SPLIT - Ratio question help

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Zw2021

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hello, I have a silly doubt but I hope it get resolved?
I am currently learning ratio and proportion and I have observed that when we multiply the ratio with an LCM we sometimes take a common LCM and use it only one. Let me give you two examples.

Question 1: In a company, three engineers earn as much as five clerks. and 8 clerks earn as much as 5 accountants. If the total earnings of 4 engineers, 6 clerks and 3 accountants is 7,86,00. Find the total earning of 1 engineer, 1 clerk and 1 accountant.

Answer: 3E=5C (E=3/5C) 8C=5A (8/5C)
4E+6C+3A = 786000
4* 3/5C + 6C +3* 8/5C = 786000
(20/3 + 6 + 24/5) C =786000
(100 + 90 + 72/ 15) = 786000 (my question refers to this line. please remember.
and the sum goes on



Question 2 . A bag contains coins of denomination Rs. 1, 2 and 5. The value of all Rs. 1, 2 and 5 Rs coins is in the ratio 3:5:4. If the total no. of coins is 252, Find the number of rs 5 coins.

Answer: no. of coin = (3/1) : (5/2): (4/5) (No of coins= net value/Denomination)
(3/1) : (5/2) : (4/5) * 10 (my question refers to this line. please remember)
30:25:8

Now my question is that in first case why do we make the denominator common but in the second case we multiply it only once and don't write 10 in the denominator?

if you need any more information regarding the question please let me know.
 
Zw2021 said:
hello, I have a silly doubt but I hope it get resolved?
I am currently learning ratio and proportion and I have observed that when we multiply the ratio with an LCM we sometimes take a common LCM and use it only one. Let me give you two examples.

Question 1: In a company, three engineers earn as much as five clerks. and 8 clerks earn as much as 5 accountants. If the total earnings of 4 engineers, 6 clerks and 3 accountants is 7,86,00. Find the total earning of 1 engineer, 1 clerk and 1 accountant.

Answer: 3E=5C (E=3/5C) 8C=5A (8/5C)
4E+6C+3A = 786000
4* 3/5C + 6C +3* 8/5C = 786000
(20/3 + 6 + 24/5) C =786000
(100 + 90 + 72/ 15) = 786000 (my question refers to this line. please remember.
and the sum goes on



Question 2 . A bag contains coins of denomination Rs. 1, 2 and 5. The value of all Rs. 1, 2 and 5 Rs coins is in the ratio 3:5:4. If the total no. of coins is 252, Find the number of rs 5 coins.

Answer: no. of coin = (3/1) : (5/2): (4/5) (No of coins= net value/Denomination)
(3/1) : (5/2) : (4/5) * 10 (my question refers to this line. please remember)
30:25:8

Now my question is that in first case why do we make the denominator common but in the second case we multiply it only once and don't write 10 in the denominator?

if you need any more information regarding the question please let me know.
Click to expand...
You need to rewrite both solutions fixing typos, etc. Otherwise it's hard to understand what you are asking about.
E.g. how do we go from
(20/3 + 6 + 24/5) C =786000
to
(100 + 90 + 72/ 15) = 786000 ?

Where did C disappear? How did 20/3, 6, 24/5 became 100, 90, 72/15???
 
You went

[MATH]3e = 5c \implies e = (5/3)c \ne (3/5)c.[/MATH] First error.

[MATH]5a = 8c \implies a = (8/5)c \ne (5/8)c.[/MATH] Second error..

[MATH]4e + 6c + 3a = 786,000 \implies 4 * (5/3)c + 6c + 3 * (8/5)c \ne 4 * (3/5)c + 6c + 3 * (5/8)c.[/MATH] Third error

So you started horribly. Errors everywhere.

Then you went [MATH]4 * (3/5)c = (12/5)c \ne (20/3)c. [/MATH] Fourth error. But it negated your first error and half the third error.

And [MATH]3 * (5/8)c = (15/8)c \ne (24/5)c.[/MATH] Fifth error. But it negated your second error and the rest of the third error.

