SAT (Stage-1) Mock Test - 17

(x² – y²)(x² – 2xy + y²) = 3
Or, (x – y)(x + y)(x² – 2xy + y²) = 3
Or, (x – y)(x + y)(x – y)² = 3
x + y = 3
We have, x + y = 3 and x – y = 1
Adding two, we get 2x = 4
i.e. x = 2
and y = 1
Value of xy = 2

Let the number of pen and that of pencil be x and y, respectively.

Then, 15x + 8y = 190

Maximum positive integral values of x and y satisfying the equation are given below:

Maximum number of pens and pencils together is 20 pencils and 2 pens.
i.e. For full money division, he can buy 2 pens and 20 pencils.
2 × 15 + 20 × 8 = 30 + 160 = 190
Total items = 20 + 2 = 22

Thickness of the wood = 2.5 cm
Now, external dimensions of the wooden box are as follows.
Length = 120 cm, width = 80 cm and height = 40 cm
Volume of the wood = (120 cm × 80 cm × 40 cm) - (115 cm × 75 cm × 35 cm)
= (384,000 - 301,875) cm³ = 82,125 cm³

Let p(x) = x³ + ax² + bx + c

Let the zeroes be z₁, z₂ and –1.

z₁z₂(-1) = -c

Or, z₁z₂ = c = Product of the other two zeroes ... (i)

Now, as one zero is –1, p(-1) = 0

Or, (-1)³ + a(-1)² + b(-1) + c = 0

Or, -1 + a – b + c = 0

Or, b + 1 – a = c … (ii)

Equating the left hand sides of equations (i) and (ii),

z₁z₂ = b + 1 – a

These four numbers have only 1 as the H.C.F.

Factors of 63 = 1 × 3 × 3 × 7

Factors of 121 = 1 × 11 × 11

Factors of 49 = 1 × 7 × 7

Factors of 841 = 1 × 29 × 29

So, the H.C.F. of 49, 63,121 and 841 = 1