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TIME SAVING PROJECT [#permalink]
22 Jan 2007, 15:34

1

This post was BOOKMARKED

please post in this thread questions that are easy but took you too much time to solve. we'll try to discuss methods of saving time: improving you arithmetic skills, finding shortcuts, representing things in a simpler way that requires less calculations.

since these things are better learnt by example - please use this thread to post such examples. we don't argue here what the answer is. please specify your solution and the way you solved the question and why it took you time. we address and discuss here only time saving strategies

(A) x / (x-1) (B) x / (x+1) (C) (x-1) / x (D) (x+1) / x (E) x

again... please try to outline the way you did it, so we can address specific pitfalls and time-wasting paths.

there are 2 ways to solve. either way is fine, and each may find a different way to be simpler than the other...

1) solve the given equation using algebra:
(x+y)/(xy)=1 => (multiply each side by xy to eliminate denominator)
x+y = xy => (substract y from each side to get all y's on the right)
x = xy-y => (take y as a common factor of expressions)
x = y(x-1) => (divide by x-1 to isolate y)
y= x/(x-1)
(answer is A)

2) simplify expression first
(x+y)/(xy) = 1/x + 1/y (from my exprience i notice that immediately. however it can be done in two simple algebraic steps)
now solve...
1/x+1/y=1 => (just pass 1/x to the right side)
1/y = 1-1/x => (represent it as a simple fraction using common denominator)
1/y = (x-1)/x => (invert both sides)
y = x/(x-1)
answer is A

for me path no. 2 is faster because i recognize immediately some patterns. some people may find path no.1 faster as it doesn't require anything more than a sequence of simple algebraic steps.

the number 1 time waster, "picking numbers", shouldn't be used here at all... although the question is put in a way that "invites" number plugin... don't go that path. the stem information is given in the form of equation - and it is usually difficult to "pick numbers" that solve a given equation. use "picking numbers" if you are given a characteristics of the numbers and need to check an equation.....

also, if you followed path 1 above, note my remarks to see why i chose certain solving operations. it is just guidelines, and there will inevitably be exceptions - but when solving an equation, try to:
a) get rid of fractions by multiplying everything with the common denominator.
b) pass all sub-expressions with the "solving variable" to one side (using addition/substraction)
c) extract the solving variable as a common factor.
d) divide to isolate the solving variable.

If the two-digit intergers M and N are positive and have the same digits, but in reverse order, which of the following CANNOT be the sum of M and N?

(1) 181 (2) 165 (3) 121 (4) 99 (5) 44

the key here is to avoid picking numbers, unless you are a picking number wizard

without picking numbers the solution goes like this:

if the digits are x and y
then M is 10x+y and N is 10y+x
M+N = 11x+11y = 11(x+y)
which means that M+N must be divisible by 11.
44,99,121,165 are divisible by 11. 181 is not. hence 181 is your answer.
answer is A.

if you wanted to pick numbers anyway, do it wisely. here is what you should do.
first, you should notice that you are not picking numbers for M and N, but for their digits (x and y).
second, you should notice that the unit's digit of M+N is the same as the unit's digit of x+y
for 44 you should look for x and y whose sum is 4. 3 and 1 would do. 31+13 = 44.
for 99, chouse 4 and 5. 45+54=99
for 121 choose something with some 1 or 11. 1 is impossible, so choose 5 and 6. 56+65 = 121.
for 165 choose something which adds to 5 or 15. choosing 2,3 won't do since 23+32=55, but choosing 6 and 9 will do: 69+96=165

having found 4 examples for B-E leaves you with the right answer A.

here is the question:
Q: If a, b, and c are digits, is (a + b + c) a multiple of 8?

(1) The three-digit number abc is a multiple of 8.
(2) a x b x c is a multiple of 8.

answer is E, of course... but how can you get to it fast?

if you try to pick numbers to look for an answer - then your search may be long and time-consuming. avoid that if possible.
however - if you guess an answer, and pick number just to verify it - then your journey will probably be short and focused.

why is that? just for starter - if you try to check independently that st1 is insufficient, then that st2 is insufficient, and then that both together are insufficient - it is 3 searches.... if you know that answer is E you search only once....

in DS questions, picking numbers can be used only to prove insufficiency. don't use it to prove sufficiency.... so if you feel, or have reasons to believe that something is insufficient - try to pick numbers: one set of numbers that answer positively to stem, and one set that gives a negative answer.

if you have reasons to believe something is insufficient try not to do it 1 by 1. start with both together first (i.e. trying to prove E is the answer)...

but again - use picking numbers only if you have a strong feel that a statement is insufficient.

enough general strategies.....

so why the answer here is E?
st1 - you probably know about a test of divisibility that involves adding digits.... that works for 3 and 9 (i.e if sum of digits is a multiple of 3 so is the original number, and vice versa)...... it doesn't work for 8.
st2 - you probably have the feel from experience, that properties of sum of numbers and properties of product of the same numbers are independent....

so it seems that neither statement provides us with real information... so it is time to (educatedly) guess that the answer might be E...

hey, don't rush to next question....we need to verify our educated guess by picking numbers.... but now, that we are focused, it would be easy:

we seek two sets of numbers a,b,c. each set must satisfy st1 AND st2.
one should have a sum divisible by 8 and one should have a sum that is not.

how to pick them wisely?

first consider st2. it says a*b*c should be divisible by 8. there is an easy way to fulfil that.... just pick one of them, say a, to be 8. now regardless of the two others st2 is going to be satisfied....

the second one is easy too.... we need 3 digit number, that starts with 8... let's just write down the first 4
800, 808, 816, 824

all of these satisfy both st1 and st2.
lets check stem:

for 808 a+b+c = 16 answer is yes
for 816 a+b+c = 15 answer is no.

