# Fun Math Problems With Answers

Fun Math Problems With Answers | mathjokes.net
These are the best and most fun math problems we can find. All of these tricky math problems are based on real math concepts and can be solved with purely math and logic.

## Fun Math Problem With Answer 1

A truck travels from X to Y.
Going uphill, it goes at 56 mph.
Going downhill, it goes at 72 mph.
On level ground, it goes at 63 mph.
If it takes 4 hours to travel from X to Y, and 5 hours to come back, what is the distance between X and Y?

When you reverse the direction,
uphill becomes downhill, downhill becomes uphill, and flat stays flat.

Suppose when going from x to y we have
u miles of uphill,
f miles of flat
d miles of downhill
(and the reverse coming back).

x to y: u / 56 + f / 63 + d / 72 = 4 hours
y to x: u / 72 + f / 63 + d / 56 = 5 hours

Subtracting top from bottom:
u/72 - u/56 + d/56 - d/72 = 1
1/72 - 1/56 = 1/8*9 - 1/7*8 = -2/7*8*9

-2/7*8*9 u + 2/7*8*9 d = 1
-2u + 2 d = 7*8*9 = 504
2 (d - u) = 504
d - u = 252

And we want
u/56 + f/63 + d/72 = 4
u/72 + f/63 + d/56 = 5

Now suppose u = 0.
Then we get d = 252
x to y takes: 0/56 + f/63 + 252/72 = 3.5 + f/63 hours
f/63 = 0.5 hour (to bring total to 4)
f = 31.5
and total distance = 283.5

For the return journey we have
0/72 + 0.5 + 252/56 = 0.5 + 4.5 = 5 hours, as required.

## Fun Math Problem With Answer 2

A father, in his will, left his money to his children in the following manner:
\$1000 to the first born and 1/10 of what then remains, then
\$2000 to the second born and 1/10 of what then remains, then
\$3000 to the third born and 1/10 of what then remains,
and so on.
When this was done each child had the same amount.

How many children were there?

There are 9 children.
Here are two solutions, working from either end of the line of children:

Let n be the number of children.
1) The last shall be first:
The last one got n * 1000 + 1/10 * 0 = a multiple of 1000.
So they all got (the same) multiple of 1000 = n * 1000.

When the 2nd to last gets his share, there is 2 * n * 1000 available
(since each gets the same, 2nd to last takes half of what's there).
Last one's share = 2nd to last one's share
n * 1000 = (n-1)* 1000 + 1/10 ( 2 * n * 1000 - (n-1)*1000)

Solve for n:
n = n - 1 + (2 n - 1 n + 1) / 10 (dividing by 1000)
0 = -1 + (n+1) / 10
n+1 = 10
n = 9

and each child gets 9000, total 81000.

1000 + (81000-1000) / 10 = 1000 + 8000 = 9000
2000 + (72000-2000) / 10 = 2000 + 7000 = 9000
and so on
8000 + (18000-8000) / 10 = 8000 + 1000 = 9000
9000 + (9000-9000) / 10 = 9000 + 0 = 9000
2) Starting from the front of the line.
Each one gets 1/n of the "pot" = n * 1000. (see above for why that is the case).

Total pot = n * (n*1000) = n^2 * 1000.

First one gets n * 1000 =
1000 + (n^2 * 1000 - 1000) / 10

Divide by 1000:
n = 1 + (n^2 - 1) / 10
10 n = 10 + n^2 - 1
n^2 - 10n + 9 = 0

(n - 9) (n - 1) = 0

n = 9 or n = 1.

The case n = 1 is not very interesting,
or is precluded by the given, "to the second born ...".

## Fun Math Problems 3

A gorilla harvests 3,000 bananas and needs to carry them 1,000 miles to the supermarket. He can only carry 1,000 at a time. Since he is a gorilla he eats 1 banana every mile he goes in any direction. He can (and will have to) leave bananas anywhere along the way. Once all his bananas have reached the end he DOES NOT need any to eat to get back.
Remember he eats 1 banana every mile he goes even if he is going back to pick up more bananas. What is the maximum number of bananas he can get to the market ?

