# Challenging word problems (equations) (1 Viewer)

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c) A man is now 5 times as old as his son. In 8 years time, he will only be 3 times as old as his son. How old is the father now?

d) Dave is presently 20 years older than his daughter Alison. If Dave’s age 12 years from now is twice Alison’s age 12 years from now, find their present ages

e) I have some 20 cent coins and 5 cent coins in a box, totalling $2.55.. There are 6 more 5 cent coins than 20 cent coins. How many of each coin do I have? Last edited: #### Qeru ##### Well-Known Member a) When the digits of a two-digit number, neither digit zero, are reversed, the number formed is 36 less than the original number. The sum of the digits of the original number could be: b) In triangle PQR, the length of each side is a whole number of centimetres. Also, PQ is 14 cm longer than PR, and QR is 30 cm longer than PR. Find the minimum possible perimeter of triangle PQR in centimetres. c) A man is now 5 times as old as his son. In 8 years time, he will only be 3 times as old as his son. How old is the father now? d) Dave is presently 20 years older than his daughter Alison. If Dave’s age 12 years from now is twice Alison’s age 12 years from now, find their present ages e) I have some 20 cent coins and 5 cent coins in a box, totalling$2.55.. There are 6 more 5 cent coins than 20 cent coins. How many of each coin do I have?
Let the number be $\bg_white 10X+Y$ where X is the 10s digit and Y is the 1s digit. For example $\bg_white 36=3(10)+6$. We know that when the digits are reversed i.e. $\bg_white 10Y+X$ the number is 36 less than the original number. So:

$\bg_white 10X+Y=10Y+X+36 \implies 9(X-Y)=36 \implies X-Y=4 \implies X=4+Y$.
note that $\bg_white 1 \leq Y. Now we can just start subbing some Y values in to get (X,Y) solutions:
Y=1 gives (5,1)
Y=2 gives (6,2)
Y=3 gives (7,3)
Y=4 gives (8,4)
Y=5 gives (9,5)

So all sums $\bg_white X+Y$ include:
$\bg_white X+Y=6,8,10,12,14$

b) $\bg_white PQ+14=QR$ and $\bg_white PR+30=QR$. Let: $\bg_white PQ=x, PR=y,QR=t$ for simplicity. Then we get:
$\bg_white x+14=t, y+30=t$. Eliminating t gives: $\bg_white x+14=y+30 \implies y=x-16$.Similar to a we want positive sols for y and x. Its pretty easy to see that the smallest positive sol is when $\bg_white x=17$ hence $\bg_white y=1$ and $\bg_white t=31$.This is true since x,y,t are all directly proportional so the smallest x value gives the smallest y and t values. So $\bg_white P_{min}=17+1+31=49cm$

someone else can do c,d,e.