Elementary Maths

samliew · 2010-10-19 03:47:23

Is it possible to solve this question without the use of algebra?

"A class has 136 pupils. 5/8 of the boys and 3/4 of the girls wear glasses. There are 93 pupils who wear glasses. How many boys are there in the class?"

bobbym · 2010-10-19 05:17:21

Hi samliew;

The problem leads to a 2 X 2 linear set of simultaneous equations. So although this might be in an Algebra class it technically belongs to Linear Algebra.

Another way of course is by computer.

There are other ways to program this but this what I like:

In pseudocode:

1) Generate all the ordered pairs that total to 136

Output:

(0,136)(1,135)(2,134) ,,, (134,2)(135,1)(136,1) where the first number is the number of boys and the second number is the number of girls.

2) Pick from the above all pairs such that 5 / 8 of the first one plus 3 / 4 of the second one = 93.

Output:

(72,64) means 72 boys and 64 girls

In a functional programming language the program would look like this:

l=Table[{b,136-b},{b,0,136}]

Select[l,5/8 First[#]+3/4 Last[#]==93&]

{72,64}

Welcome to the forum!

samliew · 2010-10-19 09:42:37

I was also wondering about solving the problem that way. Would it also be better to assume that boys = girls = 136/2 = 68, so start testing pairs starting from the middle of the ordered pairs? (if this question happened to appear in a test where external aids like calculators and excel spreadsheets are not allowed)

bobbym · 2010-10-19 10:03:18

Yes, there is a method similar to the way the program works but just a little smarter.
You do not need 137 ordered pairs to start, you only need every 8th one.
Why, because 5 / 8 of the first number in every pair must be a whole number.
You cannot have 5.75 boys. So your list of ordered pairs is only.

Only 18 possibilites. You will now have to try them all as I see no reason other than hindsight to start in the middle. The average number of tries is about 9. Each try should take you about 30 seconds so figure 5 - 6 minutes to solve it.

soroban · 2010-10-20 08:03:17

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bobbym · 2010-10-20 08:08:50

Hi guys;

The average number of guesses before getting the correct answer is 9 ( from the list of 18 ) no matter where you start from.
Starting from the middle is great when you know the answer already.

What I am saying is why is the answer biased towards the center rather then the ends, or anywhere else?

Bob · 2010-10-21 09:57:31

Hi Samliew

Here's my method:

First, note that 136 is divisible by 8 (17x 8) and the numbers of boys must be as well or 5/8 would not give a whole number.

So the number of girls must also be divisible by 8. (since number of boys + number of girls = 136).

So simplify the problem by putting the boys into 8 equally sized blocks and the girls into 8 equally sized groups.

Make a graph with the number in a block across and the number in a group up.

Your problem is solved if I can find the number of boys in a block. (This answer x 8)

The first graph below shows all the possible points that make the number in a block plus the number in a group = 17.

Next simplification. Instead of 3/4 of the girls, use the equivalent fraction 6/8.

5/8 of the boys means 5 blocks and 6/8 of the girls means 6 groups.

So 5 blocks plus 6 groups must add up to 93.

Now I want to find a second line to show possible points for this constraint.

I could guess wildly but a little thought about the numbers lets me find two pairs of coordinates quickly.

93 is divisible by 3 but is odd, so if I choose 3 for the number in a block (that's 5 x 3 = 15 boys) then I need 78 girls (that's 13 in a group).

5 x 3 + 6 x 13 = 93

So one point is (3,13)

Now choose 15 for the number in a block, then I need 3 girls in a group. 5 x 15 + 6 x 3 = 93. So second point is (15,3).

Plot these and join to make the second line. (graph 2)

The lines cross at (9,8) so there are 9 boys in a block => there are 8 x 9 = 72 boys. ( and 8 x 8 = 64 girls)

No algebra!, No guesswork!

Bob

Last edited by Bob (2010-10-21 10:04:22)

bobbym · 2010-10-22 05:53:06

Hi guys;

For that matter it can even done easier. Call x = number of boys and y = number of girls.Form the 2 equations,

Graph them:

http://www.mathsisfun.com/graph/functio … 3333333333

zoom in:

http://www.mathsisfun.com/graph/functio … 8333333333

Starting to see x = 72 and y = 64 by eyeballing, check by plugging in.

Bob · 2010-10-22 21:42:00

hi bobbym

Ok, I admit I went 'all round the houses' to do this, but I was trying to avoid algebra.

