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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. A number is doubled and 9 is added. If resultant is trebled, it becomes 75. What is that number
- 8
- 10
- 12
- 14
Answer: Option A
Explanation:
=> 3(2x+9) = 75
=> 2x+9 = 25
=> x = 8
2. Find a positive number which when increased by 17 is equal to 60 times the reciprocal of the number
- 17
- 15
- 8
- 3
Answer: Option D
Explanation:
If the number is x,
Then, x + 17 = 60/x
x2 + 17x - 60 = 0
(x + 20)(x - 3) = 0
x = 3, -20, so x = 3 (as 3 is positive)
3. The product of two numbers is 120 and the sum of their squares is 289. The sum of the number is
- 20
- 23
- 27
- 150
Answer: Option B
Explanation:
We know
\begin{aligned}
(x + y)^2 = x^2 + y^2 + 2xy
\end{aligned}
\begin{aligned}
=> (x + y)^2 = 289 + 2(120)
\end{aligned}
\begin{aligned}
=> (x + y) = \sqrt{529} = 23
\end{aligned}
4. 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
5. 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}
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