So by great good luck you ended up with the correct equation

[MATH](20/3)c + 6c + (24/5)c = 786000 \implies ((20/3) + 6 + (24/5))c = 786000.[/MATH]
From here you concluded that

[MATH]((20/3) + 6 + (24/5))c = 786000 \implies (100 + 90 + 72/15) = 786000 \implies 195 \approx 786000.[/MATH] Sixth error. You forgot c. Perhaps that was just a typo. I’ll suppose it was and what you really meant was

[MATH]((20/3) + 6 + (24/5))c = 786000 \implies (100 + 90 + 72/15)c = 786000.[/MATH] Seventh error. Let’s do it correctly.

[MATH]((20/3) + 6 + (24/5))c = 786000 \implies 15\{(20/3) + 6 + (24/5)\}c = 15 * 786000 \implies [/MATH]
[MATH](100 + 90 + 72)c = 11,790,000 \implies 262c = 11,790,000 \implies c = \dfrac{11,790,000}{262} = 45,000.[/MATH]
Let’s check.

[MATH]\therefore e = \dfrac{5}{3} * 45,000 = 75,000.[/MATH]
[MATH]a = \dfrac{8}{5} * 45,000 = 72000.[/MATH]
[MATH]4 * 75000 + 6 * 45000 + 3 * 72000 = 300000 + 270000 + 216000 = 786000 \ \checkmark.[/MATH]
So what you were asking about was an ERROR in your work. Now if you copied it from a book, I strongly suspect that you copied

(100 + 90 + 72)/15 as 100 + 90 + 72/15, which is still yet another error
 
Hey, looks like I've made a lot of typos. Yes @JeffM I've copied this from an exam paper because I did not understand the calculation. You have multiplied 15 on both the sides, 15{(20/3)+6+(24/5)}c=15∗786000

but in the answer key of the exam 15 is multiplied only on the left hand side of the equation.

I'll correct the typos now.
For the first question

3E=5C therefore E= 5/3C
8C=5A therefore A=8/5C


ACCORDING TO THE QUESTION,
4E + 6C + 3A = 786000
substituting the values of E and A

4 * 5/3 C + 6C + 3* 8/5 C = 786000

(20/3 + 6 + 24/5 ) C = 786000
Taking LCM as 15
(100 + 90 + 72)/ 15 C = 786000. -------------------- (here denominators are made common, right? also as I have mentioned earlier, the answer key does not multiply 786000 by 15) ​

262/15 C = 786000
therefore, C = 45000.

if C= 45,000 then

E= 5/3 * 45000 = 75000
A= 8/5 * 45000 = 72000






FOR QUESTION NO.2

ANSWER

Net Value 3 : 5 : 4
Denomi- 1 : 2 : 5
nation

(Net value = NO. of coins * denomination.
Therefore No. of coins = net Value / denomination) -----------(this is a formula for this sum. This shows the relation between net value, no of coins and denomination)

​

No of coins = (3/1) : (5/2) : (4/5)
Multiplying by 10
252 = (3/1 : 5/2 : 4/5) * 10 ------------------- (here too only one side of the equation is multiplied and they did not take 10 as denominator which was done in case of 15 )​
252 = 30 : 25 : 8

diving 252 in the ratio 30 : 25 : 8 we get

120 coins of Rs. 1
100 coins of Rs. 2
and 32 coins of Rs. 5




So my question is in question 1 why was a denominator taken but in question 2 the denominator was cancelled ?
 
Zw2021 said:
Hey, looks like I've made a lot of typos. Yes @JeffM I've copied this from an exam paper because I did not understand the calculation. You have multiplied 15 on both the sides, 15{(20/3)+6+(24/5)}c=15∗786000

but in the answer key of the exam 15 is multiplied only on the left hand side of the equation.

I'll correct the typos now.
For the first question

3E=5C therefore E= 5/3C
8C=5A therefore A=8/5C


ACCORDING TO THE QUESTION,
4E + 6C + 3A = 786000
substituting the values of E and A

4 * 5/3 C + 6C + 3* 8/5 C = 786000

(20/3 + 6 + 24/5 ) C = 786000
Taking LCM as 15
(100 + 90 + 72)/ 15 C = 786000. -------------------- (here denominators are made common, right? also as I have mentioned earlier, the answer key does not multiply 786000 by 15) ​

262/15 C = 786000
therefore, C = 45000.

if C= 45,000 then

E= 5/3 * 45000 = 75000
A= 8/5 * 45000 = 72000






FOR QUESTION NO.2

ANSWER

Net Value 3 : 5 : 4
Denomi- 1 : 2 : 5
nation

(Net value = NO. of coins * denomination.
Therefore No. of coins = net Value / denomination) -----------(this is a formula for this sum. This shows the relation between net value, no of coins and denomination)

