Arithmetic - GMAT Quantitative

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Question

Let be the set of all of the multiples of 3 between 29 and 50. How many subsets of can be formed?

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Answer

The multiples of 3 between 29 and 50 are 30, 33, 36, 39, 42, 45, and 48 - therefore, has seven elements total.

The number of subsets in a set can be calculated by raising 2 to the power of the number of elements. Therefore, the answer to our question is $2^{7} = 128$ .

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Question

Data sufficiency question- do not actually solve the question

Find the mean of a set of 5 numbers.

1. The sum of the numbers is 72.

2. The median of the set is 15.

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Answer

Statement 2 does not provide enough information about the mean as it can vary greatly from the median. Statement 1 is sufficient to calculate the mean, because even though it is impossible to calculate the set of numbers, the mean is calculated by dividing the sum by the total number of incidences in the set.

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Question

How much greater is the average of the integers from 500 to 700 than the average of the integers from 60 to 90?

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Answer

In this case, average is also the middle value.

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Question

Data sufficiency question- do not actually solve the question

A bag of marbles consist of a mixture of black and red marbles. What is the probability of choosing a red marble followed by a black marble?

1. The probability of choosing a black marble first is $\small \frac{1}{3}$ .

2. There are 10 black marbles in the bag.

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Answer

From statement 1, we know the probabilty of choosing the first marble. However, since the marble is not replaced, it is impossible to calculate the probability of choosing the second marble. By knowing the information in statement 2 combined with statement 1, we can calculate the total number of marbles initially present.

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Question

A certain major league baseball player gets on base 25% of the time (once every 4 times at bat).

For any game where he comes to bat 5 times, what is the probability that he will get on base either 3 or 4 times? - Hint – add the probability of 3 to the probability of 4.

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Answer

$\small n=5$

$\small p=.25$

$\small x=3$

$\small prob = .088$

$\small n=5$

$\small p=.25$

$\small x=4$

$\small prob = .015$

$\small .088 + .015 = .103$

Binomial Table

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Question

Assume that we are the immortal gods of statistics and we know the following population statistics:

average driving speed for women=50 mph with a standard deviation of 12

average driving speed for men=45 mph with a standard deviation of 11

We look down from our statistical Mount Olympus and notice that the Earth mortals have randomly sampled 60 women and 65 men in an attempt to detect a significant difference in the average driving speed.

What is the probability that the Earth mortals will properly reject the assumption (i.e. the null hypothesis) that there is no significant difference between the average driving speeds. The Earth mortals have decided to use a 2-tailed 95% confidence test.

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Answer

standard deviation of the difference between the sample means =

$\sqrt{12^{2}/60 + 11^{2}/65}=2.064$

at 95% (2-tailed) = 1.96

$1.96=(\bar{w}-\bar{m})/2.064$

the sample difference must be 4 or greater

(note: the probability of the sample difference being -4 or less is so small (4.5 standard deviations) that it will be ignored and we will only consider the probability that the difference is 4 or more.)

the probablity of the sample difference being 4 or greater (knowing that the population difference is 5) =

the table shows that .3156 lies below -.48, so, .6844 lies above -.48

In English - there is a .6844 probability that the 2 sample means will yield a sample difference that is 1.96 or more standard deviations above 0.

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Question

In a popular state lottery game, a player selects 5 numbers (on 1 ticket) out of a possible 39 numbers. There are 575,757 possible 5 number combinations.

$\frac{39!}{(5!*34!)}=575,757$

So, the odds are 575,757 to 1 against winning.

What are the odds of getting 4 of the 5 numbers correct on 1 ticket?

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Answer

575,757 must be divided by -

$\left ( \frac{5!}{(4!\cdot 1!)} \right )\cdot \left ( \frac{34!}{33!\cdot 1!} \right )=170$

$\frac{575,757}{170}=3386.8$

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Question

One card from another deck is added to a standard deck of fifty-two cards. The cards are shuffled and one card is removed.

A card is then drawn at random. What is the probability that that card is an ace?

Statement 1: The card that was added was a spade.

Statetment 2: The card that was removed was a jack.

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Answer

You need to know two things to answer this question - the rank of the added card, and the rank of the removed card. The second statement is useful but not sufficient; the first is irrelevant to the question.

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Question

Your friend at work submits a bold hypothesis. He suggests that the number of sales per day follow the pattern - Monday-10%, Tuesday-10%, Wednesday-10%, Thursday 35% and Friday 35%.

