ISEE Upper Level Quantitative Reasoning - Data Analysis and Probability

Below is a table that gives the population of Washington City for five census years.

$\begin{matrix} \textrm{\underline{Year }}& \textrm{\underline{Population}} \ 1970& 3,872\ 1980& 4,239\ 1990& 4,512\ 2000& 4,412 \ 2010& 4,832 \end{matrix}$

Which is the greater quantity?

(a) The population increase in Washington City from 1970 to 1975

(b) The population increase in Washington City from 1975 to 1980

It is impossible to tell from the information given

(a) and (b) are equal

(a) is greater

(b) is greater

Explanation

No information is given in the table about the population of Washington City in 1975. Therefore, it cannot be determined with any certainty which population change was the greater.

Below is a table that gives the population of Washington City for five census years.

$\begin{matrix} \textrm{\underline{Year }}& \textrm{\underline{Population}} \ 1970& 3,872\ 1980& 4,239\ 1990& 4,512\ 2000& 4,412 \ 2010& 4,832 \end{matrix}$

Which is the greater quantity?

(a) The population increase in Washington City from 1970 to 1975

(b) The population increase in Washington City from 1975 to 1980

It is impossible to tell from the information given

(a) and (b) are equal

(a) is greater

(b) is greater

Explanation

No information is given in the table about the population of Washington City in 1975. Therefore, it cannot be determined with any certainty which population change was the greater.

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

(a) The probability that the sum will be at least eight

(b) The probability that the sum will be at most six

(a) and (b) are equal

(a) is greater

(b) is greater

It is impossible to tell from the information given

Explanation

The following table illustrates the ways that the dice can come up. The orange cells indicate a roll of 6 or less; the pink cells indicate a roll of eight or more.

Dice_table

Since each of these outcomes are equally likely, and there are fifteen outcomes that are favorable to each of the two events, the probabilities are the same.

Mark's numeric grade in his Spanish class is determined by five equally weighted hourly tests and a final, weighted twice as much as an hourly test. The highest score possible on each is 100.

Going into finals week, Mark's hourly test scores are 92, 66, 84, 77, and 87. What must Mark score on his final, at minimum, in order to achieve a grade of 80 or better for the term?

Explanation

Mark's grade is a weighted mean in which his hourly tests have weight 1 and his final has weight 2. If we call his final, then his term average will be

$\frac{92 + 66 + 84 + 77 + 87 + 2x}{1 + 1 + 1 + 1 + 1 + 2}$ ,

which simplifies to

$\frac{2x+ 406 }{7}$ .

Since Mark wants his score to be 80 or better, we solve this inequality:

$\frac{2x+ 406 }{7} \geq 80$

$\frac{2x+ 406 }{7} \cdot 7\geq 80\cdot 7$

$2x+ 406 \geq 560$

$2x+ 406-406 \geq 560-406$

$2x \geq 154$

$2x \div 2\geq 154\div 2$

$x\geq 77$

Mark must score 77 or better on his final.

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

$A \cap B$

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

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

This is the correct choice.

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.

Venn

In the above Venn diagram, the universal set is defined as $U = \left { a, b, c, d, e, f, g, h\right }$ . Each of the eight letters is placed in its correct region. Which of the following is equal to $\overline{A\cup B}$ ?

$\overline{A\cup B} = \left { c\right }$

$\overline{A\cup B} = \left { a,c,f\right }$

$\overline{A\cup B} = \left { a,f\right }$

$\overline{A\cup B} = \left { b,c,d,e,g,h\right }$

$\overline{A\cup B} = \left { b,d,e,g,h\right }$

Explanation

$\overline{A\cup B}$ is the complement of $A\cup B$ - the set of all elements in not in $A\cup B$ .

$A\cup B$ is the union of sets and - the set of all elements in either or . Therefore, $\overline{A\cup B}$ is the set of all elements in neither nor , which can be seen from the diagram to be only one element - . Therefore,

$\overline{A\cup B} = \left { c\right }$

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 $\frac{1}{2}$ .