Regularly solving AP 10th Class Maths Model Papers Set 6 contributes to the development of problem-solving skills.

## AP SSC Maths Model Paper Set 6 with Solutions

Instructions :

- In the duration of 3 hours 15 minutes, 15 minutes of time is allotted to read the question paper.
- All answers shall be written in the answer booklet only.
- Question paper consists of 4 Sections and 33 questions.
- Internal choice is available in section – IV only.
- Answers shall be written neatly and legibly.

Section – I

(12 × 1 = 12M)

Question 1.

Which of the following point lies in Q3?

A) (3, -2)

B) (3, 2)

C) (-3, -2)

D) (- 3,2)

Solution:

C) (-3, -2)

Question 2.

Draw the graph of ax^{2} + bx + c = 0 when b^{2} – 4ac < 0.

Solution:

Question 3.

\(\frac{7}{5}\) is the zero of 7x – 5. Is it True/False ?

Solution:

False

p(x) = 7x – 5

p(\(\frac{7}{5}\)) = 7(\(\frac{7}{5}\)) – 5 = \(\frac{49}{5}\) – 5 = \(\frac{49-25}{5}\) = \(\frac{24}{5}\)

p(\(\frac{7}{5}\)) ≠ 0. So, \(\frac{7}{5}\) is not a zero of p(x).

Question 4.

From the figure, find ∠PAQ.

Solution:

∠PAQ = 180° – 120° = 60°

Question 5.

In 2, 4, 6, 8, 10 …….. of A.P. , common difference is ……

Solution:

2

Explanation:

2, 4, 6, 8, 10 …….. is an A.P

Common difference = d = a_{2} – a_{1} = 4 – 2 = 2

Question 6.

Choose the correct answer.

Statement p : Sin 45° = \(\frac{1}{\sqrt{2}}\)

Statement q : Tan 30° = \(\frac{1}{\sqrt{3}}\)

A) p true, q false

B) p false, q true

C) both p, q are true

D) both p, q are false

Solution:

C) both p, q are true

Question 7.

If the ratio of areas of two similar triangles is 2 : 1 then the ratio of their perimeters is ……..

Solution:

\(\sqrt{2}\) : 1

Question 8.

Choose the correct matching:

a) x + 2y = 4, 3x + 6y = 7, ( ) i) unique solution

b) x + 2y = 4, 3x + 6y = 12 ( ) ii) no solution

c) x + 2y = 4, 3x – y = 2 ( ) iii) infinite solutions

A) a – i, b – ii, c – iii

B) a – iii, b – ii, c – i

C) a – ii, b – iii, c – i

D) a – ii, b – i, c – iii

Solution:

C) a – ii, b – iii, c – i

Question 9.

Choose the correct matching:

A) a – (i), b -(ii), c – (iii)

B) a – (i), b – (iii), c – (ii)

C) a – (iii), b – (i), c – (ii)

D) a – (ii), b – (iii), c – (i)

Solution:

C) a – (iii), b – (i), c – (ii)

Question 10.

If A = {1, 2} and B = {3, 4} then A∪B

Solution:

A ∪ B = {1, 2} ∪ {3, 4} = {1, 2, 3, 4}

Question 11.

If a dice is thrown, what is the probability of getting an odd number less than 3?

Solution:

\(\frac{1}{6}\)

Question 12.

In Mode = l + \(\left[\frac{f_1-f_0}{2 f_1-f_0-f_2}\right]\) × h, ‘l’ represents ….

A) lower limit

B) upper limit

C) lower boundary

D) upper boundary

Solution:

C) lower boundary

Section – II

(8 × 2 = 16 M)

Note:

- Answer all the questions.
- Each question carries 2 marks.

Question 13.

What value(s) of x will make DE || AB, in the given figure AD = 8 x + 9, CD = x + 3, BE = 3x + 4, CE = x.

