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Beginners Forum (Reworked questions)
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20 Jun 2005, 05:26
Hi, I'm starting this thread, reworking on previous problems posted in this forum. I'll try and work out one old problem a day (timepermitting), giving the solution and the explanation/basics behind the question. I hope it's helpful to members since they won't have to go through the entire thread to read the solutions, and also I'll rework the wordings for some of the more ambiguous questions.



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I'll start with this one today.
A slot machine in a Las Vegas casino has an average profit of $600 for each 8 hours shift for the five days sunday through thrusday , inclusive . If the average per shift profit on Friday and saturday is 25% greater than on the other days of the week and the slot machine is in operation every hour of everyday, wheat is the total weekly profit that the casino makes fromt he slot machine? $4500 $9000 $13500 $15500 $27000
This questions tests on your ability to breakdown the long passage, assimilate the information and turn them into equations to work on.
We're told the slot machines has:
1. Average profit/shift = $600 for Sunday  Thursday
2. Average profit/shift = 25% greater on Friday and Saturday
3. Slot machine is in operation every hour of everyday
Since we know the machine works non stop for 24 hours each day, and there's a 8 hour shift, so each day there's 24/8 = 3 shifts
So, from Sunday to Thursday, there's a total of 5*3 = 15 shifts
So the total profit from the 5 days = 600*15 = $9000
We also know from the question, that the average profit/shift = (125/100)*600 = $750
And for the two days, we have 2*3 = 6 shifts
Therefore, the total profit for Friday and Saturday = 750*6 = $4500
So all we need to do now is to add up the two total profits to give the total weekly profit. That sum is $9000+$4500 = $13,500.
The answer is therefore, C.



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A leopard spots a deer from a distance of 200 meters. As the leopard starts chasing the deer, the deer also starts running. Given that the speed of the deer is 10 km/h and that of the leopard is 12 km/h, how far would have the deer run before it is caught?
A. 3 km B. 4 km C. 2 km D. 1 km E. 5 km
This question test the ability to recall the DistanceSpeedTime formula, and also to manipulate the question to formulate the equation for solving the question (by use of algebra).
In such DistanceSpeedTime question (which involves A catching up with B), there's normally one quantity that's similar.
Also, remember to make sure that your units match, otherwise your answer will not come out right.
We know from the passage:
1. The Leopard and the Deer are 200 meters apart = 0.2 km
2. THe speed of the Leopard = 12km/hr
3. The speed of the Deer = 10 km/hr
Let's assume that the Deer has travelled x km before it is caught. The Leopard would then have to travel (x+0.2)km. There is one quantity that is the same here, that is the time taken for the deer to travel that x km, and for the Leopard to travel (x+0.2km).
Since we know that Time = Distance/Speed, we can equate:
Time taken by Deer to travel x km = Time taken by Leopard to travel (x+0.2) km
x/10 = (x+0.2)/12
12x = 10x + 2
2x = 2
x = 1 km
The answer is therefore, D.



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A bicycle rider coasts down the hill, travelling 4ft in the first second. In each succeeding second, he travels 5ft farther than in the preceeding second. If the rider reaches the bottom of the hill in 11 seconds, find the total distance travelled
For this question, what I would do is to brute force the answer, instead of looking at the algebraic way of solving it. It's much faster, and doesn't require much manipulation with algebraic terms. But for the purpose of learning, we'll look at both ways.
BruteForce:
1s  4 ft
2s  9 ft
3s  14 ft
4s  19 ft
5s  24 ft
6s  29 ft
7s  34 ft
8s  39 ft
9s  44 ft
10s  49 ft
11s  54 ft
Total = 4 + 9 + 14 + 19 + 24 + 29 + 34 + 39 + 44 + 49 + 54 = 319 ft
Pros of bruteforcing
 Doesn't require much thought process. Just scribble and add
Cons of bruteforcing
 Error Prone, since adding up under time pressure can lead to mistakes
Algebraic Way:
We know for the first second, the rider travels 4 ft. Subsequent second, he travels 5 ft more than the previous second. From this, we know we're dealing with an arithmetic progression (A.P) where each term is (a+n)
(a = first term, n = arithmetic difference).
The sum of an arithmetic progression, S, is defined as: S = n/2 + {2a + (n1)d} (1)
where n is the number of terms
a is the first term
d is the difference
S = n/2 (a+l) (2)
where n is the number of terms
a = first term
l = last term
Since we only know n, a and d, we use equation (1).
S = 11/2 + {2(4) + (111)5} = 319 ft
Pros of using this method:
You can quickly work out the sum for a large number of terms (e.g. Sum of a A.P with 1000 terms)
Cons of using this method:
 Need to be able to identify an A.P
 Need to be able to recall the formula



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Given 2^199 + 2^199 = 2^X, solve for X.
To work out this problem, you need to be sure of a couple of operations involving indices. These are listed below.
1. You can only add indices with the same base and power. That is, you can add a^b + a^b. But you can't add (a^b + c^d), or (a^b + c^b) or (a^b +a^d).
2. Multiplication of two indices with the same root will result in the solution keeping the base, and the expoenents adding up together. E.g. (a^b)(a^d) = a^(b+d).
For this question, 2^199 + 2^199 can be written as (2*2^199) since we're adding two similar numbers.
Think of it as adding any two similar numbers, for instance, 4+4 would give the same as 2*4.
2*2^199 can be written as (2^1)(2^199). Using rule (2) listed above, this would give us 2^(199+1) = 2^200.
Since 2^200 = 2^x, therefore, x = 200.



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Is X a number greater than zero? (1) x^2  1 = 0 (2) x^3 + 1 = 0
For D.S, I would usually try to understand what the question is asking. This would help to narrow down quickly which statement holds. I will use this method for most of the D.S questions I attempt on.
For this question, it is asking if x is greater than zero. In effect, it's asking if x is positive.
From statement (1), we're given x^2  1 = 0. This can be factorized as (x+1)(x1) = 0, so x can be either 1 or 1. So we can't tell if x is positive, and therefore (1) is not sufficient.
Be carefult when dealing with powers of 2, since they will always give you two answers, one positive and one negative.
From statement (2), we're given x^3 + 1 = 0. This can be equated as x^3 = 1. We know x has to be 1 since no other number would result in x^3 giving 1 as a result. So statement (2) is sufficient.
The answer is therefore B.



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What is Ramâ€™s age? (I) In 15 years Ram will be twice as old as Diana would be (II) Diana was born 5 years ago
From the question, we know we must arrive at only one number, the age of Ram.
I'll start with statement (2), since it's the easiest (shortest and easily understood statement)
If Diana was born 5 yrs ago, then she would be 5 yrs old today. But that's all the information we got, so statement (2) is not sufficient.
From statement (1), we're told in 15 yrs time, Ram will be twice as old as Diana would be in 15 yrs time. This can be put in the form of an equation which gives us R+15 = 2(D+15) where R denotes Ram's age today, and D denotes Diana's age today. But we can't solve the equation, so statement (1) is not sufficient.
Using both statement (1) and statement (2), we know the value of D, Diana's age today from statement (2). We can plug this value into the equation obtained from statment (1) and solve for R.
Note: We do not need to solve for R. All we need to know, is whether R can be solved.
The answer is therefore C.
P.S: For interest, let's solve for R. R+15 = 2(D+15) => R+15 = 2(5+15) => R+15 = 40 => R = 25 yrs old today



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125 passengers commute everyday. Is the median commute per day less than 40km ?
1) 62 passengers commute within 35 kms from their house 2) 63 passengers commute within 45 kms from their house
For this question, you will need to know that the median is the middle number of a series of numbers arranged from smallest to largest.
From the statement, we know that we're dealing with 125 commuters. The median is the middle number, which is 63. Thus, the question is asking "Does the 63rd commuter travel less than 40km per day ?"
Using statment (1), all we know is 62 passengers commute within 35km from their house. But we do not know how these 62 passengers are picked (It could be any order, such as, 1st, 4th, 7th, 9th commuter)
The same problem with statement (2).
Using both (1) and (2), we can rearrange the commuters from 1st to 125th and only know that the median, which is the 63rd commuter could be travelling anywhere between 35km and 45km (since we do not know if this 63rd commuter originally came from the 62 passenger that lived within 35km of their house, or he/she came from the 63 passengers that lived within 45km of their house), and so we do not know if the 63rd commuter does less than 40km a day.
The answer is therefore E.



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During a certain season, a team won 80 percent of its first 100 matches and 50 percent of remaining matches. If the team won 70 percent of its games for entire season, what was the total number of games that team played.
A. 180 B. 170 C. 156 D. 150 E. 105
The fastest way to solve this problem is to equate the sentences, and solve for the unknown.
We're told that the team won 80% of its first 100 games. So we know the team won 80 games out of the first 100 games.
Then we're told that the team won 50% of its remaining matches. Let's assume the number of remaining games = x, then 50% of x would be 0.5x.
Finally, we're told that the team won 70% of its game for the entire season. We know that the total number of games they played = (100+x). 70% of this, is 70+0.7x.
So we can equate:
70+0.7x = 80 + 0.5x
0.2x = 10
x = 50
So the total number of games they played is 100+x = 150. The answer is therefore D.



