Ratio of money distributed among friends [closed]
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
closed as off-topic by Saad, José Carlos Santos, Cesareo, Vidyanshu Mishra, jgon Dec 4 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.
add a comment |
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
closed as off-topic by Saad, José Carlos Santos, Cesareo, Vidyanshu Mishra, jgon Dec 4 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.
What have you tried?
– Tito Eliatron
Nov 29 at 18:37
I couldn't find a logic to solve itt.
– Wardah
Nov 29 at 18:38
add a comment |
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
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
asked Nov 29 at 18:26
Wardah
12
12
closed as off-topic by Saad, José Carlos Santos, Cesareo, Vidyanshu Mishra, jgon Dec 4 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 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.
What have you tried?
– Tito Eliatron
Nov 29 at 18:37
I couldn't find a logic to solve itt.
– Wardah
Nov 29 at 18:38
add a comment |
What have you tried?
– Tito Eliatron
Nov 29 at 18:37
I couldn't find a logic to solve itt.
– Wardah
Nov 29 at 18:38
What have you tried?
– Tito Eliatron
Nov 29 at 18:37
What have you tried?
– Tito Eliatron
Nov 29 at 18:37
I couldn't find a logic to solve itt.
– Wardah
Nov 29 at 18:38
I couldn't find a logic to solve itt.
– Wardah
Nov 29 at 18:38
add a comment |
2 Answers
2
active
oldest
votes
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 = $450$.
Is there any technique to solve it with simple algebra?
– Wardah
Nov 29 at 18:57
Yes, I believe the other answer solves it in an algebraic way.
– Sean Lee
Nov 29 at 18:59
I am just a beginer. Found it very hard to understand. As it is not solved completly.
– Wardah
Nov 29 at 19:02
add a comment |
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 ?.
I am just a beginer. I tried it but failed to understand. I will be grateful if you could explain it in some depth.
– Wardah
Nov 29 at 19:03
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 = $450$.
Is there any technique to solve it with simple algebra?
– Wardah
Nov 29 at 18:57
Yes, I believe the other answer solves it in an algebraic way.
– Sean Lee
Nov 29 at 18:59
I am just a beginer. Found it very hard to understand. As it is not solved completly.
– Wardah
Nov 29 at 19:02
add a comment |
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 = $450$.
Is there any technique to solve it with simple algebra?
– Wardah
Nov 29 at 18:57
Yes, I believe the other answer solves it in an algebraic way.
– Sean Lee
Nov 29 at 18:59
I am just a beginer. Found it very hard to understand. As it is not solved completly.
– Wardah
Nov 29 at 19:02
add a comment |
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 = $450$.
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 = $450$.
answered Nov 29 at 18:48
Sean Lee
1387
1387
Is there any technique to solve it with simple algebra?
– Wardah
Nov 29 at 18:57
Yes, I believe the other answer solves it in an algebraic way.
– Sean Lee
Nov 29 at 18:59
I am just a beginer. Found it very hard to understand. As it is not solved completly.
– Wardah
Nov 29 at 19:02
add a comment |
Is there any technique to solve it with simple algebra?
– Wardah
Nov 29 at 18:57
Yes, I believe the other answer solves it in an algebraic way.
– Sean Lee
Nov 29 at 18:59
I am just a beginer. Found it very hard to understand. As it is not solved completly.
– Wardah
Nov 29 at 19:02
Is there any technique to solve it with simple algebra?
– Wardah
Nov 29 at 18:57
Is there any technique to solve it with simple algebra?
– Wardah
Nov 29 at 18:57
Yes, I believe the other answer solves it in an algebraic way.
– Sean Lee
Nov 29 at 18:59
Yes, I believe the other answer solves it in an algebraic way.
– Sean Lee
Nov 29 at 18:59
I am just a beginer. Found it very hard to understand. As it is not solved completly.
– Wardah
Nov 29 at 19:02
I am just a beginer. Found it very hard to understand. As it is not solved completly.
– Wardah
Nov 29 at 19:02
add a comment |
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 ?.
I am just a beginer. I tried it but failed to understand. I will be grateful if you could explain it in some depth.
– Wardah
Nov 29 at 19:03
add a comment |
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 ?.
I am just a beginer. I tried it but failed to understand. I will be grateful if you could explain it in some depth.
– Wardah
Nov 29 at 19:03
add a comment |
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 ?.
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 ?.
edited Nov 29 at 19:05
answered Nov 29 at 18:40
DeepSea
70.8k54487
70.8k54487
I am just a beginer. I tried it but failed to understand. I will be grateful if you could explain it in some depth.
– Wardah
Nov 29 at 19:03
add a comment |
I am just a beginer. I tried it but failed to understand. I will be grateful if you could explain it in some depth.
– Wardah
Nov 29 at 19:03
I am just a beginer. I tried it but failed to understand. I will be grateful if you could explain it in some depth.
– Wardah
Nov 29 at 19:03
I am just a beginer. I tried it but failed to understand. I will be grateful if you could explain it in some depth.
– Wardah
Nov 29 at 19:03
add a comment |
What have you tried?
– Tito Eliatron
Nov 29 at 18:37
I couldn't find a logic to solve itt.
– Wardah
Nov 29 at 18:38