Tricky Math Problem
May 17, 2020 15:59:18 #
polonois wrote:
Huh? I never came across that memory hint. I just remember what to do.Yes, remember "Please excuse my dear aunt sally."
May 17, 2020 16:00:42 #
OneShot1
Loc: Wichita, KS, USA
Shoe = 10, Boy = 5, Bouquet = 4
10 + [(5 +4+4 + 10+10) x 4] =
10 +[33 x 4] =
10 + 132 =
142 <-------
10 + [(5 +4+4 + 10+10) x 4] =
10 +[33 x 4] =
10 + 132 =
142 <-------
May 17, 2020 16:10:05 #
May 17, 2020 16:11:20 #
David Martin wrote:
43
i.e., 5 + [5+10+4] x 2 = 5 + 19 x 2 = 5 + 38 = 43
where [5+10+4] = boy wearing 2 shoes and holding 2 bouquets
i.e., 5 + [5+10+4] x 2 = 5 + 19 x 2 = 5 + 38 = 43
where [5+10+4] = boy wearing 2 shoes and holding 2 bouquets
That's what I finally got after posting the wrong number peviously. Missed the sneakers and popcorn
May 17, 2020 16:18:00 #
May 17, 2020 16:23:45 #
May 17, 2020 16:38:29 #
May 17, 2020 16:41:38 #
Leo_B
Loc: Houston suburb
Delderby wrote:
Interesting that computer spreadsheets will not agree with MDAS. Lack of brackets will regard the multiplier sign as a plus sign. My non-math logic would apply everything from left to right, with the result applied before moving onward. 
Well, that's how me and Jethro learned our ciphering back in 6th grade. Please My Dear Aunt Sally, working left to right in the process.
May 17, 2020 17:35:35 #
May 17, 2020 17:59:23 #
TheShoe
Loc: Lacey, WA
Half of a pair of shoes is worth less than half of the cost of a pair to most of us.
May 17, 2020 18:16:16 #
TheShoe
Loc: Lacey, WA
16
6A = 30; A = 5
2(A+B) = 20; B = 5
4C + B = 13; C = 2
X = A + B + 3C; X = 5 + 5 + 6; X = 16
6A = 30; A = 5
2(A+B) = 20; B = 5
4C + B = 13; C = 2
X = A + B + 3C; X = 5 + 5 + 6; X = 16
May 17, 2020 19:05:29 #
May 17, 2020 19:09:17 #
sepena
Loc: Fulltime RV Traveller
randave2001 wrote:
OK, I changed my mind. Here is what I come up with:
2 shoes = 10 therefore 1 shoe =5
Boy = 5
2 bouquets = 4 therefore 1 bouquet = 2
Final formula is:
1 shoe + (boy holding 2 bouquets and wearing 2 shoes) X 1 bouquet which would numerically be 5 + (5+4+10) X 2 or 5+19X2. Using the MDAS method of solving that would then be 5+38=43 (final answer).
2 shoes = 10 therefore 1 shoe =5
Boy = 5
2 bouquets = 4 therefore 1 bouquet = 2
Final formula is:
1 shoe + (boy holding 2 bouquets and wearing 2 shoes) X 1 bouquet which would numerically be 5 + (5+4+10) X 2 or 5+19X2. Using the MDAS method of solving that would then be 5+38=43 (final answer).
May 17, 2020 19:09:58 #
sepena
Loc: Fulltime RV Traveller
randave2001 wrote:
OK, I changed my mind. Here is what I come up with:
2 shoes = 10 therefore 1 shoe =5
Boy = 5
2 bouquets = 4 therefore 1 bouquet = 2
Final formula is:
1 shoe + (boy holding 2 bouquets and wearing 2 shoes) X 1 bouquet which would numerically be 5 + (5+4+10) X 2 or 5+19X2. Using the MDAS method of solving that would then be 5+38=43 (final answer).
2 shoes = 10 therefore 1 shoe =5
Boy = 5
2 bouquets = 4 therefore 1 bouquet = 2
Final formula is:
1 shoe + (boy holding 2 bouquets and wearing 2 shoes) X 1 bouquet which would numerically be 5 + (5+4+10) X 2 or 5+19X2. Using the MDAS method of solving that would then be 5+38=43 (final answer).
May 17, 2020 19:13:37 #
TheShoe wrote:
The boy is holding 2 bouquets and wearing 2 shoes, and MDAS still applies (enforced below with parentheses):16
6A = 30; A = 5
2(A+B) = 20; B = 5
4C + B = 13; C = 2
X = A + B + 3C; X = 5 + 5 + 6; X = 16
6A = 30; A = 5
2(A+B) = 20; B = 5
4C + B = 13; C = 2
X = A + B + 3C; X = 5 + 5 + 6; X = 16
X = A + ((B + 2A + 2C) x C) = 5 + ((5 + 10 + 4) x 2) = 5 + (19 x 2) = 5 + 38 = 43
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