How to find the missing part of a list

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

Questions 1 - 10

A geometric sequence begins as follows:

Which is the greater quantity?

(a) The seventh term of the sequence

(b)

(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 $15 \div 5 = 3$ , 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.

Define $L = \left { x ; | ; x \textrm{ is a multiple of 4 }\right }$ .

Which of the following is not a subset of the set ?

$\left { 4876, 3296, 8878, 9004, 7612\right }$

None of the other responses gives a correct answer.

$\left { 9456, 9260, 7848, 6116, 8744\right }$

$\left { 8908, 5428, 9380, 7212, 7756 \right }$

$\left { 8320, 9652, 8772, 5432, 9936\right }$

Explanation

For a set to be a subset of , all of its elements must also be elements of - that is, all of its elements must be multiples of 4. An integer is a multple of 4 if and only the number formed by its last two digits is also a multiple of 4, so all we have to do is examine the last two digits of each number in all four sets.

Of all of the numbers in the four sets listed, only 8,878 has this characteristic:

$78 \div 4 = 19 \textrm{ R }2$

8,878 is not a multiple of 4, so among the sets from which to choose,

$\left { 4876, 3296, 8878, 9004, 7612\right }$

is the only set that is not a subset of .

$\left { 2,3,4,5,6,7,8,9\right }$

The above set represents the numbered cards in a standard deck of cards. What value is missing?

Explanation

The numbered cards in a deck range from to .

Since is not in the above set, it is the missing NUMBER card from the deck.

King and Ace are incorrect choices as they are called face cards in a deck of cards.

Assorted 2

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

Figure (c)

Figure (d)

Figure (a)

Figure (b)

Explanation

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.

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

(b) is greater

(a) is greater

(a) and (b) are equal

It is impossible to tell from the information given

Explanation

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

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?

Explanation

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.

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?

Explanation

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.

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?

Explanation

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.

Mary is making a very long necklace with a variety of beads. The beads are white, blue, and black, and she strings them on the necklace, in that order. What color is the 213th bead?

black

blue

white

gray

Explanation

A number is divisible by 3 when the sum of its digits is divisble by 3. The sum of the digits of 213 equals 6, which is evenly divisible by 3.

Therefore, because 213 is a number that is evenly divisble by 3, the 213th bead is going to be the third color that Mary uses, which is black.

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?

Explanation

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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