Andrew had several shares of stock at the beginning of Monday. During

Question

Andrew had several shares of stock at the beginning of Monday. During the morning on Monday, he sold 2/3 of these shares, and in the afternoon on Monday, he sold 5/6 of his remaining shares. If he did not buy any shares during Monday, what was the ratio of the number of shares that Andrew sold Monday morning to the number of shares he sold Monday afternoon?

A) 2 to 5

B) 4 to 5

C) 5 to 4

D) 12 to 5

E) 5 to 1

A) 2 to 5

B) 4 to 5

C) 5 to 4

D) 12 to 5

E) 5 to 1

JohnGalt1 · Answer

So lets go with random number:
because denominator of two fractions are 3 and 6 respectively, for our random number lets take 18.
Now in the morning -> 18*2/3=12 shares sold
Afternoon -> (18-12)*5/6=5 shares sold
Assuming that no Shares were purchased on monday, the ratio will be 12/5. So D is correct
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KarishmaB · Answer

Morning sold - 2/3rd of Total
Left - 1/3rd of Total

Afternoon sold - (5/6) of (1/3) of Total = 5/18 of Total

Ratio = (2/3) / (5/18) = 12/5

Answer (D)

ScottTargetTestPrep · Answer

Solution:
 
Andrew sold 2/3 of his stocks on Monday morning and 5/6 x (1 - 2/3) = 5/6 x 1/3 = 5/18 of his stocks on Monday afternoon. Therefore, the ratio in question is 2/3 : 5/18 = 12 : 5.

Alternate Solution:

Let’s assume that Andrew originally had 90 shares of stock. On Monday morning, he sold 2/3 x 90 = 60 shares, and he had 30 shares remaining. On Monday afternoon he sold 5/6 x 30 = 25 shares. Thus, the ratio of the number of shares sold on Monday morning to the number of shares sold on Monday afternoon is 60 : 25, which reduces to 12 : 5.

Answer: D

Andrew had several shares of stock at the beginning of Monday. During

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