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

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Question 1158487: When the value of a is less than 0 in the equation y = a∛(x - h) + k, the graph will be...
closer to the y axis,
the same as the parent function graph,
farther from the y axis,
reflected across the x axis?

Answer by KMST(5391) About Me  (Show Source):
You can put this solution on YOUR website!
y=a%2Aroot%283%2Cx-h%29 has a graph that crosses the x-axis at x=h.
The function increases for a%3E0 , like y=root%283%2Cx-1%29: graph%28300%2C300%2C-5%2C5%2C-5%2C5%2C%28x-1%29%5E%281%2F3%29%2C-%281-x%29%5E%281%2F3%29%29 ,
and decreases for a%3C0 , like y=-2root%283%2Cx-1%29: .
The greater the value of abs%28a%29 , the steeper the curve.
We can say something similar about y=a%2Aroot%283%2Cx-h%29%2Bk , except that at x=h} the function crosses the line y=k ,
as for y=root%283%2Cx-1%29%2B2: , or y=-2root%283%2Cx-1%29:


Question 1210633: In a store selling Brand X and Brand Y shoes, some of the 100 pairs of shoes were sold. The ratio of the number of Brand X shoes to Brand Y shoes was 7/3 before the sale and became 5/1 after the sale. If an equal number of shoes were sold from both brands, how many total pairs of shoes remained in the store after the sale?
Answer by ikleyn(53930) About Me  (Show Source):
You can put this solution on YOUR website!
.
In a store selling Brand X and Brand Y shoes, some of the 100 pairs of shoes were sold.
The ratio of the number of Brand X shoes to Brand Y shoes was 7/3 before the sale
and became 5/1 after the sale.
If an equal number of shoes were sold from both brands,
how many total pairs of shoes remained in the store after the sale?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


The problem says that initially were there 100 pairs of shoes of the brands X and Y in proportion 7:3.


It means that were there 70 pairs of brand X and 30 pairs of brand Y.


OK.  Next, let x be the number of pairs sold of each brand.


Then for remaining pairs, 70-x and 30-x, we have this proportion

    %2870-x%29%2F%2830-x%29 = 5%2F1.


Solve it and find x. For it, cross multiply and then simplify

    70-x = 5(30-x),

    70 - x = 150 - 5x,

    5x - x = 150 - 70,

       4x   =   80,

        x   =   20.


So 20 pairs of each brand were solved. 
Hence, the remaining was 70-20 = 50 pairs of brand X and 30-20 = 10 pairs of brand Y,
giving the total 50+10 = 60 remaining pairs.    ANSWER

Solved.




Question 970486: The ratio of girls to boys in Liza's classroom is 5 to 4. How many girls are in her classroom if there is a total of 27 students?
Found 2 solutions by josgarithmetic, n2:
Answer by josgarithmetic(39835) About Me  (Show Source):
You can put this solution on YOUR website!
5x girls to 4x boys

