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**Class X****Mathematics – Basic****Max. Marks: 80****Duration: 3 hrs.**

**General Instructions**

__Read the following instructions very carefully and strictly follow them:__

- This question paper comprises four sections – A, B, C and D. This question. Paper carries 40 questions. All questions are compulsory.
**Section A:**Question Number 1 to 20 comprises of 20 questions of one mark each.**Section B:**Question Number 21 to 26 comprises of 6 questions of two marks each.**Section C:**Question Number 27 to 34 comprises of 8 questions of three marks each.**Section D:**Question Number 35 to 40 comprises of 6 questions of four marks each.- There is no overall choice in the question Paper. However, an internal choice has been provided in 2 questions of one mark, 2 questions of two marks, 3 questions of three marks and 3 questions of four marks. You have to attempt only one of the choices in such questions.
- In addition to this, separate instructions are given with each section and question, wherever necessary.
- Use of calculators is not permitted.

**Section - A**

**Question numbers 1 to 20 carry 1 mark each. Choose the correct option in question numbers 1 to 10.****Q.1. If a pair of linear equations is consistent, then the lines represented by them are (1 Mark)****(a) parallel ****(b) intersecting or coincident ****(c) always coincident ****(d) always intersecting****Ans.** b**Solution.**

intersecting or coincident If a pair of linear equations is consistent, then the lines represented by them are intersecting or coincident.**Q.2. The distance between the points (3, –2) and (–3, 2) is (1 Mark)(a) √52 units(b) 4√10 units(c) 2√10 units(d) 40 unitsAns.** a

Distance =

(a) 8

(b) 1/8

(c) -8

(d) -1/8

Ans.

8 cot

= 8(cot

= 8 x -1

= -8

(a) πr

(b) πl (r

(c)

(d)

Ans.

Solution.

The total surface area of a frustum-shaped glass tumbler iswhere radii r

(a) 5 x 8 x 3

(b) 15 x 2

(c) 10 x 2

(d) 5 x 2

Ans.

120 = 20 × 6

= 5 × 4 × 2 × 3

= 5 × 2

(a) 12

(b) 84

(c) 2√3

(d) –12

Ans.

The given equation is:

4x

Discriminant = b

Here, b = –6, a = 4, and c = 3

So, Discriminant = (–6)

= 36 – 48 = –12

(a) (–3, 6)

(b) (6, – 6)

(c) (6, –12)

(d) (3/2, -3)

Ans.

(0, 0) and (x, y).

So, (0 + x)/2 = 3 or, x = 6

And (0 + y)/2 = –6 or, y = –12

(a) 0

(b) many

(c) 2

(d) 1

Ans.

In the given figure, number of tangents parallel to tangent PQ is 1.

The upper limit of median class is

(a) 15

(b) 10

(c) 20

(d) 25

Ans.

Sum of frequencies (n)= 70

Middle observation = ((n/2) + 1)

= (70/2 + 1)

= 36

36

(a) 1

(b) 1/2

(c) not defined

(d) 0

Ans.

The probability of an impossible event is 0.

⇒ 4a - 4 = 0

⇒ a = 1

Ans.

= 8 : 27

Ans.

Q.16. In Figure-2, the angle of elevation of the top of a tower AC from a point B on the ground is 60°. If the height of the tower is 20 m, find the distance of the point from the foot of the tower. (1 Mark)

Ans.

√3 = 20 / AB

AB = 20 / √3

So, the required distance is 20 / √3 m.

OR

If cos A = sin 42°, then find the value of A. (1 Mark)

Ans.

= tan (90° - 50°) x tan50°

= cot 50° x tan50°

= 1 (∵ tan θ cot θ = 1)

cos A = sin42°

⇒ cos A = sin (90° - 48°)

⇒ cos A = cos 48°

⇒ A = 48°

Ans.

Probability of an event = Number of favourable outcomes / Number of all possible outcomes

Probability of getting head both the times = 1 / 4.

Ans.

Height of the cone =

= 12

Therefore, the height of the cone is 12m.

OR

Find the 11th term of the A.P. ─27, ─22, ─17, ─12, …. (1 Mark)

Ans.

⇒ 2x = -6 + 8

⇒ 2x = 2

⇒ x = 1

-27, -22, -17, -12,...

a

a

= -27 + 50

= 23

**Section - B**

**Question numbers 21 to 26 carry 2 marks each.****Q.21. Find the roots of the quadratic equation. (2 Marks)****3x ^{2} - 4√3 x + 4 = 0**

= √3x (√3x - 2) - 2 (√3x - 2)

= (√3x - 2) (√3x - 2)

So, the roots of the equation are the values of x for which

(√3x - 2)(√3x - 2) = 0

Now, √3x - 2 = 0 for x = 2 / √3

So, this root is repeated twice, one for each repeated factor √3x - 2.

Therefore, the roots of 3x

OR

Find the LCM of 150 and 200. (2 Marks)

Ans.