​

No of coins = (3/1) : (5/2) : (4/5)
Multiplying by 10
252 = (3/1 : 5/2 : 4/5) * 10 ------------------- (here too only one side of the equation is multiplied and they did not take 10 as denominator which was done in case of 15 )​
252 = 30 : 25 : 8

diving 252 in the ratio 30 : 25 : 8 we get

120 coins of Rs. 1
100 coins of Rs. 2
and 32 coins of Rs. 5




So my question is in question 1 why was a denominator taken but in question 2 the denominator was cancelled ?
Click to expand...
1. Yes, that's common denominator. Why would we do anything to the right side if we didn't change the value of the left side?
2. Ratio's components can be multiplied or divided by the same number - it will remain the same. E.g. ratio 1:2 is the same as 3:6. I think using an equality in that case is wrong.
 
Hey @lev888 I understood the first point. But can you please explain the second point again.
Are you saying that not multiplying the 15 and 10 (question 1 and 2 respectively) on both sides is wrong?
 
In beginning algebra, we tend to teach as fundamental principles things like

[MATH]a(b + c) = ab + ac.[/MATH] (1)

[MATH]a = b \implies ca = cb.[/MATH] (2)

[MATH]\dfrac{1}{a} * b = b / a.[/MATH] (3)

But there are other fundamental principles that we do not stress to beginning students. Among those are

[MATH]1 * a = a.[/MATH] (4)

[MATH]b \ne 0 \implies \dfrac{b}{b} = 1.[/MATH] (5)

I think we do not stress them because the student previously learned them in arithmetic.

Look at the following chain of logic

[MATH]\text {GIVEN: } \left ( \dfrac{20}{3} + 6 + \dfrac{24}{5} \right )c = 786000.[/MATH]
[MATH]\therefore 1 * \left ( \dfrac{20}{3} + 6 + \dfrac{24}{5} \right )c = 1 * 786000 \text { by principle 2.}[/MATH]
[MATH]\therefore 1 * \left ( \dfrac{20}{3} + 6 + \dfrac{24}{5} \right )c = 786000 \text { by principle 4}.[/MATH]
[MATH]\therefore \dfrac{15}{15} * \left ( \dfrac{20}{3} + 6 + \dfrac{24}{5} \right )c = 786000 \text { because } 1 = \dfrac{15}{15} \text { by principle 5.}[/MATH]
[MATH]\therefore \dfrac{1}{15} * \left ( \dfrac{20 * 15}{3} + 6 * 15 + \dfrac{24 *15}{5} \right )c = 7860000 \text { by principle 1.}[/MATH]
[MATH]\therefore \dfrac{1}{15} * (100 + 90 + 72)c = 786000 \text { by arithmetic simplification.}[/MATH]
[MATH]\therefore ((100 + 90 + 72)/15) = 786000 \text { by principle 3.}[/MATH]
In essence, your text multiplied (20/3 + 6 + 24/5) by 15/15, which changed nothing, but simplified it to
(100 + 90 + 72)/15 without explaining the detailed steps of what the text was doing.

Frankly, I hate this way of solving the problem. Get rid of fractions as soon as possible. It is so easy to make mistakes with fractions.

[MATH]4 * \dfrac{5}{3} * c + 6c + 3 * \dfrac{8}{5} * c = 786000 \implies[/MATH]
[MATH]15(4 * \dfrac{5}{3} * c + 6c + 3 * \dfrac{8}{5} * c) = 15 * 786000 \implies[/MATH]
[MATH]100c + 90c + 72c = 11790000 \implies [/MATH]
[MATH]262c = 11790000 \implies[/MATH]
[MATH]c = 45000 [/MATH].

Easy. Clear fractions right away.
 