5x%2B4x=27 sum of the students
9x=27
x=3 factor found


5%2A3=highlight%2815%29 girls
4%2A3=highlight%2812%29 boys

Answer by n2(91) About Me  (Show Source):
You can put this solution on YOUR website!
.
The ratio of girls to boys in Liza's classroom is 5 to 4. How many girls are in her classroom
if there is a total of 27 students?
~~~~~~~~~~~~~~~~~~~~~~~~~~


        It can be solved mentally without using equations.


According to the problem, we can group the students in sets, each containing 5 girls and 4 boys.

Each such group has 5 + 4 = 9 students.

So, the number of such groups is 27/9 = 3.

Thus, the number of girls in the class is 3*5 = 15.         ANSWER

Solved.


//////////////////////////////////


In normal teaching process, the students should learn this grouping method first,
allowing to solve this problem mentally (at 4-th or 5-th grade).

Later, in 6-th or 7-th grade, students will learn proportions and will solve such problems using proportions.

But for normal mathematical development, students should learn the mental solution way first,
using the grouping method.




Question 1210616: Emir, Cillian, and Lasha are avid collectors of vintage postage stamps. During a local exhibition, they decided to donate parts of their collections to a museum. The ratio of the number of stamps Emir and Cillian gave away was 4:7, while the ratio of the number of stamps Cillian and Lasha gave away was 3:5. The number of stamps Emir gave away was 20 fewer than the number of stamps he kept for himself. Cillian kept 3/2 as many stamps as Emir kept, and Lasha kept 4/3 as many stamps as Cillian kept. After their donations, they had a combined total of 738 stamps left in their collections. How many stamps did Lasha give away?
Found 3 solutions by MathTherapy, ikleyn, KMST:
Answer by MathTherapy(10853) About Me  (Show Source):
You can put this solution on YOUR website!
Emir, Cillian, and Lasha are avid collectors of vintage postage stamps. During a local exhibition, they decided to donate parts of their
collections to a museum. The ratio of the number of stamps Emir and Cillian gave away was 4:7, while the ratio of the number of stamps
Cillian and Lasha gave away was 3:5. The number of stamps Emir gave away was 20 fewer than the number of stamps he kept for himself.
Cillian kept 3/2 as many stamps as Emir kept, and Lasha kept 4/3 as many stamps as Cillian kept. After their donations, they had a
combined total of 738 stamps left in their collections. How many stamps did Lasha give away?
**************************************************************************
Emir:    E
Cillian: C
Lasha:  L
Ratio of E:C = 4:7
Ratio of C:L = 3:5
Since C (Cillian) obviously gave away the same amount, the ratios can be changed from E:C = 4:7 to: 4(3):7(3) = 12:21, and from C:L = 3:5 
to 3(7):5(7) = 21:35
We now have the ratio of the 3 donors as: E:C:L = 12:21:35
As such, E(Emir), C(Cillian), and L(Lasha) gave away 12x, 21x, and 35x stamps, respectively, with x being the MULTIPLICATIVE FACTOR

Since the number of stamps Emir gave away was 20 FEWER than the number of stamps he kept for himself, he kept 20 MORE than he donated, or 12x + 20
And, as Cillian kept 3%2F2 as many stamps as Emir did, the number Cillian kept was %283%2F2%29%2812x+%2B+20%29 = 18x + 30 
Now, since Lasha kept 4%2F3 as many as Cillian, Lasha kept %284%2F3%29%2818x+%2B+30%29 = 24x + 40 stamps

As they had a combined 738 stamps left in their collections, we get the following TOTAL remaining-stamps equation: 
                 12x + 20 + 18x + 30 + 24x + 40 = 738
                                                          54x + 90 = 738
                                                                   54x = 648
                        MULTIPLICATIVE factor, or x = 648%2F54 = 12
        Lasha gave away 35x stamps (see above), or 35(12) = 420 stamps

Answer by ikleyn(53930) About Me  (Show Source):
You can put this solution on YOUR website!
.
Emir, Cillian, and Lasha are avid collectors of vintage postage stamps.
During a local exhibition, they decided to donate parts of their collections to a museum.
The ratio of the number of stamps Emir and Cillian gave away was 4:7,
while the ratio of the number of stamps Cillian and Lasha gave away was 3:5.
The number of stamps Emir gave away was 20 fewer than the number of stamps he kept for himself.
Cillian kept 3/2 as many stamps as Emir kept,
and Lasha kept 4/3 as many stamps as Cillian kept.
After their donations, they had a combined total of 738 stamps left in their collections.
How many stamps did Lasha give away?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~


Let' start from this point

    "The ratio of the number of stamps Emir and Cillian gave away was 4:7, 
     while the ratio of the number of stamps Cillian and Lasha gave away was 3:5."


We want to write this statement as a long proportion of three terms.


For it, write partial proportions E:C = 4:7 and C:L = 3:5 as long three terms proportion

