# Middle School Math Quiz (no questions on probability)

2 posters

Page

**1**of**1**## Middle School Math Quiz (no questions on probability)

Relatively easy Middle School quiz problems chosen from a set in a math magazine. Please explain your answers. You need not provide big narratives, just a short explanation will do. Your explanation will help others on the forum to see the logic behind the solution.

1. It takes 6 days to cross a desert for any person. One person can carry only enough food and water to survive for 4 days. What is the minimum number of people who must start out to ensure that one person successfully gets across and others safely return to their starting point.

2. When the waiter was asked to slice the 16-in (diameter) piazza into exactly 4 slices of equal size, he decided to do so using concentric circles rather the traditional radial slices. Determine the radii of each of the 3 cuts he had to make.

3. The numbers a and b are positive integers with a < b. Solve for a and b in the following equation:

a

1. It takes 6 days to cross a desert for any person. One person can carry only enough food and water to survive for 4 days. What is the minimum number of people who must start out to ensure that one person successfully gets across and others safely return to their starting point.

2. When the waiter was asked to slice the 16-in (diameter) piazza into exactly 4 slices of equal size, he decided to do so using concentric circles rather the traditional radial slices. Determine the radii of each of the 3 cuts he had to make.

3. The numbers a and b are positive integers with a < b. Solve for a and b in the following equation:

a

^{3}+ b^{3 }= 243.**indophile**- Posts : 4338

Join date : 2011-04-29

Location : Glenn Dale, MD

## Re: Middle School Math Quiz (no questions on probability)

1. 3 people 1 returns after 1 day passing on a day's ration each to the other two

The second person return after two days. passing on a days ration to the third person who has rations now for 6 days including the 4 days of rations he carried.

So the rations will be like this

4 4 4 Day 1

_ 4 4 Day 2 ->Person 1 has turned back with 1 day's rations

_ _ 4 Day 3 -> Person 2 has runed back with 2 days rations

_ _ 3 Day 4

_ _ 2 Day 5

_ _ 1 Day 6 --> Enough to get him across for the last day

Alternatively 12 rations is what they carry(4*3)

Person 1 needs 2 rations for 1 day to and 1 day from

Person 2 needs 4 rations for 2 days to and 2 days from

Person 3 needs 6 rations for 6 days to

for a total of 12 rations (2+4+6)

2. inner most has to be 1/4 * pi * 8^2 = pi* (8/2)^2 or 4 inches

Let us say the second one is r

=> pi*r^2 - pi *4^2 = pi*4^2 => r = sqrt(2)*4

If the radius third one is r2 => pi *(r2)^2 - pi *2*(4^2) = pi*( 4^2)

=>r2 = sqrt(3)*4

So the three radii are 4, sqrt(2)*4, sqrt(3)*4

3. 243 = 729 - 486 = 9^3 - 3*9 (18) = 9^3 - 3*(6*3)*(6+3)

Using (a+b)^3 = a^3 + b^3 + 3ab(a+b)

We have a+b = 3+6 => a = 3, b = 6

The second person return after two days. passing on a days ration to the third person who has rations now for 6 days including the 4 days of rations he carried.

So the rations will be like this

4 4 4 Day 1

_ 4 4 Day 2 ->Person 1 has turned back with 1 day's rations

_ _ 4 Day 3 -> Person 2 has runed back with 2 days rations

_ _ 3 Day 4

_ _ 2 Day 5

_ _ 1 Day 6 --> Enough to get him across for the last day

Alternatively 12 rations is what they carry(4*3)

Person 1 needs 2 rations for 1 day to and 1 day from

Person 2 needs 4 rations for 2 days to and 2 days from

Person 3 needs 6 rations for 6 days to

for a total of 12 rations (2+4+6)

2. inner most has to be 1/4 * pi * 8^2 = pi* (8/2)^2 or 4 inches

Let us say the second one is r

=> pi*r^2 - pi *4^2 = pi*4^2 => r = sqrt(2)*4

If the radius third one is r2 => pi *(r2)^2 - pi *2*(4^2) = pi*( 4^2)

=>r2 = sqrt(3)*4

So the three radii are 4, sqrt(2)*4, sqrt(3)*4

3. 243 = 729 - 486 = 9^3 - 3*9 (18) = 9^3 - 3*(6*3)*(6+3)

Using (a+b)^3 = a^3 + b^3 + 3ab(a+b)

We have a+b = 3+6 => a = 3, b = 6

**Hellsangel**- Posts : 14721

Join date : 2011-04-28

## Re: Middle School Math Quiz (no questions on probability)

Yes to all. A different method was used for # 3 however.

3. 3 and 6. Since a^3 + b^3 = 243 = 3^5, and a3 + b3 = (a + b)(a2 - ab + b2), we know that (a + b) must be a power of 3 with its value being less than 243. Therefore, a+b = 3 or 9 or 27 or 81. The set of perfect cubes less than 243 contains 1, 8, 27, 64, 125, 216. Consequently, the values for a and b come from the set of {1,2,3,4,5,6}. Combining these two ideas, we find that the only solution is a=3 and b=6.

3. 3 and 6. Since a^3 + b^3 = 243 = 3^5, and a3 + b3 = (a + b)(a2 - ab + b2), we know that (a + b) must be a power of 3 with its value being less than 243. Therefore, a+b = 3 or 9 or 27 or 81. The set of perfect cubes less than 243 contains 1, 8, 27, 64, 125, 216. Consequently, the values for a and b come from the set of {1,2,3,4,5,6}. Combining these two ideas, we find that the only solution is a=3 and b=6.

**indophile**- Posts : 4338

Join date : 2011-04-29

Location : Glenn Dale, MD

Similar topics

» Middle School Math Quiz

» Middle School Math Quiz

» Middle School Math Quiz

» Middle School Math Quiz

» probability questions

» Middle School Math Quiz

» Middle School Math Quiz

» Middle School Math Quiz

» probability questions

Page

**1**of**1****Permissions in this forum:**

**cannot**reply to topics in this forum