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QUANT + DS Questions

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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by kauldheeraj » Sun Aug 30, 2009 5:43 am
Here are two questions which are not difficult , but I feel that these should not be part of any GMAT Exam since I haven't encountered any such question or of the same nature.
Need some logical explanation of these questions:

1. Each of 435 bags contains at least one of the following three items: raisins, almonds, and
peanuts. The number of bags that contain only raisins is 10 times the number of bags that contain
only peanuts. The number of bags that contain only almonds is 20 times the number of bags that
contain only raisins and peanuts. The number of bags that contain only peanuts is one-fifth the
number of bags that contain only almonds. 210 bags contain almonds. How many bags contain
only one kind of item?
(A) 256
(B) 260
(C) 316
(D) 320
(E) It cannot be determined from the given information.

Data Sufficiency:
2.A, B, C, D, and E are airline pilots with very busy travel schedules. Given that D is able to meet at
any time that B cannot meet, do the schedules of A, B, C, D, and E allow three of these five
individuals to meet together for two uninterrupted hours?
(1) Pilots A and C, who cannot meet together, are not able to end any meeting during the AM
hours of any weekday.
(2) Pilots B and E, who can never meet for longer than 2 uninterrupted hours, are only available
to meet for two straight hours starting at 10:30 PM on any weekday and not ending during the AM
hours of any weekend day.
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Source: — Quantitative Reasoning |
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by gmatcoach » Tue Sep 01, 2009 4:26 am
Problem 1:

Answer = 320

Please refer to the attached venn diagram:

R - # of bags with raisins
P - # of bags with peanuts
A - # of bags with almonds
r - # of bags with raisins only
p - # of bags with peanuts only
a - # of bags with almonds only
m - # of bags with raisins and peanuts only
n - # of bags with almonds and peanuts only
l - # of bags with almonds and raisins only
c - # of bads with almonds, raisins and peanuts

Given data:

1. r = 10p
2. a = 20m
3. p = a/5
4. A = 210 (= a+n+c+l)
5. r+a+p+l+m+n+c = 435

We have 5 equations and 7 unknowns. But we need to find a+p+r (sum of 3 unknowns) which may have a unique solution.

From equations 1,3:

a+p+r = a+a/5 + 10(a/5) = 16a/5.

From equation 5,
(r+a+p)+l+m+n+c = 16a/5 + l+m+n+c = 435
substitute for a from equation 2:
64m + l+m+n+c = 435 ==> 65m - 435 = (l+n+c)

From equation 4:
l+n+c = 210-a.

So, 65m - 435 = 210 - a = 210 - 20m ==> 45m = 225 ==> m = 5 ==> a = 100

a + p + r = 16a/6 = 320. Answer is D.
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by kauldheeraj » Tue Sep 01, 2009 8:00 am
That is wonderfully explained and it no more looks like a tough problem at all.
Thanks a lot for your response.
Any tips for Quant, coz Im finding it hard to hit the questions rightly to solve them real fast.
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