InfyTQ Algebra Quiz 1

Correct option is d

 

Sol:

 

x³+y³ = (x + y) × (x² − xy + y²)

 

Putting given values of x³+y³ and (x + y)

18 = 6 × ((x+y)² − 3xy)

18 = 6 × (36 − 3xy)

18  = 216 − 18xy

 

18xy = 198

xy = 11

 

x^{4} +y^{4}= (x^{2} + y^{2})^{2} - 2x^{2}y^{2}

= (x² + y²)² - 2 x 121

[Putting the value of xy]

= ((x + y)² - 2xy)² - 2 x 121

[Putting values of (x+y) and xy]

= (36 - 2 x 11)² - 2 x 121

= -46

The correct option is D.

Sol:

First Faulty JCB = 1200 x (2 \frac{1}{3}%)

=1200 \times \frac{7}{3 \times 100}= 28

Defective JCB = 1200 \times \frac{13}{3 \times 100}= 52

So total defective and faulty JCB are 52+28=80

Therefore non defective and non-faulty JCB are 1200-(52+28) = 1120

So the final answer to this question is Option D: 1120

Sol:

\frac{6}{4} + \frac{31}{6} \times \frac{3}{7} = \frac{104}{28}

5 + \frac{7}{3} = \frac{22}{3}

\frac{\frac{104}{28}}{\frac{22}{3}} = \frac{39}{77}

\frac{39}{77}- \frac{38}{77} = \frac{1}{77}

The correct option is c.

Sol:(\frac{3}{10})^{-8} \times (\frac{10}{3})^{-2}=(\frac{3}{10})^{x}


Now, (\frac{3}{10})^{-8} \times (\frac{3}{10})^{2}=(\frac{3}{10})^{x}


on solving,

(\frac{3}{10})^{-6}= \frac{3}{10})^{x}

Thus, x = -6.

The correct option is c.

Sol:

=>\frac{(x^{2} - y^{2})}{ (x-y)^{2}}= 4

=>\frac{(x-y)(x+y)}{(x-y)^{2}} =4

=>\frac{(x+y)}{(x-y)}=4

substitute x=8,

=>\frac{(8+y)}{(8-y)}=4

=>8+y=4(8-y)

=>8+y=32-4y

=>5y= 32 - 8

y=\frac{24}{5}

Let \frac{0.006}{x} = 0.01

Then x =\frac{0.006}{0.01}=0.6

Sol:

he gets 100B+A paise instead of 100A+B paise

 

(100B + A) - (100A + B) - 5 = (100A+B)

 

99B -99A - 5 = 100A + B

 

98B = 199A + 5

 

B = 2A +\frac{ (3A+5)}{98}

 

only an integer value of A = 31

will make B an integer

with the constraints that both A & B < 100

substituting, B = 2 x 31 + 1 = 63

 

ans: 31 Rs 63 paise

The correct option is b.

 

Sol:

 

Let initially X sugar canes were present there, and it is increasing by Y sugar cane per day, then for the first condition We get,

X+24Y = 24 x 70 ----(1)

For the 2nd condition, we have,

X+60Y = 60 x 30----(2) Now, On solving equation (1) and (2), we get

X = 1600 and

Y =\frac{ 10}{3}

Third Condition,

X+96Y = 96 x N -----(3) [N = Number of workers required]

Putting the values of X and Y in equation (3), We get

1600 + 96 x  (\frac{10}{3}) = 96 N

⇒N = (\frac{1920}{96})

N = 20.

Correct option is b.

 

Sol:

(\frac{ x }{ y})^{a-1} = (\frac{ y }{ x})^{a-3}

(\frac{ y }{ x})^{1-a} = (\frac{ y }{ x})^{a-3}

1-a=a-3

a=2

The correct option is b.

 

Sol:

 

As per the equation given,

 

(1) a+b = x

 

(2) b+c=y

 

(3) c+a = z

 

adding (1), (2) and (3)

 

a+ b + b + c + c + a = x + y + z

 

2a + 2b + 2c = x + y + z

 

2 (a + b + c) = x + y + z

 

Here, a + b + c = n

 

Hence, 2 n = x + y + z