Questions on Word Problems: Problems on percentages, ratios, and fractions answered by real tutors!

Algebra ->  Percentage-and-ratio-word-problems -> Questions on Word Problems: Problems on percentages, ratios, and fractions answered by real tutors!      Log On


   



Tutors Answer Your Questions about Percentage-and-ratio-word-problems (FREE)


Question 1210595: A donut shop sold 3/5 of their glazed donuts in the morning, In the afternoon, they sold 5/6 of the remaining glazed donuts. The ratio of the remaining glazed donuts to the remaining chocolate donuts was 2:7. If they had 42 chocolate donuts left, how many total donuts (glazed + chocolate) did the shop have at the very beginning?
Found 2 solutions by ikleyn, KMST:
Answer by ikleyn(53751) About Me  (Show Source):
You can put this solution on YOUR website!
.
A donut shop sold 3/5 of their glazed donuts in the morning.
In the afternoon, they sold 5/6 of the remaining glazed donuts.
The ratio of the remaining glazed donuts to the remaining chocolate donuts was 2:7.
If they had 42 chocolate donuts left, how many total donuts (glazed + chocolate) did the shop have
at the very beginning?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


Let 'x' be the number of glazed donuts in the chop at the very beginning.
Let 'c' be the number of remaining chocolate donuts, which number is mentioned in the problem:  c = 42.


In the morning the shop sold  %283%2F5%29x  of their glazed donuts.
Hence,  %282%2F5%29x  glazed donuts remained.


Of this  %282%2F5%29x  glazed donuts,  5%2F6  were sold afternoon.
So,  %282%2F5%29x%2A%285%2F6%29 = %281%2F3%29x  glazed donuts were sold afternoon.

Thus  %283%2F5%29x + %281%2F3%29x = %289%2F15%29x+%2B+%285%2F15%29x = %2814%2F15%29x  glazed donuts 
were sold during the day, and  %281%2F15%29x  glazed donut remained at the evening.


From the condition, we can write

    %28%281%2F15%29x%29%2Fc = 2%2F7,  or  x%2F%2815c%29 = 2%2F7,

which gives 

    x = %28%282%2A15%29%2F7%29c.

Substitute here c= 42 to get  

    x = %28%282%2A15%29%2F7%29%2A42 = 2*15*6 = 180.


So, the starting number of the glazed donuts was 180 at the very beginning.


It is what we can get from the given problem, using all given information.


But we have no any other data to determine the number of the chocolate donuts at the very beginning.


So, in the given form, the problem is DEFECTIVE and does not allow to answer the posed question.

Analyzed as far as it is possible with explanation WHY the problem is FAULT.



Answer by KMST(5345) About Me  (Show Source):
You can put this solution on YOUR website!
Not enough information.
We know how many chocolate donuts the donut shop had at the end of the day,
and could calculate how many glazed donut the donut shop had at the end of the day,
but without more information there is no way to calculate how many chocolate donuts the donut shop had at the very beginning.


Question 1210596: A high-performance drone started its flight with a full battery charge. During the first phase (Takeoff and Ascent), it used 30% of its total capacity plus an additional 8 mAh. After this phase, it entered the second phase (Data Collection), where it used 60% of the remaining charge plus an additional 12 mAh. At the end of the mission, the remaining battery charge was 136 mAh less than the charge consumed during the Data Collection phase. Find the battery capacity used during the Takeoff and Ascent phase.
Answer by greenestamps(13327) About Me  (Show Source):
You can put this solution on YOUR website!


Full charge to start:

x

Used in takeoff and ascent phase: 30% plus 8

.3x%2B8

Remaining:

x-%28.3x%2B8%29=.7x-8

Used during data collection phase: 60% of remaining plus 12

.6%28.7x-8%29%2B12=.42x%2B7.2

Remaining:

%28.7x-8%29-%28.42x%2B7.2%29=.28x-15.2

The amount remaining was 136 less than the amount used during the data collection phase:

.28x-15.2=%28.42x%2B7.2%29-136=.42x-128.8
113.6=.14x
x=113.6%2F.14=5680%2F7

Amount used during takeoff and ascent:

.3x%2B8=1760%2F7

ANSWER: 1760/7 mAh, or 251 3/7



Question 284255: The airfare of a flight from Amarillo, Tx to New Orleans cost $280, which includes a 7% sales tax. What is the cost of the flight before tax?
Answer by ikleyn(53751) About Me  (Show Source):
You can put this solution on YOUR website!
.
The airfare of a flight from Amarillo, Tx to New Orleans cost $280, which includes a 7% sales tax.
What is the cost of the flight before tax?
~~~~~~~~~~~~~~~~~~~~~~


        In the post by @mananth, the treatment is not adequate to the problem.
        Due to this reason, his solution is conceptually incorrect.

        I came to bring a correct solution.


Let the cost of the ticket be x dollars before tax.
The purchase including the 7% sales tax is (1+0.07)x = 1.07x.

We have this equation

        1.07x = 280 dollars.

Hence   x = 280%2F1.07 = 261.682243.

We round it to the closest cent.

ANSWER.   The cost of the flight before the tax is  $261.68.

Solved correctly.




Question 187282: Felipe jobs for 10 miles and then walks another 10 miles. He jobs 2 1/2 miles per hour faster than he walks, and the entire distance of 20 miles takes 6 hours. find the rate at which he walks and the rate at which he jogs.
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39792) About Me  (Show Source):
You can put this solution on YOUR website!
                     SPEED         TIME           DISTANCE
JOG                  r+2.5         10/(r+2.5)        10
WALK                 r             10/r
TOTAL                                6  

10%2F%28r%2B2%261%2F2%29%2B10%2Fr=6
.
.

Answer by MathTherapy(10806) About Me  (Show Source):
You can put this solution on YOUR website!
Felipe jobs for 10 miles and then walks another 10 miles.  He jobs 2 1/2 miles per hour faster than he walks, and
the entire distance of 20 miles takes 6 hours.  find the rate at which he walks and the rate at which he jogs.
**************************************************************
The other person's solution is rather lengthy, and in this author's opinion, a few UNNECESSARY variables were introduced. 
Furthermore, he suggests using the quadratic formula to solve his final equation, which can be easily FACTORIZED. 

Let the speed at which he walks, be S
Then, his jogging speed = S+%2B+2%261%2F2, or S + 2.5
Time taken to walk 10 miles: 10%2FS, and time taken to jog 10 miles = 10%2F%28S+%2B+2.5%29
As he takes 6 hours to walk and jog 20 miles, we get the following
TOTAL-TIME equation: 10%2FS+%2B+10%2F%28S+%2B+2.5%29+=+6
                              10(S + 2.5) + 10S = 6S(S + 2.5) ----- Multiplying by LCD, S(S + 2.5)
                                     10S+%2B+25+%2B+10S+=+6S%5E2+%2B+15S
                                              20S+%2B+25+=+6S%5E2+%2B+15S
                           6S%5E2+%2B+15S+-+20S+-+25+=+0
                                      6S%5E2+-+5S+-+25+=+0
                          6S%5E2+-+15S+%2B+10S+-+25+=+0
                        3S%282S+-+5%29+%2B+5%282S+-+5%29+=+0
                               (2S - 5)(3S + 5) = 0
                                                 2S - 5 = 0      or     3S + 5 = 0 ---- Setting FACTORS equal to 0
                                                      2S = 5      or            3S = - 5 (IGNORE)
                  Walking speed, or S = 5%2F2 = 2.5 mph

                 Jogging speed: S + 2.5 = 2.5 + 2.5 = 5 mph

                 You can do the CHECK!!


Question 1195448: Clifford had 80% fewer sweets than Ashwin. After Ashwin gave 210 sweets to Clifford, Clifford had 2/3 as many sweets as Ashwin. How many sweets did each of them have in the end?
(a) Ashwin
(b) Clifford

Answer by MathTherapy(10806) About Me  (Show Source):
You can put this solution on YOUR website!
Clifford had 80% fewer sweets than Ashwin. After Ashwin gave 210 sweets to Clifford, Clifford had 2/3 as many sweets as Ashwin. How many sweets did each of them have in the end?
(a) Ashwin
(b) Clifford
***************
Let amount Ashwin started with, be A
Since "Clifford had 80% fewer sweets than Ashwin," then Clifford started with (1 - 80%) of A, or .2A
When Ashwin gave 210 sweets to Clifford, Ashwin had A - 210 left
After receiving 210 from Ashwin, Clifford ended up with .2A + 210
With Clifford ending up with 2%2F3 as many sweets as Ashwin, we get the following FINAL-COUNT equation:
                                                     %282%2F3%29%28A+-+210%29+=+.2A+%2B+210
                                                          2(A - 210) = .6A + 630 -- Multiplying by LCD, 3
                                                             2A - 420 = .6A + 630
                                                              2A - .6A = 630 + 420
                                                                    1.4A = 1,050
Number of sweets Ashwin started with, or A+=+%221%2C050%22%2F1.4+=+%2210%2C500%22%2F14+=+highlight%28+750%29
Number of sweets Clifford started with: .2A = .2(750) = 150


Question 1210594: Luka and his friends are preparing for a local bazaar. Luka spent 2 1/2 hours making friendship bracelets, which was 3/4 of the time he spent baking cookies. The ratio of the number of cookies he baked to the number of bracelets he made was 5:2.
1. How many hours did Luka spend baking cookies?
2. If he sold 3/5 of his cookies and 1/2 of his bracelets, and he had 48 cookies left, how many bracelets did he have remaining?

Found 4 solutions by KMST, MathTherapy, ikleyn, josgarithmetic:
Answer by KMST(5345) About Me  (Show Source):
You can put this solution on YOUR website!
Luka and his friends were preparing for a local bazaar.
We do not know what his friends did to prepare for the bazaar.
We know that Luka make friendship bracelets and cookies.
We all agree that Luka spent 3 hours and 20 minutes baking cookies.
We also know that the ratio of the number of cookies Luka baked to the number of bracelets he made was 5:2.
That means that for every 5 cookies he made, he made 2 bracelets.
That means that the number of bracelets he made is highlight%282%2F5%29 of the number of cookies he made.
We assume that Luka took all the cookies and bracelets he made to the bazaar to be sold.

2. If Luka sold 3/5 of his cookies, and he had 48 cookies left, what do we know?

In that case, and if no one got any cookies for free, the fraction of the cookies Lukas made that he did not sold must be the fraction of cookies Luka had left.
That fraction is 1-3%2F5+=+5%2F5-3%2F5+=+highlight%282%2F5%29 of the cookies he made. and that amounts to 48 .
But we know that the number of bracelets Luka made is 2%2F5 of number of cookies he made.
That means that Luka made 48 bracelets.
If Luka "sold 1/2 of his bracelets" at the bazaar, then he had the other 1/2 of his bracelets remaining, and that is 1/2 of 24, or
%281%2F2%29%2A24=highlight%2812%29 bracelets remaining.

Answer by MathTherapy(10806) About Me  (Show Source):
You can put this solution on YOUR website!
Luka and his friends are preparing for a local bazaar. Luka spent 2 1/2 hours making friendship bracelets, which was
3/4 of the time he spent baking cookies. The ratio of the number of cookies he baked to the number of bracelets he
made was 5:2.
1. How many hours did Luka spend baking cookies?
2. If he sold 3/5 of his cookies and 1/2 of his bracelets, and he had 48 cookies left, how many bracelets did he have
remaining?
***********************************************
This author would like to chime in on this problem.

PART 2.

Let the multiplicative factor be x
As the ratio of the number of cookies to bracelets made was 5:2, the number of cookies and bracelets made, were 5x and 2x,
respectively
With the fraction of bracelets sold being 1%2F2, the fraction of bracelets remaining was 1+-+1%2F2 = 1%2F2
So, number of bracelets remaining was %281%2F2%292x+=+x 

Likewise, with the fraction of cookies sold being 3%2F5, the fraction of cookies remaining was 1+-+3%2F5 = 2%2F5
So, number of cookies remaining was %282%2F5%295x+=+2x
 
With the remaining number of cookies being 48, or 2x, we get the following REMAINING
NUMBER-of-cookies equation: 2x = 48
                                                                                                               
Number of bracelets remaining (same as multiplicative factor) or x+=+48%2F2+=+highlight%2824%29

OR

A 5:2 ratio of the number of cookies to number of bracelets means that there were 5 parts cookies to 2 parts bracelets

With 3%2F5 of cookies sold, number of cookies remaining = 1+-+%283%2F5%295, or %282%2F5%295, or 2 parts
With 1%2F2 of bracelets sold, number of bracelets remaining 1+-+%281%2F2%292, or %281%2F2%292, or 1 part

Now, we have 3 parts remaining and a cookies-to-bracelets ratio of 2:1
This puts the remaining cookies at 2%2F%282+%2B+1%29+=+2%2F3, and remaining bracelets at 1%2F%282+%2B+1%29+=+1%2F3
As total remaining cookies were 48, we get the Remaining number of bracelets as: 48%2F2+=+highlight%2824%29

@KMST is correct! This author misread "....and he had 48 cookies left" as, he had 48 cookies and bracelets left. Corrections
were made above, and now, 24 is the correct number of bracelets that remained, not 16, as previously stated.

Thanks to @KMST for responding, thereby bringing attention to this error!!

Answer by ikleyn(53751) About Me  (Show Source):
You can put this solution on YOUR website!
.
Luka and his friends are preparing for a local bazaar. Luka spent 2 1/2 hours making friendship bracelets,
which was 3/4 of the time he spent baking cookies. The ratio of the number of cookies he baked
to the number of bracelets he made was 5:2.
1. How many hours did Luka spend baking cookies?
2. If he sold 3/5 of his cookies and 1/2 of his bracelets, and he had 48 cookies left,
how many bracelets did he have remaining?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

                                P a r t   1

Time for making bracelets is  2 1/2 hours, or  5/2 hours.


It is 3/4 of the time baking cookies.


So, we write this equation

    %283%2F4%29x = 5%2F2  hours.


where x is the time under the question 1.

To find x, divide equation (1) by  3%2F4.  You will get

    x = %28%285%2F2%29%29%2F%28%283%2F4%29%29 = %285%2A4%29%2F%282%2A3%29 = 20%2F6 = 10%2F3 = 3 1/3 hours or 3 hours and 20 minutes.


ANSWER to question 1: Luka spent 3 1/3 hours, or 3 hours and 20 minutes, baking cookies.

Thus part (1) is completed.

            Actually, this part is pure arithmetical problem 
            and can be solved MENTALLY without using equations.


2 1/2 hours is 150 minutes.

It is 3/4 of the baking time.

So, 1/4 of the baking time is 1/3 of 150 minutes, which is 50 minutes.

Hence, the whole baking time is 4*50 = 200 minutes, which is the same 3 hours and 20 minutes as we got above.

So, solving arithmetically, you get the same answer.



Answer by josgarithmetic(39792) About Me  (Show Source):
You can put this solution on YOUR website!
First three sentences mean, for some c time to bake cookies
2%261%2F2=3c%2F4
5%2F2=%283%2F4%29c
%285%2F2%29%284%2F3%29=c
c=%285%2A2%29%2F3
c=3%261%2F3hours baking cookies


Question 1210589: A fruit seller had a collection of apples, pears, and oranges.
Initially, 1/4 of the fruits were apples.