Last edited:
JeffM said:
In essence, your text multiplied (20/3 + 6 + 24/5) by 15/15, which changed nothing, but simplified it to
(100 + 90 + 72)/15 without explaining the detailed steps of what the text was doing.
Click to expand...
You might be right, but I see it as simply adding the fractions, not multiplying by 15/15:
`(20/3 + 6 + 24/5 ) C = 786000 `
`(100/15 + 90/15 + 72/15) C = 786000 ` <-- this step was skipped
`(100 + 90 + 72)/15 C = 786000 `
 
Zw2021 said:
Hey @lev888 I understood the first point. But can you please explain the second point again.
Are you saying that not multiplying the 15 and 10 (question 1 and 2 respectively) on both sides is wrong?
Click to expand...
We are dealing with completely different concepts in 1 and 2.
1. This is an equation. I don't see any multiplying by 15 or 10 there. We simply add the fractions on the left, that's it.
2. We are given a ratio: (3/1) : (5/2) : (4/5) We can multiply each component by a number, the ratio doesn't change. Do you understand that 1:2 is the same as 3:6? They multiplied by 10 to simplify things, this step is not necessary.
You can do this:
That ratio means we can express the numbers of coins as (3/1)x, (5/2)x, (4/5)x.
Now we can set up an equation:
252 = (3/1)x + (5/2)x + (4/5)x
It's incorrect to write 252 = 30 : 25 : 8. You can't equate the number of coins on the left to a ratio on the right.
Back to the equation:
252 = (3/1)x + (5/2)x + (4/5)x
252 = (3+5/2+4/5)x
252 = (30/10+25/10+8/10)x
252 = (63/10)x
x = 40
Substitute 40 for x in the 3 expressions above, we get 120, 100, 32 coins.
 
lev888 said:
You might be right, but I see it as simply adding the fractions, not multiplying by 15/15:
`(20/3 + 6 + 24/5 ) C = 786000 `
`(100/15 + 90/15 + 72/15) C = 786000 ` <-- this step was skipped
`(100 + 90 + 72)/15 C = 786000 `
Click to expand...
@Lev I am not disagreeing with you. There are multiple chains of logic that get you to the same point. My point is that we frequently skip steps in the chain because we assume the student will fill in the missing links. For example, what axioms or theorems justify your assertion that

(20/3 + 6 + 24/5) = (100/15 + 90/15 + 72/15)

The actual chain of logic, using fundamental principles, is

[MATH]\left ( \dfrac{20}{3} 6 + \dfrac{24}{5} \right ) = \left ( \dfrac{20}{3} + 6 + \dfrac{24}{5} \right ) * 1 \text { because } a = a *1.[/MATH]
[MATH]\left ( \dfrac{20}{3} + 6 + \dfrac{24}{5} \right ) * 1 = \left ( \dfrac{20}{3} + 6 + \dfrac{24}{5} \right ) * \dfrac{3 * 5}{3 * 5} \text {because } b \ne 0 \implies \dfrac{b}{b} = 1.[/MATH]
[MATH]\left ( \dfrac{20}{3} + 6 + \dfrac{24}{5} \right ) * \dfrac{3 * 5}{3 * 5} = \left ( \dfrac{20 * \cancel 3 * 5}{\cancel 3 * 3 * 5} + \dfrac{6 * 3 * 5}{3 * 5} + \dfrac{24 * 3 * \cancel 5}{\cancel 5 * 3 * 5} \right ) \text { because of distribution.}[/MATH]
[MATH]\left ( \dfrac{20 * \cancel 3 * 5}{\cancel 3 * 3 * 5} + \dfrac{6 * 3 * 5}{3 * 5} + \dfrac{24 * 3 * \cancel 5}{\cancel 5 * 3 * 5} \right ) = \left ( \dfrac{100}{15} + \dfrac{90}{15} + \dfrac{72}{15} \right ) \text { because of arithmetic simplification.}[/MATH]
Now no one sane goes through all those steps, but when students are confused by an explanation, I have frequently found that confusion to be due to the student‘s not seeing all the steps left out. Of course, I would not do this problem the way this student’s text did in the first place. I would get rid of fractions at the beginning and avoid all of this.
 
JeffM said:
@Lev I am not disagreeing with you. There are multiple chains of logic that get you to the same point. My point is that we frequently skip steps in the chain because we assume the student will fill in the missing links. For example, what axioms or theorems justify your assertion that

(20/3 + 6 + 24/5) = (100/15 + 90/15 + 72/15)
Click to expand...
Yes, there are many ways to do it. I chose mine because 1. The solution appeared to use it and 2. The student did say: "in first case why do we make the denominator common", so it made sense to emphasize the point that we simply added the fractions by converting to common denominator, which didn't change the left side. Therefore, it was not necessary to multiply anything by anything.
 
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