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**SECTION - A**

**Questions 1 to 20 carry 1 mark each**

**Q.1. Which one of the following is not equal to ****Ans: **(a)**Q.2. A is the foot of perpendicular from a point (4, -9,7) on x-axis. The coordinates of A are (a) (4,0,0) (b) (0,-9,0) (c) (0,0,7) (d) (-4,9,-7) Ans:** Choice (a) is correct.

We know that on x-axis, y = 0 and z = 0. Therefore, the coordinates of foot of perpendicular A are (4,0,0).

(a) 0

(b) 1

(c) -1

(d) any number

Ans:

(a) 4

(b) 6

(c) 5

(d) 2

OR

If the point (3, a) lies on the line represented by the linear equation, 2x - 3y = 5, then the value of a is:

(a) 1/3

(b) 1/2

(c) 1/4

(d) 1

Ans:

Explanation: Put x = 2, y = 0 in the equation 2x + 3y = k

2(2) + 3(0) = k

4 + 0 = k

Hence, k = 4

Or

Correct option: (a)

2(3) - 3 (a) = 5

(Putting x = 3 and y = a in the given equation)

⇒ 6-3a = 5

⇒ 6 - 5 = 3a

⇒ a=1/3

(a) 75°

(b) 90°

(c) 105°

(d) 120°

Ans:

(b) 1/4

(c) 1/16

(d) -1/4

The given sequence is a G.P. with a = 8, r =1/2

The nth term a

The sixth term a

(a) 0

(b) 1

(c) 2

(d) 3

Ans:

So. √2 is a polynomial of degree 0.

(a) 60°, 80°, 100°, 120°

(c) 120°, 60°, 80°, 100°

(b) 120°, 100°, 80°, 60°

(d) 80°, 100°, 120°, 60°

Ans:

Explanation: Let the angles be 3x, 4x, 5x and 6x,

Then

3x + 4x + 5x + 6x - 360°

⇒ 18x = 360°

⇒ x = 20°

The angles are 60°, 80°, 100° and 120°

(a) 4:1

(b) 2:1

(c) 1:4

(d) 1:2

Ans:

[∵ Surface area of a spherical balloon whose radius is 7 cm

= 4π x 7 x 7 cm

Surface area of a spherical balloon whose radius is 14 cm

= 4π x 14 x 14 cm

∴ Ratio of surface areas =

(a) 8.4

(b) 0.84

(c) 80.4

(d) 0.084

Ans:

Given, sum of absolute values of deviations from mean = 84

**Questions 11 to 15 carry one mark each**

**Q.11. When a non-negative number is added to both sides of a linear equation, then its solution is .............****Ans:** When a non-negative number is added to both sides of a linear equation, then its solution is unaffected.**Q.12. Two supplementary angles are in ratio 2 : 7. Find the measures of angles.ORIn fig. if AB || CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.Ans:** 2x + 7x= 180° ⇒ x = 20°

So the angles are

2x = 2 x 20° = 40°

and 7x - 7 x 20° '= 140°

So two angles are 40° and 140°

Or

x = ∠APQ - 50° (Alternate interior angles)

50° + y = 127° (Alternate interior angles)

⇒ y = 127°-50°

= 77°

cot 5° cot 10° cot 15° ... cot 85°

= (cot 5° cot 85°)(cot 10° cot 80°)(cot 15° cot 75°)...(cot 40° cot 50°)cot 45°

= {cot 5° cot (90° - 5°) } {cot 10 cot(90° - 10°)}{cot 15° cot (90° - 15) ..{cot 40° cot(90° - 40°) } cot 45°

= (cot 5° tan 5°)(cot 10° tan 10°)(cot 15° tan 15°)...(cot 40° tan 40°) cot 45°

= (1) (1) (1). . .(1) (1) = 1

Or

**Questions 16 to 20 carry one mark each**

**Q.16. π is an irrational numberOrIn 5**True :

OR

True :

5

Find the area of a triangle whose sides are 7 cm, 24 cm and 25 cm.

