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123 - 131 Arithmetics

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123 - 131 Arithmetics

by phoenix9801 » Sat Jul 10, 2010 11:10 pm
Hi, can someone please help to explain by step-by-step instructions (in detail) in the most simplest way and clear to understand. Would Greatly appreciate it.

123-
Joanna bought only $0.15 stamps and $0.29 stamps. How many $0.15 stamps did she buy?

(1) She bought $4.40 worth of stamps

(2) She bought an equal number of $0.15 stamps and $0.29 stamps.



131-
When a player in a certain game tossed a coin a number if times, 4 more heads than tails resulted. Heads or tails resulted each time the paler tossed the coin. How many times did heads results?

(1) The player tossed the coin 24 times.

(2) The player received 3 points each time heads resulted and 1 point each time tails resulted, for a total of 52 points.


(note the answers is D , I hope some can explain it in a simplest way)
Last edited by phoenix9801 on Mon Jul 12, 2010 1:22 pm, edited 2 times in total.
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by albatross86 » Sat Jul 10, 2010 11:34 pm
Question 1

Let number of 0.15$ stamps be X and number of 0.29$ stamps be Y. What is X?

1. 0.15X + 0.29Y = 4.40

=> 15X + 29Y = 440

Now, this is an equation in two variables, where the variables may only take INTEGER values.

Here, it may be possible that the equation has only one solution.

First, find the LCM of 15 and 29 => 435. Add the lower coefficient, i.e. 15 => 435 + 15 = 450

Now 450 > 440. In such a situation, where the RHS is less than this value that we found by adding the lower coefficient to the LCM of the coefficients, we can safely assume that there is only ONE true INTEGER solution to the problem. Do not bother solving it, just realize that this is sufficient and move on.

Otherwise you could test cases, but if you can remember this simple test it's a lot faster :)

SUFFICIENT.

2. X = Y

So what, it could be any number under the sun.

INSUFFICIENT.

Pick A.

Question 2

H = T + 4
H = ?

Now, we have one equation for H in terms of T. If we can get one more INDEPENDENT (i.e. which is not just the same eqn multiplied throughout by some constant) simultaneous equation for H and T, this is sufficient to find each value.

1. H + T = 24

Two equations, two variables. We can easily find H.
SUFFICIENT

2. 3H + T = 52

Two equations, two variables. We can find H..
SUFFICIENT

Pick D
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by GMATGuruNY » Sun Jul 11, 2010 2:00 am
Joanna bought only $0.15 stamps and $0.29 stamps. How many $0.15 stamps did she buy?

(1) She bought $4.40 worth of stamps

(2) She bought an equal number of $0.15 stamps and $0.29 stamps

Statement 1:

15x + 29y = 440.

Normally, 2 variables and 1 equation would be insufficient, but we have to recognize that this problem is restricted to positive integers. We can't buy 1/2 of a stamp.

Let's start with the 15 cent stamps:

15*1 = 15
15*2 = 30
15*3 = 45
etc.

These are all multiples of 5. To spend 440 in total, the amount spent on the 29 cent stamps also will have to be a multiple of 5:

29*5 = 145
29*10 = 290
29*15 = 435

Only the 3 options above could work. Let's test out each. The amount left over will need to be divisible by 15:

440-145=295, which is not divisible by 15.
440-290=150. 150=15*10. This works: 15*10 + 29*10 = 440.
440-435=5, which is not divisible by 15.

Since only one combination works, SUFFICIENT. Joanna bought 10 of the 15 cent stamps.

Statement 2:

Without a total, Joanna could buy 1 of each kind of stamp, 2 of each, 3 of each, etc.
INSUFFICIENT.

The correct answer is A.

This is a tricky question. The GMAT writers are hoping that the average test taker will pick C. When a problem is restricted to positive integers, be careful! Play out the different scenarios before you decide whether a statement is sufficient.
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by phoenix9801 » Mon Jul 12, 2010 2:16 pm
Can you explain how did you know when to use 1,2,and 3 and 5,10,15. Is it way to get LCM to get 440 prove it ?????????
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by GMATGuruNY » Mon Jul 12, 2010 7:03 pm
phoenix9801 wrote:Can you explain how did you know when to use 1,2,and 3 and 5,10,15. Is it way to get LCM to get 440 ?????????
Joanna has to spend 440¢. She buys only 15¢ or 29¢ stamps. 15x + 29y = 440.

