# Trial and Error

Trial and error refers to the process of verifying that a certain choice is right (or wrong). We simply substitute that choice into the problem and check. Some questions can only be solved by trial and error; for others we must first decide if there isn't a faster way to arrive at the answer. In the examples to follow, we test all choices for your benefit. Once you have the right answer, there is no need to check the rest of the choices.

\[(2, 3), (3, 5), (4, 4), (6, 3), (10, 0)\]

How many of the above pairs of integers are solutions to \( 2x + 3y = 20 ?\)

(A) \(\ \ 1\)

(B) \(\ \ 2\)

(C) \(\ \ 3\)

(D) \(\ \ 4\)

(E) \(\ \ 5\)

Correct Answer: B

Solution:We try each of the pairs of integers:

For \((2, 3)\), we have \( 2 \times 2 + 3 \times 3 = 4 + 9 = 13 \neq 20 \).

For \((3, 5)\), we have \( 2 \times 3 + 3 \times 5 = 6 + 15 = 21 \neq 20 \).

For \((4, 4)\), we have \( 2 \times 4 + 3 \times 4 = 8 + 12 = 20 \). This is a solution.

For \((6, 3)\), we have \( 2 \times 6 + 3 \times 3 = 12 + 9 = 21 \neq 20 \).

For \((10, 0)\), we have \( 2 \times 10 + 3 \times 0 = 20 + 0 = 20 \). This is a solution.Thus, 2 of the pairs are solutions.

Incorrect Choices:

(A),(C),(D), and(E)

See the solution for why these choices are wrong.

\[ ab - 2a - 2b - 2 = 0 \]

Which of the following pairs of numbers \( (a, b) \) is a solution to the equation above?

\(\begin{array}{r r l}
&\text{I.}\ &(3, 8)\\

&\text{II.}\ &(4, 5)\\

&\text{III.}\ &(4.5, 4.5)\\
\end{array}\)

(A)\(\ \ \) I only

(B)\(\ \ \) II only

(C)\(\ \ \) I and II only

(D)\(\ \ \) I and III only

(E)\(\ \ \) I, II and III

When the wind blows, half of the leaves on a tree fall, and then 5 more. When the wind blows a second time, again half of the leaves fall and then 5 more. If there are no leaves remaining on the tree, how many leaves are there at the start?

(A) \(\ \ 5\)

(B) \(\ \ 10\)

(C) \(\ \ 15\)

(D) \(\ \ 30\)

(E) \(\ \ 50\)

Correct Answer: D

Solution 1:Let's analyze each answer using the trial and error approach.

(A) If there are 5 leaves at the start, when the wind blows the first time, half of the leaves fall, which is 2.5, and then 5 more, so there are \( 5 - 2.5 - 5 = - 2.5 \) leaves left. This does not make sense, so we eliminate this choice.

(B) If there are 10 leaves at the start, when the wind blows the first time, half of the leaves fall, which is 5, and then 5 more, so there are \( 10 - 5 - 5 = 0 \) leaves left. When the wind blows the second time, half of the remaining leaves fall, which is 0, and then 5 more. Thus there are \( 0 - 0 - 5 = - 5 \) leaves left. This does not make sense, so we eliminate this choice.

(C) If there are 15 leaves at the start, when the wind blows the first time, half of the leaves fall, which is 7.5, and then 5 more. Thus there are \( 15 - 7.5 - 5 = 2.5 \) leaves left. When the wind blows the second time, it blows down half of the remaining leaves, which is 1.25, and then 5 more. Thus there are \( 2.5 - 1.25 - 5 = -3.75 \) leaves left. Wrong choice.

(D) If there are 30 leaves at the start, when the wind blows the first time, half of the leaves fall, which is 5, and then 5 more. Thus there are \( 30 - 15 - 5 = 10 \) leaves left. When the wind blows the second time, half of the remaining leaves fall, which is 5, and then 5 more. Thus there are \( 10 - 5 - 5 = 0 \) leaves left. This is the correct answer.

(E) If there are 50 leaves at the start, when the wind blows the first time, half of the leaves fall, which is 25, and then 5 more. Thus there are \( 50- 25 - 5 = 20 \) leaves left. When the wind blows the second time, it blows down half of the remaining leaves, which is 10, and then 5 more. Thus there are \( 20 - 10 - 5 = 5 \) leaves left. But we are told that no leaves remain on the tree. Wrong choice.

Thus, the answer is (D).

Solution 2:We can solve this problem by working backwards.

At the end, we are left with 0 leaves.

Just before that, 5 leaves fall, so there were 5 leaves on the tree.

Just before that, half of the leaves fall, so there are \(2\cdot 5=10\) leaves on the tree. Just before that, 5 leaves fall, so there are \(10+5=15\) leaves on the tree.

And just before that, half of the leaves fall, so there are \(2\cdot 15=30\) leaves on the tree.

Incorrect Choices:

(A)

This is the number of leaves that are blown down right at the end.

(B)

This is the number of leaves that are on the tree before the second wind.

(C)

This is the number of leaves that are on the tree just after the first wind blows half of the leaves down.

(E)

This choice is offered to confused you.

There are several people in a meeting, and each pair of them shake hands. If there are a total of 210 handshakes, how many people are in the meeting?

(A)\(\ \ \) 14

(B)\(\ \ \) 15

(C)\(\ \ \) 18

(D)\(\ \ \) 20

(E)\(\ \ \) 21

Correct Answer: E

Solution:If there are \(n\) people at the meeting, each person will shake hands with \(n-1\) other people (a person cannot shake hands with himself). So, there are \(n\cdot (n-1)\) ways we can pair the people at the meeting. But, the number of handshakes isn't equal to the number of ways we can pair the people. Since the handshake between person A and person B is the same as the handshake between person B and person A, we must divide \(n\cdot (n-1)\) by 2 so as to not count each handshake twice. Let's analyze each answer choice.

(A) If there are 14 people, there will be \( \frac{14 \times 13 } { 2} = 91 \) handshakes. Wrong choice.

(B) If there are 15 people, there will be \( \frac{15 \times 14 } { 2} = 105 \) handshakes. Wrong choice.

(C) If there are 18 people, there will be \( \frac{18 \times 17 } { 2} = 153 \) handshakes. Wrong choice.

(D) If there are 20 people, there will be \( \frac{20 \times 19 } { 2} = 190 \) handshakes. Wrong choice.

(E) If there are 21 people, there will be \( \frac{21 \times 20 } { 2} = 210 \) handshakes. Correct answer.

Incorrect Choices:

(A),(B),(C), and(D)

The solution explains how to eliminate these choices.

**Cite as:**Trial and Error.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/sat-trial-and-error/