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From May 1, 1960 to May 1, 1975, the closing price of a share of stock X doubled. What was the closing price of a share of stock X on may 1, 1960?
1. From May 1, 1975 to May 1, 1984, the closing price of a share of stock X doubled.
2. From May 1, 1975 to May 1, 1984, the closing price of a share of stock X increased by $4.50
We're given in the passage:
From May 1,1960  May 1, 1975 (15 yrs) > closing price of stock X doubled. So if the closing price on May 1, 1960 is n, then the closing price on May 1, 1975 is 2n. The question is asking us if we can find a value for n.
Using statement (1), we only know that the closing price on May 1,1984 is twice that of the closing price on May 1, 1975. So if closing price on May 1, 1984 is 2b, then on May 1, 1975, the closing price is b. We have no other information to solve the problem, so (1) is insufficient.
Using statement (2), we know that the closing price on May 1, 1984 increased by $4.50 from the closing price on May 1, 1975. But that's all we know. Therefore, statement (2) is also insufficient.
Using both statement (1) and (2), we know that the $4.50 increase is equivalent to twice the closing price on May 1, 1975. We can therefore find the closing price on May 1, 1975. Since we know from the passage that the closing price on May 1, 1960 is half that on May 1, 1975, we can therefore find out the closing price on May 1 1960.
The answer is therefore C.
Note: Again, we do not need to solve for the exact closing price. Just knowing that it can be solved is sufficient to answer the question. For all D.S questions, this should be the approach in order to save time.



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What is the largest integer N such that 8 less than 5 times N is less than 43?
This question is probably just testing your ability to translate a word problem into an equation and solve for the eqaution. For this question, the equation invovled is an inequality.
So 5 times N = 5N, and 8 less than 5N = 5N8. This value is less than 43.
So 5N8 < 43
5N < 51
N < 51/5
N < 10.2
Since we're asked for an integer value, the largest value of N is therefore 10.



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For all integers p, F is the number of distinct factors of p that are divisible by 3 or 2. If p =36, what is the value of F?
This question is testing your ability to find all the factors of a number.
To solve this question, we need to find all the factors of 36, that are divisible by either 3 or 2.
Primefactorizing 36, we have 36 = 2 * 18 = 2 * 3 * 6 = 2 * 2 * 3 * 3
The factors of 36 are therefore: 1, 2, 3, 4, 6, 9, 12, 18, 36
The factors that are divisible by 3 or 2 = 2, 3, 4, 6, 9, 12, 18, 36
The number of distinct factors, F, is therefore 8.



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The temperature in degrees celsius(C) can be converted to temperature in degrees farenheit (F) by the formular F= (9/5)C+32. What is the temperature at which F=C?
This question is just testing algebraic ability.
We're given F = (9/5)C + 32
Since F = C, therefore we can replace F with C to give C = 9C/5 + 32 => 0.8C = 32 => C = 40 degress celsius. Since C = F, therefore the temperature is also 40 degrees fahrenheit.



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What were the gross revenues from ticket sales for a certain film during the second week in which it was shown.
1. Gross revenues during the second week were $1.5 million less than during the first week.
2. Gross revenues during the third week were $2.0 million less than during the first week
Again, this question asks for a definite figure  the gross revenues from ticket sales from a certain film during the second week. Therefore, we can only answer the question if we can find a way to find that value.
Using statement (1), we're told that the gross revnues during the second week is $1.5m less than the first week. We can't solve any further. So (1) is insufficient.
Using statement (2), we're told that the gross revenues for the third week is $2m less than the first week. Again, we have no other information to sovle the problem. So (2) is insufficient.
Using both statement (1) and (2), we can equate from (1):
W2 + 1.5 = W1 (A)
and we can equate from (2):
w3 + 2 = w1  (B)
By equating both (A) and (B), we still can't find a value for the gross sales during the second week. So the answer is E.



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In a certain pond, 50 fish were caught, tagged and returned to the pond. A few days later, 50 fish were caught again, of which 2 were found to have been tagged. If the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond, what is the approximate number of fish in the pond?
From the question, we know 50 fish were caught and tagged and returned to the pond. So our pond now contains 50 tagged fish. In the second catch, 2 fishes were found to be tagged, and we're told:
% of tagged fish in second catch = % of tagged fish in the pond
The % of tagged fish in second catch is simply (2/50)*100% = 4%
If we assume to total number of fishes to be x, then the % of tagged fish in the pond would be (50/x)*100%
So equating both sides of the equation:
4% = (50/x)(100)%
x = (50*100)/4 = 1250 fishes.



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Town T has 20,000 residents, 60% of whom are female. what percentage of the residents were born in town T?
1. The number of female residents who were born in town t is twice the number of male residents who were NOT born in town T
2. The number of female residents who were NOT born in Town T is twice the number of female residents who were born in Town T
From the passage, we know T has 20,000 residents of which, 60% or (60/100)*20,000 = 12,000 are female, and 8,000 (20,00012,000) are male.
We're asked what percentage of residents were born in town T. This would be found by (# of residents born in town T)/(# of residents in town T) * 100%. Since we already know the denominator, what the question is really asking is the number of residents born in town T.
From statement (1), we're told:
# of female residents born in town T = 2*(# of male residents not born in town T)
We do now have the value for both LHS and RHS, and so statement (1) is not sufficient.
Note: The question stem only gives us the total number of female and male residents, and never broke them down into those that were born in town T, and those that were born outside town T.
From statement (2), we're told:
# of female residents not born in town T = 2*(# of female residents born in town T)
Let's say the number of female residents who were born in town T is x, then the number of female residents who were not born in town T would be (12,000x). We can now equate both LHS and RHS and solve for the number of female residents who were born in town T. However, this still does not answer the question as we're lacking a figure representing the number of male residents born in town T. So statement (2) alone is not sufficient.
Using both (1) and (2), we can now use the value we found in (2), to calculate the number of male residents who were not born in town T, and therefore find the number of male residents who were born in town T. This would allow us to answer the question, and so C is the answer.



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If pq is not equal to 0, is x an interer ?
1. x = 3p2q 2. p = q
The question tells p*q != 0
Note: != denotes not equal
This could mean p*q gives us either a positive value, or a negative value. If pq is positive, p and q could be positive, or p and q could be negative. If pq is negative, either p or q is negative.
Using (1), x = 3p2q. There's a number of combinations of p and q, so statement (1) is not sufficient to answer the question.
Statement (2) alone provides no relation between x and p and q, so statement (2) by itself is also insufficient.
Using both statement (1) and statement (2), we still can't tell if x is an integer or not. Let's say p = q = integer (say 4), then x would be an integer. If p and q are irrational numbers such as 1/10, then x would not be an integer.
The answer is therefore, E.



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Updated on: 20 Jun 2005, 18:28
If x and y are positive, is y > 2? 1. x > 2y 2. x < y + 2
From statement (1), we have x > 2y => y < x/2. So y can take on various positive values dependign on what postive values x takes on. Statement (1) is therefore not sufficient.
From statement (2), we have x < y + 2 => x  y < 2. So now x and y can take on sets of value, limited by the inequality above. If x,y = 3,2, then x  y < 2 but y is not greater than 2. However, if x,y = 10,9, then x y < 2 but y is now greater than 2. So statement (2) insufficient.
Using both statement (1) and statement (2), we have y < x/2 and xy < 2. Now, we'll test for possible x,y pairs that satisfies both inequalities.
If x = 10, y < 5. (say, y = 4), then xy = 104 > 2. So x cannot be any value larger than 10.
If x = 5, y < 2.5 (say y = 2.4), then xy = 2.6 > 2. So x cannot be any value larger than 5.
If x = 3, y < 1.5 (say y = 1.4), then xy = 1.6 < 2. So x can be 3 and below.
If x = 4, y < 2 (say y = 1.9), then xy = 2.1 > 2. So x cannot be 4 and above.
If x = 3.59, y < 1.795 (say y = 1.69) then xy = 1.9 < 2. So Maximum value of x would be approximately 3.59 and y would be less than 2 for all values of x 3.59 and below.
The answer is therefore C.
Note: The question does not say x and y are integers, so we must also consider decimals (or fractions)
Originally posted by ywilfred on 20 Jun 2005, 18:13.
Last edited by ywilfred on 20 Jun 2005, 18:28, edited 2 times in total.



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If X and Y are integers and Y =/x+3/ + /4x/, does y equal 7?
1. x < 4 2. x > 3
Using statement (1), we know x can be 3 and below. This will give us a number of y values, so statement (1) is not sufficient.
Using statement (2), the same problem exists. x can be any value above 3 and this will give us a number of y values. So statement (2) is not sufficient.
Using both statements, we can get the inequality: 3 < x < 4.
Now x will have to take on either values 2, 1, 0, 1, 2 or 3.
If x = 0, y = 7.
If x = 1, y = 7
If x = 2, y = 7
If x = 3, y = 7
If x = 1, y = 7
If x= 2, y = 7
So using both statements will limit values of x that always result in y taking on the value of 7.
The answer is therefore C.



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If x and y are nonzero integers, is x ^y < y ^x
1. x = y^2 2. y > 2
For such questions, the statement is sufficient if we can answer a yes or a no to the questions "Is x^y < y^x".
Using statement (1), we have x = y^2. So if y = 1, x = 1. then x^y is not less than y^x. However, if y = 3, x = 9. Then x^y is less than x^y. So we can't answer the question posed and statement (1) is therefore not sufficient.
Using statement (2), we have y > 2, but we have no values of x to work with. So we still can't answer the question. Statement (2) is therefore not sufficient.
Using both statement (1) and statement (2), we know y is 3 and above. For all values of y that is 3 and above, y^x is always less then x^y. So we can now answer the question  "Is x^y < y^x ?" The answer is no.
The answer therefore is C.







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