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ISEE Upper Level Quantitative Reasoning › Sets

Questions 1 - 10

Which is the greater quantity?

(a) The number of odd integers such that

(b) The number of even integers such that

(a) is greater.

(b) is greater.

(a) and (b) are equal.

It is impossible to tell from the information given.

Explanation

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.

A geometric sequence begins as follows:

Which is the greater quantity?

(a) The fourth element of the sequence

(b) 30

(a) is greater

(b) is greater

(a) and (b) are equal

It is impossible to tell from the information given

Explanation

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.

An arithmetic sequence begins as follows:

Which is the greater quantity?

(a) The fourth term of the sequence

(b) 200

(a) is greater

(b) is greater

(a) and (b) are equal

It is impossible to tell from the information given

Explanation

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.

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) $\frac{1}{2}$

(a) is greater

(b) is greater

(a) and (b) are equal

It is impossible to tell from the information given.

Explanation

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 $\frac{1}{2}$ .

Which is the greater quantity?

(a) The number of odd integers such that

(b) The number of even integers such that

(a) is greater.

(b) is greater.

(a) and (b) are equal.

It is impossible to tell from the information given.

Explanation

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.

The Fibonacci sequence is formed as follows:

$F_{1} = 1$

$F_{2}= 1$

For all integers , $F_{n}= F_{n-1}+ F_{n-2}$

Which of the following is true of $F_{1,000}$ , the one-thousandth number in this sequence?

$F_{1,000} = F_{997}+ 2F_{998}$

$F_{1,000} = F_{997}+ F_{998}$

$F_{1,000} =2 F_{997}+3F_{998}$

$F_{1,000} = 2 F_{997}+ F_{998}$

$F_{1,000} = 3F_{997}+ 2F_{998}$

Explanation

To express $F_{1,000}$ , the one-thousandth term of the sequence, in terms of $F_{997}$ and $F_{998}$ alone, we note that, by definition of the sequence, each term, except for the first two, is equal to the sum of the previous two. Therefore,

$F_{999} = F_{997}+ F_{998}$

Also

$F_{1,000} = F_{998}+ F_{999}$ , and, substituting:

$F_{1,000} = F_{998}+ \left ( F_{997}+ F_{998} \right )$

and

$F_{1,000} = F_{997}+ 2 F_{998}$ ,

the correct choice.

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)

(b) is greater.

(a) is greater.

(a) and (b) are equal.

It is impossible to tell from the information given.

Explanation

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.

A geometric sequence begins as follows:

Which is the greater quantity?

(a) The fourth element of the sequence

(b) 30

(a) is greater

(b) is greater

(a) and (b) are equal

It is impossible to tell from the information given

Explanation

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.

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

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

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

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

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

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

Explanation

$A \cap B$ 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

$A \cap B$

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