Brain Teaser

#1
First person to answer the following (showing working) wins a NAAFI pie:

How many soldiers are there in a group of 27 sailors and soldiers if there are four fifths as many sailors as soldiers?

(A) 6
(B) 15
(C) 21
(D) 22


:)

No, I'm not sat in the Kremlin on a Level 1 Maths resit! This is one of the enlistment apptitude questions for the Canadian Forces (CF) and most of the people I work with couldn't even do it.

8O
 
#3
go on then...
4/5 = sailors as a fraction of group.
therefore, for every 5 soldiers, there will be 4 sailors.
27 in group, which splits to 3 lots of [4 sailors] and [5 soldiers]. (working= 4+5 =9 ---> 27/9=3)

4(sailors) * 3 = 12 sailors in total.
5(soldiers) * 3 = 15 soldiers in total.

Answer B.
 
#4
Well done that man!
Quick calculation: 4 fifths plus 5 fifths (as in 4 fifths of the OTHER half) = 9; 27/9 = 3; 4 x 3 = 12. Looks like the guy who sat the test couldn't even do it!
 
#6
4/5 the quantity of soldiers are sailors.

100 soldiers would mean 80 sailors etc ( just to explain the math)

We know that for every 5 soldiers there are 4 sailors

5 + 4 = 9
5 + 4 = 9
5 + 4 = 9

15 soldiers + 12 sailors = 27 service persons

B
 

Fugly

ADC
DirtyBAT
#8
4/5 sailors to soldiers. 9 men.

27/9 = 3 groups.

3*4 sailors = 12 sailors
3*5 soldiers = 15 soldiers.

B.
 

Fugly

ADC
DirtyBAT
#12
verticalgyro said:
Fugly said:
4/5 sailors to soldiers. 9 men.

27/9 = 3 groups.

3*4 sailors = 12 sailors
3*5 soldiers = 15 soldiers.

B.
Nice one slowpoke.

Typical Tiff, waiting until the DS answer is already revealed and then posting it in a slightly different format to make it look like your own work.

It's Wednesday afternoon, shouldn't you be out playing Golf whilst deciding which poor cnut is having to work the weekend to cover up for your planning failures?
The answer wasn't up there when I started typing, but I'm using my extra wide head dobber and I couldn't find my programmable calculator. I had to ask the cleaner in the end.
 
#13
genuine_exscaley said:
Well done that man!
Quick calculation: 4 fifths plus 5 fifths (as in 4 fifths of the OTHER half) = 9; 27/9 = 3; 4 x 3 = 12. Looks like the guy who sat the test couldn't even do it!
Nice one fella, completely wrong. :p

I thought I'd chuck in some algebra just to prove the 18-months wasn't wasted:

Being an obvious fan of meat filled pastries, I will use the analogy of pies.

The complete group (soldiers) is a whole pie (P). This means that the incomplete group is 4/5 of P. Both added both together make 27, therefore the simple algebraic equation is:

P + 4P = 27
5

We now multiply the whole equation by 5 to get rid of the fractions:

5P + 20P = 135
5

This cancels down to: 5P + 4P = 135

Therefore: 9P = 135

Therefore: P = 15

As P = Soldiers, there are 15 soldiers in the group.
 
#14
Wouldn't it be simpler to just separate those wearing DPM from those wearing dark blue and then count them ?
 

Fugly

ADC
DirtyBAT
#15
MOC_411 said:
We now multiply the whole equation by 5 to get rid of the fractions:

5P + 20P = 135
5
[pedant]

After multiplying by 5, you didn't remove the denominator of 5. Marks deducted.

:D

[/pedant]
 
#16
lacrabat said:
Wouldn't it be simpler to just separate those wearing DPM from those wearing dark blue and then count them ?
Nah. Count all the green legs and then divide by 2.

Wait a min. This count, its not at Headley Court is it? Do I get a point for showing my working?
 
#17
"Nice one fella, completely wrong. :p"

Just a question of NRTFQ. Blushing furiously but laughing at his own stupidity.
 
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