GMAT Prep 7 ques

hi vivek!
search the question bank and you will find many of your questions answered before

question bank means previous posts on this link or somewhere else.

Milk Fat

X gallons contain X/100 fat (fat = 1%)
Y gallons contain 2y/100 fat (fat =2%)
Z gallons contain 3z/100 fat (fat =3%)

total fat = x/100 + 2y/100 + 3z/100

x+y+z gallons contain 1.5*(x+y+z)/100 fat (fat = 1.5 %)

so x/100 + 2y/100 + 3z/100 = 1.5*(x+y+z)/100

multiply by 200

2x + 4y + 6z = 3(x+y+z) = 3x + 3y + 3z

x = y + 3z

Ans: a

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letter code

as the alphabets can be repeated, and their odder matters

single alphabet code: 26 ways..................................(code = X)
double alphabet code: 26*26 ways............................(code = XX)
triple alphabet code: 26*26*26 ways.........................(code = XXX)

add all three you get 26 + 26^2 + 26^3 = 18278

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yes in the same forum

DS question:
the expression can be true only if the triangular symbol(i will be writing it as #) is either a + or X

choice 1: gives us only two possibilities addition 3+5 > 3
3 x 2 > 3

for both the expression in question is true..............sufficient

choice 2:
3 x 1 = 3, 3 / 1 = 3
gives two choices, one of which does not satisfies the req

hence ans: A

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Let the growth of tree = x feet per yr

at the end of 4th year it is: 4 + 4x
6th year it is: 4 + 6x

growth in two yrs = 2x

2x/(4 + 4x) = 1/5

10x = 4 + 4x

x = 2/3

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CIRCLE

let the centre be X

PXO = 2*PRO = 70

chord is parallel to dia. perpendicular line from centre X will bisect both chords at 90 deg

similarly QOR will be 35, so QXR will be 70

so PXQ will be 180 - 70-70 = 40

minor arc pq = pi*dia*angle at centre / 360 = pi*18*40 / 360 = 2pi

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next circle

OB = OC = BC = 1
hence its equilaterla triangle
each angle = 60 deg

angle BOC = 60 = OCB

angle OAB = 1/2(angle BOC) = 30

hence angle CBA = 90

ALSO ANY TRIANGLE WITH DIAMETER AS ONE SIDE AND THIRD POINT ON CIRCUMFRENCE WILL BE A RT ANGLED TRIANGLE WITH RT ANGLE AT POINT ON CIRCUMFRENCE

AC=2

AB^2 = 2^2 - 1^2; AB = sq rt 3

area = AB*BC/2 = sq rt 3/2
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Let M, W, C represent number of men, women and children on tour

given w/c = 5/2 or C = 2W/5

from I: C/M = 5/11 or C = 5M/11

equqting C from two equations: 2W/5 = 5M/11 or M = 22W/25

no further info.................not sufficient

from II: W < 30

not sufficient alone

BOTH I and II: as W < 30
and M = 22W/25, only one value of W=25 will satisfy as people caanot be in fractions

M = 22

both together, not alone Ans:C

mental wrote:
letter code

as the alphabets can be repeated, and their odder matters

single alphabet code: 26 ways..................................(code = X)
double alphabet code: 26*26 ways............................(code = XX)
triple alphabet code: 26*26*26 ways.........................(code = XXX)

add all three you get 26 + 26^2 + 26^3 = 18278

..........................................................................

I used the same logic, but I want to add that there was no need to calculate 26 + 26^2 + 26^3 = 18278
look at units digit. if number has units digit 6, any power of this number will have units digit 6

we have 3 numbers with units digit 6. Guess, the summ of these numbers will have what number in units digit? Only 8!!!!!! 6*3 = 18
We have only 1 choise with units digit equal to 8

Hey, we saved 1 minute perhaps!!!!

mental wrote:Let M, W, C represent number of men, women and children on tour

given w/c = 5/2 or C = 2W/5

from I: C/M = 5/11 or C = 5M/11

here I made another thing - brought in correspondence with each other
ratio of W:C:M = 25:10:22
so number of men can be 22, 44, 66 and so on
INSUFF


(2)
W < 30
INSUFF

(1) and (2)
W < 30 then only W:C:M = 25:10:22, so M = 22

can any1 explain 2 ques on circles...mental if u can elborate a bit,,,,

Vivek - could you highlight the areas and i shall try to help

if there is a triangle with diameter as one side and the third point anywhere on circumfrence,
then,
the triangle is a rt angled triangle
diameter is the longest side of that triangle (hypotenuse)


let us consider the arc BC
the abgle subtended by any arc at centre is double the angle subtended by same arc on circumfrence.

mental wrote:CIRCLE

let the centre be X

PXO = 2*PRO = 70

chord is parallel to dia. perpendicular line from centre X will bisect both chords at 90 deg

similarly QOR will be 35, so QXR will be 70

so PXQ will be 180 - 70-70 = 40

minor arc pq = pi*dia*angle at centre / 360 = pi*18*40 / 360 = 2pi

.......................................................

Here, I wasnt able to how 70 u have calculated