Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 30 May 2016, 15:49

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# tough quantitative contd... from question no. 50

Author Message
Manager
Joined: 04 Apr 2010
Posts: 163
Followers: 1

Kudos [?]: 138 [6] , given: 31

tough quantitative contd... from question no. 50 [#permalink]

### Show Tags

25 Apr 2010, 17:23
6
KUDOS
2
This post was
BOOKMARKED
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 100% (00:03) wrong based on 3 sessions

### HideShow timer Statistics

50. A set of data consists of the following 5 numbers: 0,2,4,6, and 8. Which two numbers, if added to create a set of 7 numbers, will result in a new standard deviation that is close to the standard deviation for the original 5 numbers?

A). -1 and 9
B). 4 and 4
C). 3 and 5
D). 2 and 6
E). 0 and 8

Soln: SD = Sqrt(Sum(X-x)^2/N)
Since N is changing from 5 to 7. Value of Sum(X-x)^2 should change from 40(current) to 48. So that SD remains same.

so due to new numbers it adds 8. Choice D only fits here.

51. How many integers from 0 to 50, inclusive, have a remainder of 1 when divided by 3?

A.14 B.15. C.16 D.17 E.18

Soln: if we arrange this in AP, we get
4+7+10+.......+49

so 4+(n-1)3=49: n=16
C is my pick

52. If k=m(m+4)(m+5) k and m are positive integers. Which of the following could divide k evenly?

I.3 II.4 III.6

Soln: The idea is to find what are the common factors that we get in the answer.

m = 1, k = 30 which is divisible by 1,2,3,5,6,10, 15 and 30
m = 2, k = 84 which is divisible by 1,2,3,4,6, ....

As can be seen, the common factors are 1,2,3,6

So answer is 3 and 6

53. If the perimeter of square region S and the perimeter of circular region C are equal, then the ratio of the area of S to the area of C is closest to
(A) 2/3
(B) 3/4
(C) 4/3
(D) 3/2
(E) 2
Soln: and the answer would be B...here is the explanation...

Let the side of the square be s..then the perimeter of the square is 4s
Let the radius of the circle be r..then the perimeter of the circle is 2*pi*r

it is given that both these quantities are equal..therefore

4s=2*pi*r

which is then s/r=pi/2

Now the ratio of area of square to area of circle would be

s^2/pi*r^2

(1/pi)*(s/r)^2

= (1/pi)*(pi/2)^2 from the above equality relation

pi=22/7 or 3.14

the value of the above expression is approximate =0.78 which is near to answer B

54. Two people walked the same distance, one person's speed is between 25 and 45,and if he used 4 hours, the speed of another people is between 45 and 60,and if he used 2 hours, how long is the distance?
A.116 B.118 C.124 D.136 E.140

Soln: First person-- speed is between 25mph and 45mph
so for 4 hrs he can travel 100 miles if he goes at 25mph speed
and for 4 hrs he can travel 130 miles if he goes at 45 mph speed

similarly

Second person-- speed is between 45mph and 60mph
so for 2 hrs he can travel 90 miles if he goes at 45mph speed
and for 2 hrs he can travel 120 miles if he goes at 60 mph speed

So for the first person---distance traveled is greater than 100 and Less than 130

and for the second person---distance traveled is greater than 90 and less than 120

so seeing these conditions we can eliminate C, D, and E answers...

but I didn’t understand how to select between 116 and 118 as both these values are satisfying the conditions...

55. How many number of 3 digit numbers can be formed with the digits 0,1,2,3,4,5 if no digit is repeated in any number? How many of these are even and how many odd?
Soln: Odd: fix last as odd, 3 ways __ __ _3_
now, left are 5, but again leaving 0, 4 for 1st digit & again 4 for 2nd digit: _4_ _4_ _3_ =48 Odd.

100-48= 52 Even

56. How many 3-digit numerals begin with a digit that represents a prime and end with a digit that represents a prime number?