    E:C:L = 12:21:35.


The logic for it is that 4:7 is the same as 12:21  (after multiplication by 3),
while proportion 21:35 is the same as 3:5  (after dividing by 7). 


So, we can assume that Emir gave away 12x stamps, Cillian gave away 21x stamps and Lasha gave away 35x stamps.


Next, let E be the number of stamps Emir kept.

Then the number of stamps Cillian kept was %283%2F2%29E,

and the number of stamps Lasha kept was %284%2F3%29%2A%283%2F2%29E = 2E.


We are given that the combined total number of stamps left in their collection was 738.


So, we write this equation

    E + %283%2F2%29E + 2E = 738.

Hence

    2E + 3E + 4E = 738*2,  9E = 1476,  E = 1476/9 = 164.


Now, the problem says "The number of stamps Emir gave away was 20 fewer than the number of stamps he kept for himself."

It means that

    12x = 164 - 20.


From it, we find

    12x = 144,  x = 144/12 = 12.


Very good.


Now we are in position to find the number of stamps Lasha gave away.  It is  35x = 35*12 = 420.


ANSWER.  Lasha gave away 420 stamps.

Solved completely.



Answer by KMST(5391) About Me  (Show Source):
You can put this solution on YOUR website!
EDIT: My calculations are wrong. See The much clearer and direct solution by iklein.
For every 4 stamps donated by Emir, there were 7 stamps donated by Cillian.
For every 3 stamps donated by Cillian, there were 5 stamps donated by Lasha.
That means that for every 7%2A3=21 stamps given away by Cillian, there were
4%2A3=12 stamps given away by Emir, and
This is wrong --> 5%2A3=15 stamps given away by Lasha.
It should have been 5%2A7=35 .

Let us make
the number of stamps donated by Emir %22=%2212x ,
the number of stamps given by Cillian %22=%2221x , and
the number of stamps given by Lasha %22=%2215x ,

EDIT:Here is where the other solutions are better (besides having no mistakes). I insisted on calculating number of stamps kept based on x , but that unnecessarily complicates the calculations.
The number of stamps Emir gave away was 20 fewer than the number of stamps he kept for himself means Emir kept 12x%2B20 stamps. See the much clearer and direct solution by Iklein.
Cillian kept 3/2 as many stamps as Emir, Cillian and Lasha kept means
Cillian kept %283%2F2%29%2812x%2B20%29=18x%2B30 stamps.

Lasha kept 4/3 as many stamps as Cillian kept means
Lasha kept %284%2F3%29%2818x%2B30%29=24x%2B40 stamps.
After their donations, the total number of stamps Emir, Cillian and Lasha had left in their collections was
%2812x%2B20%29%2B%2818x%2B30%29%2B%2824x%2B40%29=738<-->%2812%2B18%2B24%29x%2B20%2B30%2B40=738-->54x%2B90=738-->54x=738-90-->54x=648-->x=648%2F54-->x=12

Lasha gave away15%2A12=highlight%28180%29 stamps.

In table form that looks like:
-->
After solving %2812x%2B20%29%2B%2812x%2B20%29%2B%2824x%2B40%29=738->->-> x=12 , we can substitute that value to get

We can calculate the number of stamps each one originally had, and the number each donated,
but we are only asked for the number Lasha donated: 15%2A12=180.


Question 1210613: Marek and Zaven had 900 coins altogether. Marek spent 25% fewer coins than Zaven. Zaven was left with 50% as many coins as the number he spent. Marek had 75% of the total number of coins the two friends had left in the end. How many coins did Marek have in the end?

Found 4 solutions by greenestamps, math_tutor2020, KMST, josgarithmetic:
Answer by greenestamps(13362) About Me  (Show Source):
You can put this solution on YOUR website!


After working a short way through the problem, I chose to set the problem up like this, to make the work easier by avoiding having to work with fractions or decimals.

Marek spent 25% fewer coins than Zaven -- i.e., he spent three-fourths as many as Zaven. So

Let 4x = # Zaven spent
Then 3x = # Marek spent

The number Zaven left with was half the number he spent. So

2x = # Zaven was left with; and then
4x+2x = 6x = # Zaven started with

The total number of coins the two of them had is 900, so

900-6x = # Marek started with

Then

(90-6x)-3x = 900-9x = # Marek finished with

In the end, Marek had 3/4 of the total number of coins. Here we have two (at least) ways to continue; we could either say that the number Marek finished with is 3/4 of the total 900, or we could say Marek finished with 3 times as many as Zaven. For an unknown reason, I chose the second option when I first worked the problem all the way through.

900-9x = 3(2x)
900-9x = 6x
900 = 15x
x = 900/15 = 60

The problem asks for the number Marek finished with.

ANSWER: 900-9x = 900-540 = 360


Answer by math_tutor2020(3837) About Me  (Show Source):
You can put this solution on YOUR website!

I'd go with the last method tutor KMST mentioned.
ZavenMarekTotal
Start3x900-3x900
Spent2x1.5x3.5x
Finalx900-4.5x900-3.5x


Refer to the given statement that
"Marek had 75% of the total number of coins the two friends had left in the end"
which means we'll use the last two entries of the bottom row to form this equation
