1. 6
2. 12
3. 44 years.
Let the number of years elapsed on the lease be y.
Then the number of years left on the lease is (99-y).
3y / 4 = 3(99-y) / 5
15y = 12(99-y)
15y = 1188 – 12y
27y = 1188
y = 44
4. “You forgot the spaces between ‘Johnson’ and ‘and’ and ‘and’ and ‘Smith’ ”
5. 720 times
6. 2 cakes
7. 60 legs
For any centipede to have the largest possible number of legs means the largest possible share of the available 122 legs.
Obviously this implies only one centipede. That leaves 10 heads shared between spiders and ants. Since spiders have more legs than ants, we are looking for the least possible number of spiders (1) hence 9 ants.
1 spider and 9 ants account for (8 + 54) = 62 legs. That leaves 60 legs available to the centipede.
8. Carrots $2.99, spring onions $1.99, parsnips $5.99
9. A = 8 B =2 C=1
111A+111B+111C=110B+1001C which reduces to
111A+B=890C
The most 111A can be is 999 so the most (111A + B) can be is 999 + 8 = 1007.
So C can only be 1. For example, if C was 2, 890C would be 1780.
Therefore 111A+B=890
A must be 8 for the equality to be possible,
=> B=2
10. 1009
A possible way of approaching the problem is as follows:
a. Eliminate even numbers.
b. Eliminate numbers ending in 5.
c. Eliminate numbers using the “divisible by 3” rule. Any whole number which is divisible by 3 has digits which add up to a multiple of 3
d. Of the remaining possibilities (1001, 1003, 1007, 1009, 1013, 1019, ………..) begin with the lowest and attempt to divide by successive primes beginning with 7. You can stop when you get to a square root. As 31 squared is 961 and 37 squared is 1361, that effectively means if you get to 31 in your attempt to divide by successive primes you’ve gone far enough.
1001 = 7 X 143 = 7 X 11 X 13 NOT PRIME
1003 is not divisible by 7, 11, or 13. However it is divisible by 17.
1003 = 17 X 59 NOT PRIME
1007 is not divisible by 7, 11, 13 or 17. However it is divisible by 19
1007 = 19 X 53 NOT PRIME
1009 is not divisible by 7, 11, 13, 17, 19, 23, 29 or 31. It is therefore PRIME.