dude ...thaaaaaaaaaaaaaaanx a million man..it helped me a lot i feel a lil embarassed here cuz ..i mean whn i did da questions maself bein a lil more vigiliant..i got dem...but i still am confused about dat functions one..cuz i mean.. if x=a=b ..how come f(x)= -3(x) justifies f(a+b)=f(a)+f(b) i have dis princeton review gmat manual but it dosnt have ne topic relating to dese sorta functions n am blank at it...
n yeah man my gmats on 13th sept...my mock gmat scores(p.review/gmat-prep/p.prep/kaplan) r around 640-660 ...ne word of advice?
n am like realy pissed @ verbal..sumtymes i skore like 40-45 while sumtymes its like 34-35...am pretty confident on da SC..but da problem iz ..i get stressed out by lookin at da monitor continuosly for 4 hrs n in da end am like..f*** it...
iv been practicing da tests n everything since god know when..like last sptember... am no genius unlike u guys but stil workin at it..
ill appreciate advices,if any n yeah i have sum more questions..am sure deyll b lame n easy for u experts...but again il realy apreciate da help...don kno wht i wud hv done widout u guys...!! :) god bless ya!
see, if a=b=x, when f(x) = -3x, then f(a) = -3a=-3x and f(b) = -3b=-3x f(a)+f(b) = -3x-3x = -6x now f(a+b) = -3(a+b) = -3(x+x) = -6x = f(a)+f(b)
hmm...did u solve kaplan 800?? cuz it has some sums that are really good and open up ur mind to a wider range of flaws that the incorrect statements hv..so ur task becomes all the more easier..and solve tht book even if u tend to get more than a few sums wrong cuz it will all help in the end as long as u dont repeat ur mistakes...well, wat i follow is, i m pretty fast at SC and reasonably fast at CR but freakingly slow at RC..so i compensate for this slowness by catching up on SC and CR and then taking time off the screen for say a minute after every 1/2 hr or so...cuz it really strains ur eyes... and dude, the amount of tests u give does not matter, but it is how much u learn frm each test that matters. so just try to ease off the gas for a while and have a look at all ur mistakes and solve them again..see if u repeat the same mistakes...correct them..all will be fine...all the best!!
and dude, dont wrry, keep posting all ur doubts, no matter hw silly they might seem, who knows, some of the sums might help open our eyes too!
1)hmm can u xplain the sy plain question... i have an idea dat dis question would be solved thru the equation y=mx +b but how.. what wud be the figure if v draw it...??? i mean da line"K" wud b da hypotenuse or what? btw da ans iz "A"
2)the second one iz pretty stupid i admit... well...if da triange is 2wice da area ..den is it spose to have 2ice da base..given a constant height..... am ending up wid "E" but its "C"
n hey guys tell me...dere r only two kinda triangles ...30:60:90 n da 45:45:90..right? iz dere like a 30:30:120 tri? Killer thankz!!!:
1. the question says that k has a +ve slope...slope can be determined as (y2-y1)/(x2-x1)...since we know that the triangle is formed by the line and the two axes, we know that the intersection points lie on the axes with the origin being the third point...so one point will hv x-intercept zero and the other will hv y-intercept zero....since one point has x-intercept 4, its y-intercept will be 0..so one point is (4,0) now the second point will hv x-intercept zero...so, for the slope to be +ve, its y-intercept must be -ve (using the above formula for slope) so u can rule the last 2 options out..since area is 12 and length of one side is 4, the length of other side must be 6...so answer is A...
2. see, the area of bigger triangle is twice the area of smaller triangle...and u can see that both the triangles are similar since their angles are congruent.. so both base and height of the two triangles must be in the same ratio...., now the area of smaller triangle is 1/2 x s x height..but since area of bigger triangle is twice that area, the effective area of bigger triangle is s x height..and u ll get this only when S=sqrt(2) s and HEIGHT = sqrt(2) height...therefore, area of bigger triangle = 1/2 x sqrt(2) s x sqrt(2) height = s x height = 2 x area of smaller triangle..so answer is C.
