﻿ Mathcounts Answer Keys

# Sprint 17597 and solutions

April 9, 2019

### Friday April 26 sprint.

##### Problem 1:
Work backwards. She spent 1/2 of her money at the BBQ and had $7 left. That means$7 is 1/2 of the money she had before she entered the BBQ. $7 is half of$14 so she had $14 when she entered the BBQ. She spent 1/3 of her money at Nob Hill and had$14 left. That means $14 is 2/3 of the money she had when she entered Nob Hill. 14 = 2/3 x Solve for x by mulitplying both sides by the reciprocal of 2/3 or 3/2  3/2 * 14 = 3/2 * 2/3 x  21 = x That means she had$21 when she entered Nob Hill. Since she left Nob Hill with $14, she spent$21-$14 or$7 at Nob Hill.
##### Problem 2:
Starting with a single cell, after d divisions there will be 2^d cells. For example, if the cell divides 4 times, there will be 2^4 cells. If you start with 5 cells instead of 1, there will be 5 times as many cells after 4 divisions or 5 * 2^4.

In general, if you start with n cells there are n*2^d cells after d divisions.

The cells divide every 8 hours so there are 3 divisions per day. They divide for 4 days so there are a total of 12 divisions. The answer is computed by evaluating 5*2^12.

##### Problem 3:
Ava has to wear 3 items of clothing - a pair of pants or skirt, a pair of matching shoes and a blouse.

You can represent the 3 clothing choices with 3 slots shown below She has 5 pairs of pants and 5 skirts so she has 10 choices for the first slot.

She has 10 pairs of shoes so she has 10 choices for the second slot.

She has 12 blouses so she has 12 choices for the third slot.

Populate the slots as shown: Since the choices are independent of each other, you multiply the 3 numbers and get 1200.

##### Problem 4:
The problem describes the picture as representing the top 3 steps of a 47 step stairway. Invert the idea and think of the picture as the first three steps of a 47 step stairway. The picture is the same but lets you think slightly differently.

Imagine that the stairs are made up of little rectangles that are as high as the riser and as wide as the tread. The first step would be made with just one rectangle, the second step would be made with two rectangles, the third step would be made with 3 rectangles, the 4th would be made with four rectangles and so on to the 47th steps which would be made from a stack of 47 rectangles. The first 3 steps are shown below. To compute the area you just add all the rectangles up. 1+2+3+...+47 and multiply the area of one rectangle by the sum.

Adding 47 numbers isn't easy unless you realize that you're adding an arithmetic sequence. The average of the arithmetic sequence can be found by adding the first and last term together and dividing by 2.

bar x = (1+47)/2 Multiply the average by the number of terms, 47, to get the sum, 1128.

Each rectangle is 2 cm high by 8 cm wide. The area of each rectangle is 16 cm^2 so the area of the total stair figure would be 1128*16 or 18048.

##### Problem 5:
The question asks what is the probability that Kai and Joshua get a snickers? The problem has two events. Joshua gets a snickers and Kai gets a snickers. The probability that Joshua gets a snickers is 13/18 since there are 13 snickers out of a total of 18 candies. After Joshua has taken his snickers, there are 12 left so the probability that Kai gets a snickers is 12/17. 17 because there 17 candies left after Joshua took a snickers.

The probability of two events happening is found by multiplying them together.

13/18 * 12/17 = 156/306

The probability tree of the two events is linked below. ##### Problem 6:
This problem just requires using a calculator properly.

You're given the following formula: and asked to compute f(20). That means everywhere there's an 'n', you substitute a '20' like so:

f(20) = ((1+\sqrt(5))^20 - (1-\sqrt(5))^20)/(2^20\sqrt(5))

When you enter the above expression in your calculator use the fraction button that looks like: Use the circular cursor button to escape the square root operator and the exponent operator and be sure to close your parenthesis. That is, every opening left parenthesis must be matched with a closing right parenthesis.