He sold 80 apples and 1/3 of the pears.
After this, he bought more oranges, increasing the number of oranges by 60%.
At this point, the number of pears he had left was twice the number of apples he had left.
Finally, he realized that the total number of fruits he had now was exactly 20% more than the total number he started with.
How many oranges did the fruit seller have at the very beginning?

Answer by greenestamps(13327) About Me  (Show Source):
You can put this solution on YOUR website!


The problem asks for the number of oranges he started with. So let that be our variable.

x = original number of oranges

We will need another variable.

y = original number of apples

The original number of apples is 1/4 of the total number of fruits, so the total number of fruits is 4y. This gives us

4y-(x+y) = 3y-x = original number of pears

To start, then, we have

apples: y
pears: 3y-x
oranges: x

He sells 80 apples and 1/3 of the pears, leaving...

apples: y-80
pears: (2/3)(3y-x) = 2y-(2/3)x
oranges: x

He buys more oranges, increasing the number of oranges by 60% -- i.e., multiplying the number of oranges by 160%, or 8/5.

apples: y-80
pears: (2/3)(3y-x) = 2y-(2/3)x
oranges: (8/5)x

After doing that, he has twice as many pears as apples

2y-%282%2F3%29x=2%28y-80%29
2y-%282%2F3%29x=2y-160
%282%2F3%29x=160
x=240

That is what the problem asked us to find, so we are done.

ANSWER: x = 240 oranges

Although the problem doesn't require us to do any more work, it is curious to continue to find the original numbers of apples and pears.

The total number of fruits at the end was 20% more than the number at the beginning.

at the beginning: 4y
at the end: %28y-80%29%2B%282y-%282%2F3%29x%29%2B%288%2F5%29x=3y-80-160%2B384=3y%2B144

The total number at the end was 20% greater than -- i.e. 6/5 as much -- as at the beginning:

3y%2B144=%286%2F5%29%284y%29=%2824%2F5%29y
144=%289%2F5%29y
y=%285%2F9%29144=80

The fruits he started with:
apples: y = 80
pears: 3y-x = 240-240 = 0
oranges: 240



Question 1210592: Arthur starts with a certain number of rare trading cards. First. Arthur gives 4 1/2 cards to a local collector to complete a set. After this, the ratio of the cards Arthur has left to the cards he started with is exactly 3:4. Next, Ben takes half of the remaining cards plus another half of a card. Then, Cole takes half of what Ben left behind plus another half of a card. Finally, Dan takes half of what Cole left behind plus another half of a card. After Dan is finished, there are exactly 2 cards left. How many trading cards did Arthur start with?
Answer by greenestamps(13327) About Me  (Show Source):
You can put this solution on YOUR website!


Arthur starts with a certain number of rare trading cards. First, Arthur gives 4 1/2 cards to a local collector to complete a set. After this, the ratio of the cards Arthur has left to the cards he started with is exactly 3:4. Next, Ben takes half of the remaining cards plus another half of a card. Then, Cole takes half of what Ben left behind plus another half of a card. Finally, Dan takes half of what Cole left behind plus another half of a card. After Dan is finished, there are exactly 2 cards left. How many trading cards did Arthur start with?

The second and third sentences make no sense for a couple of reasons.

(1) Rare trading cards become much less valuable if you cut them in half. Arthur giving 4 1/2 cards to another collector is nonsense in the real world.
(2) If you work the problem in the "forwards" direction, the second and third sentences together mean Arthur started with 18 cards, and after giving 4 1/2 to another collector he had 13 1/2 cards left. Working from there, the fractions of cards get even smaller, leading to nonsensical numbers (and Arthur does NOT end up with 2 cards).

So we have to ignore those first few sentences to work the problem.

And then the problem is much easier to solve by working backwards, starting with the 2 cards Arthur finished with.

In each transaction, the remaining number of cards was cut in half, and then the number of cards remaining was reduced by another one half of a card.

To work each transaction backwards -- i.e. "undo" each transaction --, we need to add one half of a card and then double the number of cards. (The opposite (inverse) of "cut in half and subtract one half" is "add one half and double").

These transactions occur three times. To find the number of cards Arthur started with, we need to start with the 2 cards he finished with and do the "undo" transaction three times.

2 cards plus half a card is 2 1/2 cards; doubled is 5 cards
5 cards plus half a card is 5 1/2 cards; doubled is 11 cards
11 cards plus half a card is 11 1/2 cards; doubled is 23 cards

Arthur started with 23 cards (and at no point did any of the rare cards have to be cut in half!)

ANSWER: 23

Note that answer again ignores, as it must, the first part of the statement of the problem that says the first thing Arthur did was give 4 1/2 cards to someone else.



Question 1210591: Three brothers—Leo, Sam, and Jax—are sharing a supply of energy bars during a hike. Leo starts the trip with a backpack full of bars. He eats 3 1/2 bars. After eating them, the ratio of the bars he has left to the bars he started with is exactly 3:4. Sam then takes the remaining bars. He eats half of them plus half a bar more. Jax takes what is left. He eats half of that amount plus half a bar more. After Jax is finished, there are exactly 2 bars left in the backpack. How many energy bars did Leo start with?
Found 2 solutions by MathTherapy, ikleyn:
Answer by MathTherapy(10806) About Me  (Show Source):
You can put this solution on YOUR website!
Three brothers—Leo, Sam, and Jax—are sharing a supply of energybars during a hike. Leo starts the trip with a backpack
full of bars. He eats 3 1/2 bars. After eating them, the ratio ofthe bars he has left to the bars he started with is
exactly 3:4. Sam then takes the remaining bars. He eats half ofthem plus half a bar more. Jax takes what is left. He eats
half of that amount plus half a bar more. After Jax is finished,there are exactly 2 bars left in the backpack. How many
energy bars did Leo start with?
*******************************************
METHOD 1
Let the multiplicative factor be x
Since the ratio of the bars he has left to the bars he started with is exactly 3:4, then the number of bars
he has left is 3x, and the number of bars he started with is 4x
With the number of eaten bars being 3%261%2F2, or 3.5, we get: 4x - 3x = 3.5, and x = 3.5
This means that the number Leo starts with is 4(3.5) = 14. 

******
CHECK
******
Original number of bars: 14
Number eaten by Leo: 3.5
Number remaining after Leo ate 3.5: 14 - 3.5 = 10.5
Number left after Sam ate 1%2F2 of remainder, plus 1%2F2 bar:%281%2F2%2910.5+-+.5+=+5.25+-+.5+=+4.75
Number left after Jax ate 1%2F2 of NEW remainder, plus1%2F2 bar: %281%2F2%294.75+-+.5+=+2.375+-+.5+=+1.875

As seen above, a 2-bar end-result didn't ensue. So, itw's then decided to apply a different method.

METHOD 2
Let the number of bars Leo started with, be B
With matrix%281%2C3%2C+3%261%2F2%2C+or%2C+3.5%29 being eaten, remainder is: B - 3.5
With Sam eating 1%2F2 of remainder, plus 1%2F2 bar,remainder becomes: %281%2F2%29%28B+-+3.5%29+-+.5 = %28B+-+3.5+-+1%29%2F2 = %28B+-4.5%29%2F2
As Jax ate 1%2F2 of remainder, plus 1%2F2 bar, then NEW remainder becomes: %281%2F2%29%28%28B+-+4.5%29%2F2%29+-+.5 = %28B+-+4.5%29%2F4+-+.5 = %28B+-+4.5+-+2%29%2F4 = %28B+-+6.5%29%2F4
Since 2 bars are now left, we get: %28B+-+6.5%29%2F4+=+2                                                          
                                                           B - 6.5 = 8 ----Cross-multiplying 
   Original number of bars, or B = 8 + 6.5 = 14.5

OR

Some like to go from the end to the beginning. In other words, from the 2-bar end result to the original number of bars.
This is illustrated below.

End-Number of bars: 2
Number of bars before Jax ate 1%2F2 bar, plus 1%2F2 of remainder: %282+%2B+.5%29%2F%281%2F2%29 = (2.5)*2 = 5
Number of bars before Sam ate 1%2F2 bar, plus 1%2F2 of remainder: %285+%2B+.5%29%2F%281%2F2%29 = (5.5)*2 = 11
Number of bars before Leo ate 3.5: 11 + 3.5 = 14.5 

******
CHECK
******
Original number of bars: 14.5
Number eaten by Leo: 3.5
Number remaining after Leo ate 3.5: 14.5 - 3.5 = 11
Number left after Sam ate 1%2F2 of remainder, plus 1%2F2 bar:%281%2F2%2911+-+.5+=+5.5+-+.5+=+5
Number left after Jax ate 1%2F2 of NEW remainder, plus1%2F2 bar: %281%2F2%295+-+.5+=+2.5+-+.5+=+2

VOILA!!!

There's an obvious CONTRADICTION, as one method produces an orignal amount of 14, while the other finds
the count to be 14.5. As a result of this. there doesn't seem to be a SOLUTION to this problem, at all!!

Answer by ikleyn(53751) About Me  (Show Source):
You can put this solution on YOUR website!
.
Three brothers—Leo, Sam, and Jax—are sharing a supply of energy bars during a hike. Leo starts the trip with a backpack
full of bars. He eats 3 1/2 bars. After eating them, the ratio of the bars he has left to the bars he started with is
exactly 3:4. Sam then takes the remaining bars. He eats half of them plus half a bar more. Jax takes what is left. He eats
half of that amount plus half a bar more. After Jax is finished, there are exactly 2 bars left in the backpack. How many
energy bars did Leo start with?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


Let X be the number of energy bars Leo did start with (the unknown value under the problem's question).

Leo eats 3.5 bars first.

The number of bars left is x-3.5.


Write the proportion as stated in the problem

    %28x-3.5%29%2Fx = 3%2F4.


From this proportion

    4x - 14 = 3x,

    4x - 3x = 14,

       x    = 14.


So, the answer to the problem's question is 14 energy bars.


You may check that this answer satisfies the problem's conditions,
Thus, all the problem's conditions are consistent and do not contradict each other.

At this point, the problem is solved completely.




Question 1210590: A fruit seller had a collection of apples, pears, and oranges. Initially, 1/4 of the fruits were apples. He sold 80 apples and 1/3 of the pears. After this, he bought more oranges, increasing the number of oranges by 60%. At this point, the number of pears he had left was twice the number of apples he had left. Finally, he realized that the total number of fruits he had now was exactly 20% more than the total number he started with. How many oranges did the fruit seller have at the very beginning?

Answer by josgarithmetic(39792) About Me  (Show Source):
You can put this solution on YOUR website!
.
.
.
.
240 oranges


Question 1155528: Don Glover borrowed $18,000 for 210 days and paid $702.80 in simple interest on the loan. What annual simple interest rate did Don paid on the loan? (Round your answer to the nearest hundredth of a percent)
Answer by ikleyn(53751) About Me  (Show Source):
You can put this solution on YOUR website!
.
Don Glover borrowed $18,000 for 210 days and paid $702.80 in simple interest on the loan.
What annual simple interest rate did Don paid on the loan?
(Round your answer to the nearest hundredth of a percent)
~~~~~~~~~~~~~~~~~~~~~~~