900-4.5x = 0.75*(900-3.5x)
That equation solves to x = 120

Plugging that value of x back into the table above produces the following:
ZavenMarekTotal
Start360540900
Spent240180420
Final120360480


Answer: 360 coins

Answer by KMST(5391) About Me  (Show Source):
You can put this solution on YOUR website!
DATA WE HAVE:
Marek and Zaven had 900 coins altogether.
That is what they had initially.

Zaven was left with 50% (half) as many coins as the number he spent.
He spent twice as many coins as he was left with.
For every 1 coin he had left, there were 2 he had spent, and 3 that he originally had.
That tells me that the number of coins Zaven initially had was a multiple of 3.

Marek spent 25% fewer coins than Zaven.
Marek spent 100%-25%=75% of what Zaven spent.
If I knew how much Zaven spent I can multiply times 0.75 to find what Marek spent.
For every 4 coins Zaven spent Marek spent 3. So I could instead divide by 4 and multiply times 3.

Marek had 75% of the total number of coins the two friends had left in the end.

Could Marek have initially had 75% of the total number of coins the two friends had left in the end? It does not seem possible.
I think what the "in the end" part applies to the whole sentence, and what was meant is:
In the end, Marek had 75% of the total number of coins the two friends had left.

AN EASY WAY TO SOLVE IT (guess and check):

Fist Guess:
What if Zaven initially had 300coins.
Then, Zaven must have spent 200coins ,
and must have had left with 300coins-200coins=100coins in the end.
Then, Marek
must have started with 900coins-300coins=600coins
must have spent 0.75%2A200coins=150coins ,
and have had left 600coins-150coins=450coins in the end.
Then, Zaven and Marek then together had 100coins%2B450coins=550coins in the end,
and that would mean that in the end,
the number of coins Marek had,
as a fraction and as a percentage of the number of coins Zaven and Marek had together, was
450coins%2F%22550+coins%22=9%2F11=%220.818181...%22=%2281.8181...%22%22%25%22

We need 75%22%25%22 as an answer, so we should try to have Marek start with a little less with respect to Zaven.

second Guess:
What if Zaven initially had 360coins.
Then, Zaven must have spent 240coins ,
and must have had left with 360coins-240coins=120coins in the end.
Then, Marek
must have started with 900coins-360coins=540coins
must have spent 0.75%2A240coins=180coins ,
and have had left 540coins-180coins=360coins in the end.
Then, Zaven and Marek then together had 120coins%2B360coins=480coins in the end,
and that would mean that in the end,
the number of coins Marek had,
as a fraction and as a percentage of the number of coins Zaven and Marek had together, was
360coins%2F%22480+coins%22=3%2F4=75%2F100=0.75=75%2F100=75%22%25%22

A HYBRID OF THE SOLUTION ABOVE AND jogsarithmetic's solution:
Let's call the number of coins Zaven had at start, what he spent, and what he had left in the end 3x, 2x, and x, respectively.

Then, we have just one variable, and the equation
%28900-4.5x%29%2F%28900-3.5x%29=0.75<-->900-4.5x=0.75%28900-3.5x%29<-->3600-18x=3%28900-3.5x%29<-->3600-18x=2700-10.5x<-->3600-2700=7.5x%29
x=900%2F7.5=120
In th e end, the number of coins Marek had was 900-4.5%2A120=900-540=highlight%28360%29

Some teachers want problems solved the way it was taught in class, and may not like alternative solutions.
Some teacher may want to give extra point f=if more that one way to solve the problem is shown.

Answer by josgarithmetic(39835) About Me  (Show Source):
You can put this solution on YOUR website!
             Marek             Zaven          TOTAL

Start        900-z               z             900

AFTER
SPENT       900-z-0.75x         z-x=0.5x

            900-z-0.75x=(0.75)(900-z-0.75x+0.5x)

That does not go directly to the quanity asked in the question.
System is system%28z-x=0.5x%2C900-z-0.75x=0.75%28900-z-0.25x%29%29
But a more efficient setup may be possible.

(*WARNING: possible fault in setup; algebraic steps produced faulty outcome)


Question 1210615: Lauri and Anže are baking cookies for a village festival. Anže baked 2/3 as many cookies as Lauri. After the morning rush, Anže had 10 fewer cookies left than the number he had originally baked. Lauri had 1/4 as many cookies left as Anže. By the end of the day, the difference between the number of cookies they each had left was 30. How many cookies did Lauri bake at first?
Found 2 solutions by ikleyn, KMST:
Answer by ikleyn(53930) About Me  (Show Source):
You can put this solution on YOUR website!
.
Lauri and Anže are baking cookies for a village festival. Anže baked 2/3 as many cookies as Lauri.
After the morning rush, Anže had 10 fewer cookies left than the number he had originally baked.
Lauri had 1/4 as many cookies left as Anže. By the end of the day, the difference
between the number of cookies they each had left was 30. How many cookies did Lauri bake at first?
~~~~~~~~~~~~~~~~~~~~~~~~~~~