A) 16 B) 80 c) 160 D) 180 E) 240

Soln: The first digit can be 2, 3, 5, or 7 (4 choices)
The second digit can be 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9 (10 choices)
The third digit can be 2, 3, 5, or 7 (4 choices)

4 * 4 * 10 = 160

57. There are three kinds of business A, B and C in a company. 25 percent of the total revenue is from business A; t percent of the total revenue is from B, the others are from C. If B is $150,000 and C is the difference of total revenue and 225,000, what is t? A.50 B.70 C.80 D.90 E.100 Soln: Let the total revenue be X. So X= A + B+ C which is X= 1/4 X+ 150 + (X-225) 4X= X+ 600 + 4X -900 Solving for X you get X= 300 And 150 is 50% of 300 so the answer is 50 % (A) 58. A business school club, Friends of Foam, is throwing a party at a local bar. Of the business school students at the bar, 40% are first year students and 60% are second year students. Of the first year students, 40% are drinking beer, 40% are drinking mixed drinks, and 20% are drinking both. Of the second year students, 30% are drinking beer, 30% are drinking mixed drinks, and 20% are drinking both. A business school student is chosen at random. If the student is drinking beer, what is the probability that he or she is also drinking mixed drinks? A. 2/5 B. 4/7 C. 10/17 D. 7/24 E. 7/10 Soln: The probability of an event A occurring is the number of outcomes that result in A divided by the total number of possible outcomes. The total number of possible outcomes is the total percent of students drinking beer. 40% of the students are first year students. 40% of those students are drinking beer. Thus, the first years drinking beer make up (40% * 40%) or 16% of the total number of students. 60% of the students are second year students. 30% of those students are drinking beer. Thus, the second years drinking beer make up (60% * 30%) or 18% of the total number of students. (16% + 18%) or 34% of the group is drinking beer. The outcomes that result in A is the total percent of students drinking beer and mixed drinks. 40% of the students are first year students. 20% of those students are drinking both beer and mixed drinks. Thus, the first years drinking both beer and mixed drinks make up (40% * 20%) or 8% of the total number of students. 60% of the students are second year students. 20% of those students are drinking both beer and mixed drinks. Thus, the second years drinking both beer and mixed drinks make up (60% * 20%) or 12% of the total number of students. (8% + 12%) or 20% of the group is drinking both beer and mixed drinks. If a student is chosen at random is drinking beer, the probability that they are also drinking mixed drinks is (20/34) or 10/17. 59. A merchant sells an item at a 20% discount, but still makes a gross profit of 20 percent of the cost. What percent of the cost would the gross profit on the item have been if it had been sold without the discount? A) 20% B) 40% C) 50% D) 60% E) 75% Soln: Lets suppose original price is 100. And if it sold at 20% discount then the price would be 80 but this 80 is 120% of the actual original price...so 66.67 is the actual price of the item now if it sold for 100 when it actually cost 66.67 then the gross profit would be 49.99% i.e. approx 50% 60. If the first digit cannot be a 0 or a 5, how many five-digit odd numbers are there? A. 42,500 B. 37,500 C. 45,000 D. 40,000 E. 50,000 Soln: This problem can be solved with the Multiplication Principle. The Multiplication Principle tells us that the number of ways independent events can occur together can be determined by multiplying together the number of possible outcomes for each event. There are 8 possibilities for the first digit (1, 2, 3, 4, 6, 7, 8, 9). There are 10 possibilities for the second digit (0, 1, 2, 3, 4, 5, 6, 7, 8, 9) There are 10 possibilities for the third digit (0, 1, 2, 3, 4, 5, 6, 7, 8, 9) There are 10 possibilities for the fourth digit (0, 1, 2, 3, 4, 5, 6, 7, 8, 9) There are 5 possibilities for the fifth digit (1, 3, 5, 7, 9) Using the Multiplication Principle: = 8 * 10 * 10 * 10 * 5 = 40,000 61.A bar is creating a new signature drink. There are five possible alcoholic