/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 69 A manufacturer of bicycles has 4... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 69

A manufacturer of bicycles has 4815 wheels, 2305 frames, and 2255 handlebars. (a) How many bicycles can be manufactured using these parts? (b) How many parts of each kind are left over? (c) Which part limits the production of bicycles?

Short Answer

Expert verified
(a) The manufacturer can make 2255 bicycles. (b) There are 305 wheels, 50 frames, and 0 handlebars left over. (c) The part that limits the production of bicycles is the handlebars.

Step by step solution

01

Determine the required quantity for each part per bicycle

To build a bicycle, you need 2 wheels, 1 frame, and 1 handlebar.
02

Calculate the number of bicycles that can be made from each part

To find out how many bicycles can be made from each part, divide the total number of each part by the required quantity of that part per bicycle. Wheels: \( \frac{4815}{2} = 2407.5 \) bicycles Frames: \( \frac{2305}{1} = 2305 \) bicycles Handlebars: \( \frac{2255}{1} = 2255 \) bicycles Note: You can't make 0.5 of a bicycle, so the number should be rounded down.
03

Find the minimum number of bicycles that can be made from all parts

To find the total number of bicycles that can be made, you need to find the minimum number of bicycles that can be made from each part. The smallest number of bicycles that can be made from all parts is 2255 (from handlebars), which is the limiting part. So, the manufacturer can make 2255 bicycles. #b) How many parts of each kind are left over?#
04

Calculate the number of parts used for making the bicycles

Multiply the limiting number of bicycles (2255) by the required quantity of each part per bicycle. Wheels used: 2255 bicycles * 2 = 4510 wheels Frames used: 2255 bicycles * 1 = 2255 frames Handlebars used: 2255 bicycles * 1 = 2255 handlebars
05

Calculate the leftover parts

Subtract the number of parts used from the initial quantity. Leftover wheels: 4815 wheels - 4510 wheels = 305 wheels Leftover frames: 2305 frames - 2255 frames = 50 frames Leftover handlebars: 2255 handlebars - 2255 handlebars = 0 handlebars So, there are 305 wheels, 50 frames, and 0 handlebars left over. #c) Which part limits the production of bicycles?#
06

Identify the limiting part

During the analysis of the problem in part a), we found out that the limiting number of bicycles is 2255, and it was determined by the number of handlebars available. Therefore, the part that limits the production of bicycles is the handlebars.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In 1865 a chemist reported that he had reacted a weighed amount of pure silver with nitric acid and had recovered all the silver as pure silver nitrate. The mass ratio of silver to silver nitrate was found to be \(0.634985 .\) Using only this ratio and the presently accepted values for the atomic weights of silver and oxygen, calculate the atomic weight of nitrogen. Compare this calculated atomic weight with the currently accepted value.

Without doing any detailed calculations (but using a periodic table to give atomic weights), rank the following samples in order of increasing number of atoms: \(3.0 \times 10^{23}\) molecules of \(\mathrm{H}_{2} \mathrm{O}_{2}, 2.0 \mathrm{~mol} \mathrm{CH}_{4}, 32 \mathrm{~g} \mathrm{O}_{2}\).

A method used by the US. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the air sample through a "bubbler" containing sodium iodide, which removes the ozone according to the following equation: \(\mathrm{O}_{3}(g)+2 \mathrm{Nal}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) \(\mathrm{O}_{2}(g)+\mathrm{I}_{2}(\mathrm{~s})+2 \mathrm{NaOH}(a q)\) (a) How many moles of sodium iodide are needed to remove \(5.95 \times 10^{-6}\) mol of \(\mathrm{O}_{3}\) ? (b) How many grams of sodium iodide are needed to remove \(1.3 \mathrm{mg}\) of \(\mathrm{O}_{3}\) ?

(a) Diamond is a natural form of pure carbon. How many moles of carbon are in a \(1.25\) -carat diamond (1 carat \(=0.200 \mathrm{~g}\) )? How many atoms are in this diamond? (b) The molecular formula of acetylsalicylic acid (aspirin), one of the most common pain relievers, is \(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\). How many moles of \(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\) are in a \(0.500-\mathrm{g}\) tablet of aspirin? How many molecules of \(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{O}_{4}\) are in this tablet?

Give the empirical formula of each of the following compounds if a sample contains (a) \(0.0130 \mathrm{~mol} \mathrm{C}, 0.0390\) mol \(\mathrm{H}\), and \(0.0065 \mathrm{~mol} \mathrm{O}\); (b) \(11.66 \mathrm{~g}\) iron and \(5.01 \mathrm{~g}\) oxygen; (c) \(40.0 \% \mathrm{C}, 67 \% \mathrm{H}\), and \(53.3 \% \mathrm{O}\) by mass
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Chemical Analysis

Read Explanation

Chemistry Branches

Read Explanation

Physical Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.