        The assignment requires the answer to the nearest hundredth of a percent.
        @mananth failed this request and rounded differently.
        See my correct calculations below.


Simple Interest = P * r * t

P = 18000

r = ?

t = 210/365 years

SI =702.80

702.80 = 18000+%2A+r+%2A+%28210%2F365%29

r = 702.80%2F%2818000%2A%28210%2F365%29%29 = 0.067862963.

Rounded to the nearest hundredth of a percent, this value is r = 6.79%.         ANSWER


So, @mananth either does not read the assignment to the end, or his code makes rounding
based on its own rules that @mananth does not control to the end - and then @mananth does not read/look at the output of his code.

My conclusion after checking this series of the @mananth solutions is

        (1)   that his work is not trustworthy,

    and

        (2)   He never worked in engineering environment, where reliability of output is critically important.

Nobody taught him to check, re-check, cross-check and double-check his output.




Question 1155532: Hector Elizondo took out a 65-day loan of $7500 at an annual interest rate of 5.5%. Find the simple interest due on the loan. (Round your answer to the nearest cent.)
Answer by ikleyn(53751) About Me  (Show Source):
You can put this solution on YOUR website!
.
Hector Elizondo took out a 65-day loan of $7500 at an annual interest rate of 5.5%.
Find the simple interest due on the loan. (Round your answer to the nearest cent.)
~~~~~~~~~~~~~~~~~~~~~~~


        The assignment requires the answer to the nearest cent.
        @mananth failed this request.
        See my correct calculations below.


Interest = P * R *T

convert days to years = 65/365

Interest = 7500 * 0.055 * (65/365)

= $73.46 to the nearest cent.         ANSWER


So, @mananth either does not read the assignment to the end, or his code makes rounding
based on its own rules that @mananth does not control to the end - and then @mananth does not read/look at the output of his code.




Question 1155533: Kristi Yang borrowed $18,000. The term of the loan was 60 days, and the annual simple interest rate was 7.1%. Find the simple interest due on the loan. (Round your answer to the nearest cent.)
Answer by ikleyn(53751) About Me  (Show Source):
You can put this solution on YOUR website!
.
Kristi Yang borrowed $18,000. The term of the loan was 60 days, and the annual simple interest rate was 7.1%.
Find the simple interest due on the loan. (Round your answer to the nearest cent.)
~~~~~~~~~~~~~~~~~~~~~~~~~~~


        The assignment requires the answer to the nearest cent.
        @mananth failed this request and rounded to the nerest dollar.
        See my correct calculations below.


SI = P * R * t
P = 18000
R = 7.1% = 0.071
T = 60 days = (60/365) years

Plug values

18000 * 0.071 * (60/365) = 210.0822

Interest = $ 210.08 to the nearest cent.         ANSWER


So, @mananth either does not read the assignment to the end, or his code makes rounding
based on its own rules that @mananth does not control to the end - and then @mananth does not read/look at the output of his code.




Question 1201771: A Pendulum Is L Meters Long. The Time, R, In Seconds That It Takes To Swing Back And Forth Once Is Given By T=2 Square Root Of L . Suppose The Pendulum Takes 5.42 Seconds To Swing Back And Forth Once. What Is Its Length? Carry Your Intermediate Computations To At Least Four Decimal Places, And Round Your Answer To The Nearest Tenth. Meters ×
Answer by ikleyn(53751) About Me  (Show Source):
You can put this solution on YOUR website!
.
A Pendulum Is L Meters Long. The Time, R, In Seconds That It Takes To Swing Back And Forth Once Is Given
By T=2 Square Root Of L . Suppose The Pendulum Takes 5.42 Seconds To Swing Back And Forth Once.
What Is Its Length? Carry Your Intermediate Computations To At Least Four Decimal Places,
And Round Your Answer To The Nearest Tenth. Meters ×
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


        Calculations in the post by @mananth are incorrect, leading to a wrong answer.
        I came to bring a correct solution.



T = 2%2Asqrt%28L%29


5.42 = 2%2Asqrt%28L%29


5.42/2 = sqrt%28L%29


2.71 = sqrt%28L%29


L = 2.71%5E2 = 7.4341 meters  (with four decimal places).


ANSWER.  The length is 7.4 meters (rounded to the nearest tenth of a meter).

Solved correctly.




