i have 2 tricky math questions?

Question

1) Gilbert, Erin, and Jennifer shared a paper route equally. Since Erin's papers were all delivered in one apartment building, she finished quickly and delivered the last 20 of Jennifer's papers. Erin calculated that she had delivered half of the total papers. How many did Gilbert deliver?

2) A speaker was so boring that one half of the audience left after a few minutes. Five minutes later, one-third of the remaining audience left. Ten minutes later, one-fourth of those remaining left, leaving only 12 people in the audience. How many people were in the audience in the beginning of the lecture?



HINT: These are Algebra questions

thank you

Answer 1

1) If they shared the delivery equally, then they would each deliver p number of papers but Erin delivered 20 of Jennifer's, so
Gilbert = p
Erin = p + 20
Jennifer = p - 20

Total = p + (p + 20) + (p - 20) ---- (1)

Since erin's delivery is 1/2 of the total papers, we have that

p + 20 = [p + (p + 20) + (p - 20)]/2 ----- (2)

From (2) we have that
2(p + 20) = p + (p + 20) + (p - 20)
2p + 40 = p + p + 20 + p - 20

Collect like terms
40 + 20 - 20 = p + p + p - 2p
40 = 3p - 2p
40 = p
p = 40

So Gilbert delivered 40 papers


2) Considering the last sentence - one-fourth of those remaining left, leaving only 12 people in the audience. Since 1/4 left leaving 12, it means that 12 is 3/4 of the audience

If 3/4 = 12
4/4 = (4/4 X 12)/(3/4)
= 4/4 X 12 X 4/3
= 16

Remember that 1/3 left remaining 16, so 16 was the 2/3 of the audience.

If 2/3 = 26
3/3 = (3/3 X 16)/(2/3)
= 3/3 X 16 X 3/2
= 24

Also, remember that this fugure was just 1/2 of the original audience
If 1/2 = 24
2/2 = (2/2 X 24)/1/2
= 2/2 X 24 X 2/1
= 48 which is the total audience

Check ...... working backwards
At ther beginning, there were 48 people in the audience;
After a few minutes, 1/2 left remaining 24

Five minutes later 1/3 of the remaining people left;
1/3 of 24 is 8, so if 8 left, we have 24 - 8 = 16 people left.

Ten minutes later 1/4 of the remaining people left;
1/4 of 16 is 4, so if 4 left, we have 16 - 4 = 12 people left!

Answer 2

Thanks for the hint.

1. Let T be the total, which is what we're wanting to find, let G, E and J be the number Gilbert, Erin and Jennifer delivered, respectively.

T = G + E + J
E = T/2 = T/3 + 20 because Erin figured she delivered half the total, which included her 1/3rd of the total plus 20 more.

If T/2 = T/3 + 20, T/2 - T/3 =20 or 6(T/2 - T/3) = 6*20
3T - 2T = 120 or T = 120

G = T/3 = 40

2. Let T be the total to begin with.

Then T/2 left after a few minutes, leaving T/2 in attendance.
Then T/6 left after five minutes, leaving 3T/6 - T/6 = T/3 .
Then T/12 left after ten minutes, leaving 4T/12 - T/12 = 3T/12 = 12, or 3 T = 12*12 or T = (12*12)/3 = 48 .

To check, 1/2 left in a few minutes leaving 24, 1/3 left after five minutes leaving 16 then 1/4 left after 10 minutes leaving 12.

Answer 3

1) Since they shared a paper route equally, they were all supposed to deliver x papers. However, Erin delivered 20 of Jennifers, so they delivered x, x+20, and x-20.

Now, we know that x+20 = 1/2 the total, and the total is 3x so:

x+20 = 1/2(3x) = 1.5x
.5x = 20
x = 40

So Gilbert delivered 40, Erin 60, and Jennifer 20.

60 is 1/2 of the total of 120


2) After the first group left, 1/2 remained

After the second group left 2/3*1/2 remained

After the last group left, 3/4*2/3*1/2 remained

So we know that 3/4*2/3*1/2*x = 12 if x = original audience, so...

6/24x = 1/4x = 12
x = 48


Hope this helps!

Answer 4

very helpful hint... :/
anyways..

1)
Let the number of papers delivered by Erin be E
Let the number of papers delivered be Gilbert be G
Let the number of papers delivered by Jennifer be J
Let the total number of papers be T

E - 20 = J ~~Eq.1~~
J = G ~~Eq.2~~
T / 2 = E ~~Eq.3~~

sub Eq3 into Eq1:
(T/2) - 20 = J
sub Eq.2 into the above equation
(T/2) - 20 = G
Since Gilbert delivered one third of all the papers, G = T/3
(T/2) - 20 = T/3
(T/2) - (T/3) = 20
(3T/6) - (2T/6) = 20
T/6 = 20
T = 120

G = T/3
120/3
G = 40 papers

Answer 5

1) 40

2) 12 + 4 + 8 + 24 = 48

Answer 6

it appears that evidently like this product is often divisible via 12 -cool! i think of it is going like this: first get rid of a few situations: a million) a b c d are under no circumstances an identical quantity as 0 is divisible via 12 2) if 3 or extra numbers are even, between the products is divisible via 12. why? 2^n -2 is often divisible via 3 for any sequence of three even numbers. (e.g. 2, 6, 8 or 2,4,6 - I plugged in values as much as one hundred) so 3 of the variations will produce a piece of three factors of three) the final last case is 3 or extra unusual numbers. yet unusual numbers minus unusual numbers provide even numbers returned, which will produce a set of things which will provide 12 returned.

Answer 7

x=number of papers each normally deliver
so
G=x
E=x+20
J=x-20
since E=(papers delivered)/2
G+J=the other half
therfore
E=G+J
x+20=2x-20
x=40=G

for the other question
x=original amount of people
since half the people left
you have x*1/2
then since 1/3 of them left, 2/3 of x*1/2 is left so
you have x*1/2*2*/3
and 3/4 of x*1/2*2/3 are left and
you have x*1/2*2/3*3/4 which is 12
simplifying you have
x/4 = 12
x=48

Answer 8

The secong one is 268, and the first one is Gilbert delivered 40.

Answer 9

1). 40

2). 48

Answer 10

sorry i have no idea about the first one, but i believe that the second one is 288. good luck with your math