Arithmetic Practice Questions — GRE General Test — Page 8

Question 141Question →

Let $a$ and $b$ be positive integers such that $\gcd(a, b) = 60$ and $\text{lcm}(a, b) = 75,600$ . Which of the following statements must be true? Select all that apply.

Select all that apply

The product

ab

is equal to

4,536,000

.

Neither

a

nor

b

can be divisible by

49

.

There are exactly

8

distinct unordered pairs of positive integers

\{a, b\}

that satisfy the given conditions.

Both

a

and

b

must be multiples of

9

.

The greatest common divisor of

(a + b)

and

60

is equal to

30

.

Show answer & explanation

Answer: The product $ab$ is equal to $4,536,000$ .; Neither $a$ nor $b$ can be divisible by $49$ .; There are exactly $8$ distinct unordered pairs of positive integers $\{a, b\}$ that satisfy the given conditions.

Answer

The statements asserting that

ab = 4,536,000

, that neither number is divisible by

49

, and that there are exactly

8

distinct unordered pairs

\{a, b\}

are all correct.

The product of the GCD and LCM of two numbers always yields their product, confirming

ab = 4,536,000

. The prime factorization of the LCM shows that

7

appears only to the first power, making divisibility by

49

impossible for either number. Finally, assigning the minimum and maximum prime exponents across the four distinct prime factors

(2, 3, 5, 7)

yields

2^4 = 16

ordered pairs, which corresponds to exactly

8

unordered pairs.

Step-by-Step Solution

1

Apply the product identity for GCD and LCM.

ab = \gcd(a, b) \times \text{lcm}(a, b) = 60 \times 75,600 = 4,536,000

.

The product of two positive integers is always equal to the product of their greatest common divisor and least common multiple.

2

Find the prime factorizations of

\gcd(a, b)

and

\text{lcm}(a, b)

to determine exponent bounds.

\gcd(a, b) = 2^2 \times 3^1 \times 5^1 \times 7^0

and

\text{lcm}(a, b) = 2^4 \times 3^3 \times 5^2 \times 7^1

.

The exponent of each prime factor in

\gcd(a, b)

gives the minimum exponent in

a

and

b

, while the exponent in

\text{lcm}(a, b)

gives the maximum exponent.

3

Evaluate prime factor constraints for prime

7

and prime

3

.

Maximum power of

7

in

a

or

b

is

7^1 = 7

, so neither can be divisible by

49

. For prime

3

, one integer has

3^1

(not divisible by

9

) and the other has

3^3 = 27

.

An integer is divisible by

49 = 7^2

only if its prime factorization contains

7^2

or higher. Similarly, a number is a multiple of

9 = 3^2

only if its prime exponent for

3

is at least

2

.

4

Calculate the total number of distinct unordered pairs

\{a, b\}

.

For each of the

4

prime factors (

2, 3, 5, 7

), there are

2

choices for which integer gets the higher exponent. Total ordered pairs

(a, b) = 2^4 = 16

. Unordered pairs

\{a, b\} = 16 / 2 = 8

.

Since

\min < \max

for all prime factors,

a \neq b

for all pairs, allowing exact division by

2

to account for symmetry.

5

Analyze

\gcd(a + b, 60)

.

Since

60 \mid a

and

60 \mid b

, it follows that

60 \mid (a + b)

, making

\gcd(a + b, 60) = 60

.

If a number

k

divides both

a

and

b

, it must divide their linear combination

a + b

.

Key Concept

Prime Factorization, GCD-LCM Identities, and Counting Valid Integer Pairs

Question 142Question →

If $|x - 4| = 9$ and $x < 0$ , what is the value of $x$ ?

Show answer & explanation

Answer: -5

Answer

The value of

x

is

-5

.

The absolute value equation

|x - 4| = 9

specifies that the distance between

x

and

4

on the real number line is equal to

9

. Moving

9

units to the left of

4

gives

4 - 9 = -5

, and moving

9

units to the right gives

4 + 9 = 13

. Because

x

is specified to be negative (

x < 0

), the correct value of

x

is

-5

.

Step-by-Step Solution

1

Set up the two linear equations representing the absolute value equation

|x - 4| = 9

.

x - 4 = 9

or

x - 4 = -9

By definition,

|A| = B

(where

B \ge 0

) implies

A = B

or

A = -B

.

2

Solve for

x

in both equations.

x = 13

or

x = -5

Adding

4

to both sides of

x - 4 = 9

gives

x = 13

, and adding

4

to both sides of

x - 4 = -9

gives

x = -5

.

3

Select the value of

x

that satisfies the given condition

x < 0

.

x = -5

The value

13

is positive, while

-5

is negative and fulfills

x < 0

.

Key Concept

Absolute Value as Distance and Solving Absolute Value Equations

Estimated Time:45s

Question 143Question →

At the beginning of the year, a cloud storage service had a certain number of active paid subscriptions. During the first six months of the year, the number of active subscriptions increased by $20\%$ . During the second six months of the year, the number of active subscriptions decreased by $15\%$ relative to the total at the middle of the year. If the service had $12,240$ active subscriptions at the end of the year, what was the number of active subscriptions at the beginning of the year?

Show answer & explanation

Answer: 12000

Answer

12000

To find the initial subscription count, model the successive percentage changes sequentially. Increasing the initial amount

N

by

20\%

yields

1.20N

. Decreasing that intermediate total by

15\%

multiplies

1.20N

by

0.85

, giving

1.02N

. Setting

1.02N = 12,240

gives

N = 12,000

.