1)hmm can u xplain the sy plain question... i have an idea dat dis question would be solved thru the equation y=mx +b but how.. what wud be the figure if v draw it...??? i mean da line"K" wud b da hypotenuse or what? btw da ans iz "A"
2)the second one iz pretty stupid i admit... well...if da triange is 2wice da area ..den is it spose to have 2ice da base..given a constant height..... am ending up wid "E" but its "C"
n hey guys tell me...dere r only two kinda triangles ...30:60:90 n da 45:45:90..right? iz dere like a 30:30:120 tri? Killer thankz!!!:
no dude...ther can be a 30-30-120 triangle also!! or for that matter, a 10-10-160, 60-40-60 etc...as long as all three add up to 180 degrees!!
no dude...ther can be a 30-30-120 triangle also!! or for that matter, a 10-10-160, 60-40-60 etc...as long as all three add up to 180 degrees!!
hmm well ma point was do dese triangles have like some sorta fixed ratio for their sides... i mean just as 30:60:90 has a:a(root3):2a and 45:45:90 has it....
hey guys..i have dis one weird lookin diagram...even after da xplanation am unable to interpret wht da question iz tryin to imply...
am realy confused now..ma gmats on 13th and suddenly my skores sseem to decline.... yestday i got like 560 on petersons n today i am stil at 580 :|
i feel like jumping off a plane....
*sigh*
dude, they have mentioned that the dist. btn G1 and G2 is 160 and each is 80m frm the wall..this implies that both are equidistant from the wall..the arcs specify the range of the spotlight..they are asking u the length of the wall that the spotlight can cover (the bold line between the two arcs)...the dist btn G1 and wall is 80m..the spotlight radius is 100m..so the top half wall height is 60m (right angled triangle)...hence, length of wall covered by spotlight = 120m..answer is A.
ahh thx ajay....da diagram looked quite scary...but da solution was quite simple...
newaz looks like am lackin at diagrams n triangles...can u xplain dis 1 too...
1. a total of 26^4 stocks can be assigned 4 letter codes and a total of 26^5 stocks can be assigned 5 letter codes..so total no. of stocks = 26^4 + 26^5 = 26^4(26+1) = 27(26^4)...answer C.
2.
let us assume kaye had 5x stamps and alberto had 3x stamps...after kaye gave 10 stamps, the ratio was 7:5 hence, (5x-10)/(3x+10) = 7/5 solve for x and u ll get x = 30 thus, kaye originally had 150 stamps and alberto had 90...so after the gift, kaye had 140 and alberto had 100...diff = 40..answer C.
3. dude...try to form a triangle at the LHS side of origin where the (-sqrt3, 1) point is...u ll get the angle between the line and x axis as 30 degrees...so the angle on the RHS between the line of (s,t) and x axis is 60 degrees...plus, u know the radius of circle is 2 (apply pythagoras thm to LHS triangle)...so u ll get the length of base of RHS triangle (which is the s-co ordinate) as 1..answer is B.
3. dude...try to form a triangle at the LHS side of origin where the (-sqrt3, 1) point is...u ll get the angle between the line and x axis as 30 degrees...so the angle on the RHS between the line of (s,t) and x axis is 60 degrees...plus, u know the radius of circle is 2 (apply pythagoras thm to LHS triangle)...so u ll get the length of base of RHS triangle (which is the s-co ordinate) as 1..answer is B.
hmm dude..please correct me if im wrong..from what iv understood...uv drawn a perpendicular from point P to x-axis n since the triangle formed wud b 90 deg with hypotenuse being the radius=2 of the circle... the base(distance) wud be 1....and since a triangle at point Q wud b identical...its base iz 1 .thus s=1.... well i stil doubt it.. :S
hmm dude..please correct me if im wrong..from what iv understood...uv drawn a perpendicular from point P to x-axis n since the triangle formed wud b 90 deg with hypotenuse being the radius=2 of the circle... the base(distance) wud be 1....and since a triangle at point Q wud b identical...its base iz 1 .thus s=1.... well i stil doubt it.. :S
thx 4 da enormous help..!!
