Fractions and Rational Numbers Alıştırma Soruları — GRE General Test

Which of the following rational numbers are strictly greater than $\frac{3}{5}$ ? Select all that apply.

Geçerli olan tümünü seçin

\frac{7}{10}

\frac{2}{3}

\frac{5}{9}

\frac{4}{7}

\frac{11}{15}

Cevabı ve açıklamayı göster

Cevap: $\frac{7}{10}$ ; $\frac{2}{3}$ ; $\frac{11}{15}$

Cevap

The fractions greater than 3/5 are 7/10, 2/3, and 11/15.

Converting 3/5 to decimal form yields 0.60. Evaluating the options: 7/10 equals 0.70, 2/3 is approximately 0.667, and 11/15 is approximately 0.733. Because each of these three values is strictly greater than 0.60, all three are correct choices.

Adım Adım Çözüm

1

Convert the benchmark fraction 3/5 to decimal form.

3/5 = 0.60.

Decimal conversion provides a clear standard for comparing rational numbers.

2

Convert each given choice to decimal form and compare it to 0.60.

7/10 = 0.70; 2/3 ≈ 0.667; 5/9 ≈ 0.556; 4/7 ≈ 0.571; 11/15 ≈ 0.733.

Direct comparison reveals which decimals exceed 0.60.

3

Select all fractions whose values exceed 0.60.

The values 0.70, 0.667, and 0.733 are all strictly greater than 0.60.

The corresponding fractions 7/10, 2/3, and 11/15 satisfy the given condition.

Anahtar Kavram

Comparing Rational Numbers and Fractions

Soru 2Soru →

A baker has a flour mixture consisting only of wheat flour and rye flour. Currently, wheat flour accounts for $\frac{2}{5}$ of the total weight of the mixture. If the baker adds $9$ pounds of wheat flour to the mixture, wheat flour will account for $\frac{1}{2}$ of the new total weight of the mixture. What was the total weight, in pounds, of the original flour mixture?

Cevabı ve açıklamayı göster

Cevap: 45

Cevap

The total weight of the original flour mixture was 45 pounds.

The initial total weight of the mixture is 45 pounds. Initially, wheat flour makes up

\frac{2}{5} \times 45 = 18

pounds. Adding 9 pounds of wheat flour increases the wheat flour to

18 + 9 = 27

pounds and the total weight to

45 + 9 = 54

pounds. The new fraction of wheat flour is

\frac{27}{54} = \frac{1}{2}

, which satisfies the given conditions.

Adım Adım Çözüm

1

Express the initial weight of wheat flour in terms of the initial total weight

W

.

Initial weight of wheat flour =

\frac{2}{5}W

.

Wheat flour represents

\frac{2}{5}

of the total mixture.

2

Formulate an equation reflecting the addition of 9 pounds of wheat flour.

\frac{\frac{2}{5}W + 9}{W + 9} = \frac{1}{2}

Adding 9 pounds of wheat flour increases both the amount of wheat flour and the total weight of the mixture by 9 pounds.

3

Solve the algebraic equation for

W

.

Cross-multiplying gives

2\left(\frac{2}{5}W + 9\right) = W + 9

, which simplifies to

\frac{4}{5}W + 18 = W + 9

. Subtracting

\frac{4}{5}W

and

9

from both sides gives

\frac{1}{5}W = 9

, so

W = 45

.

Isolating

W

gives the value of the original total weight.

Anahtar Kavram

Setting up and solving equations involving fractional parts when a quantity is added to both the part and the whole.

Soru 3Soru →

A budget planner allocates $\frac{1}{3}$ of a monthly stipend to housing expenses and $\frac{2}{5}$ of the stipend to food. If no other expenses are incurred from these two categories, what fraction of the total monthly stipend remains unallocated?

\frac{3}{8}

\frac{4}{15}

\frac{5}{8}

\frac{7}{15}

\frac{11}{15}

Cevabı ve açıklamayı göster

Cevap: $\frac{4}{15}$

Cevap

\frac{4}{15}

of the total monthly stipend remains unallocated.

To find the remaining unallocated fraction, first determine the total fraction spent by adding

\frac{1}{3}

and

\frac{2}{5}

using the common denominator

15

, yielding

\frac{5}{15} + \frac{6}{15} = \frac{11}{15}

. Subtracting this total from

1

gives

1 - \frac{11}{15} = \frac{4}{15}

. Thus, the option equal to

\frac{4}{15}

is correct.

Adım Adım Çözüm

1

Calculate the total fraction allocated to housing and food.

\frac{1}{3} + \frac{2}{5} = \frac{5}{15} + \frac{6}{15} = \frac{11}{15}

Find a common denominator (15) to combine the fractions.

2

Subtract the allocated fraction from 1 to determine the unallocated portion.

1 - \frac{11}{15} = \frac{15}{15} - \frac{11}{15} = \frac{4}{15}

The total stipend is represented by 1 whole.

Anahtar Kavram

Adding and subtracting rational numbers using common denominators

Tahmini Süre:1m 0s

Soru 4Soru →

An artist sold $\frac{1}{3}$ of her paintings on the first day of an exhibition. On the second day, she sold $\frac{2}{5}$ of the paintings that remained after the first day. On the third day, she sold $\frac{1}{4}$ of the paintings that remained after the second day. If $18$ paintings remained unsold at the end of the third day, how many paintings did the artist have originally?

45

60

72

90

108

Cevabı ve açıklamayı göster

Cevap: 60

Cevap

The artist originally had 60 paintings.

To find the original total, determine the remaining fraction of paintings after each day sequentially. After day one,

\frac{2}{3}

remain. After day two,

\frac{3}{5}

of

\frac{2}{3}

remain, which is

\frac{2}{5}

. After day three,

\frac{3}{4}

of

\frac{2}{5}

remain, which simplifies to

\frac{3}{10}

of the original total. Given that

\frac{3}{10}

of the total equals

18

, dividing

18

by

\frac{3}{10}

gives

60

.

Adım Adım Çözüm

1

Calculate the fraction of paintings remaining after the first day.

Remaining fraction after Day 1 =

1 - \frac{1}{3} = \frac{2}{3}

.

