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PSAT Math : PSAT Mathematics

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Example Questions

Example Question #4 : How To Find Consecutive Integers

In the repeating pattern 9,5,6,2,1,9,5,6,2,1......What is the 457th number in the sequence?

Possible Answers:

Correct answer:

Explanation:

There are 5 numbers in the sequnce.

How many numbers are left over if you divide 5 into 457?

There would be 2 numbers!

The second number in the sequence is 9,5,6,2,1

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Example Question #5 : How To Find Consecutive Integers

If $w,x,y,and\ z$ are consecutive, non-negative integers, how many different values of are there such that $\frac{w*x*y*z}{2}$ is a prime number?

Possible Answers:

$\infty$

Correct answer:

Explanation:

Since $w,x,y,and\ z$ are consecutive integers, we know that at least 2 of them will be even. Since we have 2 that are going to be even, we know that when we divide the product by 2 we will still have an even number. Since 2 is the only prime that is even, we must have:

$\frac{w*x*y*z}{2} = 2 \Rightarrow w*x*y*z = 4$

What we notice, however, is that for w=0 , we have the product is 0. For w=1 , we have the product is 24. We will then never have a product of 4, meaning that $\frac{w*x*y*z}{2}$ is never going to be a prime number.

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Example Question #362 : Arithmetic

Four consecutive odd integers have a sum of 32. What are the integers?

Possible Answers:

5,7,9,11

6,8,10,12

3,5,7,9

7,8,9,10

7,8,10,12

Correct answer:

5,7,9,11

Explanation:

Consecutive odd integers can be represented as x, x+2, x+4, and x+6.

We know that the sum of these integers is 32. We can add the terms together and set it equal to 32:

x + (x+2) + (x+4) + (x+6) = 32

4x + 12 = 32

4x = 20

x = 5; x+2=7; x+4 = 9; x+6 = 11

Our integers are 5, 7, 9, and 11.

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Example Question #1 : How To Find The Common Difference In Sequences

How many integers in the following infinite series are positive: 100, 91, 82, 73 . . . ?

Possible Answers:

Correct answer:

Explanation:

The difference between each number in the series is 9. You can substract nine 11 times from 100 to get 1: 100 – 9x11 = 1. Counting 100, there are 12 positive numbers in the series.

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Example Question #363 : Arithmetic

In a sequence of numbers, each term is times larger than the one before it. If the 3^rd term of the sequence is 12, and the 6^th term is 96, what is the sum of all of the terms less than 250?

Possible Answers:

192

384

378

372

381

Correct answer:

381

Explanation:

Let's call the first term in the sequence a₁ and the nth term a_n.

We are told that each term is r times larger than the one before it. Thus, we can find the next term in the sequence by multiplying by r.

a₁ = a₁

a₂ = r(a₁)

a₃ = r(a₂) = r(r(a₁)) = r²(a₁)

a₄ = r(a₃) = r(r²(a₁)) = r³(a₁)

a_n = r^(n–1)a₁

We can use this information to find r.

The problem gives us the value of the third and the sixth terms.

a₃ = r²(a₁) = 12

a₆ = r⁵(a₁) = 96

Let's solve for a₁ in terms of r and a₃.

a₁ = 12/(r²)

Let's then solve for a₁ in terms of r and a₆.

a₁ = 96/(r⁵)

Now, we can set both values equal and solve for r.

12/(r²) = 96/(r⁵)

Multiply both sides by r⁵ to get rid of the fraction.

12r⁵/r² = 96

Apply the property of exponents which states that a^b/a^c = a^b–c.

12r³ = 96

Divide by 12 on both sides.

r³ = 8

Take the cube root of both sides.

r = 2

This means that each term is two times larger than the one before it, or that each term is one half as large as the one after it.

a₂ must equal a₃ divided by 2, which equals ¹²/₂ = 6.

a₁ must equal a₂ divided by 2, which equals ⁶/₂ = 3.

Here are the first eight terms of the sequence:

3, 6, 12, 24, 48, 96, 192, 384

The question asks us to find the sum of all the terms less than 250. Only the first seven terms are less than 250. Thus the sum is equal to the following:

sum = 3 + 6 + 12 + 24 + 48 + 96 + 192 = 381

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Example Question #11 : Sequences

The first term of an arithmetic sequence is 20; the tenth term is 100. What is the sixteenth term?

