# Algebra

Quantitative Aptitude Questions and Answers section on “Algebra” with solution and explanation for competitive examinations such as CAT, MBA, SSC, Bank PO, Bank Clerical and other examinations.
1.

If x – y = 2 and $x^{2}+y^{2}=20$, Then what would be the value of $(x+y)^{2}$ :
[A]38
[B]36
[C]16
[D]12

36
$\because \left ( x-y \right )^{2}=x^{2}+y^{2}-2xy$
$=> 2^{2}=20-2xy$
$=> 2xy=20-4=16..............(1)$
$\therefore (x+y)^{2}=x^{2}+y^{2}+2xy$
= 20 + 16 = 36
So option (B) is the right answer.

2.

If $x^{2}+y^{2}-4x-4y+8=0$, Then what would be the value of x-y:
[A]4
[B]-4
[C]0
[D]8

0
$x^{2}+y^{2}-4x-4y+8=0$
$=>x^{2}-4x+4+y^{2}-4y+4=0$
$=>\left ( x-2 \right )^{2}+\left ( y-2 \right )^{2}=0$
$=> x=2, y=2$
$\therefore x-y=0.$
So option [C] is the right answer.

3.

If $x-y=\frac{x+y}{7}=\frac{xy}{4}$, then what would be the value of xy:
[A]$\frac{4}{3}$
[B]$\frac{3}{4}$
[C]$\frac{1}{4}$
[D]$\frac{1}{3}$

$\frac{4}{3}$
$\because x-y=\frac{x+y}{7}=\frac{xy}{4}=K$
$=> x-y=K.............(1)$
$=> x+y=7K............(2)$
$=> xy=4K.................(3)$
$\because \left ( x+y \right ) ^{2}-\left ( x-y \right ) ^{2}=49K^{2}-K^{2}$
$=> 4xy=48K^{2}$
$=> 16K=48K^{2}$
$=> K=\frac{1}{3}$
$\therefore xy=4K=4\times \frac{1}{3}=\frac{4}{3}$
Hence option [A] is the right answer.

4.

If $x^{2}+y^{2}-2x+6y+10=0$, Then $\left ( x^{2}+y^{2} \right )$ is equals to :
[A]4
[B]6
[C]8
[D]10

10
$x^{2}-2x+y^{2}+6y+10=0$
$=> x^{2}-2x+1+y^{2}+6y+9=0$
$=> \left ( x-1 \right )^{2}+\left ( y+3 \right )^{2}=0$
$=> x=1, y=-3$
$\therefore x^{2}+y^{2}=1+9=10$
Hence option (D) is the right answer.

5.

If $a^{2}=2$, Then (a+1) is equals to :
[A]a – 1
[B]$\frac{2}{a-1}$
[C]$\frac{a+1}{3-2a}$
[D]$\frac{a-1}{3-2a}$

$\frac{a-1}{3-2a}$
$\therefore a^{2}=2$
$=>a=\sqrt{2}$
$=>a+1=\sqrt{2}+1$
Such type of questions can be solve quickly by using alternativ method so by the $4^{th}$ alternative
$=>\frac{a-1}{3-2a}=\frac{\sqrt{2}-1}{3-2\sqrt{2}}$
$=>\frac{\sqrt{2}-1}{3-2\sqrt{2}}\times \frac{3+2\sqrt{2}}{3+2\sqrt{2}}$
$=>\frac{3\sqrt{2}-3+4-2\sqrt{2}}{9-8}$ $= 1+\sqrt{2}$
$\therefore 1+\sqrt{2} = 1+\sqrt{a}$
Hence option [D] is correct.

6.

If x is a real number, Then what would be the minimum value of $\left ( x^{2}-x-1 \right )$:
[A]$\frac{3}{4}$
[B]0
[C]1
[D]$\frac{1}{4}$

$\mathbf{\frac{3}{4}}$
Because for expression $ax^{2}+bx+c$, a>0
Minimum value is : $\frac{4ac-b^{2}}{4a}$
Because here expression is : $x^{2}-x+1$
$\therefore a=1, b=-1, c=1$
Then Minimum value is : $\frac{4\times 1\times 1-1}{4\times 1\times 1}=\frac{3}{4}$
Hence option [A] is the right answer.

7.

If a * b = 2 (a + b), then 5 * 2 is equal to :
[A]20
[B]3
[C]10
[D]14

14
Given that
$a * b = 2 (a + b)$
$\therefore 5 * 2 = 2 (5 + 2)$
$= 2\times 7 = 14$
Hence option [D] is the rigth answer.

8.

If a * b = 2a – 3b + ab, then 3 * 5 + 5 * 3 is equal to :
[A]28
[B]26
[C]22
[D]24

22
Give that a * b = 2a – 3b + ab
$=> 3 * 5 = 2\times 3 - 3\times 5 + 3\times 5 = 6$
$=> 5 * 3 = 2\times 5 - 3\times 3 + 3\times 5 = 10 - 9 + 15 = 16$
Therefore, $3 * 5 + 5 * 3 = 6 + 16 = 22$
Hence option [C] is the rigth answer.

9.

Two numbers x and y (x > y) are such that their sum is equal to three times their difference. Then value of $\frac{3xy}{2\left ( x^{2} - y^{2} \right )}$ will be :
[A]$1\frac{2}{3}$
[B]1
[C]$\frac{2}{3}$
[D]$1\frac{1}{2}$

1
$\left ( x+y \right ) = 3 \left ( x-y \right ) = 3x - 3y$
$=> 3y + y = 3x - x$
$=> 2x = 4y$
$=> x = 2y$
$=> \frac{x}{y} = \frac{2}{1}$
$\therefore x = 2, y = 1$
$\frac{3xy}{2\left ( x^{2} - y^{2} \right )} = \frac{3 \times 2 \times 1}{2 \times \left ( 4 - 1 \right )} = \frac{6}{6} = 1$
Hence option [C] is the rigth answer.

10.

Find the value of
$\left ( 1+\frac{1}{x} \right ) \left ( 1+\frac{1}{x+1} \right )\left ( 1+\frac{1}{x+2} \right )\left ( 1+\frac{1}{x+3} \right )$:
[A]$\frac{1}{x}$
[B]$\frac{x+4}{x}$
[C]$1 + \frac{1}{x+4}$
[D]$x + 4$

$\mathbf{\frac{x+4}{x}}$
Given expression
$=> \left ( 1+\frac{1}{x} \right ) \left ( 1+\frac{1}{x+1} \right )\left ( 1+\frac{1}{x+2} \right )\left ( 1+\frac{1}{x+3} \right )$
$=> \frac{x+1}{x}\times \frac{x+2}{x+1}\times \frac{x+3}{x+2}\times \frac{x+4}{x+3}$
$=> \frac{x + 4}{x}$
Hence option [B] is the rigth answer.