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# Algebra Aptitude Questions

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.