Simplification-Practice test

Aptitude Practice Paper Quantitative
1.

If (x-k) is a common factor of the equations 4x2– 9x + 5 = 0 and (p-1) x2 – 3x-1 = 0 and k <= 1, then find the value of p.

Answer: Option C

Explanation:

If (x-k) is a factor of 4x2 – 9x + 5 = 0

4k2-9k + 5 = 0 –> k = 5/4, 1

As k <= 1 k = 1

As k is also a factor of

(p-1)x2 – 3x – 1 = 0 –>(p-1)k2-3k-1 = 0

–> (p-1) 12-3-1 = 0 –> p = 5

2.

The sum and the product of the roots of the quantratic equation x2 + 20x + 3 = 0 is

Answer: Option D

Explanation:

Given equation is x2 + 20x + 3 = 0 —-(1)

Comparing (1) with general form of quantratic equation (1) in option 4, sum of the roots and the product of the roots are -20 and 3 respectively.

3.

If 1.5x=0.04y then the value of (y-x)/(y+x) is

Answer: Option B

Explanation:

x/y = 0.04/1.5 = 2/75
So (y-x)/(y+x) = (1 – x/y)/(1 + x/y) = (1 – 2/75)/ (1 + 2/75) = 73/77.

4.

If x2 + (2a + 3)x + (5-a) = 0 has both the roots as real and negative, then what is the greatest positive integral value of a?

Answer: Option D

Explanation:

Since both the roots are negative, Sum of the roots is negative and product of the roots is positive.

2a + 3 < 0 –> a > -3/2.

Also, 5 – a > 0 –> a < 5

Greatest possible integer value of a = 4

5.

The value of  (x – y)³ + (y – z)³ + (z – x)³ is equal to :
12 (x – y) (y – z) (z – x)

Answer: Option D

Explanation:

Since (x – y) + (y – z) + (z – x) = 0, so (x – y)³ + (y – z)³ + (z – x)³

= 3 (x – y) (y – z) (z – x).

3 (x – y) (y – z) (z – x) = 1/4.
12(x – y) (y – z) (z – x)

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