John is buying some burgers and fries for his friends. Burgers cost $$\$ 4.30$$ each and fries cost $$\$ 3.10$$ per order. There is no tax on the food. If he has $$\$ 50$$ and buys 6 orders of fries, what is the maximum number of burgers he can buy?

Short Answer

Expert verified
John can buy a maximum of 7 burgers for his friends.

Step by step solution

01

Find the total cost of fries.

To find the total cost of fries, we need to multiply the number of orders of fries (6) by the cost of each order, which is $3.10: \(6 * 3.10\).
02

Calculate the remaining amount of money.

Now, we need to subtract the total cost of the fries from the initial budget of $50. This will give us the remaining amount of money John can use to buy burgers.
03

Find the maximum number of burgers he can buy.

To find the maximum number of burgers John can buy, we need to divide the remaining amount of money by the cost of each burger, which is $4.30. Since John can only buy a whole number of burgers, we need to take the largest whole number quotient. Let's perform the calculations:
04

Step 1:

Total cost of fries = \(6 * 3.10 = \$ 18.60\)
05

Step 2:

Remaining amount of money = \(50 - \)18.60 = $31.40
06

Step 3:

Maximum number of burgers = \(\left\lfloor \frac{31.40}{4.30} \right\rfloor\) Dividing \(31.40 by \)4.30 gives approximately 7.30. Taking the largest whole number quotient (ignoring decimals), we get: Maximum number of burgers = 7 So, John can buy a maximum of 7 burgers for his friends.

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 Vaia!

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

If two times a number is equal to that number minus 4 , what is the number? A) \(-7\) B) \(-6\) C) \(-4\) D) \(-3\)

sessions for the long-time client package, will she be able to create this package for her clients? A) No, because the closest package that she can offer consists of three hot yoga and three zero gravity yoga sessions. B) No, because the closest package that she can offer consists of four hot yoga and four zero gravity yoga sessions. C) Yes, because she can offer five hot yoga and five zero gravity yoga sessions. D) Yes, because she can offer six hot yoga and six zero gravity yoga sessions. 13\. Cuthbert is conducting a chemistry experiment that calls for a number of chemicals to be mixed in various quantities. The one amount of which he is unsure is grams of potassium, \(p\). If Cuthbert is certain that \(\left(3 p^2+\right.\) \(14 p+24)-2\left(p^2+7 p+20\right)=0\), what is one possible value of \(3 p+6\), the exact number of grams of potassium that Cuthbert would like to use for this experiment? A) 20 B) 18 C) 12 D) 10

$$ 0.27(a+b)=0.15 a+0.35 b $$ An athletic trainer is attempting to produce a carbohydrate-electrolyte solution that is at \(27 \%\) carbohydrates by mass, which is the maximum amount of saturation allowed by her league. A supply company provides solutions that are at \(15 \%\) and \(35 \%\) carbohydrates by mass, respectively. Based on the equation above, if the trainer uses 10 quarts of the \(15 \%\) solution, how many quarts of the \(35 \%\) solution will she need? A) 180 B) 90 C) 30 D) 15 $$ f(x)=(x-b)^2-4 $$

A) NO CHANGE B) one, of the major industrial towns, C) one, of the major industrial towns D) one of the major industrial towns

If \(2 \sqrt{x}=x-3\), which of the following is the solution set for \(x\) ? A) \(\\{-1,9\\}\) B) \(\\{1,-9\\}\) C) \(\\{9\\}\) D) \(\\{1,9\\}\)

See all solutions

Recommended explanations on English Textbooks

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.

Sign-up for free