Solution:

In ∆ABC, DE || AB

∴ By Thale’s theorem,

\(\frac{C D}{A D}\) = \(\frac{C E}{C B}\)

⇒ (x + 3) (3x + 4) = x (8x + 9)

⇒ 3x^{2} + 4x + 9x + 12 = 8x^{2} + 9x

⇒ 8x^{2} + 9x – 3x^{2} – 4x – 9x – 12 = 0

⇒ 5x^{2} – 4x – 12 = 0

⇒ 5x^{2} – 10x + 6x – 12 = 0

⇒ 5x(x – 2) + 6(x – 2) = 0

⇒ (x – 2) (5x + 6) = 0

⇒ x = 2 or x = \(\frac{-6}{5}\)

Question 14.

Is 2t – 1 = 2t + 5 a linear equation in one variable?

Solution:

2t – 1 = 2t + 5

2t – 2t = 5 + 1

0 = 6

There is no variable in it. So it is not a linear equation in one variable.

Question 15.

Determine the value of \(2^{2+\log _2^3}\)

Solution:

\(2^{2+\log _2 3}\) = 2^{2} × \(2^{\log _2 3}\)

= 4 × \(\log _2 3\)

= 4 × 3

= 12

We know that

(i) a^{m + n} = a^{m} × a^{n}

(ii) \(a^{\log _a N}\) = N

Question 16.

2, 3, 5, 7, 8, 10, 15 …… is an arithmetic progression? Why ?

Solution:

Given that 2, 3, 5, 7, 8, 10, 15…………

a_{2} – a_{1} = 3 – 2 = 1

a_{3} – a_{2} = 5 – 3 = 2

a_{4} – a_{3} = 7 – 5 = 2

a_{5} – a_{4} = 8 – 7 = 1

∴ a_{2} – a_{1} ≠ a_{3} – a_{2} = a_{4} – a_{3} ≠ a_{5} – a_{4}

So it is not an arithmetic progression.

Question 17.

What is the probability for drawing out a ’Red king’ from a deck of cards?

Solution:

Total deck of cards = 52

Total possible outcomes = 52

No. of favourable outcomes to drawing out a red king = 2

∴ Probability P (E)

No. of favourable out comes

\(=\frac{\text { No. of favourable out comes }}{\text { Total possible out comes }}\)

= \(\frac{2}{52}\) = \(\frac{1}{26}\)

Question 18.

Write the formula to find median of a grouped data?

Solution:

Median = l + \(\frac{\frac{n}{2}-c f}{f}\) × h

Where l = lower limit of the median class.

h = length of the median class.

f = frequency of the median class,

cf = cumulative frequency of the class preceeding the median class.

Question 19.

A cylinder and cone have bases of equal radii and are of equal heights. Show that their volumes are

in the ratio of 3 : 1.

Solution:

It is given that the bases have equal radii and are of equal heights.

∴ Ratio of the volumes of the cylinder and the cone = πr^{2}h : \(\frac{1}{3}\)πr^{2}h

= 1 : \(\frac{1}{3}\) = 3 : 1

Question 20.

Rinki observed a ball on the ground from the balcony of the first floor of a building at an angle of depression θ. If the height of the first floor of the building is ‘x’ meters. Draw the diagram for this data

Solution:

AB = height of the building = x mt

∠XAC = Angle of depression = θ

Section – III

(8 × 4 = 32 M)

Note:

- Answer all the questions.
- Each question carries 4 marks.

Question 21.

How many multiples of 4 lie between 10 and 250?

Solution:

The list of multiples of 4 lie between 10 and 250 is,

12, 16, 20, 24, ……., 248

Here, a = 12 ; d = 4 ; a_{n} = 248

a_{n} = a + (n – 1)d

248 = 12 + (n – 1)4

248 – 12 = (n – 1)4

(n – 1)4 = 236

n – 1 = \(\frac{236}{4}\) = 59

n = 59 + 1 = 60

∴ There are 60, multiples of 4 lie between 10 and 250.

