Patterns - SSAT Upper Level Quantitative

Card 0 of 100

In the following number sequence, what number goes in place of the circle?

$20, 10, 16, 8, 14, 7, \square, \bigcirc...$

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The sequence is generated by alternately dividing by 2 and adding 6:

$20 \div 2 = 10$

$16 \div 2 = 8$

$14 \div 2 = 7$

- This number replaces the square.

$13\div 2 = 6$\frac{1}{2}$$ - This number replaces the circle.

A sequence of numbers begins:

What is the one-hundredth entry in this sequence?

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Let be the $n\mathrm{th}$ entry in the sequence. Then $f(n) = $n^{2}$ + 1$ . The one-hundredth entry is therefore

$f(100) = 100 ^{2}+1=10,001$

Define $a \bigtriangledown b = ab + b + 1$

Which of the following expressions is equal to $xy \bigtriangledown 1$ ?

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Replace with and with 1:

$xy \bigtriangledown 1 = xy \cdot 1 + 1 + 1 = xy + 2$

Find the sum of this infinite geometric series:

$4 + $\frac{7}{2}$ + $\frac{49}{16}$ + ... + 4 \cdot \left ( \frac {7}{8} \right ) ^{n}+...$

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The sum of an infinite series with first term and common ratio is

$S = $\frac{a}{1-r}$$

Set $a=4,r= $\frac{7}{8}$$ , and evaluate:

$S = $\frac{a}{1-r}$ = $\frac{4}{1-\frac{7}${8}} = $\frac{4}{\left( \frac{1}${8}\right )} = 4 \cdot 8=32$

Give the next number in the following sequence:

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The sequence is generated by alternately adding 2, then multiplying by 2:

$5 \times 2 = 10$

$12 \times 2 = 24$

$26 \times 2 = 52$

$54 \times 2 = $\underline{108}$$ , which is the correct choice.

In the following numeric sequence, what number goes in place of the circle?

$6, 18, 10, 30, 22, 66, \square, \bigcirc...$

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The sequence is generated by alternately multiplying by 3, then adding 8:

$6 \times 3 = 18$

$10 \times 3 = 30$

$22 \times 3 = 66$

- This number replaces the square.

$58 \times 3 = 174$ - This number replaces the circle.

Which of the following numbers can complete the sequence?

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Each subsequent number in this set is half the previous number, minus 1.

For example, the number after 13 is 25 because:

$13\cdot2-1=26-1=25$

Thus, the number after 25 is equal to:

$25\cdot2-1=50-1=49$

Define an operation on the real numbers as follows:

$a \bigtriangleup b = 3a - 2b$

Find the value of that makes this statement true:

$x\bigtriangleup 5 = 32$

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Replace in the defintiion:

$a \bigtriangleup b = 3a - 2b$

$x\bigtriangleup 5 =3x - 2 \cdot 5$

$x\bigtriangleup 5 =3x - 10$

Now, set this equal to 32 and solve for :

$x\bigtriangleup 5 = 32$

$3x \div 3= 42 \div 3$

In the following number sequence, what number goes in place of the circle?

$1, 9, 25, 49, 81, 121, \bigcirc...$

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The sequence comprises the squares of the odd integers, in order:

...

The next number, which replaces the circle, is

.

Define an operation on the set of real numbers as follows:

$a; \S; b = $a^{2}$ + b ^{2}$

Evaluate:

$-4; \S; \left ( -5 \right )$

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$a; \S; b = $a^{2}$ + b ^{2}$

$-4; \S; \left ( -5 \right ) = \left ( -4 \right $)^{2}$ + \left ( -5\right ) ^{2}$

$= \left ( -4 \right )\times \left ( -4 \right ) + \left ( -5\right ) \times \left ( -5 \right )$

What number replaces the circle?

$1, 1, 2, 4, 6, 18, 21, 84, 88, \square, \bigcirc...$

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The sequence is formed by alternately multiplying by a number, then adding the same number; the number incrementally increases every other term.

$1 \times 1 = 1$

$2 \times 2 = 4$

$6 \times 3 = 18$

$21 \times4 = 84$

$88 \times 5 = 440$ , the number which replaces the square.

, the number which replaces the circle.

$\left \lfloor N \right \rfloor$ is defined as the greatest integer less than or equal to .