Step-by-Step Solution

1

Express the mid-year subscription total in terms of the initial subscription count

N

.

Mid-year subscriptions

= N \times (1 + 0.20) = 1.20N

A

20\%

increase scales the original quantity by a factor of

1.20

.

2

Express the end-of-year subscription total in terms of

N

after a

15\%

decrease from the mid-year total.

End-of-year subscriptions

= 1.20N \times (1 - 0.15) = 1.20N \times 0.85 = 1.02N

A

15\%

decrease reduces the new base quantity to

85\%

of its mid-year value.

3

Set the algebraic expression for the final total equal to the given numerical total and solve for

N

.

1.02N = 12,240 \implies N = 12,000

Dividing the final amount by the net multiplier

1.02

yields the original starting value.

Key Concept

Successive Percentage Changes and Base Shift

Estimated Time:1m 30s

Question 144Question →

A specialty baking recipe consists of three ingredients: flour, sugar, and butter. By weight, $\frac{1}{2}$ of the recipe is flour and $\frac{1}{3}$ of the recipe is sugar, with the remaining portion consisting entirely of butter. Which of the following statements about the ingredients in the recipe must be true? Select all such statements.

Select all that apply

The butter accounts for

\frac{1}{6}

of the total weight of the recipe.

The ratio of the weight of sugar to the weight of butter is

2 : 1

.

The combined weight of flour and sugar accounts for

\frac{2}{5}

of the total weight of the recipe.

The weight of sugar is

50\%

less than the weight of flour.

The ratio of the weight of flour to the combined weight of the remaining ingredients is

1 : 2

.

Show answer & explanation

Answer: The butter accounts for $\frac{1}{6}$ of the total weight of the recipe.; The ratio of the weight of sugar to the weight of butter is $2 : 1$ .

Answer

The statement that butter accounts for 1/6 of the total weight of the recipe and the statement that the ratio of the weight of sugar to the weight of butter is 2:1 are both correct.

The statement asserting that butter accounts for 1/6 of the total weight is correct because subtracting the combined fraction of flour and sugar (5/6) from the total (1) yields 1/6. The statement asserting that the ratio of sugar to butter is 2:1 is correct because dividing 1/3 by 1/6 equals 2.

Step-by-Step Solution

1

Calculate the fraction of the recipe that is butter.

Butter fraction =

1 - \left(\frac{1}{2} + \frac{1}{3}\right) = 1 - \frac{5}{6} = \frac{1}{6}

.

The total of all constituent fractions must sum to 1.

2

Calculate the ratio of sugar to butter.

\text{Ratio} = \frac{1/3}{1/6} = \frac{1}{3} \times \frac{6}{1} = \frac{2}{1} = 2 : 1

.

Divide the fraction representing sugar by the fraction representing butter.

3

Evaluate the statement regarding combined flour and sugar weight.

Combined fraction =

\frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6} \neq \frac{2}{5}

.

Adding fractions requires finding a common denominator (6).

4

Evaluate the statement regarding percentage difference between sugar and flour.

Percent less =

\frac{1/2 - 1/3}{1/2} \times 100\% = \frac{1/6}{1/2} \times 100\% = \frac{1}{3} \times 100\% = 33.33\% \neq 50\%

.

Percent decrease must use the original reference quantity (flour) as the denominator.

5

Evaluate the ratio of flour to the combined remaining ingredients.

Remaining ingredients fraction =

1/3 + 1/6 = 1/2

. Ratio of flour (1/2) to remaining (1/2) =

1 : 1 \neq 1 : 2

.

Compare the part (flour) to the other part (sugar + butter), not to the whole.

Key Concept

Operations on rational numbers, fraction subtraction/addition using common denominators, part-to-part ratio comparisons, and percentage change base selection.

Question 145Question →

How many positive integers $n$ less than $1,000$ are divisible by $12$ , leave a remainder of $4$ when divided by $5$ , and are not divisible by $9$ ?

11

12

13

15

17

Show answer & explanation

Answer: 12

Answer

12

The problem asks for positive integers

n < 1,000

that satisfy three conditions:

n = 12k

,

n \equiv 4 \pmod 5

, and

9 \nmid n

. Expressing

n = 12k

and substituting into the modular equation yields

2k \equiv 4 \pmod 5

, which simplifies to

k \equiv 2 \pmod 5

. For

k \le 83

, setting

k = 5m + 2

yields

17

values for

m \in \{0, 1, \dots, 16\}

. Furthermore,

12k

is divisible by

9

if and only if

k

is a multiple of

3

, which corresponds to

m \equiv 2 \pmod 3

. The values

m = 2, 5, 8, 11, 14

(

5

values) must be excluded. Subtracting

5

from

17

leaves exactly

12

valid integers.

Step-by-Step Solution

1

Express

n

in terms of its divisibility by

12

and set the upper bound.

n = 12k

for a positive integer

k

. Since

n < 1,000

, we have

12k < 1,000

, which implies

k \le 83

.

Any positive integer divisible by

12

must be a positive integer multiple of

12

.

2

Apply the modular remainder condition modulo

5

.

12k \equiv 4 \pmod 5 \implies 2k \equiv 4 \pmod 5 \implies k \equiv 2 \pmod 5

.

