HELP!

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HELP!

by fizzanasir » Thu Oct 13, 2011 11:11 pm
A tourist purchased a total of $1500 worth of traveler's checks in $10 and $50 denominations. During the trip the tourist cashed 7 checks and then lost all the ret. If the number of $10 checks cashed was one more or one less than the number of $50 checks cashed. What is the minimum possible value o the checks that were lost?

a) $ 1,430
b) $ 1,310
c) $ 1,290
d) $ 1,270
e) $ 1,150
Source: — Problem Solving |

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by shankar.ashwin » Thu Oct 13, 2011 11:22 pm
We aim to spend the maximum in the 7 notes that were cashed inorder to find the min value of the checks lost.

So total is 7 and a difference of 10$ and 50$ is 1.

So highest such combination would be to spend 4-$50 and 3- $10 notes

4* 50 = 200 (and) 3 * 10 = 30

So min amt would be 1500 - 230 = 1270 D IMO

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by fizzanasir » Fri Oct 14, 2011 1:35 am
i have a few more questions if you could help me, it would be great :)

Today Rose is twice as old as Sam and Sam is 3 years younger than Tina. If Rose, Sam and Tina are all alive 4 years from today, which o the following must be true on that day?

I) Rose is twice as old as Sam
II) Sam is 3 years younger than Tina
III) Rose is older than Tina

A) I only
B) II only
C) III only
D) I and II
E) II and III



Q2. If each of the following fractions were written as a repeating decimal, which would have the longest sequence of different digits?

A) 2/11
B) 1/3
C) 41/99
D) 2/3
E) 23/77

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by shankar.ashwin » Fri Oct 14, 2011 2:14 am
fizzanasir wrote:i have a few more questions if you could help me, it would be great :)

Today Rose is twice as old as Sam and Sam is 3 years younger than Tina. If Rose, Sam and Tina are all alive 4 years from today, which o the following must be true on that day?

I) Rose is twice as old as Sam
II) Sam is 3 years younger than Tina
III) Rose is older than Tina

A) I only
B) II only
C) III only
D) I and II
E) II and III
Today, R=2S and S=T-3 --> R = 2S = 2T-6

I prefer to pick numbers, you may use algebra as well.

If T = 6 -> S = T -3 -> S = 3 (I pick T = 6, because thats the case where I would have a possibility of T and R being of the same age)
and, R = 2S = 6

Four years from now, (add 4 to R,S and T)

T = 10, S = 7 and R = 10

You can clearly see only statement 2 is true.

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by shankar.ashwin » Fri Oct 14, 2011 2:25 am
fizzanasir wrote: Q2. If each of the following fractions were written as a repeating decimal, which would have the longest sequence of different digits?

A) 2/11
B) 1/3
C) 41/99
D) 2/3
E) 23/77
I would manually divide the numbers and find. 1/3 and 2/3 is 0.33 and 0.66 so it would contain only 1 digit,

So skip them, try dividng the other 3 options you can find E follows no pattern and go on) 0.2987...

On problems like these make sure you start from option E as answers tend to be towards the end keeping in mind people would spend a lot of time solving all options

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by fizzanasir » Fri Oct 14, 2011 3:06 am
how do you know which one of the recurring digits are the longest? like i did all division manually, but i couldnt decide which one is the answer, they are all recurring

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by shankar.ashwin » Fri Oct 14, 2011 12:08 pm
They ask for the longest sequence of different digits and not the longest sequence itself.

Say, 2/11 - 0.181818 is a seq with only 2 digits and
41/99 - .414141 is a sequence with only 2 digits again.

P.S
Found a shortcut for this on the forum
Express any fraction such that the denominator is 9,99,999 or so on.

For eg,

2/11 = 18/99 = 0.181818 (The decimal would be repetition of the numerator)
41/99 is already in that form.
1/3 = 33/99 = 0.3333
2/3 = 66/99 = 0.6666 and so on.
Not something very important, but thought it would be helpful somewhere :)