# Veritas Prep Daily GMAT Challenge – Critical Reasoning and Algebra!:

by on June 28th, 2010

Each day this week, we will be challenging the Beat The GMAT community with a Verbal and Quant question that we consider especially difficult. The first 10 people to provide the correct answer to either question in the comments section will win a Veritas Prep GMAT prep book.  Make sure to submit your answer(s) by 5pm PDT today to be eligible for the contest.  Winners will be announced in tomorrow’s Veritas Prep post.  Good luck!

## Critical Reasoning Challenge Problem

Zebra mussels, a nuisance when they clog the intake pipes of nuclear power plants and water plants along the Great Lakes, have some redeeming qualities. Since the mussels feed voraciously on algae that they filter from the water that passes by them, bags of zebra mussels suspended in the discharge streams of chemical plants significantly improve water quality, even removing some hazardous waste.

Which one of the following is most strongly supported on the basis of the statements above, if they are true?

(A)  Zebra mussels arrived in the Great Lakes on transatlantic freighters and, since they have no natural enemies there, are rapidly displacing the native species of clams.
(B)  If the mussels spread to areas of the Mississippi River where native clams provide the basis for a cultured pearl industry, that industry will collapse, since the mussels are unsuitable for such use and would displace the clams.
(C)  There is no mechanical means available for clearing intake pipes by scraping the mussels from them.
(D)  The algae on which the mussels feed would, if not consumed by the mussels, themselves clog the intake pipes of nuclear power plants and water plants.
(E)  Any hazardous waste the mussels remove from chemical-plant discharge will remain in the mussels, if they do not transform it, and they then must be regarded as hazardous waste.

## Algebra Challenge Problem

If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually did.  How many more miles would he have covered than he actually did had he driven 2 hours longer and at an average rate of 10 miles per hour faster on that day?

(A)  100
(B)  120
(C)  140
(D)  150
(E)  160

Free Trial Class This Thursday: Join Brian Galvin, Veritas Prep’s Director of Academic Programs and one of our highest-rated GMAT instructors, this Thursday July 1 for a 3-hour Combinatorics & Probability GMAT lesson.  This is lesson 12 of Veritas Prep’s 14 lesson Complete Course! RSVP TODAY.

This week only – 50% off Veritas Prep GMAT books: Just for this week, take 50% off Veritas Prep GMAT books. Use discount code VPBooks10.

• Verbal : E
Quant : D

• Critical Reasoning Challenge Problem : E
Algebra Challenge Problem : D

• Hi Paras,

Congratulations on answering both challenge problems correctly! At your convenience, can you please send me a private message (VP_Marisa) with the address that you would like the books shipped to?

Thanks!
Marisa Peck

• CR : E
Quant: D

• Critical reasoning : E
Quant : D

• Hi Vibhavender Y,

Congratulations on answering both challenge problems correctly! At your convenience, can you please send me a private message (VP_Marisa) with the address that you would like the books shipped to?

Thanks!
Marisa Peck

• CR: E
Algebra: D

• Hi Shin,

Congratulations on answering both challenge problems correctly! At your convenience, can you please send me a private message (VP_Marisa) with the address that you would like the books shipped to?

Thanks!
Marisa Peck

• CR : E
Quant: D

• sorry,
CR:D
Algebra: D

• lol, glazed eyes + typo
back to my original answer
CR : E
Quant: D

• CR: E
Alzebra

• Alzebra : D

• Critical Reasoning Problem
Ans E

Algebra Problem
Ans D

Explanation:
From the given data (S+5) * ( T +1) = D + 70 where S is the speed of motorist drives for the time T .
on solving we get
5T+S=65------> (1)
another equation is 10T+2S+20 = 2(5T+S) + 20
= 2*65 +20 using (1)
=150

• CR: E

Quant:

1st clue:
(T+1) (S+5) = D+70
ST+5T+S+5=D+70
ST+5T+S=D+65
2ST+10T+S=2D+130 - Multiply by 2 on both sides

2nd clue:
(T+2)(S+10)=D
ST+10T+2S+20=D
ST+10T+2S=D-20

Subtract 1st clue from 2nd clue

ST=2D+130-D+20
ST=D+150

So answer is D. 150.

• cr E
quant D

• Hi Vinodh,

Congratulations on answering both challenge problems correctly! At your convenience, can you please send me a private message (VP_Marisa) with the address that you would like the books shipped to?