Since

12 \equiv 2 \pmod 5

, simplifying the congruence gives

k \equiv 2 \pmod 5

.

3

Determine the candidate set of values for

k

.

k

can be written as

k = 5m + 2

. For

1 \le k \le 83

, the non-negative integer

m

ranges from

0

to

16

inclusive, yielding

17

possible values.

The maximum value

5m + 2 \le 83

gives

5m \le 81 \implies m \le 16

.

4

Apply the condition that

n

is not divisible by

9

.

12k = 3 \times 4 \times k

is divisible by

9

if and only if

k

is divisible by

3

.

5m + 2 \equiv 2m + 2 \pmod 3

, so

2m + 2 \equiv 0 \pmod 3 \implies m \equiv 2 \pmod 3

.

Since

12

already contains one factor of

3

,

12k

needs

k

to supply another factor of

3

to be divisible by

9

.

5

Filter out the excluded values of

m

and calculate the final count.

Values of

m \in \{0, 1, \dots, 16\}

with

m \equiv 2 \pmod 3

are

m \in \{2, 5, 8, 11, 14\}

(

5

values). The number of valid integers is

17 - 5 = 12

.

Subtracting the excluded multiples of

9

from the candidate set leaves the exact number of valid integers.

Key Concept

Modular arithmetic, linear congruences, and counting integer multiples with divisibility constraints

Estimated Time:2m 0s

Question 146Question →

A supercomputer processor operates at a standard speed of $1.25 \times 10^9$ calculations per second. During a maintenance diagnostic, the processor's operational speed is reduced by $96\%$ . Operating exclusively at this reduced speed, how many seconds will it take the processor to complete a workload of $1.8 \times 10^{13}$ calculations?

3.6 \times 10^4

1.5 \times 10^4

3.6 \times 10^5

1.44 \times 10^5

3.6 \times 10^6

Show answer & explanation

Answer: $3.6 \times 10^5$

Answer

3.6 \times 10^5

seconds

The correct answer is derived by first finding the reduced processing speed:

4\%

of

1.25 \times 10^9

equals

0.04 \times 1.25 \times 10^9 = 5.0 \times 10^7

calculations per second. Dividing the target workload of

1.8 \times 10^{13}

by

5.0 \times 10^7

yields

\frac{1.8}{5.0} \times 10^{13-7} = 0.36 \times 10^6 = 3.6 \times 10^5

seconds.

Step-by-Step Solution

1

Determine the reduced operational speed percentage.

Remaining speed percentage is

100\% - 96\% = 4\% = 0.04

.

A

96\%

reduction means the processor operates at

4\%

of its original capacity.

2

Calculate the reduced operational speed in scientific notation.

Reduced speed

= 0.04 \times (1.25 \times 10^9) = 0.05 \times 10^9 = 5.0 \times 10^7

calculations per second.

Multiplying the decimal coefficient

0.04

by

1.25

gives

0.05

, which adjusts to

5.0 \times 10^7

in scientific notation.

3

Divide total workload by reduced speed to compute time in seconds.

Time

= \frac{1.8 \times 10^{13}}{5.0 \times 10^7} = \left(\frac{1.8}{5.0}\right) \times 10^{13 - 7} = 0.36 \times 10^6

seconds.

Workload divided by rate gives total duration. Exponents are subtracted when dividing powers of ten.

4

Convert the final result to standard scientific notation.

0.36 \times 10^6 = 3.6 \times 10^5

seconds.

Moving the decimal point one place to the right requires decreasing the exponent of 10 by 1.

Key Concept

Decimals and Scientific Notation Operations

Estimated Time:2m 0s

Question 147Question →

A bakery uses flour and sugar in a ratio of $5:2$ by weight to make a cake batter. If a baker needs to prepare a batch using a total of $28$ pounds of flour and sugar combined, how many pounds of flour are needed?

8

pounds

14

pounds

20

pounds

35

pounds

70

pounds

Show answer & explanation

Answer: $20$ pounds

Answer

20

pounds

To find the amount of flour, determine the fraction of the total mixture that consists of flour. The ratio of flour to sugar is

5:2

, meaning flour makes up

5

out of

5 + 2 = 7

total parts. Multiplying this fraction by the total weight gives

\frac{5}{7} \times 28 = 20

pounds.

Step-by-Step Solution

1

Determine the total number of ratio parts

Total ratio parts =

5 + 2 = 7

The ratio

5:2

means that for every

5

parts of flour, there are

2

parts of sugar, giving a total of

7

parts.

2

Calculate the weight of one ratio part

Weight per part =

28 \text{ pounds} \div 7 = 4 \text{ pounds}

Dividing the total weight by the total number of parts yields the weight of a single part.

3

Multiply the weight per part by the number of flour parts

Flour weight =

5 \times 4 \text{ pounds} = 20 \text{ pounds}

Since flour represents

5

of the

7

parts, multiplying

5

by

4

gives the total weight of flour required.

Key Concept

Part-to-Whole Ratios

Estimated Time:45s

Question 148Question →

If $2^x \times 4^3 = 2^9$ , what is the value of $x$ ?

Show answer & explanation

Answer: 3

Answer

The value of

x

is 3.

By rewriting

4^3

as

(2^2)^3 = 2^6

, the equation becomes

2^x \cdot 2^6 = 2^9

. Applying the product rule gives

2^{x+6} = 2^9

, which simplifies to

x + 6 = 9

, yielding

x = 3

.