Question 1204130: At 4.45 pm, James, Leo and Noah are at the starting point of a 400-metre circular path. James takes 8 minutes to walk 1 round, Leo takes 180 seconds to run I round while Noah cycles 2 rounds in 3 minutes. Find the time when all three of them will next meet at the starting point.
Answer by ikleyn(53751) About Me  (Show Source):
You can put this solution on YOUR website!
.
At 4.45 pm, James, Leo and Noah are at the starting point of a 400-metre circular path.
James takes 8 minutes to walk 1 round, Leo takes 180 seconds to run I round while Noah cycles 2 rounds in 3 minutes.
Find the time when all three of them will next meet at the starting point.
~~~~~~~~~~~~~~~~~~~~~

James takes 8 minutes to walk 1 round,

Leo takes 3 minutes to run 1 round.


So, if to consider only these two persons, they will be at the starting point simultaneously
next time in  LCM(8,3) = 24 minutes and not earlier.


24 is a multiple of 3, so Noah also will be at the starting point together with James and Leo at this time moment.


So, 24 minutes is the closest time when three friends will meet at the staring point.


The clock will display the time 4.45 pm + 24 minutes = 5.09 pm.    ANSWER

Solved.

Thus, I solved the problem mentally without using fractions, working with integer numbers, ONLY.

Compare with the solution by @mananth.




Question 1165330: in 2015 the median age at first marriage of U.S. women (27.1) was what percent less than median age at first marriage of U.S. men (29.2 years)
Answer by CPhill(2189) About Me  (Show Source):
You can put this solution on YOUR website!
This is a percentage decrease calculation.
The formula for percentage less than (or percentage decrease) is:
$$\text{Percentage Less} = \frac{\text{Reference Value} - \text{Comparison Value}}{\text{Reference Value}} \times 100\%$$
In this case:
* Reference Value (Men's age) $= 29.2$ years
* Comparison Value (Women's age) $= 27.1$ years
$$\text{Percentage Less} = \frac{29.2 - 27.1}{29.2} \times 100\%$$
```python?code_reference&code_event_index=2
# Given values
men_age = 29.2
women_age = 27.1
# Calculate the percentage difference
percentage_less = ((men_age - women_age) / men_age) * 100
print(f"Percentage less: {percentage_less}")
```
```text?code_stdout&code_event_index=2
Percentage less: 7.191780821917801
```
The median age at first marriage of U.S. women (27.1 years) was approximately **7.19% less** than the median age for U.S. men (29.2 years) in 2015.
### Calculation Steps:
1. **Find the Difference:**
$$29.2 \text{ years} - 27.1 \text{ years} = 2.1 \text{ years}$$
2. **Calculate the Percentage Decrease (using the men's age as the reference):**
$$\text{Percentage Less} = \frac{\text{Difference}}{\text{Men's Age}} \times 100\%$$
$$\text{Percentage Less} = \frac{2.1}{29.2} \times 100\%$$
$$\text{Percentage Less} \approx 0.071917 \times 100\%$$
$$\mathbf{\approx 7.19\%}$$


Question 730075: last month a store manager decided to decrease the prices of all items by 10%. this month he increased the prices by 10%. What would be the price for a pair of pants that had cost $750 before prices were decreased last month??

Please include your solution... thank you in advance!! :)

Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39792) About Me  (Show Source):
Answer by ikleyn(53751) About Me  (Show Source):
You can put this solution on YOUR website!
.
last month a store manager decided to decrease the prices of all items by 10%. this month he increased the prices by 10%.
What would be the price for a pair of pants that had cost $750 before prices were decreased last month??
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


Last month the price decreased from 750 dollars to 750 - 0.1*750 = 750*(1-0.1) = 750*0.9 = 675 dollars.


This month the price increased from 675 dollars to 675 + 0.1*675 = 675*(1+0.1) = 675*1.1 = 742.50 dollars.


    
                           Notice here 
    that the price did not return to 750 dollars, the original level !



You can compose these long calculations into one short composition for two steps

      750 dollars  ----->  750*0.9*1.1 = 750*0.99 = 742.50 dollars


which gives you the same answer.


Notice this factor 0.99, which prevents the price to return to the original level ! !

Solved, with explanations.


/\/\/\/\/\/\/\/\/\/\/\/


Ignore the post by @lynnlo, since it gives wrong answer and is irrelevant.




Question 731816: Isabelle bought 12 wallet-sized photos for $36. Use a ratio table to determine how much she will pay for 5 more photos.
I drew the ratio table but cannot figure out what to do next. Please help me. Thanks!

Found 2 solutions by timofer, ikleyn:
Answer by timofer(155) About Me  (Show Source):
You can put this solution on YOUR website!
12 photos for 36 dollars
5 more photos for additional x dollars

12%2F36=5%2Fx
1%2F3=5%2Fx
3%2F1=x%2F5
15=x-------------pay 15 dollars more

Answer by ikleyn(53751) About Me  (Show Source):
You can put this solution on YOUR website!
.
Isabelle bought 12 wallet-sized photos for $36. Use a ratio table to determine how much she will pay for 5 more photos.
I drew the ratio table but cannot figure out what to do next. Please help me. Thanks!
~~~~~~~~~~~~~~~~~~~~~~~~~~~~


        I will provide two solutions.

        First solution is of the 5-th grade level student, who just solves problems on division and multiplication
        of integer numbers, but does not know proportions yet.

        Second solution is of the level of the 6-th grade level student, who just learned proportions.

        You select the one of these two solutions, which suits better your needs and your level.


                First solution of the 5-th grade level student

12 photos cost &36 dollars - hence, one photo costs  36/12 = 3 dollars.

Then 5 more than 12 photos costs (12+5)*3 = 17*3 = 51 dollars.

It is your ANSWER :  51 dollars.


                Second solution of the 6-th grade level student

Make a proportion

    36%2F12 = x%2F%2812%2B7%29,


where 'x' is what you want to get.


Simplify the proportion

    36%2F12 = x%2F17.


Now from the proportion find  x = %2836%2A17%29%2F12 = 3*17 = 51  dollars.


It is the same value, as in my first solution above.

Solved in two different ways for your better understanding.

--------------------

By the way, the meaning of the proportion is that the unit price for one photo is the same in both cases.




Question 731830: A television that has a square screen is advertised as having a 38-inch diagonal screen. Which of the following represents the width of the television to the nearest inch?
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39792) About Me  (Show Source):
You can put this solution on YOUR website!
The shape is a square so length and width are the same size.
d for the diagonal length, d=38.
x%5E2%2Bx%5E2=d%5E2
2x%5E2=d%5E2
2x%5E2=38%5E2
x%5E2=%2838%2F2%29%2838%29
x%5E2=19%2A38
x%5E2=2%2A19%2A19
highlight%28x=19%2Asqrt%282%29%29

"the following,..." for choices of answers was not shown. They must include 19sqrt%282%29, which is the
choice needing to be picked.

Answer by ikleyn(53751) About Me  (Show Source):
You can put this solution on YOUR website!
.
A television that has a square screen is advertised as having a 38-inch diagonal screen.
Which of the following represents the width of the television to the nearest inch?
~~~~~~~~~~~~~~~~~~~~~~~~~

Your post asks "which of the following", but does not provide what should follow.

So, I qualify your post as self-contradictory with the intention to deceive a reader,
and recommend you to throw it to the closest garbage bin as soon as you can, i.e. immediately.

Do not post this kind of messages to this forum.




Question 733675: a store is having a sale were all merchindise is 1/4 off . a women spends $36. how much would it have been at regular price

Answer by ikleyn(53751) About Me  (Show Source):
You can put this solution on YOUR website!
.
a store is having a sale were all merchindise is 1/4 off.
a women spends $36. how much would it have been at regular price
~~~~~~~~~~~~~~~~~~~~~~~~~~

So, in this problem, $36 is 3/4 of the regular price.


Then  1/4  of the regular price is  36/3 = 12 dollars;  
hence, the regular price is 4*12 = 48 dollars.


ANSWER.  The regular price is $48.

Solved.

The answer in the post by @lynnlo is incorrect.





Question 734916: how do you solve 16 m^2 = ____cm^2
Answer by ikleyn(53751) About Me  (Show Source):
You can put this solution on YOUR website!
.

1 meter = 100 centimeters.


1 square meter = 100^2 cm^2 = 10,000 cm^2.


16 m^2 = 16*10,000 = 160,000 cm^2.



ANSWER.  16 m^2 = 160,000 cm^2.

It is not simple - - - it is SUPER-simple conversion.




Question 740765: In a school of 600 students the number of boys is 360.when 60 new students are admitted in the school the ratio of the number of boys to the number of girls became 13:9. find the number of newly admitted boys..?

Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13327) About Me  (Show Source):
You can put this solution on YOUR website!


The original number of students was 600. After 60 new students are admitted, the total number of students is 660.

The resulting ratio of boys to girls is 13:9.

Let 13x = new number of boys
Let 9x = new number of girls

13x+9x = 660
22x = 660
x = 30

The new number of boys is 13x = 390.

The original number of boys was 360.

The number of new boys admitted was 390-360 = 30.

ANSWER: 30 new boys


Answer by ikleyn(53751) About Me  (Show Source):
You can put this solution on YOUR website!
.
In a school of 600 students the number of boys is 360. when 60 new students are admitted in the school
the ratio of the number of boys to the number of girls became 13:9. find the number of newly admitted boys ?
~~~~~~~~~~~~~~~~~~~~~~~~~

Originally, were there 600 students in the school, 360 boys and 600-360 = 240 girls.

60 new students were admitted.

Let x be the number of boys among these 60; then the number of girls among these 60 was (60-x).


Now the number of boys is 360+x; the number of girls is 240 + (60-x) = 300-x.


We make the ratio according to the problem

    %28360%2Bx%29%2F%28300-x%29 = 13%2F9.


To find 'x', cross multiply, then simplify

    9*(360+x) = 13*(300-x),

    9*360 + 9x = 13*300 - 13x

    9x + 13x = 13*300 - 9*360

       21x   =      660

         x   =       660/21 = 30.


ANSWER.  30 boys were newly admitted.


CHECK.  %28360%2B30%29%2F%28300-30%29 = 390%2F270 = 13%2F9.   ! correct !

Solved.




Question 646752: If in the army there is one officer for every 16 privates, how many officers are there in a regiment consisting of 1,105 officers and privates?
Answer by ikleyn(53751) About Me  (Show Source):
You can put this solution on YOUR website!
.
If in the army there is one officer for every 16 privates, how many officers are there in a regiment
consisting of 1,105 officers and privates?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


        The solution in the post by @Theo is incorrect and goes far out of the target.
        I came to bring a correct solution.
        I will give two solutions: one arithmetic and other algebraic.

                Arithmetic solution


According to the problem we can group all people in groups, each containing 16 privates and 1 officer.
So, each group is 17 people. The number of groups is 1105 divided by 17, which is

    1105%2F17 = 65.


So, there 65 groups and, correspondingly, 65 officers.


ANSWER.  65 officers and  1105 - 65 = 1040 privates.


                Algebraic solution


Let x be the number of officers.  Then the number of privates is 16x, according to the problem.


Write an equation for the total

    x + 16x = 1105.


Simplify and find x

    17x = 1105,  x = 1105/17 = 65.


We get the same answer:  65 officers  and  1105 - 65 = 1040 privates.

So, the problem is solved in two ways for your better understanding.

The algebra solution is, actually, the same Arithmetic solution, retold using formulas instead of words.




Question 1210397: Andrew bought some apples and pears. The ratio of the number of apples bought to the number of pears bought was 7:4. He spent $61.20. He paid $22.80 more for the apples than the pears. Each apple was $0.30 more than each pears.
(a) How much did he spend on the pears?
(b) How many pears did he buy?

Found 4 solutions by MathTherapy, greenestamps, ikleyn, josgarithmetic:
Answer by MathTherapy(10806) About Me  (Show Source):
You can put this solution on YOUR website!
Andrew bought some apples and pears. The ratio of the number of apples bought to the number of pears bought was 7:4. He spent $61.20. He paid $22.80 more for the apples than the pears. Each apple was $0.30 more than each pears.
(a) How much did he spend on the pears?
(b) How many pears did he buy?

Let multiplicative factor be x
As ratio of apples to pears is 7:4, number of apples and pears purchased = 7x and 4x, respectively

Let amount spent on pears be p
Since $22.80 more was spent on apples than pears, then amount spent on apples = p + $22.80
A total of $61.20 was spent, so we then have: p + p + 22.8 = 61.2
                                                        2p = 61.2 - 22.8
                                                        2p = 38.4
                        p, or amount spent on pears = highlight_green%2838.4%2F2+=+highlight%28%22%2419.20%22%29%29 

Since $19.20 was spent on “4x” pears, then cost of each pear = 19.2%2F%284x%29+=+4.8%2Fx
As $19.20 was spent on pears, $61.20 - $19.20, or $42 was spent on “7x” apples, and so, each apple cost 42%2F%287x%29+=+6%2Fx
Now, since each apple was $0.30 more than each pear, we get: 6%2Fx+=+.3+%2B+4.8%2Fx  
                                                             6 = .3x + 4.8 ---- Multiplying by LCD, x 
                                                       6 - 4.8 = .3x
                                                           1.2 = .3x                                                             
                            x, or multiplicative factor = 1.2%2F.3+=+12%2F3+=++4

Number of pears purchased: 4x = 4(4) = 16

Answer by greenestamps(13327) About Me  (Show Source):
You can put this solution on YOUR website!


There are many ways you could set up this problem for solving. I looked briefly at a few different ways before choosing a method that looked as if it would be relatively easy. I would hope that other tutors might come to you with different solutions.

He spent a total of $61.20; and he spent $22.80 more for the apples than for the pears.

let a = cost of the apples
let p = cost of the pears

a%2Bp=61.20 the total cost was $61.20
a-p=22.80 the cost of the apples was $22.80 more than the cost of the pears

Solve the pair of equations by adding the two equations to solve for a and then use that number to find p.

2a=84.00
a=42.00
p=42.00-22.80=19.20

The total cost of the apples was $42; the total cost of the pears was $19.20.

The ratio of the number of apples to the number of pears was 7:4. So

let 7x = number of apples
then 4x = number of pears

Find the cost of x apples and the cost of x pears:

7x=42
x=6
The cost of x apples is $6.

4x=19.20
x=4.80
The cost of x pears is $4.80.

The cost of x apples is $6 - $4.80 = $1.20 more than the cost of x pears. Each apple costs $0.30 more than each pear, so x = $1.20/$0.30 = 4.

The cost of x=4 apples is $6, so the cost of each apple is $6/4 = $1.50.

The cost of x=4 pears is $4.80, so the cost of each pear is $4.80/4 = $1.20.

Check the results we have with the original given information:

He bought 7x = 28 apples at $1.50 each for a total of 28($1.50) = $42.00.
He bought 4x = 16 pears at $1.20 each for a total of 16($1.20) = $19.20.
The total he spent was $42.00+$19.20 = $61.20.

Looks good. Now answer the questions that were asked.

(a) How much did he spend on the pears?
ANSWER: $19.20

(b) How many pears did he buy?
ANSWER: 16


Answer by ikleyn(53751) About Me  (Show Source):
You can put this solution on YOUR website!
.
Andrew bought some apples and pears. The ratio of the number of apples bought to the number of pears bought was 7:4.
He spent $61.20. He paid $22.80 more for the apples than the pears. Each apple was $0.30 more than each pears.
(a) How much did he spend on the pears?
(b) How many pears did he buy?
~~~~~~~~~~~~~~~~~~~~~~~~~

Let A be the amount he spent for apples.

Let P be the amount he spent for pears.


From the problem, we have these two equations

    A + P = 61.20,    (1)

    A - P = 22.80.    (2)


To find P, subtract equation (2) from equation (1).  You will get

    2P = 61.20 - 22.80

    2P = 38.4,  P = 38.4/2 = 19.2.


Thus, Andrew spent $19.20 for pears.  It is the answer for question (a).


From this, we conclude that Andrew spent  $61.20 - $19.20 = $42.00 for apples.



                   Now we start solving (b).



Andrew has 7n apples and 4n pears.  The number 'n' is unknown, and we want to find it.


The price for one apple is  42%2F%287n%29 = 6%2Fn.


The price for one pear is  19.2%2F%284n%29 = 4.8%2Fn.


The difference equation for price is

    6%2Fn - 4.8%2Fn = 0.3  dollars.


Simplify and find 'n'

    1.2%2Fn = 0.3,

    n = 1.2%2F0.3 = 4.


So, Andrew bought  7n = 7*4 = 28 apples and 4n = 4*4 = 16 pears.


It is the answer to question (b).

All questions are answered and the problem is solved completely.



Answer by josgarithmetic(39792) About Me  (Show Source):
You can put this solution on YOUR website!
This is not very neat and much likely not the best way, but this should work.
          PRICE    QUANTY    COST

Apple     p+0.3     7x      7x(p+0.3)

Pear       p        4x      4x*p

Totals                      61.20

and too, the 22.8 dollars difference
system%287x%28p%2B0.3%29-4xp=22.8%2C7x%28p%2B0.3%29%2B4xp=61.2%29


Question 1210396: Hakim spent 3/8 of his money on 3 cupcakes and 8 muffins. Then he spent 4/5 of the remaining money on 15 pieces of waffle. Each cupcake cost 2/3 as much as a muffin. Each piece of waffle cost $0.20 more than a cupcake. What was the cost of a muffin?
Found 3 solutions by MathTherapy, mccravyedwin, greenestamps:
Answer by MathTherapy(10806) About Me  (Show Source):
You can put this solution on YOUR website!
Hakim spent 3/8 of his money on 3 cupcakes and 8 muffins. Then he spent 4/5 of the remaining money on 15 pieces of waffle. Each cupcake cost 2/3 as much as a muffin. Each piece of waffle cost $0.20 more than a cupcake. What was the cost of a muffin?

