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Simple Ratios Shortcut - Why does it work?
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I am preparing for a competitive examination which consists of quantitative aptitude. We came across a question on Ratios as follows:
Two numbers are in the ratio 4 : 5. If 7 is added to each, the ratio
becomes 5 : 6. Find the numbers.
Now to solve such questions, which have a given ratio, a number added to both the numbers in the ratio and the final ratio, our teacher suggested a simple short cut to save some time in the examination. It is as follows:
If the difference between the antecedent and the consequent of the ratio of the original numbers (here, ratio of original numbers = 4 : 5, difference = 1) is same as the difference between the antecedent and the consequent of the ratio of the final numbers (here, 5 : 6, difference = 1), then compare the difference (1) with the number added to the original numbers(here, 7).
Using that, proportionately find the number corresponding to the original ratio points.
E.g. here, difference = 1 and number added = 7.
So "if 7 corresponds to 1 ratio point" says he, "4 ratio points (the antecedent of original ratio) must correspond to 28 and 5 to 35. And in this way we get the original numbers as 28 and 35."
Can somebody please explain why/how this short cut works? Also would this work only in this specific case, or could it be applied to some other cases also, albeit with some modification? Also, he used the term "ratio point" to denote both the difference between the ante. and conse. of the ratio as well as the ante. and conse. of the ratios. Is the term legit?
ratio
asked Jul 26 at 18:07
Neil
32
add a comment |Â
up vote
0
down vote
favorite
I am preparing for a competitive examination which consists of quantitative aptitude. We came across a question on Ratios as follows:
Two numbers are in the ratio 4 : 5. If 7 is added to each, the ratio
becomes 5 : 6. Find the numbers.
Now to solve such questions, which have a given ratio, a number added to both the numbers in the ratio and the final ratio, our teacher suggested a simple short cut to save some time in the examination. It is as follows:
If the difference between the antecedent and the consequent of the ratio of the original numbers (here, ratio of original numbers = 4 : 5, difference = 1) is same as the difference between the antecedent and the consequent of the ratio of the final numbers (here, 5 : 6, difference = 1), then compare the difference (1) with the number added to the original numbers(here, 7).
Using that, proportionately find the number corresponding to the original ratio points.
E.g. here, difference = 1 and number added = 7.
So "if 7 corresponds to 1 ratio point" says he, "4 ratio points (the antecedent of original ratio) must correspond to 28 and 5 to 35. And in this way we get the original numbers as 28 and 35."
Can somebody please explain why/how this short cut works? Also would this work only in this specific case, or could it be applied to some other cases also, albeit with some modification? Also, he used the term "ratio point" to denote both the difference between the ante. and conse. of the ratio as well as the ante. and conse. of the ratios. Is the term legit?
ratio
asked Jul 26 at 18:07
Neil
32
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am preparing for a competitive examination which consists of quantitative aptitude. We came across a question on Ratios as follows:
Two numbers are in the ratio 4 : 5. If 7 is added to each, the ratio
becomes 5 : 6. Find the numbers.
Now to solve such questions, which have a given ratio, a number added to both the numbers in the ratio and the final ratio, our teacher suggested a simple short cut to save some time in the examination. It is as follows:
If the difference between the antecedent and the consequent of the ratio of the original numbers (here, ratio of original numbers = 4 : 5, difference = 1) is same as the difference between the antecedent and the consequent of the ratio of the final numbers (here, 5 : 6, difference = 1), then compare the difference (1) with the number added to the original numbers(here, 7).
Using that, proportionately find the number corresponding to the original ratio points.
E.g. here, difference = 1 and number added = 7.
So "if 7 corresponds to 1 ratio point" says he, "4 ratio points (the antecedent of original ratio) must correspond to 28 and 5 to 35. And in this way we get the original numbers as 28 and 35."
