of the point which is 5 units to the right of the y-axis and 4 units below the x-axis.

Solution:

Right of y-axis:$$x=5$$

Below x-axis:$$y=-4$$

Therefore,$$\boxed{(5,-4)}$$

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Q25. Find the area of a triangle whose base is 16 cm and height is 9 cm.

Solution:$$A=\frac{1}{2}\times b\times h$$$$=\frac{1}{2}\times16\times9$$$$=8\times9$$$$\boxed{A=72\text{ cm}^2}$$

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SECTION – C

Short Answer Questions

6 × 3 = 18 Marks

Q26. Rationalise:

$$\frac{4}{\sqrt{5}}$$

Solution:$$\frac{4}{\sqrt{5}}\times\frac{\sqrt{5}}{\sqrt{5}}$$$$=\frac{4\sqrt{5}}{5}$$$$\boxed{\frac{4\sqrt{5}}{5}}$$

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Q27. If

$$p(x)=x^2-5x+6$$

find $p(2)$ and $p(3)$.

Solution:

For $x=2$,$$p(2)=2^2-5(2)+6$$$$=4-10+6$$$$\boxed{p(2)=0}$$

For $x=3$,$$p(3)=3^2-5(3)+6$$$$=9-15+6$$$$\boxed{p(3)=0}$$

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Q28. Find three solutions of:

$$2x+y=6$$

Solution:

Putting $x=0$,$$2(0)+y=6$$$$y=6$$

Therefore,$$\boxed{(0,6)}$$

Putting $x=1$,$$2(1)+y=6$$$$y=4$$

Therefore,$$\boxed{(1,4)}$$

Putting $x=3$,$$2(3)+y=6$$$$y=0$$

Therefore,$$\boxed{(3,0)}$$

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Q29. In triangle ABC,

$$\angle A=55^\circ,\qquad \angle B=65^\circ$$

Find $\angle C$.

Solution:$$\angle A+\angle B+\angle C=180^\circ$$$$55^\circ+65^\circ+\angle C=180^\circ$$$$120^\circ+\angle C=180^\circ$$$$\boxed{\angle C=60^\circ}$$

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Q30. In a parallelogram ABCD, if

$$\angle A=70^\circ$$

find the remaining three angles.

Solution:

Opposite angles of a parallelogram are equal.$$\angle C=\angle A$$$$\boxed{\angle C=70^\circ}$$

Adjacent angles are supplementary.$$\angle A+\angle B=180^\circ$$$$70^\circ+\angle B=180^\circ$$$$\boxed{\angle B=110^\circ}$$

Similarly,$$\boxed{\angle D=110^\circ}$$

Therefore,$$\boxed{\angle A=70^\circ,\;\angle B=110^\circ,\;\angle C=70^\circ,\;\angle D=110^\circ}$$

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Q31. Find the mean, median and mode of:

$$4,6,5,4,7,8,4,6,5$$

Solution:

Arrange the data:$$4,4,4,5,5,6,6,7,8$$

Mean

$$\text{Mean} = \frac{4+6+5+4+7+8+4+6+5}{9}$$$$=\frac{49}{9}$$$$\boxed{\text{Mean}=\frac{49}{9}\approx5.44}$$

Median

There are 9 observations.$$\text{Median}=5^\text{th}\text{ observation}$$$$\boxed{\text{Median}=5}$$

Mode

4 occurs three times.$$\boxed{\text{Mode}=4}$$

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SECTION – D

Long Answer Questions

4 × 5 = 20 Marks

Q32. Factorise completely:

$$x^3-3x^2-4x+12$$

Solution:

Grouping the terms:$$x^3-3x^2-4x+12$$$$=x^2(x-3)-4(x-3)$$$$=(x-3)(x^2-4)$$

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

we get:$$x^2-4=(x-2)(x+2)$$

Therefore,$$\boxed{x^3-3x^2-4x+12=(x-3)(x-2)(x+2)}$$

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

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Q33. 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:

Since$$AB\parallel CD$$

we have$$\angle ABO=\angle CDO$$

Also,$$\angle BAO=\angle DCO$$

Opposite sides of a parallelogram are equal:$$AB=CD$$

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

by ASA congruence.

