Testbook Logo
Exams SuperCoaching Test Series Skill Academy
More
English
English
Hindi
Telugu
Bengali
Marathi
Malayalam
Tamil
Download Ratios, Rates, Proportional Relationships and Units MCQs Free PDF

Ratios, Rates, Proportional Relationships and Units MCQ Quiz - Objective Question with Answer for Ratios, Rates, Proportional Relationships and Units - Download Free PDF

Last updated on Mar 19, 2025

Latest Ratios, Rates, Proportional Relationships and Units MCQ Objective Questions

Ratios, Rates, Proportional Relationships and Units Question 1:

The density of a certain mineral is \(2,500\) kg/m³, and a cube of this mineral weighs \(10,000\) kg. Calculate the side length of the cube in meters, to the nearest hundredth.

  1. 1.59
  2. 2.20
  3. 1.25
  4. 2.30

Answer (Detailed Solution Below)

Option 1 : 1.59

Ratios, Rates, Proportional Relationships and Units Question 1 Detailed Solution

To find the side length of the cube, begin with the density equation: \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \). Given the density \(2,500\) kg/m³ and mass \(10,000\) kg, solve for \(V\): \( 2,500 = \frac{10,000}{V} \). Solving gives \( V = \frac{10,000}{2,500} = 4 \) cubic meters. Since the cube has volume \( V = s^3 \), we have \( s^3 = 4 \). Solve for \(s\): \( s = \sqrt[3]{4} \approx 1.5874 \). Rounding gives \(1.59\).

Ratios, Rates, Proportional Relationships and Units Question 2:

A cube has a density of \(250\) kilograms per cubic meter and a mass of \(500\) kilograms. To the nearest hundredth, what is the length of one edge of the cube in meters?

  1. 2.50
  2. 2.51
  3. 1.26
  4. 2.53

Answer (Detailed Solution Below)

Option 3 : 1.26

Ratios, Rates, Proportional Relationships and Units Question 2 Detailed Solution

To find the length of one edge of the cube, we need to determine the volume of the cube first. The density of the cube is given as \(250\) kg/m3, and the mass is \(500\) kg. Using the formula for density, \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \), we can calculate the volume of the cube as follows: \( 250 = \frac{500}{V} \). Solving for \(V\), we get \( V = \frac{500}{250} = 2 \) cubic meters. Since the cube's volume is \(2\) cubic meters, and volume \( V = s^3 \) for a cube, we have \( s^3 = 2 \). Taking the cube root of both sides gives \( s = \sqrt[3]{2} \approx 1.26 \). 

Ratios, Rates, Proportional Relationships and Units Question 3:

A cubic tank holds \(1000\) liters of water and has a density of water at \(1000\) kg/m³. If the mass of the water is \(1000\) kilograms, what is the length of one side of the tank in meters?

  1. 0.99
  2. 1.00
  3. 1.01
  4. 1.02

Answer (Detailed Solution Below)

Option 2 : 1.00

Ratios, Rates, Proportional Relationships and Units Question 3 Detailed Solution

To determine the length of one side of the tank, we first note that the density of water is \(1000\) kg/m³, and the mass is also \(1000\) kg. The volume formula for density is \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \). Given the density and mass are both \(1000\), the volume \( V \) is \( \frac{1000}{1000} = 1 \) cubic meter. Since the tank is cubic, \( V = s^3 \). Therefore, \( s^3 = 1 \). Taking the cube root of both sides gives \( s = \sqrt[3]{1} = 1 \). Therefore, the length of one edge of the tank is \(1.00\) meters. Option \(2\) is the correct answer as it directly matches the calculated result.

Ratios, Rates, Proportional Relationships and Units Question 4:

A metal block with a density of \(8,900\) kg/m³ has a mass of \(1,780\) kg and is shaped like a cube. What is the length of one side of the cube in meters, rounded to the nearest hundredth?

  1. 0.58
  2. 0.59
  3. 0.60
  4. 0.61

Answer (Detailed Solution Below)

Option 1 : 0.58

Ratios, Rates, Proportional Relationships and Units Question 4 Detailed Solution

First, calculate the volume of the metal block using its density and mass. The formula is \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \). Plug in the values: \( 8,900 = \frac{1,780}{V} \). Solve for \(V\): \( V = \frac{1,780}{8,900} \approx 0.2 \) cubic meters. Since the block is a cube, use the cube volume formula \( V = s^3 \). Therefore, \( s^3 = 0.2 \). Solve for \(s\) by taking the cube root: \( s = \sqrt[3]{0.2} \approx 0.5848 \). Round to the nearest hundredth, giving \(0.58\). 

