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Question Detail
Difference between a two-digit number and the number obtained by interchanging the two digits is 36, what is the difference between two numbers
- 2
- 4
- 8
- 12
Answer: Option B
Explanation:
Let the ten digit be x, unit digit is y.
Then (10x + y) - (10y + x) = 36
=> 9x - 9y = 36
=> x - y = 4.
1. if the sum of \begin{aligned} \frac{1}{2} \end{aligned} and \begin{aligned} \frac{1}{5} \end{aligned} of a number exceeds \begin{aligned} \frac{1}{3} \end{aligned} of the number by \begin{aligned} 7\frac {1}{3} \end{aligned}, then number is
- 15
- 20
- 25
- 30
Answer: Option B
Explanation:
Seems a bit complicated, isnt'it, but trust me if we think on this question with a cool mind then it is quite simple...
Let the number is x,
then, \begin{aligned} (\frac{1}{2}x + \frac{1}{5}x) - \frac{1}{3}x = \frac{22}{3} \end{aligned}
\begin{aligned}
=> \frac{11x}{30} = \frac{22}{3}
\end{aligned}
\begin{aligned}
=> x = 20
\end{aligned}
2. Find the number, when 15 is subtracted from 7 times the number, the result is 10 more than twice of the number
- 5
- 15
- 7.5
- 4
Answer: Option A
Explanation:
Let the number be x.
7x -15 = 2x + 10 => 5x = 25 => x = 5
3. The sum of the squares of three numbers is 138, while the sum of their products taken two at a time is 131. Their sum is
- 15
- 20
- 25
- 35
Answer: Option B
Explanation:
Let the numbers be a, b and c.
Then,
\begin{aligned}
a^2 + b^2 + c^2 = 138
\end{aligned}
and (ab + bc + ca) = 131
\begin{aligned}
(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)
\end{aligned}
= 138 + 2 x 131 = 400
\begin{aligned}
=> (a + b + c) = \sqrt{400} = 20.
\end{aligned}
4. find the number, difference between number and its 3/5 is 50.
- 120
- 123
- 124
- 125
Answer: Option D
Explanation:
Let the number = x,
Then, x-(3/5)x = 50,
=> (2/5)x = 50 => 2x = 50*5,
=> x = 125
5. Sum of a rational number and its reciprocal is 13/6. Find the number
- 2
- 3/2
- 4/2
- 5/2
Answer: Option B
Explanation:
\begin{aligned} => x + \frac{1}{x} = \frac{13}{6} \end{aligned}
\begin{aligned} => \frac{x^2+1}{x} = \frac{13}{6} \end{aligned}
\begin{aligned} => 6x^2-13x+6 = 0 \end{aligned}
\begin{aligned} => 6x^2-9x-4x+6 = 0 \end{aligned}
\begin{aligned} => (3x-2)(2x-3) \end{aligned}
\begin{aligned} => x = 2/3 or 3/2 \end{aligned}
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