Let amount spent, be D, cost of a muffin, M, and cost of a cupcake. C

For the first purchase, we get: %283%2F8%29D+=+3C+%2B+8M
                                   3D = 24C + 64M ---- Multiplying by LCD, 8 ---- eq (i)

For the second purchase, we get:%284%2F5%29%281++-++3%2F8%29+%2A+D+=+15W
                                %284%2F5%29%285%2F8%29D+=+15W
                                %28cross%284%29%2Fcross%285%29%29%28cross%285%29%2F2cross%288%29%29D+=+15W
                                D%2F2 = 15W ---- eq (ii)

Each cupcake cost 2%2F3 as much as a muffin, so: C = %282%2F3%29M+=+2M%2F3 ---- eq (iii)

Each piece of waffle cost $0.20 more than a cupcake, so: W = .2+%2B+2M%2F3 ---- eq (iv)

D%2F2+=+15W ------ eq (ii)
D%2F2+=+15%28.2+%2B+2M%2F3%29 ---- Substituting .2+%2B+2M%2F3 for W, in eq (ii) 
D%2F2+=+3+%2B+10M
D = 6 + 20M ---- Multiplying by LCD, 2 ---- eq (v)


 3D = 24C + 64M ------ eq (i)
3D+=+24%282M%2F3%29+%2B+64M --- Substituting 2M%2F3 for C, in eq (i) 
3D = 8(2M) + 64M
3D = 16M + 64M
3D = 80M ---- eq (vi)


   D =  6 + 20M ---- eq (v)
  3D =      80M ---- eq (vi)
  3D = 18 + 60M ---- Multiplying eq (v) by 3 ---- eq (vii)
    0 = - 18 + 20M  --- Subtracting eq (vii) from eq (vi)
- 20M = - 18
Cost of a muffin, or highlight_green%28M+=+%28-+18%29%2F%28-+20%29+=+9%2F10+=+highlight%28%22%240.90%22%29%29

Answer by mccravyedwin(421) About Me  (Show Source):
You can put this solution on YOUR website!
Most tutors, including me, often ignore where it says this:

"Also, if possible, provide a 'check' at the end, so check if the values 
you computed in fact are correct".

Nearly always, we skip this check.

If we would read and heed that, then perhaps we might use a different 
method, to make it easier to check.

Hakim spent 3/8 of his money on 3 cupcakes and 8 muffins. Then he spent 4/5
of the remaining money on 15 pieces of waffle. Each cupcake cost 2/3 as much as
a muffin. Each piece of waffle cost $0.20 more than a cupcake. What was the 
cost of a muffin?

Let T = total money Hakim had at the beginning.
Let C = cost of a cupcake
Let M = cost of a muffin
Let W = cost of a piece of waffle

Hakim spent 3/8 of his money on 3 cupcakes and 8 muffins. 
expr%283%2F8%29T+=+3C+%2B+8M

Then he spent 4/5 of the remaining money

which was T+-+%283C+%2B+8M%29

on 15 pieces of waffle. 

expr%284%2F5%29%28T-%283C%2B8M%29%5E%22%22%29=15W

Each cupcake cost 2/3 as much as a muffin. 

C=expr%282%2F3%29M

Each piece of waffle cost $0.20 more than a cupcake. 

W+=+C+%2B+0.20



Go to any of the online solvers for systems of equations

https://www.wolframalpha.com/
https://www.symbolab.com/solver/system-of-equations-calculator
https://cowpi.com/math/systemsolver/4x4.html
https://www.wolframalpha.com/calculators/system-equation-calculator
There are others also.

Type in 

(3/8)T = 3C + 8M,  (4/5)(T-(3C+8M))=15W, C = (2/3)M, W = C+0.20

Press ENTER, get

C = 0.6, M = 0.09, T = 24, W = 0.8 

which we interpret as

C = $0.60, M = $0.90, T = $24.00, W = $0.80.  

What was the cost of a muffin?

$0.90   <--- solved, but not checked.

Now let's check:

Hakim spent 3/8 of his money on 3 cupcakes and 8 muffins.
(3/8)x$24.00 = $9.00

3x$0.60 = $1.80, 8x$0.90 = $7.20, $1.80 + $7.20 = $9.00. 

That checks.

So his remaining money was $24.00 - $9.00 = $15.00

Then he spent 4/5 of the remaining money 

which was (4/5)x$15.00 = $12.00

on 15 pieces of waffle.

15x$0.80 = $12.00 and, indeed, that checks.   

Each cupcake cost 2/3 as much as a muffin. 

$0.60 = (2/3)($0.90)
$0.60 = $0.60, so that checks.

Each piece of waffle cost $0.20 more than a cupcake.

$0.80 = $0.60 + $0.20
$0.80 = $0.80, so that checks. 

Now, as you see, everything checks.

Edwin

Answer by greenestamps(13327) About Me  (Show Source):
You can put this solution on YOUR website!


You could possibly do a bunch of work "in the background" to set up this problem using a single variable, but it appears to me that would get quite messy. So let's not try to be clever and just go with separate variables for the costs of each of the three items.

He spent 3/8 of his money on the first purchase and 4/5 of the remaining amount on the second purchase. After spending 3/8 of his money on the first purchase, he had 5/8 of it left, so when he spent 4/5 of his remaining money on the second purchase that was (4/5)(5/8) = 4/8 of his original money.

The cost of each cupcake is 2/3 the cost of a muffin:
m = cost of a muffin
c = cost of a cupcake
c = (2/3)m

The cost of each piece of waffle is $0.20 more than the cost of a cupcake:
w = cost of a piece of waffle
w = c+0.20 = (2/3)m+0.20

Let x be his original amount of money

The cost of 3 cupcakes and 8 muffins was 3/8 of his original money:

3%28%282%2F3%29m%29%2B8m=%283%2F8%29x
2m%2B8m=%283%2F8%29x
10m=%283%2F8%29x [1]

The cost of 15 pieces of waffle was 4/8 of his original money:

15%28%282%2F3%29m%2B0.20%29=%284%2F8%29x
10m%2B3=%284%2F8%29x [2]

The given information was such that this approach to solving the problem leads to two equations which are easy to solve. Comparing [1] and [2],

3=%281%2F8%29x
24=x

The amount of money he started with was $24.

From [1], the cost of 10 muffins was 3/8 of $24, or $9. So the cost of each muffin is $9/10 = $0.90.

ANSWER: $0.90



Question 1170245: 1.a) Deduce that the total surface area S of a cylinder closed at both ends with height, H and base radius R is given by: S=2πR(R+H) where π is a constant.
b) Find the volume of S given that H =15.0m and R =5.0m (leaving your answer in terms of π)
c) Calculate the amount of water the tank can hold leaving your answer in terms of π.
d) Calculate the height (h) of a cuboid tank of cross-sectional area 25 metre cube which has the same capacity as the cylinder tank in (c) above.


Found 2 solutions by ikleyn, CPhill:
Answer by ikleyn(53751) About Me  (Show Source):
You can put this solution on YOUR website!
.

In your post,  in the last question,  you write

        ". . . cross-sectional area 25 metre cube . . . ".


Dear missis or mister writer,  the area is  NEVER  measures in  "metre cube ".

The appropriate unit for the area is  " square meter ".


As I noticed from your posts, you make this error systematically.


It tells me a lot about mathematical qualification of a person who created this problem.



Answer by CPhill(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Let's break down this problem step by step.
**1.a) Deduce the Total Surface Area of a Closed Cylinder**
* **Area of the Top and Bottom Circles:** Each circle has an area of πR². Since there are two circles (top and bottom), their combined area is 2πR².
* **Lateral Surface Area (Curved Surface):** Imagine unrolling the side of the cylinder. This forms a rectangle with height H and width equal to the circumference of the base (2πR). Therefore, the lateral surface area is 2πRH.
* **Total Surface Area (S):** To find the total surface area, add the areas of the top, bottom, and lateral surfaces:
S = 2πR² + 2πRH
S = 2πR(R + H)
**1.b) Find the Volume of the Cylinder**
* **Volume Formula:** The volume (V) of a cylinder is given by V = πR²H.
* **Given Values:** H = 15.0 m and R = 5.0 m.
* **Calculation:**
V = π(5.0 m)²(15.0 m)
V = π(25 m²)(15.0 m)
V = 375π m³
**1.c) Calculate the Amount of Water the Tank Can Hold**
The amount of water the tank can hold is equal to its volume.
* **Answer:** The tank can hold 375π m³ of water.
**1.d) Calculate the Height of a Cuboid Tank with the Same Capacity**
* **Cuboid Volume Formula:** The volume of a cuboid is given by V = Area of base × height.
* **Given Information:**
* Volume of the cuboid = Volume of the cylinder = 375π m³
* Cross-sectional area (base area) of the cuboid = 25 m²
* **Calculation:**
375π m³ = 25 m² × h
h = (375π m³) / (25 m²)
h = 15π m
**Answers:**
a) S = 2πR(R + H)
b) V = 375π m³
c) 375π m³
d) h = 15π m


Question 1168192: Kameron paid $3.60 for a book. It was marked down 50% of its original price, and he also used a coupon to save an additional 40% off of the sale price. What was the original price?
Write the answer rounded to the nearest cent. Do not include a $ sign.