Ratios, Rates, Proportional Relationships and Units Question 5:

The density of a brick is 1800 kilograms per cubic meter. If the brick has dimensions of 0.2 meters by 0.1 meters by 0.05 meters, what is its mass in kilograms?

  1. 1.8
  2. 1.7
  3. 1.9
  4. 2.0

Answer (Detailed Solution Below)

Option 1 : 1.8

Ratios, Rates, Proportional Relationships and Units Question 5 Detailed Solution

The volume \( V \) of the brick can be calculated as \( V = 0.2 \times 0.1 \times 0.05 = 0.001 \) cubic meters. The mass \( M \) can be calculated using the formula \( M = \text{Density} \times \text{Volume} \). So, \( M = 1800 \times 0.001 = 1.8 \) kilograms. Thus, the mass of the brick is 1.8 kilograms, matching option 1.

Top Ratios, Rates, Proportional Relationships and Units MCQ Objective Questions

Ratios, Rates, Proportional Relationships and Units Question 6

Download Solution PDF

A trivia tournament organizer wanted to study the relationship between the number of points a team scores in a trivia round and the number of hours that a team practices each week. For the study, the organizer selected 55 teams at random from all trivia teams in a certain tournament. The table displays the information for the 40 teams in the sample that practiced for at least 3 hours per week. 

  Number of points per round
Hours practiced  6 to 13 points 14 or more points Total
3 to 5 hours 6 4 10
More than 5 hours 4 26 30
Total 10 30 40


Which of the following is the largest population to which the results of the study can be generalized?

  1. All trivia teams in the tournament that scored 14 or more points in the round 
  2. The 55 trivia teams in the sample 
  3. The 40 trivia teams in the sample that practiced for at least 3 hours per week 
  4. All trivia teams in the tournament 

Answer (Detailed Solution Below)

Option 4 : All trivia teams in the tournament 

Ratios, Rates, Proportional Relationships and Units Question 6 Detailed Solution

Download Solution PDF
Choice D is correct. It's given that the organizer selected 55 teams at random from all trivia teams in the tournament. A table is also given displaying the information for the 40 teams in the sample that practiced for at least 3 hours per week. Selecting a sample of a reasonable size at random to use for a survey allows the results from that survey to be applied to the population from which the sample was selected, but not beyond this population. Thus, only the sampling method information is necessary to determine the largest population to which the results of the study can be generalized. Since the organizer selected the sample at random from all trivia teams in the tournament, the largest population to which the results of the study can be generalized is all trivia teams in the tournament. 
Choice A is incorrect. The sample was selected at random from all trivia teams in the tournament, not just from the teams that scored an average of 14 or more points per round. 
Choice B is incorrect. If a study uses a sample selected at random from a population, the results of the study can be generalized to the population, not just the sample. 
Choice C is incorrect. If a study uses a sample selected at random from a population, the results of the study can be generalized to the population, not just a subset of the sample. 
Download Solution PDF

Ratios, Rates, Proportional Relationships and Units Question 7:

The ratio \(\frac{c}{d} = 6\) and \(\frac{72c}{kd} = 6\). Find \(k\).

  1. 12
  2. 6
  3. 72
  4. 36

Answer (Detailed Solution Below)

Option 3 : 72

Ratios, Rates, Proportional Relationships and Units Question 7 Detailed Solution

The ratio \(\frac{c}{d} = 6\) implies \(c = 6d\). Substituting into the equation \(\frac{72c}{kd} = 6\), we have \(\frac{72(6d)}{kd} = 6\). Simplifying gives \(\frac{432d}{kd} = 6\). Canceling \(d\) results in \(\frac{432}{k} = 6\). Solving for \(k\), multiply both sides by \(k\) to get \(432 = 6k\). Dividing both sides by 6, \(k = 72\). Therefore, \(k\) is 72.

Ratios, Rates, Proportional Relationships and Units Question 8:

If the ratio of \(a\) to \(b\) is 3 and the ratio of \(9a\) to \(mb\) is also 3, what is the value of \(m\)?