Step-by-Step Solution

1

Convert all terms to base 2

4^3 = (2^2)^3 = 2^{2 \times 3} = 2^6

To apply exponent rules, expressions should share a common base.

2

Apply product rule of exponents to the left side of the equation

2^x \times 2^6 = 2^{x+6}

When multiplying exponential terms with the same base, add their exponents:

a^m \cdot a^n = a^{m+n}

.

3

Equate the exponents from both sides

x + 6 = 9

Since the bases are identical and non-zero, the exponents must be equal.

4

Solve the linear equation for

x

x = 3

Subtract 6 from both sides of the equation.

Key Concept

Combining exponential terms with equal bases using power rules (

a^m \cdot a^n = a^{m+n}

and

(a^m)^n = a^{mn}

).

Question 149Question →

An agricultural research station recorded the annual grain harvest of a farm over a four-year period. In the second year, the harvest increased by $20\%$ compared to the first year. In the third year, the harvest increased by $25\%$ compared to the second year. In the fourth year, due to unfavorable weather, the harvest decreased by $20\%$ compared to the third year. If the grain harvest in the fourth year was $3,600$ metric tons, what was the grain harvest, in metric tons, in the first year?

2,400

2,880

3,000

3,060

4,500

Show answer & explanation

Answer: 3,000

Answer

3,000 metric tons

To find the initial amount, represent the successive percentage changes as sequential decimal multipliers. A

20\%

increase corresponds to a multiplier of

1.20

, a

25\%

increase corresponds to

1.25

, and a

20\%

decrease corresponds to

0.80

. The net change factor is

1.20 \times 1.25 \times 0.80 = 1.20

. Since the fourth-year harvest equals

1.20

times the first-year harvest, dividing

3,600

by

1.20

yields

3,000

metric tons.

Step-by-Step Solution

1

Define variables and write multiplier expressions for each year's percent change

Let

x

be the harvest in the first year. Year 2 harvest

= 1.20x

; Year 3 harvest

= 1.25 \times (1.20x)

; Year 4 harvest

= 0.80 \times (1.25 \times 1.20x)

.

Percentage increases and decreases must be applied sequentially to the immediately preceding period's value.

2

Calculate the combined percentage multiplier for the four-year period

Combined multiplier

= 1.20 \times 1.25 \times 0.80 = 1.20

.

Multiplying the decimal factors yields the overall factor relating the first year's harvest to the fourth year's harvest.

3

Set up and solve the equation for the first year's harvest (

x

)

1.20x = 3,600 \implies x = \frac{3,600}{1.20} = 3,000

.

Dividing the fourth year's total harvest by the combined multiplier gives the initial first year harvest.

Key Concept

Successive Percentage Changes and Base Values

Question 150Question →

If $k$ is a positive integer such that $k$ is divisible by $6$ and $k + 1$ is divisible by $5$ , what is the remainder when $k^2 + 5k$ is divided by $30$ ?

0

4

6

14

24

Show answer & explanation

Answer: $6$

Answer

The remainder when

k^2 + 5k

is divided by

30

is

6

.

The correct answer is

6

. Since

k

is a multiple of

6

and leaves a remainder of

4

when divided by

5

, the general form of

k

modulo

30

is

24

. Evaluating

k^2 + 5k

modulo

30

yields

24^2 + 5(24) = 576 + 120 = 696

, which gives a remainder of

6

when divided by

30

.

Step-by-Step Solution

1

Determine the congruence class of

k

modulo 6 and modulo 5.

k \equiv 0 \pmod 6

and

k \equiv 4 \pmod 5

.

Given that

k

is divisible by 6, its remainder modulo 6 is 0. Since

k + 1

is divisible by 5,

k + 1 \equiv 0 \pmod 5

, which implies

k \equiv 4 \pmod 5

.

2

Find the smallest positive integer value of

k

modulo 30 satisfying both conditions.

k \equiv 24 \pmod{30}

.

The multiples of 6 are 0, 6, 12, 18, 24, 30, ... Among these, 24 gives a remainder of 4 when divided by 5. Since 5 and 6 are coprime,

k \equiv 24 \pmod{30}

.

3

Substitute

k \equiv 24 \pmod{30}

into the expression

k^2 + 5k

and compute the remainder modulo 30.

The remainder is

6

.

k^2 + 5k = k(k+5) \equiv 24(24+5) = 24(29) \pmod{30}

. Using modular arithmetic,

24 \equiv -6 \pmod{30}

and

29 \equiv -1 \pmod{30}

, so

(-6)(-1) = 6 \pmod{30}

.

Key Concept

Chinese Remainder Theorem and Modular Arithmetic Properties

Estimated Time:1m 30s

Question 151Question →

Let $p = 0.00036 \times 10^{-2}$ and $q = 9.0 \times 10^{-7}$ . Which of the following statements are true? Select all that apply.

Select all that apply

p + q = 4.5 \times 10^{-6}

\frac{p}{q} = 4

\sqrt{p \times q} = 1.8 \times 10^{-6}

p - q = 2.7 \times 10^{-7}

\sqrt{p^2 + q^2} = p + q

Show answer & explanation

Answer: $p + q = 4.5 \times 10^{-6}$ ; $\frac{p}{q} = 4$ ; $\sqrt{p \times q} = 1.8 \times 10^{-6}$

Answer

The correct statements are the ones asserting

p + q = 4.5 \times 10^{-6}

,

\frac{p}{q} = 4

, and

\sqrt{p \times q} = 1.8 \times 10^{-6}

.

