Ratio of money distributed among friends [closed]












0












$begingroup$


Some money (dollars) is divided among friends A,B and C in the ratio of 5:6:9



After A gives fifty dollars to his mother, the ratio becomes 3:4:6



Find the amount of money A has after giving fifty dollars to his mother.






ratio







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$endgroup$



closed as off-topic by Saad, José Carlos Santos, Cesareo, Vidyanshu Mishra, jgon Dec 4 '18 at 14:33


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, José Carlos Santos, Cesareo, Vidyanshu Mishra, jgon

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    What have you tried?
    $endgroup$
    – Tito Eliatron
    Nov 29 '18 at 18:37










  • $begingroup$
    I couldn't find a logic to solve itt.
    $endgroup$
    – Wardah
    Nov 29 '18 at 18:38
















0












$begingroup$


Some money (dollars) is divided among friends A,B and C in the ratio of 5:6:9



After A gives fifty dollars to his mother, the ratio becomes 3:4:6



Find the amount of money A has after giving fifty dollars to his mother.






ratio







share|cite|improve this question









$endgroup$



closed as off-topic by Saad, José Carlos Santos, Cesareo, Vidyanshu Mishra, jgon Dec 4 '18 at 14:33


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, José Carlos Santos, Cesareo, Vidyanshu Mishra, jgon

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    What have you tried?
    $endgroup$
    – Tito Eliatron
    Nov 29 '18 at 18:37










  • $begingroup$
    I couldn't find a logic to solve itt.
    $endgroup$
    – Wardah
    Nov 29 '18 at 18:38














0












0








0





$begingroup$


Some money (dollars) is divided among friends A,B and C in the ratio of 5:6:9



After A gives fifty dollars to his mother, the ratio becomes 3:4:6



Find the amount of money A has after giving fifty dollars to his mother.






ratio







share|cite|improve this question









$endgroup$




Some money (dollars) is divided among friends A,B and C in the ratio of 5:6:9



After A gives fifty dollars to his mother, the ratio becomes 3:4:6



Find the amount of money A has after giving fifty dollars to his mother.






ratio




ratio






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 29 '18 at 18:26









WardahWardah

32




32




closed as off-topic by Saad, José Carlos Santos, Cesareo, Vidyanshu Mishra, jgon Dec 4 '18 at 14:33


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, José Carlos Santos, Cesareo, Vidyanshu Mishra, jgon

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by Saad, José Carlos Santos, Cesareo, Vidyanshu Mishra, jgon Dec 4 '18 at 14:33


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, José Carlos Santos, Cesareo, Vidyanshu Mishra, jgon

If this question can be reworded to fit the rules in the help center, please edit the question.












  • $begingroup$
    What have you tried?
    $endgroup$
    – Tito Eliatron
    Nov 29 '18 at 18:37










  • $begingroup$
    I couldn't find a logic to solve itt.
    $endgroup$
    – Wardah
    Nov 29 '18 at 18:38


















  • $begingroup$
    What have you tried?
    $endgroup$
    – Tito Eliatron
    Nov 29 '18 at 18:37










  • $begingroup$
    I couldn't find a logic to solve itt.
    $endgroup$
    – Wardah
    Nov 29 '18 at 18:38
















$begingroup$
What have you tried?
$endgroup$
– Tito Eliatron
Nov 29 '18 at 18:37




$begingroup$
What have you tried?
$endgroup$
– Tito Eliatron
Nov 29 '18 at 18:37












$begingroup$
I couldn't find a logic to solve itt.
$endgroup$
– Wardah
Nov 29 '18 at 18:38




$begingroup$
I couldn't find a logic to solve itt.
$endgroup$
– Wardah
Nov 29 '18 at 18:38










2 Answers
2






active

oldest

votes


















0












$begingroup$

The trick here is to observe that ratios are invariant to multiplication, i.e. 5:6:9 expresses the same proportions as 10:12:18 (Just by multiplying every number by 2).



So, given that we have the ratios 5:6:9 and 3:4:6, these can be rewritten as 10:12:18 and 9:12:18 respectively:



$$
begin{align}
text{Before} quad 5:6:9 &iff 10:12:18 \
text{After} quad 3:4:6 &iff 9:12:18
end{align}
$$



Now, the problem states that A has given $$50$ to his mother, and from the above ratios, we see that this has decreased the ratio from 10 to 9 for A (the ratio between the other two has not changed since they did not receive/give away any money).



This indicates that one unit of the ratio corresponds to $$50$.