        This problem is good to solve it by the backward method.
        Doing this way, the solution to the problem becomes simpler.


Let x be the number of cookies Lauri had left at the end of the day.
Then, according to the problem, Anze had left 4x cookies at the end of the day.


Next, according to the problem, 

    4x - x = 30,  or  3x = 30,  which implies  x = 30/3 = 10.


So, at the end of the day, Lauri had 10 cookies left, while Anze had 4*10 = 40 cookies left.


Hence, initially Anze did bake 40 + 10 = 50 cookies.


These 50 cookies are 2/3 of the number of cookies that Lauri initially baked.


Hence, Lauri baked initially  %283%2F2%29%2A50 = 3*25 = 75  cookies.


ANSWER.  Lauri did baked 75 cookies at first.

Solved.



Answer by KMST(5391) About Me  (Show Source):
You can put this solution on YOUR website!
x=number of cookies Lauri baked.
%282%2F3%29x=number of cookies Anže baked
If we assume that "after the morning rush" and "by the end of the day" are the same time, and the time when
Anže had 10 fewer cookies left than the number he had originally baked, and
Lauri had 1/4 as many cookies left as Anže,
we have enough data to solve the problem.
At the end, Anže had %282%2F3%29x-10 cookies left,
while Lauri had %281%2F4%29%28%282%2F3%29x-10%29 cookies left.
The difference is:
cookies, and we are told that
%283%2F4%29%28%282%2F3%29x-10%29=30Multiplying both side times 4%2F3 , we get
%282%2F3%29x-10=40-->%282%2F3%29x=50-->x=50%283%2F2%29-->x=highlight%2875%29


Question 1210614: Leo spent 25% of his savings on 3 gaming mice and 5 charging cables on Friday. The cost of one gaming mouse was three times the cost of one charging cable. On Saturday, he spent 4/5 of his remaining savings to buy some more gaming mice and another 8 charging cables. How many gaming mice did he buy on Saturday?
Answer by KMST(5391) About Me  (Show Source):
You can put this solution on YOUR website!
Something is wrong with this problem,

We know that one gaming mouse cost as much as 3 charging cables on Friday.
Let's call the cost of one charging cable on Friday one blip.
Then the cost of a gaming mouse on Friday was 3 blips.
On Friday Leo spent money on 3 gaming mice and 5 charging cables.
The amount Leo spent on Friday, in blips, is
3%2A%283blips%29%2B5%281blip%29=9blips%2B5blips=14blips
Those 14+blips Leo spent on Friday were 25%22%25%22=1%2F4 of his savings.
That means that all of Leo's savings, in blips were 4%2A14blips=56blips
On Saturday, Leo's remaining savings, were 56blips-14blips=42blips
On Saturday Leo spent 4%2F5 of his remaining savings.
In blips, the amount Leo spent on Saturday was %284%2F5%29%2A42=168%2F5=33.6
There was a typo somewhere, or else the prices changed, because with whole number prices, the total spent would always be a whole number.


Question 228010: $15000 is deposited every year in an account yeilding 6% interest compounded annually, how much money will have been saved after 10 years?
Answer by ikleyn(53930) About Me  (Show Source):
You can put this solution on YOUR website!
.
$15000 is deposited every year in an account yeilding 6% interest compounded annually,
how much money will have been saved after 10 years?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


        This my post is written as a reaction to two previous posts by @rfer and @JWG,
        that do not solve the problem or solve it incorrectly.

        This problem is a standard typical problem on ordinary annuity,  of the beginner level.