ingredients in the drink: rum, vodka, gin, peach schnapps, or whiskey. There are five possible non-alcoholic ingredients: cranberry juice, orange juice, pineapple juice, limejuice, or lemon juice. If the bar uses two alcoholic ingredients and two non-alcoholic ingredients, how many different drinks are possible? A. 100 B. 25 C. 50 D. 75 E. 3600 Soln: The first step in this problem is to calculate the number of ways of selecting two alcoholic and two non-alcoholic ingredients. Since order of arrangement does not matter, this is a combination problem. The number of combinations of n objects taken r at a time is C(n,r) = n!/(r!(n-r!)) The number of combinations of alcoholic ingredients is C(5,2) = 5!/(2!(3!)) C(5,2) = 120/(2(6)) C(5,2) = 10 The number of combinations of non-alcoholic ingredients is C(5,2) = 5!/(2!(3!)) C(5,2) = 120/(2(6)) C(5,2) = 10 The number of ways these ingredients can be combined into a drink can be determined by the Multiplication Principle. The Multiplication Principle tells us that the number of ways independent events can occur together can be determined by multiplying together the number of possible outcomes for each event. The number of possible drinks is = 10 * 10 = 100 62. The sum of the even numbers between 1 and n is 79*80, where n is an odd number. N=? Soln: The sum of numbers between 1 and n is = (n(n+1))/2 1+2+3+.....+n=(n(n+1))/2 {formula} we are looking for the sum of the even numbers between 1 and n, which means: 2+4+6+.....+(n-1) n is ODD =1*2+2*2+2*3+......+2*((n-1)/2) =2*(1+2+3+.....+*((n-1)/2)) from the formula we obtain : =2*(((n-1)/2)*((n-1)/2+1))/2 =((n-1)/2)*((n+1)/2) =79*80 => (n-1)*(n+1)=158*160 => n=159 63. A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together? (A) 3510 (B) 2620 (C) 1404 (D) 700 (E) 635 Soln: 4W 2M == 5C4.(8C2-1) = 5.(27) = 135 3W 3M == 5C3.(8C3-6) = 10.50 = 500 total 635 Max. number of possibilities considering we can choose any man 8c2 * 5C4 + 8C3*5c3 = 700. consider it this way.... from my previous reply max possible ways considering we can chose any man = 700 now we know that 2 man could not be together... now think opposite... how many ways are possible to have these two man always chosen together... since they are always chosen together... For chosing 2 men and 4 women for the committee there is only 1 way of chosing 2 men for the committee since we know only two specific have to be chosen and there are 5C4 ways of choosing women 1*5C4 = 5 For choosing 3 men and 3 women for the committee there are exactly 6C1 ways choosing 3 men for the committee since we know two specific have to be chosen so from the remaining 6 men we have to chose 1 and there are 5C4 ways of choosing women 6C1*5C3 = 60 So total number of unfavorable cases = 5 + 60 = 65 Now since we want to exclude these 65 cases... final answer is 700-65 = 635 64. In how many ways can the letters of the word 'MISSISIPPI' be arranged? a) 1260 b) 12000 c) 12600 d) 14800 e) 26800 Soln: Total # of alphabets = 10 so ways to arrange them = 10! Then there will be duplicates because 1st S is no different than 2nd S. we have 4 Is 3 S and 2 Ps Hence # of arrangements = 10!/4!*3!*2! 65. Goldenrod and No Hope are in a horse race with 6 contestants. How many different arrangements of finishes are there if No Hope always finishes before Goldenrod and if all of the horses finish the race? (A) 720 (B) 360 (C) 120 (D) 24 (E) 21 Soln: two horses A and B, in a race of 6 horses... A has to finish before B if A finishes 1... B could be in any of other 5 positions in 5 ways and other horses finish in 4! Ways, so total ways 5*4! if A finishes 2... B could be in any of the last 4 positions in 4 ways. But the other positions could be filled in 4! ways, so the total ways 4*4! if A finishes 3rd... B could be in any of last 3 positions in 3 ways, but the other positions could be filled in 4! ways, so total ways 3*4! if A finishes 4th... B could be in any of last 2 positions in 2 ways, but the other positions could be filled in 4! ways, so total ways... 2 * 4! if A finishes 5th .. B has to be 6th and the top 4 positions could be filled in 4! ways.. A cannot finish 6th, since he has to be ahead of B therefore total number of ways 5*4! + 4*4! + 3*4! + 2*4! + 4! = 120 + 96 + 72 + 48 + 24 = 360 66. On how many ways can the letters of the word "COMPUTER" be arranged? 