Thanks!
Marisa Peck

• let his normal speed be x and he travels for y hrs
so normally he cover xy hrs
now he travel for 1 more hr so time =y+ with speed x+5
so distance travelled (x+5)(y+1)=xy+5y+x+5
now given (xy+5y+x+5 )-xy=70
5y+x=65 --- eq1

now if he drive 2 hrs longer so time travelled y+2 with speed x+10
so distance travelled xx+10(y+2)=xy +10 y+2x+20
distance travelled more=(xy +10 y+2x+20)-xy=10y+2x+20=2(5y+x)+20=2*65+20=150

• critical reasoning D

points to be taken care of:
1)musels feed on algae so this will be removed from water and hazardous waste
option A)transatlantic freighters its not mentioned anywhere
b)we dont know effect of mussel on pearl industry
c)we are not sure wether there are means or not
d)not neccesarily it will clog but yes it will definately detoriate water condition
e)most logical as hazardous waste remove will remain in musse if not transformed as algae will be feed upon and nothing has been done about hazardous waste so it should be transformed

• sorry critical reasoning answer E

• Verbal: D
Quants: D

• Critical Reasoning E
Algebra D

• Verbal:D
a),B) and C) are out of scope
E) Any hazardous is not supported by the passage,the stimulus says some hazardous;hence E is wrong
Quant:D
let tot dis be 100 with speed s and time t,i.e st=100;new dis is (s+5)(t+1);st+5+5t+s=170--->5t+s=65;reqd (s+10)(t+2)-st=10t+2s+20;using 5t+s=65--->10t+2s=130 ans is 150

• Verbal

i feel E is correct
a,b,c felt r out of scope
d is providing additonal information

Algebra

D-distance
s=speed
t=time

D=st (1)
(formulae)

given
d+70=(s+5)(t+1)
d+70=st+s+5t+5 -(2)
sub (1) in (2)
st+70=st+s+5t+5
s+5t=65 - (3)

Question is (t+2)(s+10)? that is how much extra distance it will cover
(t+2)(s+10)
=ts+10t+2s+20
=d+2(5t+s)+20

sub eq (3)
=d+2*(65)+20
=d+150

Extra distance travelled
(d+150)-d=150

• CR : E

Quants: D

• The critical reasoning is a strangely worded question, and one could make the argument that there's no good irrefutable answer, but I think D is the best choice here. The consensus on the board is that E is the answer, but I don't think it's quite right. Here's why: if we were to alter the wording of answer choice E to say that some of the toxins ingested by the mussels were excreted as waste, it wouldn't necessarily undermine the central argument - removing some toxins but leaving others would still have a net positive effect. The key, I think, is that the mussels improve water quality by consuming ALGAE, which means that if the algae themselves don't pose a problem, using the mussels to eat them would be rather pointless. The only answer choice that mentions the potentially problematic impact of the algae is D. The reason I hesitate is that the algae could be harmful without necessarily clogging the pipes, but I still suspect it's the best of the lot.

• Verbal : B
Quant : E

• Verbal - E
Quant - D

• CR: E
Quant: D

• Algebra challenge problem D

• Verbal:D
Quant:D

• My answers are

CR : D
PS: D

• Verbal: D
Quant: D

• Quant : 150 =>D

Cr -> E

• Quant Answer is - 150 (Choice D)

here's the explanation

Let 'S' be the original speed and 'T' be the original time that the biker drives.
Distance covered 'D' would then be equals Speed 'S' * Time 'T'

1. D = ST

The modified speed on a particular day is -> S + 5
The modified time on that particular day is -> T + 1

So new distance travelled on that day is

2. D + 70 = (S+5) * (T + 1)

from 1 and also solving Right Hand Side of 2 we get

ST + 70 = ST + 5T + S + 5

or

3. 65 = 5T + S

now let the new extra distance traveled by the biker is 'X'

so new equation becomes

4. D + X = (S + 10) * (T + 2)

from 1 and also solving Right Hand Side of 4 we get

ST + X = ST + 10T + 2S + 20
OR
5. X = 10T + 2S + 20

Now multiply statement 3 above by 2 to get

6. 130 = 10T + 2S

substitute statement 6 in statement 5 above to get the answer 150.

------------------------------

Verbal

I feel the answer is choice - D

Zebra mussles feed on algae and algae might themselves choke the nuclear plant. so D is my choice.

HTH

Thank you,

Gaurav Malhotra

• My answers are
1) CR -D.
2) Algebra - D.

CR -- 1) D
Algebra -- 2) D

• CR - E
Explanation: If we agree that mussels will remove the hazardous waste from the water, there is no evidence that they will transform the waste, and the waste will still remain in the mussels. Therefore, there is possibility that the mussels will become hazardous waste themselves.

Algebra - D
Let x be the original speed and y be the original time.
xy + 70 = (x+1)(y+5)
xy + 70 = xy + 5x + y + 5
5x + y = 65
From second piece of infomration,
(x+2)(y+10) = xy + 10x + 2y + 20
substituting 5x+y = 65,
xy + 2(5x + y) + 20 = xy + 150
Hence, the answer is 150 miles.

