Answer:

Class 10 maths term 2 important questions with solutions standard

Math Final Term Board Exam Class 10

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Q1. Let a and b be two positive integers such that a = p³q⁴ and b = p²q³ , where p and q are prime numbers. If HCF(a,b) = p^mqⁿ and LCM(a,b) = p^rq^s, then (m+n)(r+s)=
(a) 15
(b) 30
(c) 35
(d) 72

Solution:  HCF(a,b) = p^mqⁿ  = p²q³  

Therefore, m = 2 and n = 3 

LCM(a,b) = p^rq^s  = p³q⁴ 

Therefore, r = 3 and s = 4 

Now (m+n)(r+s) = (2 + 3) (3 + 4) = 5 x 7 = 35 

Correct Answer : (c) 35 

Q2. Let p be a prime number. The quadratic equation having its roots as factors of p is
(a) x² –px +p=0
(b) x²–(p+1)x +p=0
(c) x²+(p+1)x +p=0
(d) x² –px+p+1=0

Solutions: (b) x²– (p+1)x +p = 0

Q3. If α and β are the zeros of a polynomial f(x) = px² – 2x + 3p and α + β = αβ, then p is
(a)-2/3
(b) 2/3
(c) 1/3
(d) -1/3

Solutions: Given that α + β = αβ 

Here α + β = - b/a 

                 = - (-2) / p = 2/p 

αβ = c/a = 3p/p = 3 

Now α + β = αβ  

=> 2/p = 3 

=> p = 2/3 

Correct Answer : (b) 2/3 

Q4. If the system of equations 3x + y = 1 and (2k-1)x + (k-1)y = 2k+1 is inconsistent, then k =
(a) -1
(b) 0
(c) 1
(d) 2

Solution: 

3x + y = 1   ................. (i) 

(2k-1)x + (k-1)y = 2k+1  ............ (ii) 

a₁/a₂ = 3/(2k-1) 

b₁/b₂ = 1/(k-1) 

c1/c2 = 1/2k+1 

For a pair of linear equation to be inconsistent 

a₁/a₂ = b₁/b₂ 

3/(2k-1)  = 1/(k-1) 

3(k - 1) = 2k - 1

3k - 2 = 2k -1 

3k - 2k = - 1 + 2 

k = 1 

Correct Answer: (c) 1 

Q5. If the vertices of a parallelogram PQRS taken in order are P(3,4), Q(-2,3) and R(-3,-2), then the coordinates of its fourth vertex S are
(a) (-2,-1)
(b) (-2,-3)
(c) (2,-1)
(d) (1,2)

Solution: 

Now the mid-point of PR where PR and QS are two diagonals pf ||gm PQRS. 

P(3,4) and R(-3,-2) 

x = (3 + -3 )/ 2 = 0 

y = (4 + - 2) / 2 = 1

The coordinates of mid point of PR is O(0, 1) 

Mind that diagonals PR and OS bisect each other at point O(0, 1)

 Q(-2,3) and let S(x, y)  and O(0, 1) is also mid point of QS 

Then, using mid-point formula 

x = (-2 + x) / 2 

0 = - 2 + x 

gives x = 2 

y = (3 + y)/ 2 

1 = (3 + y)/ 2  

1 x 2 = 3 + y 

y = 2 - 3 

y = - 1 

The coordinates of  S(2, - 1)

Correct Answer: (c) (2, - 1)