You and he then record the number of sales for the following week: Monday-120, Tuesday-85, Wednesday-105, Thursday-325 and Friday-365.

After viewing the observed data, your friend expresses serious concern regarding his hypothesis.

You can help; you can tell him the probability of the observed data occuring if the hypothesis is true. Hint - Excel ChiTest.

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Answer

Use Excel ChiTest to get the .063 probability. If you are old-fashioned, you can also obtain the Chi-Squared number (8.928) by using ChiInv; but, it is not needed

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Question

A certain tutor boasts that his 2 week training program will increase a student's score on a 2400 point exam by at least 100 points (4.167%). A 10 student 'before-and-after' study was conducted to validate the claim. The following results were obtained - the 3 columns represent the before, afer and increase numbers for each of the 10 students:

1300 1340 40
1670 1790 120
1500 1710 210
1360 1660 300
1580 1730 150
1160 1320 160
1910 2100 190
1410 1490 80
1710 1880 170
1990 2060 70

Assume the null Hypothesis:

'The average increase is less than 100 points'

What is the highest level of significance (p-value) at which the null hypothesis will be rejected?

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Answer

Using Excel, the average increase (column 3) is 149 and Standard Deviation of the increases is 76.37

$76.37/\sqrt{10}=24.15$

Using Excel - a t value of 2.03 for 9 degrees of freedom = .036

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Question

A test for a new drug was conducted. In the control or placebo group, 7 of 210 participants experienced positive results. The group that took the drug experienced 27 out of 374 positive resluts.

The placebo group had a sucess rate of .0333 and the drug group had a success rate of .0722. The difference is .0389 and the overall percentage (for both groups combined) is .0582

At what level is the difference of .0389 significant? Asked another way - what is the p-value for .0389?

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Answer

The standard error of the difference (.0389) =

$\sqrt{(.0582 * (1-.0582)) * (1/374 + 1/210)} = .02$

test statistic -

from the table (or excel NormsDist) - Z=1.9245 translates to .9729

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Question

A type 1 error (False Alarm or 'Convicting the innocent man') occurs when we incorrectly reject a true null hypothesis.

A type 2 error (failure to detect) occurs when we fail to reject a false null hypothesis.

Which one of the following 5 statements is false?

Note - only 1 of the statements is false.

A) For a given sample size (n=100), decreasing the significane level (from .05 to .01) will decrease the chance of a type 1 error.

B) For a given sample size (n=100), increasing the significane level (from .01 to .05) will decrease the chance of a type 2 error.

C) The ability to correctly detect a false null hypothesis is called the 'Power' of a test.

D) Increasing sample size (from 100 to 120) will always decrease the chance of both a type 1 error and a type 2 error.

E) None of the above statements are true.

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Answer

Statements A, B, C, and D are all true - so -

The only false statement is E (the statement that declares that A and B and C and D are all false)

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Question

Four coins - a fair penny, a loaded penny, a fair nickel, and a loaded nickel - are tossed. What is the probability that all four will come up heads?

Statement 1: Yesterday, out of 100 tosses, the loaded penny came up heads 70 times.

Statement 2: Yesterday, out of 100 tosses, the loaded nickel came up heads 40 times.

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Answer

While experiments such as repeated tossings can give an idea of the probability that a coin will come up heads or tails, they do not provide a definitive answer, so neither statement is helpful here.

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Question

Jerry is a Cardinal fan and he and his family live on a street with 9 other families that are all Cardinal fans. One block to the north, there are 11 families and they are all Cub fans. These 21 households all buy their lawn fertilizer from Ben's Lawn and Garden Shop. Jerry suspects that Ben (who is originally from Chicago) is a Cub fan and that he provides better fertilizer to the Cub fans than to the Cardinal fans, while charging the same price for all.

Last Saturday everyone in town mowed their lawn. At 2:00 AM Sunday morning, Jerry snuck around town and weighed all of the grass clippings for the 21 households in question.

The weights (in lbs) of the grass clippings for the 10 Cardinal homes were:

82, 85, 90, 74, 80, 89, 75, 81, 93, 75

The weights (in lbs) of the grass clippings for the 11 Cub homes were:

90, 87, 93, 75, 88, 96, 90, 82, 95, 97, 78

The Cardinal average was 82.4; the Cub average was 88.27.

At what level is the 5.87 lb difference significant? - Asked another way - what is the p value for the 5.87 lb difference.

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Answer

Cub variance = 53.218; Cardinal variance = 45.378.

