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

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Question

An arithmetic sequence begins

Which is the greater quantity?

(a) The fiftieth term of the sequence

(b)

Answer

The common difference of the sequence is , so the fiftieth term will be:

$6 + (50-1) \times 9 = 6 + 49 \times 9 = 447$ ,

which is less than 450.

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Question

An arithmetic sequence begins:

Which is the greater quantity?

(a)

(b) The tenth term of the sequence

Answer

The common difference of the sequence is . The tenth term of the sequence is $100 + (10-1) (-11) = 100 + 9 \cdot (-11) = 100 - 99 = 1$ .

This makes (b) greater.

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Question

A geometric sequence begins as follows:

Which is the greater quantity?

(a) The seventh term of the sequence

(b)

Answer

The common ratio of the sequence is $15 \div 5 = 3$ , so the next four terms of the sequence are:

$45 \times 3 = 135$

$135\times 3 = 405$

$405\times 3 = 1,215$

$1,215 \times 3 = 3,645$ , the seventh term, which is greater than 3,000.

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Question

The Fibonacci sequence begins

with each subsequent term being the sum of the previous two.

Which is the greater quantity?

(a) The product of the fifth and eighth terms of the Fibonacci sequence

(b) The product of the sixth and seventh terms of the Fibonacci sequence

Answer

By beginning with and taking the sum of the previous two terms to get each successive term, we can generate the Fibonacci sequence:

(a) The fifth and eighth terms are 5 and 21; their product is $5 \times 21 = 105$

(b) The sixth and seventh terms are 8 and 13; their product is $8 \times 13 = 104$

(a) is greater.

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Question

The Fibonacci sequence begins

with each subsequent term being the sum of the previous two.

Which is the greater quantity?

(a) The product of the eighth and tenth terms of the Fibonacci sequence

(b) The square of the ninth term of the Fibonacci sequence

Answer

By beginning with and taking the sum of the previous two terms to get each successive term, we can generate the Fibonacci sequence:

(a) The eighth and tenth terms are 21 and 55; their product is $21\times 55 = 1,155$

(b) The ninth term is 34; its square is $34 ^{2} = 34 \times 34 = 1,156$ .

(b) is greater

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Question

Five candidates are running for school board. A voter chooses two of the candidates.

Which is the greater quantity?

(a) The number of ways a voter can mark his or her ballot

(b)

Answer

This is a choice of two out of five without regard to order - that is, the number of combinations of two out of a set of five. This is

$C(5,2) = \frac{5!}{(5-2)!2!}= \frac{5!}{3!; 2!} = \frac{120}{6 \cdot 2} = 10$

There are ten possible ways to choose, and .

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Question

Seven students are running for student council. A student can vote for any three. Steve wants to vote for his girlfriend Marsha.

Which is the greater quantity?

(a) The number of ways Steve can fill out his ballot so that he can vote for Marsha

(b)

Answer

Since one of Steve's votes has already been decided, this is a choice of two out of the remaining six without regard to order - that is, the number of combinations of two out of a set of six. This is

$C (6,2) = \frac{6!}{(6-2)!; 2!}= \frac{6!}{4!; 2!} = \frac{720}{24 \cdot 2} = 15$

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Question

Eight students are running for student council. A student can vote for any two. Phil only wants to vote for male candidates; Sandee only wants to vote for female candidates.

Which is the greater quantity?

(a) The number of ways Phil can fill out his ballot so that he can vote for two male candidates

(b) The number of ways Sandee can fill out her ballot so that she can vote for two female candidates

Answer

This question is impossible to answer, because it is not known how many candidates are male and how many are female.

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Question

$1,1,2,3,5,8,\square$

What is the next number in the series above?

Answer

First, you must determine the pattern in the series. The pattern is adding the two preceding terms to give you the next term (ex: ). Therefore, apply that rule to calculate , which gives you 13 as the next term.

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Question

Refer to the above diagram. The top row gives a sequence of figures. Which figure on the bottom row comes next?

Answer

The square with the diagonal line alternates between the square on the left and the square on the right. Therefore, the next figure in the sequence will have its diagonal line in the rightmost square, eliminating Figures (b) and (d) and leaving Figures (a) and (c).

Also, the shaded square moves one position to the right from figure to figure, so in the next figure, the shaded square must be the one at the extreme right. Figure (c) matches that description.

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Question

What are the next two numbers of this sequence?

Answer

The sequence is formed by alternately adding and adding to each term to get the next term.

$\begin{matrix} 10 + 4 = 14\ 14 + 5 = 19\ 19+4= 23\ 23+5=28\ 28+4 = 32\ 32+5=37\ 37+4=41\ 41+5=46 \end{matrix}$

and are the next two numbers.

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Question

Define two sets as follows:

$A = \left { a, c, d, e, f, h, j, l, p \right }$

$B = \left { c, e, f, h, j, o, p, s, z \right }$

Which of the following is a subset of $A \cap B$ ?

Answer

We demonstrate that all of the choices are subsets of $A \cap B$ .

