Sample Paper: Math Class ten
Mathematics Class Tenth
sample paper math class ten term-1 mcqs-3
Question (21) In the figure given below AD is the median of triangle ABC, find the coordinates of the point B.
(a) (– 13, –13)
(b) (– 11, 10)
(c) (– 10, – 10)
(d) (– 15, – 15)
Answer : (d) (– 15, – 15)
Explanation :
We know that, the coordinates of the point P(x, y) which divides the line segment joining points A (x1, y1) and B(x2, y2) internally, in the ratio 1 : 1 are `((x_1+x_2)/2, (y_1+y_2)/2)`
In the given figure C is the mid-point of BD, i.e. C divides BD in the ratio of 1:1
Thus, let the B (x1, y1)
Therefore, coordinates of point C
⇒ – 10 = x1 + 5
⇒ – 10 – 5 = x1
⇒ – 15 = x1
⇒ x1 = – 15
Similarly,
–5 = `(y_1+5)/2`
⇒ – 5 × 2 = y1 + 5
⇒ – 10 = y1 + 5
⇒ – 10 – 5 = y1
⇒ – 15 = y1
⇒ y1 = – 15
Thus, coordinate of point B is (–15, – 15)
Thus, option (d) (–15, – 15) is the correct answer.
Shortcut Method
Since, D is the median, this means D divides BC in 1:1 ratio.
Here, difference between the coordinates of C and D = –10
Thus, difference between B and D will also be equal to –10
Thus, coordinates of B = coordinates of C + (–10)
Or, coordinates of B = coordinates of C – 10
Thus, B = C ( –5 – 10, – 5 – 10)
⇒ The coordinates of B = –15, – 15
Thus, option (d) (–15, – 15) is the correct answer.
Question (22) The cooridinates of point B, if B is the midpoint of the line segment AC in the given figure is
(a) (2.5, 4)
(b) (7, 5.2)
(c) (5.2, 7)
(d) (4, 2.5)
Answer : (a) (2.5, 4)
Explanation :
Since, OA = 8 units and it lies on the y-axis, thus coordinates of A = (0, 8)
i.e. A (0, 8)
Similarly, OC = 5 units and it lies on x-axis,
Thus, coordinates of C = (5, 0)
i.e. C (5, 0)
Now, the point B is the mid-point of A( 0, 8) and C (5, 0)
We know that, the coordinates of the point P(x, y) which divides the line segment joining points A (x1, y1) and B(x2, y2) internally, in the ratio 1 : 1 are `((x_1+x_2)/2, (y_1+y_2)/2)`
In the given figure B is the mid-point of AC, i.e. B divides AC in the ratio of 1:1
Thus, let the B (x, y)
Therefore, coordinates of point B
Similarly, y = 8/2 = 4
Thus, coordinates of given mid-point B = 2.5, 4
Thus, option (a) (2.5, 4) is the correct answer.
Question (23) The values of x and y in the above rectangle are
(a) 4, 1
(b) 1, 4
(c) 8, 3
(d) 3, 8
Answer : (b) (1, 4)
Explanation :
Since, the given figure is a rectangle, thus,
AD = BC and AB = CD
Thus, x + 3 y = 13 - - - - (i)
And, 3 x + y = 7 - - - - - (ii)
Now equation (i) × 3 – equation (ii)
(x + 3 y ) 3 – (3 x + y) = 13 × 3 – 7
⇒ y = 4
Now, after substituting the value of y = 4 in equation (i), we get
x + 3 × 4 = 13
⇒ x + 12 = 13
⇒ x = 13 – 12
⇒ x = 1
Thus, x = 1 and y = 4
Thus, option (b) 1, 4 is the correct answer.
Question (24) The value of the given expression
(3 – cot 30o) (√3 + 1) = 2 A
Hence A can be replaced by
(a) cot 60o
(b) tan 60o
(c) 2 tan 30o
(d) 2 tan 60o
Answer : (b) tan 60o
Explanation :
Given,
⇒ A = tan 60o
[∵ tan 60o = √ 3]
Thus, A can be replaced by tan 60o
Thus, option (b) tan 60o is the correct answer.
