Practice Model question set 1 - quantitative aptitude Online Quiz (set-1) For All Competitive Exams

Given equation,

$√{{2x}/{3 - x}} - √{{3 - x}/{2x}} = 3/2$

Let $√{{2x}/{3 - x}}$ = a

∴ a - $1/a = 3/2$

⇒ $2(a^2 - 1)$ = 3a

⇒ $2a^2$ - 3a - 2 = 0

⇒ $2a^2$ - 4a + a - 2 = 0

⇒ 2a(a - 2) + 1(a - 2) = 0

⇒ (2a + 1) (a - 2) = 0

If a - 2 = 0

Now, put a = 2

⇒ $√{{2x}/{3 - x}}$ = 2

Squaring both sides, then we get

⇒ 2x = 4(3 - x)

⇒ 6x = 12 ⇒ x = 2

If 2a + 1 = 0,

⇒ a = -$1/2, a ≠ {-1}/2$

x = 2 is the root of equation.

Given that, f(x) is a polynomial with constant term 10 and all coefficient are integer. Let $k, k_1, k_2, ...... k_{n - 1}$ be roots of nth degree polynomial.

Now (x - k) is a factor of f(x), where k is an integer.

Product of roots = ${\text"Constant term"}/{\text"Coefficient of" x^n}$

i.e., $k. k_1. k_2......k_{n - 1} = {10}/1$ Coefficient of $x^n$ is 1

⇒ $k. k_1. k_2....k_{n -1}$ = 10 = 5.2.1....1.

Therefore, the possible value of k is 5.

Let &alpha, β be the roots of the equation $lx^2$ + mx + m = 0

Given ${α}/{β} = p/q$

Now α + β (sum of roots) = ${- m}/l$

and α β (product of roots) = $m/l$

Consider $√{p/q} + √{q/p} + √{m/l}$

Using (1)

= $√{{α}/{β}} = √{{β}/{α}} + √{m/l}$

= ${α + β}/{√{α β}} + √{m/l}$

= ${- m/l}/{√{m/l}} + √{m/l} = - √{m/l} + √{m/l}$ = 0

∴ option (d) is correct.

Given equation

$x^2$ - 15x + r = 0

Sum of roots = 15

p + q = 15 .....(i)

and p - q = 1 ......(ii)

From equation (i) and (ii) we have

p = 8, q = 7

Now, $p^2$ - 15 p + r = 0

$(8)^2$ - 15 (8) + 4 = 0

∴ r = 56

Let the roots of the equation be α and β

∴ α + β = $5/k$ and α β = $6/k$

Given, ${α}/{β} = 2/3$

⇒ α = $2/3$ β

∴ $2/3 β + β = 5/k and 2/3 β^2 = 6/k$

⇒ $5/3 β = 5/k and β^2 = 9/k$

β = $3/k and β^2 = 9/k$

⇒ $9/{k^2} = 9/k$

⇒ k = 1 and k ≠ 0. It is not satisfy the given condition.

Let the roots of equation,

$ax^2 + bx + c = 0$ are - α and $- 1/{α}$.

∴ (- α)$(- 1/{α}) = c/a$

⇒ 1 = $c/a$ ⇒ c = a

Given, $1/{a + b + x} = 1/a + 1/b + 1/x$

∴ $1/{a + b + x} - 1/x = 1/a + 1/b ⇒ {-(a + b)}/{(a + b + x)x} ={(a + b)}/{ab}$

⇒ $x^2$ + (a + b)x + ab = 0 ⇒ (x + a) (x + b) = 0

Hence, x = -a, -b.

Let the roots of equation be α and 2 α

Sum of root

- (2 α + α) = ${3a - 1}/{a^2 - 5a + 3}$

-3 α = ${3a - 1}/{a^2 - 5a + 3}$....

Squaring Both the side

$9α^2 = {(3a - 1)^2}/{(a^2 - 5a + 3)^2}$ ......(i)

Product of root

2 α × α = $2/{a^2 - 5a + 3}$

$α^2 = 1/{a^2 - 5a + 3}$

Dividing equation (i) from (ii)

${9 α^2}/{α^2} = {(3a - 1)^2}/{(a^2 - 5a + 3)^2} × {a^2 - 5a + 3}/1$

9 = ${(3a - 1)^2}/{(a^2 - 5a + 3)}$

$9a^2 - 45a + 27 = 9a^2 + 1 - 6a$

26 = 39a

a = $2/3$

Given, $log_{10} (x^2 - 6x + 45) = 2$

⇒ $(x^2 - 6x + 45) = 10^2 = 100$

⇒ $x^2$ - 6x - 55 = 0

⇒ $x^2$ - 11x + 5x - 55 = 0

⇒ x(x - 11) + 5(x - 11) = 0

⇒ (x + 5) (x - 11) = 0

∴ x = 11, -5.

Note: Let the quardatic equation be $ax^2$ + b + c = 0.

To find roots of this equation quickly, we find two factors of 'b' such that their sum is equal to b and their product is equal to the product of the coefficient of x2 and the constant term 'c'.

Let two such factors be α and β.

The α + β = b and α β = ca

In the second step, we divide these factors by the coefficient of $x^2$ ,

ie be 'a'.

In the next step, we change the signs of the outcome. These are the

roots of the equation.