Selling

\frac{1}{3}

of the total leaves

\frac{2}{3}

of the original total.

2

Calculate the fraction of paintings remaining after the second day.

Remaining fraction after Day 2 =

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

.

Selling

\frac{2}{5}

of the remaining paintings leaves

\frac{3}{5}

of that remaining portion.

3

Calculate the fraction of paintings remaining after the third day.

Remaining fraction after Day 3 =

\frac{2}{5} \times \left(1 - \frac{1}{4}\right) = \frac{2}{5} \times \frac{3}{4} = \frac{6}{20} = \frac{3}{10}

.

Selling

\frac{1}{4}

of the remaining paintings leaves

\frac{3}{4}

of the second day's remaining portion.

4

Set up the equation with the given remaining quantity and solve for the total number of paintings

N

.

\frac{3}{10} N = 18 \implies N = 18 \times \frac{10}{3} = 60

.

The final remaining fraction represents

18

paintings.

Anahtar Kavram

Sequential Fraction of Remainder Problems

Soru 5Soru →

At a certain technology company, $\frac{4}{9}$ of the employees work in the sales department and $\frac{1}{3}$ of the employees work in the engineering department. If all of the remaining employees work in the administration department, what fraction of the total employees works in the administration department?

\frac{2}{9}

\frac{7}{9}

\frac{5}{12}

\frac{4}{9}

\frac{1}{9}

Cevabı ve açıklamayı göster

Cevap: $\frac{2}{9}$

Cevap

\frac{2}{9}

Expressing

\frac{1}{3}

as

\frac{3}{9}

allows direct addition with

\frac{4}{9}

, giving

\frac{7}{9}

for sales and engineering combined. Subtracting

\frac{7}{9}

from 1 leaves

\frac{2}{9}

for administration.

Adım Adım Çözüm

1

Find the total fraction of employees working in sales and engineering.

\frac{4}{9} + \frac{1}{3} = \frac{4}{9} + \frac{3}{9} = \frac{7}{9}

Convert

\frac{1}{3}

to an equivalent fraction with a common denominator of 9 before adding.

2

Subtract the combined fraction from 1 to find the remaining fraction.

1 - \frac{7}{9} = \frac{9}{9} - \frac{7}{9} = \frac{2}{9}

The sum of all departmental fractions must equal 1 whole.

Anahtar Kavram

Adding fractions with unlike denominators and determining the remaining part of a whole.

Soru 6Soru →

If $x$ and $y$ are non-zero rational numbers such that $\frac{x}{y} < -\frac{1}{2}$ and $x + y > 0$ , which of the following statements must be true? Select all that apply.

Geçerli olan tümünü seçin

xy < 0

\frac{1}{x} + \frac{1}{y} < 0

x > y

|x| > |y|

x^2 + y^2 > (x+y)^2

Cevabı ve açıklamayı göster

Cevap: $xy < 0$ ; $\frac{1}{x} + \frac{1}{y} < 0$ ; $x^2 + y^2 > (x+y)^2$

Cevap

The statements

xy < 0

,

\frac{1}{x} + \frac{1}{y} < 0

, and

x^2 + y^2 > (x+y)^2

must be true.

Because

\frac{x}{y} < -\frac{1}{2}

,

x

and

y

must carry opposite algebraic signs, establishing that

xy < 0

. Combining the reciprocals into

\frac{x+y}{xy}

places a positive numerator over a negative denominator, ensuring the sum of reciprocals is strictly negative. Expanding

(x+y)^2 = x^2 + y^2 + 2xy

shows that adding the negative term

2xy

makes

(x+y)^2 < x^2 + y^2

.

Adım Adım Çözüm

1

Determine the sign of the product

xy

xy < 0

The quotient of two non-zero real numbers is negative if and only if they have opposite signs. Since

\frac{x}{y} < -\frac{1}{2} < 0

,

x

and

y

must have opposite signs, so their product is negative.

2

Evaluate the sum of reciprocals

\frac{1}{x} + \frac{1}{y}

\frac{1}{x} + \frac{1}{y} < 0

Finding a common denominator gives

\frac{1}{x} + \frac{1}{y} = \frac{x+y}{xy}

. Given

x+y > 0

(positive) and

xy < 0

(negative), dividing a positive number by a negative number yields a negative result.

3

Compare

x^2 + y^2

with

(x+y)^2

x^2 + y^2 > (x+y)^2

Using algebraic expansion,

(x+y)^2 = x^2 + y^2 + 2xy

. Because

xy < 0

,

2xy

is negative. Subtracting a positive value (or adding a negative value) to

x^2 + y^2

results in a smaller quantity.

4

Test counterexamples for

x > y

and

|x| > |y|

Neither statement is required to be true.

Let

x = -2

and

y = 3

. Both are rational numbers. Check conditions:

\frac{-2}{3} < -\frac{1}{2}

holds, and

-2 + 3 = 1 > 0

holds. For this counterexample,

x = -2 < 3 = y

and

|-2| = 2 < 3 = |y|

.

Anahtar Kavram

Signs and inequalities of rational numbers and reciprocal operations

Soru 7Soru →

A chemical storage vessel contains a mixture composed of three liquid components: Component $X$ , Component $Y$ , and Component $Z$ . Initially, Component $X$ accounts for $\frac{3}{8}$ of the total mixture volume, and Component $Y$ accounts for $\frac{5}{12}$ of the total mixture volume, with Component $Z$ occupying the remainder of the volume.

During a processing stage, $\frac{1}{3}$ of Component $X$ is extracted and $\frac{2}{5}$ of Component $Y$ is extracted, while Component $Z$ remains completely unchanged. If the total volume of the mixture remaining in the vessel after processing is 170 milliliters, what was the total volume, in milliliters, of the mixture in the vessel before processing?

Cevabı ve açıklamayı göster

Cevap: 240

Cevap

The initial total volume of the mixture was 240 milliliters.