Hence,$$\boxed{AO=OC}$$

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

Therefore, the diagonals of a parallelogram bisect each other.

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Q34. A cylinder has radius 7 cm and height 10 cm. Find:

1. Curved Surface Area
2. Total Surface Area
3. Volume

Take$$\pi=\frac{22}{7}$$

Solution:

Given,$$r=7\text{ cm}$$$$h=10\text{ cm}$$

1. Curved Surface Area

$$CSA=2\pi rh$$$$=2\times\frac{22}{7}\times7\times10$$$$=2\times22\times10$$$$\boxed{CSA=440\text{ cm}^2}$$

2. Total Surface Area

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

3. Volume

$$V=\pi r^2h$$$$=\frac{22}{7}\times7^2\times10$$$$=\frac{22}{7}\times49\times10$$$$=22\times7\times10$$$$\boxed{V=1540\text{ cm}^3}$$

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Q35. The marks obtained by 10 students are:

$$11,14,16,12,14,18,14,15,10,16$$

Find:

1. Mean
2. Median
3. Mode
4. Range

Solution:

Arrange the data:$$10,11,12,14,14,14,15,16,16,18$$

Mean

$$\text{Sum}=11+14+16+12+14+18+14+15+10+16$$$$=140$$$$\text{Mean}=\frac{140}{10}$$$$\boxed{\text{Mean}=14}$$

Median

There are 10 observations.$$\text{Median} = \frac{5^\text{th}+6^\text{th}}{2}$$$$=\frac{14+14}{2}$$$$\boxed{\text{Median}=14}$$

Mode

14 occurs three times.$$\boxed{\text{Mode}=14}$$

Range

$$\text{Range}=18-10$$$$\boxed{\text{Range}=8}$$

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SECTION – E

Case-Based Questions

3 × 4 = 12 Marks

Q36. Coordinate Geometry Case Study

A rectangular park is represented by the points:$$A(1,2),\quad B(7,2),\quad C(7,6),\quad D(1,6)$$

(a) Find the abscissa of A.

Solution:

The x-coordinate is the abscissa.$$\boxed{2}$$

(b) Find the ordinate of C.

Solution:

The y-coordinate is the ordinate.$$\boxed{6}$$

(c) Find AB and BC.

Solution:$$AB=7-1$$$$\boxed{AB=6\text{ units}}$$

Similarly,$$BC=6-2$$$$\boxed{BC=4\text{ units}}$$

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Q37. Circular Garden Case Study

A circular garden has radius 14 m.

(a) Write the formula for its circumference.

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

(b) Find its circumference.

$$C=2\times\frac{22}{7}\times14$$$$=44\times1$$$$\boxed{C=88\text{ m}}$$

(c) Find its area.

$$A=\pi r^2$$$$=\frac{22}{7}\times14\times14$$$$=22\times2\times14$$$$\boxed{A=616\text{ m}^2}$$

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Q38. Probability Case Study

A bag contains 10 balls numbered from 1 to 10. One ball is selected randomly.

(a) How many total outcomes are possible?

$$\boxed{10}$$

(b) Find the probability of getting an odd number.

Odd numbers are:$$1,3,5,7,9$$

Number of favourable outcomes:$$5$$

Total outcomes:$$10$$

Therefore,$$P(O)=\frac{5}{10}$$$$\boxed{P(O)=\frac{1}{2}}$$

(c) Find the probability of getting a number greater than 6.

Numbers greater than 6 are:$$7,8,9,10$$

Favourable outcomes:$$4$$

Therefore,$$P=\frac{4}{10}$$$$\boxed{P=\frac{2}{5}}$$

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ANSWER KEY – 2020 MODEL PAPER

Q.No.	Answer	Q.No.	Answer
1	C	11	C
2	C	12	C
3	C	13	A
4	B	14	C
5	C	15	B
6	C	16	D
7	A	17	B
8	B	18	C
9	A	19	B
10	B	20	B