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vivek.kapoor83
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- Joined: Mon Aug 04, 2008 1:56 am
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hi vivek!
search the question bank and you will find many of your questions answered before
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GMAT Prep 7 ques
Source: Beat The GMAT — Problem Solving |
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
............................................................
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
..........................................................................
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
............................................................
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
..........................................................................
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
........................................................
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
.......................................................
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
........................................................
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
.......................................................
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
.......................................................
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
.......................................................
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
...................................................
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
...................................................
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
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
I used the same logic, but I want to add that there was no need to calculate 26 + 26^2 + 26^3 = 18278mental 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
..........................................................................
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!!!!
here I made another thing - brought in correspondence with each othermental 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
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
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.
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