To find the initial total volume, first find the fraction of the initial mixture that is Component Z:

1 - (3/8 + 5/12) = 5/24

. Next, compute the remaining amounts of each component relative to the initial total volume

V

: Component X has

(1 - 1/3) \times 3/8 = 1/4 = 6/24

remaining; Component Y has

(1 - 2/5) \times 5/12 = 1/4 = 6/24

remaining; Component Z retains its

5/24

. Summing these yields

6/24 + 6/24 + 5/24 = 17/24

of the original volume. Setting

(17/24)V = 170

mL gives

V = 170 \times (24/17) = 240

mL.

Adım Adım Çözüm

1

Find the initial fraction of the mixture representing Component

Z

.

Component

Z = 1 - \left(\frac{3}{8} + \frac{5}{12}\right) = 1 - \left(\frac{9}{24} + \frac{10}{24}\right) = 1 - \frac{19}{24} = \frac{5}{24}

.

The sum of all component fractions must equal 1.

2

Calculate the remaining fraction for each component relative to the initial total volume

V

.

Remaining

X = \left(1 - \frac{1}{3}\right) \times \frac{3}{8}V = \frac{2}{3} \times \frac{3}{8}V = \frac{1}{4}V = \frac{6}{24}V

.
Remaining

Y = \left(1 - \frac{2}{5}\right) \times \frac{5}{12}V = \frac{3}{5} \times \frac{5}{12}V = \frac{1}{4}V = \frac{6}{24}V

.
Remaining

Z = \frac{5}{24}V

.

When a fraction of a component is removed, the remaining fraction of that component is multiplied by its original portion of the total mixture.

3

Sum the remaining component fractions to find the total remaining volume as a fraction of

V

.

Total Remaining Fraction

= \frac{6}{24}V + \frac{6}{24}V + \frac{5}{24}V = \frac{17}{24}V

.

Combining the remaining amounts gives the fraction of the initial mixture left.

4

Solve for the initial total volume

V

using the given final volume of 170 mL.

\frac{17}{24}V = 170 \implies V = 170 \times \frac{24}{17} = 10 \times 24 = 240 \text{ mL}.

Multiplying the final volume by the reciprocal of the remaining fraction yields the original volume.

Anahtar Kavram

Multi-step fraction operations and solving for original whole amounts

Soru 8Soru →

A logistics company received a shipment of identical freight containers. On Monday, the crew unloaded $\frac{2}{7}$ of the total shipment. On Tuesday, they unloaded $\frac{3}{5}$ of the remaining containers. On Wednesday, they unloaded $\frac{1}{2}$ of the containers that remained after Tuesday's work. If 30 containers remained unloaded at the end of Wednesday, what was the total number of containers in the original shipment?

80

105

140

210

525

Cevabı ve açıklamayı göster

Cevap: 210

Cevap

The total number of containers in the original shipment was 210.

The correct answer is 210. Working forward, after Monday

\frac{5}{7}

of the initial shipment

N

remains. On Tuesday,

\frac{2}{5}

of that remainder stays unloaded, which equals

\frac{2}{5} \times \frac{5}{7}N = \frac{2}{7}N

. On Wednesday,

\frac{1}{2}

of that remainder stays unloaded, yielding

\frac{1}{2} \times \frac{2}{7}N = \frac{1}{7}N

. Setting

\frac{1}{7}N = 30

gives

N = 210

.

Adım Adım Çözüm

1

Determine the fraction of containers remaining after Monday.

Since

\frac{2}{7}

of the total shipment

N

was unloaded on Monday, the fraction remaining is

1 - \frac{2}{7} = \frac{5}{7}

of

N

.

Subtracting the fraction unloaded on Monday from 1 gives the remaining fraction.

2

Calculate the fraction of containers remaining after Tuesday.

On Tuesday,

\frac{3}{5}

of the remaining

\frac{5}{7}N

was unloaded, which is

\frac{3}{5} \times \frac{5}{7}N = \frac{3}{7}N

. The remaining fraction after Tuesday is

\frac{5}{7}N - \frac{3}{7}N = \frac{2}{7}N

.

Alternatively, if

\frac{3}{5}

of the remainder was unloaded, then

1 - \frac{3}{5} = \frac{2}{5}

of the remainder was left:

\frac{2}{5} \times \frac{5}{7}N = \frac{2}{7}N

.

3

Calculate the fraction of containers remaining after Wednesday.

On Wednesday,

\frac{1}{2}

of the remaining

\frac{2}{7}N

was unloaded, leaving

1 - \frac{1}{2} = \frac{1}{2}

of that remainder. Thus, the final fraction remaining is

\frac{1}{2} \times \frac{2}{7}N = \frac{1}{7}N

.

Multiplying the remaining fraction after Tuesday by the fraction left unhandled on Wednesday yields the overall fraction of the original shipment remaining.

4

Solve for the total initial number of containers

N

.

Set

\frac{1}{7}N = 30

, which gives

N = 30 \times 7 = 210

.

Equating the calculated final remaining fraction to the given numerical count allows solving for the total original quantity.

Anahtar Kavram

Sequential Fraction of Remaining Quantities

Soru 9Soru →

A container originally holds a liquid mixture in which $\frac{2}{5}$ of the total volume is pure juice. After $6$ liters of pure juice are added to the container, pure juice makes up $\frac{1}{2}$ of the new total volume. How many liters of mixture were originally in the container?

14

20

24

30

36

Cevabı ve açıklamayı göster

Cevap: 30

Cevap

30 liters

The original mixture volume

V

contains

\frac{2}{5}V

liters of pure juice. When

6

liters of pure juice are added, the new volume of juice becomes

\frac{2}{5}V + 6

and the new total volume becomes

V + 6

. Setting

\frac{2}{5}V + 6 = \frac{1}{2}(V + 6)

leads to

\frac{1}{10}V = 3

, giving

V = 30

liters.

Adım Adım Çözüm

1

Define variables for the original volume and pure juice volume.

Let

V

represent the original volume of the mixture in liters. The original volume of pure juice is

\frac{2}{5}V

.

Expressing the initial quantity of juice in terms of the unknown total volume establishes the baseline algebraic relationship.

2

Set up an equation incorporating the added quantity of pure juice.

New juice volume

= \frac{2}{5}V + 6

, and new total volume

= V + 6

. The equation is

\frac{2}{5}V + 6 = \frac{1}{2}(V + 6)

.