$A \cap B$ is the intersection of and - that is, the set of all elements of both sets. Therefore,

$A \cap B = \left { c,e,f,h,j,p\right }$

$\left { c,e,f,h,j,p\right }$ itself is one of the choices; it is a subset of itself. The empty set $\varnothing$ is a subset of every set. The other two sets listed comprise only elements from $A \cap B$ , making them subsets of $A \cap B$ .

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Question

Let the universal set be the set of all positive integers. Also, define two sets as follows:

$A = \left { 8, 16, 24, 32, 40, 48, 56...\right }$

$B = \left { 1, 4, 9, 16, 25, 36, 49,...\right }$

Which of the following is an element of the set $A \cap B$ ?

Answer

We are looking for an element that is in the intersection of and - in other words, we are looking for an element that appears in both sets.

is the set of all multiples of 8. We can eliminate two choices as not being in by demonstrating that dividing each by 8 yields a remainder:

$484 \div 8 = 60 \textrm{ R }4$

$900 \div 8 = 112 \textrm{ R }4$

is the set of all perfect square integers. We can eliminate two additional choices as not being perfect squares by showing that each is between two consecutive perfect squares:

$18^{2}= 324< 352 < 361 = 19^{2}$

$18^{2}= 324< 336 < 361 = 19^{2}$

This eliminates 352 and 336. However,

$256 = 16 ^{2}$ .

It is also a multiple of 8:

$256 \div 8 = 32$

Therefore, $256 \in A \cap B$ .

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Question

Define two sets as follows:

$A = \left { 8, 16, 24, 32, 40, 48, 56...\right }$

$B = \left { 1, 4, 9, 16, 25, 36, 49,...\right }$

Which of the following is not an element of the set $A \cup B$ ?

Answer

$A \cup B$ is the union of and , the set of all elements that appear in either set_._ Therefore, we are looking to eliminate the elements in and those in to find the element in neither set.

is the set of all multiples of 8. We can eliminate two choices as mulitples of 8:

$272 \div 8 = 34$ , so $272 \in A$

$344\div 8 = 43$ , so $344 \in A$

is the set of all perfect square integers. We can eliminate two additional choices as perfect squares:

$\sqrt{225}=15$ , so $225 \in B$

$\sqrt{441}=21$ , so $441 \in B$

All four of the above are therefore elements of $A \cup B$ .

420, however is in neither set:

$420 \div 8 = 52 \textrm{ R } 4$ , so $420 \notin A$

and

$20 = \sqrt{400}<\sqrt{420} < \sqrt{441} = 21$ , so $420 \notin B$

Therefore, $420 \notin A \cup B$ , making this the correct choice.

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Question

Find the missing part of the list:

$\small \small 1, 3, 9, 27,...,243,729$

Answer

To find the next number in the list, multiply the previous number by $\small 3$ .

$27\times3=81$

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Question

Seven students are running for student council; each member of the student body will vote for three. Derreck does not want to vote for Anne, whom he does not like. How many ways can he cast a ballot so as not to include Anne among his choices?

Answer

Derreck is choosing three students from a field of six (seven minus Anne) without respect to order, making this a combination. He has ways to choose. This is:

$C(6,3)= \frac{6!}{(6-3)! ; 3! } = \frac{6!}{3! ; 3! } = \frac{720 }{6 \times 6}= 20$

Derreck has 20 ways to fill the ballot.

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Question

Ten students are running for Senior Class President. Each member of the student body will choose four candidates, and mark them 1-4 in order of preference.

How many ways are there to fill out the ballot?

Answer

Four candidates are being selected from ten, with order being important; this means that we are looking for the number of permutations of four chosen from a set of ten. This is

$P (10,4) = \frac{10!}{(10-4)!}= \frac{10!}{6!} = 10 \times 9 \times 8 \times 7 = 5,040$

There are 5,040 ways to complete the ballot.

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Question

Ten students are running for Senior Class President. Each member of the student body will choose five candidates, and mark them 1-5 in order of preference.

Roy wants Mike to win. How many ways can Roy fill out the ballot so that Mike is his first choice?

Answer

Since Mike is already chosen, Roy is in essence choosing four candidates from nine, with order being important. This is a permutation of four elements out of nine. The number of these is

$P (9,4) = \frac{9!}{(9-4)!} = \frac{9!}{5!} = 9 \times 8 \times 7 \times 6 = 3,024$

Roy can fill out the ballot 3,024 times and have Mike be his first choice.

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Question

The junior class elections have four students running for President, five running for Vice-President, four running for Secretary-Treasurer, and seven running for Student Council Representative. How many ways can a student fill out a ballot?

Answer

These are four independent events, so by the multiplication principle, the ballot can be filled out $4 \times 5 \times 4\times 7 = 560$ ways.

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Question

The sophomore class elections have six students running for President, five running for Vice-President, and six running for Secretary-Treasurer. How many ways can a student fill out a ballot if he is allowed to select one name per office?

Answer

These are three independent events, so by the multiplication principle, the ballot can be filled out $6 \times 5 \times 6 = 180$ ways.

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