Quantitative Aptitude
SURDS AND INDICES MCQs
Surds & Indices, Indices And Surds, Power
Givan \(\left(\frac{a}{b}\right)^{x-1} = \left(\frac{b}{a}\right)^{x-3}\)
\(\left(\frac{a}{b}\right)^{x-1} = \left(\frac{a}{b}\right)^{-(x-3)}= \left(\frac{a}{b}\right)^{(3-x)}\)
x - 1 = 3 - x
2x = 4
x = 2.
xz = y2 \(\Leftrightarrow \) 10(0.48z) = 10(2 x 0.70) = 101.40
0.48z = 1.40
z = \(\frac{140}{48}\) = \(\frac{35}{12}\) = 2.9 (approx.)
5a = 3125 \(\Leftrightarrow \) 5a = 55
a = 5.
So, 5(a - 3) = 5(5 - 3) = 52 = 25.
3x - y = 27 = 33 \(\Leftrightarrow \) x - y = 3 ....(i)
3x + y = 243 = 35 \(\Leftrightarrow \) x + y = 5 ....(ii)
On solving (i) and (ii), we get x = 4.
(256)0.16 x (256)0.09 = (256)(0.16 + 0.09)
= (256)0.25
= \((256)^{\left(\frac{25}{100}\right)}\)
= \((256)^{\left(\frac{1}{4}\right)}\)
= \((4^{4})^{\left(\frac{1}{4}\right)}\)
= \(4^{4\left(\frac{1}{4}\right)}\)
= 41
= 4
(10)150 ÷ (10)146 = \(\frac{10^{150}}{10^{146}}\)
= 10150 - 146
= 104
= 10000.
Given Exp. = \(\frac{1}{\left(1+\frac{x^{b}}{x^{a}}+\frac{x^{c}}{x^{a}}\right)} +\frac{1}{\left(1+\frac{x^{a}}{x^{b}}+\frac{x^{c}}{x^{b}}\right)}+\frac{1}{\left(1+\frac{x^{b}}{x^{c}}+\frac{x^{a}}{x^{c}}\right)}\)
= \(\frac{x^{a}}{(x^{a}+x^{b}+x^{c})}+\frac{x^{b}}{(x^{a}+x^{b}+x^{c})}+\frac{x^{c}}{(x^{a}+x^{b}+x^{c})}\)
= \(\frac{(x^{a}+x^{b}+x^{c})}{(x^{a}+x^{b}+x^{c})}\)
= 1
Let (25)7.5 x (5)2.5 ÷ (125)1.5 = 5x.
Then, \(\frac{(5^{2})^{7.5}\times (5)^{2.5}}{(5^{3})^{1.5}}\) =5x
\(\frac{5^{(2\times7.5)}\times 5^{2.5}}{5^{(3\times1.5)}}\) = 5x
\(\frac{5^{15}\times5^{2.5}}{5^{4.5}}\) = 5x
5x = 5(15 + 2.5 - 4.5)
5x = 513
Therefore x = 13.
(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.
The given expression is (0.04)^-1.5. We can simplify this expression using the following formula:
a^(-n) = 1/(a^n)
where a is a non-zero real number and n is a positive integer.
Using this formula, we get:
(0.04)^-1.5 = 1/(0.04)^1.5
Now, we can simplify the expression inside the parentheses using the following formula:
a^n = (a^m)^(n/m)
where a is a non-zero real number and m and n are integers.
Using this formula with a=0.04, n=3, and m=2, we get:
(0.04)^1.5 = (0.04^2)^(3/2) = 0.0016^(3/2) = 0.000064
Therefore,
(0.04)^-1.5 = 1/(0.04)^1.5 = 1/0.000064 = 15625/1 = 125
Hence, the correct answer is option B, 125.
In summary, we used the formula a^(-n) = 1/(a^n) to convert the negative exponent to a positive exponent, and then used the formula a^n = (a^m)^(n/m) to simplify the expression inside the parentheses. Finally, we obtained the answer by taking the reciprocal of the simplified expression.
If you think the solution is wrong then please provide your own solution below in the comments section .
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.