In a contest run by a store, each customer whose purchase exceeds $100 is allowed to draw a discount coupon from a jar. at the beginning of the contest, the jar contains 30 slips for a 5% discount, 15 slips for an x% discount, and 5 slips for a 15% discount. if the expected value of the first draw from the jar is 6.6%, the value of x is __ . at one point in the contest, the jar contains 4 slips for a 5% discount, y slips for an x% discount, and 2 slips for a 15% discount. if the expected value on the next draw is 8%, the value of y is __ .

Answers:
x = 7%
y = 2 slips

Explanation:

The expected value is the result of the sum of each value times its probabilities:

Expeted value = probability 1 × value 1 + probability 2 × value 2 + probability 3 × value 3 +

Case 1: at the beginning of the contest:
total number of slips: 30 + 15 + 5 = 50

probability 1 = 30/50

value 1 = 5%

probability 2 = (15/50)
value 2 = x%

probability 3 = (5/50)

value 3 = 15%

⇒  Expected value = 6.6% = (30/50) 5% + (15/50)x% + (5/50)15%
⇒ (15/50)x% = 6.6% - (30/50)5% - (5/50)15%
⇒ (15/50) x% = 2.1% 
⇒ x% = (50 / 15) 2.1% = 7%

7%

2) Case 2: at one point, ...

Yet, the equation for the expected value is:

Expeted value = probability 1 × value 1 + probability 2 × value 2 + probability 3 × value 3 +

Only the probabilities have changed, but the discounts are the same. This is x% is the same value found above: 7%.

The total number of slips now is 4 + y + 2 = 6 + y

And the expected value becomes:
8% = [ 4 / (6+y) ] 5% + [ y / (6 + y) ] 7% + [2 / (6 + y)] 15%

From which you obtain:

Mulitplying by 6+ y: 8% [6 + y] = 4×5% + y×7% + 2×15%

⇒ 8% y + 8%×6 = 4×5% + y 7% + 2×15%

⇒ 8% y - 7%y = 4×5% + 2×15% - 6×8%
⇒  0.01y = 0.2 + 0.3 - 0.48 = 0.02
⇒ y = 2

Let's name these two digits as a (tens digit) and b (units digit) and then rephrase the problem statement using a and b.

The part of the sentence "three times the tens digit is 2 less than twice the units digits" translates into:

3a = 2b - 2

Considering that both a and b have to be integer numbers less than 10 and that a will have to be an even number

(since 2b - 2 is always even and for 3a to be even, a will have to be even),

the only possible options that will satisfy this equation are: 47 and 24.

The second part of the problem statement does not make much sense.

What exactly is twice the number obtained by reversing the digits?

In case of 47, twice the number obtained by reversing the digits is 2*74 = 148.

In case of 24, twice the number obtained by reversing the digits is 2*42 = 84.