1. M must always occur at the third place 2. Vowels occupy the even positions. Soln: For 1. 7*6*1*5*4*3*2*1=5,040 For 2.) I think It should be 4 * 720 there are 4 even positions to be filled by three even numbers. in 5*3*4*2*3*1*2*1 It is assumed that Last even place is NOT filled by a vowel. There can be total 4 ways to do that. Hence 4 * 720 67. A shipment of 10 TV sets includes 3 that are defective. In how many ways can a hotel purchase 4 of these sets and receive at least two of the defective sets? Soln: There are 10 TV sets; we have to choose 4 at a time. So we can do that by 10C4 ways. We have 7 good TV’s and 3 defective. Now we have to choose 4 TV sets with at least 2 defective. We can do that by 2 defective 2 good 3 defective 1 good That stands to 3C2*7C2 + 3C3*7C1 (shows the count) If they had asked probability for the same question then 3C2*7C2 + 3C3*7C1 / 10C4. 68. A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people? A. 420 B. 2520 C. 168 D. 90 E. 105 Soln: 1st team could be any of 2 guys... there would be 4 teams (a team of A&B is same as a team of B&A)... possible ways 8C2 / 4. 2nd team could be any of remaining 6 guys. There would be 3 teams (a team of A&B is same as a team of B&A)... possible ways 6C2 / 3 3rd team could be any of remaining 2 guys... there would be 2 teams (a team of A&B is same as a team of B&A). Possible ways 4C2 / 2 4th team could be any of remaining 2 guys... there would be 1 such teams... possible ways 2C2 / 1 total number of ways... 8C2*6C2*4C2*2C2 ------------------- 4 * 3 * 2 * 1 = 8*7*6*5*4*3*2*1 -------------------- 4*3*2*1*2*2*2*2 = 105 (ANSWER)... Another method: say you have 8 people ABCDEFGH now u can pair A with 7 others in 7 ways. Remaining now 6 players. Pick one and u can pair him with the remaining 5 in 5 ways. Now you have 4 players. Pick one and u can pair him with the remaining in 3 ways. Now you have 2 players left. You can pair them in 1 way so total ways is 7*5*3*1 = 105 ways i.e. E 69. In how many ways can 5 people sit around a circular table if one should not have the same neighbors in any two arrangements? Soln: The ways of arranging 5 people in a circle = (5-1)! = 4! For a person seated with 2 neighbors, the number of ways of that happening is 2: AXB or BXA, where X is the person in question. So, for each person, we have two such arrangements in 4!. Since we don't want to repeat arrangement, we divide 4!/2 to get 12 70. There are 4 copies of 5 different books. In how many ways can they be arranged on a shelf? A) 20!/4! B) 20!/5(4!) C) 20!/(4!)^5 D) 20! E) 5! Soln: 4 copies each of 5 types. Total = 20 books. Total ways to arrange = 20! Taking out repeat combos = 20!/(4! * 4! * 4! * 4! * 4!) – each book will have 4 copies that are duplicate. So we have to divide 20! By the repeated copies. 71. In how many ways can 5 rings be worn on the four fingers of the right hand? Soln: 5 rings, 4 fingers 1st ring can be worn on any of the 4 fingers => 4 possibilities 2nd ring can be worn on any of the 4 fingers => 4 possibilities 3rd ring can be worn on any of the 4 fingers => 4 possibilities 4th ring can be worn on any of the 4 fingers => 4 possibilities 5th ring can be worn on any of the 4 fingers => 4 possibilities Total possibilities = 4*4*4*4*4 = 4^5. 72. If both 5^2 and 3^3 are factors of n x (2^5) x (6^2) x (7^3), what is the smallest possible positive value of n? Soln: Write down n x (2^5) x (6^2) x (7^3) as = n x (2^5) x (3^2) x (2^2) x (7^3), = n x (2^7) x (3^2) x (7^3) now at a minimum 5^2 and a 3 is missing from this to make it completely divisible by 5^2 x 3^3 Hence answer = 5^2 x 3 = 75 73. Obtain the sum of all positive integers up to 1000, which are divisible by 5 and not divisible by 2. (1) 10050 (2) 5050 (3) 5000 (4) 50000 Soln: Consider 5 15 25 ... 995 l = a + (n-1)*d l = 995 = last term a = 5 = first term d = 10 = difference 995 = 5 + (n-1)*10 thus n = 100 = # of terms consider 5 10 15 20.... 