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**Class 10 maths term 2 important questions with solutions standard**

**Math Final Term Board Exam Class 10**

**Q1. Let a and b be two positive integers such that a = p ^{3}q^{4} and b = p^{2}q^{3} , where p and q are prime numbers. If HCF(a,b) = p^{m}q^{n} and LCM(a,b) = p^{r}q^{s}, then (m+n)(r+s)=**

(a) 15

(b) 30

(c) 35

(d) 72

**Solution:** HCF(a,b) = p^{m}q^{n } = p^{2}q^{3 }

Therefore, m = 2 and n = 3

LCM(a,b) = p^{r}q^{s } = p^{3}q^{4 }

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) x^{2} –px +p=0

(b) x^{2}–(p+1)x +p=0

(c) x^{2}+(p+1)x +p=0

(d) x^{2} –px+p+1=0

**Solutions:** (b) x^{2}– (p+1)x +p = 0

**Q3. If α and β are the zeros of a polynomial f(x) = px ^{2} – 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)

a_{1}/a_{2} = 3/(2k-1)

b_{1}/b_{2} = 1/(k-1)

c1/c2 = 1/2k+1

For a pair of linear equation to be inconsistent

a_{1}/a_{2} = b_{1}/b_{2}

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)

3 Questions Found on same Topics.

Hey! 3 Questions Found on same Chapter.

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