Lakshya Education MCQs

Question:

If m and n are whole numbers such that mn = 121, the value of (m - 1)n + 1 is:

Options:
A.1
B.10
C.121
D.1000
Answer: Option D

We know that 112 = 121.

Putting m = 11 and n = 2, we get:

(m - 1)n + 1 = (11 - 1)(2 + 1) = 103 = 1000.

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More Questions on This Topic :

Question 1.

\(\frac{1}{1+a^{(n-m)}}+\frac{1}{1+a^{(m-n)}} = ?\)

Options:
  1.    0
  2.    \(\frac{1}{2}\)
  3.    1
  4.    am + n
Answer: Option C

\(\frac{1}{1+a^{(n-m)}}+\frac{1}{1+a^{(m-n)}} = \frac{1}{\left(1+\frac{a^{n}}{a^{m}}\right)} + \frac{1}{\left(1+\frac{a^{m}}{a^{n}}\right)}\)

\(\frac{a^{m}}{\left(a^{m}+a^{n}\right)}+\frac{a^{n}}{\left(a^{m}+a^{n}\right)}\)

= \(\frac{{\left(a^{m}+a^{n}\right)}}{{\left(a^{m}+a^{n}\right)}}\)

= 1.

Question 2.

\(\frac{\left(243\right)^{\frac{n}{5}}\times3^{2n+1}}{9^{n}\times3^{n-1}}= ?\)

Options:
  1.    1
  2.    2
  3.    9
  4.    3n
Answer: Option C

Given Expression = \(\frac{\left(243\right)^{\frac{n}{5}}\times3^{2n+1}}{9^{n}\times3^{n-1}}\)

= \(\frac{\left(3^{5}\right)^{\left(\frac{n}{5}\right)} \times3^{2n+1}}{\left(3^{2}\right)^{n}\times3^{n-1}}\)

= \(\frac{\left(3^{5\times(\frac{n}{5})}\times3^{2n-1}\right)}{\left(3^{2n}\times3^{n-1}\right)}\)

= \(\frac{3^{n}\times3^{2n-+1}}{3^{2n}\times3^{n-1}}\)

= \(\frac{3^{(n+2n+1)}}{3^{(2n+n-1)}}\)

= \(\frac{3^{3n+1}}{3^{3n-1}}\)

= 3(3n + 1 - 3n + 1)   

= 32   = 9.

 

Question 3.

(0.04)-1.5 = ?

Options:
  1.    25
  2.    125
  3.    250
  4.    625
Answer: Option B

(0.04)-1.5 =  \(\left(\frac{4}{100}\right)^{-1.5}\)

\(\left(\frac{1}{25}\right)^{-(\frac{3}{2})}\)

= \(\left(25\right)^{\left(\frac{3}{2}\right)}\)

= \(\left(5^{2}\right)^{\left(\frac{3}{2}\right)}\)

= \(\left(5\right)^{2\times\left(\frac{3}{2}\right)}\)

= 53

= 125.

Question 4.

\(\left(\frac{x^{b}}{x^{c}}\right)^{(b+c-a)}.\left(\frac{x^{c}}{x^{a}}\right)^{(c+a-b)}. \left(\frac{x^{a}}{x^{b}}\right)^{(a+b-c)} = ?\)

Options:
  1.    xabc
  2.    1
  3.    xab + bc + ca
  4.    xa + b + c
Answer: Option B

Given Exp. = \(x^{(b-c)(b+c-a)}.x^{(c-a)(c+a-b)}.x^{(a-b)(a+b-c)}\) 

= \( x^{(b-c)(b+c)-a(b-c)}. x^{(c-a)(c+a)-b(c-a)}. x^{(a-b)(a+b)-c(a-b)}\)

= \(x^{(b^{2}-c^{2}+c^{2}-a^{2}+a^{2}-b^{2})} . x^{-a(b-c)-b(c-a)-c(a-b)}\)

= \(\left(x^{0}\times x^{0}\right)\)

= \(\left(1\times1\right) = 1\)

 

 

Question 5.

If x = 3 + 22, then the value of  \(\left(x-\frac{1}{x}\right)\) is :

Options:
  1.    1
  2.    2
  3.    22
  4.    33
Answer: Option B

\(\left(x-\frac{1}{x}\right)^{2} = x+\frac{1}{x}-2\)

= (3 + 22) + \(\frac{1}{(3+22)}-2\)

= (3 + 22) + \(\frac{1}{(3+22)}-2\times \frac{{(3-22)}}{{(3-22)}}-2\)

= (3 + 22) + (3 - 22) - 2

   = 4.

Therefore \(\left(x-\frac{1}{x}\right) = 2.\)