Can somebody please explain why/how this short cut works? Also would this work only in this specific case, or could it be applied to some other cases also, albeit with some modification? Also, he used the term "ratio point" to denote both the difference between the ante. and conse. of the ratio as well as the ante. and conse. of the ratios. Is the term legit?
ratio
asked Jul 26 at 18:07
Neil
32
I am preparing for a competitive examination which consists of quantitative aptitude. We came across a question on Ratios as follows:
Two numbers are in the ratio 4 : 5. If 7 is added to each, the ratio
becomes 5 : 6. Find the numbers.
Now to solve such questions, which have a given ratio, a number added to both the numbers in the ratio and the final ratio, our teacher suggested a simple short cut to save some time in the examination. It is as follows:
If the difference between the antecedent and the consequent of the ratio of the original numbers (here, ratio of original numbers = 4 : 5, difference = 1) is same as the difference between the antecedent and the consequent of the ratio of the final numbers (here, 5 : 6, difference = 1), then compare the difference (1) with the number added to the original numbers(here, 7).
Using that, proportionately find the number corresponding to the original ratio points.
E.g. here, difference = 1 and number added = 7.
So "if 7 corresponds to 1 ratio point" says he, "4 ratio points (the antecedent of original ratio) must correspond to 28 and 5 to 35. And in this way we get the original numbers as 28 and 35."
Can somebody please explain why/how this short cut works? Also would this work only in this specific case, or could it be applied to some other cases also, albeit with some modification? Also, he used the term "ratio point" to denote both the difference between the ante. and conse. of the ratio as well as the ante. and conse. of the ratios. Is the term legit?
ratio
asked Jul 26 at 18:07
Neil
32
asked Jul 26 at 18:07
Neil
32
asked Jul 26 at 18:07
Neil
32
asked Jul 26 at 18:07
Neil
32
32
add a comment |Â
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
0
down vote
accepted
So if your first numbers are $N$ and $M$ with $N ne M$ and $frac NM = frac nm$.
An if you add $wne 0$ to each number you get the ration $frac N+wM+w =frac pq$.
The shortcut says if $d= p-n= q-m$ then $N = n*frac wd$ and $M = m*frac wd$.
Why?
Well $frac NM = frac nm$ which means there is a $k$ so that $N =nk$ and $M= mk$.
And $frac nk + wmk + w = frac pq$
So $q(nk+w) = p(mk + w)$ so
$k(qn - pm) = w(p-q)$
Now $d = p-n= q-m$ so $p-q = n -m$. Let $v=p-q = n-m$
$k(q(m+v) - (q+v)m) = wv$
$k(qm + qv - qm - mv) = wv$
$k*v(q-m) = wv$
$k*v*d = wv$
$k= frac wd$.
The only assumption is $vne 0$ and $dne 0$. Which is fair. If $v = n-m =0$ then $n=m$ and $N=M$ which we assumed was not the case. If $d = 0$ then we have $frac NM = frac nm = frac N+wM+w $ so $N(M+w) = M(N+w)$ so $Nw = Mw$ but as $wne 0$ then $frac NM = frac ww = 1$ and $N=M$ which we assumed was not the case.
answered Jul 26 at 19:16
fleablood
60.3k22575
Perfect. This is it.
– Neil
Jul 26 at 21:00
add a comment |Â
up vote
2
down vote
Your numbers are $4k$ and $5k$
When you add $7$ to them you get $4k+7$ and $5k+7$
The new ration is $$frac 4k+75k+7=frac 56$$
Solve for $k$ and you get $k=7$
Thus your numbers are $28$ and $35$
answered Jul 26 at 18:33
Mohammad Riazi-Kermani
27.3k41851
add a comment |Â
up vote
0
down vote
Because you pass from $displaystyle4xover 5x$ to $displaystyle5xover 6x$ by adding $x$ to both numerator and denominator.
answered Jul 26 at 18:13
Aretino
21.7k21342
add a comment |Â
up vote
0
down vote
Yes, your statement is true.
$$frac nk+d(n+1)k+d=frac (n+1)k(n+2)k implies d=k $$
answered Jul 26 at 18:40
Mohammad Riazi-Kermani
27.3k41851
@fleablood offers a more generalized solution
– Neil
Jul 26 at 21:01
Thanks for the comment, my solution is also correct but shorter in details.
