Computation brain training will help you sharpen your math skills.
Below you will find several varieties of computation puzzles including Computation Cross,
Computation Square, and Computation word math puzzles.
COMPUTATION CROSS 14
Rules:
These computation brain training puzzles are completed just like regular crosswords, only
letters are replaced by numbers, and the numbers are entered one number per square
The clues are the sum of the boxes given.
For example the first clue(1 across) for CROSS1 , is 13.
Looking at the diagram we see there are two squares for 1across.
So their sum of those two squares must equal 13.
For computation cross1 the answer for clue 1across is 6 in square one, followed by 7 in
square two, (because 6+7 =13).
Computation cross1

Clues Computation cross1 Clues:

Computation cross2

Clues:

Computation cross3

Clues:

Computation cross4

Clues:

COMPUTATION SQUARE 12
Rules:
These puzzles consist of a five x five board ( 5 x5).
Only the numbers 19 are used, and only one digit per square is allowed.
A digit may appear more than once in a solution.
A prime number is one that is divisible by ONLY the number 1 and itself ( 3, 7, 11, are prime).
Computation Square1


Computation Square2


Computation Brain Training: WORD MATH PUZZLES
MAY DAY
Thirteen children from Miss Kendrick's homeroom class have been selected to participate
in this year's May Day festivities.
They must choose a class leader to captain the project.
Miss Kendricks has the children form a circle and count clockwise, with each count of
thirteen being reached that child will be eliminated from the game, and the count restarts to the
left of the last eliminated child.
Little Cassie Wilson volunteers to begin the count. Because she volunteered Miss Kendricks
allows Cassie to pick which child the count begins with.
Call Cassie child A, and the rest of the children in the circle BM.
With which child should Cassie choose the count to begin to ensure she is the last one
left, and hence wins the game?
(You can find the answer with explanation at the bottom of this page)
THE MOPED INCIDENT:
Jack had to go to the city  15 miles away from home. So he hopped on his trusty moped ( at a top
speed of 20mph), and zoomed into town. Jack picked up his fandangle and prepared for the
return trip leaving the city at 2pm. But after riding only twothirds of the way home, Jack's moped
got a flat tire and he was forced to walk the rest of the way.
Jack arrived home at 330pm, and was immediately assailed by his wife who
demanded to know what took him so long. When he explained about the moped incident,
she only looked at him with consternation and said he should have walked faster.
Exactly how fast did Jack walk?
(You can find the answer with explanation at the bottom of this page)
MOPED INCIDENT PART DEUX
A few weeks later Jack and his wife went on a holiday trip using their Chrysler station wagon,
as the wife refused to be seen on the back of Jack's moped.
The only thing his spouse was grateful for was that Jack always adhered to the posted speed limits
on both local and state highways.
Assuming that the posted limits, respectively were 35mph and 55mph, and their trip took 10 hours
(in which they traveled 470 miles), How much of their travel was on local highways
and how much on interstates?
(You can find the answer with explanation at the bottom of this page)
Answer to MAYDAY :
Cassie should pick child G to start.
Then the order of elimination will be: F, G, I, L, C, K, H, J, B, D, E, M.
Answer to MOPED :
The key is to use the formula which gives the relationship between speed, distance, and time.
Speed is given as the distance traveled over time. Mathematically we write this using symbols :
Speed (s) = distance (d) over (divided by /) time (t), or s = d / t.
Now lets apply this to our particular situation. We are given s =20mph, we know the initial time
he left was 2pm, but we need to calculate the distance(d).
We are told Jack traveled 2/3 of the way home, and we know the total distance was 15 miles
(from home to city), thus Jack traveled (15 divided by 2/3) or 10 miles , when the tire gave out.
Thus it took Jack how long to travel 10 miles? Well if he was traveling at 20mph, which means
in one hour he would have gone 20miles, then Jack must have taken 1/2 hour to travel 10 miles.
Which you can verify by substituting t = d / s, or 10mile/20mph = 1/2 hour.
Further Jack left at 2pm, therefore traveled 1/2 hour and blew his tire.
He must have blown his tire at 2:30pm.
Jack now walks the rest of the way ( which is 15 miles  10 miles) = 5 miles.
We now know he walked from 2:30 pm until he arrived home at 3:30pm, therefore he walked 5 miles
in one hour. ( When we ask for rate, we really mean speed), and we know speed =
d / t, or substituting for the variables , we have Speed(s) = 5miles / 1 hour = 5 mph.
Jack walked home at the rate of 5 miles per hour.
MOPED INCIDENT PART DEUX (ANSWER):
Again, we make use of the formula s = d / t.
We are given t = 10 hours, and d = 470 miles.
Therefore we have s = 47 Mph.
Since this is well over 35mph, but under 55 mph, we can conclude they probably spent more
time on the interstate, than the local roads.
The difference in speeds was 20mph, and the average of 47mph, was 8 under the 55mph,
but 12 over the 35 mph speed limit. Using this ratio (8/12) we can figure how much time was
spent driving at the 35mph limit and the 55 mph limit.
Because we have 8/12 or 4/6 , we can see that they spent 4 hours driving at 35mph and 6 hours
driving at 55mph, and then we can plug in our formula as a check.
we have d = s x t, or
d1= 35 x 4 hours = 140 miles.
d2 = 55 x 6 hours = 330 miles.