Maya and Leo are digitizing a set of archival files. Working together

Question

Maya and Leo are digitizing a set of archival files. Working together at their respective constant rates, they can finish the job in 20 fewer hours than Leo would need to finish the job alone. How many hours would Maya need to finish the job alone, working at her own constant rate?

(1) Working alone, Maya would need 15 fewer hours than Leo would need to finish the job alone.
(2) The two of them can finish the job together in 10 hours.

Adit_ · Answer

We have:
   = 1
(x) + ] = 1

Statement 1:

Maya works 15 fewer hours than Leo.
Which means she works x+5 hours only.
On substituting we find the value of x.
Substitute to find Maya's working hours.
SUFFICIENT.

Statement 2:

Clearly Sufficient.
As we are given the value of x directly, we just need to substitute and find Maya's working hours.

Answer:  Option D

DropletMaverick · Answer

we need to look for rate of each person so 
combined rate =1/10
now if we know combined rate we can easily calculate the rate of Leo =1/30
and hence the rate of Maya will be =1/15
so i think 
 Answer : C

BongBideshini · Answer

Let's say Leo took t hours;
Together they took t - 20

Statement (1)  
Maya’s time = t - 15

Using the work-rate equation:
1/t + 1/(t - 15) = 1/(t - 20)

Multiply through by t(t - 15)(t - 20):
 
(t - 15)(t - 20) + t(t - 20) = t(t - 15)
t^2 - 35t + 300 + t^2 - 20t = t^2 - 15t
2t^2 - 55t + 300 = t^2 - 15t
t^2 - 40t + 300 = 0
(t - 30)(t - 10) = 0
 
t = 30 or t = 10
But t = 10 is impossible, because together time would be -10
so, t = 30
Maya’s time = t - 15 = 15 hours.
 Statement (1) is sufficient.

Statement (2):  
Together they finish in 10 hours.

t-20 = 10
t=30
Using the work-rate equation:
1/M + 1/30 = 1/10
1/M = 1/10 - 1/30 = 1/15
M = 15
 Statement (2) is sufficient.

Answer: D

nemoexplicabo · Answer

Let M be Maya’s time alone and L be Leo’s time alone (hours). “Together in 20 fewer hours than Leo alone” means their combined time is L − 20, so rates add: 1/M + 1/L = 1/(L − 20). 
 Statement (1) gives M = L − 15. Substitute into the rate equation: 1/(L − 15) + 1/L = 1/(L − 20). Multiply everything by L(L − 15)(L − 20) and you get: L(L − 20) + (L − 15)(L − 20) = L(L − 15). Simplify to L^2 − 20L + L^2 − 35L + 300 = L^2 − 15L, so L^2 − 40L + 300 = 0. That quadratic factors nicely as (L − 30)(L − 10) = 0, and Leo cannot take 10 hours if working with Maya would then take −10 hours, so L = 30. Then M = 15. Statement (1) is sufficient. 
 Statement (2) says their together time is 10, so L − 20 = 10, giving L = 30 immediately. Plug into the stem rate equation: 1/M + 1/30 = 1/10, so 1/M = 1/10 − 1/30 = 1/15 and M = 15. Statement (2) is sufficient.  
Answer: D.
Takeaway: in Work/Rates Data Sufficiency, the “difference in time” lives in the denominators, not the rates. Once you write the clean rate equation, it collapses fast.

Maya and Leo are digitizing a set of archival files. Working together

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Maya and Leo are digitizing a set of archival files. Working together

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