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Number N is randomly selected from a set of all primes betwe

by BTGmoderatorAT » Sun Nov 05, 2017 2:53 pm
Number N is randomly selected from a set of all primes between 10 and 40, inclusive. Number K is selected from a set of all multiples of 5 between 10 and 40 inclusive. What is the probability that N+K is odd?

(A) 1/2
(B) 2/3
(C) 3/4
(D) 4/7
(E) 5/8

Is there a strategic approach to this question? Can any experts help?
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Answer

by EconomistGMATTutor » Mon Nov 06, 2017 7:41 am
Hello ardz24.

First, you have to notice that N is always an odd number. Then, if you want that N+K is odd, then K has to be even (because N is odd). (it doesn't matter how we select N).

Now, K has to be selected from {10, 15, 20, 25, 30, 35, 40}. (7 options).

So, the favorable cases are k=10, 20, 30, 40. (4 options).

The probability that N+K is odd is equal to:

$$\frac{Favorable\ options}{Total\ options}=\frac{4}{7}.$$

So, the answer should be D.

I hope this can help you.

I'm available if you'd like any follow up.

Regards.
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by Brent@GMATPrepNow » Fri Nov 10, 2017 9:50 am
ardz24 wrote:Number N is randomly selected from a set of all primes between 10 and 40, inclusive. Number K is selected from a set of all multiples of 5 between 10 and 40 inclusive. What is the probability that N+K is odd?

(A) 1/2
(B) 2/3
(C) 3/4
(D) 4/7
(E) 5/8
Notice that there are 2 ways that the sum N+K can be ODD
1) N is ODD and K is EVEN
2) N is EVEN and K is ODD

However, N cannot be EVEN, since all of the primes between 10 and 40 are ODD (11, 13, 17, 19, . . . . 31, 37)
So, case 2 (above) is impossible

So, P(N+K is odd) = P(N is odd AND K is even)
= P(N is odd) x P(K is even)

------ASIDE-------
Possible values of N: (11, 13, 17, 19, . . . . 31, 37)
So, P(N is odd) = 1

Possible values of K: (10, 15, 20, 25, 30, 35, 40)
So, P(K is even) = 4/7
-----------------------------------------

So, P(N+K is odd) = P(N is odd AND K is even)
= P(N is odd) x P(K is even)
= 1 x 4/7
= 4/7
= D

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by Scott@TargetTestPrep » Tue Jan 09, 2018 10:12 am
ardz24 wrote:Number N is randomly selected from a set of all primes between 10 and 40, inclusive. Number K is selected from a set of all multiples of 5 between 10 and 40 inclusive. What is the probability that N+K is odd?

(A) 1/2
(B) 2/3
(C) 3/4
(D) 4/7
(E) 5/8
In order for N + K to be odd, we need an even + odd or odd + even.

Since N is a prime between 10 and 40, N must be odd.

Thus, we need to determine how many even multiples of 5 there are from 10 to 40 inclusive. We have 10, 20, 30, and 40. We also have (40 - 10)/5 + 1 = 7 total multiples of 5 from 10 to 40 inclusive.

Thus, the probability that N is odd and K is even is 1 x 4/7 = 4/7.

Answer: D

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by [email protected] » Tue Jan 09, 2018 12:10 pm
Hi ardz24,

We're told that number N is randomly selected from a set of all PRIMES between 10 and 40, inclusive and number K is selected from a set of all multiples of 5 between 10 and 40 inclusive. We're asked for the probability that N+K is ODD. For the SUM to be ODD, one of the numbers must be EVEN and the other must be ODD. Since all of the primes from 10 to 40 are ODD, K MUST be EVEN... sowe're ultimately asked for the probability that K is EVEN...

Multiples of 5 from 10 and 40: 10, 15, 20, 25, 30, 35, 40 ---7 numbers
Even numbers in this set: 10, 20, 30, 40 -- 4 numbers

The probability that K is EVEN (and thus, N+K is ODD) = 4/7

Final Answer: D

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