Question 517322: i choose three different number cards from a stack. no number is odd. each number is less than 20. the sum of the two of the numbers equals the third number.The sum of all three numbers is 24.what number is on each card.
Answer by bucky(2189)   (Show Source):
You can put this solution on YOUR website! The answer to this problem must meet the following criteria:
.
(1) The three cards that are drawn all have even numbers.
(2) Each card must have a value less than 20
(3) The two cards with the smaller values must have a sum that equals the value of the third card.
(4) The total of all three numbers drawn must be 24
.
Assume that the number on each card is unique. This means that there are no duplicate cards ... there will be only one card for each number. Also assume that there is no card with the number zero, and all cards have positive numbers.
.
Let's look at criteria 1 & 2. Since there are only even numbers and the values of each card must be less than 20 we can meet these criteria by realizing that the possible cards that were drawn were: 2, 4, 6, 8, 10, 12, 14, 16, and 18 ... all even and all less than 20.
.
Now let's look at criterion 3. Also let's represent the three unknown cards with the letters A, B, and C in which A and B are the smaller values and C is the largest of the three. From criterion 3 we can write the equation:
.
A + B = C
.
Let's now consider criterion 4. The sum of all three cards equals 24. In equation form this is:
.
A + B + C = 24
.
From criterion 3 we know that C = A + B. So we can substitute A + B for C in the equation from criterion 4. This substitution leads to:
.
A + B + A + B = 24
.
Combining like terms gives:
.
2A + 2B = 24
.
Dividing both sides (all terms) by 2 reduces this equation to:
.
A + B = 12
.
One thing we learn from this comes from knowing that A + B = C and therefore C must equal 12. The largest card is 12. And since A + B = 12, the two remaining smaller cards must be two cards whose sum is 12. Obviously, those two cards are 10 and 2. Do the three cards 2, 10, and 12 meet the original 4 criteria. They are all even, they are all less than 20, when added the smaller two cards (2 and 10) have a sum equal to the third card (12), and the sum of all three cards (2, 10, and 12) is 24. So that's the answer. The three cards are 2, 10, and 12. BUT NOT SO FAST...
.
There is also another possibility in which A + B = 12. The two cards 4 and 8 also will work. Combined with the third card of 12, we have three cards 4, 8, and 12 that also meet all the criteria. They are all even, all less than 20, the total of the two smaller cards (4 + 8) equals the largest card 12, and the sum of all three (4 + 8 + 12) is 24.
.
There are two correct answers to this problem.
.
Hope this helps you to see how you can work your way through this problem. You can also try other various combinations from the cards 2, 4, 6, 8, 10, 12, 14, 16, and 18 and you will find that no other combinations will simultaneously satisfy criteria 3 and 4. That can only be done by 2, 10, and 12 or by 4, 8, and 12.
.
|
|
|