Since now he has 9 units, he has $9 times $50 = $450$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Is there any technique to solve it with simple algebra?
    $endgroup$
    – Wardah
    Nov 29 '18 at 18:57










  • $begingroup$
    Yes, I believe the other answer solves it in an algebraic way.
    $endgroup$
    – Sean Lee
    Nov 29 '18 at 18:59










  • $begingroup$
    I am just a beginer. Found it very hard to understand. As it is not solved completly.
    $endgroup$
    – Wardah
    Nov 29 '18 at 19:02



















0












$begingroup$

Hint: Let $x$ be the amount $A$ has to begin with. Then: $dfrac{x}{5} = dfrac{b}{6}, dfrac{x-50}{3} = dfrac{b}{4}$. Can you solve this system for $x$ ? $b =$ amount $B$ has. To be more specific, isolate $b$, and get $dfrac{6x}{5} = dfrac{4(x-50)}{3}$. Can you take it from here ?.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I am just a beginer. I tried it but failed to understand. I will be grateful if you could explain it in some depth.
    $endgroup$
    – Wardah
    Nov 29 '18 at 19:03


















2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

The trick here is to observe that ratios are invariant to multiplication, i.e. 5:6:9 expresses the same proportions as 10:12:18 (Just by multiplying every number by 2).



So, given that we have the ratios 5:6:9 and 3:4:6, these can be rewritten as 10:12:18 and 9:12:18 respectively:



$$
begin{align}
text{Before} quad 5:6:9 &iff 10:12:18 \
text{After} quad 3:4:6 &iff 9:12:18
end{align}
$$



Now, the problem states that A has given $$50$ to his mother, and from the above ratios, we see that this has decreased the ratio from 10 to 9 for A (the ratio between the other two has not changed since they did not receive/give away any money).



This indicates that one unit of the ratio corresponds to $$50$.



Since now he has 9 units, he has $9 times $50 = $450$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Is there any technique to solve it with simple algebra?
    $endgroup$
    – Wardah
    Nov 29 '18 at 18:57










  • $begingroup$
    Yes, I believe the other answer solves it in an algebraic way.
    $endgroup$
    – Sean Lee
    Nov 29 '18 at 18:59










  • $begingroup$
    I am just a beginer. Found it very hard to understand. As it is not solved completly.
    $endgroup$
    – Wardah
    Nov 29 '18 at 19:02
















0












$begingroup$

The trick here is to observe that ratios are invariant to multiplication, i.e. 5:6:9 expresses the same proportions as 10:12:18 (Just by multiplying every number by 2).



So, given that we have the ratios 5:6:9 and 3:4:6, these can be rewritten as 10:12:18 and 9:12:18 respectively:



$$
begin{align}
text{Before} quad 5:6:9 &iff 10:12:18 \
text{After} quad 3:4:6 &iff 9:12:18
end{align}
$$



Now, the problem states that A has given $$50$ to his mother, and from the above ratios, we see that this has decreased the ratio from 10 to 9 for A (the ratio between the other two has not changed since they did not receive/give away any money).



This indicates that one unit of the ratio corresponds to $$50$.



Since now he has 9 units, he has $9 times $50 = $450$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Is there any technique to solve it with simple algebra?
    $endgroup$
    – Wardah
    Nov 29 '18 at 18:57










  • $begingroup$
    Yes, I believe the other answer solves it in an algebraic way.
    $endgroup$
    – Sean Lee
    Nov 29 '18 at 18:59










  • $begingroup$
    I am just a beginer. Found it very hard to understand. As it is not solved completly.
    $endgroup$
    – Wardah
    Nov 29 '18 at 19:02














0












0








0





$begingroup$

The trick here is to observe that ratios are invariant to multiplication, i.e. 5:6:9 expresses the same proportions as 10:12:18 (Just by multiplying every number by 2).



So, given that we have the ratios 5:6:9 and 3:4:6, these can be rewritten as 10:12:18 and 9:12:18 respectively:



$$
begin{align}
text{Before} quad 5:6:9 &iff 10:12:18 \
text{After} quad 3:4:6 &iff 9:12:18
end{align}
$$



Now, the problem states that A has given $$50$ to his mother, and from the above ratios, we see that this has decreased the ratio from 10 to 9 for A (the ratio between the other two has not changed since they did not receive/give away any money).



This indicates that one unit of the ratio corresponds to $$50$.



Since now he has 9 units, he has $9 times $50 = $450$.






share|cite|improve this answer









$endgroup$



The trick here is to observe that ratios are invariant to multiplication, i.e. 5:6:9 expresses the same proportions as 10:12:18 (Just by multiplying every number by 2).



So, given that we have the ratios 5:6:9 and 3:4:6, these can be rewritten as 10:12:18 and 9:12:18 respectively:



$$
begin{align}
text{Before} quad 5:6:9 &iff 10:12:18 \
text{After} quad 3:4:6 &iff 9:12:18
end{align}
$$



Now, the problem states that A has given $$50$ to his mother, and from the above ratios, we see that this has decreased the ratio from 10 to 9 for A (the ratio between the other two has not changed since they did not receive/give away any money).



This indicates that one unit of the ratio corresponds to $$50$.



