yr 7 homework?

Question

this is my brothers homework, he is 12 years old. im 20 and i cant do it... please could someone help, i feel quite dumb. :S thanks.
Q1: You have to place four A's in a 4 by 4 square with one letter in each row, in each column, and in each of the two main diagnols. How many different arrangements are there?
Q2: Two packages, each shaped like a cuboid, are tied with three loops of string, one in each of the three directions. For package A the three loops have lenghts 40cm, 60cm, 60cm. while for package B the three loops have lenghts 40cm, 60cm, 80cm. Which package has the greater volume?
Any help VERY much appreciated! I havent done maths in about 4years so am totally stuck!

yeh i thought q 2 wud be B as well, bt then i thought.... thats just too easy compared to the other question? lol. i guess it is then. thanks

Answer 1

there are 4 squares that the A can be in in the first row. Once you have put this A in any square, there are 3 squares left for the second row, if it is not going to be in the same column as the first A.
Using the same logic, there are 2 columns left that the A can go in on the 3rd row. And you are only left with one column in the 4th row.

Therefore there are 4*3*2*1 ways, ie 24 ways.

I am quite good at maths but cannot find a way to mathematically prove how many have one A in each diagonal that works for all cases! So, the only thing I can think of is to just draw it and count them:
Axxx Axxx Axxx Axxx Axxx Axxx
xAxx xAxx xxAx xxAx xxxA xxxA
xxAx xxxA xxxA xAxx xAxx xxAx
xxxA xxAx xAxx xxxA xxAx xAxx
(3rd and 5th)

xAxx xAxx xAxx xAxx xAxx xAxx
Axxx Axxx xxAx xxAx xxxA xxxA
xxAx xxxA Axxx Axxx Axxx xxAx
xxxA xxAx xxxA xxxA xxAx Axxx
(3rd, 4th, 6th)

and so on!!
I hope this helps your brother!!!

(have just worked out the other question and thought I would work it out for you properly, and now disagree with the other answers!!!!

In this, L is length, B is breadth and H is height:

First package:
L+L+H+H=40 =2L+2H
B+B+H+H=60=2B+2H
L+L+B+B=60=2L+2B

length and height are the same.
(60=2B+2H=2L+2B;
60-2B=2H=2L)

40=2L+2H
40=2L+2L
40=4L
10=L
10=H

60=2B+2H
60=2B+2*10
60=2B+20
60-20=2B
40=2B
20=B

Volume = L*B*H
Volume = 10*20*10
Volume = 2,000cm³


Second package
L+L+H+H=40=2L+2H
B+B+H+H=60=2B+2H
L+L+B+B=80=2L+2B

40=2L+2H
60=2B+2H
2H=60-2B=40-2L

60-2B=40-2L
60-40=2B-2L
20=2B-2L

80=2B+2L
20=2B-2L
80-2L=2B
20+2L=2B
80-2L=20+2L
80-20=2L+2L
60=4L
15=L

80=2L+2B
80=30+2B
50=2B
25=B

40=2L+2H
40=30+2H
10=2H
5=H

volume = 15*25*5
=1875cm³

I hope this explains things to your brother!

Answer 2

Q2
Its not quite the way queenbee says but the answer B has the larger volume is true.

the loops are 2longsides+2shortsides; 2 longsides+2tops; 2 shortsides+2tops

remember top is same as base.


I'm still thinking about Q1

Answer 3

Question 2 is "B". = 40X60X80 = 19200. That's obviously more than A.

Answer 4

Wow... not even GCSE maths was that difficult...