That is, the prime factorisation of 6n would contain the prime 5. This is not possible

∵ 6

So, the prime numbers in the factorisation of 6n are 2 and 3.

So, the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorisation of 6

So, there is no natural number n for which 6n ends with the digit zero.

We have,

150 = 5

and, 200 = 5

Here, 2

So, LCM(150, 200) = 2

Ans.

tan (A + B) = √3

or tan (A + B) = tan60°

or A + B = 60°......(1)

Again, we have

tan (A - B) = 1 / √3

or tan (A - B) = tan30°

or A - B = 30°....(2)

On adding equations (1) and (2), we get

2A = 90°

or A = 45°

On putting this value of A in equation (1), we get

B = 15°

Ans.

BC / ZY = 6 / 12

= 1 / 2

AC / XY = 2√3 / 4√3

= 1 / 2

AB / XZ = 3 / 6

= 1 / 2

Ratios of the corresponding sides of the given pair of triangles are equal.

i.e., BC / YZ = AC / XY = AB / XZ = 1 / 2.

Therefore, by SSS similarity critarion, ΔABC ~ ΔXZY.

The corresponding angles are equal in ΔABC and ΔXZY. i.e.,

∠A = ∠X = 80°,

∠B = ∠Z = 60°

and

∠C = ∠Y

In ΔABC,

∠A + ∠B + ∠C = 180°

⇒ 80° + 60° + ∠C = 180°

⇒ ∠C = 180° - 140°

⇒ ∠C = 40°

⇒ ∠Y = 40°

Ans.

Number of good bulbs = 98

Total number of outcomes = 98 +14 = 112

Probability of getting a good bulbs = Number of favourable outcomes / Total number of outcomes

= 98 = 112

= 7 / 8.

OR

The following distribution shows the transport expenditure of 100 employees: (2 Marks)

Find the mode of the distribution.

Ans.

Here, we observe that class marks and frequencies are small quantities.

So, we use direct method to compute the mean and proceed as below.

Mean,

= 230 / 10

= 23

Therefore, mean for the following distribution is 23.

From the given data, we have

l = 400, f

Mode =

= 480

Mode of the given data is 480.

**Section - C**

**Question number 27 to 34 carry 3 marks each.****Q.27. A quadrilateral ABCD is drawn to circumscribe a circle. Prove that: AB + CD = AD + BC. (3 Marks)Ans.**

In the given figure, quadrilateral ABCD is circumscribing the given circle and its sides are touching the circle at P, Q, R and S.

We have to prove that

AB + CD = AD + BC

We know that lengths of tangents drawn from a point to a circle are equal.

Therefore, from figure, we have

DR = DS, CR = CQ, AS = AP, BP = BQ

Now,

L.H.S. = AB + CD = (AP + BP) + (CR + DR)

= (AS + BQ) + (CQ + DS)

= AS + DS + BQ + CQ

= AD + BC

= R.H.S.

OR

Solve for x and y: (3 Marks)

According to question, y - x = 26......(1)

and y = 3x + 4....(2)

Substituting the value of y from equation (2) in equation (1), we get

3x + 4 - x = 26

or 2x = 26 - 4

or 2x = 22

or x = 11

Putting this value of x in equation (1), we get

y - 11 = 26

or y = 26 + 11 = 37

Hence, the numbers are 11 and 37.

Let 1 / x = p and 1 / y = q

then given equations can be written as:

2p + 3q = 13

2p + 3q - 13 = 0......(1)

and

5p - 4q = -2

5p - 4q + 2 = 0......(2)

Using cross-multiplication method, we get

⇒ p / - 46 = q / -69 = 1 / -23

⇒ p / - 46 = 1 / -23 and q / - 69 = 1 / -23

⇒ p = -46 / -23 and q = -69 / -23

⇒ p = 2 and q = 3

⇒ 1 / x = 2 and 1 / y = 3

⇒ x = 1 / 2 and y = 1 / 3

Ans.

So we can find integers r and s(≠ 0) such that √3 = r / s.

Suppose r and s have a common factor other than 1.

Then we divide r and s by the common factor and get

√3 = a / b

where a and b are coprime.

So, √3b = a

Squaring on both sides, we get

3b

Therefore,

a

So, we can write a = 3c for some integer c.

Now,

3b

⇒ 3b

⇒ b

This means that b

Therefore,

a and b have at least 3 as a common factor.

But this contradicts the fact that a and b are coprime.

So, our assumption that √3 is a rational is wrong.

Hence, √3 is an irrational number.

(i) Write the coordinates of the points A, B, C and D, using the 10 × 10 grid as coordinate axes.

(ii) Find whether ABCD is a parallelogram or not.

Ans.

Now, using distance formula, we will find the length of each side of the quadrilateral ABCD.

We see that sides AB, BC, CD and DA are equal in lengths, Therefore, quadrilateral ABCD is a parallelogram.

Ans.

We have to find a

We know that sum of first n terms of an AP is given by

S = n / 2 [2a + n - 1 d]

So,

1050 = 7(20 + 13d)

or d = 10

We know that

an = a + (n - 1)d

So, a

= 10 + 20 x 10

= 210.

