Q2: If 2 different representatives are to be selected at random from a group of 10 employees and if p is the probability that both representatives selected will be women, is p > ? (1) More than of the 10 employees are women. (2) The probability that both representatives selected will be men is less than .
A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient. B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient. D. EACH statement ALONE is sufficient. E. Statements (1) and (2) TOGETHER are NOT sufficient.
Q13: If n is a positive integer, what is the remainder when 3^(8n+3) + 2 is divided by 5?
0
1
2
3
4
Q19: Each of the 30 boxes in a certain shipment weighs either 10 pounds or 20 pounds, and 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?
4
6
10
20
24
Q25: If the sequence x1, x2, x3, , xn, is such that x1 = 3 and xn+1= 2xn - 1 for n 1, then x20 - x19 =
May be the right mentors and guidance can get you a 770 in GMAT. To find out, you could be part of the event "GLOBAL MBA DREAM" in Hyderabad where an ISB representative will be answering all admissions related questions face to face along with an Oxford alumni giving an academic session called "GMAT Made Easy". If you want you can register for the event, free of charge:
#2, Panchsheela Towers 2nd Floor Park Lane, Yellammagutta Secunderabad H.O. Hyderabad - 500003 Landmark: Beside Taruni Phone: 040-66499389 416, Amrutha Estates Himayathnagar Hyderabad - 500 029. Landmark: Above Pick 'N' Move Phone: 040 - 66254600 #204, 2nd floor, Mekins Mayank Maheswari Plaza Ameerpet Hyderabad-16 Landmark: Close to Green park Hotel Phone:040-66364600
Q2: If 2 different representatives are to be selected at random from a group of 10 employees and if p is the probability that both representatives selected will be women, is p > ? (1) More than of the 10 employees are women. (2) The probability that both representatives selected will be men is less than .
Q13: If n is a positive integer, what is the remainder when 3^(8n+3) + 2 is divided by 5?
0
1
2
3
4
Q19: Each of the 30 boxes in a certain shipment weighs either 10 pounds or 20 pounds, and 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?
4
6
10
20
24
Q25: If the sequence x1, x2, x3, , xn, is such that x1 = 3 and xn+1= 2xn 1 for n 1, then x20 x19 =
2^19
2^20
2^21
2^20 - 1
2^21 - 1
Ashish,
The first and last question has already been answered by someone on page 156. Please have a look.
Q13. Is the answer 5)-remainder 4? Q19. Is the answer 4) - 20 boxes must be removed? Lets assume 10 pound boxes = x So 20 pound boxes will be = 30-x (10x+20(30-x))/30 = 18 => x=6 (10 pound boxes) => so 30-6=24 (20 pound boxes) Now take second part. We know that 10 pound boxes are 6 in number, however we have to find out how many boxes of 20 pounds are there to make the average 14. and so how many 20 pound boxes are removed. Lets assume 20 pound boxes now to be = x So (10*6+20*x) / (6+x)=14 => x=4 So 20 pounds boxes to be removed are = 24 - 4 => 20
Last month 15 homes were sold in Town X. The average (arithmetic mean) sale price of the homes was $150,000 and the median sale price was $130,000. Which of the following statements must be true?
I. At least one of the homes was sold for more than $165,000. II. At least one of the homes was sold for more than $130,000 and less than $150,000. III. At least one of the homes was sold for less than $130,000.
A) I only B) II only C) III only D) I & II E) I & III ------------------------------------------------------------------------ A set of 15 different integers has a median of 25 and a range of 25. What is the greatest possible integer that could be in this set?
A) 32 B) 37 C) 40 D) 43 E) 50 ----------------------------------------------------------------------
Last month 15 homes were sold in Town X. The average (arithmetic mean) sale price of the homes was $150,000 and the median sale price was $130,000. Which of the following statements must be true?
I. At least one of the homes was sold for more than $165,000. II. At least one of the homes was sold for more than $130,000 and less than $150,000. III. At least one of the homes was sold for less than $130,000.
A) I only B) II only C) III only D) I & II E) I & III
------------------------------------------------------------------------ A set of 15 different integers has a median of 25 and a range of 25. What is the greatest possible integer that could be in this set?
Median is 130000 ie it is the middle value of 15 values or 8th value in the series. so the 7 values behind the median can be all equal to 130000 or less.So we cannot say At least one of the homes was sold for less than $130,000. 3rd statement is out. Total value of all homes = 15*150000=2250000 Total value of all 8 homes ie 1st to the 8th= 8*130000=1040000 Total value of remeining 7 homes= 2250000-1040000=1210000 Value of 1 home=1210000/7= 172857 so statement 2 is also out and statement 1 is true. So ans A:-o
2. I think D
median is 25 means 8th term in the series is 25 .so the 1st to 7th no: in this series can be either 25 or less than that.But we have to maximize the range ie 1st term shiuld be as small as possible ie 1st term = 25-7=18 So last term of series with 1st term 18 and range 25 is 43. Ans D;)
I have a doubt in Question (2).. why did you subtract 7 from 25 to get the smallest integer? are you assuming that the integers are consecutive?
Why can't the integers NOT be consecutive..? then the smallest integer cannot be determined right?