The standard deviation of the difference (5.87) is:

$\sqrt{88.27/11 + 82.4/10} = 3.062$

TDIST of 1.918 with 19 Degrees of Freedom = .035

So, we would reject the null hypothesis (the hypothesis that claims that the means are equal) at 95% confidence (p=.05) and not reject at 99% (p=.01)

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Question

A marble is selected at random from a box of red, yellow, and blue marbles. What is the probability that the marble is yellow?

There are ten blue marbles in the box.

There are eight red marbles in the box.

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Answer

To determine the probability that the marble is yellow we need to know two things: the number of yellow marbles, and the number of marbles total. The first quantity divided by the last quantity is our probablility.

But the two given statements together only tell us that eighteen marbles are not yellow. This is not enough information. For example, if there are two yellow marbles, the probability of drawing a yellow marble is $\frac{2}{20} = 0.1$ . But if there are twenty-two yellow marbles, the probability of drawing a yellow marble is $\frac{22}{40} = 0.55$

Therefore, the answer is that both statements together are insufficient to answer the question.

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Question

Several decks of playing cards are shuffled together. One card is drawn, shown, and put aside. Another card is dealt. What is the probability that the dealt card is red, assuming the first card is known?

The card removed before the deal was red.

The cards were shuffled again between the draw and the deal.

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Answer

To answer this question you need to know two things: the number of red cards left and the number of total cards left. The second statement is irrelevant, as a reshuffle does not change the composition of the deck. The first statement tells you that there is one fewer red card than black cards, but it does not tell you how many of each there are, as you do not know how many decks of cards there were.

And that information, which is not given, affects the answer. For example, if there were four decks, there were 103 red cards out of 207; if there were six decks, there were 155 red cards out of 311. The probabilities would be, respectively,

$\small \frac{103}{207}\approx0.4976$

and

$\small \small \frac{155}{211}\approx0.4984$ ,

a small difference, but nonetheless, a difference.

The correct answer is that both statements together are insufficient to answer the question.

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Question

Data sufficiency question

What is the probability of choosing a red marble at random from a bag filled with marbles?

1. There are only red and black marbles in the bag

2. $\frac{1}{3}$ of the marbles are black

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Answer

From statement 1, we learn that there are only two different colored marbles in the bag. From statement 2, we learn that $\frac{1}{3}$ are black which tells us that $\frac{2}{3}$ are red. Without statement 1, it is impossible to determine if there is another color of marble in the bag.

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Question

A pair of dice - one fair, one loaded, but each with the usual numbers 1-6 on their faces - are rolled. What is the probability that one die will show a 5 and the other will show a 6?

Statement 1: The probability of rolling a 5 on the loaded die is $\frac{1}{5}$ .

Statement 2: The probability of rolling a 6 on the loaded die is $\frac{1}{8}$ .

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Answer

The probability of rolling a 5 and the probability of rolling a 6 on the fair die are both $\frac{1}{6}$ .

Let the probabilities of rolling a 5 and a 6 on the loaded die be and , respectively.

The probability of rolling a 6 on the fair die and a 5 on the loaded die is $\frac{1}{6} \cdot P$ .

The probability of rolling a 5 on the fair die and a 6 on the loaded die is $\frac{1}{6} \cdot Q$ .

Therefore, the probability of rolling a 5-6 one way or the other is their sum:

$\frac{1}{6} \cdot P + \frac{1}{6} \cdot Q$

which is dependent on both probabilites given in the two statements.

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Question

Two fair dice are rolled. What is the probability that each die (not the sum of the dice) will come up a composite number?

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Answer

There are two composite numbers on each die, 4 and 6. On one die, the probability of rolling a 4 or a 6 is $\frac{2}{6} = \frac{1}{3}$ ; the probabililty of rolling two composite numbers is $\frac{1}{3} \cdot \frac{1}{3} = \frac{1}{9}$ .

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Question

Four cards are removed from a standard deck of fifty-two cards. Did the probability that a randomly drawn card will be a club increase, decrease, or stay the same?

Statement 1: The four cards that were removed were all kings.

Statement 2: One card of each suit was removed.

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Answer

Thirteen of the fifty-two cards in a standard deck - one-fourth - are clubs, so the probability of drawing a club is $\frac{1}{4}$ .

Either statement tells us that one card of each suit was removed (there is one king of each suit), leaving twelve clubs out of forty-eight cards - one fourth of them. This keeps the probability of drawing a club at $\frac{1}{4}$ .

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