– Mohammad Riazi-Kermani
Jul 26 at 21:11
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
So if your first numbers are $N$ and $M$ with $N ne M$ and $frac NM = frac nm$.
An if you add $wne 0$ to each number you get the ration $frac N+wM+w =frac pq$.
The shortcut says if $d= p-n= q-m$ then $N = n*frac wd$ and $M = m*frac wd$.
Why?
Well $frac NM = frac nm$ which means there is a $k$ so that $N =nk$ and $M= mk$.
And $frac nk + wmk + w = frac pq$
So $q(nk+w) = p(mk + w)$ so
$k(qn - pm) = w(p-q)$
Now $d = p-n= q-m$ so $p-q = n -m$. Let $v=p-q = n-m$
$k(q(m+v) - (q+v)m) = wv$
$k(qm + qv - qm - mv) = wv$
$k*v(q-m) = wv$
$k*v*d = wv$
$k= frac wd$.
The only assumption is $vne 0$ and $dne 0$. Which is fair. If $v = n-m =0$ then $n=m$ and $N=M$ which we assumed was not the case. If $d = 0$ then we have $frac NM = frac nm = frac N+wM+w $ so $N(M+w) = M(N+w)$ so $Nw = Mw$ but as $wne 0$ then $frac NM = frac ww = 1$ and $N=M$ which we assumed was not the case.
answered Jul 26 at 19:16
fleablood
60.3k22575
Perfect. This is it.
– Neil
Jul 26 at 21:00
add a comment |Â
up vote
0
down vote
accepted
So if your first numbers are $N$ and $M$ with $N ne M$ and $frac NM = frac nm$.
An if you add $wne 0$ to each number you get the ration $frac N+wM+w =frac pq$.
The shortcut says if $d= p-n= q-m$ then $N = n*frac wd$ and $M = m*frac wd$.
Why?
Well $frac NM = frac nm$ which means there is a $k$ so that $N =nk$ and $M= mk$.
And $frac nk + wmk + w = frac pq$
So $q(nk+w) = p(mk + w)$ so
$k(qn - pm) = w(p-q)$
Now $d = p-n= q-m$ so $p-q = n -m$. Let $v=p-q = n-m$
$k(q(m+v) - (q+v)m) = wv$
$k(qm + qv - qm - mv) = wv$
$k*v(q-m) = wv$
$k*v*d = wv$
$k= frac wd$.
The only assumption is $vne 0$ and $dne 0$. Which is fair. If $v = n-m =0$ then $n=m$ and $N=M$ which we assumed was not the case. If $d = 0$ then we have $frac NM = frac nm = frac N+wM+w $ so $N(M+w) = M(N+w)$ so $Nw = Mw$ but as $wne 0$ then $frac NM = frac ww = 1$ and $N=M$ which we assumed was not the case.
answered Jul 26 at 19:16
fleablood
60.3k22575
Perfect. This is it.
– Neil
Jul 26 at 21:00
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
So if your first numbers are $N$ and $M$ with $N ne M$ and $frac NM = frac nm$.
An if you add $wne 0$ to each number you get the ration $frac N+wM+w =frac pq$.
The shortcut says if $d= p-n= q-m$ then $N = n*frac wd$ and $M = m*frac wd$.
Why?
Well $frac NM = frac nm$ which means there is a $k$ so that $N =nk$ and $M= mk$.
And $frac nk + wmk + w = frac pq$
So $q(nk+w) = p(mk + w)$ so
$k(qn - pm) = w(p-q)$
Now $d = p-n= q-m$ so $p-q = n -m$. Let $v=p-q = n-m$
$k(q(m+v) - (q+v)m) = wv$
$k(qm + qv - qm - mv) = wv$
$k*v(q-m) = wv$
$k*v*d = wv$
$k= frac wd$.
