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Aptitude - Problems on Numbers - Discussion

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

Let the number be x.

Then, x + 17 = 60
x

x2 + 17x - 60 = 0

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

x = 3.

Discussion:
21 comments Page 2 of 3.
Hariteja said:   1 decade ago
Check from the options. It will be easy.
(1)
Professor tajinder said:   1 decade ago
@Revathy

See we have (x-3)(x+20)=0

Now x-3=0
x=3

x+20=0
x=-20

But we have to find the postive no, so the answer is 3.

Hope you will undertand it.
(1)
Sirisha said:   1 decade ago
You substitute the options in the place of x. Then we get LHS=RHS.
Sindhu said:   1 decade ago
Reciprocal means if you take x reciprocal of x is 1/x.
Pankaj said:   1 decade ago
I'm not understand reciprocal means in this question?
Joe said:   1 decade ago
Let the number be x but since we have 3,17,10 and 20 as the numbers but we don't know which is the correct one the here is the Equation,

x+17 = 60/x.
So am using 3 as my first number.

3(3+17) = 60/3*3.
3(20) = 20*3.

60 = 60.

Hence the Answer is 3.
Rakesh said:   1 decade ago
Find a positive number which when increased by 17 is equal to 60 times the reciprocal of the number.

But here you're not taking reciprocal of number to solve your just taking x.
Rajeev said:   1 decade ago
@Ravi we have to subdivide the equation.

We have x2 + 17x - 60 = 0,

Now, x2+20x-3x-60 = 0 (we can write 17x as (20x-3x)).

x(x+20)-3(x+20).

Then (x-3)(x+20) = 0.
Raam said:   6 years ago
What if the number of times is 59 rather than 60? Please tell me.
Ravi said:   1 decade ago
How we got (x-3)(x+20)?
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