Sample Paper: Math Class ten

Sample Paper: Math MCQs-2

Question (11) If sin θ₁ = 1/2 and cos θ₂ = 1/2, the the value of 2(θ₂ – θ₁) is

(a) 90^o

(b) 60^o

(c) 30^o

(d) 0^o

Answer : (b) 60^o

Explanation :

Given,

sin θ₁ = 1/2

⇒ sin θ ₁ = sin 30^o

[∵ sin 30^o = 1/2]

⇒ θ₁ = 30^o

And, as given again, cos θ₂ = 1/2

⇒ cos θ ₂ = cos 60^o

[∵ cos 60^o = 1/2]

⇒ θ₂ = 60^o

Now, 2(θ₂ – θ₁)

= 2( 60^o – 30^o)

= 2 (30^o)

= 2 × 30^o

= 60^o

Thus, the option (b) 60^o is the correct answer.

Question (12) If 174 x + 326 y = 471 and 326 x + 174 y = 529, then the value of x + y is

(a) 1

(b) 2

(c) 500

(d) 1000

Answer : (b) 2

Explanation :

Given,

174 x + 326 y = 471 - - - - (i)

326 x + 174 y = 529 - - - - (ii)

Thus, x + y = ?

After adding equation (i) and (ii) we get,

174 x + 326 y + 326 x + 174 y = 471 + 529

⇒ 500 x + 500 y = 1000

⇒ 500 ( x + y ) = 1000

⇒ x + y = 1000/500

⇒ x + y = 2

Thus, option (b) 2 is the correct answer.

Question (13) The smallest number by which 1/17 should be multiplied so that its decimal expansion terminates after one decimal place is

(a) 10/17

(b) 17/100

(c) 17/10

(d) 17

Answer : (c) 17/10

Explanation :

The given number 1/17 is when multiplied by 17/10, we get

`1/17xx17/10 = 1/10 = 0.1`

Thus, option (c) 17/10 is the correct answer.

Question (14) Two natural numbers having HCF = 1 are called

(a) Composite numbers

(b) Prime numbers

(c) Co-prime numbers

(d) Odd numbers

Answer : (b) Prime numbers

Explanation :

Numbers which have HCF equal to 1 are called Prime numbers

Thus, (b) Prime numbers is the correct answer.

Question (15) Observe this figure and find ∠ P.

(a) 40^o

(b) 60^o

(c) 80^o

(d) 140^o

Answer : (a) 40^o

Explanation :

In the given triangle ∠ P = ?

In the given triangles ABC and PQR

`(AB)/(RQ)=(AC)/(RP)=(BC)/(PQ)=1/2`

∴ △ ABC ∼ △ RPQ

Thus, ∠ C = ∠ P

We know that the sum of all angles of a triangle = 180^o

Thus, in △ ABC,

∠ A + ∠ B + ∠ C = 180^o

⇒ 80^o + 60^o + ∠ C = 180^o

⇒ 140^o + ∠ C = 180^o

⇒ ∠ C = 180^o – 140^o

⇒ ∠ C = 40^o

Now, since ∠ C = ∠ P

Thus, ∠ P = 40^o

Thus, option (a) 40^o is the correct answer.

Question (16) From the given figure, the value of cot 30^o is

(a) 1/2

(b) √3/2

(c) 1/√3

(d) √3

Answer : (d) √3

In this given triangle ABD,

AB = a

BD = a/2

∠ DAB = 30^o

Thus, cot 30^o = ?

Using Pythagoras Theorem

(AB)² = (AD)² + (BD)²

Now,

Thus, option (d) √3 is the correct answer.

Question (17) If the points P (6, 1), Q (8, 2), R (9, 4) and S (x, 3) are the vertices of a parallelogram, taken in order, find the value of x.

(a) 5

(b) 6

(c) 7

(d) 8

Answer : (d) 7

Explanation :

Let the figure of given parallelogram is as follows:

We know that, the diagonals of a parallelogram bisects each other.

