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Question 1188238: Justin had some cards. He gave Alan 1/10 of the cards and bought another 45
new cards. Then he used 1/3 of the new total number of cards and bought 40 new
cards. He had 160 cards in the end. How many cards did he have at first?
Found 5 solutions by Boreal, MathTherapy, Octo-pie7, ikleyn, greenestamps:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! He had x at first.
Then he had 0.9x, then 0.9x + 45
then he had (1/3)(0.9x+45)+40=160
so (1/3)(0.9x+45)=120
0.9x+45=360, multiplying by 3
0.9x=315
x=350 cards to begin with
315 after giving away 10%
then 360 after getting 45
1/3 of those is 120, and another 40 is 160.
he had 350 to begin with.
Answer by MathTherapy(10555)  (Show Source):
Answer by Octo-pie7(11)  (Show Source):
You can put this solution on YOUR website! Let new cards be x
(x - x/10) + 45 = 9x/10 + 45
1/3 of (9x/10 + 45) = 9x/30 + 45
(9x/10 + 45) - (9x/30 + 15) + 40 = 160
=> 9x/10 - 9x/30 + 70 = 160
=> (27x - 9x)/30 = 90
x = (90 * 30)/18
= 15 * 10
= 150 cards
ratio method:
He has 160 cards in the end.
He bought 40 new cards.
So, He had (160 - 40) = 120 cards.
He used 1/3 of the new total no. of cards.
Then,
He used : He not used
= 1 : 2
60 : 120
Then no. of total new cards is 120 + 60 = 180
He bought 45 new cards .
Before buying 45 cards he had,
(180 - 45) = 135 cards.
He gave Alan 1/10 of the cards.
Justin : Alan
9 : 1
135 : 15
So, At first Justin had
(135 + 15) = 150 cards.
Answer by ikleyn(52817)  (Show Source):
You can put this solution on YOUR website! .
The simplest way to solve such problems is a BACKWARD method,
when you MENTALLY start from the end and move to the beginning,
making inverse operations to the described, step by step.
Doing this way, the problem can be solved in your head mentally, with NO equations.
Pen or pencil together with piece of paper are good assistants.
Many problems of this kind are designed, created and intended SPECIALLY for you
to learn and apply this method, and this given problem is ESPECIALLY GOOD for you to start with.
See the lesson
- Solving problems on the remaining amount
in this site and find there several similar solved problems.
Start reading this lesson from the very beginning.
Happy learning ( ! )
Answer by greenestamps(13203)  (Show Source):
You can put this solution on YOUR website!
Yes, this particular problem is probably solved most easily by working backward.
He finished with 160 cards, after buying 40 more, so before buying those 40 he had 120.
The 120 cards was after he used 1/3 of what he had. This part is a bit tricky to do mentally. Since the 120 was after he used 1/3 of what he had before, the 120 is 2/3 of what he had before; therefore the number he had before was (3/2) times 120, which is 180.
That's the first couple of steps; I'll leave it to you to go the last couple.
If you are going to solve it "forwards" using formal algebra, take a moment to see what is happening in the problem before deciding how to set the problem up.
The algebraic solution from one of the other tutors starts immediately by letting x be the initial number of cards; this leads quickly to equations involving fractions, which are always a bit harder to work with.
Instead, noticing that the first thing that happens is he give away 1/10 of his cards, start with the initial number of cards being 10x. That gives us....
initial number: 10x
after giving 1/10 away: 9x
after buying 45 more: 9x+45
after using 1/3 of those (2/3 of 9x+45 are left): 6x+30
after buying 40 more: 6x+70
At that point he had 160 cards:
6x+70=160
6x=90
x=90/6=15
ANSWER: The number he started with was 10x=150
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