no dude..i have not mentioned that the triangles are similar at all...see, u drop a line perpendicular to X axis on the LHS..let us name that point S(-sqrt3,0)...in triangle OPS, PS=1, OS=sqrt3, tanPOS = 1/sqrt3..hence POS=30 degrees...now, POQ=90 degrees given...drop a perpendicular from Q to X axis and name tht point T(sqrt3,0)...hence, QOT=60 degrees in triangle QOT, OQ=2, QOT=60 degrees, OT=cos-1QOT x OQ = 1/2 x 2 = 1...hope u hv understood now..
no dude..i have not mentioned that the triangles are similar at all...see, u drop a line perpendicular to X axis on the LHS..let us name that point S(-sqrt3,0)...in triangle OPS, PS=1, OS=sqrt3, tanPOS = 1/sqrt3..hence POS=30 degrees...now, POQ=90 degrees given...drop a perpendicular from Q to X axis and name tht point T(sqrt3,0)...hence, QOT=60 degrees in triangle QOT, OQ=2, QOT=60 degrees, OT=cos-1QOT x OQ = 1/2 x 2 = 1...hope u hv understood now..
Hi Simpler explainationis fo rthis is slope of the line OP i -1/sqrt3 so slope of OQ is sqrt 3 reason is OP and OQ are perpendicular Slope OQ is t/s = sqrt 3 other eqtion t sqare+s sqare =4 (circle equation) radius u will get the answer using bot
ATTN all!!! ZERO is a multiple of any number and it is included in the list of multiples of a number...i verified this while giving the Full length test of Kaplan Premier Edition in the book....
Each of the 30 boxes in a certain shipment with either 10 pounds or 20 pounds, and the average (arithmetic mean) weight of the boxes in the shipment is 18 pounds. If the average weight of the boxes in the shipment is to be reduced to 14 pounds by removing some of the 20-pound boxes, how many 20-pound boxes must be removed? A. 4 B. 6 C. 10 D. 20 E. 24
Each of the 30 boxes in a certain shipment with either 10 pounds or 20 pounds, and the average (arithmetic mean) weight of the boxes in the shipment is 18 pounds. If the average weight of the boxes in the shipment is to be reduced to 14 pounds by removing some of the 20-pound boxes, how many 20-pound boxes must be removed? A. 4 B. 6 C. 10 D. 20 E. 24
i feel the answer is B - 6...because, if the avg weight is 18, then total weight of consignment = 30*18 = 540..if u want to reduce avg to 14, then total weight shud be 30*14 = 420..so u hv to reduce 120 pounds...so 6 20 pound boxes shud be removed...
Each of the 30 boxes in a certain shipment with either 10 pounds or 20 pounds, and the average (arithmetic mean) weight of the boxes in the shipment is 18 pounds. If the average weight of the boxes in the shipment is to be reduced to 14 pounds by removing some of the 20-pound boxes, how many 20-pound boxes must be removed? A. 4 B. 6 C. 10 D. 20 E. 24
30 boxes average is 18 . so total weight is 540 . Now assume x number of 20 pound boxes are removd .
equation is 540-20*x= (30-x)*14 solving you get x=20 so answer is D
i feel the answer is B - 6...because, if the avg weight is 18, then total weight of consignment = 30*18 = 540..if u want to reduce avg to 14, then total weight shud be 30*14 = 420..so u hv to reduce 120 pounds...so 6 20 pound boxes shud be removed...
Ha here it is if you are removing certain numbrer of boxes how can it still be 30*14? it shld be (30-x)*14. is nt it?
Hi All, Note: I had posted a similar question earlier. Please note however that the questions in this attachment are different from the earlier ones. Attached are some of the questions from GMATPrep that I got wrong. The options marked with the bold black dot in the attachment are mine and are the wrong answers. The options marked by a square blue frame are correct as these are the correct answers marked by the software. However the software does not explain why a particular choice is correct and others are wrong. I will be very grateful if some of you can help justify why the correct choice is actually correct! Many thanks. - TheToastedBread
If x is the product of the positive integers from 1 to 8, inclusive, and if i, k, m, and p are positive integers such that x = 2i3k5m7p, then i + k + m + p =
