How to find the missing part of a list - ISEE Upper Level Quantitative Reasoning

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Question

A geometric sequence begins as follows:

Which is the greater quantity?

(a) The nine hundred ninety-ninth term of the sequence

(b) The one-thousandth term of the sequence

Answer

The common ratio is $\frac{20}{-10} = -2$ , which means that the terms will alternate in sign. The first term is negative; therefore, all of the odd-numbered terms will be negative and all of the even-numbered terms will be positive. This makes the one thousandth term greater than the nine hundred ninety-ninth.

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Question

Define the universal set, , as follows:

$U = \left { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10\right }$

Also, let

$A = \left { 1, 2, 3, 6, 9\right }$ and $B = \left { 3, 5, 7, 9, 10\right }$

Which of these sets represents the complement of $A \cup B$ ?

Answer

$A \cup B$ , the union of and is the set of all elements in , , or both. Merge the two sets to get:

$A \cup B = \left { 1, 2, 3, 5, 6, 7, 9, 10\right }$

The complement of a set is the set of all elements in the universal set that are not in the set. The only elements in not in $A \cup B$ are 4 and 8, so

$\overline{ \left (A \cup B \right )}= \left { 4,8\right }$

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Question

A class has 25 students. If 60% of them are boys, how many students are girls?

Answer

If 60% of the students are boys, 40% are girls (100 – 60 = 40). Multiply 25 by 40% (25 * 0.4 = 10); therefore, 10 students are girls.

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Question

We can divide the natural numbers into four sets:

$A = \left { 1, 5, 9, 13, 17, ... \right }$

$B= \left { 2, 6, 10, 14, 18...\right }$

$C = \left { 3, 7, 11, 15, 19...\right }$

$D = \left { 4, 8, 12, 16, 20...\right }$

In which of these sets would 197 be a member?

Answer

The sets are divided according to the remainder obtained when each element is divided by 4. 197 divided by 4 yields a remainder of 1; all of the elements of match this description, so $197 \in A$ .

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Question

Which of the following is an example of two sets and such that $A \cap B = \varnothing$ ?

Answer

$\varnothing$ refers to the empty set, the set with no elements; $A \cap B = \varnothing$ if and only if the two sets have no elements in common. In four of these cases, and share an element, which in each of these four choices is underlined:

$A = \left { \Delta , \Gamma , \Theta , \underline{\Lambda} , \Sigma \right }$

$B = \left { \Omega , \Psi , \Phi , \Pi , \underline{\Lambda} \right }$

$A = \left { \textrm{Lincoln, Washington, Jefferson, Jackson, \underline{Madison}}\right }$

$B = \left { \textrm{ Monroe, Kennedy, \underline{Madison}, McKinley , Taft} \right }$

$A = \left { a,v,\underline{r}, w,s,y\right }$

$B = \left { k, p, \underline{r}, q,x\right }$

$A = \left { \underline{2}, 8, 4, 1, 10\right }$

$B = \left { 5, 9, 7, \underline{2}, 0\right }$

$A = \left { \textrm{John, Mary, Fred, Stan, Harry } \right }$ and $B = \left { \textrm{Greg, Larry, Phil, Andy, Joanne }\right }$ do not have an element in common, so this is the right choice.

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Question

Define set $A = \left { 1, 3, 5, 6, 7, 8, 10 \right }$ . How could we define set so that $A \cap B = \left { 1, 5, 6, 8\right }$ ?

Answer

$A \cap B$ is the set of all elements in both and .