QR : RS = 2 : 1

thus,

Or

Also, (0, -1), (1, 1) ∈ R, ∴ f(o) = 0 + b = b, f(1) = a + b

Thus, - 1 =f(0) = b and 1 = f(1) = a + b ⇒ a = 1 - b = 1 -(-1) = 2

Hence, f(x) = 2 x - 1.

Let the angles of the quadrilateral are 3x, 5x, 9x and 13x,

We know that, sum of angles of a quadrilateral = 360°

∴ 3x + 5x + 9x +13x = 360°

⇒ 30x = 360° ⇒ x = 12°

∴ Angles of the quadrilateral are

3x = 3 x 12 = 36°,

5x = 5 x 12 = 60°,

9x = 9 x 12 =108°

and 13x = 13 x 12 =156°

Ans:

Area of triangle is 1/2 x b x h

**SECTION - B**

**Question number 21 to 26, carry 2 marks each.**

**Q.21. Find the value of****Ans:****Q.22. Solve the following equation :****Ans: **The given equation is**Q.23. A die is thrown 300 times and the outcomes are noted as given belowIf a die is thrown at random, then find the probability of getting prime number(i) 1(ii) 2(iii) 3(iv) 5Ans:** Here, total number of trials = 300

(i) We know that, 1 is not a prime number, so we cannot determine the probability of prime number.

(ii) P (getting 2 on a die)

(iii) P (getting 3 on a die)

(iv) P (getting 5 on a die)

Ans:

AD = BC [Given]

∠BAD = ∠ABC [Given]

AB = AB [Common]

ΔBAD ≌ ΔABC [By SAS congruence rule]

∴ ∠BDA = ∠ACB [By cpct]

Or

The base (non-equal side) of an isosceles triangle is 8 cm. Find its area, if the perimeter of the triangle is 32 cm.

Ans:

In ΔABC, we have

Since each rose bed occupies a spate of 6 sq. m.

Required number of rose beds in the field =

Or

Here, perimeter of an isosceles triangle = 32 cm

Base or non-equal side of the triangle = 8 cm

Let each equal side of the isosceles triangle be x cm.

∴ x + x + 8 = 32

⇒ 2x = 32 - 8 = 24

⇒ x = 12 cm

Area of the triangle by using Heron’s formula

Q.26. A line cuts x-axis at point A and y-axis at point B. The point (2,3) divides AB in the ratio of 3: 2.

...(1)

It is given that the line (1) cuts the x-axis, then by putting y = 0 in (1), we have x = a

∴ The coordinates of the point A are A(a, 0).

Similarly, the line (1) cuts the y-axis then by putting x = 0 in (1), we have

y=b

∴ The coordinates of the point B are 5(0, b).

Let C be a point C(2, 3) which divides AB in the ratio 3 : 2.

If C(2, 3) divides AB in the ratio 3 : 2, i.e., AC: CB = 3 : 2, then

∴ The required equation of the line AB is

⇒ x + y = 5

Or

Ans:

According to the question, we have

On solving these equations, we get

Thus, Q divides PR in the ratio 1/2 : 1 or 1: 2.

**SECTION - C**

**Question 27 to 34, carry 3 marks each****Q.27. Factorise: 8x ^{3} + 20x^{2} + 6x - 9**

By hit and trial method, we find that

= -27 + 45 - 9 - 9 = 0

∴ (2x + 3) is a factor of p(x).

Now, by long division method, we obtain

∴ p(x) = (2x + 3) (4x

= (2x + 3) (4x

= (2x + 3) {2x (2x + 3) - 1(2x + 3)}

= (2x + 3) (2x + 3) (2x - 1)

Or

Let p(x) = x

Since (x + 1) and (x + 2) are factors of p(x)

∴ Put x = -1 and -2, we obtain

p(-1) = (-1)

⇒ -1 + 3 + 3α + β = 0

⇒ 3α + β = -2 ...(i)

p(-2) = (-2)

⇒ -8 + 12 + 6a + β = 0

⇒ 6α + β = - 4 ...(ii)

Subtracting (i) from (ii). we have

Hence, values of α and β are α = -1/2 and β =0.

x

When n = 1, P(1) - x

∴ P(1) is true.