Possible Answers:

128

156

$153 \frac{1}{3}$

$162\frac{2}{9}$

148

Correct answer:

$153 \frac{1}{3}$

Explanation:

Given terms $a_{m}, a_{n}$ of an arithmetic sequence, m < n , the common difference is

$d = \frac{a_{n} - a_{m}}{n - m}$

Setting $m = 1, n = 10, a_{1} = 20, a_{10} = 100$ , the common difference is found to be

$d = \frac{100 - 20}{10-1} = \frac{80}{9} = 8 \frac{8}{9}$

The $n\textup{th}$ term of the sequence $a_{n}$ is

$a_{n} = a_{1} + (n-1)d$

Setting $a_{1} = 20, n = 16, d = 8 \frac{8}{9}$ , this term is found to be

$a_{16} = 20 + (16-1) \cdot 8\frac{8}{9}$

$a_{16} = 20 + 15 \cdot 8\frac{8}{9}$

$a_{16} = 20 + 133 \frac{1}{3}$

$a_{16} = 153 \frac{1}{3}$

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Example Question #151 : Integers

An arithmetic sequence has as its first term 17. Its tenth term is . What is the sum of its second and ninth terms?

Possible Answers:

$\frac{9X- 153}{10}$

$\frac{10X + 170}{9}$

$\frac{10X - 170}{9}$

$\frac{9X + 153}{10}$

None of the other choices gives the correct response.

Correct answer:

None of the other choices gives the correct response.

Explanation:

It is not necessary to determine the common difference or any other terms to answer this question. If we let be the common difference, then

$a_{2} = a _{1}+ d = 17 + d$

and

$a_{10} = a _{9}+ d$ , or

$a _{9} = a_{10} - d = X- d$

The sum of the second and ninth terms is given by the expression

$a_{2} +a _{9} = 17 + d+ X - d = X + 17$

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Example Question #12 : Sequences

The first and fourth terms of an arithmetic sequence are, respectively,

$a _{1} = x^{2}- x+ 17$

and

$a _{4} = 3 x^{2}- 5x+ 81$ .

Which of the following is the common difference of the sequence?

Possible Answers:

$\frac{2 }{3} x^{2} - \frac{4}{3}x- \frac{64}{3}$

$\frac{2 }{3} x^{2} - \frac{4}{3}x + \frac{64}{3}$

$\frac{1}{2}x^{2}- x - 16$

$\frac{1}{2}x^{2}- x + 16$

$x^{2}- 2x + 32$

Correct answer:

$\frac{2 }{3} x^{2} - \frac{4}{3}x + \frac{64}{3}$

Explanation:

Given terms $a_{m}, a_{n}$ of an arithmetic sequence, m < n , the common difference is

$d = \frac{a_{n} - a_{m}}{n-m}$ .

Set m = 1, n = 4 :

$d = \frac{a_{4} - a_{1}}{4-1}$

$d = \frac{\left (3 x^{2}- 5x+ 81 \right ) - \left (x^{2}- x+ 17 \right )}{3}$

$d = \frac{3 x^{2}- x^{2} - 5x+ x + 81 - 17 }{3}$

$d = \frac{2 x^{2} - 4x + 64 }{3}$

$d = \frac{2 }{3} x^{2} - \frac{4}{3}x + \frac{64}{3}$

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Example Question #3 : How To Find The Common Difference In Sequences

Which of the following are not natural numbers?

I. 1

II. 0

III. 349010

IV. -2

V. 1/4

Possible Answers:

II, III, IV, V

II, IV, V

I, IV, V

I, V

IV, V

Correct answer:

II, IV, V

Explanation:

Natural numbers are defined as whole numbers 1 and above. II, IV, V are not natural numbers.

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Example Question #1 : How To Find The Answer To An Arithmetic Sequence

-27, -24, -21, -18…

In the sequence above, each term after the first is 3 greater than the preceding term. Which of the following could not be a value in the sequence?

Possible Answers:

657

461

501

126

Correct answer:

461

Explanation:

All of the values in the sequence must be a multiple of 3. All answers are multiples of 3 except 461 so 461 cannot be part of the sequence.

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PSAT Math : PSAT Mathematics

Study concepts, example questions & explanations for PSAT Math

All PSAT Math Resources

Example Questions

Example Question #4 : How To Find Consecutive Integers

Example Question #5 : How To Find Consecutive Integers

Example Question #362 : Arithmetic

Example Question #1 : How To Find The Common Difference In Sequences

Example Question #363 : Arithmetic

Example Question #11 : Sequences

Example Question #151 : Integers

Example Question #12 : Sequences

Example Question #3 : How To Find The Common Difference In Sequences

Example Question #1 : How To Find The Answer To An Arithmetic Sequence

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