Question 22.

Solve the following pair of equations:

\(\frac{2}{x}\) + \(\frac{3}{y}\) = 13 ; \(\frac{5}{x}\) – \(\frac{4}{y}\) = -2

Solution:

Observe the given pair of equations. They are not linear equations.

We have 2(\(\frac{1}{x}\)) + 3(\(\frac{1}{y}\)) = 13 …….. (1)

5(\(\frac{1}{x}\)) – 4(\(\frac{1}{y}\)) = -2 …….. (2)

If we substitute \(\frac{1}{x}\) = p and \(\frac{1}{y}\) = q, we get a pair of linear equations:

2p + 3q = 13 ………… (3)

5p – 4q = -2 ………. (4)

[Coefficients of q are 3 and 4 and their L.C.M. is 12. Using the elimination method:]

[‘q’ has opposite sign, so we add the two equations.]

p = \(\frac{46}{23}\) = 2

Substitute the value of p in equation (3)

2(2) + 3q = 13 ; 3q = 13 – 4 = 9

q = \(\frac{46}{23}\)

But, \(\frac{1}{x}\) = p = 2 ⇒ x = \(\frac{1}{2}\)

\(\frac{1}{y}\) = q = 3 ⇒ y = \(\frac{1}{3}\)

Question 23.

Find two consecutive positive integers, sum of whose squares is 613.

Solution:

Let the first positive integer be x

Then the consecutive positive integer is x + 1

Sum of the squares = x^{2} + (x + 1)^{2}

= x^{2} + x^{2} + 2x + 1

= 2x^{2} + 2x + 1

Given sum of the squares = 613

∴ 2x^{2} + 2x + 1 = 613

2x^{2} + 2x + 1 – 613 = 0

2x^{2} + 2x – 612 = 0

x^{2} + x – 306 = 0

x^{2} + 18x – 17x – 306 = 0

x(x + 18) – 17 (x + 18) = 0

(x + 18)(x – 17)

x + 18 = 0 or x – 17 = 0

x + 18 = 0 gives x = -18

x – 17 = 0 gives x = 17

For x = 17 its consecutive positive integer is x + 1 = 17 + 1 = 18.

Two consecutive positive integers are 17, 18.

Question 24.

Find all the zeroes of 2x^{4} – 3x^{3} – 3x^{2} + 6x – 2, if you know that two of its zeroes are \(\sqrt{2}\) and –\(\sqrt{2}\).

Solution:

Since two of the zeroes are \(\sqrt{2}\) and –\(\sqrt{2}\), therefore we can divide by

Now, by splitting -3x, we factorize 2x^{2} – 3x + 1 as (2x – 1) (x – 1). So, its zeroes are given by x = \(\frac{1}{2}\) and x = 1. Therefore, the zeroes of the given polynomial are \(\sqrt{2}\), –\(\sqrt{2}\) and 1 and \(\frac{1}{2}\).

Question 25.

Prove that the points A (-7, -3), B (5, 10), C (15, 8) and D (3, – 5) taken in order are the vertices of a

parallelogram.

Solution:

Let A(-7, -3), B(5, 10), C(15, 8) and D(3, -5) be the given points.

Opposite sides AB = CD, BC = DA

Diagonals AC ≠ BD

Therefore ABCD is a parallelogram.

The given points taken in order form a parallelogram.

Question 26.

A heap of rice is in the form of a cone of diameter 12 m. and height 8 m. Find its volume? How much canvas cloth is required to cover the heap?

(Use π = 3.14)

Solution:

Diameter of the conic heap of rice = 12 m

∴ Its radius(r) = –\(\frac{12}{2}\)m = 6 m

height (h) = 8 m

Its volume (v) = \(\frac{1}{3}\)πr^{2}h

Required canvas cloth to cover the heap

= Curved surface area of heap

= πrl = 3.14 × 6 × 10 m^{2}

= 188.4 m^{2}

Question 27.

