Skip over navigation
Answer: $4$ more five-pence coins than two-pence coins
Forming an equation
Let the number of five-pence coins be $x$
So the number of two-pence coins is $50 - x$
Then $5x + 2(50 - x) = 181$, that is $3x = 81$, that is $x = 27$.
So there are $27$ five-pence coins and $23$ two-pence coins, so there are $4$ more five-pence coins than two-pence coins.
Starting from 25 of each coin
25$\times$2p + 25$\times$5p = 175p
Swap $n\times$2p for $n\times$5p to get 181p
175p $-n\times$2p + $n\times$5p = 181p
$\Rightarrow$ 175p + $n($5p$-$2p$)$ = 181p
$\Rightarrow$ 175p + $n\times$3p = 181p
$\Rightarrow n\times$3p = 3p
$\Rightarrow n$ = 2
After swapping 2 2ps for 2 5ps there are 4 more 5ps than 2ps
Fifty Coins
Age 11 to 14
ShortChallenge Level 
Answer: $4$ more five-pence coins than two-pence coins
Forming an equation
Let the number of five-pence coins be $x$
So the number of two-pence coins is $50 - x$
Then $5x + 2(50 - x) = 181$, that is $3x = 81$, that is $x = 27$.
So there are $27$ five-pence coins and $23$ two-pence coins, so there are $4$ more five-pence coins than two-pence coins.
Starting from 25 of each coin
25$\times$2p + 25$\times$5p = 175p
Swap $n\times$2p for $n\times$5p to get 181p
175p $-n\times$2p + $n\times$5p = 181p
$\Rightarrow$ 175p + $n($5p$-$2p$)$ = 181p
$\Rightarrow$ 175p + $n\times$3p = 181p
$\Rightarrow n\times$3p = 3p
$\Rightarrow n$ = 2
After swapping 2 2ps for 2 5ps there are 4 more 5ps than 2ps
This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.
You may also like
Starting Fibonacci
What is the first term of a Fibonacci sequence whose second term is 4 and fifth term is 22?