We can test each set and determine which elements are shared by that set and :

If $B = \left { 1, 2, 5, 6, 9, 12, 13, 15\right }$ :

then

$A \cap B$

$= \left { \underline{1}, 3, \underline{5}, \underline{6}, 7, 8, 10 \right } \cap \left { \underline{1}, 2,\underline{ 5}, \underline{6}, 9, 12, 13, 15\right }$

$= \left { 1,5,6\right }$

If $B = \left { 1, 2, 4, 5, 6, 10, 11, 12\right }$ :

then

$A \cap B$

$= \left { \underline{1}, 3, \underline{5}, \underline{6}, 7, 8, \underline{10} \right } \cap \left { \underline{1}, 2, 4, \underline{5}, \underline{6}, \underline{10}, 11, 12\right }$

$= \left { 1,5,6, 10\right }$

If $B = \left { 1,5, 10\right }$ :

then

$A \cap B$

$= \left { \underline{1}, 3, \underline{5},6, 7, 8, \underline{10} \right } \cap \left { \underline{1}, \underline{5}, \underline{10}\right }$

$= \left { 1,5, 10\right }$

If $B = \left {5, 6,8, 9, 16\right }$ :

then

$A \cap B$

$= \left { 1, 3, \underline{5}, \underline{6}, 7, \underline{8}, 10 \right } \cap\left {\underline{5}, \underline{6},\underline{8},9, 16\right }$

$= \left { 5,6,8\right }$

If $B = \left { 1, 2, 5, 6, 8, 9, 11, 12\right }$ :

then

$A \cap B$

$= \left { \underline{1}, 3, \underline{5}, \underline{6}, 7, \underline{8}, 10 \right } \cap \left { \underline{1}, 2, \underline{5}, \underline{6}, \underline{8}, 9, 11, 12\right }$

$= \left { 1,5,6, 8\right }$

This is the correct choice.

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Question

A pair of fair dice are tossed. Which is the greater quantity?

(a) The probability that the product of the numbers will be at least .

(b) $\frac {1}{3}$

Answer

Out of a possible thirty-six rolls, the following result in a product of or greater:

This is ten equally probable rolls out of thirty-six, resulting in a probability of

$\frac {10}{36} = \frac{5}{18}$ .

Since $\frac {1}{3} = \frac {6} {18} > \frac {5} {18}$ , (b) is the greater quantity.

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Question

Which is the greater quantity?

(a) The number of odd integers such that

(b) The number of even integers such that

Answer

This question can be most easily answered by matching each element in the set in (a) with the next consecutive integer, which is in the set in (b):

$1 \rightarrow 2$

$3 \rightarrow 4$

$5 \rightarrow 6$

...

$197 \rightarrow 198$

Every element in the second set has a match, but there is an unmatched element in the first set. Therefore (a) is the greater quantity.

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Question

An arithmetic sequence begins as follows:

Which is the greater quantity?

(a) The fourth term of the sequence

(b) 200

Answer

The common difference of the sequence is , so the next two terms of the sequence are:

215 is the fourth term. This makes (a) greater.

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Question

Which quantity is greater?

(a) The number of even integers that satisfy the inequality

(b) The number of multiples of 4 that satisfy the inequality

Answer

The easiest way to answer this is to try to match each element in the first set to one the second set as follows:

$2 \rightarrow 4$

$4 \rightarrow 8$

$6 \rightarrow 12$

...

$998 \rightarrow 1,996$

In other words, each element in the set in (a) is paired with the element in the set in (b) that is its double. Since there is a one-to-one correspondence, the two sets are of equal aize, and (a) and (b) are equal quantities.

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Question

Which quantity is greater?

(a) The number of even integers that satisfy the inequality $0 < x \leq 1,000$

(b) The number of odd integers that satisfy the inequality $0 < x \leq 1,000$

Answer

The easiest way to answer this is to try to match each element in the first set to one in the second set as follows:

$2 \rightarrow 1$

$4 \rightarrow3$

$6\rightarrow 5$

...

$1,000 \rightarrow 999$

Since there is a one-to-one correspondence between the elements of the two sets, (a) and (b) are equal.

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Question

A pair of fair dice are tossed. Which is the greater quantity?

(a) The probability that the product of the two dice will be an even number

(b) $\frac{1}{2}$

Answer

The product of the two dice will be an odd number only if both dice are odd; the rolls favorable to that event are:

$\left { (1,1) , (1,3), (1,5), (3,1) , (3,3), (3,5),(5,1) , (5,3), (5,5) \right }$ ,

or nine out of the thirty-six possible rolls. This makes twenty-seven of the thirty-six equally probable rolls favorable to getting an even product. Since half of 36 is 18, the probability of getting an even product is greater than $\frac{1}{2}$ .