**ORDraw a circle of radius 2.5 cm. Take a point P at a distance of 8 cm from its centre. Construct a pair of tangents from the point P to the circle. (3 Marks)Ans.**

cosec A – sin A sec A – cos A =

= R.H.S.

OR

In Figure – 5 ABCD is a square with side 7 cm. A circle is drawn circumscribing the square. Find the area of the shaded region. (3 Marks)

Ans.

Diameter of the smaller circle = 7 cm

Radius of the smaller circle = 7 / 2 cm

Area of the smaller circle

Area of shaded region

= Area of the smaller circle + 2 x Area of segment OCB

= 66.5cm

ABCD is a square with side 7 cm. Then,

Length of the diagonal of square = 7√2 cm

Diameter of circle = Diagonal of square

⇒ BD = 7√2 cm

Radius of circle = 7√2 / 2 cm

Area of shaded region = Area of circle - Area of the square

= 77 - 49

= 28 cm

Therefore, the area of the shaded region is 28 cm

**Section - D**

**Question numbers 35 to 40 carry 4 marks each.Q.35. Find other zeroes of the polynomial (4 Marks)**

The two zeroes of p(x) are

Therefore, (x -

Also, (x -

and so x

Now,

3x

= (x

= (x

Equating (x

Hence, the zeroes of the given polynomial are : √2, -√2, - 2 / 3 and 2.

The given polynomial is g(x) = x

Here, divisor is x

Divide g(x) = x

So, Quotient = x - 1 and Remainder = -2x + 3.

The division algorithm states that

Dividend = Divisor x Qoutient + Remainder

RHS = Divisor x Qoutient + Remainder

= (x

= x

= x

Thus, the division alogorithm is verified.

Ans.

In ΔABC, we have AB

tan45° = AB / BC

or 1 = AB / BC

or BC = AB = 75...... (1)

Now,

In ΔABD, we have

tan30° = AB / BD

or 1 / √3 = 75√3

or BD = 75√3...... (2)

From (1) and (2), we get

CD = BC + BD = 75 + 75√3 = 75 (1+ √3)

Therefore, the distance between the two ships is 75(1 + √3) m.

OR

In Figure-6, in an equilateral triangle ABC, AD ⊥ BC, BE ⊥ AC and CF ⊥ AB. Prove that 4(AD

(i) Join BE and CD.

(ii) Draw DM ⊥ AC and EN ⊥ AB.

and

Therefore,

Similarly,

But area (ΔBDE) = area(ΔDEC)...(3)

Therefore, from (1), (2) and (3), we get

Hence, proved.

ABC is an equilateral triangle.

Therefore, AB = BC = AC

Now,

Using Pythagoras theorem in ΔADC, we get

AC

......(1)

Similarly, using Pythagoras theorem in ΔAEB, we get

3AB

Again, using Pythagoras theorem in ΔAFC, we get

3AB2 = 4CF2......(3)

On adding equations (1), (2) and (3) we get

3AB

or, 9AB

or, 4 AD

Hence, proved.

Ans.

Radius of upper end of the frustum = r = 20 cm

Radius of lower end of the frustum = r2 = 8 cm

Capacity of container = Volume of the frustum

= 9125cm

**ORA rectangular park is to be designed whose breadth is 3 m less than its length. Its area is to be 4 square metres more than the area of a park that has already been made in the shape of an isosceles triangle with its base as the breadth of the rectangular park and of altitude 12 m. Find the length and breadth of the park. (4 Marks)Ans.** Let A and B be the time taken by the smaller and the larger taps respectively to fill the tank.

Since both the taps together can fill the tank in

Or,

Tap with larger diameter takes 10 hours less than smaller one to fill the tank.

So, A - 10 = B

Or, ....(2)

By placing the value of 1 / B from 2 in to 1, we get

Or, 75A - 375 = 4A

Or, 4A

Or, 4A

Or, 4A

Or, 4A (A - 25) -15 (A - 25) = 0

Or, (A - 25)(4a - 15) = 0.

So, B = 25 - 10 = 15 hours

Let L be the length of the rectangle.

So, breadth of the rectangle = L - 3

Area of the rectangle = L L - 3.....1

Base of the isosceles triangle = L - 3

Altitude of the isosceles triangle = 12 m

Area of the isosceles triangle = 1 / 2 (12) (L - 3) .....(2)

Given that

L(L - 3) = 1 / 2 (12) (L - 3) + 4

Or, L

Or, L2 - 9L + 14 = 0

Or, L

Or, L(L - 7) + 2(L - 7) = 0

Or, (L - 7)(L - 2) = 0

So, L = 7m (L ≠ 2 ∵ L - 3 is negative.)

Breadth = 7m - 3m = 4m

Length and breadth of the rectangle are 7m and 4m respectively.

Ans.

Now, plot (10, 7), (20, 21), …, (80, 100) on the graph.

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