Question 1
If $$5^{a} = 9^{b} = 2025$$, then the value of $$\frac{ab}{a + b}$$ = _____.
CMAT Logarithms, Surds and Indices Questions
CMAT Logarithms, Surds and Indices Questions
Solution
It is given,
$$5^a=9^b=2025$$
$$5^a=2025$$
Applying log on both the sides, we get
$$\log_55^a=\log_52025$$
$$a=\log_52025$$
Similarly, we get $$b=\log_92025$$
$$\dfrac{ab}{a+b}=\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}$$
$$\dfrac{1}{a}=\log_{2025}5$$
$$\dfrac{1}{b}=\log_{2025}9$$
$$\dfrac{1}{a}+\dfrac{1}{b}=\log_{2025}45$$
$$\dfrac{ab}{a+b}=\log_{45}2025\ =\ 2$$
The answer is option C.
Question 2
Arrange the following numbers in increasing order :
A. $$\sqrt{59} - \sqrt{51}$$
B. $$\sqrt{37} - \sqrt{29}$$
C. $$\sqrt{87} - \sqrt{79}$$
D. $$\sqrt{79} - \sqrt{71}$$
Choose the correct answer from the options given below:
Solution
8 = $$\left(\sqrt{59}-\sqrt{51}\right)\left(\ \sqrt{\ 59}+\sqrt{51}\right)$$
8 = $$\left(\sqrt{37}-\sqrt{29}\right)\left(\ \sqrt{37}+\sqrt{29}\right)$$
We know that, $$\sqrt{59}+\sqrt{51}$$ is greater than $$\sqrt{37}+\sqrt{29}$$
This implies, $$\sqrt{37}-\sqrt{29}$$ is greater than $$\sqrt{59}-\sqrt{51}$$ as the product is constant.
Similarly, $$\sqrt{59}-\sqrt{51}$$ is greater than $$\sqrt{79}-\sqrt{71}$$ and $$\sqrt{79}-\sqrt{71}$$ is greater than $$\sqrt{87}-\sqrt{79}$$.
The correct order is B > A > D > C
The answer is option D.
Question 3
Given below are two statements:
Statement I: $$(243)^{0.16} \times (9)^{0.1} = 0.3$$
Statement II: If $$3.105 \times 10^{P} = 0.00239 + 0.000715$$, then $$P = -3$$
In the light of the above statements choose the most appropriate answer from the options given below:
Solution
Statement I:
$$\left(243\right)^{0.16}\times\left(9\right)^{0.1}=3^{0.8}\times3^{0.2}=3$$
Therefore, statement I is incorrect.
Statement II:
$$3.105\times10^P=0.00239+0.000715$$
$$3.105\times10^P=0.003105$$
$$3105\times10^P=3.105$$
P = -3
Therefore, statement II is correct.
The answer is option D.