  1. 9
  2. 3
  3. 27
  4. 1

Answer (Detailed Solution Below)

Option 1 : 9

Ratios, Rates, Proportional Relationships and Units Question 8 Detailed Solution

Given that the ratio \(\frac{a}{b} = 3\), we can express this as \(a = 3b\). We also have \(\frac{9a}{mb} = 3\). Substituting \(a = 3b\) into this equation gives \(\frac{9(3b)}{mb} = 3\). Simplifying, we have \(\frac{27b}{mb} = 3\). Dividing both sides by \(b\) results in \(\frac{27}{m} = 3\). Solving for \(m\), we multiply both sides by \(m\) to get \(27 = 3m\). Dividing both sides by 3, we find \(m = 9\). Therefore, the value of \(m\) is 9.

Ratios, Rates, Proportional Relationships and Units Question 9:

If \(\frac{p}{q} = 5\) and \(\frac{35p}{tq} = 5\), what is the value of \(t\)?

  1. 7
  2. 35
  3. 5
  4. 1

Answer (Detailed Solution Below)

Option 2 : 35

Ratios, Rates, Proportional Relationships and Units Question 9 Detailed Solution

The given ratio \(\frac{p}{q} = 5\) implies that \(p = 5q\). Substituting this into the second equation \(\frac{35p}{tq} = 5\) yields \(\frac{35(5q)}{tq} = 5\). Simplifying, we get \(\frac{175q}{tq} = 5\). Canceling \(q\) gives \(\frac{175}{t} = 5\). Solving for \(t\), we multiply both sides by \(t\) to obtain \(175 = 5t\). Dividing both sides by 5 results in \(t = 35\). Thus, the value of \(t\) is 35.

Ratios, Rates, Proportional Relationships and Units Question 10:

A factory produces widgets at a constant rate of 30 widgets per minute. How many widgets does the factory produce in a day, assuming it operates 8 hours a day?

  1. 14,400
  2. 24,000
  3. 28,800
  4. 32,000

Answer (Detailed Solution Below)

Option 1 : 14,400

Ratios, Rates, Proportional Relationships and Units Question 10 Detailed Solution

To find out how many widgets the factory produces in a day, we first calculate how many widgets it produces in one hour. At 30 widgets per minute, in one hour (60 minutes), the factory produces 30 × 60 = 1,800 widgets. Since the factory operates for 8 hours a day, the total production in a day is 1,800 × 8 = 14,400 widgets. Therefore, the correct answer is option 1, 14,400 widgets. The other options are incorrect as they do not correctly calculate the widgets produced in a day based on the given rate and operating hours.

Ratios, Rates, Proportional Relationships and Units Question 11:

There are five equal glasses containing milk in the ratio 3 : 4 : 5 : 6 : 7. How many glasses are at least 50% full of milk if the total volume of milk in the glasses is 60% of the total volume of the glasses?

  1. 2
  2. 3
  3. 4
  4. 5

Answer (Detailed Solution Below)

Option 2 : 3

Ratios, Rates, Proportional Relationships and Units Question 11 Detailed Solution

Let the quantities of milk in the glasses be 3x ml, 4x ml, 5x ml, 6x ml and 7x ml. Let the volume of each glass be 100 ml. Total volume of the glasses = 500 ml

3x + 4x + 5x + 6x + 7x = 0.6(500)

⇒ x = 12

The quantities of milk in the glasses are 36 ml, 48 ml, 60 ml, 72 ml and 84 ml, i.e., 3 glasses are filled with milk to at least 50% of their capacity.

Ratios, Rates, Proportional Relationships and Units Question 12:

The density of a certain mineral is \(2,500\) kg/m³, and a cube of this mineral weighs \(10,000\) kg. Calculate the side length of the cube in meters, to the nearest hundredth.

  1. 1.59
  2. 2.20
  3. 1.25
  4. 2.30

Answer (Detailed Solution Below)

Option 1 : 1.59

Ratios, Rates, Proportional Relationships and Units Question 12 Detailed Solution

To find the side length of the cube, begin with the density equation: \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \). Given the density \(2,500\) kg/m³ and mass \(10,000\) kg, solve for \(V\): \( 2,500 = \frac{10,000}{V} \). Solving gives \( V = \frac{10,000}{2,500} = 4 \) cubic meters. Since the cube has volume \( V = s^3 \), we have \( s^3 = 4 \). Solve for \(s\): \( s = \sqrt[3]{4} \approx 1.5874 \). Rounding gives \(1.59\).