Converting

p

to

3.6 \times 10^{-6}

and

q

to

0.9 \times 10^{-6}

shows that their sum is

4.5 \times 10^{-6}

, their ratio is

4

, and the square root of their product

\sqrt{3.24 \times 10^{-12}}

is

1.8 \times 10^{-6}

.

Step-by-Step Solution

1

Convert

p

into standard scientific notation.

p = 0.00036 \times 10^{-2} = 3.6 \times 10^{-4} \times 10^{-2} = 3.6 \times 10^{-6}

.

Expressing numbers in consistent scientific notation enables direct arithmetic operations.

2

Align

q

to the same power of ten for comparison.

q = 9.0 \times 10^{-7} = 0.9 \times 10^{-6}

.

Writing terms with matching exponents allows straightforward addition and subtraction.

3

Test each statement using the simplified values.

p + q = (3.6 + 0.9) \times 10^{-6} = 4.5 \times 10^{-6}

(True);

\frac{p}{q} = \frac{3.6 \times 10^{-6}}{0.9 \times 10^{-6}} = 4

(True);

\sqrt{p \times q} = \sqrt{3.24 \times 10^{-12}} = 1.8 \times 10^{-6}

(True);

p - q = 2.7 \times 10^{-6} \neq 2.7 \times 10^{-7}

(False);

\sqrt{p^2 + q^2} \neq p + q

(False).

Evaluating each given equation determines all correct options.

Key Concept

Decimal arithmetic and scientific notation require aligning powers of ten for addition/subtraction and proper application of radical properties.

Question 152Question →

If $k = 4.5 \times 10^{-8}$ , what is the value of $\frac{(0.000009)^2 \times 400,000}{k}$ expressed in scientific notation?

7.2 \times 10^2

7.2 \times 10^4

7.2 \times 10^{-14}

7.2 \times 10^{10}

7.2 \times 10^{-2}

Show answer & explanation

Answer: $7.2 \times 10^2$

Answer

7.2 \times 10^2

Converting

0.000009

to

9 \times 10^{-6}

and squaring yields

81 \times 10^{-12}

. Multiplying this by

400,000 = 4 \times 10^5

produces

324 \times 10^{-7}

. Dividing by

4.5 \times 10^{-8}

gives

72 \times 10^1

, which in proper scientific notation (

a \times 10^n

with

1 \le a < 10

) is

7.2 \times 10^2

.

Step-by-Step Solution

1

Convert the components of the numerator into scientific notation

0.000009 = 9 \times 10^{-6}

and

400,000 = 4 \times 10^5

Converting all numbers to scientific notation simplifies subsequent exponent operations.

2

Square the first decimal term

(9 \times 10^{-6})^2 = 9^2 \times (10^{-6})^2 = 81 \times 10^{-12}

Applying the exponent to both the coefficient and the power of 10 gives

81 \times 10^{-12}

.

3

Multiply the terms in the numerator

(81 \times 10^{-12}) \times (4 \times 10^5) = (81 \times 4) \times 10^{-12 + 5} = 324 \times 10^{-7}

Multiply coefficients and add powers of 10.

4

Divide the numerator by the denominator

k = 4.5 \times 10^{-8}

\frac{324 \times 10^{-7}}{4.5 \times 10^{-8}} = \left(\frac{324}{4.5}\right) \times 10^{-7 - (-8)} = 72 \times 10^1

Divide coefficients and subtract the denominator exponent from the numerator exponent.

5

Convert the final result to standard scientific notation format

a \times 10^n

where

1 \le a < 10

72 \times 10^1 = 7.2 \times 10^2

Shift the decimal point one place to the left and increase the exponent of 10 by 1.

Key Concept

Decimals and Scientific Notation Operations

Estimated Time:2m 0s

Question 153Question →

At the start of 2024, a technology company allocated a total research budget of $B$ dollars between Department X and Department Y, such that Department X received $60\%$ of the total budget and Department Y received the remaining $40\%$ . In 2025, Department X's budget was increased by $25\%$ , while Department Y's budget was decreased by $20\%$ . In 2026, the combined budget of both departments was reduced by a uniform $10\%$ . If the total combined budget of the two departments in 2026 was $\$288,900$ , what was the initial total research budget $B$ at the start of 2024?

\$270,000

\$297,000

\$300,000

\$321,000

\$340,000

Show answer & explanation

Answer: $\$300,000$

Answer

\$300,000

The initial budget

B

is divided into

0.60B

for Department X and

0.40B

for Department Y. In 2025, Department X's budget becomes

0.60B \times 1.25 = 0.75B

and Department Y's budget becomes

0.40B \times 0.80 = 0.32B

, yielding a combined 2025 budget of

0.75B + 0.32B = 1.07B

. In 2026, a

10\%

reduction results in a combined budget of

1.07B \times 0.90 = 0.963B

. Setting

0.963B = 288,900

and dividing gives

B = \$300,000

.

Step-by-Step Solution

1

Express the 2024 budget allocations for each department in terms of

B

.

Department X received

0.60B

and Department Y received

0.40B

.

Department X was allocated

60\%

of the total initial budget

B

, leaving

40\%

for Department Y.

2

Calculate the budget of each department in 2025 after their respective percentage changes.