995 995 = 5 + (n-1)*5 => n = 199 Another approach... Just add up 995 + 985 + 975 + 965 + 955 + 945 = 5820, so it has to be greater than 5050, and the only possible choices left are 1) & 4) Also, series is 5 15 25.... 985 995 # of terms = 100 sum = (100/2)*(2*5 + (100-1)*10) = 50*1000 = 50000 74. If the probability of rain on any given day in city x is 50% what is the probability it with rain on exactly 3 days in a five day period? 8/125 2/25 5/16 8/25 3/4 Soln: Use binomial theorem to solve the problem.... p = 1/2 q = 1/2 # of favorable cases = 3 = r # of unfavorable cases = 5-3 = 2 total cases = 5 = n P(probability of r out of n) = nCr*p^r*q^(n-r) 75. The Full House Casino is running a new promotion. Each person visiting the casino has the opportunity to play the Trip Aces game. In Trip Aces, a player is randomly dealt three cards, without replacement, from a deck of 8 cards. If a player receives 3 aces, they will receive a free trip to one of 10 vacation destinations. If the deck of 8 cards contains 3 aces, what is the probability that a player will win a trip? A. 1/336 B. 1/120 C. 1/56 D. 1/720 E. 1/1440 The probability of an event A occurring is the number of outcomes that result in A divided by the total number of possible outcomes. There is only one result that results in a win: receiving three aces. Since the order of arrangement does not matter, the number of possible ways to receive 3 cards is a combination problem. The number of combinations of n objects taken r at a time is C(n,r) = n!/(r!(n-r)!) C(8,3) = 8!/(3!(8-3)!) C(8,3) = 8!/(3!(5!)) C(8,3) = 40320/(6(120)) C(8,3) = 40320/720 C(8,3) = 56 The number of possible outcomes is 56. Thus, the probability of being dealt 3 aces is 1/56. 76. The Full House Casino is running a new promotion. Each person visiting the casino has the opportunity to play the Trip Aces game. In Trip Aces, a player is randomly dealt three cards, without replacement, from a deck of 8 cards. If a player receives 3 aces, they will receive a free trip to one of 10 vacation destinations. If the deck of 8 cards contains 3 aces, what is the probability that a player will win a trip? A. 1/336 B. 1/120 C. 1/56 D. 1/720 E. 1/1440 Soln: Since each draw doesn't replace the cards: Prob. of getting an ace in the first draw = 3/8 getting in the second, after first draw is ace = 2/7 getting in the third after the first two draws are aces = 1/6 thus total probability for these mutually independent events = 3/8*2/7*1/6 = 1/56 77. Find the probability that a 4 person committee chosen at random from a group consisting of 6 men, 7 women, and 5 children contains A) exactly 1 woman B) at least 1 woman C) at most 1 woman Soln: A.) 7C1* 11C3/ 18C4 B) 1 - (11C4/18C4) C) (11C4/18C4) + (7C1*11C3/18C4) 78. A rental car service facility has 10 foreign cars and 15 domestic cars waiting to be serviced on a particular Saturday morning. Because there are so few mechanics, only 6 can be serviced. (a) If the 6 cars are chosen at random, what is the probability that 3 of the cars selected are domestic and the other 3 are foreign? (b) If the 6 cars are chosen at random, what is the probability that at most one domestic car is selected? Soln: A) 10C3*15C3/25C6 B) Probability of no domestic car + Probability of 1 domestic car = 10C6/25C6 + 15C1 *10C5/25C6 79. How many positive integers less than 5,000 are evenly divisible by neither 15 nor 21? A. 4,514 B. 4,475 C. 4,521 D. 4,428 E. 4,349 Soln: We first determine the number of integers less than 5,000 that are evenly divisible by 15. This can be found by dividing 4,999 by 15: = 4,999/15 = 333 integers Now we will determine the number of integers evenly divisible by 21: = 4,999/21 = 238 integers some numbers will be evenly divisible by BOTH 15 and 21. The least common multiple of 15 and 21 is 105. This means that every number that is evenly divisible by 105 will be divisible by BOTH 15 and 21. Now we will determine the number of integers evenly divisible by 105: = 4,999/105 = 47 integers Therefore the positive integers less than 5000 that are not evenly divisible by 15 or 21 are 4999-(333+238-47)=4475 80) Find the least positive integer with four different prime factors, each greater than 