Adding pure juice increases both the pure juice component and the total volume of the mixture by

6

liters.

3

Solve the linear equation for

V

.

Expanding the right side gives

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

. Subtracting

\frac{2}{5}V

from both sides gives

6 - 3 = \left(\frac{1}{2} - \frac{2}{5}\right)V \Rightarrow 3 = \frac{1}{10}V \Rightarrow V = 30

.

Finding a common denominator for the fractions

\frac{1}{2} = \frac{5}{10}

and

\frac{2}{5} = \frac{4}{10}

allows solving for

V

directly.

Anahtar Kavram

Solving linear equations involving fractions of whole quantities and mixture relationships.

Tahmini Süre:1m 0s

Soru 10Soru →

A storage tank is initially $\frac{2}{3}$ full of liquid. After $\frac{1}{4}$ of the liquid inside the tank is drained out, $15$ gallons of liquid are added to the tank. If the tank is then $\frac{5}{6}$ full, what is the total capacity of the tank, in gallons?

30

36

45

60

90

Cevabı ve açıklamayı göster

Cevap: 45

Cevap

45 gallons

The correct answer of 45 gallons is found by calculating the liquid remaining after draining

\frac{1}{4}

of the initial

\frac{2}{3} C

, which leaves

\frac{1}{2} C

. Adding 15 gallons brings the total volume to

\frac{5}{6} C

. Solving

\frac{5}{6} C - \frac{1}{2} C = 15

gives

\frac{1}{3} C = 15

, so

C = 45

.

Adım Adım Çözüm

1

Determine the fraction of liquid in the tank after draining.

The initial amount of liquid is

\frac{2}{3} C

, where

C

is the total capacity. Draining

\frac{1}{4}

of this liquid leaves

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

of the initial liquid. Thus, the remaining liquid is

\frac{3}{4} \times \frac{2}{3} C = \frac{1}{2} C

.

The fraction drained applies to the existing liquid volume, not the total container capacity.

2

Set up an equation incorporating the added liquid and final tank fraction.

\frac{1}{2} C + 15 = \frac{5}{6} C

Adding

15

gallons to the remaining liquid yields a volume equal to

\frac{5}{6}

of the total capacity.

3

Solve for the total capacity

C

.

Subtracting

\frac{1}{2} C

(which is

\frac{3}{6} C

) from both sides gives

15 = \frac{5}{6} C - \frac{3}{6} C = \frac{2}{6} C = \frac{1}{3} C

. Multiplying both sides by

3

yields

C = 45

.

Isolating

C

determines the overall volume of the tank.

Anahtar Kavram

Multi-step operations with fractions and fractional parts of a whole

Tahmini Süre:1m 30s

Soru 11Soru →

A laboratory vessel contains a liquid solution consisting of water and ethanol, where water makes up $\frac{3}{8}$ of the total volume. First, $\frac{1}{5}$ of the total volume of the solution is drained and replaced with an equal volume of pure ethanol. Next, $\frac{1}{4}$ of the resulting mixture is evaporated, removing water and ethanol in proportion to their presence. Finally, pure water is added to fill the vessel back to its original total volume. What fraction of the final solution is water? Express your answer as a decimal.

Cevabı ve açıklamayı göster

Cevap: 0.475

Cevap

The fraction of the final solution that is water is 0.475 (or 19/40).

By following the multi-step fractional changes to the liquid volume, the remaining water prior to refilling is

\frac{9}{40}

of the original capacity. Refilling the missing

\frac{1}{4}

(or

\frac{10}{40}

) volume with pure water yields

\frac{19}{40} = 0.475

of the total solution as water.

Adım Adım Çözüm

1

Track water content after the initial replacement.

Water fraction becomes

\frac{3}{10}

of the original volume.

Let the original total volume be

V

. Initially, water volume is

\frac{3}{8}V

. Removing

\frac{1}{5}

of the solution leaves

\frac{4}{5}

of the original solution, so the water volume becomes

\frac{4}{5} \times \frac{3}{8}V = \frac{3}{10}V

. Replacing the removed volume with pure ethanol brings total volume back to

V

, with water occupying

\frac{3}{10}V

.

2

Track water content after evaporation.

Water volume becomes

\frac{9}{40}V

and total solution volume becomes

\frac{3}{4}V

.

Evaporating

\frac{1}{4}

of the solution leaves

\frac{3}{4}

of the mixture intact. The remaining water volume is

\frac{3}{4} \times \frac{3}{10}V = \frac{9}{40}V

.

3

Calculate the final water fraction after refilling with pure water.

Final water volume is

\frac{19}{40}V = 0.475V

.

To restore the total volume from

\frac{3}{4}V

back to

V

, an amount equal to

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

of pure water is added. Adding this to the existing water gives

\frac{9}{40}V + \frac{1}{4}V = \frac{9}{40}V + \frac{10}{40}V = \frac{19}{40}V

. Dividing by total volume

V

yields

\frac{19}{40} = 0.475

.

Anahtar Kavram

Sequential fractional reduction and component tracking

Tahmini Süre:2m 30s

Soru 12Soru →

If $a$ and $b$ are positive rational numbers such that $a < b < 1$ , which of the following expressions must be strictly greater than $1$ ? Select all that apply.

Geçerli olan tümünü seçin

\frac{b}{a}

\frac{a+1}{b}

\frac{a}{b^2}

\frac{b+1}{a+1}

\frac{a+b}{2b}

Cevabı ve açıklamayı göster

Cevap: $\frac{b}{a}$ ; $\frac{a+1}{b}$ ; $\frac{b+1}{a+1}$

Cevap

The expressions that must be strictly greater than 1 are the ratio of b to a, the fraction (a + 1) over b, and the fraction (b + 1) over (a + 1).

The correct options are those involving ratios where the numerator is demonstrably greater than the denominator. The ratio of b to a compares two positive numbers where b is larger, giving a quotient greater than 1. The fraction with numerator (a + 1) has a value over 1 while denominator b is under 1, ensuring the quotient exceeds 1. The fraction comparing (b + 1) to (a + 1) has a larger numerator because adding 1 preserves the inequality b > a.

