At a monthly meeting, \(\dfrac25\) of the attendees were males and \(\dfrac78\) of the male attendees arrived on time.

Vincen wrote: ↑

Wed Oct 28, 2020 8:29 am

At a monthly meeting, \(\dfrac25\) of the attendees were males and \(\dfrac78\) of the male attendees arrived on time. If \(\dfrac9{10}\) of the female attendees arrived on time, what fraction of the attendees at the monthly meeting did not arrive on time?

(A) \(\dfrac{11}{100}\)

(B) \(\dfrac3{25}\)

(C) \(\dfrac7{50}\)

(D) \(\dfrac3{20}\)

(E) \(\dfrac4{25}\)

Answer: A

Source: Official Guide

Solution:

We can let the total number of attendees = 800. (This is a convenient number that allows the calculations to be easier.)

Thus, there are 2/5 x 800 = 320 males and 800 - 320 = 480 females.

Since 7/8 of the males arrived on time, 7/8 x 320 = 280 males arrived on time.

Since 9/10 of the female attendees arrived on time, 9/10 x 480 = 432 females arrived on time.

Thus, 800 - (280 + 432) = 88 attendees did not arrive on time.

Therefore, the fraction of the attendees who did not arrive on time is 88/800 = 11/100.

Answer: A