Since now he has 9 units, he has $9 times $50 = $450$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 29 '18 at 18:48









Sean LeeSean Lee

381111




381111












  • $begingroup$
    Is there any technique to solve it with simple algebra?
    $endgroup$
    – Wardah
    Nov 29 '18 at 18:57










  • $begingroup$
    Yes, I believe the other answer solves it in an algebraic way.
    $endgroup$
    – Sean Lee
    Nov 29 '18 at 18:59










  • $begingroup$
    I am just a beginer. Found it very hard to understand. As it is not solved completly.
    $endgroup$
    – Wardah
    Nov 29 '18 at 19:02


















  • $begingroup$
    Is there any technique to solve it with simple algebra?
    $endgroup$
    – Wardah
    Nov 29 '18 at 18:57










  • $begingroup$
    Yes, I believe the other answer solves it in an algebraic way.
    $endgroup$
    – Sean Lee
    Nov 29 '18 at 18:59










  • $begingroup$
    I am just a beginer. Found it very hard to understand. As it is not solved completly.
    $endgroup$
    – Wardah
    Nov 29 '18 at 19:02
















$begingroup$
Is there any technique to solve it with simple algebra?
$endgroup$
– Wardah
Nov 29 '18 at 18:57




$begingroup$
Is there any technique to solve it with simple algebra?
$endgroup$
– Wardah
Nov 29 '18 at 18:57












$begingroup$
Yes, I believe the other answer solves it in an algebraic way.
$endgroup$
– Sean Lee
Nov 29 '18 at 18:59




$begingroup$
Yes, I believe the other answer solves it in an algebraic way.
$endgroup$
– Sean Lee
Nov 29 '18 at 18:59












$begingroup$
I am just a beginer. Found it very hard to understand. As it is not solved completly.
$endgroup$
– Wardah
Nov 29 '18 at 19:02




$begingroup$
I am just a beginer. Found it very hard to understand. As it is not solved completly.
$endgroup$
– Wardah
Nov 29 '18 at 19:02











0












$begingroup$

Hint: Let $x$ be the amount $A$ has to begin with. Then: $dfrac{x}{5} = dfrac{b}{6}, dfrac{x-50}{3} = dfrac{b}{4}$. Can you solve this system for $x$ ? $b =$ amount $B$ has. To be more specific, isolate $b$, and get $dfrac{6x}{5} = dfrac{4(x-50)}{3}$. Can you take it from here ?.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I am just a beginer. I tried it but failed to understand. I will be grateful if you could explain it in some depth.
    $endgroup$
    – Wardah
    Nov 29 '18 at 19:03
















0












$begingroup$

Hint: Let $x$ be the amount $A$ has to begin with. Then: $dfrac{x}{5} = dfrac{b}{6}, dfrac{x-50}{3} = dfrac{b}{4}$. Can you solve this system for $x$ ? $b =$ amount $B$ has. To be more specific, isolate $b$, and get $dfrac{6x}{5} = dfrac{4(x-50)}{3}$. Can you take it from here ?.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I am just a beginer. I tried it but failed to understand. I will be grateful if you could explain it in some depth.
    $endgroup$
    – Wardah
    Nov 29 '18 at 19:03














0












0








0





$begingroup$

Hint: Let $x$ be the amount $A$ has to begin with. Then: $dfrac{x}{5} = dfrac{b}{6}, dfrac{x-50}{3} = dfrac{b}{4}$. Can you solve this system for $x$ ? $b =$ amount $B$ has. To be more specific, isolate $b$, and get $dfrac{6x}{5} = dfrac{4(x-50)}{3}$. Can you take it from here ?.






share|cite|improve this answer











$endgroup$



Hint: Let $x$ be the amount $A$ has to begin with. Then: $dfrac{x}{5} = dfrac{b}{6}, dfrac{x-50}{3} = dfrac{b}{4}$. Can you solve this system for $x$ ? $b =$ amount $B$ has. To be more specific, isolate $b$, and get $dfrac{6x}{5} = dfrac{4(x-50)}{3}$. Can you take it from here ?.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 29 '18 at 19:05

























answered Nov 29 '18 at 18:40









DeepSeaDeepSea

71.2k54487




71.2k54487












  • $begingroup$
    I am just a beginer. I tried it but failed to understand. I will be grateful if you could explain it in some depth.
    $endgroup$
    – Wardah
    Nov 29 '18 at 19:03


















  • $begingroup$
    I am just a beginer. I tried it but failed to understand. I will be grateful if you could explain it in some depth.
    $endgroup$
    – Wardah
    Nov 29 '18 at 19:03
















$begingroup$
I am just a beginer. I tried it but failed to understand. I will be grateful if you could explain it in some depth.
$endgroup$
– Wardah
Nov 29 '18 at 19:03




$begingroup$
I am just a beginer. I tried it but failed to understand. I will be grateful if you could explain it in some depth.
$endgroup$
– Wardah
Nov 29 '18 at 19:03



-

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