Sample Paper: Math Class ten

Sample Paper: Math MCQs-1

Question (1) The ratio of HCF and LCM of even prime number and the least composite number is:

(a) 1 : 3

(b) 1 : 1

(c) 1 : 2

(d) 2 : 1

Answer : (c) 1 : 2

Explanation

Prime Number: Numbers which have only two factors 1 and number itself are called prime numbers.

Example: 1, 3 , 5 , 7 , 11, . . .

Even Prime Number: 2 is only the even prime number which has only two factors 1 and number itself, i.e. 2.

Composite Number : Numbers which have more than two factors, 1, number itself and other number(s), are called Composite Numbers.

Example : (i) 4. This is the least composite number which has three factors, 1, 2 and 4.

(ii) 6. Six (6) is also a composite number which has more than two factors, i.e. 1, 2, 3 and 6

Now, in the given question, the ratio of even prime number and least composite number is to be calculated.

Here, the even prime number = 2

And least composite number = 4

Thus, LCM of these two numbers, 2 and 4

The prime factor of 2 = 1 × 2

The prime factor of 4 = 1 × 2 × 2

Thus the LCM of given numbers 2 and 4 = 1 × 2 × 2 = 4

And the HCF of these two numbers, 2 and 4

The prime factor of 2 = 1 × 2

The prime factor of 4 = 1 × 2 × 2

Thus, HCF = 2

Now, the ratio of HCF and LCM

= 2 : 4 = 1 : 2

Thus, answer (c) 1 : 2 is the correct option.

Question (2) For what value of p, the pair of equations 6x – 3y + 10 = 10 and 2x + ky + 9 = 0 have no solution:

(a) 12

(b) 6

(c) 1

(d) – 1

Answer: (d) – 1

Explanation :

If for a linear equation, `(a_1)/(a_2)=(b_1)/(b_2)!=(c_1)/(c_2)` then it forms parallel lines and equations have no solution.

For the given equations 6x – 3y + 10 = 10 and 2x + ky + 9 = 0

a₁ = 6,

a₂ = 2,

b₁ = –3,

b₂ = k,

c₁ = 10

and c₂ = 9

Thus, `6/2=(-3)/k !=10/9`

⇒ 3 k = – 3

⇒ k = – 3 / 3

⇒ k = – 1

Thus, for the value of k = – 1 the given equations have no solution.

Hence, option (d) – 1 is the right option.

Question (3) The coordinates of the centroid of a triangle whose vertices are (–7, 4), (3, – 5) and (10, – 2) is :

(a) (1, 2)

(b) (1, 3)

(c) (– 1, 2)

(d) (2, – 1)

Answer (d) (2, – 1)

Explantion :

We know that centroid of a triangle in which coordinates of the vertices are given,

Here given, vertices of the triangle are (–7, 4), (3, – 5) and (10, – 2)

Thus, x₁ = –7, x₂ = 3 and x₃ = 10

And, y₁ = 4, y₂ = – 5 and y₃ = –2

Thus, centroid of the given triangle

Thus, centroid of the given triangle = (2, – 1)

Thus, option (d) is the correct answer.

Question (4) △ TAP ∼ △ ROD. If TM and RN are altitudes of △ TAP and △ ROD respectively and TM : RN = 4 : 9 then AP² : OD² =

(a) 2 : 3

(b) 16 : 81

(c) 81 : 16

(d) 4 : 9

Answer : (b) 16 : 81

Explanation :

Given,

And, △ TAP ∼ △ ROD

TM and RN are altitudes of △ TAP and △ ROD respectively

And TM : RN = 4 : 9

Thus, AP² : OD² = ?

We know that if two triangles are similar then its sides will be in proportion.

Thus, TM: RN = AP : OD = TA : RO = TP : RD

Now, since TM : RN = 4 : 9 (as given in the question)

Thus, AP : OD = 4 : 9

Thus, (AP)² : (OD)² = 4² : 9²

⇒ (AP)² : (OD)² = 16 : 81

Thus, option (b) 16 : 81 is the correct answer.

Question (5) The lines x = 0 and x = ? 5 has/ have

(a) No solution

(b) Infinite number of solutions

(c) Two solutions

(d) Unique solution

Answer : (a) No solution

Explanation :

As given, lines are x = 0 and x = – 5

Thus, these lines can be drawn as follows.

Since, given lines are parallel, thus it has no solution.

Hence, option (a) No solution is the correct answer.

Question (6) For what value of k, do the equations 3x – y + 8 = 0 and 6x – ky = – 16 represent coincident lines.

(a) 1/2

(b) – 1/2

(c) 2

(d) – 2

Answer : (c) 2

Explanation :

Given, equations are

3x – y + 8 = 0 - - - - (1)

6x – ky = – 16

⇒ 6x – ky + 16 = 0 - - - (2)

We know that for coincident lines,

`a_1/a_2 = b_1/b_2=c_1/c_2`

For the given equations,

a₁ = 3

a₂ = 6

b₁ = – 1

And, b₂ = – k

Thus,

⇒ – 3k = – 6

⇒ k = 6/3 = 2

Thus, k = 2

Thus, for the value of k = 2, the given lines will be coincident.

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

Question (7) If sin P = p and sec P = q, then the value of tanP is

(a) p/q

(b) p . q

(c) q/p

(d) 1/pq

Answer (b) p . q

Explanation :

Given, sinP = p

And, secP = q

Thus, cos P = 1/q

Now, `tanP=(sinP)/(cosP)`

`=>tan P = p/(1/q)`

⇒ tan P = p . q

Thus, option (b) p . q is the correct answer.

Question (8) The HCF of two numbers is 7 and their product is 16170, then their LCM is

(a) 210

(b) 490

(c) 350

(d) 2310

Answer : (d) 2310

Explanation :

Given, HCF = 7 and their product = 16170

Thus, LCM = ?

Here, LCM × HCF = 16170

⇒ LCM × 7 = 16170

⇒ LCM = 16170/7

⇒ LCM = 2310

Thus, option (d) 2310 is the correct answer.

Question (9) If 2 tan² β + sec² β = 2, then β is

(a) 0^o

(b) 15^o

(c) 45^o

(d) 60^o

Answer : (d) 60^o

Explanation

Given,

2 tan² β + sec² β = 2

Then, β = ?

2 tan² β + sec² β = 2

⇒ tan²β + tan² β + sec² β = 2

⇒ – 1 + tan² β = 2

⇒ tan² β = 2 + 1

⇒ tan² β = 3

⇒ tan β = √3

⇒ tan β = tan 60^o

⇒ β = 60^o

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

Question (10) If 2 sin 3 θ = 1, thenθ =?

(a) 10^o

(b) 15^o

(c) 20^o

(d) 30^o

Answer : (a) 10^o

Explanation :

Given, 2 sin 3 θ = 1

⇒ sin 3 θ = 1/2

⇒ sin 3 θ = sin 30^o

[∵ sin 30^o = 1/2]

⇒ 3 θ = 30^o

⇒ θ = 30^o/3

⇒ θ = 10^o

Thus, option (a) 10^o is the correct answer.

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