Aptitude - Problems on Numbers - Discussion

Discussion Forum : Problems on Numbers - General Questions (Q.No. 10)
10.
Find a positive number which when increased by 17 is equal to 60 times the reciprocal of the number.
3
10
17
20
Explanation:

Let the number be x.

 Then, x + 17 = 60 x

x2 + 17x - 60 = 0

(x + 20)(x - 3) = 0

x = 3.

Discussion:
19 comments Page 1 of 2.

Anitha Karre said:   4 years ago
In given options, suppose a positive number is 10.

Which is added by 17. i.e10+17=27 by a given question which is equal to 60. So see in the units place of both numbers (27,60) '7' & '0' which is not equal.

Suppose 3 is +ve number added by 17 gives 20 which is equal to 60/3=20.
So the positive number is 3.

Raam said:   6 years ago
What if the number of times is 59 rather than 60? Please tell me.

Debrupa said:   6 years ago
Reciprocal of x = 1/x.

Here, the equation should be ->
x+17=60*(1/x)
=> x=3,-20.

But, as we have to calculate +ve no only, so answer will be 3.
(3)

Balag said:   8 years ago
5 is wrong. Because 12 * 5 = 60. Not -60. So it is wrong.
(2)

Steven Than said:   8 years ago
Yes, I agree @Kabali.

(2)

Denzel said:   8 years ago
How did you get x2 + 17x - 60 ?
(2)

Kabali said:   8 years ago
x2 + 12x + 5x - 60 = 0 also solution.
x(x + 12) - 5(x + 12) = 0.
Hence x = 5 is also the answer.
(2)

Kausik said:   9 years ago
How you come to the assumption of x2+20x-3x-60 = 0 (we can write 17x as (20x-3x))?

20-3 = 17 is one of the assumptions for the result of 17, but how this can be narrowed to the right assumption?
(3)

Naresh said:   9 years ago