The total spent (440) has a units digit of 0. 29 is a strange number. There can't be too many ways we can add a multiple of 15 to a multiple of 29 and get a sum with a units digit of 0.

Multiples of 15 end in 5 or 0: 15, 30, 45, 60, etc.

If we use a multiple of 15 that ends in 5, we need to use a multiple of 29 that ends in 5 as well, so that when we add the units digits, we'll get 5 + 5 = 0. To get a multiple of 29 that ends in 5, we need to multiply 29 by a multiple of 5 that ends in 5:

5 * 29 = 145
15 * 29 = 435

If we use a multiple of 15 that ends in 0, we need to use a multiple of 29 that ends in 0 as well, so that when we add the units digits, we'll get 0 + 0 = 0. To get a multiple of 29 that ends in 0, we need to multiply 29 by a multiple of 10:

10 * 29 = 290
20 * 290 = 580 (too big, because the total spent is 440)

So for the number of 29$ stamps, the only viable options are:

5 * 29 = 145
15 * 29 = 435
10 * 29 = 290

145 doesn't work because 440 - 145 = 295, which is not divisible by 15.
435 doesn't work because 440 - 435 = 5, which is not divisible by 15.
290 works because 440 - 290 = 150, which is divisible by 15: 150/15 = 10.

So the only viable option is 10*15 + 10*29 = 440.

So Joanna bought 10 of each kind of stamp.

Does the explanation above help?
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by barcebal » Tue Jul 13, 2010 9:38 am
Now, this is an equation in two variables, where the variables may only take INTEGER values.

Here, it may be possible that the equation has only one solution.

First, find the LCM of 15 and 29 => 435. Add the lower coefficient, i.e. 15 => 435 + 15 = 450

Now 450 > 440. In such a situation, where the RHS is less than this value that we found by adding the lower coefficient to the LCM of the coefficients, we can safely assume that there is only ONE true INTEGER solution to the problem. Do not bother solving it, just realize that this is sufficient and move on.

Otherwise you could test cases, but if you can remember this simple test it's a lot faster Smile
What is the RHS?
How did you come by this rule?
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by albatross86 » Thu Jul 15, 2010 3:21 am
Hey,

After reviewing the rule I posted, I have found some extraneous cases which would not work, so I apologize and request you to ignore it.

It is supposed to be used as a negation test, i.e. if the value we calculated is greater than the RHS( Right-hand side) then there may be more than 1 integer solutions, but if it is less, there is only ONE or NO integer solutions.

Please use Mitch's method which is much more GMAT friendly!
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by neonaan » Wed Aug 04, 2010 7:26 pm
Hi All.
I saw Mitch's answer but the approach to this is even convenient if the answer chosen is C.

Explanation: 0.15 x + 0.29 y =4.40 (from 1)

from 2) Joanna bought equal number of 0.15c stamps as 0.29c stamps.

using 2 & 1, we can state this as 0.15x + 0.29x = 4.40 => 0.44x =4.40 => x=10

0.15*10 + 0.29 *10 =4.40

Please throw light on this issue if this approach is incorrect.
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by pradeepkaushal9518 » Thu Aug 05, 2010 6:19 am
both option A and C can be explained to be correct so what to do in that case?
as we know 2 variables requires 2 equations.so C can be the answer why should i do lot of things to prove that A is correct.
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by Ian Stewart » Thu Aug 05, 2010 12:33 pm
pradeepkaushal9518 wrote:both option A and C can be explained to be correct so what to do in that case?
as we know 2 variables requires 2 equations.so C can be the answer why should i do lot of things to prove that A is correct.
Answer choice C on GMAT Data Sufficiency questions reads:

C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

If statement 1 is sufficient alone, answer choice C is not correct, because of what I highlighted in red. So you can never have a question where answers A and C can both legitimately be 'explained to be correct'. Because of this, you always need to check each statement individually before proceeding to combine them.
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