Evaluate the expression for :

$\left \lfloor 2x +0.5 \right \rfloor + \left \lfloor 3x +0.2\right \rfloor$

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Substitute:

$\left \lfloor 2x +0.5 \right \rfloor + \left \lfloor 3x +0.2\right \rfloor$

$= \left \lfloor 2 \cdot 0.7 +0.5 \right \rfloor + \left \lfloor 3 \cdot 0.7 +0.2\right \rfloor$

$= \left \lfloor 1.4 +0.5 \right \rfloor + \left \lfloor 2.1 +0.2\right \rfloor$

$= \left \lfloor 1.9\right \rfloor + \left \lfloor 2.3\right \rfloor$

$\left \lceil N\right \rceil$ is defined as the least integer greater than or equal to .

Evaluate for :

$\left \lceil 4 x -0.4 \right \rceil + 5x+ 0.3$

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Substitute :

$\left \lceil 4 x -0.4 \right \rceil + 5x+ 0.3$

$= \left \lceil 4 \cdot 0.8 -0.4 \right \rceil + 5 \cdot 0.8 + 0.3$

$= \left \lceil 3.2 -0.4 \right \rceil + 4 + 0.3$

$= \left \lceil 2.8 \right \rceil + 4 + 0.3$

What number replaces the circle in this sequence?

$\bigcirc , \square , 8, 6, 12, 10, 20, 18, 36....$

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This sequence is alternately generated by subtracting two and multiplying by two.

We can therefore find the numbers that replace the square and the circle by reversing the pattern - alternately dividing by two and adding two beginning at 36:

$36 \div 2 = 18$

$20 \div 2 =10$

$12 \div 2 = 6$

$8 \div 2 = 4$ - this replaces the circle

- this replaces the square

Anya, Brian, Clark, and Donna represented Central High in a math contest. The team score is the sum of the three highest scores; Anya outscored Clark, Brian outscored Donna; Donna outscored Clark. Whose scores were added to determine the team score?

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Let be the scores by Anya, Brian, Clark, and Donna, respectively.

Three inequalities can be deduced from these statements:

Anya outscored Clark:

Brian outscored Donna:

Donna outscored Clark:

The first and third statements can be combined to arrive at:

so Brian and Donna both outscored Clark. Since Anya outscored Clark also, Clark finished last among the four, and the team score was the sum of Anya's, Brian's and Donna's scores.

Define a sequence of numbers as follows:

For all integers ,

Evaluate .

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Using the definition of this particular sequence we will plug in to find, then use that to find the next term and so one and so forth.

$= $a_{1}$+4$

$= $a_{2}$+9$

$= $a_{3}$+16$

33 is the correct choice.

Define a sequence of numbers as follows:

For all integers ,

Evaluate .

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Using the definition of this particular sequence we will plug in to find, then use that to find the next term and so one and so forth.

$= \left $(a_{1}$+2 \right $)^{2}$$

$= \left (4+2 \right $)^{2}$$

$= $6^{2}$$

$= \left $(a_{2}$+3 \right $)^{2}$$

$= \left (36+3 \right $)^{2}$$

$= $39^{2}$$

Give the sum of the infinite geometric series whose first two terms are 6 and 5, in that order.

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The sum of an infinite geometric series with initial term and common ratio is:

$S = $$\frac{a_{1}$$}{1-r}$ .

The initial term is and the common ratio is $r = $$\frac{a_{2}$$$}{a_{1}$} = $\frac{5}{6}$$ ; therefore,

$S = $\frac{6}{1- \frac{5}${6}} = $\frac{6}{\frac{1}${6}} = 6 \cdot 6 = 36$

Give the sum of the infinite geometric series that begins

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The sum of an infinite geometric series with initial term and common ratio is:

$S = $$\frac{a_{1}$$}{1-r}$ .

The initial term is and the common ratio is $r = $$\frac{a_{2}$$$}{a_{1}$} = $\frac{-6}{7}$ = -$\frac{6}{7}$$ ; therefore,

$S = $$\frac{a_{1}$$}{1-r} = $\frac{7}{1-\left ( - \frac{6}${7}\right)} =\frac{7}{\frac{13}${7}} = 7 \cdot $\frac{7}{13}$ = $\frac{49}{13}$ = 3$\frac{10}{13}$$

Examine the above figure. In the top row, the cubes of the whole numbers are written in ascending order. In each successive row, each entry is the difference of the two entries above it - five of those entries have been calculated for you.

What is the fifth entry in the third row?

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The fifth entry in the third row is the difference of the sixth and fifth entries in the second row.

The sixth entry in the second row is the difference of 216 and 125:

The fifth entry in the second row is the difference of 343 and 216:

Now subtract these two: