Originally posted by @iamatiger Thanks 😀
Got any other tricky ones?
I think the key was trying to solve for messenger_speed = k*army_speed.
At first I tried messenger_speed = k+army_speed
But there is no solution for k then, because the army speed does not cancel out.
There's another tricky one on the same page but it looked so complicated I never even tried it.All the others were pretty basic.
Feel free to try this.I got lost half way through reading the question!!
Suppose you were told that half the apples you can see could not be seen and 2/3rds of those not seen can be seen and that you could see 6 dozen apples more than cannot be seen and half the apples that cannot be seen could be seen and 3/4's of the apples that can be seen couldn't be seen and that you would then miss out in seeing half a dozen than you see how many apples do you have?
Originally posted by @venda There's another tricky one on the same page but it looked so complicated I never even tried it.All the others were pretty basic.
Feel free to try this.I got lost half way through reading the question!!
Suppose you were told that half the apples you can see could not be seen and 2/3rds of those not seen can be seen and that you could see 6 dozen apples more ...[text shortened]... nd that you would then miss out in seeing half a dozen than you see how many apples do you have?
half the apples you can see could not be seen Does the book have any hints on what that means?
Also
you would then miss out in seeing half a dozen than you see Is there a word missing in that bit?
Originally posted by @iamatiger [b]half the apples you can see could not be seen Does the book have any hints on what that means?
Also
you would then miss out in seeing half a dozen than you see Is there a word missing in that bit?[/b]
I presume it means that if you could see 12 apples for instance six of them were hidden from view and yes , I think it should read more than you see
Originally posted by @venda I presume it means that if you could see 12 apples for instance six of them were hidden from view and yes , I think it should read [b]more than you see[/b]
Originally posted by @iamatiger I give up, I dont know what it means at all.
I'm not surprised!!.
Anyway in the book it says:-
You can see 144 apples. and couldn't see 432 apples.Therefore,there is a total of 576 apples that you have.
I realise now that's just the answer and there's no explanation.Sorry
I think you'll enjoy this one Tiger.I saw it in a newspaper ages ago and spent ages trying solve it.It's just basic algebra but I had to enlist the help of the internet to solve it.I'm sure you'll be able to do better.it's an age thing!!:-
Five men of different ages under fifty are celebrating their
joint birthday in a pub.Alan's age is the average of their 5
ages.When Bob is three times Alan's age today,the sum of
Alan's and Colin's ages will equal the sum of their five
ages today.Furthermore , Derek will then be 3 times his age
today.and Eric will be one year more than double Bob's age
today.The landlord checked that the youngest of the five was
old enough to buy alcohol.Who is the youngest and how old is
he?
> Dave
That was fun, thanks!
Alan's age is the average of their 5 ages:
{1} 4a = b+c+d+e
When Bob is three times Alan's age today:
in 3a - b years...
the sum of Alan's and Colin's ages will equal the sum of their five ages today:
3a-b+a+3a-b+c=5a
{2} c=2b-2a
Derek will then be 3 times his age today:
3a-b+d=3d
{3} d=(3a-b)/2
Eric will be one year more than double Bob's age today:
3a-b+e=2b+1
{4} e=3b-3a+1
Substitute the expressions for c,d and e in {2},{3},{4} into {1} and rearrange to get:
15a = 11b+2
Some experimentation shows that if both are aged between 18 and 50 then the only possibility is a=28, b=38.
Expressions {2},{3},{4} then give: c=20, d=23, e=31
Originally posted by @iamatiger That was fun, thanks!
[b]Alan's age is the average of their 5 ages:
{1} 4a = b+c+d+e
When Bob is three times Alan's age today:
in 3a - b years...
the sum of Alan's and Colin's ages will equal the sum of their five ages today:
3a-b+a+3a-b+c=5a
{2} c=2b-2a
Derek will then be 3 times his age today:
3a-b+d=3d
{3} d=(3a-b)/2
...[text shortened]... 28, b=38.
Expressions {2},{3},{4} then give: c=20, d=23, e=31