Use the standard formula for the Future value of the ordinary annuity


    FV = P%2A%28%28%281%2Br%29%5En-1%29%2Fr%29,    (1)


where  FV is the future value of the account;  P is the annual payment (deposit); 
r is the annual compounding percentage rate presented as a decimal; 
n is the number of deposits (= the number of years, in this case).


Under the given conditions, P = 15000;  r = 0.06;  n = 10.  So, according to the formula (1), 
you get at the end of the 10-th year


    FV = 15000%2A%28%28%281%2B0.06%29%5E10-1%29%2F0.06%29 = $197,711.92.    ANSWER


Note that you deposit only  10*$15000 = $150,000.  The rest is what the account earns/accumulates in 10 years.

Solved completely.




Question 1210604: Two athletes, Leo and Jasper, were responsible for distributing premium energy bars to the teams during a weekend tournament. Leo distributed 540 energy bars on the first day. Jasper distributed a quantity equal to exactly 4/9 as many bars as Leo. Jasper had a remaining stock of bars that was 15% greater than the number of bars he just distributed. Leo had a remaining stock that was 2.25 times the size of Jasper’s remaining stock. Calculate the total number of energy bars Leo and Jasper had in their combined inventory before the tournament began.
Found 3 solutions by math_tutor2020, greenestamps, josgarithmetic:
Answer by math_tutor2020(3837) About Me  (Show Source):
You can put this solution on YOUR website!

Both tutors josgarithmetic and greenestamps have typos in their scratch work which lead to an incorrect final answer.


The first equation josgarithmetic wrote, J-240=1.15*240, should solve to J = 516.
This then has L-540=2.25(J-240) become L = 1161.
Therefore the total should be J+L = 516+1161 = 1677


greenestamps made an error when computing 2.25 times 276 = 581
Instead it should be 2.25 times 276 = 621
So the updated calculation at the end should be
540+240+276+621 = 1677

Answer by greenestamps(13362) About Me  (Show Source):
You can put this solution on YOUR website!


Directly from the problem description....

Number sold by Leo: 540

Number sold by Jasper: 4/9 of 540 = 240

Number Jasper has left: 1.15 times 240 = 276

Number Leo has left: 2.25 times 276 = 581

Total combined original inventory: 540+240+276+581 = 1637

ANSWER: 1637


Answer by josgarithmetic(39835) About Me  (Show Source):
You can put this solution on YOUR website!
Combined beginning energy bar, E.
L%2BJ=E
                                       Leo             Jasper
Remain distribution day 1            L-540             J-(4/9)540
                                     L-540             J-240


From Description
J-240=1.15%2A240
J=1.15%2A240%2B240
J=2.25%2A240----THIS STEP A MISTAKE. REST IS ALL WRONG.
J=540
and
L-540=2.25%28J-240%29
L-540=2.25%28540-240%29
L=540%2B2.25%2A300
L=1215

E=L%2BJ
E=1215%2B540
cross%28E=1755%29


Question 1210597: Liam and Noah shared a collection of 540 vintage stamps. Liam sold 2/5 fewer stamps than Noah. After the sale, Noah found that the number of stamps he had remaining was exactly 2/3 of the number of stamps he had sold. If Liam’s remaining stamps accounted for 4/7 of the total stamps the two boys possessed at the end, how many stamps did Liam have at first?
Found 4 solutions by MathTherapy, KMST, josgarithmetic, greenestamps:
Answer by MathTherapy(10853) About Me  (Show Source):
Answer by KMST(5391) About Me  (Show Source):
You can put this solution on YOUR website!
The question is unclear, and I cannot find an interpretation that would yield a reasonable result (a whole number for all numbers of stamps involved in those transactions).

Liam and Noah shared a collection of 540 vintage stamps."
Apparently Noah has a number N of stamps, and Liam has a number L of stamps, and L%2BN=540

The third sentence seems clear, and we can translate it into an equation that leads us to
highlight%28x=%283%2F5%29N%29 for the number of stamps Noah sold.
The second sentence is a puzzle:
"Liam sold 2/5 fewer stamps than Noah." What could that mean?
A) I would think it could mean that the difference x-y is %282%2F5%29x
B) It could mean that the number of stamps Liam sold is y=%282%2F5%29x
C) It could mean that while Noah sold 3%2F5 of his stamps,
the ratio of the number of stamps Liam sold to the number he originally had is
3%2F5-2%2F5=1%2F5

I tried all three approaches and failed to find a whole number for all numbers of stamps involved.
However, if you do not calculate all numbers involved, you can find that Noah could start with N=324 stamps and Liam with L=216 stamps, from there on all the numbers of stamps would be fractional, but if Liam sells 1%2F5 of his stamps the non-integer numbers of remaining Liam's stamps and total stamps left are in exactly a 4%2F7 ratio.