If the reason for the original question was to help someone who has an aversion to algebra your

might put them off.

You could argue that 'blocks' and 'groups' are just xs and ys really but they don't look like them.

I once got a remedial (at maths; she was very skilled at avoiding work!) child to construct formulas for an investigation but using jelly babies*. Instead of T = 3x + 17 she was quite happy to write T = 3J + 17 because she thought the Js were just jelly babies. I only told her at the end that she'd been doing algebra!

* In the UK jelly = jello and a jelly baby is a small sweet made of jelly. They're quite popular and the first thing kids like to do is 'bite the heads off'. What that tells us about children I would rather not speculate. But at the end of the lesson you can tidy up by eating the visual aids. That's got to be worth it!

Bob

bobbym · 2010-10-22 21:45:15

No problem Bob. I understand completely. I believe that you solved the problem in the best manner.
He should be satisfied now.

By the way I ate the heads too and I almost turned out okay.

Lance Caelan · 2010-12-27 18:18:01

1) A polygon has n sides. If one of its interior angles is 84 deg, and the other interior angles are each equal to 156 deg, find n.

2a) If a product of two positive integers is a prime number, what must be the value of the smaller integer?
2b) Given that x and y are positive integers, solve x^squared - y^squared = 13

3) Given that 950k is a perfect cube, find the smallest possible value of k.

Please include workings. Very grateful for your help.

________________________
Homework help

John E. Franklin · 2010-12-27 18:33:39

This is inresponse to the 5/8 boys, 3/4 girls wearing lunettes (glasses).
Without algebra, this is my step toward one of many possible solutions, perhaps.
50/80 + 30/40 = 80 pupils, 13 short of 93 pupils.
13 = 5 + 5 + 3, so just add those in and you are done.
(50 + 5 + 5)/(80 + 8 + 8) + 33/44 = 93 pupils
Hence boys = 80 + 8 + 8 = 96 and girls = 44. QED
woops made a mistake above, I forgot the that we must have 136 students total.
And 44 + 96 is 140, so this is wrong.
This is a fun problem, reminds me of pythagorean triples due to the whole number nature requirement.
So to finish properly, take the
60/96 and the 33/44 and adjust them up and down each by 15, three 5's or five 3's.
(60 - 5- 5 -5)/(96-8-8-8) + (33 + 3 + 3 + 3 + 3 + 3)/(44 + 4 + 4 + 4 + 4 + 4)
Then we have:
44 boys/72 + 48 girls/64, which is better because 72 + 64 = 136 students!!! Yeah!!!!

Last edited by John E. Franklin (2010-12-28 03:57:15)

bobbym · 2010-12-27 19:05:57

Hi Lance Caelan;

It would be a lot easier to find your questions if you would make a new thread rather than tacking them on to already answered questions.
You see John was busy giving his solution to the earlier question of this thread. I almost did not see yours.

2a) If a product of two positive integers is a prime number, what must be the value of the smaller integer?

The smaller integer is a 1. A prime number is a number that only has itself and 1 as a factor.

2b) Given that x and y are positive integers, solve x^squared - y^squared = 13

This is done by inspection. the first few squares are 1,4,9,16... We only have to try the first 3. So it is easy to see that 2^2 + 3^2 = 13.

3) Given that 950k is a perfect cube, find the smallest possible value of k.

The answer to this is also very easy. k = 7220. To do this here is a hint, factor 950. Please try to figure it yourself.

Math Is Fun Forum

#1 2010-10-19 03:47:23

Elementary Maths

#2 2010-10-19 05:17:21

Re: Elementary Maths

#3 2010-10-19 09:42:37

Re: Elementary Maths

#4 2010-10-19 10:03:18

Re: Elementary Maths

#5 2010-10-20 08:03:17

Re: Elementary Maths

#6 2010-10-20 08:08:50

Re: Elementary Maths

#7 2010-10-21 09:57:31

Re: Elementary Maths

#8 2010-10-22 05:53:06

Re: Elementary Maths

#9 2010-10-22 21:42:00

Re: Elementary Maths

#10 2010-10-22 21:45:15

Re: Elementary Maths

#11 2010-12-27 18:18:01

Re: Elementary Maths

#12 2010-12-27 18:33:39

Re: Elementary Maths

#13 2010-12-27 19:05:57

Re: Elementary Maths

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