Today is MY DAY !!!
Doing all my usual classic mistakes !
May be the blazing Saudi sun has taken the toll on me !!!
Thnks Bhavin, for pointing out the mistakes.
Hi,
can u provide the working? Cant figure out much with solution
Hi,
can u provide the working? Cant figure out much with solution
yash, you want working for which problem??
Hi Bhavin,
Actually for both .........:)
Thanx !
Hi All,
I need some help on the attached question. This is from GMATPrep and I got it wrong. The option marked with the bold black dot in the attachment is my wrong answer. The option marked by a square blue frame is 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 someone can help justify why the correct choice is actually correct. Thanks in advance.
Cheers!
I need some help on the attached question. This is from GMATPrep and I got it wrong. The option marked with the bold black dot in the attachment is my wrong answer. The option marked by a square blue frame is 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 someone can help justify why the correct choice is actually correct. Thanks in advance.
Cheers!
in a certain game played with red chips and blue chips, each red chip has a point value
of X and each blue chip has a point value of Y, where X>Y and X and Y are positive
integers. If a player has 5 red chips and 3 blue chips, what is the average (arithmetic
mean ) point value of the 8 chips that the player has?
(1) The average point value of one red chip and one blue chip is 5.
(2) The average point value of the 8 chips that the player has is an integer.
________________________
An equilateral triangle that has an area of 9 3^1/2 is inscribed in a circle. What is the
area of the circle?
A . 6pi B. 9pi C. 12 pi D. 9pi 3^1/2 E. 18pi 3^1/2
_______________________________-
If A@B=A^2+B^2 for all real numbers A and B. then (A@B)@C=
A. A^2+B^2+C^2 B. 2A^2+2B^2+C^2
C. A^4+B^4+C^2 D. A^4+B^4+2A^2B^2+C^2
_________________________________
of X and each blue chip has a point value of Y, where X>Y and X and Y are positive
integers. If a player has 5 red chips and 3 blue chips, what is the average (arithmetic
mean ) point value of the 8 chips that the player has?
(1) The average point value of one red chip and one blue chip is 5.
(2) The average point value of the 8 chips that the player has is an integer.
________________________
An equilateral triangle that has an area of 9 3^1/2 is inscribed in a circle. What is the
area of the circle?
A . 6pi B. 9pi C. 12 pi D. 9pi 3^1/2 E. 18pi 3^1/2
_______________________________-
If A@B=A^2+B^2 for all real numbers A and B. then (A@B)@C=
A. A^2+B^2+C^2 B. 2A^2+2B^2+C^2
C. A^4+B^4+C^2 D. A^4+B^4+2A^2B^2+C^2
_________________________________
[LEFT]in a certain game played with red chips and blue chips, each red chip has a point value[/LEFT]
[LEFT]of X and each blue chip has a point value of Y, where X>Y and X and Y are positive
integers. If a player has 5 red chips and 3 blue chips, what is the average (arithmetic
mean ) point value of the 8 chips that the player has?
(1) The average point value of one red chip and one blue chip is 5.
(2) The average point value of the 8 chips that the player has is an integer.
________________________[/LEFT]
[LEFT]_________________________________[/LEFT]
[LEFT]Hi Yash,[/LEFT]
[LEFT]isn't it a DS question? :smile:[/LEFT]
[LEFT]Anyway, answer to this question is (C) - both statements taken together.[/LEFT]
[LEFT]From first statement we can find out that X+Y = 10 only[/LEFT]
[LEFT]And second statement says that average of all 8 chips is an integer.[/LEFT]
[LEFT]Now, using first statement, we can say that sum of 3 red chips and 3 blue chips will be 30 [ 3(x+y) = 30]
Therefore, we need to find out the sum of remaining 2 red chips. Note that X>Y, means X can have value 5 If we plug in these values of X to find the sum of remaining 2 redchips, we can see that only X= 9 gives us an average of all 8 chips as an integer.[/LEFT]
[LEFT]==> [3(x+y) + 18]/8 ==> [30 + 18]/8 ==> 48/8 = 6[/LEFT]
[LEFT]Therefore using both the statement togehter we can find out the required average.[/LEFT]
[LEFT]________________________[/LEFT]
[LEFT]An equilateral triangle that has an area of 9 3^1/2 is inscribed in a circle. What is the
area of the circle?
A . 6pi B. 9pi C. 12 pi D. 9pi 3^1/2 E. 18pi 3^1/2
_______________________________-[/LEFT]
If
[LEFT]A@B=A^2+B^2 for all real numbers A and B. then (A@B)@C=
A. A^2+B^2+C^2 B. 2A^2+2B^2+C^2
[LEFT]C. A^4+B^4+C^2 D. A^4+B^4+2A^2B^2+C^2
_________________________________[/LEFT]
[/LEFT]
[LEFT]1. Answer is (C) - 12pi
Solution: (3^1/2 * a^2)/ 4 = 9 * 3^1/2
a^2 = 36
a = 6
Because equilateral triangle is inscribed in a circle, radius of circle "r" will be equal to 2/3 of the altitude of the triangle.[/LEFT]
[LEFT]==> r = 2/3 [ ( 3^1/2 a)/2]
==> r = 2 * 3^1/2[/LEFT]
[LEFT]Therefore area of the circle with radius r is 12pi.[/LEFT]
[LEFT]2. answer is (D)... it is the simple expansion of (a^2 + b^2)^2 [/LEFT]
let me know the OAs plz...
Hi Bhavin ,
Two more problems:
1)Sum of 50 positive integers is 2550. What is he sum of even integers from 102 to 200 inclusive?