Let us assume that P(k) is true for any natural number k.

i.e., P(k) : x

i.e., x

Now, we have to prove that P(k + 1) is also true.

i.e., P(k + 1) : x

Now, x

[using (2)]

which is divisible by (x - y).

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, P(n) : x

Prove that ΔABE ≌ ΔACD.

Ans:

To prove ΔABE ≌ ΔACD

Proof In ΔABE and ΔACE , we have

AD = AE [given]

BD = CE [given]

and ∠ADB = ∠AEC

∴ ΔABD = ΔACE [by SAS congruence rule]

Hence, AB = AC [by CPCT]

Now, in ΔABE and ΔACD, we have

ΔACD, we have

AB = AC [proved above]

BE = CD

[∵ BD = CE ⇒ BD + DE = CE + ED ⇒ BE = CD]

and ∠ABE = ∠ACD

[∵ angles opposite to equal sides are equal]

∴ ΔABE ≌ ΔACD [by SAS congruence rule]

Hence proved

Ans:

Let PB = QC = DR =x

If BQ = y , then RC = y (∴ BC = DC)

By Pythagoras theorem,

...(i)

...(ii)

From (i) and (ii) PQ = QR

∴ ∠QPR = ∠PRQ = 45°

⇒

Now, AD ⊥ BC and let AB = BC = CA = x

⇒

In rt ∠ed ∠BDA, ∠D = 90°

AB

⇒

Thus, the length of each string is 10√3 m.

**Q.32. The Mean and Standard Deviation of a group of 100 observations were found to be 20 and 3, respectively. Later on, it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the Mean and Standard Deviation if the incorrect observations are omitted.Ans:** We have n = 100, and

We know that:

⇒

⇒ ...(1)

= 2000-(21 +21 + 18) [using (1)]

= 2000 - 60

= 1940

...(2)

Now,

⇒

⇒ ...(3)

= 40900 - [(21 )

= 40900 - [441 + 441 + 324]

= 40900 - (1206)

= 3969

[using (2)]

Hence, the new Mean and Standard Deviation are 20 and 3.036.

(i) number line

(ii) cartesian plane?

Also, show it by the graph.

Ans:

⇒ 2x + 2y + 3 = 2y ⇒ 2x = - 3 ⇒ x = -3/2

which is a linear equation in one variable. (i) On number line, the solution of the given equation is unique (one solution).

i.e.

(ii) In cartesian plane,

The number of solutions are infinite as solutions are in the form where m can be any real number.

(a) a number less than 14

(b) a number which is a perfect square

(c) a prime number less than 29

OR

Three coins are tossed simultaneously 400 times and following frequencies of the outcomes were recorded

(i) Find the probability of getting no head

(ii) Find the probability of getting one head

(iii) Find the probability of getting exactly two heads

Ans:

Total number of cards marked with number 2 to 101 = 100

(a) Cards marked with number < 14 are with number 2 to 13 .

⇒ Total number of cards having a number < 14 = 12

∴ P (that a number

(b) Cards with number as a perfect square from 2 to 101 are :

4,9,16,25,36,49, 64,81,100

⇒ Total number of cards having a perfect square number = 9

⇒ P(that a perfect square number)

(c) Cards with a prime number < 29 are :

2, 3, 5,7, 11, 13, 17, 19, 23 .