2. Soln: 3*5*7*11 = 1155 81) From the even numbers between 1 and 9, two different even numbers are to be chosen at random. What is the probability that their sum will be 8? Soln: Initially you have 4 even numbers (2,4,6,8) you can get the sum of 8 in two ways => 2 + 6 or 6 + 2 so the first time you pick a number you can pick either 2 or 8 - a total of 2 choices out of 8 => 1/2 after you have picked your first number and since you have already picked 1 number you are left with only 2 options => either (4,6,8) or (2,4,8) and you have to pick either 6 from the first set or 2 from the second and the probability of this is 1/3. Since these two events have to happen together we multiply them. ½ * 1/3 = 1/6 82) 5 is placed to the right of two – digit number, forming a new three – digit number. The new number is 392 more than the original two-digit number. What was the original two-digit number? Soln: If the original number has x as the tens digit and y as the ones digit (x and y are integers less than 10) then we can set up the equation: 100x + 10y + 5 = 10x + y + 392 90x + 9y = 387 9(10x+y) = 387 10x + y = 43 ==> x = 4, y = 3 the original number is 43, the new number is 435 83) If one number is chosen at random from the first 1000 positive integers, what is the probability that the number chosen is multiple of both 2 and 8? Soln: Any multiple of 8 is also a multiple of 2 so we need to find the multiples of 8 from 0 to 1000 the first one is 8 and the last one is 1000 ==> ((1000-8)/8) + 1 = 125 ==> p (picking a multiple of 2 & 8) = 125/1000 = 1/8 84) A serial set consists of N bulbs. The serial set lights up only if all the N bulbs are in working condition. Even if one of the bulbs fails then the entire set fails. The probability of a bulb failing is x. What is the probability of the serial set failing? Soln: Probability of x to fail. Probability of a bulb not failing = 1-x probability that none of the N bulbs fail, hence serial set not failing = (1-x)^N probability of serial set failing = 1-(1-x)^N 85) Brad flips a two-sided coin 8 times. What is the probability that he gets tails on at least 7 of the 8 flips? 1/32 1/16 1/8 7/8 none of the above Soln: Number of ways 7 tails can turn up = 8C7 the probability of those is 1/2 each Since the question asks for at least 7, we need to find the prob of all 8 tails - the number of ways is 8C8 = 1 Add the two probabilities 8C7*(1/2)^8 = 8/2^8 -- for getting 7 Prob of getting 8 tails = 1/2^8 Total prob = 8/2^8+1/2^8 = 9/2^8 Ans is E. 86. A photographer will arrange 6 people of 6 different heights for photograph by placing them in two rows of three so that each person in the first row is standing in front of someone in the second row. The heights of the people within each row must increase from left to right, and each person in the second row must be taller than the person standing in front of him or her. How many such arrangements of the 6 people are possible? A. 5 B. 6 C. 9 D. 24 E. 36 Soln: 5 ways 123 124 125 134 135 456 356 346 256 246 87. As a part of a game, four people each much secretly chose an integer between 1 and 4 inclusive. What is the approximate likelihood that all four people will chose different numbers? Soln: The probability that the first person will pick unique number is 1 (obviously) then the probability for the second is 3/4 since one number is already picked by the first, then similarly the probabilities for the 3rd and 4th are 1/2 and 1/4 respectively. Their product 3/4*1/2*1/4 = 3/32 88. Which of the sets of numbers can be used as the lengths of the sides of a triangle? I. [5,7,12] II. [2,4,10] III. [5,7,9] A. I only B. III only C. I and II only D. I and III only E. II and III only Soln: For any side of a triangle. Its length must be greater than the difference between the other two sides, but less than the sum of the other two sides. Answer is B 89. A clothing manufacturer has determined that she can sell 100 suits a week at a selling price of 200$ each. For each rise of 4$in the selling price she will sell 2 less suits a week. If she sells the suits for x$ each, how many dollars a week will she receive from sales of the suits?