When he has more than 2000 bananas, it costs him 5 bananas / mile to
transport them, since he makes 3 trips forward and 2 trips back.

When he has more than 1000 bananas, it costs him 3 bananas / mile, for 2
trips forward and 1 trip back.

And when he has 1000 or fewer, it's 1 banana / mile.

I think the way to do it is: see how far he can get 2000 bananas, then 1000
bananas, and then make one final trip.

200 miles @ 5 bananas / mile = 1000 bananas cost.

So if the first stopping point is at 200 miles,
he would carry 1000, eat 200, drop 600, and eat 200.
Then he would do that again.
On the third trip, he doesn't eat the second 200, so at the 200 mile
marker he has 600 + 600 + 800 = 2000 bananas.

If the second leg is 333 miles, it works out as:
carry 1000, eat 333, drop 334, eat 333.
carry 1000, eat 333, drop 667.

Now he has 1001 bananas, and it's just as well to throw one away. To
make it come out exact, we'd need to use fractional miles and fractional
bananas.

Now we have 1000 bananas at the 533 mile marker.

For the last leg, he just carries the last 1000 bananas,
eating 467 along the way, and arriving with 533 bananas.

If the initial legs are shorter, it works out the same.
If the initial legs are longer, he does more miles at higher banana /
mile rates, ending up with fewer bananas.

## Fun Math Problems 4

Consider twin sisters with two different retirement savings plan:
Plan 1: Tera begins a retirement account at age 20. She stars with \$2000 and then saves \$2000 per year at 7% interest compounded annually for 10 years. Then she stops contributing to the account but keeps her savings invested at the same rate.

Plan 2: Janna doesn't save any money in her twenties. When she turns 30 she starts with \$2000 and then saves \$2000 per year at 7% interest compounded annually for 35 years.

Which one has more at age 65 ?

At the end of the 10 years, the first one has
2000 x (1.07^10 + 1.07^9 + 1.07^8 + ... + 1.07^2 + 1.07) =
29567.20

This then is multiplied by 1.07^35 = 10.68 to yield 315,676.62

The other one gets
2000 x (1.07 + 1.07^2 + 1.07^3 + ... + 1.07^35) = 2000 x 137.4 = 274,800

She doesn't catch up.

## Fun Math Problems 5

Measuring Cups
Diana has three-cup, nine-cup, and twenty-cup containers, each with no markings. with these containers she can measure various amounts. using the three containers how many whole cups of water from one to twenty, inclusive, can she measure ?

C3 = 3-cup cup, C9 = 9-cup cup, C20 = 20-cup cup.

1: Fill C9, pour into C20, fill C9 pour into C20, fill C3. Since C20 has
2 cups remaining, pour from C3 into C20, leaving 1 in C3.

2: Fill C20, pour 9 into C9. Empty C9, pour 9 into C9, leaving 2 in C20.

3: Fill C3

4: Follow directions for 1, pour it into C9, then fill C3
(there might be a quicker way)

5: Follow directions for 2, pour into C9, fill C3
(there might be a quicker way)

6: Fill C9, pour into C3, leaving 6.

7: C9 -> C20, C9 -> C20 (leaves space of 2 in C20).
Fill C9, pour 2 into remaining space in C20, leaving 7.

8: C20 -> C9, leaving 11. -> C3 leaving 8

9: C9

10: Follow directions for 1, then fill C9

11: Follow directions for 2, then fill C9

12: C9 + C3

13: Follow directions for 1, put it in C20,

14: Follow directions for 5, put it in C20, then fill C9

15: C3 -> C20 twice, then C9

16: C9 -> C20 twice. Then C3 -> C20, leaving 1 in C3.
C9 -> C3 leaves 7 in C9.
Empty C20, pour 7 into C20, then add another C9.

17: C20 -> C3 leaving 17

18: C9 -> C20 twice

19: C9 -> C20 twice, leaving 2. C3 -> C20, leaving 1 in C3. Empty C20.
Then two C9's into C20 + 1 from C3.

20: C20

That's it.