# TCS NQT Ninja and Digital aptitude questions - 29

### 1. The perimeter of a equilateral triangle and regular hexagon are equal.  Find out the ratio of their areas?

a. 3:2
b. 2:3
c. 1:6
d. 6:1
Correct Option: b
Explanation:
Let the side of the equilateral triangle = $a$ units and side of the regular hexagon is $b$ units.
Given that,  $3a = 6b$ $\Rightarrow \dfrac{a}{b} = \dfrac{2}{1}$
Now ratio of the areas of equilateral triangle and hexagon = $\dfrac{{\sqrt 3 }}{4}{a^2} : \dfrac{{3\sqrt 3 }}{2}{b^2}$
$\Rightarrow \dfrac{{\sqrt 3 }}{4}{\left( 2 \right)^2} : \dfrac{{3\sqrt 3 }}{2}{\left( 1 \right)^2}$
$\Rightarrow 2:3$

### 2. What is the remainder of (32^31^301) when it is divided by 9?

a. 3
b. 5
c. 2
d. 1
Correct option: b
Explanation:
See solved example 6 here
$\dfrac{{{{32}^{{{31}^{301}}}}}}{9}$ = $\dfrac{{{5^{{{31}^{301}}}}}}{9}$
Euler totient theorem says that ${\left[ {\dfrac{{{a^{\phi (n)}}}}{n}} \right]_{{\mathop{\rm Re}\nolimits} m}} = 1$
$\phi (n) = n\left( {1 - \dfrac{1}{a}} \right)\left( {1 - \dfrac{1}{b}} \right)...$ here $n = {a^p}.{b^q}...$
Now $\phi (9) = 9\left( {1 - \dfrac{1}{3}} \right) = 6$
Therefore, ${5^6}$ when divided by 9 remainder 1.
Now $\dfrac{{{{31}^{301}}}}{6} = {1^{301}} = 1$
So ${{{31}^{301}}}$ can be written as 6k + 1
$\Rightarrow {5^{{{31}^{301}}}} = {\left( {{5^6}} \right)^K}{.5^1}$
$\dfrac{{{5^{{{31}^{301}}}}}}{9} = \dfrac{{{{\left( {{5^6}} \right)}^K}{{.5}^1}}}{9} = \dfrac{{{1^K}.5}}{9} = 5$

### 3. Which of the following numbers must be added to 5678 to give a reminder 35 when divided by 460?

a. 980
b. 797
c. 955
d. 618
Correct option: b
Explanation:
Let $x$ be the number to be added to 5678.
When you divide 5678 + $x$ by 460 the remainder = 35.
Therefore, 5678 + $x$ = 460k + 35 here $k$ is some quotient.
$\Rightarrow$ 5643 + $x$ should exactly divisible by 460.
Now from the given options x = 797.

Total = 34.