Soln: Let y be the number of $4 increases she makes, and let S be the number of suits she sells. Then X = 200 + 4y ==> y = x/4 - 50 S = 100 - 2y ==> S = 100 - 2[x/4 - 50] = 100 - x/2 + 100 = 200 - x/2 so the answer is that the number of suits she'll sell is 200 - x/2 90. A certain portfolio consisted of 5 stocks, priced at$20, $35,$40, $45 and$70, respectively. On a given day, the price of one stock increased by 15%, while the price of another decreased by 35% and the prices of the remaining three remained constant. If the average price of a stock in the portfolio rose by approximately 2%, which of the following could be the prices of the shares that remained constant?

A) 20, 35, 70

B) 20, 45, 70

C) 20, 35, 40

D) 35, 40, 70

E) 35, 40, 45

Soln: Add the 5 prices together:
20 + 35 + 40 + 45 + 70 = 210

2% of that is 210 x .02 = 4.20

Let x be the stock that rises and y be the stock that falls.

.15x -.35y = 4.20 ==> x = (7/3)y + 27

This tells us that the difference between x and y has to be at least 27. Since the answer choices list the ones that DONT change, we need to look for an answer choice in which the numbers NOT listed have a difference of at least 27.

91. if -2=<x=<2 and 3<=y<=8, which of the following represents the range of all possible values of y-x?
(A) 5<=y-x<=6
(B) 1<=y-x<=5
(C) 1<=y-x<=6
(D) 1<=y-x<=10
(E) 1<=y-x<=10

Soln: you can easily solve this by subtracting the two inequalities. To do this they need to be in the opposite direction; when you subtract them preserve the sign of the inequality from which you are subtracting.

3 < y < 8
multiply the second one by (-1) to reverse the sign
2 > x > -2
Subtract them to get
3 - 2 < y - x < 8 - (-2)
1 < y - x < 10

92. Of a group of 260 people who purchased stocks, 61 purchased A, 88 purchased B, 56 purchased C, 75 purchased D, 60 purchased E. what is the greatest possible number of the people who purchased both B and D?

A:40 B:50 C:60 D:75 E:80

since they have asked us to find out the greatest possible number buying both B as well as D, the answer has to be the smallest no between the two which is 75...as all the guys purchasing D can also buy B and only 75 out of 88 purchasing B can simultaneously purchase D as well....

93. There are 30 people and 3 clubs M, S, and Z in a company. 10 people joined M, 12 people joined S and 5 people joined Z. If the members of M did not join any other club, at most, how many people of the company did not join any club?

A: 4 B: 5 C: 6 D: 7 E: 8

Soln: total no of people = 30
no joining M = 10
no joining S = 12
no joining Z = 5
question asked - AT MOST how many people did not join any group?

solution: now since none of the members of M joined any other group, the no of people left = 30-10(for M)=20
since the question says at most how many did not join any group, lets assume the all people who join Z also join S. so no of people joining group S and Z are 12 (note that there will be 5 people in group S who have also joined Z)

therefore no of people not joining any group = 20-12=8
Hence e

94. Find the numbers of ways in which 4 boys and 4 girls can be seated alternatively.

1) in a row
2) in a row and there is a boy named John and a girl named Susan amongst the group who cannot be put in adjacent seats
3) around a table