The only assumption is $vne 0$ and $dne 0$. Which is fair. If $v = n-m =0$ then $n=m$ and $N=M$ which we assumed was not the case. If $d = 0$ then we have $frac NM = frac nm = frac N+wM+w $ so $N(M+w) = M(N+w)$ so $Nw = Mw$ but as $wne 0$ then $frac NM = frac ww = 1$ and $N=M$ which we assumed was not the case.
answered Jul 26 at 19:16
fleablood
60.3k22575
So if your first numbers are $N$ and $M$ with $N ne M$ and $frac NM = frac nm$.
An if you add $wne 0$ to each number you get the ration $frac N+wM+w =frac pq$.
The shortcut says if $d= p-n= q-m$ then $N = n*frac wd$ and $M = m*frac wd$.
Why?
Well $frac NM = frac nm$ which means there is a $k$ so that $N =nk$ and $M= mk$.
And $frac nk + wmk + w = frac pq$
So $q(nk+w) = p(mk + w)$ so
$k(qn - pm) = w(p-q)$
Now $d = p-n= q-m$ so $p-q = n -m$. Let $v=p-q = n-m$
$k(q(m+v) - (q+v)m) = wv$
$k(qm + qv - qm - mv) = wv$
$k*v(q-m) = wv$
$k*v*d = wv$
$k= frac wd$.
The only assumption is $vne 0$ and $dne 0$. Which is fair. If $v = n-m =0$ then $n=m$ and $N=M$ which we assumed was not the case. If $d = 0$ then we have $frac NM = frac nm = frac N+wM+w $ so $N(M+w) = M(N+w)$ so $Nw = Mw$ but as $wne 0$ then $frac NM = frac ww = 1$ and $N=M$ which we assumed was not the case.
answered Jul 26 at 19:16
fleablood
60.3k22575
edited Jul 26 at 19:50
answered Jul 26 at 19:16
fleablood
60.3k22575
answered Jul 26 at 19:16
fleablood
60.3k22575
answered Jul 26 at 19:16
fleablood
60.3k22575
60.3k22575
Perfect. This is it.
– Neil
Jul 26 at 21:00
add a comment |Â
Perfect. This is it.
– Neil
Jul 26 at 21:00
Perfect. This is it.
– Neil
Jul 26 at 21:00
Perfect. This is it.
– Neil
Jul 26 at 21:00
add a comment |Â
up vote
2
down vote
Your numbers are $4k$ and $5k$
When you add $7$ to them you get $4k+7$ and $5k+7$
The new ration is $$frac 4k+75k+7=frac 56$$
Solve for $k$ and you get $k=7$
Thus your numbers are $28$ and $35$
answered Jul 26 at 18:33
Mohammad Riazi-Kermani
27.3k41851
add a comment |Â
up vote
2
down vote
Your numbers are $4k$ and $5k$
When you add $7$ to them you get $4k+7$ and $5k+7$
The new ration is $$frac 4k+75k+7=frac 56$$
Solve for $k$ and you get $k=7$
Thus your numbers are $28$ and $35$
answered Jul 26 at 18:33
Mohammad Riazi-Kermani
27.3k41851
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Your numbers are $4k$ and $5k$
When you add $7$ to them you get $4k+7$ and $5k+7$
The new ration is $$frac 4k+75k+7=frac 56$$
Solve for $k$ and you get $k=7$
Thus your numbers are $28$ and $35$
answered Jul 26 at 18:33
Mohammad Riazi-Kermani
27.3k41851
Your numbers are $4k$ and $5k$
When you add $7$ to them you get $4k+7$ and $5k+7$
The new ration is $$frac 4k+75k+7=frac 56$$
Solve for $k$ and you get $k=7$
Thus your numbers are $28$ and $35$
answered Jul 26 at 18:33
Mohammad Riazi-Kermani
27.3k41851
answered Jul 26 at 18:33
Mohammad Riazi-Kermani
27.3k41851
answered Jul 26 at 18:33
Mohammad Riazi-Kermani
27.3k41851
answered Jul 26 at 18:33
Mohammad Riazi-Kermani
27.3k41851
27.3k41851
add a comment |Â
add a comment |Â
up vote
0
down vote
Because you pass from $displaystyle4xover 5x$ to $displaystyle5xover 6x$ by adding $x$ to both numerator and denominator.
