OK, here are the answers:
1) Newton's score is 17.
Archimedes=11, Euclid=21 and Pythagoras=15.
Use algebra. A, E, N and P.
(A+E+N+P)/4=16
(A+P)/2=13
(A+E)/2=16
(E+P)/2=18
So.....
A+E+N+P=64
A+P=26
A+E=32
E+P=36
Using trial and error, you can come up with A, E and P. Substitute that into A+E+N+P=64 equation to get N=17.
2) 40 pieces long.
The dimensions are 40x25=1000.
Take away the two corner pieces from each sum (4 corner pieces in all)
38x23=874
874 is 1000-126 outside edges.
Trial and error and algebra.
3) Pythagoras.
The SAGA TROPHY is an anagram of PYTHAGORAS.
4) £6125.
Trial and error again using a calculator.
£125 (last 3 digits)x 49= £6125.
5) £72.80
30% of £80= £24
£80+£24=£104 in December.
30% of £104=£31.20
It reduced by this so.....
£104-£31.20=£72.80 which is January's price.
6) 26 gerbils.
Trial and error again.
Liz has 10 gerbils.
Gives Andy one, Liz has 9, so Andy will have 27, which is 3 times as many.
Gives Liz two, Andy has 24, Liz will have 12, which is twice as much.
7) The number is 9.
Call the number x.
2x+63=x squared
trial and error gives x to be 9, so that 81=81.
EIGHT IS GREAT:
888+88+8+8+8
ODD ONES OUT:
a) 361. All the rest are divisible by 9.
b) a sixth. It is the only fraction that has not got an exact decimal.
c) 1212. All the rest are palindromes (can be read the same backwards as forwards).
NINES IN A HUNDRED:
There are 20 9's in a hundred.
9, 19, 29, 39, 49, 59, 69, 79, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99
DIVISIBLE BY 3:
Yes it is. The number is 12111.
1432 means
1000+
400+
30+
2
SO.........
eleven thousand, eleven hundred and eleven means...
11000+
1100+
11
Which equals 12111 which IS divisible by 3, 4037 times.
FIVE TWOS:
There maybe more than one answer for each, but this is just one solution:
1= 2-2+2- (2/2)
2=2+2+2-2-2
3=(22/22)+2
4=2+(2/2)+(2/2)
5=2+2+2- (2/2)
6=2+2+2+2-2
7=2x2x2-(2/2)
8=2x2x2x (2/2)
9=2x2x2+ (2/2)
CATS AND MICE:
No more cats are needed. If 3 cats can catch 10 mice in 10 minutes, it means that they can catch 1 mouse every minute.
60 mice in 60 minutes means the cats have to catch 1 mouse every minute, so it is the same, hence they already have the required amount of cats.
COMPLETE THE SEQUENCE:
a) 14. The sequence goes: +1,+2, +3, +4
b) 3. The sequence is the triangular numbers, or +2, +3, +4, +5
c) 11. The sequence goes: +2, -3, +4, -5, +6, -7, +8 or the alternate numbers have a pattern of their own.
The alternate numbers go: 6, 5, 4, 3 and
8, 9, 10, hence 11.
d) 63. The sequence goes: x3, x3, x3, x3.
MY OWN PUZZLES:
1) 528.
Let x be Johns number.
Let y be Barrys number.
x-y=17
x= 2y+1
Therefore, 2y+1-y=17
y+1=17
y=16.
x-16=17
So, x=33
33x16=528.
2) 40g.
40g.
This question looks complicated and you might have been sitting there for ages working out all the facts, BUT the weight of Edam DOES NOT change, no matter how high or steep the slope is, or how fast it goes.
3) 27
Let x be the number first thought of.
x squared= (x/3) cubed
x squared= (x/3) * (x/3) * (x/3)
x squared= x cubed/27.
27x sqaured=x cubed
x=27
4) The page number is 77.
Let x be the page number.
Let y be the number of words on that page.
y=13x
(y/11)-14=x
(13x/11)-14=x
(13x/11)-(154/11)=x
13x-154=11x
13x=11x+154
2x=154
x=77.
(The number of words on that page is 77 x 13 =1001.)
5) 6 degrees.
Total clock face is 360 degrees.
Clock is split into 60 seconds/minutes/segments.
Each segment is equal.
Each segment must be 360/60=6 degrees.
There you have it. How well did you do that time?
Easy??? Thought not. The last puzzles were much easier than the first ones, weren't they! |