I think it would be beneficial for the purpose of this thread if you specify what part of the solution was most time consuming so we can address that specifically...anyway...

here is the original question:

If x>y^2>z^4 which of the following could be true?

I)x>y>z
II)z>y>x
III)x>z>y

a) I only
b) I & II
c) I & III
d) II & III
e) I, II, & III

the answer is E.

here is how i think it is the fastest to solve it:

the stem gives us an ordering of exponents of x,y,z and asks for possible orderings of x,y,z themselves.
here are few things that comes to mind immediately on understanding the question:
a) the questions asking about something that "could be true". an example is the ultimate proof of such thing - but don't rush immediately to try numbers... think a bit more...
b) we know that for positive numbers: a>b <a>b^n for any n. however you should note immediately that the question compare numbers with different exponents so the above rule would be hard to use... any rule that "immediately comse to mind" but is not fully applicable is a source for traps and troubles... so make a mental note not to fall to one of these.
c) another thing that may come to mind is that x<x>1 and x>x^2 for 0<x<1>y^2 is it possible that y>x?

of course it is... remember point c we made above.... when taking numbers less then 1 this thing can happen. a canonic example you probably encountered in your prep is 1/2 and 1/3.

if you got to this point and you are under time pressure - just mark E and move on.

but hey - you got this far, why not finding the ultimate proof - gicing an example for each statement.

following your intuition for st.1 you get x=1000 y=10 z=1 as a good example. for st.3 is just a varioation as we already noted so use: x=1000 y=-10 and z=-1

st.2 we already have x=1/3 y=1/2 to complete the picture just add z=2/3 to get a good example.

Last year Department Store X had a sales total for December that was 4 times the average (arithmetic mean) of the monthly sales totals for January through November. The sales total for December was what fraction of the sales total for the year?

Last year Department Store X had a sales total for December that was 4 times the average (arithmetic mean) of the monthly sales totals for January through November. The sales total for December was what fraction of the sales total for the year?

(1) 1/4 (2) 4/15 (3) 1/3 (4) 4/11 (5) 4/5

this is a classic for "picking numbers", although it might not look like that in the first place.

how do i recognize that:
a) it is a question with many parameters that are not known (i.e. the actual sales data for jan through dec)
b) there is a simple constraint or characteristic that combines them all (i.e. dec value is 4 times the average of the rest)
c) there is one answer that fits all possible numbers ( you know that it is a GMAT question and there is not an option like "data is not sufficient to determine" - if there was one, like in DS question - i won't use picking numbers here).

since a,b,c holds the best way is to look for one example that fits the constraint of (b). check for the answer, and know that it fits all other possibilities, and hence it is the right answer.

picking numbers here is easy. let jan through nov sales be 1 (for each month). then according to the constraint, dec sales must be 4.
so we have total yearly sales of 15. so the required fraction is 4/15

i guess you meant the number 0.756 (and not 0.0756)

you should know 3/4 is 0.75 (this is basic, you should either be already familiar with or memorize the conversions of simple fractions to their digital representation).

soo the difference is clearly 0.006
yet again immediately convert it to simple fraction 6/1000
now simplify it (divide by 2 as a common factor) to get 3/500
answer is D

Some Tips on Standard Deviation Sums particularly DS
1) If you add or subtract a particular constant from all the terms the sd deviation remains same and the mean is either increased or decreased by constant
2) If you multiply all terms by x then SD =x times old SD and mean = x times old mean.
3) For comparing the SD for two sets any information about mean ,median,mode and range are insufficient unless you can determine the individual terms from the given data.

If the two-digit intergers M and N are positive and have the same digits, but in reverse order, which of the following CANNOT be the sum of M and N?

(1) 181
(2) 165
(3) 121
(4) 99
(5) 44

without picking numbers the solution goes like this:

if the digits are x and y
then M is 10x+y and N is 10y+x M+N = 11x+11y = 11(x+y)
which means that M+N must be divisible by 11.
44,99,121,165 are divisible by 11. 181 is not. hence 181 is your answer.
answer is A.

if the digits are x and y then M is 10x+y and N is 10y+x

if the tens digit of a number is k and the unit digit is m then the number is 10k+m...

so if M and N are two digit numbers with reverse digits x and y,
then one of them has x as tens digit and y as unit digit - then the number is 10x+y. the other number must have y as tens digit and x as units digit. so it is 10y+x

for 808 a+b+c = 16 answer is yes for 816 a+b+c = 15 answer is no.

I have a question. Here, are we considering 0 as a multiple of 8? That is what made me pick diff. set of nos. I think this question was discussed before. I'm not sure of the answer, though. Should we always consider 0 as a multiple of any no.?

for 808 a+b+c = 16 answer is yes for 816 a+b+c = 15 answer is no.

I have a question. Here, are we considering 0 as a multiple of 8? That is what made me pick diff. set of nos. I think this question was discussed before. I'm not sure of the answer, though. Should we always consider 0 as a multiple of any no.?

Thanks

0 is certainly, under any circumstance, a multiple of 8.
(it is not a positive multiple of 8 though.... but no one said anything about positive)

with no options to choose from it is hard to show some important shortcuts that can be made (like rough estimation digit checking etc...)
we are left with pure computation.... and the worst thing you can do is to write it down and do paper&pencil long multiplication....

here is what i do:
987=1000-13
use the formula (a-b)^2= a^2-2ab+b^2
1000^2 is 1M
13^2 is 169
and 2*13*1000=26000
so the answer is 974,169

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