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Analyzing structure word problem: pet store (2 of 2)

Sal solves a word problem about the unknown number of bears, cats, and dogs in a pet store. This is part 2 where Sal uses an algebraic reasoning. Created by Sal Khan.

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  • female robot ada style avatar for user katherinekc10
    When he works out b/c+d+b, couldn't he have plugged in numbers that fit the formula c>d>b?

    I used c=3, d=2, and b=1.

    b/c+d+b = 1/3+2+1 = 1/6

    1/6<1/3

    I have used this way before and it seems to work and be easier.
    Is his way a more full-proof way or is my way fine?
    (19 votes)
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    • starky seedling style avatar for user Cary Wang
      Well, this is an algebra problem. Sal expects you to solve it using algebraic expressions. That's the point of this video. Though your way of solving is very useful (I use it too), it is not an algebraic solution. So no, your way is not fine, at least in this case.
      (3 votes)
  • female robot grace style avatar for user Rey #FilmmakerForLife #EstelioVeleth.
    what does analytic mean?
    (5 votes)
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  • blobby green style avatar for user Ben Holt
    If the values are negative then this ratio is larger than one third.
    (2 votes)
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  • blobby green style avatar for user Ahmed Nazeer
    How did Sal know that d = 1/3 of the rectangle in his visual analysis?
    Couldn't d be a little less than 1/3 but greater than b?
    Could anyone explain? -- NAZ
    (2 votes)
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  • mr pants teal style avatar for user Tronax
    i think there is a bit easier way to think about it, in general. ill show it here.

    we are given that c > d > b. so, its clear that c > b and d > b
    1. if c is greater than b, than we could say that c = b + k, where k is positive number
    2. if d is greater than b, than we could say that d = b + j, where j is positive number

    now we can substitude c and d: b/(c+b+d) as b/(b+k + b + b+j)
    rearrange and sum the b's: b/(b+k + b + b+j) = b/(3b + k + j)

    and now its simple, we know that b/3b > b/(3b + k + j) because adding positive numbers to denominator lowers the output.
    b/3b = 1/3, so we get
    1/3 > b/(3b + k + j)
    substitude b and d back into denominator, and here is our solution:
    1/3 > b/(c+b+d)
    (3 votes)
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  • male robot johnny style avatar for user Mohamed Ibrahim
    the last reasoning that C+b is greater than 2b, it seems very reasonable and easy, but can we express that very simple reasoning by math or sometimes we have to write the logic in words ?
    (1 vote)
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    • female robot grace style avatar for user loumast17
      Do you mean x+d . 2b? Either way, it kind of already is an algebraic argument using inequalities. Basically we are given c>d>b from the start, so we can use that. so we can then use c+1 > b+1 and c+2>b+2 and keep goign for any number and eventually get c+b > b+b
      (2 votes)
  • blobby green style avatar for user Ahmed Nasret
    kindly judge this
    b/(b+c+d) = 1/ ( (d+c)/b) +1 )
    b,c and d should be integers thus min b is 1, min d is 2 and min c is 3
    hence: the denominator expression ((b+c)/d)+1 min value is 6
    the whole expression max value is 1/6, means the expression in subject must be lesser than or equal 1/6 which in all cases lesser than 1/3
    whatever, Sal's way is better, because it covers the case of non-integers values of a,b and c. such proposal only allow us to put min values and tackle the way like i did.
    (1 vote)
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  • aqualine seed style avatar for user Ben Miller
    I am a little confused about one of the practice problems. A>B>C, and it asks the relationship between B+C and 2A-B. The analytical solution says B+C<2A-B, but when I try it with real sets of numbers I get different results. Take {A,B,C} to be the sets {3,2,1} and {4,3,2} they produce different equality relationships.
    (1 vote)
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  • piceratops ultimate style avatar for user Agus
    i made it using infinity... much more complicated than Sal but here it goes.

    "b tends to infinity,
    as d>b, d tends to infinity+1,
    and as c>d>b, c tends to infinity+2
    so the denominator will have a bigger infinity than 3x(infinity)

    thus b/(b+c+d) will always be <1/3."

    is this reasoning correct? am i commiting conceptual mistakes?

    thanks in advance
    (1 vote)
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  • female robot grace style avatar for user Bella
    if a>d>b was 1000000002>1000000001>1000000000, then wouldn't 1000000000/3000000003=1/3?
    (0 votes)
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Video transcript

In the last video, we made a visual argument as to why this expression has to be less than 1/3, and this expression we already figured out is the fraction that are bears. Now we will make an algebraic argument, or I could call it an analytic argument. And to make this argument, I'm going to leave this expression-- we know this is the fraction that are bears-- and I'm going to write this 1/3 in a form that looks a lot like this, and then based on the information we have, we can directly compare them. So how can I write 1/3? Maybe with the b as a numerator. Well, 1/3 is the same thing as b over 3b, which is the exact same thing as b over b plus b plus b. So now, this is looking pretty similar. The only difference between this expression right over here, b over c plus d plus b and b over b plus b plus b is that our denominators are different. And the only difference in our denominators, this denominator has a c plus d here, while this has a b plus b over here. Now, we have to ask ourselves a question. What is larger? Is c plus d larger than b plus b? And I encourage you to pause that and think about that for a second. Well, yes. We already see right over here. It was given to us that c is greater than d that is greater than b, so both c and d are greater than b. So c plus d is definitely going to be greater than b plus b. So this denominator right over here is greater, so this has a larger denominator. This right over here has a smaller denominator. And since we know this has a larger denominator, this has a smaller denominator, they have the exact same numerator-- they both have b as a numerator-- we know that this whole thing must be a smaller quantity. If you have the same numerator but one expression has a larger denominator, it must be smaller. Wait, so how does that work? Well, just remember. I mean, just imagine. You have the same numerator, what's going to be bigger, a over 7 or a over 5? Well here, you're dividing a by 7. You're dividing into many more chunks than over here, so this right over here is smaller. This right over here is larger. So this is the larger. This right over here is smaller. So the same numerator, the larger the denominator, the smaller the quantity is going to be. So going back to the original question, this is the smaller quantity, and this right over here, 1/3, is the larger quantity.