IR Question with Quant Twist

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IR Question with Quant Twist

by srcc25anu » Tue Apr 30, 2013 4:16 am
f(x) is a function such that f(x) = s(x) + p(x) , where s(x) is the sum of all the digits of x and p(x) is the product of all the digits of x .

Identify one x-value that, entered into function f(x) , would yield an appropriate result. Select only one answer in each column.

Options for x-value:
A. 48
B. 39
C. 24
D. 256
E. 68

Options for result:
A. 48
B. 39
C. 24
D. 256
E. 68

[spoiler]OA: B and B[/spoiler]

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by Lifetron » Wed May 01, 2013 12:25 am
I can't get the twist here. It is jus trial n error, right?

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by srcc25anu » Wed May 01, 2013 10:55 am
gughanbose wrote:I can't get the twist here. It is jus trial n error, right?
Twist was meant to say that although this is an IR question, I am posting it on Quant forum.

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by Matt@VeritasPrep » Wed May 01, 2013 8:14 pm
Where's this question from? If I didn't know better I could swear I wrote it :D

To give a bit of a hint to get us started, f(x) is just the sum of two different operations: one that takes the sum of x and another that takes the product of x. f(57), for example, would look like

f(57) = (5+7) + (5*7) = 12 + 35 = 47

because the sum of the digits of 57 is 12 and the product of the digits is 35. (That s(x) and p(x) stuff is likely there just to perplex.) Beyond that, this question just seems to test how quickly you can compute various options, how shrewdly you can size up which ones to test first, and how brave you are about picking a certain set of answers (that last remark will make sense once you reveal the OA).

If we get bored with this one, let me know - I've got a slew of other q's that test the same concept. Very often these questions test the implications of a number's remainder when divided by 3 or the notion of a digital sum, though this one doesn't seem to.

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by srcc25anu » Wed May 01, 2013 8:19 pm
this was from veritas practice questions.

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by Matt@VeritasPrep » Wed May 01, 2013 8:46 pm
srcc25anu wrote:this was from veritas practice questions.
Ha! My memory hasn't completely failed - I did write this. (Though what I was thinking when I wrote it is still unknown ...)