Question

Ram and Mohan are friends. Each has some money. If Ram gives Rs.30 to Mohan, then Mohan will have twice the money left with Ram. But if Mohan gives Rs.10 to Ram, then Ram will have thrice as much as is left with Mohan. How much money does each have?
A. Rs.62, Rs.34
B. Rs.6, Rs.2
C. Rs.170, Rs.124
D. Rs.43, Rs.26

Answer 1

Hint: Here we assume the initial money with both Ram and Mohan as two different variables. Using the conditions given in the question we subtract the money from the giving person’s initial amount and add that money to the receiving person’s initial amount. Applying the conditions of twice and thrice we make two equations and solve the value of two variables by substitution method.
* If a is twice of b then we can write \[a = 2b\] and if a is thrice of b then we can write \[a = 3b\].

Complete step-by-step answer:
Let us assume Ram has x and Mohan has y
We take two cases separately.
CASE 1:
Ram gives Rs.30 to Mohan and then Mohan will have twice the money left with Ram.
Here Ram gives the money so we subtract the amount traded from Ram’s initial money and Mohan receives the money so we add the amount traded to Mohan’s initial money.
Substituting the values of Ram’s initial money as x and Mohan’s initial money as y, we can write
Money Ram has\[ = (x - 30)\]
Money Mohan has \[ = (y + 30)\]
No we are given that Mohan will have twice the money left with Ram. So we can write
\[(y + 30) = 2(x - 30)\]
Solve the equation by opening the brackets on RHS of the equation.
\[ \Rightarrow y + 30 = 2x - 60\]
Shift all constants to one side of the equation.
\[ \Rightarrow 30 + 60 = 2x - y\]
\[ \Rightarrow 2x - y = 90\] … (1)
CASE 2:
Mohan gives Rs.10 to Ram, then Ram will have thrice as much as is left with Mohan.
Here Mohan gives the money so we subtract the amount traded from Mohan’s initial money and Ram receives the money so we add the amount traded to Ram’s initial money.
Substituting the values of Ram’s initial money as x and Mohan’s initial money as y, we can write
Money Ram has\[ = (x + 10)\]
Money Mohan has \[ = (y - 10)\]
Now we are given that Ram will have twice the money left with Mohan. So we can write
\[(x + 10) = 3(y - 10)\]
Solve the equation by opening the brackets on the RHS of the equation.
\[ \Rightarrow x + 10 = 3y - 30\]
Shift all constants to one side of the equation.
\[ \Rightarrow 30 + 10 = 3y - x\]
\[ \Rightarrow 3y - x = 40\] … (2)
Now we have two equations from equation (1) and equation (2),
\[2x - y = 90\] and \[3y - x = 40\]
We use substitution method to find the values of x and y.
We have equation (1) as\[2x - y = 90\]. Shift all the values except y to one side of the equation.
\[ \Rightarrow 2x - 90 = y\]
Substitute this value of y in equation (2)
\[ \Rightarrow 3(2x - 90) - x = 40\]
Open the bracket on LHS.
\[ \Rightarrow 6x - 270 - x = 40\]
Shift all constant values to one side of the equation.
\[ \Rightarrow 5x = 40 + 270\]
\[ \Rightarrow 5x = 310\]
Divide both sides by 5
\[ \Rightarrow \dfrac{{5x}}{5} = \dfrac{{310}}{5}\]
Cancel the same factor from numerator and denominator.
\[ \Rightarrow x = 62\]
Now we substitute the value of \[x = 62\]in equation (1)
\[ \Rightarrow 2(62) - y = 90\]
Open the brackets on LHS
\[ \Rightarrow 124 - y = 90\]
Shift all constant values to one side of the equation.
\[ \Rightarrow 124 - 90 = y\]
\[ \Rightarrow y = 34\]
So, the value of x is 62 and the value of y is 34.
Therefore, Money Ram has is Rs.62 and money Mohan has is Rs.34

So, option A is correct.

Note: Students might get confused while writing the equations from the word problem as they might multiply the value of twice or thrice on the wrong side of the equation. Keep in mind we multiply the number 2 for twice and 3 for thrice to that side of the equation which has the word twice or thrice attached to it.
Also, keep in mind we always change signs from positive to negative and vice versa when shifting a value from one side to another side.