Rs 10.18 is divided among 45 children. 1 boy gets 22p and...
- Thread starter Shaon
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Hello, Shaon!
Please check the wording of the problem.
. . There seem to be two typos.
\(\displaystyle \text{100 paisa }\:=\:\text{1 rupee}\)
\(\displaystyle \text{Let }\:\begin{array}{ccc}B &=& \text{number of boys} \\ G &=&\text{number of girls} \end{array}\)
\(\displaystyle \text{We have: }\:B + G \:=\:45\) .[1]
. . \(\displaystyle \text{and: }\:22B + 28G \:=\:1018 \quad\Rightarrow\quad 11B + 14G \:=\:509\) .[2]
\(\displaystyle \begin{array}{cccc}\text{Multiply [1] by -11:} & \text{-}11B - 11G &=&\text{-}495 \\ \text{Add [2]:} & 11B + 14G &=& 509 \end{array}\)
. . \(\displaystyle \text{and we have: }\:3G \:=\:14 \quad\Rightarrow\quad G \:=\:\frac{14}{3}\) .??
Please check the wording of the problem.
. . There seem to be two typos.
\(\displaystyle \text{100 paisa }\:=\:\text{1 rupee}\)
\(\displaystyle \text{Let }\:\begin{array}{ccc}B &=& \text{number of boys} \\ G &=&\text{number of girls} \end{array}\)
\(\displaystyle \text{We have: }\:B + G \:=\:45\) .[1]
. . \(\displaystyle \text{and: }\:22B + 28G \:=\:1018 \quad\Rightarrow\quad 11B + 14G \:=\:509\) .[2]
\(\displaystyle \begin{array}{cccc}\text{Multiply [1] by -11:} & \text{-}11B - 11G &=&\text{-}495 \\ \text{Add [2]:} & 11B + 14G &=& 509 \end{array}\)
. . \(\displaystyle \text{and we have: }\:3G \:=\:14 \quad\Rightarrow\quad G \:=\:\frac{14}{3}\) .??
Re:
My mother tried in algebraic method, which is not yet taught to us in my school, I need to do in unitary method or arithmetic method.
R = Rupees, P = Paise
1 Re = 100 paise
stapel said:How does an "R" relate to a "p"?Shaon said:
What have you tried? How far did you get? Where are you stuck?
Please be complete. Thank you!
Eliz.
My mother tried in algebraic method, which is not yet taught to us in my school, I need to do in unitary method or arithmetic method.
R = Rupees, P = Paise
1 Re = 100 paise
soroban said:
It was a typo error,
The problem is :
Rs 10.80 is divided among 45 children; I boy gets 22p and 1 girl gets 28p. How many boys and girls are there?
My mother tried to solve in algebraic method taking variables and x for no. of boys and y for no. of girls and the end result she got is 30 boys and 15 girls but she is unable to explain to me as Algebraic method is not yet taught to me in my class; I need to do in arithmetic method.
Hello, Shaon!
Now, it can be solved . . .
\(\displaystyle \text{Let }\:\begin{array}{ccc}B &=& \text{number of boys} \\ G &=&\text{number of girls} \end{array}\)
\(\displaystyle \text{We have: }\:B + G \:=\:45\) .[1]
. . \(\displaystyle \text{and: }\:22B + 28G \:=\:1080 \quad\Rightarrow\quad 11B + 14G \:=\:540\) .[2]
\(\displaystyle \begin{array}{cccc}\text{Multiply [1] by -11:} & \text{-}11B - 11G &=&\text{-}495 \\ \text{Add [2]:} & 11B + 14G &=& 540 \end{array}\)
. . \(\displaystyle \text{and we have: }\:3G \:=\:45 \quad\Rightarrow\quad \boxed{G \:=\:15}\)
Substitute into [1]: .\(\displaystyle B + 15 \:=\:45\quad\Rightarrow\quad\boxed{ B \:=\:30}\)
Therefore, there are 30 boys and 15 girls.
Now, it can be solved . . .