Adım Adım Çözüm

1

Analyze the ratio of b to a

Since

b > a > 0

,

\frac{b}{a} > 1

is always true.

Dividing any positive number by a smaller positive number results in a value greater than 1.

2

Analyze the fraction (a + 1) over b

Since

a > 0

,

a + 1 > 1

. Since

b < 1

,

a + 1 > b

, which implies

\frac{a+1}{b} > 1

.

The numerator is greater than 1 and the denominator is less than 1, so the fraction exceeds 1.

3

Test counterexamples for the fraction a over b squared

For

a = \frac{1}{8}

and

b = \frac{1}{2}

,

\frac{a}{b^2} = \frac{1/8}{1/4} = \frac{1}{2} \le 1

.

The expression is not guaranteed to be greater than 1 for all valid values of a and b.

4

Analyze the fraction (b + 1) over (a + 1)

Since

b > a

, adding 1 to both quantities yields

b + 1 > a + 1 > 0

, so

\frac{b+1}{a+1} > 1

.

Adding the same positive quantity to two numbers maintains their relative order.

5

Analyze the average fraction (a + b) over 2b

Since

a < b

,

a + b < 2b

, meaning

\frac{a+b}{2b} < 1

.

The numerator is the sum of a and b, which is strictly less than twice b.

Anahtar Kavram

Properties of Rational Number Inequalities and Fraction Comparison

Tahmini Süre:1m 30s

Soru 13Soru →

A community library received a donation of $120$ new books. On Monday, $\frac{3}{8}$ of the donated books were cataloged. On Tuesday, $\frac{1}{3}$ of the remaining uncataloged books were cataloged. How many of the donated books remain to be cataloged?

Cevabı ve açıklamayı göster

Cevap: 50

Cevap

50 books remain to be cataloged.

First, find the number of books cataloged on Monday: 3/8 of 120 is 45 books. Subtracting 45 from 120 leaves 75 books uncataloged. On Tuesday, 1/3 of those remaining 75 books were cataloged, which is 25 books. Subtracting 25 from 75 gives 50 books remaining to be cataloged.

Adım Adım Çözüm

1

Calculate the number of books cataloged on Monday.

45 books

Multiply the initial total of 120 books by the fraction cataloged on Monday, which is 3/8.

2

Calculate the uncataloged books remaining after Monday.

75 books

Subtract the 45 cataloged books from the initial 120 books.

3

Calculate the number of books cataloged on Tuesday.

25 books

Multiply the remaining 75 books by the fraction cataloged on Tuesday, which is 1/3.

4

Calculate the final number of uncataloged books.

50 books

Subtract the 25 books cataloged on Tuesday from the 75 uncataloged books remaining after Monday.

Anahtar Kavram

Fraction of a Remaining Quantity

Tahmini Süre:50s

Soru 14Soru →

A university research lab receives a multi-year grant. In the first quarter, the lab spends $\frac{2}{7}$ of the total grant on laboratory equipment. In the second quarter, the lab spends $\frac{3}{5}$ of the remaining funds on research staff salaries. At the beginning of the third quarter, the lab receives a supplementary addition equal to $\frac{1}{4}$ of the funds remaining at the end of the second quarter. If the total amount of money in the grant immediately following the third-quarter addition is $\$45,000$ , what was the initial dollar amount of the research grant?

$84,000

$126,000

$157,500

$210,000

$315,000

Cevabı ve açıklamayı göster

Cevap: $126,000

Cevap

$126,000

The correct answer of

126,000 is found by tracking the fraction of the initial grant

G

remaining after each step. Spending

\frac{2}{7}

leaves

\frac{5}{7}G

. Spending

\frac{3}{5}

of that remainder leaves

\frac{2}{5} \times \frac{5}{7}G = \frac{2}{7}G

. Adding

\frac{1}{4}

of

\frac{2}{7}G

increases the balance by

\frac{1}{14}G

, resulting in a final total of

\frac{2}{7}G + \frac{1}{14}G = \frac{5}{14}G

. Setting

\frac{5}{14}G = 45,000

yields

G = 126,000$.

Adım Adım Çözüm

1

Determine the remaining fraction after the first quarter.

Fraction remaining =

1 - \frac{2}{7} = \frac{5}{7}

of the initial grant

G

.

The lab spent

\frac{2}{7}

of the total grant.

2

Calculate the remaining fraction after the second quarter.

Fraction remaining =

\left(1 - \frac{3}{5}\right) \times \frac{5}{7}G = \frac{2}{5} \times \frac{5}{7}G = \frac{2}{7}G

.

The lab spent

\frac{3}{5}

of the remaining funds, leaving

\frac{2}{5}

of those funds.

3

Account for the third-quarter supplementary addition.

Final fraction =

\frac{2}{7}G + \frac{1}{4}\left(\frac{2}{7}G\right) = \frac{2}{7}G + \frac{1}{14}G = \frac{4}{14}G + \frac{1}{14}G = \frac{5}{14}G

.

An additional

\frac{1}{4}

of the second-quarter remaining funds was added to the grant.

4

Solve for the initial grant amount

G

.

\frac{5}{14}G = 45,000 \implies G = 45,000 \times \frac{14}{5} = 9,000 \times 14 = 126,000

.

The final remaining balance of

\$45,000

corresponds to

\frac{5}{14}

of the original grant.

Anahtar Kavram

Multi-step sequential fraction operations and base tracking in word problems

Soru 15Soru →

An industrial chemical reservoir is emptied by three pumps, $P_1$ , $P_2$ , and $P_3$ , operating independently at different constant rates. Working alone, $P_1$ can empty the full reservoir in $6$ hours, $P_2$ can empty it in $8$ hours, and $P_3$ can empty it in $12$ hours. Initially, the reservoir is completely full. First, $P_1$ and $P_2$ work together for $2$ hours. Then $P_1$ is turned off, and $P_3$ is turned on to work alongside $P_2$ . Additionally, while $P_2$ and $P_3$ are working together, a defect causes liquid to leak out of the bottom of the reservoir at a constant rate equal to $\frac{1}{6}$ of the combined emptying rate of $P_2$ and $P_3$ . How many additional hours will it take to completely empty the remaining liquid from the reservoir?

