Question

Andrea can clean a house 4 times as fast as Denise. When they work together, Andrea and Denise can clean a large house in 8 hours. How many hours would it take Denise to clean the house by herself?

20

15

10

5

Answer 1

@lalaly

Answer 2

@CGGURUMANJUNATH

Answer 3

ANDREA'S one day work =1/4

Answer 4

ok but that is not one the answer choices! @CGGURUMANJUNATH

Answer 5

it is a bit confusing.

Answer 6

@jdoe0001

Answer 7

these are the answers
20

15

10

5

Answer 8

Sorry i found my mistake give me a min

Answer 9

ok

Answer 10

so let's see how much they did in 1hr each \(\large \begin{array}{cccccc}
denise&&andrea&&total\\
\hline\\
\cfrac{1}{d}&+&\cfrac{1}{{\color{blue}{ a}}}&=&\cfrac{1}{8}&\implies \cfrac{1}{d}+\cfrac{1}{{\color{blue}{ 4d}}}=\cfrac{1}{8}
\end{array}\)

Answer 11

@lalaly did you find the mistake!

Answer 12

ok then what should I do next? @jdoe0001

Answer 13

well, solve for "d", which is how long "denise" takes

Answer 14

the idea being that we dunno how long they each really take in 1hr

andrea can do in 1hr 1/(nth) of the work
and that (nth) figure, will depend on her speed, so her work rate is 1/(nth) => 1/a

same goes for "denise", she'll do 1/(nth) of the work in 1hr, thus 1/d
since "a" and "d" are two distinct values
because andrea is faster than denise

but we do know that since they finish the whole house in 8hrs
they can only do 1/8th of the work in 1hr

Answer 15

ok and then @jdoe0001

Answer 16

and then you'd solve for "d"

Answer 17

since denise's rate is 1/d in 1hr
"d" is how long she really takes by herself

Answer 18

so the answer would be B @jdoe0001

Answer 19

well, what did you get?

Answer 20

Wait for the 4d what do I multiply the 4 with?? @jdoe0001

Answer 21

you may want to multiply both sides by "8d" see what that gives you

Answer 22

what i got was 8d/8d + 8d/32d = 1/64d @jdoe0001

Answer 23

\(\bf {\color{blue}{ 8d\times }}\cfrac{1}{d}+{\color{blue}{ 8d\times }}\cfrac{1}{4d}={\color{blue}{ 8d\times }}\cfrac{1}{8}\implies
\cfrac{8d}{d}+\cfrac{8d}{4d}=\cfrac{8d}{8}\)

can we cancel out anything terms there from the fractions?

Answer 24

the d! @jdoe0001

Answer 25

\(\bf \cfrac{8\cancel{d}}{\cancel{d}}+\cfrac{\cancel{8d}}{\cancel{4d}}=\cfrac{\cancel{8}d}{\cancel{8}}\)

Answer 26

so.. what do you think we're left with?

Answer 27

8 =d

Answer 28

@jdoe0001

Answer 29

well... \(\bf \cfrac{8d}{4d}\implies \cfrac{8}{4}\cdot \cfrac{d}{d}\) so... what would our \(\cfrac{8}{4}\) turn into?

Answer 30

the 1st fraction gives 8, yes
the 2nd gives something else

Answer 31

2

Answer 32

@jdoe0001

Answer 33

ahhhh so we end up with \(\bf {\color{blue}{ 8d\times }}\cfrac{1}{d}+{\color{blue}{ 8d\times }}\cfrac{1}{4d}={\color{blue}{ 8d\times }}\cfrac{1}{8}\implies
\cfrac{8d}{d}+\cfrac{8d}{4d}=\cfrac{8d}{8}\implies 8+2=d\)

now you can see how long "denise" takes by herself :)

Answer 34

ITS 10! OH OK! @jdoe0001

Answer 35

hmmm lemme rethink that one for a sec...

Answer 36

somehow it has a bad taste =)

Answer 37

oh ok! @jdoe0001 can you help me on another one!

