Put -ac = ab + bc ; -ab = ac+ bc and -bc =ab + ac

you get : 1/(a2-bc

(ab + bc + ca)/[(a+b+c)(abc)]

put, ab+bc+ca=0

you get value = 0/[(a+b+c)(abc)] = 0

Sol: n(x+y) = 20 + 15 = 35

Total boys = nx = 20

Þ nx = 2 x 10 = 4 x 5 = 10 x 2

Similarly,

Total girls = ny = 15

Þ 3 x 5

The possible value of n may be n = 5

x = 4 and y = 5

OR,

Hcf of 20 and 15 is 5

Þ No. of group will be n =5

then Total girls = ny = 15

Þ 3 x 5

The possible value of n may be n = 5

x = 4 and y = 5

Q. The pair of equations y=0 and y= -5 has

1.one solution 2.two solutions 3.infinitely many solutions 4.no solution

Ans: o.x + y = 0

0.x + y = -5

therefore, a1/a2 = b1/b2 ≠ c1/c2

Þ The pair of equations y=0 and y= -5 has no solution

Q. If secA + tanA = 1/x find the value of SecA and tanA

Q. There are 20 cars and motorcycle in a parking lot. If there are 56 wheels together find the no of cars and motorcycles

Ans: Le t the no. of car is x then no. of motorcycle will be (20-x)

A/Q, 4x + 2(20-x) = 56

Þ x = 8

the no. of car is x = 8 then no. of motorcycle will be (20-x) = 20-8 = 12

Q. If constant term in Quardatic polynomial is zero ,then prove that one of its zero is zero

Sol: let the quadratic polynomial be P(x) = a x2 + bx + c

Þ P(x) = a x2 + bx [since the constant term is zero]

Now P(0) = 0 + 0 = 0

Thus, 0 is one zero of p(x).

Q. If p(x) = ax2 + bx +c and a + c = b, then one of the zeroes is

(a) b/a (b) c/a (c) -c/a (d) -b/a

Sol: c/a

Q. if asin2q + bcos2 q = c , then sow that cot2q = (c-a)/(b-c)

ab + ac

ab + ac

(ab + bc + ca)/[(a+b+c)(abc)]

put, ab+bc+ca=0

you get value = 0/[(a+b+c)(abc)] = 0

Q. A class of 20 boys and 15 girls is divided into n groups so that each group has x boys and y girls. Find x ,y and n.

Sol: n(x+y) = 20 + 15 = 35

Total boys = nx = 20

Þ nx = 2 x 10 = 4 x 5 = 10 x 2

Similarly,

Total girls = ny = 15

Þ 3 x 5

The possible value of n may be n = 5

x = 4 and y = 5

OR,

Hcf of 20 and 15 is 5

Þ No. of group will be n =5

then Total girls = ny = 15

Þ 3 x 5

The possible value of n may be n = 5

x = 4 and y = 5

Q. The pair of equations y=0 and y= -5 has

1.one solution 2.two solutions 3.infinitely many solutions 4.no solution

Ans: o.x + y = 0

0.x + y = -5

therefore, a1/a2 = b1/b2 ≠ c1/c2

Þ The pair of equations y=0 and y= -5 has no solution

Q. If secA + tanA = 1/x find the value of SecA and tanA

Q. There are 20 cars and motorcycle in a parking lot. If there are 56 wheels together find the no of cars and motorcycles

Ans: Le t the no. of car is x then no. of motorcycle will be (20-x)

A/Q, 4x + 2(20-x) = 56

Þ x = 8

the no. of car is x = 8 then no. of motorcycle will be (20-x) = 20-8 = 12

Q. If constant term in Quardatic polynomial is zero ,then prove that one of its zero is zero

Sol: let the quadratic polynomial be P(x) = a x2 + bx + c

Þ P(x) = a x2 + bx [since the constant term is zero]

Now P(0) = 0 + 0 = 0

Thus, 0 is one zero of p(x).

Q. If p(x) = ax2 + bx +c and a + c = b, then one of the zeroes is

(a) b/a (b) c/a (c) -c/a (d) -b/a

Sol: c/a

Q. if asin2q + bcos2 q = c , then sow that cot2q = (c-a)/(b-c)