Show that

Solution:

Question 28.

A die is thrown once. Find the probability of getting

i) a prime Number

ii) a number lying between 2 and 6

iii) an odd number

iv) multiple of 3.

Solution:

When a die is thrown, sample space S = {1, 2, 3, 4, 5, 6}

i) Let A denotes the event of getting a prime number. Then A = {2, 3, 5}

So, required probability = P(A)

= \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\)

ii) Let B denote the event of getting a number between 2 and 6.

Then, B = {3, 4, 5}

∴ Required probability = P(B) = \(\frac{n(B)}{n(S)}\)

\(\frac{3}{6}\) = \(\frac{1}{2}\)

iii) Let C denotes the event of getting an odd number. Then C = {1, 3, 5}

So, required probability, P(C) = \(\frac{\mathrm{n}(\mathrm{C})}{\mathrm{n}(\mathrm{S})}\)

\(\frac{3}{6}\) = \(\frac{1}{2}\)

iv) Let D denotes the event of getting multiple of 3. Then D = {3, 6}

No. of favorable outcomes = 2

So, required probability,

p(D) = \(\frac{\mathrm{n}(\mathrm{D})}{\mathrm{n}(\mathrm{S})}\)

Probability = P (D) = \(\frac{2}{6}\) = \(\frac{1}{3}\)

Section-IV

(5 × 8 = 40 M)

Note:

- Answer all the questions.
- Each question carries 8 marks.
- There is an internal choice for each question.

Question 29.

A = {x : x set of even prime}

B = {x : x is a natural number < 12}

C = {x : x is a multiple of 4 less than or equal to 12}

D = {x : x is a factors of 12}

Find

(i) A ∪ B

(ii) B ∩ C

(iii) C – D

(iv) A – D

(OR)

If log \(\left(\frac{x+y}{3}\right)\) = \(\frac{1}{2}\)(log x + log y), then find the value of \(\frac{x}{y}\).

Solution:

A = {x : x is set of even prime} = {2}

B = {x : x is a natural number < 12}

= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}

C = {x : x is a multiple of 4 less than or equal to 12}

= {4, 8, 12}

D = {x : x is a factor of 12} = {1, 2, 3, 4, 6, 12}

(i) A∪B

A∪B= {2} ∪ {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}

= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}

(ii) B ∩ C

B ∩ C = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11} ∩ {4, 8, 12} = {4, 8}

(iii) C – D

C – D = {4, 8, 12} – {1, 2, 3, 4, 6, 12} = {8}

(iv) A – D = {2} – {1, 2, 3, 4, 6, 12} = { } = φ

(OR)

Question 30.

For which acute angle ‘θ’

(OR)

A motor boat whose speed is 18 km/h in still water. It takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.

Solution:

(OR)

Let the speed of the stream be x km/h

Therefore, the speed of the boat upstream = (18 – x) km/h and the speed of the boat downstream = (18 + x) km/h.

The time taken to go upstream distance 24

\(\frac{\text { distance }}{\text { speed }}\) = \(\frac{24}{18-x}\) hours.

Similarly, the time taken to go down-stream

= \(\frac{24}{18+x}\) hours

According to the question,

= \(\frac{24}{18-x}\) – \(\frac{24}{18+x}\) = 1

i.e., 24(18 + x) – 24(18 – x)

= (18 – x)(18 + x)

i.e., x^{2} + 48x – 324 = 0

Using the quadratic formula, we get

x = \(\frac{-48 \pm \sqrt{48^2+1296}}{2}\) = \(\frac{-48 \pm \sqrt{3600}}{2}\)

= \(\frac{-48 \pm 60}{2}\) = 6 (or) -54

Since x is the speed of the stream, it cannot be negative. So, we ignore the root x = – 54. Therefore, x = 6 gives the speed of the stream as 6 km/hr.

Question 31.

The mean pocket allowance is 18/-. Find the missing frequencies.

