## 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

We know that 112 = 121.

Putting m = 11 and n = 2, we get:

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

Earn Reward Points by submitting Detailed Explaination for this Question

## 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

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

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

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

= 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

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

(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

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
$$\left(x-\frac{1}{x}\right)^{2} = x+\frac{1}{x}-2$$
= (3 + 22) + $$\frac{1}{(3+22)}-2$$
= (3 + 22) + $$\frac{1}{(3+22)}-2\times \frac{{(3-22)}}{{(3-22)}}-2$$
Therefore $$\left(x-\frac{1}{x}\right) = 2.$$