A:
1) 4! * 4! * 2
2) 4! * 4! * 2 - number of ways with John and Susan sitting together
= (4! * 4! * 2) - (7 * 3! * 3! *2)

The way that JS arrangements are found is by
bracketing J and S and considering it to be a single entity. So a
possible arrangement is (B=boy, G=girl)

(JS) B G B G B G number of arrangements is 7 x 3! x 3! = 252
(SJ) G B G B G B number of arrangements is 7 x 3! x 3! = 252

3) Fix one boy and arrange the other 3 boys in 3! ways. Arrange the 4
girls in 4! ways in the gaps between the boys.

Total arrangements = 3! x 4!

= 6 x 24

= 144

95. From a group of 3 boys and 3 girls, 4 children are to be randomly selected. What is the probability that equal numbers of boys and girls will be selected?

A. 1/10

B. 4/9

C. 1/2

D. 3/5

E. 2/3

Soln: Total number of ways of selecting 4 children = 6C4 = 15

with equal boys and girls. => 2 boys and 2 girls. => 3C2 * 3C2 = 9.

Hence p = 9/15 = 3/5

96. What is the reminder when 9^1 + 9^2 + 9^3 + ...... + 9^9 is divided by 6?

(1) 0
(2) 3
(3) 4
(4) 2
(5) None of these

Soln: Remainder of 9*odd /6 is 3
remainder of 9*even/6 is 0

9^1 + 9^2 + 9^3 + ...... + 9^9=9*(1+9+9^2+.....9^8)

1+9+9^2+.....9^8 is odd.

Thus we obtain 3 as a remainder when we divide 9*(1+9+9^2+.....9^8) by 6.

Another way: We get the value as 9*odd/6 = 3*odd/2 Since 3*odd = odd; odd/2 = XXXX.5
so something divided by 6, gives XXXX.5, hence remainder is 6*0.5 = 3

97. Find the value of 1.1! + 2.2! + 3.3! + ......+n.n!

(1) n! +1
(2) (n+1)!
(3) (n+1)!-1
(4) (n+1)!+1
(5) None of these

Soln: 1.1! + 2.2! + 3.3! + ......+n.n!
=1.1! + (3-1)2! + (4-1)3! +......+ ((n+1)-1) n!
=1.1!+3!-2!+4!-3!+.......+(n+1)!-n!

So it is (n+1)! -1 (Answer choice 4)

98. The numbers x and y are three-digit positive integers, and x + y is a four-digit integer. The tens digit of x equals 7 and the tens digit of y equals 5. If x < y, which of the following must be true?

I. The units digit of x + y is greater than the units digit of either x or y.
II. The tens digit of x + y equals 2.
III. The hundreds digit of y is at least 5.

A. II only

B. III only

C. I and II

D. I and III

E. II and III

Soln: x= abc
y= def

x = a7c
y= b5f

x > y and x+y = wxyz.

I. The units digit of x + y is greater than the units digit of either x or y.

It can carryover one digit. False

II. The tens digit of x + y equals 2.

It can be 2 or 3. False

III. The hundreds digit of y is at least 5.

a+b+1 >= 10
a >b so a at least 5. True.

Ans: b

99. Among 5 children there are 2 siblings. In how many ways can the children be seated in a row so that the siblings do not sit together?
(A) 38
(B) 46
(C) 72
(D) 86
(E) 102
Soln: The total number of ways 5 of them can sit is 120
when the siblings sit together they can be counted as one entity
therefore the number of ways that they sit together is 4!=24, but since
the two siblings can sit in two different ways e.g. AB and BA we multiply 24 by 2 to get the total number of ways in which the 5 children can sit together with the siblings sitting together - 48
In other words 4P4*2P2
the rest is obvious 120-48=72

-----------i will continue from question no. 101 in my next post... dont forget to give me kudos
_________________

Consider me giving KUDOS, if you find my post helpful.
If at first you don't succeed, you're running about average. ~Anonymous

tough quantitative contd... from question no. 50   [#permalink] 25 Apr 2010, 17:23
Display posts from previous: Sort by