1. I think A
Median is 130000 ie it is the middle value of 15 values or 8th value in the series. so the 7 values behind the median can be all equal to 130000 or less.So we cannot say At least one of the homes was sold for less than $130,000. 3rd statement is out. Total value of all homes = 15*150000=2250000 Total value of all 8 homes ie 1st to the 8th= 8*130000=1040000 Total value of remeining 7 homes= 2250000-1040000=1210000 Value of 1 home=1210000/7= 172857 so statement 2 is also out and statement 1 is true. So ans A:-o
2. I think D
median is 25 means 8th term in the series is 25 .so the 1st to 7th no: in this series can be either 25 or less than that.But we have to maximize the range ie 1st term shiuld be as small as possible ie 1st term = 25-7=18 So last term of series with 1st term 18 and range 25 is 43. Ans D;)
but in the same way... can't i say 25 is the largest of the smallest possible integer that will yield the largest integer on the right hand side?
can we say that all the numbers on the left hand side of 25 need not be less than 25.. and can be 25 instead? (which means the first 8 numbers in the set are each "25").
then the highest integer will be 25 + 25 = 50 right?
The first and last question has already been answered by someone on page 156. Please have a look.
Q13. Is the answer 5)-remainder 4? Q19. Is the answer 4) - 20 boxes must be removed? Lets assume 10 pound boxes = x So 20 pound boxes will be = 30-x (10x+20(30-x))/30 = 18 => x=6 (10 pound boxes) => so 30-6=24 (20 pound boxes) Now take second part. We know that 10 pound boxes are 6 in number, however we have to find out how many boxes of 20 pounds are there to make the average 14. and so how many 20 pound boxes are removed. Lets assume 20 pound boxes now to be = x So (10*6+20*x) / (6+x)=14 => x=4 So 20 pounds boxes to be removed are = 24 - 4 => 20
I solved the second part without any calculation. Thought that it might be useful for u guys...
The new avg shud be 14. ie the no of 20 pound boxes should be less than the no of 10 pound boxes (
but in the same way... can't i say 25 is the largest of the smallest possible integer that will yield the largest integer on the right hand side?
can we say that all the numbers on the left hand side of 25 need not be less than 25.. and can be 25 instead? (which means the first 8 numbers in the set are each "25").
then the highest integer will be 25 + 25 = 50 right?
Read the question carefully ! It says DIFFERENT positive integers. That means every preceeding number should be lesser than 25.
1) The annual interest rate earned by an investment increased by 10 percent from last year to this year. If the annual interest rate earned by the investment this year was 1 percent, what was the annual interest rate last year? A. 1% B. 1.1% C. 9.1% D. 10% E. 10.8% 2) Of the 32 students in a certain class, 15 are in a music club and 20 are in a dance club.If 10 of the students are not in either club, how many of the students are in only one of the two clubs? A. 5 B. 9 C.10 D. 17 E. 25 3) Each side of a certain parallelogram has length 6. if the area of the parallelogram is 18. which of the following is the measure of one of its angles? A. 30 B. 45 C. 60 D. 90 E.120
1) The annual interest rate earned by an investment increased by 10 percent from last year to this year. If the annual interest rate earned by the investment this year was 1 percent, what was the annual interest rate last year? A. 1% B. 1.1% C. 9.1% D. 10% E. 10.8% 2) Of the 32 students in a certain class, 15 are in a music club and 20 are in a dance club.If 10 of the students are not in either club, how many of the students are in only one of the two clubs? A. 5 B. 9 C.10 D. 17 E. 25 3) Each side of a certain parallelogram has length 6. if the area of the parallelogram is 18. which of the following is the measure of one of its angles? A. 30 B. 45 C. 60 D. 90 E.120
Q13 3 has a cyclicity 4 when divided by 5 3^8n +3 %5 =2 2+2 % 5=4
Q14 Since the average has come down to 14 from 18 4*30 is the deficit and this can be accounted for by removing 6 20kg boxes (6*20 =4*30)
Hi Pro, That was the catch... I too earlier solved in the same way.... The new avg is not for 30 boxes rather total boxes left after removing the required nos. Correct answer is 20.
1. Choice C should have been 0.91%, not 9.1%. If the interest this yr was 1%, which is a 10% hike over last yr, it shud be 1/1.1 = .91%
2. Let the no of students in both clubs = x No of students in atleast one of the 2 clubs = 32-10 = 22 = (15-x) + x + (20-x).
ie 22 = 35-x ==> x=13
Therefore students in only one of the clubs = 15-x + 20-x = 9.
Choice B.
3. Area of the parallellogram = base*ht ie 18 = 6*h ==> h=3. If u draw a perpendicular to the base, let the angle opp to perpendicular be x.
Sin x = 3/9 ==> x=Sin-1 (1/3).
x No answer choice corresponds.
Ashish can u tell me how u got the 3rd answer ?
Consider a parallelogram ABCD , let AB be the base and CD be the side parallel to AB .
Drop a perpendicular from D to AB and name it X......and the height of this perprndicular is 3(=18/6)
So triangle AXD is a right triangle and we know AD=6(side of parallelogram) and DX=3(ht of parallelo) sin A = DX/AD=3/6=1/2.......... so A = sin -1(1/2)=30 degree
On a certain scale of intensity, each increment of 10 in magnitude represents a tenfold increase in intensity. On this scale, an intensity corresponding to a magnitude of 165 is how many times an intensity corresponding to a magnitude of 125?
On a certain scale of intensity, each increment of 10 in magnitude represents a tenfold increase in intensity. On this scale, an intensity corresponding to a magnitude of 165 is how many times an intensity corresponding to a magnitude of 125? A. 40 B. 100 C. 400 D. 1000 E. 10000
Ans E. Every change of 10 in magnitude makes the intensity 10 times. I1=10^(a/10). I2=10^(b/10).... a,b are magnitude changes
On a certain scale of intensity, each increment of 10 in magnitude represents a tenfold increase in intensity. On this scale, an intensity corresponding to a magnitude of 165 is how many times an intensity corresponding to a magnitude of 125? A. 40 B. 100 C. 400 D. 1000 E. 10000