Question (25) It tan θ = 1, then the value of (cosec4 θ + sec4 θ) is
(a) 8
(b) 4
(c) 2 &radi;2
(d) 1/4
Answer : (a) 8
Explanation :
Given, tan θ =1
⇒ tan θ = tan 45o
⇒ θ = 45o
Thus, (cosec4 θ + sec4 θ)
= cosec (45o)4 + sec (45o)4
= (√2)4 + (√2)4
= 4 + 4 = 8
Thus, option (a) 8 is the correct answer.
Question (26) If α and β are the zeroes of the quadratic polynomial x2 + x – 2 then the value of α2 β + β2 α is :
(a) –2
(b) –1
(c) 0
(d) 2
Answer : (d) 2
Explanation :
Given, x2 + x – 2
And α and β are zeroes of this polynomial,
Then, α2 β + β2 = ?
We know that, α + β = –b/a = –1/1
⇒ α + β = – 1
And, we know that, α β = c/a = –2/1
⇒ α β = – 2
Thus, α2 β + β2
= &alpha β (α + β)
= –2 ( – 1) = 2
⇒ α2 β + β2 = 2
Thus, option (d) 2 is the correct answer.
Question (27) If the HCF of 14 and 105 is expressed as 14 m – 105, then the value of m is
(a) 2
(b) 4
(c) 6
(d) 8
Answer : (d) 8
Explanation :
The prime factorization of
14 = 2 × 7
And, 105 = 3 × 5 × 7
Thus, HCF of 14 and 105 = 7
Now, as given the HCF of 14 and 105 is expressed as 14 m – 105
Thus, m = ?
This, means the HCF of 14 and 105 = 14 m – 105
⇒ 14 m – 105 = 7
⇒ 14 m = 7 + 105
⇒ 14 m = 112
⇒ m = 112/14
⇒ 8
Thus, option (d) 8 is the correct answer.
(28) The decimal expansion of `51/(3xx2^2xx5^3)` will be
(a) Non-Terminating and repeating
(b) Terminating
(c) Non-terminating and non-repeating
(d) Non-Terminating
Answer : (b) Terminating
Explanation :
The given expression `51/(3xx2^2xx5^3)`
Can be written as `17/(2^2xx5^3)`
We know that, if a rational number x = p/q, such that prime factorisation of q is of the form 2n 5m, where n and m are non-negative integers, then x has a decimal expansion which terminates.
Since, in the given expression the prime factorisation of q is in the form of 2n 5m, thus this given expression has a terminating decimal expansion.
Thus, option (b) Terminating is the correct answer.
Question (29) The greatest number which divides 70 and 125, leaving remainder 5 and 8 respectively is :
(a) 1750
(b) 875
(c) 65
(d) 13
Answer : (d) 13
Explanation :
The given numbers are 70 and 125
And to find the greatest number, i.e. HCF which divides the given number leaving remainder 5 and 8 respectively.
Thus, the numbers can be modified as
70 – 5 = 65
And, 125 – 8 = 117
The prime factorization of 65 = 5 × 13
And, prime factorization of 117 = 3 × 3 × 13
Thus, HCF = 13
Thus, the greatest number is 13 which divides the 70 and 125 leaving the remainder 5 and 8 respectively.
Thus, option (d) 13 is the answer.
Question (30) If α and β are the zeroes of the polynomial 5x2 – 7x + 2, then the sum of their reciprocals is:
(a) 7/2
(b) 7/5
(d) 2/5
(d) 14/25
Answer : (a) 7/2
Explanation :
Given polynomial P(x) = 5x2 – 7x + 2
And zeros of the polynomial are α and β
Now, we know that, α + β = – b/a
= –( –7 )/5
⇒ α + β = 7/5
α β = c/a
⇒ α β = 2/5
Now, sum of reciprocal of α and β
After substituting the values of α + β and α β, we get
Thus, sum of reciprocal of α and β = 7/2
Thus, option (a) 7/2 is the correct answer.
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