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

Examine the sequence:

$5, 6, 8, 11, 15, 20, 26, \square, \bigcirc,...$

What number goes into the circle?

Explanation

Each element is obtained by adding a number to the previous one; the number added increases by 1 each time:

- this number replaces the square

- this number replaces the circle

Which of the following can be the next number after 47 in the sequence below?

${2, 5, 11, 23, 47...}\textup{\ ?}$

Explanation

Each number in the set is the value of the preceding number multiplied by 2, plus 1:

Therefore, the next number in the set will be 47 times 2, plus 1. This gives us:

Given a set $A = \left { 1,3, 6, 7, 8\right }$ , which of the following sets could we assign to such that $A \cup B = \left { 1,2, 3, 4, 5, 6, 7, 8, 9, 10\right }$ ?

$\left { 2, 9,5, 4, 10, 8\right }$

$\left { 6,5,4, 9, 10\right }$

$\left { 2, 4, 9, 10 \right }$

$\left {1, 2, 4, 5, 6, 9, 10, 12\right }$

None of the choices answer the question correctly.

Explanation

$A \cup B$ is the union of the sets, i.e. the set of all elements in , , or both.

For $A \cup B = \left { 1,2, 3, 4, 5, 6, 7, 8, 9, 10\right }$ , must include all elements in $\left { 1,2, 3, 4, 5, 6, 7, 8, 9, 10\right }$ that are not in , or $\left { 2, 4, 5, 9, 10\right }$ . This allows us to eliminate $\left { 2, 4, 9, 10 \right }$ , which excludes 5, and $\left { 6,5,4, 9, 10\right }$ , which excludes 2.

Also, cannot include any element not in $\left { 1,2, 3, 4, 5, 6, 7, 8, 9, 10\right }$ . This allows us to eliminate $\left {1, 2, 4, 5, 6, 9, 10, 12\right }$ , which includes 12.

This leaves $\left { 2, 9,5, 4, 10, 8\right }$ .

If $B = \left { 2, 9,5, 4, 10, 8\right }$ , then

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

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

This is the correct choice.

Find the missing part of the list:

Explanation

The sequence pattern follows the rule:
$x_n=3x_{n-1}-1$

Therefore,

Nine students are running for student council; each member of the student body will choose four.

The candidates include two of Eileen's sisters, Maureen and Colleen. Eileen wants to vote for one or both of them. How many ways can Eileen fill out her ballot so that one or both of her sisters is among her choices?

Explanation

If Eileen wants to choose both sisters, then she will choose two out of the other seven candidates. This is the number of combinations of two out of seven:

$C (7,2) = \frac{7!}{(7-2)!; 2!} = \frac{7!}{5!; 2!} = \frac{5,040}{120 \times 2} = 21$

If Eileen wants to choose one sister, then she will choose three out of the other eight candidates. This is the number of combinations of three out of eight:

$C (8,3) = \frac{8!}{(8-3)!; 3!} = \frac{8!}{5!; 3!} = \frac{40,320}{120 \times 6} = 56$

Since there are also two ways Eileen can choose one sister, by the multiplication principle, she can fill out the ballot $56\times 2 = 112$ ways that include one sister.

Add these two:

Jack noticed that a group of prime numbers followed a pattern in which they increased by 2, then 4, in an alternating pattern:

(11 plus 2 is 13, 13 plus 4 is 17, 17 plus 2 is 19, 19 plus 4 is 23.)

Jack believes that this pattern applies for all prime numbers, not just this group.

Which of the following is the best counter argument to his belief?

The next prime number, 29, breaks the pattern because it is 6 greater than 23.

7 is 4 less than 11.

19 is not a prime number.

Jack made a mistake on some of the addition in the set.

Explanation

While Jack's pattern applies to the group of number that he selected, it will not apply for the next prime number, 29, because it is 6 greater than the previous prime number of 23. Therefore, the correct answer is:

The next prime number, 29, breaks the pattern because it is 6 greater than 23.

Define set $A = \left { 1, 2, 4, 8, 10, a, g, j, r, w \right }$ . Which of the following sets could we define to be set so that $A \cap B = \varnothing$ ?

$B = \left { 3, 6, 7, 9, 12, b, e, h, s, t\right }$

$B = \left { 5,6,7,11,17, f,h,i,k,r\right }$

$B = \left { 3, 5, 6, 7, 9, e, g, t, u, x\right }$

$B =\left { 3, 5, 7, 10, 13, e, f, h, k, y \right }$

$B= \left { 3,5, 6,8,14,d,f,s,t,x\right }$

Explanation

For $A \cap B = \varnothing$ to be a true statement, sets and cannot have any elements in common. We can eliminate four of these choices of by noting that in each, there is one element (underlined) also in :

$\left { 3,5, 6,\underline{8},14,d,f,s,t,x\right }$

$\left { 3, 5, 7, \underline{10}, 13, e, f, h, k, y \right }$

$\left { 3, 5, 6, 7, 9, e, \underline{g}, t, u, x\right }$

$\left { 5,6,7,11,17, f,h,i,k,\underline{r}\right }$

However, $B = \left { 3, 6, 7, 9, 12, b, e, h, s, t\right }$ shares no elements with $A = \left { 1, 2, 4, 8, 10, a, g, j, r, w \right }$ , so their intersection is $\varnothing$ by definition. This is the correct choice.

The Department of Motor Vehicles wants to make all of the state's license numbers conform to two rules:

Rule 1: The number must comprise two letters (A-Z) followed by four numerals (0-9).

Rule 2: Numerals can be repeated but not letters.

Which of the following expressions is equal to the number of possible license numbers that would conform to this rule?

$26 \times 25 \times 10 ^{4}$

$25 ^{2} \times 10 ^{4}$

$26 \times 25 \times P (10,4 )$

$25 ^{2} \times P (10,4 )$

Explanation

The first letter can be any one of the 26 letters, but after this choice is made, since the letter cannot be repeated, there are 25 choices for the second letter. Each of the next four characters can each be one of the 10 numerals, with repetition allowed. By the multiplication principle, the number of possible license numbers is

$26 \times 25 \times 10 \times 10 \times 10 \times 10 = 26\times 25 \times 10 ^{4}$

A pair of fair dice are tossed. What is the probability that the product of the numbers of the faces is greater than or equal to ?

$\frac {2}{9}$

$\frac {1}{4}$

$\frac {1}{3}$

$\frac {1}{6}$

$\frac {5}{18}$

Explanation

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

$\left{ (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6) \right}$

This is eight out of thirty-six, making the probability

$\frac{8}{36} = \frac {2}{9}$ .

Define $f (x) = \frac{1}{x ^{2}+ 9 }$

What is the natural domain of ?

$(-\infty , \infty)$

$(3 , \infty)$

$(-\infty ,3)$

$(-\infty , 3) \cup (3 , \infty)$

Explanation

The only possible restriction of the domain here is the denominator $x ^{2}+ 9$ , which cannot be equal to 0. We can find any such values of as follows: