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.



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