5100 7550 10100 15500 20100
2) Amys grade was the 90th percentile of he 80 grades for her class. Of the 100 grades from another class, 19 were higher than amys and the rest were lower. If no other grade was same as amys grade then amys grade was what percntile of the grades of he two classes combined.
72nd 80th 81st 85th 92nd.
Hi Bhavin,
Actually for both .........:)
Thanx !
hi yash...
for the first sum...3 methods have already been posted ...i'll just clarify....
for A.P, sum of n terms is given by Sn = n/2 ...a= first term, d = common difference and n= no of terms..
Here, n=50 , a=102 , d=2
Hence S50 = 7550
Also, Sn = n/2 (T1 + Tn) where T1 and Tn are first and last term resp...
So, S50 = 25 ( 102+200)..
either which ways, ans is 7550..
2) 90 percentile implies 10% crowd is above him..hence 10 % of 80 = 8
and 19 from other class are above him.
2 classes combined, 8+19 = 27 people are above him and total strength is 80+100 = 180
27 is 15% of 180 i.e 15% of the crowd is above him, hence 85 percentile..
hope this solves your doubt..
Hi All,
I need some help on the attached question. This is from GMATPrep and I got it wrong. The option marked with the bold black dot in the attachment is my wrong answer. The option marked by a square blue frame is 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 someone can help justify why the correct choice is actually correct. Thanks in advance.
Cheers!
hi...here is my attempt for the exlanation..4 letters can be assigned to 4 envelopes in 4! ways...
now, 1 has to be correct address and other 3 incorrect...
any 1 of the 4 letters can be assigned to correct address in 4 ways..
now consider the remaining 3 letters...they should not be at the correct address...its a case of de arrangement..
simply consider it this way..let A, B, C, D be 4 letters..
if A is assigned to correct envelope...try out the ways B, C and D should not be in correct envelope..
For B- 2 incorrect places
For C -1 incorrect place..
D would occupy the last incorrect place..
total possibilities meeting the above criteria is 4*2*1=8
p = 8/24 = 1/3...
Hi All,
I need some help on the attached question. This is from GMATPrep and I got it wrong. The option marked with the bold black dot in the attachment is my wrong answer. The option marked by a square blue frame is 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 someone can help justify why the correct choice is actually correct. Thanks in advance.
Cheers!
Total ways four letters can be put in four envelopes is 4!
Total ways in which exactly one letter is in the right envelope is
4C1*1*3!(1-1/1!+1/2!-1/3!)=4*2=8
4C1 to choose one envelope and put the right letter in it.
3!(1-1/1!+1/2!-1/3!) is the number of ways to put 3 letters in the remaining 3 envelopes such that no envelope has its intended letter.
This is called derangement and the general formula for this is
n!(1-1/1!+1/2!-1/3!+......(-1)^n/n!)
8/24=1/3
Hi guys,
-4 (|-20|-5)=
A. -100 B. -60 C. 60 D. 75 E. 100
Acc to me the ans. is C , can someone confirm.....
-4 (|-20|-5)=
A. -100 B. -60 C. 60 D. 75 E. 100
Acc to me the ans. is C , can someone confirm.....
Hi guys,
-4 (|-20|-5)=
A. 100 B. 60 C. 60 D. 75 E. 100
Acc to me the ans. is C , can someone confirm.....
If the question is correct, then my answer is also 60.
HG
yes i too think its C..
m and n are the x and y coordinates, respectively, of a point in the coordinate
plane. If the points (m, n) and
both lie on the line defined by the equation
, what is the value of p? 
1 
2 5 8 is c right???
m and n are the x and y coordinates, respectively, of a point in the coordinate
plane. If the points (m, n) and
by the equation
is c right???
Yeah absolutely. substituting (m,n) in the equation, we get
m = (n/2) - (2/5) --- 1
substituting (m + p, n+4)
m + p = ((n+4)/2) - (2/5) --2
substituting value of m from (1) in (2)
(n/2) - (2/5) + p = (n/2) + (4/2) -(2/5)
hence p = 2
m and n are the x and y coordinates, respectively, of a point in the coordinate
plane. If the points (m, n) and
by the equation
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is c right???
absolutely..just use slope formula..
x=y/2 + 2/5 => y = 2x + 4/5 or m=slope= 2
So, n+4-n/ m+p-m = 2 => p = 2
. If y 3 and 3x/y is a prime integer greater than 2, which of the following must be true?
. x = y
. y = 1
. x and y are prime integers.
(A) None
(B) only
(C) only
(D) only
(E) and
I think ans is B but provided solution ia A........can anyone confirm
. x = y
. y = 1
. x and y are prime integers.
(A) None
(B) only
(C) only
(D) only
(E) and
I think ans is B but provided solution ia A........can anyone confirm
. If y 3 and 3x/y is a prime integer greater than 2, which of the following must be true?
. x = y
. y = 1
. x and y are prime integers.
(A) None
(B) only
(C) only
(D) only
(E) and
I think ans is B but provided solution ia A........can anyone confirm
yes...ans is A...
3x/y is a prime integer greater than 2..so x/y need not be an integer...and also x need not be equal to be y, it is also not necessary for x and y to be prime integers...Hence A..
If in doubt, just consider, x=15 and y = 9...then 3x/y is prime (5) and x not equal to y..
there are many more egs..
yes...ans is A...
3x/y is a prime integer greater than 2..so x/y need not be an integer...and also x need not be equal to be y, it is also not necessary for x and y to be prime integers...Hence A..
If in doubt, just consider, x=15 and y = 9...then 3x/y is prime (5) and x not equal to y..
there are many more egs..
Thanks buddy.........for ur speedy help.