Department X:

0.60B \times (1 + 0.25) = 0.75B

; Department Y:

0.40B \times (1 - 0.20) = 0.32B

.

Department X increased by

25\%

(

1.25

multiplier) and Department Y decreased by

20\%

(

0.80

multiplier).

3

Determine the combined budget for 2025 and apply the 2026 reduction.

Combined 2025 budget =

0.75B + 0.32B = 1.07B

. Combined 2026 budget =

1.07B \times (1 - 0.10) = 0.963B

.

The two department budgets sum to

1.07B

, which then undergoes a uniform

10\%

reduction (

0.90

multiplier).

4

Solve for the initial budget

B

using the given 2026 total of

\$288,900

.

0.963B = 288,900 \implies B = \frac{288,900}{0.963} = 300,000

.

Dividing the 2026 total by the net multiplier

0.963

gives the original budget

B

.

Key Concept

Successive percentage changes with initial weighted allocations

Estimated Time:2m 0s

Question 154Question →

A high-precision instrument measures the mass of a single micro-particle $P$ as $4.8 \times 10^{-8}$ grams and a single micro-particle $Q$ as $8.0 \times 10^{-9}$ grams. A sample consists of a combination of $P$ and $Q$ particles in a ratio of $3$ particles of $P$ for every $5$ particles of $Q$ . If the total mass of the sample is $9.2 \times 10^{-5}$ grams, what is the total number of particles in the sample?

Show answer & explanation

Answer: 4000

Answer

The total number of particles in the sample is 4,000.

Aligning powers of 10 shows that one P particle weighs

4.8 \times 10^{-8}

g and one Q particle weighs

0.8 \times 10^{-8}

g. A combined unit of 3 P particles and 5 Q particles has a mass of

3(4.8 \times 10^{-8}) + 5(0.8 \times 10^{-8}) = 18.4 \times 10^{-8} = 1.84 \times 10^{-7}

g. Dividing the total sample mass of

9.2 \times 10^{-5}

g by

1.84 \times 10^{-7}

g yields 500 units. Since each unit contains 8 particles (3 + 5), the total number of particles is

500 \times 8 = 4,000

.

Step-by-Step Solution

1

Convert the mass of particle Q so that it shares the same exponent (

10^{-8}

) as particle P.

Mass of single particle Q =

0.8 \times 10^{-8}

grams.

Aligning powers of 10 is necessary before performing addition of masses.

2

Find the combined mass of a fundamental ratio group consisting of 3 particles of P and 5 particles of Q.

Mass of one ratio group =

3(4.8 \times 10^{-8}) + 5(0.8 \times 10^{-8}) = 14.4 \times 10^{-8} + 4.0 \times 10^{-8} = 18.4 \times 10^{-8} = 1.84 \times 10^{-7}

grams.

Determines the mass contributed by each set of 8 particles.

3

Divide the total mass of the sample by the mass of a single ratio group.

Number of groups =

\frac{9.2 \times 10^{-5}}{1.84 \times 10^{-7}} = \frac{9.2}{1.84} \times 10^2 = 5 \times 10^2 = 500

groups.

Determines how many full ratio sets of particles make up the sample.

4

Multiply the number of groups by the total number of particles contained in each group (

3 + 5 = 8

).

Total particles =

500 \times 8 = 4000

.

Yields the total count of individual particles in the sample.

Key Concept

Operations with scientific notation, decimal place value alignment, and weighted proportional sums.

Question 155Question →

During an operational quality review, a manufacturing plant evaluated its annual defect rate over a three-year period. In Year 1, the defect rate was $d\%$ , where $d > 0$ . In Year 2, the defect rate decreased by $20\%$ from Year 1. In Year 3, the defect rate increased by $25\%$ from Year 2. Which of the following statements must be true? Select all that apply.

Select all that apply

The defect rate in Year 3 is equal to the defect rate in Year 1.

The defect rate in Year 2 is equal to

80\%

of the defect rate in Year 1.

The net percentage change in the defect rate from Year 1 to Year 3 is an increase of

5\%

.

The defect rate in Year 1 is

20\%

greater than the defect rate in Year 2.

The ratio of the defect rate in Year 3 to the defect rate in Year 2 is

4 : 5

.

Show answer & explanation

Answer: The defect rate in Year 3 is equal to the defect rate in Year 1.; The defect rate in Year 2 is equal to $80\%$ of the defect rate in Year 1.

Answer

The statements confirming that the defect rate in Year 3 equals the defect rate in Year 1, and that the defect rate in Year 2 is 80% of Year 1, are correct.

The statement asserting that Year 3 defect rate equals Year 1 defect rate is true because

0.80 \times 1.25 = 1.00

. The statement asserting that Year 2 defect rate is 80% of Year 1 is true because a 20% reduction leaves 80% of the initial value.

Step-by-Step Solution

1

Express the Year 2 defect rate in terms of Year 1.

d_2 = d_1 \times (1 - 0.20) = 0.80 d_1

A

20\%

decrease from Year 1 corresponds to multiplying the base value

d_1

by

0.80

.

2

Express the Year 3 defect rate in terms of Year 2 and Year 1.

d_3 = d_2 \times (1 + 0.25) = (0.80 d_1) \times 1.25 = 1.00 d_1

A

25\%

increase over Year 2 multiplies

d_2

by

1.25

. Substituting

d_2 = 0.80 d_1

shows

d_3 = d_1

.

3

Evaluate each given choice against the derived relationships.

