Alice, Betty and Chitra had $276$ cards. Betty had three times as many cards as Alice. After Chitra gave $30$ cards to Betty, Betty had twice as many cards as Chitra. How many cards did Chitra have at first

Full solution

Q. Alice, Betty and Chitra had

276

cards. Betty had three times as many cards as Alice. After Chitra gave

30

cards to Betty, Betty had twice as many cards as Chitra. How many cards did Chitra have at first

Define Variables: Let's call the number of cards Alice has $A$ , Betty has $B$ , and Chitra has $C$ . We know that $B = 3A$ and the total is $A + B + C = 276$ .
Betty's Cards Transfer: After Chitra gives $30$ cards to Betty, Betty has $B + 30$ cards. $\newline$ Now, Betty has twice as many cards as Chitra, so $B + 30 = 2(C - 30)$ .
Equation Substitution: Substitute $B$ with $3A$ in the equation $B + 30 = 2(C - 30)$ to get $3A + 30 = 2(C - 30)$ .
Simplify Equations: Now we have two equations: $3A + 30 = 2(C - 30)$ and $A + 3A + C = 276$ . Simplify the second equation to get $4A + C = 276$ .
Find $C$ in terms of $A$ : Solve the first equation for $C$ : $3A + 30 = 2C - 60$ , so $3A + 90 = 2C$ . $\newline$ Divide by $2$ to get $C = 1.5A + 45$ .
Substitute C in Equation: Substitute $C$ in the second equation: $4A + (1.5A + 45) = 276$ . Simplify to get $5.5A + 45 = 276$ .
Solve for $A$ : Subtract $45$ from both sides: $5.5A = 231$ .
Find $B$ : Divide by $5.5$ to find $A$ : $A = \frac{231}{5.5}$ . $A = 42$ .
Find $C$ : Now find $B$ : $B = 3A = 3 \times 42.$ $\newline$ $B = 126.$
Final Calculation: Finally, find $C$ using $A + B + C = 276$ : $42 + 126 + C = 276$ . $\newline$ $168 + C = 276$ .
Final Calculation: Finally, find $C$ using $A + B + C = 276$ : $42 + 126 + C = 276$ . $168 + C = 276$ .Subtract $168$ from both sides: $C = 276 - 168$ . $C = 108$ .