Wipro Algebra Quiz – 1

p^{3}+q^{3}=(p+q)(p^{2}-pq +q^{2})

or, 9=3(p^{2}-pq+q^{2})

or, p^{2}-pq +q^{2}=3

or, p^{2}+q^{2}=3+pq ....(1)

Again (p+q)^{2}=3^{2}=9

or p^{2}+q^{2}=9-2pq........(2)

From (1) &(2), 3+pq =9-2pq

or, pq =2

Now p^{4}+q^{4}=(p^{2}+q^{2})^2-2.p^{2}.q^{2}

=(3+pq)^{2}-2(pq)^{2}

=(3+2)^{2}-2.(2^{2})

=25-8

=17

 

Answer: (A)

Given, e+ \frac{1}{2e} = 2

Multiplying through by 2e,

2e² + 1 = 4e …………………………………………………………………………………….(1)

Now,

8e^{3} + \frac {1}{e^{3}} = (2e)^{3} + (\frac{1}{e})^{3}

(Using the identity a³ + b³ = (a+b) (a² - ab + b²) for a=2e and b= \frac{1}{e})

= [(2e + \frac{1}{e})] [(2e)^{2} - 2e.\frac{1}{e} + (\frac{1}{e})^{2}]

= [\frac{(2e^{2} + 1)}{e}] [4e^{2} - 2 + \frac{1}{e²}]

= [\frac{(2e^{2} + 1)}{e}] [\frac{(4e^{4} - 2e^{2}+ 1)}{e^{2}}]

= [\frac{(2e² + 1)}{e}] [\frac{(2e^{2})² + 2. 2e².1 + 1² - 6e²)}{e^{2}}]

= [\frac{(2e² + 1)}{e}] [\frac{(2e² + 1)² - 6e²)}{e²}]

Replacing 2e² + 1  from  equation(1),

= (\frac{4e}{e}) [\frac{(4e)² - 6e²)}{e²}] = 4\frac{(16e² - 6e²)}{e²}

= 4 (\frac{10e²}{e²})

= 4 \times 10 = 40

Hence, the value of 8e³ + \frac{1}{e^{3}} = 40

Answer: (C)
Explanation:

Put the value on the equation: 4−6e−e²

If e=4

Then  4-24-16 = -36

If e=6

Then 4-36-36 = -68

If e=-3

Then 4+18-9 = 13

If e=7

Then 4-42-49 = -87

Hence, Maximum value of e = -3.

Answer: (D)


Here we are seeing a Pythagorean Triplet, where 5 units and 12 units can be the respective lengths of the sides of a triangle and 13 units is the length of hypotenuse.

In the equation 5√f+12√f=13√f,thevalueof√xshouldbe2.5√f+12√f=13√f,thevalueof √f should be 2.

Here we go,

.’. √f=2√f=2

By squaring both sides,

f = 4

OR

Put the value on the equation:

If f=4

Then we get 169 that is square of 13.

Hence, f=4.

E^{2} +E^{3}

= 2^{2} +2^{3}

= 12

Explanation:

f^{3} + \frac{1}{f^{3}}

= (f +\frac{1}{f})(f^{2} - 1 +\frac{1}{f^{2}})

= (f +\frac{1}{f}) \times 0

= 0

Answer: 

p^{\frac{1}{3}}+ q^{\frac{1}{3}} – r^{\frac{1}{3}} =0

=> p^{\frac{1}{3}}+ q^{\frac{1}{3}} = r^{\frac{1}{3}}

cubing both sides

=> (p^{\frac{1}{3}}+ q^{\frac{1}{3}})^{3} = r

=> {p+q+3(p^\frac{1}{3}q^\frac{1}{3})(p^\frac{1}{3}+q^\frac{1}{3})} = r

=>p+q+3p^{\frac{1}{3}}q^{\frac{1}{3}}r^{\frac{1}{3}} = r

=>p+q-r= -3p^{\frac{1}{3}}q^{\frac{1}{3}}r^{\frac{1}{3}}

Again cubing both sides

=> (p + q – r)^{3} = - 27pqr

so (p + q – r)^{3} + 27pqr = 0

Answer: (A)
Explanation: Let the number be x. Then,
\frac{7}{23} of \frac{11x}{9} = \frac{7}{21} of \frac{3x}{8} + 12
\frac{7}{23} of  \frac{11x}{9} = \frac{7}{21} of \frac{ 3x}{8} + 12
\frac{77x}{207} = \frac{21x}{168} + 12
x[(\frac{77}{207}) - (\frac{21}{168})] = 12

\frac{(12936-4347)x}{34776}  = 12

\frac{8589x}{34776} = 12

x(0.25) = 12

x = 48

Answer: (B)
Explanation:

\frac{2h}{1+\frac{1}{1+\frac{h}{1-h}}}= 1

\frac{2h}{1+\frac{1}{\frac{1-h+h}{1-h}}} = 1

\frac{2h}{ 1+1-h}= 1

\frac{2h}{2-h} = 1

2h = 2-h

3h=2

h=\frac{2}{3}

Answer: (B)
Explanation: We have
(1-\frac{1}{7})(1-\frac{1}{8})(1-\frac{1}{9})…(1-\frac{1}{100})
= (\frac{6}{7})(\frac{7}{8})(\frac{8}{9})…(\frac{99}{100})
= \frac{6}{100}

= \frac{3}{50}