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

 

x⁴ + y⁴ = (x² + y²)2 - 2x²y²

= (x² + y²)² - 2*121

[Putting the value of xy]

= ((x + y)² - 2xy)² - 2*121

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

= (36 - 2*11)² - 2*121

= -46

The correct option is D.

Sol:

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

=1200 \times 2 \frac{7}{3 \times 100}= 28

Defective JCB = 1200 \times 2 \frac{13}{3 \times 100}= 53

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: (3/10)^-8*(10/3)^-2=(3/10)^x

Now, (3/10)^-8*(3/10)²=(3/10)^x

on solving,

(3/10)^-6=(3/10)^x

(3/10)^-6=(3/10)^x

Thus, x=-6.

The correct option is c.

Sol:

=>(x² -y² )/(x-y)² =4

=>(x-y)(x+y)/(x-y)² =4

=>(x+y)/(x-y)=4

substitute x=8,

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

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

=>8+y=32-4y

=>5y= 32 - 8

=>y=24/5

Let .006/x= .01;     Then x =.006/0.01=.6= .6/1 = 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 + (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*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+24*y = 24*70 ----(1)

For the 2nd condition, we have,

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

X = 1600 and

Y = 10 /3

Third Condition,

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

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

1600 + 96 * (10/3) = 96 * N

⇒N = (1920/96)

N = 20.

Correct option is b.

 

Sol:

( x / y )^(a-1) = ( y / x )^(a-3)

(y/x)^(1-a)=(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