Answer by josgarithmetic(39792) About Me  (Show Source):
You can put this solution on YOUR website!
g, the original price

%280.5%29g, the sale price

0.5g-0.4%2A0.5g=3.60, what he paid

You can finish.


Question 1168190: Bummer! Kai lives in California and of course, he has to pay sales tax! The sales tax rate is 8.75%. Now, does he have enough money to bring the backpack of his dream home?
(Original price: $88.95, 40% off and Kai has $55.00 saved up so far.)

Answer by MathLover1(20855) About Me  (Show Source):
You can put this solution on YOUR website!



Original price: $88.95
40% off is 0.40%2A88.95=35.58
88.95-35.58=53.37.....new price

if the sales tax rate is 8.75%=0.0875, then he needs to pay
53.37%2B+0.0875%2A53.37=58.04....new price with sales tax included
if Kai has $55.00+saved up so far he does not have enough money to bring the backpack of his dream home


Question 1169039: Picture frame has its length 8cm longer than its width. It has an inner 1-cm boundary such that a maximum 660cm^2- picture may fit into it. Find the dimension of this frame.
Answer by ikleyn(53751) About Me  (Show Source):
You can put this solution on YOUR website!
.
Picture frame has its length 8cm longer than its width. It has an inner 1-cm boundary
such that a maximum 660cm^2- picture may fit into it. Find the dimension of this frame.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


        The problem formulation as it is given in the post is  INCORRECT.

        The correct formulation, in my version,  is  THIS


    Picture frame has its length 8cm longer than its width. It has an inner 1-cm boundary 
    such that the area of the picture itself is 660cm^2. Find the dimension of this frame.

                    Below is my solution for this edited formulation.

Since the width of the frame is uniform and since the outer length of the frame is 8 cm longer 
than its outer width, the length of the picture itself, without considering the frame, is
8 cm longer than the picture width.


So, if x is the picture width, in centimeters, then the picture length is (x+8) cm,
and we have this equation

    x*(x+8) = 660 cm^2.     (1)


Now you can guess mentally the solution to this equation: it is x= 22 cm for the picture width, 
giving  22+8 = 30 cm  for the picture length.


After this guessing, note that the left side of the equation (1) is the monotonic function,
so the guessed solution is UNIQUE.


Alternatively, you can solve this quadratic equation (1) formally, using factoring or the quadratic formula.


In either case, you will get the same solution for x.


Thus the picture is 22 x 30 centimeters.


It implies that the outer dimensions of the frame are  24 x 32 cm.    


ANSWER.  The outer dimensions of the frame are  24 x 32 cm.    

Solved.




Older solutions: 1..45, 46..90, 91..135, 136..180, 181..225, 226..270, 271..315, 316..360, 361..405, 406..450, 451..495, 496..540, 541..585, 586..630, 631..675, 676..720, 721..765, 766..810, 811..855, 856..900, 901..945, 946..990, 991..1035, 1036..1080, 1081..1125, 1126..1170, 1171..1215, 1216..1260, 1261..1305, 1306..1350, 1351..1395, 1396..1440, 1441..1485, 1486..1530, 1531..1575, 1576..1620, 1621..1665, 1666..1710, 1711..1755, 1756..1800, 1801..1845, 1846..1890, 1891..1935, 1936..1980, 1981..2025, 2026..2070, 2071..2115, 2116..2160, 2161..2205, 2206..2250, 2251..2295, 2296..2340, 2341..2385, 2386..2430, 2431..2475, 2476..2520, 2521..2565, 2566..2610, 2611..2655, 2656..2700, 2701..2745, 2746..2790, 2791..2835, 2836..2880, 2881..2925, 2926..2970, 2971..3015, 3016..3060, 3061..3105, 3106..3150, 3151..3195, 3196..3240, 3241..3285, 3286..3330, 3331..3375, 3376..3420, 3421..3465, 3466..3510, 3511..3555, 3556..3600, 3601..3645, 3646..3690, 3691..3735, 3736..3780, 3781..3825, 3826..3870, 3871..3915, 3916..3960, 3961..4005, 4006..4050, 4051..4095, 4096..4140, 4141..4185, 4186..4230, 4231..4275, 4276..4320, 4321..4365, 4366..4410, 4411..4455, 4456..4500, 4501..4545, 4546..4590, 4591..4635, 4636..4680, 4681..4725, 4726..4770, 4771..4815, 4816..4860, 4861..4905, 4906..4950, 4951..4995, 4996..5040, 5041..5085, 5086..5130, 5131..5175, 5176..5220, 5221..5265, 5266..5310, 5311..5355, 5356..5400, 5401..5445, 5446..5490, 5491..5535, 5536..5580, 5581..5625, 5626..5670, 5671..5715, 5716..5760, 5761..5805, 5806..5850, 5851..5895, 5896..5940, 5941..5985, 5986..6030, 6031..6075, 6076..6120, 6121..6165, 6166..6210, 6211..6255, 6256..6300, 6301..6345, 6346..6390, 6391..6435, 6436..6480, 6481..6525, 6526..6570, 6571..6615, 6616..6660, 6661..6705, 6706..6750, 6751..6795, 6796..6840, 6841..6885, 6886..6930, 6931..6975, 6976..7020, 7021..7065, 7066..7110, 7111..7155, 7156..7200, 7201..7245, 7246..7290, 7291..7335, 7336..7380, 7381..7425, 7426..7470, 7471..7515, 7516..7560, 7561..7605, 7606..7650, 7651..7695, 7696..7740, 7741..7785, 7786..7830, 7831..7875, 7876..7920, 7921..7965, 7966..8010, 8011..8055, 8056..8100, 8101..8145, 8146..8190, 8191..8235, 8236..8280, 8281..8325, 8326..8370, 8371..8415, 8416..8460, 8461..8505, 8506..8550, 8551..8595, 8596..8640, 8641..8685, 8686..8730, 8731..8775, 8776..8820, 8821..8865, 8866..8910, 8911..8955, 8956..9000, 9001..9045, 9046..9090, 9091..9135, 9136..9180, 9181..9225, 9226..9270, 9271..9315, 9316..9360, 9361..9405, 9406..9450, 9451..9495, 9496..9540, 9541..9585, 9586..9630, 9631..9675, 9676..9720, 9721..9765, 9766..9810, 9811..9855, 9856..9900, 9901..9945, 9946..9990, 9991..10035, 10036..10080, 10081..10125, 10126..10170, 10171..10215, 10216..10260, 10261..10305, 10306..10350, 10351..10395, 10396..10440, 10441..10485, 10486..10530, 10531..10575, 10576..10620, 10621..10665, 10666..10710, 10711..10755, 10756..10800, 10801..10845, 10846..10890, 10891..10935, 10936..10980, 10981..11025, 11026..11070, 11071..11115, 11116..11160, 11161..11205, 11206..11250, 11251..11295, 11296..11340, 11341..11385, 11386..11430, 11431..11475, 11476..11520, 11521..11565, 11566..11610, 11611..11655, 11656..11700, 11701..11745, 11746..11790, 11791..11835, 11836..11880, 11881..11925, 11926..11970, 11971..12015, 12016..12060, 12061..12105, 12106..12150, 12151..12195, 12196..12240, 12241..12285, 12286..12330, 12331..12375, 12376..12420, 12421..12465, 12466..12510, 12511..12555, 12556..12600, 12601..12645, 12646..12690, 12691..12735, 12736..12780, 12781..12825, 12826..12870, 12871..12915, 12916..12960, 12961..13005, 13006..13050