Sample Paper: Math Class ten

sample paper math class ten term-1 mcqs-4

Question (31) If one of the zeroes of the quadratic polynomial x² + 3x + k is 2, then the value of k is :

(a) 10

(b) – 10

(c) – 7

(d) – 2

Answer : (b) – 10

Explanation :

Given polynomial P(x) = x² + 3x + k

Given, one of the zeros = 2

Then value of k = ?

P(2) = 2² + 3 × 2 + k = 0

⇒ 4 + 6 + k = 0

⇒ 10 + k = 0

⇒ k = – 10

Thus, option (b) – 10 is the correct answer.

Question (32) If x = a, y = b be the solution of the equations x – y = 2 and x + y = 4 then the value of a & b respectively are:

(a) 3 and 5

(b) 5 and 3

(c) 3 and 1

(d) – 1 and – 3

Answer : (c) 3 and 1

Explanation :

Given, x = a & y = b

And, x – y = 2

⇒ a – b = 2 - - - (i)

And, x + y = 4

⇒ a + b = 4 - - - - (ii)

Now, after adding equation (i) and (ii), i.e. equation (i) + equation (ii), we get

a – b + a + b = 2 + 4

⇒ 2a = 6

⇒ a = 6/2 = 3

⇒ a = 3

Now, after substituting the value of a = 3 in equation (i), we get

3 – b = 2

⇒ 3 – 2 = b

⇒ 1 = b

⇒ b = 1

Thus, value of a & b = 3, 1

Thus, option (c) 3 and 1 is the correct answer.

Question (33) ABCD is a Cyclic Quadrilateral, the measures of ∠ A and ∠ B are:

(a) 64^o , 80^o

(b) 30^o, 14^o

(c) 100^o, 70^o

(d) 116^o, 100^o

Answer : (a) 64^o , 80^o

Explanation :

Given, ABCD is a cyclic quadrilateral, in which

∠ A = 2y + 4^o

∠ B = 6x–4^o

∠ C = 4y – 4^o

And, ∠ D = 7x + 2^o

Thus, ∠ A and ∠ B = ?

We know that, in a cyclic quadrilateral, the sum of opposite angles are 180^o

Thus, in the given cyclic quadrilateral,

∠ A + ∠ C = 180^o

⇒ 2y + 4^o + 4y – 4^o = 180^o

⇒ 6y = 180^o

⇒ y = 180^o/6

⇒ y = 30^o

Now, after putting the value of y = 30^o in the expression given for the ∠A, we get

∠ A = 2y + 4^o

⇒ ∠ A = 2 × 30^o + 4^o

⇒ ∠ A = 60^o + 4^o

⇒ ∠ A = 64^o

Now, ∠ B + ∠ D = 180^o

⇒ 6x – 4^o + 7x + 2^o = 180^o

⇒ 13 x – 2^o = 180^o

⇒ 13 x = 180^o + 2^o

⇒ 13 x = 182^o

⇒ x = 182^o/13

⇒ x = 14

Now, after substituting the value of x = 14 in ∠B = 6 x – 4^o, we get

⇒ ∠ B = 6 × 14 – 4^o

⇒ ∠ B = 84 – 4

⇒ ∠ B = 80^o

Thus, ∠ A and ∠ B = 64^o & 80^o

Thus, option (a) 64^o & 80^o is the correct answer.

Question (34) In the following figure, LM is parallel to BC and LN is parallel to CD, then which of the following relation is true?

Answer : (d) `(AM)/(LC) = (AN)/(AD)`

Explanation :

As per question, in the given figure,

ML || BC

And LN ||DC

Now, in Δ ABC,

ML || BC

Thus, from BPT (Basic proportionality Theorem) (Thales Theorem), which says If a line is drawn parallel to one side of a triangle to intersect the other two distinct sides in distinct points, the other two sides are divided in the same ratio.

Similarly, in Δ ADC

As given, LN || DC

Thus, from BPT (Basic proportionality Theorem) (Thales Theorem),

Thus, option (d) is the correct answer.

