
A and B have 460 coins altogether. If \[\dfrac{3}{4}\] of $A$'s number of coins is equal to \[\dfrac{2}{5}\] of $B$'s number of coins, find how many coins must B give to A, so that they both have equal number of coins.
Answer
581.4k+ views
Hint: First we take a variable as $A$'s number of coins and another for $B$'s number of coins.
Then applying the two conditions we will get two equations.
By solving the equations we will get $A$'s number of coins and $B$'s number of coins.
Complete step by step answer:
It is given that A and B have 460 coins altogether.
Let, $A$'s number of coins be $x$ & $B$'s number of coins be $y$.
Again, the number of total coins is 460.
Thus, \[x + y{\rm{ }} = 460\]
Rearrange the equation and mark it as (1)
\[y = 460{\rm{ }} - {\rm{ }}x\] …… (1)
Again it is given that \[\dfrac{3}{4}\] of $A$'s number of coins is equal to \[\dfrac{2}{5}\] of $B$'s number of coins
That is \[\dfrac{{3x}}{4} = \dfrac{{2y}}{5}\]
By cross multiplication of the above equation we have,
\[3x \times 5 = 2y \times 4\\\]
On simplifying the above equation, we get,
\[15x = 8y\]
Rearrange the equation and mark it as (2) we get,
\[x = \dfrac{{8y}}{{15}}.....(2)\]
Let us put the value of x from (2) in (1) we get
\[\dfrac{{8y}}{{15}} + y = 460\]
Let us simplify it in the left hand side we get,
\[\dfrac{{8y + 15y}}{{15}} = 460\]
\[23y = 460 \times 15\]
Let us solve the above equation for \[y\]we get,
\[y = \dfrac{{460 \times 15}}{{23}}\]
\[y = 20 \times 15 = 300\] On simplifying the above equation,
\[y = 300\]
Substitute \[y\]in equation (1) we get,
\[\;x = 460 - 300 = 160.\]
Therefore, we have found \[\;x = 160,{\rm{ }}y = 300\]
Hence $A$'s number of coins & $B$'s number of coins are 160 & 300 respectively.
So, B has \[\left( {300 - 160} \right){\rm{ }} = 140\] more coins than A.
If we distribute these 140 coins equally between A and B then the number of coins of A and B will be the same.
Then B has to give A \[\dfrac{{140}}{2} = 70\] coins, to make the number of coins equal.
Hence, B must give A 70 coins, so that they both have equal numbers of coins.
Note:
In these kinds of questions, the logic lies in the given information. Grab that logic to form equations. If we have two equations with two variables then by method of substitution we get the solution. To find who has got more number of coins we should subtract the total number of coins each have.
Then applying the two conditions we will get two equations.
By solving the equations we will get $A$'s number of coins and $B$'s number of coins.
Complete step by step answer:
It is given that A and B have 460 coins altogether.
Let, $A$'s number of coins be $x$ & $B$'s number of coins be $y$.
Again, the number of total coins is 460.
Thus, \[x + y{\rm{ }} = 460\]
Rearrange the equation and mark it as (1)
\[y = 460{\rm{ }} - {\rm{ }}x\] …… (1)
Again it is given that \[\dfrac{3}{4}\] of $A$'s number of coins is equal to \[\dfrac{2}{5}\] of $B$'s number of coins
That is \[\dfrac{{3x}}{4} = \dfrac{{2y}}{5}\]
By cross multiplication of the above equation we have,
\[3x \times 5 = 2y \times 4\\\]
On simplifying the above equation, we get,
\[15x = 8y\]
Rearrange the equation and mark it as (2) we get,
\[x = \dfrac{{8y}}{{15}}.....(2)\]
Let us put the value of x from (2) in (1) we get
\[\dfrac{{8y}}{{15}} + y = 460\]
Let us simplify it in the left hand side we get,
\[\dfrac{{8y + 15y}}{{15}} = 460\]
\[23y = 460 \times 15\]
Let us solve the above equation for \[y\]we get,
\[y = \dfrac{{460 \times 15}}{{23}}\]
\[y = 20 \times 15 = 300\] On simplifying the above equation,
\[y = 300\]
Substitute \[y\]in equation (1) we get,
\[\;x = 460 - 300 = 160.\]
Therefore, we have found \[\;x = 160,{\rm{ }}y = 300\]
Hence $A$'s number of coins & $B$'s number of coins are 160 & 300 respectively.
So, B has \[\left( {300 - 160} \right){\rm{ }} = 140\] more coins than A.
If we distribute these 140 coins equally between A and B then the number of coins of A and B will be the same.
Then B has to give A \[\dfrac{{140}}{2} = 70\] coins, to make the number of coins equal.
Hence, B must give A 70 coins, so that they both have equal numbers of coins.
Note:
In these kinds of questions, the logic lies in the given information. Grab that logic to form equations. If we have two equations with two variables then by method of substitution we get the solution. To find who has got more number of coins we should subtract the total number of coins each have.
Recently Updated Pages
Master Class 8 Social Science: Engaging Questions & Answers for Success

Master Class 8 Maths: Engaging Questions & Answers for Success

Master Class 8 Science: Engaging Questions & Answers for Success

Class 8 Question and Answer - Your Ultimate Solutions Guide

Master Class 8 English: Engaging Questions & Answers for Success

Why are manures considered better than fertilizers class 11 biology CBSE

Trending doubts
What is BLO What is the full form of BLO class 8 social science CBSE

Citizens of India can vote at the age of A 18 years class 8 social science CBSE

Full form of STD, ISD and PCO

Right to vote is a AFundamental Right BFundamental class 8 social science CBSE

What is the difference between rai and mustard see class 8 biology CBSE

Summary of the poem Where the Mind is Without Fear class 8 english CBSE