\frac{12}{7}

hours

2

hours

\frac{10}{9}

hours

\frac{12}{5}

hours

\frac{144}{49}

hours

Cevabı ve açıklamayı göster

Cevap: $\frac{12}{7}$ hours

Cevap

\frac{12}{7}

hours

The option stating '

\frac{12}{7}

hours' is correct because during the first 2 hours,

P_1

and

P_2

empty

\frac{7}{12}

of the reservoir, leaving

\frac{5}{12}

remaining. Then,

P_2

and

P_3

have a combined rate of

\frac{5}{24}

, and the defect adds an extra

\frac{5}{144}

per hour, giving a total emptying rate of

\frac{35}{144}

reservoirs per hour. Dividing

\frac{5}{12}

by

\frac{35}{144}

yields

\frac{12}{7}

hours.

Adım Adım Çözüm

1

Calculate the combined emptying rate of

P_1

and

P_2

and determine the fraction emptied in the first

2

hours.

Rate of

P_1 = \frac{1}{6}

reservoir/hr, Rate of

P_2 = \frac{1}{8}

reservoir/hr. Combined rate = \frac{1}{6} + \frac{1}{8} = \frac{7}{24} reservoir/hr. In

2

hours, amount emptied =

2 \times \frac{7}{24} = \frac{7}{12}

of the reservoir.

Determines how much work was completed during the initial stage.

2

Find the remaining fraction of liquid in the reservoir.

Remaining fraction =

1 - \frac{7}{12} = \frac{5}{12}

of the reservoir.

Establishes the remaining volume that must be emptied in the second stage.

3

Calculate the combined rate of

P_2

,

P_3

, and the defect rate.

Combined rate of

P_2

and

P_3 = \frac{1}{8} + \frac{1}{12} = \frac{5}{24}

reservoir/hr. Defect rate = \frac{1}{6} \times \frac{5}{24} = \frac{5}{144} reservoir/hr. Total emptying rate = \frac{5}{24} + \frac{5}{144} = \frac{30 + 5}{144} = \frac{35}{144} reservoir/hr.

Combines all simultaneous emptying processes during the second stage.

4

Divide the remaining fraction by the total emptying rate to find the additional time needed.

Time = \frac{5/12}{35/144} = \frac{5}{12} \times \frac{144}{35} = \frac{5}{35} \times \frac{144}{12} = \frac{1}{7} \times 12 = \frac{12}{7} hours.

Applies the rate formula

\text{Time} = \frac{\text{Work}}{\text{Rate}}

.

Anahtar Kavram

Compound Fraction Work Rates and Rational Operations

Soru 16Soru →

Let $a$ and $b$ be rational numbers such that $-1 < a < 0 < b < 1$ . Which of the following expressions MUST be strictly greater than $1$ ? Select all such expressions.

Geçerli olan tümünü seçin

\frac{b - a}{b}

\frac{1 + b}{1 + a}

\frac{a + b}{b}

\frac{b - a}{1 - a}

\frac{1}{b - a}

Cevabı ve açıklamayı göster

Cevap: $\frac{b - a}{b}$ ; $\frac{1 + b}{1 + a}$

Cevap

The expressions

\frac{b - a}{b}

and

\frac{1 + b}{1 + a}

MUST be strictly greater than 1.

The expression representing the difference

(b - a)

divided by

b

simplifies to

1 + \frac{-a}{b}

. Because

-a

and

b

are both positive, this term is strictly greater than 1. Similarly, the expression

\frac{1 + b}{1 + a}

divides a numerator greater than 1 by a positive denominator less than 1, which always yields a value strictly greater than 1.

Adım Adım Çözüm

1

Analyze the expression

\frac{b - a}{b}

by splitting the fraction.

\frac{b - a}{b} = \frac{b}{b} - \frac{a}{b} = 1 - \frac{a}{b}

.

Since

a < 0

and

b > 0

, the quotient

\frac{a}{b}

is strictly negative. Subtracting a negative number from 1 is equivalent to adding a positive number, so

1 - \frac{a}{b} > 1

.

2

Analyze the expression

\frac{1 + b}{1 + a}

by establishing bounds for the numerator and denominator.

Since

b > 0

,

1 + b > 1

. Since

-1 < a < 0

, adding 1 yields

0 < 1 + a < 1

.

Dividing any real number greater than 1 by a positive real number less than 1 produces a value strictly greater than 1.

3

Evaluate the remaining options with counterexamples or algebraic bounds to verify they are not guaranteed to be greater than 1.

The expression

\frac{a + b}{b} = 1 + \frac{a}{b} < 1

; the expression

\frac{b - a}{1 - a} < 1

because

b < 1 \implies b - a < 1 - a

; and

\frac{1}{b - a}

can be less than 1 when

b - a > 1

(e.g.,

b = 0.8, a = -0.5 \implies b - a = 1.3

).

Demonstrating that an expression is less than 1 or can be less than 1 eliminates it from being strictly greater than 1 for all valid rational numbers

a

and

b

.

Anahtar Kavram

Properties and Inequalities of Rational Numbers

Soru 17Soru →

An investment fund allocates its total portfolio assets among three asset classes: stocks, bonds, and real estate. Initially, $\frac{5}{12}$ of the total portfolio is invested in stocks and $\frac{1}{3}$ of the total portfolio is invested in bonds, with the remaining fraction invested in real estate. If $\frac{2}{5}$ of the stock portfolio and $\frac{1}{4}$ of the bond portfolio are subsequently liquidated, what fraction of the fund's remaining total portfolio is invested in real estate?

\frac{1}{4}

\frac{3}{10}

\frac{1}{3}

\frac{4}{9}

\frac{5}{12}

Cevabı ve açıklamayı göster

Cevap: $\frac{1}{3}$

Cevap

The fraction of the fund's remaining total portfolio invested in real estate is

\frac{1}{3}

.