The numbers of stamps left are %284%2F5%29L=4L%2F5 for Liam
and %282%2F5%29%28540-L%29=%281080-2L%29%2F5 for Noah
The total unsold is
4L%2F5%2B%281080-2L%29%2F5=%281080%2B2L%29%2F5
The ratio of Liam's unsold to total unsold is 4L%2F5%29%2F%28%281080%2B2L%29%2F5%29=4L%2F%281080%2B2L%29
All we have to do is solve 4L%2F%281080%2B2L%29=4%2F7 without thinking about all the transactions.
4L%2F%281080%2B2L%29=4%2F7-->L%2F%281080%2B2L%29=1%2F7-->7L%2F%281080%2B2L%29=1-->7L=1080%2B2L%29-->5L=1080%29-->L=1080%2F5%29-->highlight%28L=216%29

Answer by josgarithmetic(39835) About Me  (Show Source):
You can put this solution on YOUR website!
Only checked the description and question carefully, followed stepwise and made two equations but
have not yet tried to solve them.
                Noah          Liam              TOTAL

START             n           540-n              540


SOLD SOME x       n-x         (540-n)-x+2x/5
Noah

Noah found that n-x=%282%2F3%29x.

Another described post-sale condition
%28540-n%29-x%2B%282%2F5%29x=%284%2F7%29%28n-x%2B%28540-n%29-x%2B%282%2F5%29x%29

Answer by greenestamps(13362) About Me  (Show Source):
You can put this solution on YOUR website!


FAULTY DATA!!

Choose a variable:
x = number Noah sold

The number Noah had left was 2/3 the number he sold:
(2/3)x = number Noah had left

Liam sold "2/5 fewer" than Noah -- i.e., he sold 3/5 AS MANY as Noah:
(3/5)x = number Liam sold

Then...

x + (3/5)x = (8/5)x = total sold

540-(8/5)x = total remaining

Liam's remaining stamps were 4/7 of the total remaining, so Noah's remaining stamps were 3/7 of the total remaining:

%282%2F3%29x=%283%2F7%29%28540-%288%2F5%29x%29

multiply by 21 to clear fractions

14x=9%28540-%288%2F5%29x%29
14x=4860-%2872%2F5%29x
70x=24300-72x
142x=24300

x = 24300/142 = 171 9/71

Obviously the answer should be a positive integer....



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 3 solutions by greenestamps, ikleyn, KMST:
Answer by greenestamps(13362) About Me  (Show Source):
You can put this solution on YOUR website!


At the end, there were 42 chocolate donuts left, and the ratio of glazed to chocolate was 2:7. Use the proportion 2:7 = x:42 to find that there were 12 glazed donuts left.

In the morning, 3/5 of the glazed donuts were sold, so the fraction remaining was 2/5.

In the afternoon, 5/6 of the remaining glazed donuts were sold, so 1/6 of them remained. The fraction of glazed donuts left was then 1/6 of 2/5, or 1/15.

The 12 remaining glazed donuts were 1/15 of the original number, so the original number of glazed donuts was 15*12 = 180.

But there is no information given that allows us to determine the original number of chocolate donuts.

ANSWER: not enough information


Answer by ikleyn(53930) 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(5391) 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(13362) 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(53930) 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(39835) 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(10853) 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(10853) 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(5391) 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(10853) 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(53930) 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(39835) 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(13362) 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(13362) 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(10853) 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(53930) 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(39835) 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(53930) 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(53930) 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(53930) 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(53930) 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(53930) 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(2264) 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):**
$$\text{Percentage Less} = \frac{\text{Difference}}{\text{Men's Age}} \times 100\%$$
$$\text{Percentage Less} = \frac{2.1}{29.2} \times 100\%$$
$$\text{Percentage Less} \approx 0.071917 \times 100\%$$
$$\mathbf{\approx 7.19\%}$$


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