Red toys = 40%, Green toys = 60%
Large toys = 50%, small toys = 50%
Small red toys = 10% implies Large Red toys = 30%, Small Green Toys = 40%, and Large Green Toys = 20%
Green and Large toys = 40
Let the number of toys = T and 40 toys are green and large.
Then 20% of T = 40 implies T = 200
30% toys are Red and Large, so 30% of 200 = 60

The correct answer is B.

sanju09 wrote:
But [spoiler]A[/spoiler] is not [spoiler]60[/spoiler]

Sorry, that was a typo, its B. Thanks for pointing that. Edited the previous post.

sanju09 wrote: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

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