My local bus company has recently expanded and no longer has enough room for all of its buses.
Twelve of their buses have to be stored outside.
If they decide to increase their garage space by 40%, this will give them enough room for all of their current buses, plus enough room to store another twelve in the future.
How many buses does the company currently own?
[Ref: ZWOL] © Kevin Stone
Direct Link: www.brainbashers.com?ZWOL
Hint: A tricky puzzle that will need a little algebra (or clever thinking).
They have enough room for 60 of these, expanding the 60 capacity by 40% will give them enough room for 84, which we know is 12 more spaces than they currently need.
If they have B buses and S spaces before the expansion, they have enough room for:
S = B - 12 
After the expansion they have more spaces, and enough room for:
S + 0.4 x S = B + 12 
Rewriting  as:
1.4S = B + 12
and again as
B = 1.4S - 12 
We can rewrite  as:
B = S + 12 
Making  =  we have:
1.4S - 12 = S + 12
Subtracting S from both sides gives:
0.4S - 12 = 12
Adding 12 to both sides gives:
0.4S = 24
Multiplying by 10 and dividing by 4 on both sides gives:
S = 60
Using S = 60 in  gives B = 72.
Within the BrainBashers school, the science department has three disciplines.
In total, 280 students study chemistry, 254 students study physics and 280 students study biology.
97 students study both chemistry and physics, 138 students study both physics and biology, 152 students study both chemistry and biology.
73 students study all three disciplines.
Can you determine how many students there are in the science department? The answer is well below 814.
[Ref: ZZYU] © Kevin Stone
Direct Link: www.brainbashers.com?ZZYU
Hint: How many students study chemistry and physics, but not biology?
The answer is most easily seen if three intersecting circles are drawn, and the numbers inside each section worked out.
Three countrymen met at a cattle market. "Look here," said Hodge to Jakes, "I'll give you six of my pigs for one of your horses, and then you'll have twice as many animals here as I've got."
"If that's your way of doing business," said Durrant to Hodge, "I'll give you fourteen of my sheep for a horse, and then you'll have three times as many animals as I."
"Well, I'll go better than that," said Jakes to Durrant; "I'll give you four cows for a horse, and then you'll have six times as many animals as I've got here."
No doubt this was a very primitive way of bartering animals, but it is an interesting little puzzle to discover just how many animals Jakes, Hodge, and Durrant must have taken to the cattle market.
[Ref: ZUCK] At A Cattle Market. Amusements In Mathematics by Henry Ernest Dudeney (1917).
Direct Link: www.brainbashers.com?ZUCK
Hint: A little algebra might help.
Jakes must have taken 7 animals to market, Hodge must have taken 11, and Durrant must have taken 21. There were thus 39 animals altogether.
This can be solved with a little algebra, where J - Jakes, H = Hodge and D = Durrant:
From the clues:
2 x (H - 6 + 1) = J + 6 - 1 
3 x (D - 14 + 1) = H + 14 - 1 
6 x (J -4 + 1) = D + 4 - 1 
These can be rearranged to give:
2H - J = 15 
3D - H = 52 
6J - D = 21 
We can now use 2 x  +  to give:
6D - J = 119 
We can now use  + 6 x  to give:
35J = 245
J = 7
We can then use J = 7 in  and  to give H = 11, and D = 21.
In the following sum the digits 0 to 9 have all been used, O = Odd, E = Even, zero is even and the top row's digits add to 9. Can you determine each digit?
[Ref: ZVZB] © Kevin Stone
Direct Link: www.brainbashers.com?ZVZB
Hint: The largest possible numbers could start with 6 and 8, therefore the first digit of the answer is 1.
E + E = E
O + O = E
E + O = O
To discuss individual letters it's easiest to represent the sum as:
A B C
D E F +
G H I J
The largest values for A and D are 6 and 8, which makes G = 1.
Since column 2 is E + O = O there can be no carry from column 1 (since E + O + 1 is always even). Therefore C and F are 3 and 5 (but we don't yet know which is which), therefore J = 8.
There can't be a carry from column 2 (as A + D is even) therefore E can't be 9 as this would force a carry.
Therefore I = 9. Hence B can't be 0. Therefore H = 0.
The last remaining odd number makes E = 7. Making B = 2.
Therefore A and D are 4 and 6 (but we don't yet know which is which).
Since the top row's digits have to add to 9 the top number must be 423.
This makes the sum 423 + 675 = 1098.