Question

please solve 19 and 20

Answer 1

Question no. 19 )

After seeing Question u can easily understand that a , b , c should be = 1 then only a + bc = b + ca = c + ab would happen but They are equal to 502 so...a = b = c is necessary .

${a}^{2} + a = 502$
(as a =b = c)

$a = \frac{ - 1 + - \sqrt{2009} }{2}$
Minimum value of a + b + c = 3a when
a =
$\frac{ - 1 - \sqrt{2009} }{ 2}$
minimum value of a + b+c =
$\frac{3}{2} \times \frac{ - 1 - \sqrt{2009} }{1}$

so option C is correct

Question no. 20
let. x and X1 be the Roots of first equation then

X + X1 =
$\frac{2008 \times 2010}{ {2009}^{2} }$

$x \times x1 = \frac{ - 1}{ {2009}^{2} }$

U could Solve both equation to get x and X1 but...in exam we don't have so much time so u can Correctly guess that X = 1 , Let me show u

X + X1 = (2008×2010)/2009^2

X + X 1 = {(2009-1)+(2009+1)}/2009^2

X + X1 = (2009^2 - 1)/2009^2

X + X1 = 1 - 1/2009^2

In question it is given X(i.e is alpha) is larger so . X = 1


in Second equation

B+B1 = -2008/2009
B×B1 = -1/2009

Here also u can easily see
B + B1 = -2009 + 1

B is smaller so B must be -2009


We need to find X-B
1-(-2009) = 2010


here X = alpha according to ur question

Answer 2

Answer:

Cost of 50 pencils  = Rs 75.

No of pencils be x Rs .

 x = 100 pencils.