answered Jul 26 at 18:13
Aretino
21.7k21342
add a comment |Â
up vote
0
down vote
Because you pass from $displaystyle4xover 5x$ to $displaystyle5xover 6x$ by adding $x$ to both numerator and denominator.
answered Jul 26 at 18:13
Aretino
21.7k21342
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Because you pass from $displaystyle4xover 5x$ to $displaystyle5xover 6x$ by adding $x$ to both numerator and denominator.
answered Jul 26 at 18:13
Aretino
21.7k21342
Because you pass from $displaystyle4xover 5x$ to $displaystyle5xover 6x$ by adding $x$ to both numerator and denominator.
answered Jul 26 at 18:13
Aretino
21.7k21342
answered Jul 26 at 18:13
Aretino
21.7k21342
answered Jul 26 at 18:13
Aretino
21.7k21342
answered Jul 26 at 18:13
Aretino
21.7k21342
21.7k21342
add a comment |Â
add a comment |Â
up vote
0
down vote
Yes, your statement is true.
$$frac nk+d(n+1)k+d=frac (n+1)k(n+2)k implies d=k $$
answered Jul 26 at 18:40
Mohammad Riazi-Kermani
27.3k41851
@fleablood offers a more generalized solution
– Neil
Jul 26 at 21:01
Thanks for the comment, my solution is also correct but shorter in details.
– Mohammad Riazi-Kermani
Jul 26 at 21:11
add a comment |Â
up vote
0
down vote
Yes, your statement is true.
$$frac nk+d(n+1)k+d=frac (n+1)k(n+2)k implies d=k $$
answered Jul 26 at 18:40
Mohammad Riazi-Kermani
27.3k41851
@fleablood offers a more generalized solution
– Neil
Jul 26 at 21:01
Thanks for the comment, my solution is also correct but shorter in details.
– Mohammad Riazi-Kermani
Jul 26 at 21:11
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Yes, your statement is true.
$$frac nk+d(n+1)k+d=frac (n+1)k(n+2)k implies d=k $$
answered Jul 26 at 18:40
Mohammad Riazi-Kermani
27.3k41851
Yes, your statement is true.
$$frac nk+d(n+1)k+d=frac (n+1)k(n+2)k implies d=k $$
answered Jul 26 at 18:40
Mohammad Riazi-Kermani
27.3k41851
answered Jul 26 at 18:40
Mohammad Riazi-Kermani
27.3k41851
answered Jul 26 at 18:40
Mohammad Riazi-Kermani
27.3k41851
answered Jul 26 at 18:40
Mohammad Riazi-Kermani
27.3k41851
27.3k41851
@fleablood offers a more generalized solution
– Neil
Jul 26 at 21:01
Thanks for the comment, my solution is also correct but shorter in details.
– Mohammad Riazi-Kermani
Jul 26 at 21:11
add a comment |Â
@fleablood offers a more generalized solution
– Neil
Jul 26 at 21:01
Thanks for the comment, my solution is also correct but shorter in details.
– Mohammad Riazi-Kermani
Jul 26 at 21:11
@fleablood offers a more generalized solution
– Neil
Jul 26 at 21:01
@fleablood offers a more generalized solution
– Neil
Jul 26 at 21:01
Thanks for the comment, my solution is also correct but shorter in details.
– Mohammad Riazi-Kermani
Jul 26 at 21:11
Thanks for the comment, my solution is also correct but shorter in details.
– Mohammad Riazi-Kermani
Jul 26 at 21:11
add a comment |Â
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