\(\displaystyle \text{Let }\:\begin{array}{ccc}B &=& \text{number of boys} \\ G &=&\text{number of girls} \end{array}\)
\(\displaystyle \text{We have: }\:B + G \:=\:45\) .[1]
. . \(\displaystyle \text{and: }\:22B + 28G \:=\:1080 \quad\Rightarrow\quad 11B + 14G \:=\:540\) .[2]
\(\displaystyle \begin{array}{cccc}\text{Multiply [1] by -11:} & \text{-}11B - 11G &=&\text{-}495 \\ \text{Add [2]:} & 11B + 14G &=& 540 \end{array}\)
. . \(\displaystyle \text{and we have: }\:3G \:=\:45 \quad\Rightarrow\quad \boxed{G \:=\:15}\)
Substitute into [1]: .\(\displaystyle B + 15 \:=\:45\quad\Rightarrow\quad\boxed{ B \:=\:30}\)
Therefore, there are 30 boys and 15 girls.
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You have 1080 paise.Shaon said:
Have the boys and girls pair off. (There may be a few "extra" of one or the other, but ignore that possibility for now. For 44/2 = 22 pairs, with one "somebody" left over.) How many paise does each pair get? What is this total for the twenty-two pairs? (Hint: The value should be "too high".)
Since girls get more than boys, kick out a girl and add a boy. That is, subtract 28 and add 22. What is the net difference?
For each swap (a boy in, for a girl out), you will have this same net reduction in the total.
Suppose the extra kid (from the originally-assumed 22 pairs, plus "somebody") was a girl. What is your total? Is it possible, by swaps, to reduce the total to the required 1080 paise? If so, then find the solution. If not, then assume that the "somebody" was a boy, find that total, and repeat the process. :wink:
Eliz.
Rs 10.80 is divided among 45 children. 1 boy gets 22p and 1 girl gets 28p. What is total number of boys and girls?
Just for your information, there is an alternate method of getting your answer.
1--From the stated problem, you can write 22B + 28G = 1080.
2--This reduces to 11B + 14G = 540
3--Dividing through by the lowest coefficient yields 1B + G + 3G/11 = 49 += 1/11.
4--(3G - 1)/11 must be an integer as does (12G - 4)/11
5--Divide through by 11 again yielding G + G/11 -4/11
6--(G - 4)/11 must be an integer k making G = 11k + 4
7--Substituting back into (2) yields B = 44 - 14k
8------k....0....1....2....3....4
.......B...44...30...16....2....-
.......G....4...15...26...37...48
Sum........48...45...42...39
9--Introducing the other piece of information from the problem statement, "B + G = 45", the answer is clearly 30 boys and 15 girls.
The method might take a bit more time but it does yield the proper solution for a given sum of boys and girls, in addition to solutions for varying sums of boys and girls. Just thought you might be interested in an alternate approach to problems of this type.
um
Just for your information, there is an alternate method of getting your answer.
1--From the stated problem, you can write 22B + 28G = 1080.
2--This reduces to 11B + 14G = 540
3--Dividing through by the lowest coefficient yields 1B + G + 3G/11 = 49 += 1/11.
4--(3G - 1)/11 must be an integer as does (12G - 4)/11
5--Divide through by 11 again yielding G + G/11 -4/11
6--(G - 4)/11 must be an integer k making G = 11k + 4
7--Substituting back into (2) yields B = 44 - 14k
8------k....0....1....2....3....4
.......B...44...30...16....2....-
.......G....4...15...26...37...48
Sum........48...45...42...39
9--Introducing the other piece of information from the problem statement, "B + G = 45", the answer is clearly 30 boys and 15 girls.
The method might take a bit more time but it does yield the proper solution for a given sum of boys and girls, in addition to solutions for varying sums of boys and girls. Just thought you might be interested in an alternate approach to problems of this type.
um
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- Feb 4, 2004
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Shaon said:
TchrWill said:
Yes, this can be solved in the alternative algebraic manner, as the poster's mother had already shown. The poster has specifically asked for "arithmetic" or "unitary" methods.soroban said:
I've shown an arithmetic method. Would somebody like to show the unitary method, perhaps explaining how to do this exercise? I mean, sure, the complete worked solution has been posted but, since it uses methods he cannot use, how does this help the student in the long term? And if he has to "show work", then cheating doesn't even help in the short term! :?
Eliz.
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Re:
I don't even know what a "unitary" method is.
Did a google search and learned that there is a unitary method for solving Schrodinger's Equation!!
http://stinet.dtic.mil/oai/oai?verb=get ... =ADA323262
Wow!!!
stapel said:
I don't even know what a "unitary" method is.
Did a google search and learned that there is a unitary method for solving Schrodinger's Equation!!
http://stinet.dtic.mil/oai/oai?verb=get ... =ADA323262
Wow!!!