Let the original total portfolio be

T

. Initially, stocks are

\frac{5}{12}T

, bonds are

\frac{4}{12}T

, and real estate is

1 - \frac{9}{12}T = \frac{3}{12}T

. After liquidating

\frac{2}{5}

of stocks and

\frac{1}{4}

of bonds, the remaining stock value is

\frac{3}{5} \times \frac{5}{12}T = \frac{3}{12}T

and the remaining bond value is

\frac{3}{4} \times \frac{4}{12}T = \frac{3}{12}T

. Real estate remains

\frac{3}{12}T

. The new total portfolio is

\frac{3}{12}T + \frac{3}{12}T + \frac{3}{12}T = \frac{9}{12}T = \frac{3}{4}T

. Therefore, real estate represents

\frac{3/12 T}{9/12 T} = \frac{3}{9} = \frac{1}{3}

of the remaining total portfolio.

Adım Adım Çözüm

1

Determine the initial fraction invested in real estate.

Real estate accounts for

1 - \left(\frac{5}{12} + \frac{1}{3}\right) = 1 - \frac{9}{12} = \frac{3}{12} = \frac{1}{4}

of the original total portfolio.

The sum of all three asset fractions must equal 1.

2

Calculate the remaining fractions for stocks and bonds after liquidation.

Remaining stocks =

\left(1 - \frac{2}{5}\right) \times \frac{5}{12} = \frac{3}{5} \times \frac{5}{12} = \frac{3}{12}

. Remaining bonds =

\left(1 - \frac{1}{4}\right) \times \frac{1}{3} = \frac{3}{4} \times \frac{4}{12} = \frac{3}{12}

.

Multiplying the unliquidated fraction by each asset's initial fraction gives its remaining fraction relative to the original total.

3

Calculate the total remaining portfolio fraction.

Total remaining portfolio =

\frac{3}{12} \text{ (stocks)} + \frac{3}{12} \text{ (bonds)} + \frac{3}{12} \text{ (real estate)} = \frac{9}{12} = \frac{3}{4}

of the original total portfolio.

Sum the remaining portions of all three asset categories.

4

Compute the ratio of real estate to the remaining total portfolio.

\frac{\text{Real Estate}}{\text{Remaining Total}} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}.

The question asks for the fraction relative to the new reduced total portfolio.

Anahtar Kavram

Sequential Fraction Allocation and Reduced Base Ratios

Tahmini Süre:1m 30s

Soru 18Soru →

Let

a

,

b

, and

c

be positive integers, and define three rational numbers

x

,

y

, and

z

as follows:

x = \frac{a}{a+b}, \quad y = \frac{b}{b+c}, \quad \text{and} \quad z = \frac{c}{c+a}

Which of the following statements MUST be true for all such positive integers

a

,

b

, and

c

? Select all that apply.

Geçerli olan tümünü seçin

The sum

x + y + z

satisfies the inequality

1 < x + y + z < 2

.

The product

(1 - x)(1 - y)(1 - z)

is identically equal to

xyz

.

The product

xyz

cannot exceed

\frac{1}{8}

.

The sum

x + y + z

can be equal to an integer.

If

x + y + z = \frac{3}{2}

, then it must be that

x = y = z = \frac{1}{2}

.

Cevabı ve açıklamayı göster

Cevap: The sum $x + y + z$ satisfies the inequality $1 < x + y + z < 2$ .; The product $(1 - x)(1 - y)(1 - z)$ is identically equal to $xyz$ .; The product $xyz$ cannot exceed $\frac{1}{8}$ .

Cevap

The statements asserting that

1 < x + y + z < 2

, that

(1 - x)(1 - y)(1 - z) = xyz

, and that

xyz \le \frac{1}{8}

are all true.

The statement bounding the sum between 1 and 2 is correct because adjusting denominators to a common sum

a+b+c

reveals that the sum is strictly greater than 1 and strictly less than 2. The identity statement is correct because subtracting each fraction from 1 shifts the numerators cyclically without changing the overall product. The statement placing an upper bound of

\frac{1}{8}

on the product is correct by applying the AM-GM inequality to each denominator term.

Adım Adım Çözüm

1

Establish strict bounds on the sum

x + y + z

.

Since

a, b, c > 0

, we have

\frac{a}{a+b} > \frac{a}{a+b+c}

. Summing all three terms gives

x + y + z > \frac{a+b+c}{a+b+c} = 1

. To find the upper bound, rewrite

\frac{a}{a+b} < \frac{a+c}{a+b+c}

. Summing all three terms gives

x + y + z < \frac{(a+c)+(a+b)+(b+c)}{a+b+c} = 2

. Thus,

1 < x + y + z < 2

.

This proves that the sum of the three fractions must always lie strictly between 1 and 2.

2

Compute the product

(1-x)(1-y)(1-z)

algebraically.

1 - x = 1 - \frac{a}{a+b} = \frac{b}{a+b}

,

1 - y = \frac{c}{b+c}

, and

1 - z = \frac{a}{c+a}

. The product is

\frac{b \cdot c \cdot a}{(a+b)(b+c)(c+a)} = \frac{abc}{(a+b)(b+c)(c+a)} = xyz

.

This proves the structural algebraic identity between

(1-x)(1-y)(1-z)

and

xyz

.

3

Apply the AM-GM inequality to find the upper bound of

xyz

.

Since

a+b \ge 2\sqrt{ab}

,

b+c \ge 2\sqrt{bc}

, and

c+a \ge 2\sqrt{ca}

, their product satisfies

(a+b)(b+c)(c+a) \ge 8\sqrt{a^2b^2c^2} = 8abc

. Taking the reciprocal gives

xyz = \frac{abc}{(a+b)(b+c)(c+a)} \le \frac{abc}{8abc} = \frac{1}{8}

.

This confirms that

xyz

has a maximum possible value of

\frac{1}{8}

(achieved when

a=b=c

).

4

Test counterexamples for the remaining statements.

Since

1 < x + y + z < 2

, no integer value is possible, ruling out integer sums. Additionally, substituting

a=1, b=2, c=1

gives

x=\frac{1}{3}, y=\frac{2}{3}, z=\frac{1}{2}

, whose sum is

\frac{3}{2}

despite

x, y, z

not being equal to

\frac{1}{2}

.