Question (35) In the given figure, `(PS)/(SQ) = (PT)/(TR)` and ∠ PST = ∠ PRT, then Δ PQR is a/ an:

(a) Equilateral Triangle

(b) Isosceles Triangle

(c) Scalene Triangle

(d) Right Triangle

Answer : (b) Isosceles Triangle

Explanation :

As per question, in the given figure,

PS/SQ = PT/TR

Thus, from the converge of Basic Proportionality Theorem which says If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

ST||QR

Now, since ST||QR and SQ is the transversal,

∴ ∠SQR = ∠PSR (Because corresponding angles are equal)

And, ∠PSR = ∠SQR = ∠TRQ

⇒ ∠SQR = ∠TRQ

⇒ ∠PQR = ∠ PRQ

Hence, the given triangle PQR is an isosceles triangle.

Thus, option (b) Isosceles Triangle is the correct answer.

Question (36) In Δ ABC, AB = 6 &rad; 3 cm, AC = 12 cm and BC = 6 cm, the angle B is:

(a) 120^o

(b) 60^o

(c) 90^o

(d) 45^o

Answer : (c) 90^o

Explanation:

Let the given figure can be drawn as follows:

Here, AB = 6&radi;3, and BC = 6 cm

Thus, to prove AC = 12 cm

Now, (AB)² + (BC)²

= (6 √ 3 cm)² + (6 cm)²

= 36 × 3 cm² + 36 cm²

= 108 cm² + 36 cm²

= 144 cm²

Now, (AC) ² = (12 cm)²

= 144 cm²

Thus, (AC)² = (AB)² + (BC)²

Since, this follows the rule of Pythagoras Theorem,

Thus, the given triangle is a right angled triangle.

Thus, ∠ B = 90^o

Hence, option (c) 90² is the correct answer.

Question (37) In the given figure DE is parallel to BC, then the value of x is :

(a) 10 cm

(b) 8 cm

(c) 6 cm

(d) 4 cm

Answer : (c) 6 cm

Explanation :

Given,

AD = 3 cm & DB = 4 cm

Thus, AB = AD + DB = 3 + 4

⇒ AB = 7 cm

BC = 14 cm

DE||BC

Now, since, DE || BC, thus according Basic Proportionality Theorem,

Δ ABC ∼ Δ ADE

After cross multiplication, we get,

x × 7 cm = 3 cm × 14 cm

⇒ 7 x cm = 42 cm

⇒ x = 42/7 = 6 cm

Thus, option (c) 6 cm is the correct answer.

Question (38) Corresponding sides of two similar triangles are in the ratio of 2 : 3. If the area of a small triangle is 48 sq. cm, then the area of a large triangle is :

(a) 230 sq. cm

(b) 106 sq. cm

(c) 107 sq. cm

(d) 108 sq. cm

Answer : (d) 108 sq. cm

Explanation :

Given, two triangles are similar.

Ratio of their corresponding sides = 2 : 3

And, area of small triangle = 48 sq cm

Thus, area of larger triangle = ?

We know form one of the theorem of areas of similar triangle that, The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Now, let the area of larger triangle = A_L

Thus, Area of smaller triangle : Area of larger triangle = (side of smaller triangle)² : (side of larger triangle)²

⇒ A_L = 12 × 9 cm²

⇒ Area of larger triangle = 108 cm²

Thus, option (d) 108 sq. cm is the correct answer.

Question (39) If (1 – cos² θ) cosec² θ = A, then the value of A is :

(a) 0

(b) 1

(c) 2

(d) 3

Answer : (b) 1

Explanation :

Given, (1 – cos² θ) cosec² θ A

Thus, value of A = ?

(1 – cos² θ) cosec² θ

= sin² θ × cosec² θ

`=cancelsin^2 theta xx 1/(cancelsin^2 theta)`

= 1 = A

Thus, option (b) 1 is the correct answer.

Question (40) After simplification we get the value of the given expression for acute angle θ as tan θ . sin θ + cos θ = A. Then the value of A is

(a) sin θ

(b) cos θ

(c) sec θ

(d) cosec θ

Answer : (c) sec θ

Explanation :

Given,

tan θ . sin θ + cos θ = A

⇒ sec θ = A

⇒ A = sec θ

Thus, option (c) sec θ is the correct answer.

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