This makes (a) the greater quantity

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Question

A pair of fair dice are rolled. Which is the greater quantity?

(a) The probability that at least one die comes up 5 or 6

(b) $\frac{1}{2}$

Answer

For the roll to be unfavorable to the event that at least one of the dice is 5 or 6, both dice would have to be 1, 2, 3, or 4. There are $4 \times 4 = 16$ ways out of 36 that this can happen, so there are ways for either or both of the two dice to be 5 or 6. Since half of 36 is 18, the probability of this event is greater than $\frac{1}{2}$ .

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Question

Which of the following is the greater quantity?

(a) The sum of the even integers from to

(b) Twice the sum of the integers from to

Answer

The quantities are equal. This can be proved as follows:

The sum of the integers from to is

.

(b) is twice this:

$2\left (1 + 2+ 3 +...+ 500 \right )$

$= 2 \times 1 + 2 \times 2+ 2 \times 3 +...+ 2 \times 500$

This is the same value as (a), the sum of the even integers from to .

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Question

A geometric sequence begins as follows:

Which is the greater quantity?

(a) The fourth term of the sequence

(b) The sixth term of the sequence

Answer

The common ratio of the geometric sequence is

$-14 \div 7 = -2$

The next four terms of the sequence are:

$-14 \times\left ( -2 \right ) = 28$

$28 \times\left ( -2 \right ) = -56$ - the fourth term

$-56 \times\left ( -2 \right ) = 112$

$112\times\left ( -2 \right ) = -224$ - the sixth term

, so the fourth term, which is (a), is greater

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Question

A geometric sequence begins as follows:

Which is the greater quantity?

(a) The fourth element of the sequence

(b) 30

Answer

The common ratio of the sequence is

$\frac{125}{250} = \frac{1}{2}$

The next two terms of the sequence can be found as follows:

$125 \times \frac{1}{2} = 62 \frac{1}{2}$

$62 \frac{1}{2}\times \frac{1}{2} = 31 \frac{1}{4}$

This is the fourth term, which is greater than 30.

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Question

An arithmetic sequence begins as follows:

Which of the following is the greater quantity?

(a) The tenth element of the sequence

(b) 70

Answer

The common difference of the sequence is ; then tenth element is

,

which is greater than 70.

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Question

A standard deck of cards is altered by removing the red queens and the three of the jacks. A card is drawn at random from this altered deck.

Which is the greater quantity?

(a) The probability of drawing a red card

(b) The probability of drawing a black card

Answer

The three removed jacks must either comprise two red cards and one black card, or one red card and two black cards. Since the two red queens are also removed, the net result is either the removal of four red cards and one black card, or three red cards and two black cards. Either way, more black cards than red cards are left in the altered deck, and (b) must be greater.

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Question

An arithmetic sequence begins as follows:

$4 \frac{1}{2} , 6 \frac{3}{4},...$ '

Which is the greater quantity?

(a) The fifth number in the sequence

(b)

Answer

The common difference of the sequence is

$6 \frac{3}{4} - 4 \frac{1}{2} = 6 \frac{3}{4} - 4 \frac{2}{4} = 2\frac{1}{4}$ .

The fifth number in the sequence is

$4 \frac{1}{2} + (5-1) \times 2\frac{1}{4} =4 \frac{1}{2} + 4 \times 2\frac{1}{4} = 4 \frac{1}{2}+ 9 = 13\frac{1}{2}$ .

This makes (b) greater.

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Question

A geometric sequence begins as follows:

Which is the greater quantity?

(a) The fifth term of the sequence

(b)

Answer

The common ratio of the sequence is

$0.02 \div 0.1 = 0.2$ .

The next three terms of the sequence are:

$0.02 \times 0.2 = 0.004$

$0.004 \times 0.2 = 0.0008$

$0.0008 \times 0.2 =0.00016$ , the fifth term.

, making (a) greater.

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