• Q1 : E
Q2 : D

• CR : D
Algebra : D

• Its D and D

• CR - E

A,B,C are out of the question.
E is out since they never mentioned that Algae could clog their pipes.
D is the best answer.

Math - D
Scenario 1
Biker rides extra 70 miles in the extra hour - speed = 70 mph ;
70mph was 5mph more than original speed => original speed = 65mph

Scenario 2
Biker rides extra 2 hours at extra 10pmh - total speed = 65 + 10 = 75mph
Extra distance = 75mph x 2h = 150 miles

• Please feel free to comment !
Thanks

• *typo* CR : E

A,B,C are out of the question.
D is out since they never mentioned that Algae could clog their pipes.
E is the best answer.

(D) The algae on which the mussels feed would, if not consumed by the mussels, themselves clog the intake pipes of nuclear power plants and water plants.

• CR: E
Choice A,B and C are out of scope
Choice D mentions that algae if not consumed would themselves clog the intake pipe. But, the premise mentioned that algae are nothing but water impurity which can be removed by Zebra mussels. So, this is not relevant.
Choice E is the best answer.

DS: D

On a normal day:
Distance(d) = Speed(s)*Time(t) i.e. d = st

On the given day:
d+70 = (s+5)*(t+1)
=> st +70 = st +5t+s+5
=> 5t+s = 65

On the day whose distance is to be calculated, let the extra distance traveled be 'x':
d+x = (s+10)(t+2)
=>st + x = st+10t+2s+20
=>x = 20 + 2(5t+s)
=>x = 20 + 2(65) = 150
Hence choice D is the best answer.

• CR : D ... everything else seemed out of scope. D seemed best out of the pathetic options. You better have good reasoning for the right answer choice, Brain Galvin! Jeez you made think about mussel excreta what is wrong with you folks!

Math : D . Easy question!

• I understand it's late-)) but what not to try?
CR: D
Math: 150

Math explained:
we have 3 equations
1) unknown initial hours, rate and distance
2) h+1 - altered hours, r+5 - altered rate, d+70 - altered distance
3) h+2 - altered hours 2, r+10 - altered rate 2, d+70 +add distance (call it y, f.e)

so we have
1) hr = d
2) (h+1)*(r+5) = d+70
3) (h+2)*(r+10) = d+70+y
we need to find y =)

so we have 70+80=150 miles more =))
2) hr + 5h + r + 5 = hr +70
3) hr + 10h + 2r + 20 = hr + 70 + y
--
2) 5h+r=65
3) 10h+2r= 50+y (lets divide this equation by 2)
---
2) 5h +r= 65
3) 5h+r = 25 +y/2

therefore
65=25+y/2
y/2=40
y=80

70+80= 150

• =))
CR: D
Math: D

• B: For CR ( I have a gut feel that B is not out of scope
Reasoning: Zebra mussels's Redeeming quality is to remove algae even though they clog nuclear plan intakes. When Zebra mussel's are
spread to Missipi rever, they would not have similar redeeming effect because Zebra mussels are not known to be suitable to Pearl industry and would displace clams.

d = st
d+70 = (s+5)(t+1)
st + 5t + 5 + s = d + 70
5t + s = 65 (since d = st) --> (1)

Requred d = (s+10)(t+1)
st + 10t + 2s + 20 = d + 2*(5t + s) + 20
= d + 2*65 + 20 (from (1))
= d + 150 (Thus D).
D: For Algebra

• CR: Answer is E
Algebra: Answer is D
Let Speed be S; Distance be S; Time be T
As per the formula, Speed = Distance/Time ==> S = D/T
Clue 1:
S+5 = (D+70)/(T+1) ==> ST+5T+S+5 = D+70 ==> ST+5T+S-65 = D ==> 65 = 5T+S
Clue 2:
S+10 = (D+x)/(T+2) ==> ST+10T+2S+20 = D+x ==> 10T+2S+20 = x ==> 130+20 = x ==> x = 150

• Thanks everyone for your comments. The winners for this article's challenges have been announced in today's (Tuesday, June 29) article "Sentence Correction and Geometry" challenge. Please submit your answers by 5pm PST to be eligible for the prizes. Good luck!

• I feel there's a much easier way to see the quant problem.

The clue basically states that the motorist drove 1 hour extra and covered 70 miles.
This simply means that for that 1 hour, he was going 70 mph.
Since that was 5 mph faster than before, he was going at 65 mph.

Then all we have to do is 65 mph + 10 = 75 mph for 2 hours = 150 miles extra.

• CR E