In a certain production lot, 40 percent of the toys are red

hi dave13

Refer to the first post for the detailed solution. You can prepare a matrix as shown in the above solution. From the matrix it is clear that 20% toys are green and large and it is also given that 40 toys are green and large. Hence we have -

20% of total toys = 40 => total toys = 40*(100/20)=200

We can let the total number of toys = n.

So 0.4n toys are red and 0.6n are green.

0.5n are large and 0.5n are small.

0.1n are red and small.

40 toys are green and large.

We can create the equation:

Total = large + green - both green and large + neither green nor large

n = 0.5n + 0.6n - 40 + 0.1n

n = -40 + 1.2n

40 = 0.2n

200 = n

Since 0.4n of the toys are red and 0.1n of the toys are red and small, then 0.3n of the toys must be red and large. Since n = 200, the number of toys that are red and large is 0.3(200) = 60.

(Note: Alternatively, we can say:

Since 0.5n of the toys are large and 40 toys are green and large, then 0.5n - 40 toys must be red and large. Since n = 200, the number of toys that are red and large is 0.5(200) - 40 = 60.)

Answer: B

use 2x2 matrix to solve

-------red------green----total
smal---10------40-------50
large--30------20--------50
total--40------60--------100

30/50 * 100= 60%
IMO B

For those who don't understand where the 200 comes from - read carefully: the given numbers are % and # of toys... I missed that too so...

In a certain production lot, 40 percent of the toys are red and the remaining toys are green. Half of toys are small and half are large. If 10 percent of the toys are red and small, and 40 toys are green and large. How many of the toys are red and large?

A. 40
B. 60
C. 70
D. 80
E. 90
Red : 40%
Red and small : 10%, so the remaining are red and large, which is 30%

Green: 60%
Green and small : 40%, green and large: 20%(as 50% large, 50% small)
Given, 20% = 40, so total toys = 200, red and large toys= 60.
B is the answer

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