⇒ Total number of cards having a prime number < 29 = 9

⇒ P (that a prime number < 20)

OR

Given : Total numbers of outcomes = 400

Now, outcomes for getting no head

= 400 - (103 + 124 + 98)

= 75

**SECTION - D**

**Question 35 to 40 carry 4 marks each****Q.35. Simplify: Or find the values of a and b.Ans:**

Or

On comparing the rational and irrational parts, we have a = 5 and b = 2

cos A cos 2A cos 4A cos 8A=

Ans.

L.H.S. = cos A cos 2A cos 4A cos 8A

[Multiplying and dividing by 2 sin A]

[∵ 2 sinθ cosθ = sin 2θ]

[Multiplying and dividing by 2]

[Multiplying and dividing by 2]

[Multiplying and dividing by 2]

We know that

cos 3A = 4 cos

⇒ 4 cos

⇒ ...(1)

L.H.S. = cos

[using (1)]

[∵ cos (720° + 3A) = cos 3A; cos (360° + 3A) = cos 3A]

= 3/4 [cos A + cos (240° + A) + cos (120° + A)] + 3/4 cos 3A

[∵ cos (180° + A) = - cos A]

= R.H.S.

Or

Calculate the mean, median and mode for the following data.

23, 25, 28, 25, 16, 23, 17, 22, 25, 25

Ans.

We know that mode is the value, which contain highest frequency.

Here, we see in the given table that 30 is the highest frequency and its corresponding marks is 25.

Hence, mode of the given data is 25.

Firstly, write the given data in ascending order.

16, 17, 22, 23, 23, 25, 25, 25, 25, 28

We can represent above data in the following table

Here, n = 10 [even]

∴

= 1/2 [Value of 5th term + Value of 6th term]

= 1/2(23 + 25) =24

Mode = Value of variable which occurs most frequently = 25

Ans.

Sum of three sides XY + YZ + ZX = 11 cm

Required: To construct ΔXYZ

Steps of construction:

1. Draw a line segment say PQ = 11 cm

(∵ XY + YZ + ZX = 11 cm)

2. Draw ∠KPQ = 60° (∵ ∠Y = 30°)

And ∠LQP = 90° (∵ ∠Z = 90°)

3. Bisect the ∠KPQ and ∠LQP . Let these intersect at point X.

4. Let MN intersect PQ at Y and RS intersect l PQ at Z

5. Join XY and XZ.

Then XYZ is the required triangle

Ans.

= 2(lb + bh + hl)

= 2 (15 x 12 + 12 x 5 + 5 x 15)

= 2 (180 + 60 + 75) = 630 cm

Extra area of cardboard required for overlapping = 5/100 x 630 = 31.50 cm

∴ Total area of cardboard required for each smaller box = 630 + 31.50 — 661.50 cm

Total surface area of bigger box

= 2 ( lb + bh + hl)

= 2 (25 x 20 + 20 x 5 + 5 x 25)

= 2 (500 + 100 + 125) = 2(725) = 1450 cm

Extra area of cardboard required for overlapping = 5/100 x 1450 = 72.50 cm

∴ Total area of cardboard required for each bigger box

= 1450 cm

Total area of cardboard required for 250 such boxes each smaller and bigger

= 250 x (661.50 + 1522.50) = 250 x 2184 = 546000 cm

Now, total cost of cardboard at the rate of INR 4 for 1000 cm

Ans.

∴ Number of possible arrangements of all letters taken all at a time = 11!/2!2!2!

∴ Number of words formed of the letters of the word =

There are 7 consonants in the given word (i.e., M, M, T, T, H, C, S).

When the consonants occur together, let us treat them as one letter (MMTTHCS).

Now there are 5 letters (i.e., MMTTHCS, A, A, E, I), out of which two A’s are alike and rest are distinct.

∴ Number of possible arrangements of these 5 letters taken all at a time = 5!/2!

Corresponding to each such arrangement, the 7 consonants can be arranged amongst themselves in 7!/2!2! ways. [∵ M and T occur two times]

∴ Number of arrangements in which consonants are together =

∴ Number of words in which consonants occur together =

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