Answer 38

hmmm something tells me that 10 is Andrea's =(

Answer 39

hehhe

Answer 40

let's see

andrea is 4 times as fast as denise
so whatever denise speed is, andrea's has to be 4 times that much

Answer 41

lemme quickly rewrite it

Answer 42

ok

Answer 43

are those the only choices you have? I seem to be getting 40

Answer 44

Yeah those are the only ones!

Answer 45

hmmm you see, it can't be 10hrs shoot

if both take 8hrs together, and andrea is faster
andrea would take about 10hrs

denise would take 4 times 10, or 40

Answer 46

each one will take longer than 8hrs
because would be working alone

Answer 47

the only way I can see one of the choices, is if andrea is only 2 times as fast

Answer 48

so.... it can't be 10 for denise
because that'd meab 1/4 of 10 for andrea

meaning 2.5hrs so andrea would finish the whole house in 2.5hrs
while working with denise, will take 8hrs lol heheh

so that'd imply andrea is a speeding bullet by herself, but with denise she's bogged down =)

Answer 49

but I'd say based on the info provided, andrea being 4 times as fast
I'd only get 40hrs for denise

Answer 50

@jdoe0001
last one please

Two bricklayers, Sam and Joe, are working on your house. Sam can complete the work in 5 hours, while Joe can complete the work in 3 hours. How many hours does the bricklaying take if they work together?

15/8

15/9

6/8

4/3

Answer 51

@mary.rojas

Answer 52

the one all the way at the bottom! @mary.rojas

Answer 53

the one that says about two bricklayers @jdoe0001

Answer 54

same layout pretty much, Sam in 1 hr can do 1/5 of the work
Joe 1hr can do 1/3 of the work

that's their work rate
in 1 hr they'll do 1/t of the work

if they put their RATE together, how long will it take?
well \(\large \large \begin{array}{cccccc}
sam&&joe&&total\\
\hline\\
\cfrac{1}{5}&+&\cfrac{1}{3}&=&\cfrac{1}{t}
\end{array}\)

they'll finish together at "t" time

Answer 55

now if we multiply both sides by "15t" then we end up with \(\bf
15t\times \cfrac{1}{5}+15t\times \cfrac{1}{3}=15t\times \cfrac{1}{t}\implies \cfrac{15t}{5}+\cfrac{15t}{3}=\cfrac{15t}{t}
\\ \quad \\
3t+5t=15\)

Answer 56

5t=15 t=3

Answer 57

@jdoe0001

Answer 58

well, 3t+5t = 8t so 3t+5t=15 => 8t = 15

then divide both sides by 8 :)

Answer 59

15/8 @jdoe0001

Answer 60

15/8 \(\checkmark\)

Answer 61

@jdoe0001 What is the simplified form of 4x^2-25/2x-5

2x - 5, with the restriction x≠-5/2
2x + 5, with the restriction x≠5/2
2x - 5, with the restriction x ≠ 5/ 2
2x + 5, with the restriction x ≠ -5/ 2

Answer 62

I got 2x+5 but I am not sure of the restriction. I believe it is the second one!

Answer 63

@jdoe0001

Answer 64

Is it B? @jdoe0001

Answer 65

is a rational
so restrictions will apply
no to make it undefined
meaning the denominator cannot be 0
meaning "x" shouldn't take anything that makes the denominator 0

so if you set the denominator to 0 and solve for "x", that'd give it out

Answer 66

2x-5 = 0 => x = 5/2, thus \(\bf x\ne \frac{5}{2}\)

Answer 67

Much simpler way of working the Andrea and Denise problem:

Denise's rate is 1 house in \(d \text{ hours} = \dfrac{1}{d}\)
Andrea's rate is 4x Denise's rate, or \(4*\dfrac{1}{d} = \dfrac{4}{d}\)

Their combined rate is \(\large\frac{1}{d}+\frac{4}{d} = \frac{5}{d}\)

1 house in 8 hours working together means

\[\frac{1}{\frac{5}{d}} = 8\]or\[\frac{d}{5}=8\]\[d=40\]So Denise takes 40 hours to do 1 house at her speedy pace of 1/40 house/hour. Andrea works 4 times faster, so she does 4 houses in 40 hours or 1 house in 10 hours, giving her a rate of 1/10 house/hour.

I agree that none of the available answers are correct for the problem as stated.