We know that, middle of a line whose coordinates of two ends are given can be calculated using formula

`(x_1+x_2)/2, (y_1+y_2)/2`

Thus, M is the middle point of PQ = middle point of QS

⇒ 2 (8 + x) = 15 × 2

⇒ 16 + 2x = 30

⇒ 2x = 30 – 16

⇒ 2x = 14

⇒ x = 14/2 = 7

Thus, x = 7

Thus, option (c) 7 is the correct answer.

Question (18) P is at (–4, 6) and Q is at (– 1, – 3). M is a point on the x-axis which is equidistant from P and Q. What are the coordinates of M ? M = ( , )

(a) (– 42, 6)

(b) (0, – 7)

(c) (– 7, 0)

(d) (6, – 42)

Answer : (c) (– 7, 0)

Explanation :

Given, there are two points P (– 4, 6) and Q (– 1, – 3)

And, M is other point at x-axis which is at equidistance from P and Q

This means, PM = QM

Since, M lies on x-axis, thus y-coordinate for M = 0, i.e. y = 0 for the point M.

Now, from distant formula we know that, distance between two points whose coordinates are (x₁, y₁) and (x₂, y₂)

`=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

Distance between P and M, i.e. PM

And distance between Q and M i.e. QM

Now, since PM = QM (as given in the question), thus, equation (i) = equation (ii)

Thus,

`sqrt(x^2+8x+52)=sqrt(x^2+2x+10)`

After squaring both sides, we get

x² + 8x + 52 = x² + 2x + 10

⇒ x² + 8x + 52 – (x² + 2x + 10)

⇒ x² + 8x + 52 – x² – 2x – 10 = 0

⇒ 6x + 42 = 0

⇒ 6x = – 42

⇒ x = – 42/6

∴ x = – 7

Now, since given point is at x axis, thus, y-coordinate = 0

Thus, coordinate of given point M = – 7, 0

i.e. M (–7, 0)

Thus, option (c) (–7, 0) is the correct answer.

Question (19) Two poles of 6 m and 11 m stand vertically on a plane ground. If the distance between their feet is 12 m, find the distance between their tops.

(a) 21 m

(b) 13 m

(c) 12 m

(d) 9 m

Answer : (b) 13 m

Explanation :

As per statistics given in the question, figure can be drawn as follows:

Here, twos poles, AB = 6 m and CD = 11 m are stand vertically.

Distance between their feet A and C = 12 m

Thus, distance between their top D and B, i.e. DB = ?

Now, since, AC || EB, and AB || CD

Thus, AC = EB = 12 m

And, AB = CE = 6 m

DE = AC – CE = 11 m – 6 m

⇒ DE = 5 m

Now, in triangle, DEB

∠ E = 90^o

Thus, from Pythagoras Theorem,

DB² = DE² + EB²

= (5 m)² + (12 m)²

= 25 m² + 144 m²

= 169 m²

`=>DB = sqrt (169m^2)`

⇒ DB = 13 m

Thus, distance between the top of the given poles = 13 m

Hence, option (b) 13 m is the correct answer.

Question (20) Radha is playing with a toy robot that moves 80 m due East and then 150 m due North. How far is the toy from the starting point?

(a) 110 m

(b) 150 m

(c) 120 m

(d) 170 m

Answer : (d) 170 m

Explanation :

As given statistics figure can be drawn as follows:

As given, a toy moves 80 m east, i.e. from A to B. After that it moves from 150 m north, i.e. from B to C.

Thus, the distance between A to C, i.e. AC = ?

Now, in the given figure, ∠ B = 90^o

Thus, in right angled triangle ABC, according to Pythagoras Theorem,

AC² = AB² + BC²

= (80 m)² + (150 m)²

= 6400 m² + 22500 m²

= 28900 m²

`=>AC = sqrt(28900 m^2)`

⇒ AC = 170 m

Thus, option (d) 170 m is the correct answer.

MCQs Test

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