Year 3 equals Year 1 (True). Year 2 is

80\%

of Year 1 (True). Net change is

0\%

not

5\%

(False). Year 1 is

25\%

greater than Year 2, not

20\%

(False). Ratio of Year 3 to Year 2 is

5:4

, not

4:5

(False).

Only the first two statements hold true under rigorous percentage calculations.

Key Concept

Successive Percent Change and Base Value Shifts

Estimated Time:1m 30s

Question 156Question →

An agricultural research station monitored the water volume in a reservoir over a three-month period. In the first month, the reservoir's water volume decreased by $x\%$ . In the second month, the remaining water volume decreased by $(x + 10)\%$ . In the third month, heavy rainfall increased the water volume remaining at the end of the second month by $50\%$ . If the final water volume at the end of the third month was exactly $84\%$ of the initial water volume prior to the first month, what is the value of $x$ ?

10

15

20

25

30

Show answer & explanation

Answer: $20$

Answer

The value of

x

is

20

.

The correct answer is

20

. Applying successive percent change multipliers to the initial volume

V_0

gives

V_3 = 1.50 \times (1 - \frac{x+10}{100}) \times (1 - \frac{x}{100}) V_0

. Setting this equal to

0.84 V_0

yields the quadratic equation

k^2 - 1.90k + 0.34 = 0

where

k = \frac{x}{100}

. Solving gives

k = 0.20

, which means

x = 20

.

Step-by-Step Solution

1

Express the successive volume changes as percentage multipliers relative to the initial volume

V_0

.

After month 1:

V_1 = V_0 \left(1 - \frac{x}{100}\right)

. After month 2:

V_2 = V_1 \left(1 - \frac{x+10}{100}\right) = V_0 \left(1 - \frac{x}{100}\right)\left(0.90 - \frac{x}{100}\right)

. After month 3:

V_3 = V_2 \times 1.50 = 1.50 V_0 \left(1 - \frac{x}{100}\right)\left(0.90 - \frac{x}{100}\right)

.

Successive percentage changes must be applied sequentially to the updated volume at each stage.

2

Set the expression for final volume

V_3

equal to

0.84 V_0

and simplify the equation.

1.50 \left(1 - \frac{x}{100}\right)\left(0.90 - \frac{x}{100}\right) = 0.84 \implies \left(1 - \frac{x}{100}\right)\left(0.90 - \frac{x}{100}\right) = \frac{0.84}{1.50} = 0.56

.

The problem states that the final volume is

84\%

of the initial volume

V_0

.

3

Substitute

k = \frac{x}{100}

and formulate a quadratic equation.

(1 - k)(0.90 - k) = 0.56 \implies k^2 - 1.90k + 0.90 = 0.56 \implies k^2 - 1.90k + 0.34 = 0

. Multiply by

100

:

100k^2 - 190k + 34 = 0 \implies 50k^2 - 95k + 17 = 0

.

Converting to a standard quadratic form

ak^2 + bk + c = 0

allows solving for

k

.

4

Solve the quadratic equation for

k

.

k = \frac{95 \pm \sqrt{(-95)^2 - 4(50)(17)}}{2(50)} = \frac{95 \pm \sqrt{9025 - 3400}}{100} = \frac{95 \pm \sqrt{5625}}{100} = \frac{95 \pm 75}{100}

. Thus

k = 0.20

or

k = 1.70

.

The quadratic formula provides the roots for

k

.

5

Select the valid physical root and determine

x

.

Since a percentage decrease in water volume cannot exceed

100\%

(

k \le 1

), we discard

k = 1.70

. Therefore

k = 0.20

, which corresponds to

x = 20

.

A initial decrease of

170\%

is physically impossible in this context.

Key Concept

Multi-step percentage change and successive base updates leading to non-linear equations

Estimated Time:3m 0s

Question 157Question →

At the beginning of a dry season, a municipal water reservoir contains an initial volume of $V$ gallons of water. During the first month, $\frac{1}{4}$ of the initial volume is released for agricultural irrigation. During the second month, $\frac{1}{5}$ of the remaining volume evaporates, after which $30,000$ gallons of water are pumped into the reservoir from an underground aquifer. If the final volume of water in the reservoir is equal to $\frac{3}{4}$ of the initial volume $V$ , what was the initial volume $V$ , in gallons?

75,000

150,000

200,000

250,000

600,000

Show answer & explanation

Answer: 200,000

Answer

200,000 gallons

Subtracting the first month's release of

\frac{1}{4}V

leaves

\frac{3}{4}V

. Evaporating

\frac{1}{5}

of this remaining amount leaves

\frac{4}{5} \times \frac{3}{4}V = \frac{3}{5}V

. Adding

30,000

gallons gives a final volume of

\frac{3}{5}V + 30,000

. Setting this equal to

\frac{3}{4}V

results in

\frac{3}{4}V - \frac{3}{5}V = \frac{3}{20}V = 30,000

, which solves to

V = 200,000

.

Step-by-Step Solution

1

Determine the volume remaining after the first month's irrigation release.

The volume remaining is

V - \frac{1}{4}V = \frac{3}{4}V

.

Releasing

\frac{1}{4}

of the initial volume leaves

1 - \frac{1}{4} = \frac{3}{4}

of the initial volume.

2

Calculate the volume remaining after the second month's evaporation.

