Carly had some chocolates. She gave 178 chocolates to her neighbour and 25 of the remainder to her daughter. With the remaining chocolates, she gave 13 of the chocolates to her son and then had 256 chocolates left. How many chocolates had she at first?

Question

Troy Sutton · Accepted Answer

Step 1 Let Carly had x chocolates. She gave 178 chocolates to her neighbour. Remaining are x−178 and as 25 of (x−178) were given her daughter, remaining were 1−25=5−25=35 of (x−178) i.e. 3(x−178)5 are left Now 13 of 3(x−178)5 were given to son remaining were 1−13=3−13=23 of 3(x−178)5 i.e. 23×3(x−178)5 are left. Hence 23×3(x−178)5=256 Hence, x−178=256×32×53 or x−178=128⧸256×⧸31⧸2×5⧸3=640 and x=640+178=818 Carly had 818 chocolates

trasahed · Accepted Answer

Step 1  Breaking the question down into its component parts:  Carly had some chocolates: →?  She gave 178 to her neighbour: →[?−178] left (remainder)  25 of the remainder to her daughter:  →[?−178]−25[?−178]=35[?−178] left  With the remaining she gave 13 to her son:  →35[?−178]−13(35[?−178])=25[?−178] left  then she had 256 chocolates left: →25[?−178]=256  Step 2  Cross multiply (same as multiply both sides by 52  [?−178]=52×256  Add 178 to both sides  ?=[52×256]+178  ?=818

Carly had some chocolates. She gave 178 chocolates to her

Answered question

Answer & Explanation

New Questions in Pre-Algebra