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Tough word problem - dividing $
did it the long way:
4+ (x-4)/2 + 4 + [(x - 8 - (x-4)/2)] /3 + 32 = x
40 + (x-4)/2 + x-12/6 = x
40 + x/2 - 2 + x/6 -2 = x
36 = x/3
x= 108
Bob = 4 + [108 - 8 - (108-4)/2] / 3 = 20
4+ (x-4)/2 + 4 + [(x - 8 - (x-4)/2)] /3 + 32 = x
40 + (x-4)/2 + x-12/6 = x
40 + x/2 - 2 + x/6 -2 = x
36 = x/3
x= 108
Bob = 4 + [108 - 8 - (108-4)/2] / 3 = 20
i love this question, i solved it correctly but i guess my method is too long, so i have adjusted to the simpler ones posted on here. thanks guys at least am on track to making it big.
i want to know it all i am ready to learn it all
logitech wrote:A) 4 + (x-4)/2
B) 4 + 1/3 [(X-12)/2]
C) 2/3 [(X-12)/2] = 32 ; (x-12) = 3 x 32
SO 4 + 1/3 [(X-12)/2] = 4+ 1/3 ( 3x32/2) = 20
I did not understand how (B) got 4 + 1/3 [(X-12)/2] .. Please help me understand
Here's what I thought
A)(x+4)/2 : Remaining (x-4)/2
//B first gets 4 from the Remaining amount and then one-third of (Remaining-4)
B)1/3 [(x-12)/2] <--this is one-third of (Remaining-4)
C)32
now A+B+C = x
I got x = 96 and B as 14, which is not even in the option.
I know my calculation is wrong for B .. Please suggest ..
duh!! got it .. anywaz thanks ..
Ann -A
Bob -B
Chole -C
Let the Total of money = X
As per the condition, A's total = 4+ (x-4)/2
Amount remaining = x - (x+4)/2 = (x-4)/2
B's total = 4+ ([ (x-4)/2 -4] /3 = 4+ (x-12)/6 = (x+12)/6
Amount remaining = (x-4)/2 - [(x+12)/6] = (x-12)/3 =32 (given)
X= 108
Therefore, B's share = 20
Bob -B
Chole -C
Let the Total of money = X
As per the condition, A's total = 4+ (x-4)/2
Amount remaining = x - (x+4)/2 = (x-4)/2
B's total = 4+ ([ (x-4)/2 -4] /3 = 4+ (x-12)/6 = (x+12)/6
Amount remaining = (x-4)/2 - [(x+12)/6] = (x-12)/3 =32 (given)
X= 108
Therefore, B's share = 20
Your equations all seem way to complicated. Here's how I solved it.
Chloe's $32 is 2/3 of "what remains," because Bob took 1/3 of "what remained" + 4.
So I did 32 = 2/3x and solved for x. x = 48. So Bob got 1/3 of that, plus 4 = 20.
My brain was frozen and I panicked, as usual on this question. But when I took a deep breath, slapped myself in the face a couple of times, and then looked at it more reaonably, I saw this and it was really easy to solve. It's strange, few of these are hard to solve, it's just that the test questions and format is such that you don't see how to set it up. once you see what it's saying, they're usually rather simple.
Chloe's $32 is 2/3 of "what remains," because Bob took 1/3 of "what remained" + 4.
So I did 32 = 2/3x and solved for x. x = 48. So Bob got 1/3 of that, plus 4 = 20.
My brain was frozen and I panicked, as usual on this question. But when I took a deep breath, slapped myself in the face a couple of times, and then looked at it more reaonably, I saw this and it was really easy to solve. It's strange, few of these are hard to solve, it's just that the test questions and format is such that you don't see how to set it up. once you see what it's saying, they're usually rather simple.
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Deependra1
- Senior | Next Rank: 100 Posts
- Posts: 53
- Thanked: 1 times
huy danaj i solved this is problem but it took me atleast three minutes to solve this question.my approach was as follows:i started the question by back solving supposing for option D bob received 26 dollars then the amount would have been 4$+22$ and before 1/3rd of amount 66 remaining the amount would have been 66+4=70$ and the earlier amount would have been 144$ as 4$+half of amount remaining that is 144-4=140/2=70$ so this way the amount of chloe remaining is not 32$ but in option A that is 20$ the amount would have been 4$+16$ which is 20$ and 1/3rd of 48 would give 16$ so the earlier amount would have been 4$+52$ which is 1/2 of 104 so the original amount would be 108.thus the amount with chloe comes to be 108-56-20 to be 32 which proves the answer to be A option as money with bob
A = 4+1/2(X-4). What remains after A's share is X-A = (X-4)/2anirudhbhalotia wrote:logitech wrote:A) 4 + (x-4)/2
B) 4 + 1/3 [(X-12)/2]
C) 2/3 [(X-12)/2] = 32 ; (x-12) = 3 x 32
SO 4 + 1/3 [(X-12)/2] = 4+ 1/3 ( 3x32/2) = 20
If X is the total amount.
Ann : 4 + (x-4)/2 (Ann receives $4 plus one-half of what remains)
Bob : 4 + 1/3(x- Share of Ann) (Bob receives $4 plus one-third of what remains)
Bob: 4+1/3[x-(4+(x-4)/2)]
Isn't it?
How did you get 4 + 1/3 [(X-12)/2] ?
B = 4+1/3((X-4)/2 - 4) = 4+1/3(X-12)/2. What remains after B's share is A-B = (X-12)/3
C = 32 = (X-12)/3 ; X = 108
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jmckenna14
- Newbie | Next Rank: 10 Posts
- Posts: 6
Nice work, Logitech
My solution is below. Note that we don't need to consider Ann's portion in the solution. We can just let K be the money remaining after Ann has received her portion and go from there.
Our equation will use the fact that, once we remove Bob's portion, we have $32 for Chloe.
So, we get K - Bob's $ = 32
The equation is K-4 - (K-4)/3 = 32 <<<< how did you come up with this equation? I am having a hard time trying to translate the word problem into alegbraic form. Any rules of thumb you use with these type of problems?
Solve for K (K=52) and then determine Bob's portion ($20).
The answer is, indeed, A
My solution is below. Note that we don't need to consider Ann's portion in the solution. We can just let K be the money remaining after Ann has received her portion and go from there.
Our equation will use the fact that, once we remove Bob's portion, we have $32 for Chloe.
So, we get K - Bob's $ = 32
The equation is K-4 - (K-4)/3 = 32 <<<< how did you come up with this equation? I am having a hard time trying to translate the word problem into alegbraic form. Any rules of thumb you use with these type of problems?
Solve for K (K=52) and then determine Bob's portion ($20).
The answer is, indeed, A
