Question Detail
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
1. If one third of one fourth of number is 15, then three tenth of number is
- 34
- 44
- 54
- 64
Answer: Option C
Explanation:
Let the number is x,
\begin{aligned}
\frac{1}{3} of\frac{1}{4} * x = 15
\end{aligned}
\begin{aligned}
=> x = 15\times 12 = 180
\end{aligned}
\begin{aligned}
=> so \frac{3}{10} \times x = \frac{3}{10} \times 180 = 54
\end{aligned}
2. Three times the first of three consecutive odd integers is 3 more than twice the third. The third integer is
- 12
- 13
- 15
- 17
Answer: Option C
Explanation:
Let the three integers be x, x+2 and x+4.
Then, 3x = 2(x+4)+3,
x= 11
Therefore, third integer x+4 = 15
3. 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}
4. 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}
5. Sum of two numbers is 25 and their difference is 13. Find their product.
- 104
- 108
- 114
- 124
Answer: Option C
Explanation:
Friends, this sort of question is quite important in competitive exams, whenever any question come which have relation between sum, product and difference, this formula do the magic:
\begin{aligned}
=> (x+y)^2 = (x-y)^2 + 4xy
\end{aligned}
\begin{aligned}
<=> (25)^2 = (13)^2 + 4xy
\end{aligned}
\begin{aligned}
<=> 4xy = (25)^2 - (13)^2
\end{aligned}
\begin{aligned}
<=> xy = \frac{456}{4} = 114
\end{aligned}