This disproves the statements claiming the sum can be an integer or that a sum of

\frac{3}{2}

requires all three fractions to be equal.

Anahtar Kavram

Bounding and algebraic properties of cyclic rational fractions

Tahmini Süre:3m 0s

Soru 19Soru →

An asset management firm divides its total portfolio into three distinct funds: Fund X, Fund Y, and Fund Z. Initially, $\frac{1}{3}$ of the total portfolio value is in Fund X, and $\frac{2}{5}$ is in Fund Y, with the remaining fraction in Fund Z. Over the course of one year, the value of Fund X increases by $\frac{1}{5}$ , the value of Fund Y decreases by $\frac{1}{4}$ , and the value of Fund Z increases by $\frac{1}{2}$ . At the end of the year, what fraction of the portfolio's total value is contained in Fund Z?

\frac{4}{11}

\frac{2}{5}

\frac{4}{7}

\frac{4}{15}

\frac{15}{16}

Cevabı ve açıklamayı göster

Cevap: $\frac{4}{11}$

Cevap

The fraction of the portfolio's total final value contained in Fund Z is

\frac{4}{11}

.

Fund Z grows to

\frac{2}{5}

of the original portfolio value, while the overall portfolio value increases to

\frac{11}{10}

of the original total. Comparing Fund Z's final value to the new total yields

\frac{2/5}{11/10} = \frac{4}{11}

.

Adım Adım Çözüm

1

Calculate the initial fraction of the total portfolio allocated to Fund Z.

The initial allocation for Fund Z is

1 - \left(\frac{1}{3} + \frac{2}{5}\right) = 1 - \frac{11}{15} = \frac{4}{15}

.

The sum of all initial fund allocations must equal 1.

2

Determine the final value of each fund relative to the initial total portfolio value

V

.

Fund X:

\frac{1}{3}V \times \left(1 + \frac{1}{5}\right) = \frac{2}{5}V

. Fund Y:

\frac{2}{5}V \times \left(1 - \frac{1}{4}\right) = \frac{3}{10}V

. Fund Z:

\frac{4}{15}V \times \left(1 + \frac{1}{2}\right) = \frac{2}{5}V

.

Multiply each initial fund share by its respective growth or reduction factor.

3

Calculate the new total portfolio value relative to the initial value

V

.

Total final value =

\frac{2}{5}V + \frac{3}{10}V + \frac{2}{5}V = \frac{4}{10}V + \frac{3}{10}V + \frac{4}{10}V = \frac{11}{10}V

.

Sum the final values of all three individual funds.

4

Compute the final fraction of the portfolio contained in Fund Z.

Fraction =

\frac{\frac{2}{5}V}{\frac{11}{10}V} = \frac{2}{5} \times \frac{10}{11} = \frac{4}{11}

.

Divide the final value of Fund Z by the new total portfolio value.

Anahtar Kavram

Rational number operations and shifting base values in multi-step fraction problems

Tahmini Süre:2m 0s

Soru 20Soru →

A laboratory container holds a solution composed solely of alcohol, acid, and water. Initially, alcohol accounts for $\frac{3}{8}$ of the solution's total volume, and acid accounts for $\frac{1}{4}$ of the total volume. In a two-step process, a chemist first removes $\frac{1}{3}$ of the alcohol present and $\frac{1}{2}$ of the acid present, with no water removed. Next, the chemist adds pure water until water accounts for $\frac{3}{5}$ of the solution's new total volume. What fraction of the final solution's total volume is alcohol?

\frac{4}{15}

\frac{1}{4}

\frac{4}{9}

\frac{2}{5}

\frac{2}{3}

Cevabı ve açıklamayı göster

Cevap: $\frac{4}{15}$

Cevap

The fraction of the final solution's total volume that is alcohol is

\frac{4}{15}

.

The correct fraction

\frac{4}{15}

is derived by calculating the remaining alcohol volume as

\frac{1}{4}

of the initial total volume

V

, and determining the final total volume

V_{\text{final}} = \frac{15}{16}V

using the constant non-water volume of

\frac{3}{8}V

. Dividing

\frac{1}{4}V

by

\frac{15}{16}V

yields

\frac{4}{15}

.

Adım Adım Çözüm

1

Determine initial volume fractions for all three components.

Let

V

be the initial total volume. Alcohol is

\frac{3}{8}V

, acid is

\frac{1}{4}V = \frac{2}{8}V

, and water is

1 - (\frac{3}{8} + \frac{2}{8}) = \frac{3}{8}V

.

Establishing the initial baseline volumes allows accurate tracking through the multi-step changes.

2

Calculate component volumes after the removal step.

Remaining alcohol =

\frac{3}{8}V \times (1 - \frac{1}{3}) = \frac{1}{4}V

. Remaining acid =

\frac{1}{4}V \times (1 - \frac{1}{2}) = \frac{1}{8}V

. Water remains

\frac{3}{8}V

.

Removing specified fractions of individual components changes their absolute volumes.

3

Calculate non-water volume and final total volume after water is added.

Total non-water volume =

\frac{1}{4}V + \frac{1}{8}V = \frac{3}{8}V

. Since water becomes

\frac{3}{5}

of the final volume

V_{\text{final}}

, non-water is

1 - \frac{3}{5} = \frac{2}{5}

of

V_{\text{final}}

. Thus,

\frac{2}{5} V_{\text{final}} = \frac{3}{8}V \implies V_{\text{final}} = \frac{5}{2} \times \frac{3}{8}V = \frac{15}{16}V

.

Adding only water keeps the non-water volume constant, providing a fixed reference point to find the new total volume.

4

Compute the final fraction of alcohol in the solution.

\text{Alcohol Fraction} = \frac{\text{Alcohol Volume}}{V_{\text{final}}} = \frac{\frac{1}{4}V}{\frac{15}{16}V} = \frac{1}{4} \times \frac{16}{15} = \frac{4}{15}.

The required quantity is the part-to-whole ratio of remaining alcohol to the final total volume.

Anahtar Kavram

Multi-step fraction operations involving component-wise removals and fixed non-water volumes in liquid mixtures.

Tahmini Süre:2m 30s

Fractions and Rational Numbers

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