Help me again please
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Help me again please
1. In a safari park there are elephants and penguins. Altogether there are 24 heads and 64 feet. How many elephants are there?
2. Find two numbers a sum of 28 where one number in three times the other.
3. Find two numbers a sum of 84 where one number is 16 more than the answer.
I dont have the answers as they are sample questions. Can you please show me how you worked the answers. Thanks
2. Find two numbers a sum of 28 where one number in three times the other.
3. Find two numbers a sum of 84 where one number is 16 more than the answer.
I dont have the answers as they are sample questions. Can you please show me how you worked the answers. Thanks
Impossible is Nothing.
Re: Help me again please
They can all be done with algebra, but I'll try without.
Elephants and penguins have one head apiece, so there are 24 animals. If they were all penguins, there would be 48 feet. The extra 16 feet must belong to 8 elephants, and the other 16 animals are penguins.sherry_d wrote:1. In a safari park there are elephants and penguins. Altogether there are 24 heads and 64 feet. How many elephants are there?
This means the sum is 4 times the other number, which must be 28/4 = 7, and the first number is 21.sherry_d wrote:2. Find two numbers a sum of 28 where one number in three times the other.
This means 84-16 = 68 must be twice the smaller number, which must be 34, and the larger one is 50.sherry_d wrote:3. Find two numbers a sum of 84 where one number is 16 more than the answer.
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First one:
64 =4e+2p (64 is total of feet, e's have 4 feet and p's have 2)
24 =e+p (24 is total of heads and e plus p must equal that number)
so, changing sides and signs etc, gives you
2p=64-4e and p=24-e
so
2(24-e) =64-4e
48 -2e=64-4e
48+2e=64
2e=64-48
2e=16
e=8
8 elephants. If you'd been asked about the penguins, you'd have substituted for e instead of p.
64 =4e+2p (64 is total of feet, e's have 4 feet and p's have 2)
24 =e+p (24 is total of heads and e plus p must equal that number)
so, changing sides and signs etc, gives you
2p=64-4e and p=24-e
so
2(24-e) =64-4e
48 -2e=64-4e
48+2e=64
2e=64-48
2e=16
e=8
8 elephants. If you'd been asked about the penguins, you'd have substituted for e instead of p.
Thank you all, I struggle a bit with algebra. Are there some resource you can point me towards? DD hasnt done them yet so I have to teach her the basics so anything that can help her or me. The 11+ books I have hardly touch the equations. WP you put me to shame, you make it seem so easy and yet I have my eyes out right now trying to make head and tail of it all
Impossible is Nothing.
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Have a look here for algebra explanations. I found this site good for many maths things.
http://www.mathsisfun.com/algebra/index.html
http://www.mathsisfun.com/algebra/index.html
Some worksheets here
http://www.math-drills.com/algebra.shtml
(Solving linear equations section)
I always get reluctant children to shop
E.g.
I started the day with £25 but now I only have £5 and 5 books. How much did each book cost?
5b + 5 = 25
Children can see that they spent £20 so the books must have cost £4 each.
If you look at the worksheets then you can call the letters whatever you want, Doctor Who characters has worked well in the past, but whatever takes your child's fancy.
3C + 5 = £20
I now have three cybermen and £5. How much were the cybermen?
http://www.math-drills.com/algebra.shtml
(Solving linear equations section)
I always get reluctant children to shop
E.g.
I started the day with £25 but now I only have £5 and 5 books. How much did each book cost?
5b + 5 = 25
Children can see that they spent £20 so the books must have cost £4 each.
If you look at the worksheets then you can call the letters whatever you want, Doctor Who characters has worked well in the past, but whatever takes your child's fancy.
3C + 5 = £20
I now have three cybermen and £5. How much were the cybermen?