Question

Dear Mitch,

When are you going to post the answers to all the Thanksgiving problems?

Bill B.

Answer

Dear Bill,

Now!

Here are the questions with the answers and explanations below each of them.

Hope this helps,

Happy holiday,

Mitch

1) At a Thanksgiving get-together, six friends meet and each person shakes hands once with each of the other people. How many handshakes take place altogether?

A) 32

B) 30

C) 21

D) 18

E) 15

ANSWER: 15, Choice E.

Here's how I recommend doing this very famous kind of question:

First, realize that this question is all about counting. Every time two people shake hands, that is ONE handshake, not two. And when one person (say person A) shakes another's hand (say person B), it's a mutual and mathematically equal experience, so there is no opportunity for Person B to then shake the hand of Person A (we're just talking mathematically).

So, in a horizontal sequence, draw a blank for each person. Here there are 6 people, so you draw six blanks. Like this:

_ _ _ _ _ _

Then, ask yourself how many people can the first one (the left-most blank representing a person) shake hands with, without repeating anyone and without counting himself?) Answer 5. Put a 5 in his blank.

Next, how many can the next person shake hands with, but not counting the ones who have already shaken hands with him/her... Answer: 4.

Then 3, then 2, then 1, and finally, the last person has been counted in every other person's tally of handshakes, so he/she gets a zero in his blank.

__5__ __4 __ __3__ __2 __ __1__ __0__

And finally, you add all those little numbers up:

5 + 4 + 3 + 2 + 1 + 0 = 15

And that's it!

2) The Wayondee-Tutmyers have a big family, and every member travels a long distance to spend Thanksgiving together. (Even those who live close by go as far away as they can before turning around to begin traveling to the event, so that they too can brag about the big journey they made 'just to be with family').

Last year, including the host, the very special Grandma Wellinthky-Wayondee-Tutmyers, 34 people showed up. (So, before we get to the challenging part, at this point the math is 1 host plus 33 guests.)

Of the 34 people present each person had either pie for dessert, or ice cream, or both pie *and* ice cream, but no one escaped without having to ingest one or both of these homemade treats.

Of the people present that night for the festivities at Grandma Wellinthky-Wayondee-Tutmyers, 22 people had pie, and 19 people had ice cream.

How many of the people present had ice cream but NOT pie?

ANSWER: 12

You figure out how many people were counted above the total number that were there that night. There were 34 people there, and there were 22 + 19 counted. So, 41 people counted even though there were only 34 people there. Since 41 is 7 more than 34, it means that 7 people were counted twice. So seven people had both ice cream and pie.

We can begin to put the numbers in columns like this:

__ Ice cream only ice cream and pie Pie only__

............................................... 7 .............................................

If 19 had ice cream, there must be 19 – 7, or 12 left in the ice cream only column

And since 22 had pie, there must be 22 – 7, or 15 in the pie only column.

__ Ice cream only ice cream and pie Pie only__

...12 ...........................................7.......................................... 15

**And from there you probably can get the idea.**

**If you'd like more explanation, by all means just write in!**

3) Joey B. drives to his grandmother's Thanksgiving dinner at the rate of 40 miles per hour. Later, returning home from Grandma's, the miracle of escaping without her bothering him about 'meeting a nice girl and settling down' fills him with such relief and excitement that he rushes home at the rate of 60 miles per hour.

Of the time he spent driving that night, going and coming, his car's average rate of movement was most likely:

A) 48 miles per hour

B) 50 miles per hour

C) 55 miles per hour

D) 59 miles per hour

E) 60 miles per hour

ANSWER: 48 miles per hour.

Why? Because we're dealing with weighted averages, and so there are more miles being driven at the slower rate than the faster rate, so the average can never be the exact midpoint (50 m.p.h.) as it would be if the driver did not have to go the same distance at both rates. Think: since he is spending more time driving at the slower rate, as the trip takes longer when driving slower, if you think of miles per minute instead of miles per hour, you will see that there are more minutes of slow driving than there are of fast driving. And so the answer to these questions is always the choice that is a little bit slower than the dead center un-weighted average. (For longer trips, the difference could be more, i.e. even slower than just a 'little bit' under the dead center point).