Ratios, Rates, Proportional Relationships and Units Question 13:

A cube has a density of \(250\) kilograms per cubic meter and a mass of \(500\) kilograms. To the nearest hundredth, what is the length of one edge of the cube in meters?

  1. 2.50
  2. 2.51
  3. 1.26
  4. 2.53

Answer (Detailed Solution Below)

Option 3 : 1.26

Ratios, Rates, Proportional Relationships and Units Question 13 Detailed Solution

To find the length of one edge of the cube, we need to determine the volume of the cube first. The density of the cube is given as \(250\) kg/m3, and the mass is \(500\) kg. Using the formula for density, \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \), we can calculate the volume of the cube as follows: \( 250 = \frac{500}{V} \). Solving for \(V\), we get \( V = \frac{500}{250} = 2 \) cubic meters. Since the cube's volume is \(2\) cubic meters, and volume \( V = s^3 \) for a cube, we have \( s^3 = 2 \). Taking the cube root of both sides gives \( s = \sqrt[3]{2} \approx 1.26 \). 

Ratios, Rates, Proportional Relationships and Units Question 14:

A cubic tank holds \(1000\) liters of water and has a density of water at \(1000\) kg/m³. If the mass of the water is \(1000\) kilograms, what is the length of one side of the tank in meters?

  1. 0.99
  2. 1.00
  3. 1.01
  4. 1.02

Answer (Detailed Solution Below)

Option 2 : 1.00

Ratios, Rates, Proportional Relationships and Units Question 14 Detailed Solution

To determine the length of one side of the tank, we first note that the density of water is \(1000\) kg/m³, and the mass is also \(1000\) kg. The volume formula for density is \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \). Given the density and mass are both \(1000\), the volume \( V \) is \( \frac{1000}{1000} = 1 \) cubic meter. Since the tank is cubic, \( V = s^3 \). Therefore, \( s^3 = 1 \). Taking the cube root of both sides gives \( s = \sqrt[3]{1} = 1 \). Therefore, the length of one edge of the tank is \(1.00\) meters. Option \(2\) is the correct answer as it directly matches the calculated result.

Ratios, Rates, Proportional Relationships and Units Question 15:

If the ratio of \(p\) to \(q\) is equivalent to 10 to \(r\), and \(p = 90\), what is \(q\) in terms of \(r\)?

  1. 9r
  2. 10r
  3. 12r
  4. 11r

Answer (Detailed Solution Below)

Option 1 : 9r

Ratios, Rates, Proportional Relationships and Units Question 15 Detailed Solution

The equation \(\frac{p}{q} = \frac{10}{r}\) can be rewritten with \(p = 90\) as \(\frac{90}{q} = \frac{10}{r}\). Cross-multiply to get \(10q = 90r\). Solving for \(q\), divide both sides by 10 to find \(q = 9r\). Hence, \(q\) in terms of \(r\) is \(9r\), making option 1 correct.

Exam Preparation
Simplified

Learn, practice, analyse and improve
Get Started for Free Download_on_the_App_Store_Badge_US-UK_RGB_blk_4SVG_092917 - www.guacandrollcantina.com
Trusted by 7.2 Crore+ Students

Related MCQ

Percentages MCQOne-variable data: Distributions and Measures of Center and Spread MCQTwo-variable data: Models MCQTwo-variable data: Scatterplots MCQProbability and Conditional Probability MCQInference from Sample Statistics and Margin of Error MCQEvaluating Statistical Claims: Observational Studies and Experiments MCQ
Exams
Test Series
SuperCoaching
Previous Year Papers
Latest Updates
MCQ Questions
Testbook Products & Resources for Students
Testbook USA
Books
Testbook Edu Solutions Pvt. Ltd.
1st & 2nd Floor, Zion Building,
Plot No. 273, Sector 10, Kharghar,
Navi Mumbai - 410210
[email protected]
Toll Free:1800 833 0800
Office Hours: 10 AM to 7 PM (all 7 days)
Company
About usCareers We are hiringTeach Online on TestbookPartnersMediaSitemap
Products
Test SeriesTestbook PassOnline CoursesOnline VideosPracticeBlogRefer & EarnBooks
Our Apps
Follow us on
Copyright © 2014-2022 Testbook Edu Solutions Pvt. Ltd.: All rights reserved
User PolicyTermsPrivacy
Get Free Access Now
Hot Links： teen patti wala game all teen patti game teen patti master app