Daily pocket allowance | 11-13 | 13-15 | 15-17 | 17-19 | 19-21 | 21-23 | 23-25 |

Number of Children | 7 | 6 | 9 | 13 | f | 5 | 4 |

(OR)

Find the co-ordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).

Solution:

Given that, mean pocket allowance is ₹ 18.

We will find the mean of the given data and equalize it with 18 to find the value of ‘f’.

(OR)

Let P and Q be the points of trisection of AB is AP = PQ = QB.

Therefore P divides AB internally in the ratio 1 : 2

Now, Q also divides AB internally in the ratio 2 : 1

We have x_{1} = 4, y_{1} = -1, x_{2} = -2, y_{2} = -3.

m_{1} = 2; m_{2} = 1

Therefore the coordinates of the point of trisection of the line joining are p(2, \(\frac{-5}{3}\)) and Q(0, \(\frac{-7}{3}\))

Question 32.

State and prove Pythagoras Theorem.

(OR)

The angle of elevation of a jet plane from a point A on the ground is 60°. After a flight of 15 seconds,

the angle of elevation changes to 30°. If the jet plane is flying at a constant height of 100\(\sqrt{3}\) meter, find the speed of the jet plane. (\(\sqrt{3}\) = 1.732)

Solution:

Statement: In a right triangle, the square of hypotenuse is equal to the sum of the squares of the other two sides.

Given : ∆ABC is the right triangle right angled at B.

RTP: AC^{2} = AB^{2} + BC^{2}

Construction: Draw BD ⊥ AC

Proof: ∆ADB ~ ∆ABC

(OR)

Let D is the initial position of the jet plane which moves to C in 15 seconds.

A is the point on the ground from where we are observing. Then

From (1) & (2)

Distance that jet flied

= CD = BE = AB – AE

= 4500 m – 1500 m = 3000 m

So, speed of the flight

= \(\frac{3000 \mathrm{~m}}{15 \text { seconds }}\) = 200 m/s

Question 33.

Draw the graph of p(x) = x^{2} – x – 12 and find its zeroes. Justify your answer.

(OR)

Draw a circle of radius 6 cm. From a point 10 cm away from its centre construct the pair of tangents to the circle and measure their lengths. Verify by using Pythogaras theorem.

Solution:

p(x) = x^{2} – x – 12

y = x^{2} – x – 12

-3 and 4 are zeroes of the quadratic polynomial because (-3, 0) and (4, 0) are intersection points of X axis.

Justification: x^{2} – x – 12 = 0 ⇒ x^{2} – 4x + 3x – 12 ⇒ (x – 4) (x + 3) = 0

x – 4 = 0 and x + 3 = 0 x = 4, x = -3 zeroes of p(x) = -3, 4.

(OR)

Construction:

- Draw a circle of radius 6 cm. Let its centre be ‘O’
- Let a point P is at a distance of 10 cm from the centre ‘O’
- Draw the perpendicular bisector of PO which cuts it at the mid point M.
- Taking M as centre with radius PM or MO, we draw a circle that intersects the given circle at two points A and B.
- Join PA and PB. These are the required tangents of the circle.

Proof:

From the figure, PO = 10 cm, OA = 6 cm

Also ∆POA is a right angled triangle.

By Pythagoras theorem,

PO^{2} = PA^{2} + OA^{2} ⇒ PA^{2} = PO^{2} – OA^{2} = 10^{2} – 6^{2} = 100 – 36 = 64

∴ PA = \(\sqrt{64}\) = 8 cm

In OPB, ∠OBP = 90°. OB = 6 cm, OP = 10 cm.

By Pythagoras theorem OP^{2} = OB^{2} + BP^{2}

⇒ BP^{2} = OP^{2} – OB^{2} ⇒ = 10^{2} – 6^{2}

= 100 – 36 = 64 ⇒ BP = \(\sqrt{64}\)

= 8 cm