The other group:

1. If 10 pounds of stuffing cost d dollars, how many pounds of stuffing can be purchased for 3 dollars?

(A) 30d

(B) (3d)/10

(C) 30/d

(D) d/30

(E) (10d)/3

ANSWER: C, 30/d

How?

Like this:

Think of the slang: Instead of asking how much something *costs*, people often ask, "How much ** is** that?"

And, in math, "is" is represented by an equal sign, as in 3 + 4 *is* 7, or 3 + 4 = 7.

So, we have:

10 = d dollars

But, more helpful than knowing how much a person can get for* d *dollars, we'd like to know how much one gets for* one* dollar.

So, since the way to turn anything into a one is by dividing it by itself, we divide each side by d to get "something = one dollar".

Like this:

10/d = 1 dollar

And to find out how many can be bought for 3 dollars, as the questions asks, we multiply that one dollar by 3, and do the same to the other side, as math requires.

And we get 30/d = 3 dollars.

2. If p pounds of sweet potato pie costs s cents, 10 pounds of that pie will cost

(A) (ps)/10 cents

(B) 10ps cents

(C) (10s)/p cents

(D) (10p)/s cents

(E) (s + p +10) cents

ANSWER: C, (10s)/p cents

Similar to above,

P = s cents

p/p = s/p cents

1 = s/p cents

So,

10 = (10s)/p cents

3. One week before Thanksgiving last year a gourmet food store reduced the price of their "family-sized" turkey dinner by half the regular price, and then, 3 days later, when the chef became concerned that he had prepared more than they were likely to sell, had the owner reduce the sale price by 10%. The final price is what percent of the original price?

(A) 5%

(B) 10%

(C) 25%

(D) 40%

(E) 45%

ANSWER: E, 45%.

When dealing with percentage questions where they do not give you a number, choose 100. First reduction: 100 becomes 50.

Second reduction: 10% of 50 = 5, so 50 goes to 45.

45 out of the original 100 is 45%.

4. On the night before Thanksgiving, Stanley S., who, although 22-years-old, still lives with his parents, sneaked into the kitchen and ate one-fourth of the pumpkin pie his mother had spent all day preparing. The following morning, he woke up early (having cleverly set his alarm clock), tip-toed downstairs, slipped back into the kitchen, and helped himself to one-half of what was left of the pie. What fraction of the entire pie did Stanley eat before the holiday dinner?

(A) 1/2

(B) 7/8

(C) 3/8

(D) 5/8

(E) 3/4

ANSWER: Choice D, 5/8.

Why? Draw a circle and divide it up into 8 equal slices. Then take away one-fourth and you are left with 6 slices. Then take away half of the remaining 6 and you get 3 slices. 3 of the original 8 *remaining*. He** ate** 5/8ths of the pie.

Now, the answer to the super-challenging balance-scale problem.

(NOTE: For this one the actual problem was in picture-form, which was the only way that made sense, and to see that picture you will have to pull it down from the original question, which is simply entitled the 'super-challenge thanksgiving problem' (or something like that).

So, in conjunction with the image of the four different balance scales, the question was:

**A series of weighings is shown. **

**All items on the scales are small toys to be used as decorations for a Thanksgiving party. There are four different kinds of toys - toy 'drumsticks', toy 'turkey platters', toy 'Pilgrim hats', and toy figures of a 'Native American'.**

**Going from top to bottom, the first three scales (scale A, scale B, and scale C) are balanced and complete. But the fourth scale (scale D) is not.**

**After taking a careful look at the scales below and analyzing the information they provide, try to problem-solve to figure out how many toy turkey platters must be put on the empty side of scale D in order to form a balance with the left-hand side of that scale, which, as depicted in the illustration, is weighed down by just one toy drumstick?**

ANSWER: Five turkey platters.

Explanation:

In order to make the two left sides of scale A and B equal, you add one turkey platter to each side of scale B. That means that one Native American equals two turkey platters + one hat. So 2 Native Americans = 4 turkey platters + 2 hats,

And that means that 4 turkey platters + 2 hats = 3 hats.

Taking away 2 hats from both sides, we see that the weight of 4 turkey platters equals the weight of one hat.

So, now you can exchange the one hat in scale B with 4 turkey platters and can balance the drumstick in D with 5 turkey platters.

I know this one was challenging, so I hope this explanation helps!

Happy holiday,

Mitch