Ratios: Given X : 3 : (9/2) = (15/4) : (9/2) : Y find the value of X and of Y?

[zafar ali]

Ratios: Given X : 3 : (9/2) = (15/4) : (9/2) : Y find the value of X and of Y?
How i can solve it? Is anyone have solution?

 

[Steven G]

What do you multiply 3 by to get 9/2?
Then multiply this number by x to get 9/2. Now solve for x
Then multiply 9/2 by this number to get y

 

[Harry_the_cat]

What do you multiply 3 by to get 9/2?
Then multiply this number by x to get 9/2. Now solve for x. Do you mean 15/4 here?
Then multiply 9/2 by this number to get y

see comment

 

[Harry_the_cat]

You could break it into 2 problems:
X : 3 = (15/4) : (9/2)
and
3 : (9/2) = (9/2) : Y

The first one can be written as:
$\displaystyle \frac{X}{3} =\frac{\frac{15}{4}}{\frac{9}{2}}$

$\displaystyle \frac{X}{3} = \frac{15}{4}*\frac{2}{9}$

$\displaystyle X = \frac{15}{4}*\frac{2}{9}*3$

Over to you to finish it.

 

[Steven G]

see comment

Yep. Thanks

 

[Denis]

Yep. Thanks

One of these days, you'll do something correctly!

 

[HallsofIvy]

What "x : 3: 9/2= 15/4: 9/2: y" means is that you can go from x to 15/4, from 3 to 9/2, and from 9/2 to y by multiplying each number on the left by the same number. Since the "middle", "3 to 9/2", are known we can set that the multiplier is (9/2)/3= 3/2. x times 3/2= 15/4 so x= (15/4)/(3/2)= (15/4)(2/3)= 5/2. And y= (3/2)(9/2)= 27/4.

 

[Steven G]

One of these days, you'll do something correctly!

One of these days, you'll do something correctly!

I do something correctly all the time. By giving you an angry face.
???????????? Do you see a pattern?

 

[engineerali]

What "x : 3: 9/2= 15/4: 9/2: y" means is that you can go from x to 15/4, from 3 to 9/2, and from 9/2 to y by multiplying each number on the left by the same number. Since the "middle", "3 to 9/2", are known we can set that the multiplier is (9/2)/3= 3/2. x times 3/2= 15/4 so x= (15/4)/(3/2)= (15/4)(2/3)= 5/2. And y= (3/2)(9/2)= 27/4.

You have solved it and I get it as Engineer but for students its not the ideal way (how they invented what you did) . Here is how it should work(For students).
First remove "y" and its corresponding number to find "x" then remove "x" and its corresponding number to get "y".
When we remove "y" we also remove "9/2" corresponding number as it is equivalent so equation becomes
x:3=15/4:9/2
x÷3=15/4÷9/2
x/3=15/4×2/9
x/3=5/6(after solving above right hand side)
x=5/6×3
x=5/2 or 2-1/2 (in proper fraction)
Now again
When we remove "x" we also remove "15/4" corresponding number as it is equivalent so equation becomes
3:9/2=9/2:y (ratio means ÷)
3÷9/2=9/2÷y
3×2/9=9/2×1/y
6/9=9/2y
Cross multiply
6(2y)=9×9
12y=81
÷ both sides by 12
y= 81/12(table of 3 used to cut fractions. We use HCF to find common too)
y=27/4 or 6-3/4(proper fraction).
Now isnt it v easy to teach but i was student like you did and trust me only 1% can understand what you did. And you showed why its done with how its done. To me its harsh on kids lol
I hope you agree
Adios

 

[Dr.Peterson]

Your method is essentially the same as Harry's post #4, except not explained as clearly, and not leaving work for the student to do (which could be done better than you did).

There are many ways to solve this; exposure to multiple methods can be beneficial. I could list a couple more very different approaches, but would not claim that any one of them is the only way to teach it.

We have no idea what level the OP is at, or what method would be most helpful. I wouldn't assume Zafar is in your 1%, or is not. This is part of the reason we want students to show some work, in order to get an idea of what methods they are learning, and where they are misunderstanding.

By the way, never write "6-3/4", which makes it look like a subtraction; a mixed number (6 3/4) is an addition.

 

[engineerali]

³

Your method is essentially the same as Harry's post #4, except not explained as clearly, and not leaving work for the student to do (which could be done better than you did).

There are many ways to solve this; exposure to multiple methods can be beneficial. I could list a couple more very different approaches, but would not claim that any one of them is the only way to teach it.

We have no idea what level the OP is at, or what method would be most helpful. I wouldn't assume Zafar is in your 1%, or is not. This is part of the reason we want students to show some work, in order to get an idea of what methods they are learning, and where they are misunderstanding.

By the way, never write "6-3/4", which makes it look like a subtraction; a mixed number (6 3/4) is an addition.

If you see i wrote proper fraction even though 6-3/4 might look odd but i used improper fraction besides it i.e. 27/4 then wrote "or" its done intentionally to not confuse answer "-" here was obviously meant for space as you presumed.
I admired his approached but this Question is from Oxford book D1 book 1 taught in Grade 7.
Now Dr. Let me tell you as what he did was that dividing the knowns meant the corresponding numbers or variables when divided or multiplied gave same answer depending on whether they are on LHS or RHS. That is art of balancing equation.
If you see Zafar was new member and posted once or twice then didn't come on forum. I just elaborated on what needed to be done. To me I provided complete solution. Obviously explanation here is impossible despite Harry doing it but i took it little further(I somehow skipped Harry's reply).
Considering Grade 7 and Trust me these questions can be solved in more different ways but method will be same to balance both sides and taking corresponding values.
Thanks for grilling me.
P.S:- The "." at end of sentence is a Dot not a decimal point
Good Day