The volume remaining after evaporation is

\left(1 - \frac{1}{5}\right) \times \frac{3}{4}V = \frac{4}{5} \times \frac{3}{4}V = \frac{3}{5}V

.

Evaporating

\frac{1}{5}

of the remaining water leaves

\frac{4}{5}

of that remaining amount.

3

Formulate the linear equation including the pumped volume and solve for

V

.

\frac{3}{5}V + 30,000 = \frac{3}{4}V \implies 30,000 = \frac{3}{4}V - \frac{3}{5}V = \frac{15 - 12}{20}V = \frac{3}{20}V \implies V = \frac{30,000 \times 20}{3} = 200,000$.

Equating the expression for the final water volume to

\frac{3}{4}V

yields a single linear equation in

V

.

Key Concept

Sequential fractional reduction of remaining amounts and solving linear equations involving rational numbers.

Question 158Question →

What is the value of $\sqrt{4^5 + 4^5 + 4^5 + 4^5}$ ?

16

64

128

256

1024

Show answer & explanation

Answer: 64

Answer

64

Combining the four terms inside the radical gives

4 \times 4^5 = 4^6

. Taking the square root yields

\sqrt{4^6} = 4^3

, which evaluates to 64.

Step-by-Step Solution

1

Rewrite the sum inside the radical as multiplication

4^5 + 4^5 + 4^5 + 4^5 = 4 \times 4^5

Adding four identical terms is equivalent to multiplying the term by 4.

2

Apply the product rule for exponents

4^1 \times 4^5 = 4^{1+5} = 4^6

When multiplying exponential expressions with the same base, add their exponents.

3

Simplify the radical expression

\sqrt{4^6} = (4^6)^{1/2} = 4^3 = 64

Taking the square root of a power is equivalent to dividing the exponent by 2.

Key Concept

Exponent Addition and Radical Laws

Question 159Question →

Printer X prints at a constant rate of $60$ pages per minute, and Printer Y prints at a constant rate of $40$ pages per minute. If both printers operate simultaneously at their respective constant rates, which of the following statements must be true? Select all that apply.

Select all that apply

Working together, the two printers print

100

pages in

1

minute.

Printer X prints

60\%

of the total number of pages produced when both printers operate for the same duration.

The ratio of the time taken by Printer Y to print a job to the time taken by Printer X to print the same job is

2:3

.

Working together, the two printers require

5

minutes to print

500

pages.

The combined printing rate of the two printers working together is

50

pages per minute.

Show answer & explanation

Answer: Working together, the two printers print $100$ pages in $1$ minute.; Printer X prints $60\%$ of the total number of pages produced when both printers operate for the same duration.; Working together, the two printers require $5$ minutes to print $500$ pages.

Answer

The correct statements are that working together the printers print 100 pages in 1 minute, Printer X prints 60% of the total pages, and working together they require 5 minutes to print 500 pages.

The combined rate is

60 + 40 = 100

pages per minute. In

1

minute,

100

pages are printed. In

5

minutes,

5 \times 100 = 500

pages are printed. Out of every

100

pages, Printer X produces

60

, which corresponds to

60\%

.

Step-by-Step Solution

1

Calculate the combined rate of both printers

Combined rate =

60 + 40 = 100

pages per minute

When two entities work together simultaneously, their work rates add directly.

2

Evaluate the proportion of total output printed by Printer X

Proportion =

\frac{60}{60 + 40} = \frac{60}{100} = 60\%

The fraction of work done by one printer is its rate divided by the total combined rate.

3

Determine the time ratio between Printer Y and Printer X for a fixed job

Time ratio Y : X =

\frac{1/40}{1/60} = \frac{60}{40} = 3:2

Time required for a fixed task is inversely proportional to the work rate.

4

Calculate time needed for 500 pages working together

Time =

\frac{500\text{ pages}}{100\text{ pages/min}} = 5\text{ minutes}

Time equals total work divided by combined rate.

Key Concept

Additive work rates and inverse relationship between rate and time in ratio problems

Estimated Time:1m 0s

Question 160Question →

A short-pulse laser emits a single pulse lasting $8.4 \times 10^{-9}$ seconds. A high-speed optical sensor completes one measurement cycle every $1.4 \times 10^{-11}$ seconds. How many measurement cycles does the sensor complete during the duration of a single laser pulse?

Show answer & explanation

Answer: 600

Answer

600

Dividing the pulse duration (

8.4 \times 10^{-9}

seconds) by the sensor cycle time (

1.4 \times 10^{-11}

seconds) yields

\frac{8.4}{1.4} \times 10^{-9 - (-11)} = 6 \times 10^2 = 600

complete cycles.

Step-by-Step Solution

1

Set up the division expression for the total number of cycles.

\frac{8.4 \times 10^{-9}\text{ seconds}}{1.4 \times 10^{-11}\text{ seconds}}

To find how many cycle intervals fit within the total pulse duration.

2

Divide the decimal coefficients.

\frac{8.4}{1.4} = 6

Separating the numerical coefficients from the powers of ten.

3

Apply exponent rules to divide powers of 10.

10^{-9 - (-11)} = 10^{-9 + 11} = 10^2 = 100

Dividing powers with the same base requires subtracting the denominator exponent from the numerator exponent.

4

Combine results to find total cycles.

6 \times 100 = 600

Multiplying coefficient quotient by the simplified power of ten.

Key Concept

Division of Numbers in Scientific Notation and Exponent Rules

Arithmetic

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept