Last month 15 homes were sold in Town X. The avg. 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 $130,0000 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 and II
e) I and III
the main deal with 'at least one' problems - which come up more often on probability than on other problem types - is that they're very difficult to treat directly. instead, when you see 'at least one', you should treat the OPPOSITE situation - i.e., none.
so, because 'at least one' and 'none' are opposites, the following statements are exactly equivalent:
* there must be at least one
* it's IMPOSSIBLE to have NONE
the second is the easier way to think about it.
so:
in this problem, you should consider the case in which NONE of the homes was sold for whatever price is mentioned in the problem, and see whether it's IMPOSSIBLE.
NOTE: I AM NOT GOING TO WRITE THE THOUSANDS. so, '130' means $130,000. you'll thank me; this problem will be much easier to read.
(preface)
the median of 15 values is the value that comes 8th in the list. therefore, the first seven values are 130 or less, the 8th value is 130, and the 9th-15th values are 130 or more.
also, Sum = Average x Number of data points, so the sum of all the prices is 15 x 150 = 2250.
(i)
let's consider the case in which NONE of the homes was sold for more than 165.
the MAXIMUM sum of prices in this case would be 8(130) + 7(165), which is the case if all of the first 8 values are 130 each (the biggest they can be) and values 9-15 are 165 each.
that's a total of 1040 + 1155 = 2195.
not high enough.
therefore, it's IMPOSSIBLE to have NO prices over 165, so this statement must be true.
(ii)
let's try to create a list with NO such house prices.
let the first 7 prices be, say, 100 each.
the 8th is 130.
so the first 8 have a sum of 830, meaning that the highest 7 have a sum of 2250 - 830 = 1420.
there are all kinds of ways to do that with no values between 130 and 150, but the simplest is to make all seven of them equal to 1420 / 7, which is greater than 200.
so (ii) doesn't have to be true.
(iii)
let's try to create a list with NO such house prices.
this would mean that the first 8 prices are all 130.
so, the last 7 prices sum to 2250 - 8(130) = 1210.
that's an average of 1210/7, which is a shade over 170. you could let all 7 of the high prices equal that value, and it would work.
therefore, (iii) doesn't have to be true.
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Here's a solution without much calculating that proves that (1) must be true.
It makes sense that if the median price is below the average price, the the average of seven highest prices should be further from 150 than the average of the seven lowest prices, which we know is no more than 130. Since the average of the seven highest prices must therefore be at least 170, at least one of them must be more than 165.
Friday, September 10, 2010
The average of 100 numbers is 6, and the standard deviation is D, where D is positive. When added which of the following numbers, the new deviation will be less than D?
A. -6 and 0
B. 0 and 0
C. 0 and 6
D. 0 and 12
E. 6 and 6
standard deviation is a measure of how spread out the data are. in other words, the standard deviation gives you a rough idea of how far away the numbers are from the average (mean).
hence the name: standard (= typical) deviation (= difference from the norm). notice that standard deviation is not simply the average of how far away the data points are from the average, but that notion is more than sufficient for conceptual understanding.
in any case, you have no idea how big the standard deviation is. it could be something like 0.0000001, meaning that every number in the set is extremely close to the average value of 6. therefore, the only way to be absolutely sure you're reducing the standard deviation is to insert two numbers each of which is exactly the average. since these points are both 0 units away from the average, they will lower the standard deviation.
Answer - E
A. -6 and 0
B. 0 and 0
C. 0 and 6
D. 0 and 12
E. 6 and 6
standard deviation is a measure of how spread out the data are. in other words, the standard deviation gives you a rough idea of how far away the numbers are from the average (mean).
hence the name: standard (= typical) deviation (= difference from the norm). notice that standard deviation is not simply the average of how far away the data points are from the average, but that notion is more than sufficient for conceptual understanding.
in any case, you have no idea how big the standard deviation is. it could be something like 0.0000001, meaning that every number in the set is extremely close to the average value of 6. therefore, the only way to be absolutely sure you're reducing the standard deviation is to insert two numbers each of which is exactly the average. since these points are both 0 units away from the average, they will lower the standard deviation.
Answer - E
Set A, B, C have some elements in common. if 16 elements are in both A and B, 17 elements are in both A and C, and 18 elements are in both B and C, how many elements do all three of the sets A, B, and C have in common?
1). of the 16 elements that are in both A and B, 9 elements are also in C
2). A has 25 elements, B has 30 elements, and C has 35 elements
Ans - A
1). of the 16 elements that are in both A and B, 9 elements are also in C
2). A has 25 elements, B has 30 elements, and C has 35 elements
Ans - A
If n and t are positive integers, what is the greatest prime factor of nt?
1). The greatest common factor of n and t is 5
2). The least common multiple of n and t is 105
(1) this only tells is that the greatest number that is in both factorizations - those of n and t - is 5. but there could be a larger factor that is part of only one of the factorizations. for instance:
- it's possible that n = t = 5. then the greatest prime factor of nt is 5.
- it's possible that n = 5 and t = 35. then the greatest prime factor of nt is 7.
insufficient.
(2) the least common multiple contains every factor of t or n at least once. (it has to; if, say, t had a factor that wasn't contained in it, then it would fail to be a multiple of t.) so, the biggest prime factor of this # will also be the biggest prime factor of the product nt.
sufficient.
try a few combinations of n and t if you aren't convinced.
answer = b
1). The greatest common factor of n and t is 5
2). The least common multiple of n and t is 105
(1) this only tells is that the greatest number that is in both factorizations - those of n and t - is 5. but there could be a larger factor that is part of only one of the factorizations. for instance:
- it's possible that n = t = 5. then the greatest prime factor of nt is 5.
- it's possible that n = 5 and t = 35. then the greatest prime factor of nt is 7.
insufficient.
(2) the least common multiple contains every factor of t or n at least once. (it has to; if, say, t had a factor that wasn't contained in it, then it would fail to be a multiple of t.) so, the biggest prime factor of this # will also be the biggest prime factor of the product nt.
sufficient.
try a few combinations of n and t if you aren't convinced.
answer = b
Is the integer k divisible by 4?
(1) 8k is divisible by 16.
(2) 9k is divisible by 12.
------------------------------------------------------------------------------------
For statement 2 :
* the prime factorization of 9k is 3 x 3 x (whatever goes into k)
* this factorization must contain 4, which is two 2's
* both of the 2's must be part of k, because the rest of the factorization is 3's
* therefore, k is divisible by 4
(1) 8k is divisible by 16.
(2) 9k is divisible by 12.
best way to go, if you don't get the theory right away: start running through an exhaustive list of possibilities for each statement, and see if either
(a) you can prove the statement is insufficient (in which case you're done), or
(b) a pattern emerges that will allow you to conclude that the statement is sufficient.
statement (1): just consider an exhaustive list of multiples of 16
if 8k = 16, then k = 2 --> NO
if 8k = 32, then k = 4 --> YES
insufficient
statement (2): just consider an exhaustive list of multiples of 12
if 9k = 12, then k isn't an integer
if 9k = 24, then k isn't an integer
if 9k = 36, then k = 4 --> YES
if 9k = 48, then k isn't an integer
if 9k = 60, then k isn't an integer
if 9k = 72, then k = 8 --> YES
a pattern has emerged (continue a couple more rounds if you like)
sufficient
(a) you can prove the statement is insufficient (in which case you're done), or
(b) a pattern emerges that will allow you to conclude that the statement is sufficient.
statement (1): just consider an exhaustive list of multiples of 16
if 8k = 16, then k = 2 --> NO
if 8k = 32, then k = 4 --> YES
insufficient
statement (2): just consider an exhaustive list of multiples of 12
if 9k = 12, then k isn't an integer
if 9k = 24, then k isn't an integer
if 9k = 36, then k = 4 --> YES
if 9k = 48, then k isn't an integer
if 9k = 60, then k isn't an integer
if 9k = 72, then k = 8 --> YES
a pattern has emerged (continue a couple more rounds if you like)
sufficient
------------------------------------------------------------------------------------
For statement 2 :
* the prime factorization of 9k is 3 x 3 x (whatever goes into k)
* this factorization must contain 4, which is two 2's
* both of the 2's must be part of k, because the rest of the factorization is 3's
* therefore, k is divisible by 4
If x is positive, which of the following could be correct ordering of x2,1/x,2x
a) x2 '<' 2x '<' 1/x
b) x2 '<' 1/x '<' 2x
c) 2x '<' x2 '<' 1/x
i) None
ii) a
iii) c
iv) a and b
v) a,b,c
The most obvious thing is to say that 1/x is the smallest, but none of them start with 1/x... so I'm suspecting a trick. That means I want to try a fraction between zero and one, because I know those are "tricky."
if x = 1/2, then:
x^2 = 1/4
1/x = 2
2x = 1
so the order would be x^2 < 2x < 1/x. So (a) is possible. Eliminate i and iii. (Looks like you got this far)
if x = 3/4, then:
x^2 = 9/16
1/x = 4/3
2x = 3/2
so the order would be x^2 < 1/x < 2x. So (b) is possible. Eliminate ii. (Looks like this is where you had trouble.)
Don't forget to try more than one number - 1/2 is the obvious starting number, but you've got to make sure that other fractions wouldn't do something different...
And (c) isn't going to work b/c the only way to make x^2 smaller than 1/x is for x^2 to be between zero and one. And if that's true, x^2 will be smaller than 2x, because x gets smaller when you square it but larger when you multiply it by 2.
a) x2 '<' 2x '<' 1/x
b) x2 '<' 1/x '<' 2x
c) 2x '<' x2 '<' 1/x
i) None
ii) a
iii) c
iv) a and b
v) a,b,c
The most obvious thing is to say that 1/x is the smallest, but none of them start with 1/x... so I'm suspecting a trick. That means I want to try a fraction between zero and one, because I know those are "tricky."
if x = 1/2, then:
x^2 = 1/4
1/x = 2
2x = 1
so the order would be x^2 < 2x < 1/x. So (a) is possible. Eliminate i and iii. (Looks like you got this far)
if x = 3/4, then:
x^2 = 9/16
1/x = 4/3
2x = 3/2
so the order would be x^2 < 1/x < 2x. So (b) is possible. Eliminate ii. (Looks like this is where you had trouble.)
Don't forget to try more than one number - 1/2 is the obvious starting number, but you've got to make sure that other fractions wouldn't do something different...
And (c) isn't going to work b/c the only way to make x^2 smaller than 1/x is for x^2 to be between zero and one. And if that's true, x^2 will be smaller than 2x, because x gets smaller when you square it but larger when you multiply it by 2.
Machine A produces pencils at a constant rate of 9,000 pencils per hour, and machine B produces pencils at a constant rate of 7,000 pencils per hour. If the two machines together must produce 100,000 pencils and if each machine can operate for at most 8 hours, what is the least amount of time, in hours, that machine B must operate?
a) 4
b) 4 and 2/3
c) 5 and 1/3
d) 6
e) 6 and 1/4
Ans : A
a) 4
b) 4 and 2/3
c) 5 and 1/3
d) 6
e) 6 and 1/4
Ans : A
jack and mark both receive hourly wage increases of 6%. after the wage increase, jack's hourly wage was how many more dollars per hour than mark?
1. before the wage increases, jack's hourly rate was 5 more than mark's
2. before the wage increases, the ratio of jack's hourly wage to mark's hourly wage was 4 to 3
(1)
before wage increases:
jack = m + 5
mark = m
after wage increases:
jack = 1.06(m + 5)
mark = 1.06m
difference = 1.06(5)
= 5.30, but you're wasting your time if you calculate 1.06(5) (all you have to notice is that it's a constant; m drops out)
sufficient
(2)
before wage increases:
jack = 4x
mark = 3x
after wage increases:
jack = 1.06(4x) = 4.24x
mark = 1.06(3x) = 3.18x
difference = 1.06x
depends on x
insufficient
answer = a
1. before the wage increases, jack's hourly rate was 5 more than mark's
2. before the wage increases, the ratio of jack's hourly wage to mark's hourly wage was 4 to 3
(1)
before wage increases:
jack = m + 5
mark = m
after wage increases:
jack = 1.06(m + 5)
mark = 1.06m
difference = 1.06(5)
= 5.30, but you're wasting your time if you calculate 1.06(5) (all you have to notice is that it's a constant; m drops out)
sufficient
(2)
before wage increases:
jack = 4x
mark = 3x
after wage increases:
jack = 1.06(4x) = 4.24x
mark = 1.06(3x) = 3.18x
difference = 1.06x
depends on x
insufficient
answer = a
In a certain year, the difference between Mary's and Jim's annual salaries was twice the difference between Mary's and Kate's annual salaries. If Mary's annual salary was the highest of the 3 people, what was the average (arithmetic mean) annual salary of the 3 people that year?
(1) Jim's annual salary was $30,000 that year.
(2) Kate's annual salary was $40,000 that year.
Ans is B
(1) Jim's annual salary was $30,000 that year.
(2) Kate's annual salary was $40,000 that year.
Ans is B
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A certain city with a population of 132,000 is to be divided into 11 voting districts, and no district is to have a population that is more than 10 percent greater than the population of any other district. What is the minimum possible population that the least populated district could be?
a) 10,700
b) 10,800
c) 10,900
d) 11,000
e) 11,100
The important insight for this problem is to realize that the way to minimize the population in the least populous voting district is to have the maximum permitted population in each of the other districts. So you have the least populous district with population, p, and then each of the remaining 10 districts would have a population 10% greater than the least populous district or 1.1p. We know the total city population is 132,000 so:
p+10*(1.1p)=132000
12p=132,000
p=11,000
a) 10,700
b) 10,800
c) 10,900
d) 11,000
e) 11,100
The important insight for this problem is to realize that the way to minimize the population in the least populous voting district is to have the maximum permitted population in each of the other districts. So you have the least populous district with population, p, and then each of the remaining 10 districts would have a population 10% greater than the least populous district or 1.1p. We know the total city population is 132,000 so:
p+10*(1.1p)=132000
12p=132,000
p=11,000
| |
Which of the following inequalities has a solution set, when graphed on the number line, is a single line segment of finite length?
A) x^4 '>'= 1
B) x^3 '>'= 27
C) x^2 '>'= 16
D) 2 '<'= |x| '<'= 5
E) 2 '<'= 3x+4 '<'= 6
A,B & C lare non liner graphs....
(D) 2 <= |x| <= 5
|x| >= 2
x >= 2 or x <= -2
x <= 5 or x >= -5
These are two segments. One is from -5 to -2. The other from 2 to 5.
(E) 2 '<'= 3x+4' <'= 6
-2 '<'= 3x '<'= 2
-2/3 '<'= x '<'= 2/3
This is one continuous segment.
A) x^4 '>'= 1
B) x^3 '>'= 27
C) x^2 '>'= 16
D) 2 '<'= |x| '<'= 5
E) 2 '<'= 3x+4 '<'= 6
A,B & C lare non liner graphs....
(D) 2 <= |x| <= 5
|x| >= 2
x >= 2 or x <= -2
x <= 5 or x >= -5
These are two segments. One is from -5 to -2. The other from 2 to 5.
(E) 2 '<'= 3x+4' <'= 6
-2 '<'= 3x '<'= 2
-2/3 '<'= x '<'= 2/3
This is one continuous segment.
When a certain tree was first planted, it was 4 feet tall, and the heigth of the tree increased by a constant amount each year for the next 6 years. At the end of the 6th year, the tree was 1/5 taller than it was at the end of the 4th year. By how many feet did the height of the tree increased each year?
Answers: 3/10, 2/5, 1/2, 2/3, 6/5
Yr0 = 4
Yr1 = 4+x
Yr2 = 4+x+x=4+2x
Yr3 = 4+x+x+x=4+3x
Yr4 = 4+x+x+x+x=4+4x
Yr5 = 4+x+x+x+x+x=4+5x
Yr6 = 4+x+x+x+x+x+x=4+6x
We are told the amount at the end of Year 6 is 6/5 of the amount at the end of year 4. Thus we can write:
4+6x = 6/5 (4+4x)
5(4+6x) = 6(4+4x)
20+30x = 24+24x
6x=4
x=2/3
Answers: 3/10, 2/5, 1/2, 2/3, 6/5
Yr0 = 4
Yr1 = 4+x
Yr2 = 4+x+x=4+2x
Yr3 = 4+x+x+x=4+3x
Yr4 = 4+x+x+x+x=4+4x
Yr5 = 4+x+x+x+x+x=4+5x
Yr6 = 4+x+x+x+x+x+x=4+6x
We are told the amount at the end of Year 6 is 6/5 of the amount at the end of year 4. Thus we can write:
4+6x = 6/5 (4+4x)
5(4+6x) = 6(4+4x)
20+30x = 24+24x
6x=4
x=2/3
If x is a positive number less than 10, is z greater than the average (arithmetic mean) of x and 10?
1) on the number line, z is closer to 10 than it is to x.
2) z = 5x
1) on the number line, z is closer to 10 than it is to x.
this means that it is the the right (larger) than the average (m) and so (1) is sufficient. We can rule out B,C,E.
2) z=5x
This one is trying to trap people who don't read the problem carefully. X is a positive number less than 10, not necessarly an integer. consider x=.1. Then the mean of x and 10 is 5.05. if z=5x, then z is .5 and is smaller than the mean. Now consider x=2. The mean of x and 10 is then 6. z=5x=10 and is larger than the mean. So 2 is not sufficient and the answer is A.
1) on the number line, z is closer to 10 than it is to x.
2) z = 5x
1) on the number line, z is closer to 10 than it is to x.
this means that it is the the right (larger) than the average (m) and so (1) is sufficient. We can rule out B,C,E.
2) z=5x
This one is trying to trap people who don't read the problem carefully. X is a positive number less than 10, not necessarly an integer. consider x=.1. Then the mean of x and 10 is 5.05. if z=5x, then z is .5 and is smaller than the mean. Now consider x=2. The mean of x and 10 is then 6. z=5x=10 and is larger than the mean. So 2 is not sufficient and the answer is A.
The points A, B, C, and D are on a number line, not necessarily in that order. If the distance between A and B is 18 and the distance between C and D is 8, what is the distance between B and D?
1) The distance between C and A is the same as the distance between C and B.
2) A is to the left of D on the number line.
Answer: E
1) The distance between C and A is the same as the distance between C and B.
2) A is to the left of D on the number line.
Answer: E
All of the stocks on the over-the-counter market are designated by either a 4-letter or a 5-letter code that is created by using the 26 letters of the alphabet. Which of the following gives the maximum number of different stocks that can be designated with these codes?
A. 2(26^5)
B. 26(26^4)
C. 27(26^4)
D. 26(26^5)
E. 27(26^5)
OA is C.
A. 2(26^5)
B. 26(26^4)
C. 27(26^4)
D. 26(26^5)
E. 27(26^5)
OA is C.
On his trip from Alba to Benton, Julia drove the first x miles at an average rate of 50 miles per hour and the remaining distance at an average rate of 60 miles per hour. How long did it take Julio to drive the first x miles ?
1> On his trip, Julio drove for a total of 10 hrs and drove a total of 530 miles
2> On his trip,it took Julio 4 more hours to drive the first x miles than to drive the remaining distance.
OA is A
1> On his trip, Julio drove for a total of 10 hrs and drove a total of 530 miles
2> On his trip,it took Julio 4 more hours to drive the first x miles than to drive the remaining distance.
OA is A
A fast food company plans to build 4 new restaurants. If there are 12 sites that satisfy the company's criteria for location of new restaurant, in how many different ways can company select the 4 sites needed for the new restaurant if the order of selection doesn't matter.
A. 48
B. 288
C. 495
D. 990
E. 11880
Combination problem
Anwer is C :
A. 48
B. 288
C. 495
D. 990
E. 11880
Combination problem
Anwer is C :
One kilogram of a certain coffee brand consists of x kilograms of Type I coffee and y kilograms of Type II coffee. The cost of the brand is C dollars per kilogram, where C = 6.5x + 8.5y. Is x '<' 0.8?
1) y '>'= 0.15
2) C '>'= 7.30
1) y '>'= 0.15
2) C '>'= 7.30
| |
x+y=1
Statement 1 implies that x could be less, equal or more than 0.8; not sufficient.
Statement 2 certainly is.
The correct answer is B
Statement 1 implies that x could be less, equal or more than 0.8; not sufficient.
Statement 2 certainly is.
The correct answer is B
Strategies for Ratio
Fact - if a data set can be split into two groups, both of which have at least the ratio a:b for some 2 characteristics, then the entire data set has at least the ratio a:b for those 2 characteristics.
The corollary for above:
if a data set can be split into two groups, and one of the groups has a ratio HIGHER than the overall ratio for some 2 characteristics, then the other group has a ratio LOWER than the overall ratio for those 2 characteristics - and vice versa.
The corollary for above:
if a data set can be split into two groups, and one of the groups has a ratio HIGHER than the overall ratio for some 2 characteristics, then the other group has a ratio LOWER than the overall ratio for those 2 characteristics - and vice versa.
An employee of Company Z is an employee of either Division X or Division Y, but not both. If each division has some part-time employees, is the ratio of the number of full-time employees to the number of part-time employees greater for division X than for company Z?
1) The ratio of the number of full-time employees to the number of part-time employees is less for Division Y than for company Z.
2) More than half of the full-time employees of Company Z are employees of Division X, and more than half of the part-time employees of Company Z are employees of Division Y.
here's a fact that you should know. i can furnish a proof if you reallyreallyreally want me to, but it should be clear:
if a data set can be split into two groups, both of which have at least the ratio a:b for some 2 characteristics, then the entire data set has at least the ratio a:b for those 2 characteristics.
in other words, if the ratio of FT to PT employees is at least, say, 3:1 in both divisions, then the overall ratio of FT to PT employees must also be 3:1.
here's a corollary:
if a data set can be split into two groups, and one of the groups has a ratio HIGHER than the overall ratio for some 2 characteristics, then the other group has a ratio LOWER than the overall ratio for those 2 characteristics - and vice versa.
this follows logically from the above statement, because it violates the first result (and common sense) if both divisions' ratios are somehow higher (or both lower) than the overall ratio.
--
statement (1)
this statement must be true, because if div. y has a lower ratio, then div. x must have a higher ratio to balance things out (see the corollary above).
so, sufficient.
if you want actual inequalities to prove this, i would be glad to provide them, but you should be able to conceptualize this result so that you have a fighting chance of completing the problem within the allotted time.
--
statement (2)
because FT and PT are mutually exclusive, this statement implies that div. x has more FT employees, but fewer PT employees, than does div. y.
therefore, the ratios are (higher / lower) for div. x and (lower / higher) for div. y, so the overall ratio must be higher for div. x.
sufficient
answer = d
1) The ratio of the number of full-time employees to the number of part-time employees is less for Division Y than for company Z.
2) More than half of the full-time employees of Company Z are employees of Division X, and more than half of the part-time employees of Company Z are employees of Division Y.
here's a fact that you should know. i can furnish a proof if you reallyreallyreally want me to, but it should be clear:
if a data set can be split into two groups, both of which have at least the ratio a:b for some 2 characteristics, then the entire data set has at least the ratio a:b for those 2 characteristics.
in other words, if the ratio of FT to PT employees is at least, say, 3:1 in both divisions, then the overall ratio of FT to PT employees must also be 3:1.
here's a corollary:
if a data set can be split into two groups, and one of the groups has a ratio HIGHER than the overall ratio for some 2 characteristics, then the other group has a ratio LOWER than the overall ratio for those 2 characteristics - and vice versa.
this follows logically from the above statement, because it violates the first result (and common sense) if both divisions' ratios are somehow higher (or both lower) than the overall ratio.
--
statement (1)
this statement must be true, because if div. y has a lower ratio, then div. x must have a higher ratio to balance things out (see the corollary above).
so, sufficient.
if you want actual inequalities to prove this, i would be glad to provide them, but you should be able to conceptualize this result so that you have a fighting chance of completing the problem within the allotted time.
--
statement (2)
because FT and PT are mutually exclusive, this statement implies that div. x has more FT employees, but fewer PT employees, than does div. y.
therefore, the ratios are (higher / lower) for div. x and (lower / higher) for div. y, so the overall ratio must be higher for div. x.
sufficient
answer = d
The function f is defined for all positive integers n by the following rule: f(n) is the number of positive integers each of which is less than n and has no positive factor in common with n other than 1. If p is prime, then f(p) =
(A) P-1
(B) P-2
(C) (P+1)/2
(D) (P-1)/2
(E) 2
If P is prime it's only factors are P and 1. So no number below it will have a common factor with it except 1. Therefore answer should just be P-1.
for eg
if p=2, then f(p) = 1 and 1 does not have a common factor with 2
---------------------------------------------------------------------------------
f(n) has n-1 positive integers less than n maximum value of f(n) = n-1
Now in the case of a prime number all the positive integers each of which is less than prime number will have no positive factor in common with prime number. Thus f(p) attains its maximum value of p-1 when p is a prime number
Alternatively take a prime 5
4 is not a factor
3 is not a factor
2 is not a factor
1 is not a factor
f(5) is 4 A may be true; B is ruled out; C is ruled out, D is ruled out, E is ruled out
only A can be the answer
(A) P-1
(B) P-2
(C) (P+1)/2
(D) (P-1)/2
(E) 2
If P is prime it's only factors are P and 1. So no number below it will have a common factor with it except 1. Therefore answer should just be P-1.
for eg
if p=2, then f(p) = 1 and 1 does not have a common factor with 2
---------------------------------------------------------------------------------
f(n) has n-1 positive integers less than n maximum value of f(n) = n-1
Now in the case of a prime number all the positive integers each of which is less than prime number will have no positive factor in common with prime number. Thus f(p) attains its maximum value of p-1 when p is a prime number
Alternatively take a prime 5
4 is not a factor
3 is not a factor
2 is not a factor
1 is not a factor
f(5) is 4 A may be true; B is ruled out; C is ruled out, D is ruled out, E is ruled out
only A can be the answer
At a two-day seminar, 90 percent of those registered attended the seminar on the first day. What percent of those registered did not attend the seminar on either day?
(1) A total of 1,000 people registered for the two-day seminar
(2) Of those registered, 80 percent attended the seminar on the second day.
Circle #1 is first day attendants, circle #2 is second day attendants, the overlap is attendants for both days, and Neither is the space outside the diagram.
The information from the original problem plus statement (1) give you the first circle (900). Insufficient to solve for Neither.
The information from the original problem plus statement (2) give you the second circle (800). Also Insufficient to solve for Neither.
Together, you still have no idea how many individuals addended Both days. Since the proper formula for such a situation is
Total = Group 1 + Group 2 - Both + Neither, in our case we have
1000 = 900 + 800 - Both + Neither.
Since we are still left with two variables (B and N), the correct answer is E.
(1) A total of 1,000 people registered for the two-day seminar
(2) Of those registered, 80 percent attended the seminar on the second day.
Circle #1 is first day attendants, circle #2 is second day attendants, the overlap is attendants for both days, and Neither is the space outside the diagram.
The information from the original problem plus statement (1) give you the first circle (900). Insufficient to solve for Neither.
The information from the original problem plus statement (2) give you the second circle (800). Also Insufficient to solve for Neither.
Together, you still have no idea how many individuals addended Both days. Since the proper formula for such a situation is
Total = Group 1 + Group 2 - Both + Neither, in our case we have
1000 = 900 + 800 - Both + Neither.
Since we are still left with two variables (B and N), the correct answer is E.
If n and y are positive integers and 450Y = n^3, which of the following must be an integer?
I. y/(3 x 2^2 x 5)
II. y/(3^2 x 2 x 5)
III. y/(3 x 2 x 5^2)
a. None
b. I only
c. II only
d. III only
e. I, II, and III
obtain all prime factors of 450 = 5x5x3x3x2
Therefore 5x5x3x3x2 x y = n^3
in order to make the right hand side a cube, we are missing one 5, one 3, and two 2s in order to make it a perfect cube
thus y/ (3)(5) (2 ^ 2) should be an integer
Answer choice B is the only choice
I. y/(3 x 2^2 x 5)
II. y/(3^2 x 2 x 5)
III. y/(3 x 2 x 5^2)
a. None
b. I only
c. II only
d. III only
e. I, II, and III
obtain all prime factors of 450 = 5x5x3x3x2
Therefore 5x5x3x3x2 x y = n^3
in order to make the right hand side a cube, we are missing one 5, one 3, and two 2s in order to make it a perfect cube
thus y/ (3)(5) (2 ^ 2) should be an integer
Answer choice B is the only choice
If x does not = -y, is (x-y)/(x+y) > 1?
(1) x > 0
(2) y < 0
(x-y)/(x+y) > 1?
means either:
a) if (x+y) is positive, then the question becomes
x-y > x+y? (direction of inequality sign stays the same as you cross-multiply)
-y > y? (subtract x -- no change to sign)
0 > 2y? (add y -- no change to sign)
0 > y? (divide by 2 -- no change to sign)
=> is y negative?
b) if (x+y) is negative, then the question becomes
x-y <>
same chain of algebra
0 <>
=> is y positive?
Even with the two conditions together, we never know for sure the sign of x+y, so we can't determine which case we're in and therefore which question is being asked. Anwer : E
(1) x > 0
(2) y < 0
(x-y)/(x+y) > 1?
means either:
a) if (x+y) is positive, then the question becomes
x-y > x+y? (direction of inequality sign stays the same as you cross-multiply)
-y > y? (subtract x -- no change to sign)
0 > 2y? (add y -- no change to sign)
0 > y? (divide by 2 -- no change to sign)
=> is y negative?
b) if (x+y) is negative, then the question becomes
x-y <>
same chain of algebra
0 <>
=> is y positive?
Even with the two conditions together, we never know for sure the sign of x+y, so we can't determine which case we're in and therefore which question is being asked. Anwer : E
Three grades of milk are 1 percent, 2 percent, and 3 percent fat by volume. If x gallons of the 1 percent grade, y gallons of the 2% grade, and z gallons of the 3% grade are mixed to give x+y+z gallons of a 1.5 percent grade, what is x in terms of y and z?
a. y + 3z
b. y+z / 4
c. 2y + 3z
d. 3y + z
e. 3y + 4.5z
Answer : A
a. y + 3z
b. y+z / 4
c. 2y + 3z
d. 3y + z
e. 3y + 4.5z
Answer : A
Question: From a group of 4 married couples, a team of 3 to be selected such that only one person from a couple is selected.
A) 16
b) 24
c) 26
d) 30
e) 32
Let's consider the case where couple #1 is on the committee. The third person can be chosen from any of the remaining six. So there are 6C1 = 6 ways of choosing the committee where both members of couple 1 are on the committee. The same reasoning holds for couples 2,3 and 4. So there are 6*4 = 24 "bad" ways of choosing the committee.
If there are 8C3 = 56 total ways of choosing the committee, and 24 "bad" ways of choosing the committee where a couple is on the committee, then that leaves us with 32 ways of choosing the committee where both members of the committee are NOT on the board.
A) 16
b) 24
c) 26
d) 30
e) 32
Let's consider the case where couple #1 is on the committee. The third person can be chosen from any of the remaining six. So there are 6C1 = 6 ways of choosing the committee where both members of couple 1 are on the committee. The same reasoning holds for couples 2,3 and 4. So there are 6*4 = 24 "bad" ways of choosing the committee.
If there are 8C3 = 56 total ways of choosing the committee, and 24 "bad" ways of choosing the committee where a couple is on the committee, then that leaves us with 32 ways of choosing the committee where both members of the committee are NOT on the board.
Baseball's World Series matches 2 teams against each other in a best-of-seven series. The first team to win four games wins the series and no subsequent games are played. If you have no special information about either of the teams, what is the probability that the World Series will consist of fewer than 7 games?
(A) 12.5%
(B) 25%
(C) 31.25%
(D) 68.75%
(E) 75%
OE - In order to determine the probability that the World Series will last fewer than 7 games, we can first
determine the probability that the World Series WILL last exactly 7 games and then subtract this value from 1.
In order for the World Series to last exactly 7 games, the first 6 games of the series must results in 3 wins and 3
losses for each team.
Let's analyze one way this could happen:
Game 1 -----Game 2 -----Game 3 -----Game 4 -----Game 5 -----Game 6
T1 Wins-----T1 Wins-----T1 Wins-----T1 Loses-----T1 Loses-----T1 Loses
There are many other ways this could happen. Using the permutation formula, there are 6!/(3!)(3!) = 20 ways for the
two teams to split the first 6 games (3 wins for each).
There are then 2 possible outcomes to break the tie in Game 7. Thus, there are a total of 20 × 2 = 40 ways for the
World Series to last the full 7 games.
The probability that any one of these 40 ways occurs can be calculated from the fact that the probability of a team
winning a game equals the probability of a team losing a game = 1/2.
Given that 7 distinct events must happen in any 7 game series, and that each of these events has a probability of
1/2, the probability that any one particular 7 game series occurs is
(1/2) ^ 7 = 1/ 128
Since there are 40 possible different 7 game series, the probability that the World Series will last exactly 7
games is:
40 * 1/128 = 40/ 128 = .3125 = 31.255%
Thus the probability that the World Series will last fewer than 7 games is:
100% - 31.25% = 68.75%.
The correct answer is D.
(A) 12.5%
(B) 25%
(C) 31.25%
(D) 68.75%
(E) 75%
OE - In order to determine the probability that the World Series will last fewer than 7 games, we can first
determine the probability that the World Series WILL last exactly 7 games and then subtract this value from 1.
In order for the World Series to last exactly 7 games, the first 6 games of the series must results in 3 wins and 3
losses for each team.
Let's analyze one way this could happen:
Game 1 -----Game 2 -----Game 3 -----Game 4 -----Game 5 -----Game 6
T1 Wins-----T1 Wins-----T1 Wins-----T1 Loses-----T1 Loses-----T1 Loses
There are many other ways this could happen. Using the permutation formula, there are 6!/(3!)(3!) = 20 ways for the
two teams to split the first 6 games (3 wins for each).
There are then 2 possible outcomes to break the tie in Game 7. Thus, there are a total of 20 × 2 = 40 ways for the
World Series to last the full 7 games.
The probability that any one of these 40 ways occurs can be calculated from the fact that the probability of a team
winning a game equals the probability of a team losing a game = 1/2.
Given that 7 distinct events must happen in any 7 game series, and that each of these events has a probability of
1/2, the probability that any one particular 7 game series occurs is
(1/2) ^ 7 = 1/ 128
Since there are 40 possible different 7 game series, the probability that the World Series will last exactly 7
games is:
40 * 1/128 = 40/ 128 = .3125 = 31.255%
Thus the probability that the World Series will last fewer than 7 games is:
100% - 31.25% = 68.75%.
The correct answer is D.
On his drive to work, Bob listens to one of three radio stations, A,B, or C. He first turns to A. If A is playing a
song he likes, he listens to it; if not, he turns to B. If B is playing a song he likes, he listens to it; if not,
he turns to C. If C is playing a song he likes, he listens to it; if not, he turns off the radio. For each station
, the probability is 0.30 that any given moment the station is playing a song Leo likes. On his drive to work, what
is the probability that Bob will hear a song he likes?
A) 0.027
B) 0.090
C) 0.417
D) 0.657
E) 0.900
Probability Leo listens to A = (Probability that A is playing a song of Leo's choice ) = 0.3
Probability Leo listens to B = (Probability that A is not playing a song of Leo's choice ) * (Probability that B is
playing a song of Leo's choice ) = (1-0.3) * 0.3 = 0.7*0.3
Probobility Leo listens to C = (Probability that A is not playing a song of Leo's choice ) * (Probability that B is
not playing a song of Leo's choice ) * (Probability that C is playing a song of Leo's choice ) = 0.7 * 0.7 * 0.3
Total Probability = 0.3 + 0.7*0.3 + 0.7*0.7*0.3 = 0.657
song he likes, he listens to it; if not, he turns to B. If B is playing a song he likes, he listens to it; if not,
he turns to C. If C is playing a song he likes, he listens to it; if not, he turns off the radio. For each station
, the probability is 0.30 that any given moment the station is playing a song Leo likes. On his drive to work, what
is the probability that Bob will hear a song he likes?
A) 0.027
B) 0.090
C) 0.417
D) 0.657
E) 0.900
Probability Leo listens to A = (Probability that A is playing a song of Leo's choice ) = 0.3
Probability Leo listens to B = (Probability that A is not playing a song of Leo's choice ) * (Probability that B is
playing a song of Leo's choice ) = (1-0.3) * 0.3 = 0.7*0.3
Probobility Leo listens to C = (Probability that A is not playing a song of Leo's choice ) * (Probability that B is
not playing a song of Leo's choice ) * (Probability that C is playing a song of Leo's choice ) = 0.7 * 0.7 * 0.3
Total Probability = 0.3 + 0.7*0.3 + 0.7*0.7*0.3 = 0.657
Two canoe riders must be selected from each of two groups of campers. One group consists of three men and one
woman, and the other group consists of two women and one man. What is the probability that two men and two women
will be selected?
(A) 1/6
(B) 1/4
(C) 2/7
(D) 1/3
(E) 1/2
1st Group: 3 Men, 1 Woman
2nd Group: 1 Man, 2 Women
Total number of ways of selection: 4C2 * 3C2 = 18
No of ways of selecting two men and two women:
1. 2 men and 0 woman from 1st group and 0 men and 2 women from 2nd group = 3C2 = 3
2. 1 man and 1 woman from 1st group and 1 man and 1 woman from 2nd group = 3C1* 2C1 = 6 Thus Total ways from the
above = 3 + 6 = 9
Therefore Probability = 9/18 = 1/2
woman, and the other group consists of two women and one man. What is the probability that two men and two women
will be selected?
(A) 1/6
(B) 1/4
(C) 2/7
(D) 1/3
(E) 1/2
1st Group: 3 Men, 1 Woman
2nd Group: 1 Man, 2 Women
Total number of ways of selection: 4C2 * 3C2 = 18
No of ways of selecting two men and two women:
1. 2 men and 0 woman from 1st group and 0 men and 2 women from 2nd group = 3C2 = 3
2. 1 man and 1 woman from 1st group and 1 man and 1 woman from 2nd group = 3C1* 2C1 = 6 Thus Total ways from the
above = 3 + 6 = 9
Therefore Probability = 9/18 = 1/2
Which of the following sets includes ALL of the solutions of x that will satisfy the equation:
|x-2| - |x-3| = |x-5|
a) {-6,-5,0,1,7,8}
b) {-4,-2,0,10/3,4,5}
c) {-4,0,1,4,5,6}
d) {-1.10/3,3,5,6,8}
e) {-2,-1,1,3,4,5}
OE - One way to solve equations with absolute values is to solve for x over a series of intervals. In each interval of x, the sign of the expressions within each pair of absolute value indicators does not change. In the equation Ix-2I - Ix-3I = Ix-5I , there are 4 intervals of interest: x less than 2 - In this interval, the value inside each of the three absolute value expressions is negative. 2 less than x less than 3 - In this interval, the value inside the first absolute value expression is positive, while the value inside the other two absolute value expressions is negative. 3 less than x less than 5 - In this interval, the value inside the first two absolute value expressions is positive, while the value inside the last absolute value expression is negative. 5 less than x - In this interval, the value inside each of the three absolute value expressions is positive. Use each interval for x to rewrite the equation so that it can be evaluated without absolute value signs. For the first interval, x less than 2, we can solve the equation by rewriting each of the expressions inside the absolute value signs as negative (and thereby remove the absolute value signs): - x + 2 - ( -x + 3 ) = - x + 5 - x + 2 + x - 3 = - x + 5 x = 6 Notice that the solution x = 6 is NOT a valid solution since it lies outside the interval x less than 2. (Remember, we are solving the equation for x SUCH THAT x is within the interval of interest). For the second interval 2 less than x less than 3, we can solve the equation by rewriting the expression inside the first absolute value sign as positive and by rewriting the expressions inside the other absolute values signs as negative: x - 2 - ( -x + 3 ) = - x + 5 x - 2 + x - 3 = - x + 5 3x = 10 x = 10/3 Notice, again, that the solution x = 10/3 is NOT a valid solution since it lies outside the interval 2 less than x less than 3 For the third interval 3 less than x less than 5, we can solve the equation by rewriting the expressions inside the first two absolute value signs as positive and by rewriting the expression inside the last absolute value sign as negative: x - 2 - ( x - 3 ) = - x + 5 x - 2 - x + 3 = - x + 5 x = 4 The solution x = 4 is a valid solution since it lies within the interval 3 less than x less than 5 Finally, for the fourth interval 5 less than x, we can solve the equation by rewriting each of the expressions inside the absolute value signs as positive: x - 2 - ( x - 3 ) = x - 5 x - 2 - x + 3 = x - 5 x = 6 The solution x = 6 is a valid solution since it lies within the interval 5 less than x. We conclude that the only two solutions of the original equation are x = 4 and x = 6. Only answer choice C contains all of the solutions, both 4 and 6, as part of its set. Therefore, C is the correct answer
|x-2| - |x-3| = |x-5|
a) {-6,-5,0,1,7,8}
b) {-4,-2,0,10/3,4,5}
c) {-4,0,1,4,5,6}
d) {-1.10/3,3,5,6,8}
e) {-2,-1,1,3,4,5}
OE - One way to solve equations with absolute values is to solve for x over a series of intervals. In each interval of x, the sign of the expressions within each pair of absolute value indicators does not change. In the equation Ix-2I - Ix-3I = Ix-5I , there are 4 intervals of interest: x less than 2 - In this interval, the value inside each of the three absolute value expressions is negative. 2 less than x less than 3 - In this interval, the value inside the first absolute value expression is positive, while the value inside the other two absolute value expressions is negative. 3 less than x less than 5 - In this interval, the value inside the first two absolute value expressions is positive, while the value inside the last absolute value expression is negative. 5 less than x - In this interval, the value inside each of the three absolute value expressions is positive. Use each interval for x to rewrite the equation so that it can be evaluated without absolute value signs. For the first interval, x less than 2, we can solve the equation by rewriting each of the expressions inside the absolute value signs as negative (and thereby remove the absolute value signs): - x + 2 - ( -x + 3 ) = - x + 5 - x + 2 + x - 3 = - x + 5 x = 6 Notice that the solution x = 6 is NOT a valid solution since it lies outside the interval x less than 2. (Remember, we are solving the equation for x SUCH THAT x is within the interval of interest). For the second interval 2 less than x less than 3, we can solve the equation by rewriting the expression inside the first absolute value sign as positive and by rewriting the expressions inside the other absolute values signs as negative: x - 2 - ( -x + 3 ) = - x + 5 x - 2 + x - 3 = - x + 5 3x = 10 x = 10/3 Notice, again, that the solution x = 10/3 is NOT a valid solution since it lies outside the interval 2 less than x less than 3 For the third interval 3 less than x less than 5, we can solve the equation by rewriting the expressions inside the first two absolute value signs as positive and by rewriting the expression inside the last absolute value sign as negative: x - 2 - ( x - 3 ) = - x + 5 x - 2 - x + 3 = - x + 5 x = 4 The solution x = 4 is a valid solution since it lies within the interval 3 less than x less than 5 Finally, for the fourth interval 5 less than x, we can solve the equation by rewriting each of the expressions inside the absolute value signs as positive: x - 2 - ( x - 3 ) = x - 5 x - 2 - x + 3 = x - 5 x = 6 The solution x = 6 is a valid solution since it lies within the interval 5 less than x. We conclude that the only two solutions of the original equation are x = 4 and x = 6. Only answer choice C contains all of the solutions, both 4 and 6, as part of its set. Therefore, C is the correct answer
Equal amounts of water were poured into two empty jars of different capacities, which made one jar 1/4 full and the
other jar 1/3 full. If the water in the jar with lesser capacity is then poured into the jar with the greater
capacity, what fraction of the larger jar will be filled with water?
A. 1/7
B. 2/7
C. 1/2
D. 7/12
E. 2/3
The jar that is 1/3 full is smaller
water in jar 1 = water in jar 2
jar 1 is now twice as full (1/4)*2 = 1/2
or
1/4 + 1/4 = 1/2
other jar 1/3 full. If the water in the jar with lesser capacity is then poured into the jar with the greater
capacity, what fraction of the larger jar will be filled with water?
A. 1/7
B. 2/7
C. 1/2
D. 7/12
E. 2/3
The jar that is 1/3 full is smaller
water in jar 1 = water in jar 2
jar 1 is now twice as full (1/4)*2 = 1/2
or
1/4 + 1/4 = 1/2
This year Henry will save a certain amount of his income, and he will spend the rest. Next year Henry will have no income, but for each dollar that he saves this year, he will have 1 + r dollars available to spend. In terms of r, what fraction of his income should Henry save this year so that next year the amount he was available to spend will be equal to half the amount that he spends this year?
(A) 1/(r+2)
(B) 1/2(r+2)
(C) 1/(3r+2)
(D) 1/(r+3)
(E) 1/(2r+3)
Let total income = x
Let saved = y
=> spent = x-y
Total dollars available that he can spend next year = y(1+r)
Given y(1+r) = (x-y)/2
=> y = x/(2r+3)
Fraction (y/x) = 1/(2r+3)
Hence E
(A) 1/(r+2)
(B) 1/2(r+2)
(C) 1/(3r+2)
(D) 1/(r+3)
(E) 1/(2r+3)
Let total income = x
Let saved = y
=> spent = x-y
Total dollars available that he can spend next year = y(1+r)
Given y(1+r) = (x-y)/2
=> y = x/(2r+3)
Fraction (y/x) = 1/(2r+3)
Hence E
Louie takes out a three-month loan of $1000. The lender charges him 10% interest per month compunded monthly. The
terms of the loan state that Louie must repay the loan in three equal monthly payments. To the nearest dollar, how
much does Louie have to pay each month?
(A) 333
(B) 383
(C) 402
(D) 433
(E) 483
Answer - C
terms of the loan state that Louie must repay the loan in three equal monthly payments. To the nearest dollar, how
much does Louie have to pay each month?
(A) 333
(B) 383
(C) 402
(D) 433
(E) 483
Answer - C
On a certain day, Tim invest $1,000 at 10 percent annual interest, compound annually, and Lana invested $ 2,000 at 5 percent annual interest, compound annually. The total amount of interest earned by Tim's investment in the first 2 years was how much greater than the total amount of interest earned by Lana's investment in the first 2 years?
A. 5
B. 15
C. 50
D. 100
E. 105
Amount = P[1+(r/100)] ^ t
Tim's investment Amount = 1000(1+10%)^2=1210
Interest earned by Tim = Amount - P = 1210 - 1000 = 210
Lana's investment Amount = 2000(1+5%)^2=2205
Interest earned by Lana = Amount - P = 2205 - 2000 = 205
Hence 210-205=5
A. 5
B. 15
C. 50
D. 100
E. 105
Amount = P[1+(r/100)] ^ t
Tim's investment Amount = 1000(1+10%)^2=1210
Interest earned by Tim = Amount - P = 1210 - 1000 = 210
Lana's investment Amount = 2000(1+5%)^2=2205
Interest earned by Lana = Amount - P = 2205 - 2000 = 205
Hence 210-205=5
The vertical position of an object can be approximated at any given time by the function: p(t) = rt – 5t2 + b where p(t) is the vertical position in meters, t is the time in seconds, and r and b are constants. After 2 seconds, the position of an object is 41 meters, and after 5 seconds the position is 26 meters. What is the position of the object, in meters, after 4 seconds?
(A) 24
(B) 26
(C) 39
(D) 41
(E) 45
(A) 24
(B) 26
(C) 39
(D) 41
(E) 45
Stephanie, Regine, and Brian ran a 20 mile race. Stephanie and Regine's combined times exceeded Brian's time by exactly 2 hours. If nobody ran faster than 8 miles per hour, who could have won the race?
I. Stephanie II. Regine III. Brian
(A) I only
(B) II only
(C) III only
(D) I or II only
(E) I, II, or III
Let the race time of Stephanie = S
Let the race time of Regine = R
Let the race time of Brian = B
It is given that -- Stephanie and Regine's combined times exceed Brian's time by 2 hours.
Hence S + R = B + 2 ------- (1)
To win the race, an individual's time must be less than 1/3 of the the combined times of all the runners.
Therefore for Brian to win the race (=> that Brian would have the lowest time) his time would need to be less than
1/3 of the combined times for all the runners.
=> B is less than 1/3 of (S+R+B)
=>3B<> is less than (S+R+B)
=> 2B is less than (S+R)
Making use of (1) in above
2B<> is less than (S+R)
=> 2B<> is less than B+2
=> B<2
Thus to win the race Brian's time must be less than 2 hours. which is impossible as fastest Brian run is 8 miles/hr
=> that the least amount of time in which he can complete the 20 mile race is 2.5 hrs.
Hence Stephanie and Regine as possible winners. Since the question gives us same information about Stephanie and
Regine, we cannot state either one as a possible winner.
Hence, the correct answer is D
I. Stephanie II. Regine III. Brian
(A) I only
(B) II only
(C) III only
(D) I or II only
(E) I, II, or III
Let the race time of Stephanie = S
Let the race time of Regine = R
Let the race time of Brian = B
It is given that -- Stephanie and Regine's combined times exceed Brian's time by 2 hours.
Hence S + R = B + 2 ------- (1)
To win the race, an individual's time must be less than 1/3 of the the combined times of all the runners.
Therefore for Brian to win the race (=> that Brian would have the lowest time) his time would need to be less than
1/3 of the combined times for all the runners.
=> B is less than 1/3 of (S+R+B)
=>3B<> is less than (S+R+B)
=> 2B is less than (S+R)
Making use of (1) in above
2B<> is less than (S+R)
=> 2B<> is less than B+2
=> B<2
Thus to win the race Brian's time must be less than 2 hours. which is impossible as fastest Brian run is 8 miles/hr
=> that the least amount of time in which he can complete the 20 mile race is 2.5 hrs.
Hence Stephanie and Regine as possible winners. Since the question gives us same information about Stephanie and
Regine, we cannot state either one as a possible winner.
Hence, the correct answer is D
A hiker walking at a constant rate of 4 miles per hour is passed by a cyclist traveling in the same direction along the same path at a constant rate of 20 miles per hour. The cyclist stops to wait for the hiker 5 minutes after passing her, while the hiker continue to walk at her constant rate. How many minutes must the cyclist wait until the hiker catches up?
A. 20/3
B. 15
C. 20
D. 25
E. 80/3
Hiker's relative speed = 16 m/h
Hiker traveled in 5 min: 16 * 5/60 = 4/3 miles.
Time taken by hiker to cover 4/3 miles: 4/(4/3) = 1/3 hrs = 20 minutes
A. 20/3
B. 15
C. 20
D. 25
E. 80/3
Hiker's relative speed = 16 m/h
Hiker traveled in 5 min: 16 * 5/60 = 4/3 miles.
Time taken by hiker to cover 4/3 miles: 4/(4/3) = 1/3 hrs = 20 minutes
A paint crew gets a rush order to paint 80 houses in a new development. They paint the first y houses at a rate of x houses per week. Realizing that they'll be late at this rate, they bring in some more painters and paint the rest of the houses at the rate of 1.25x houses per week. The total time it takes them to paint all the houses under this scenario is what fraction of the time it would have taken if they had painted all the houses at their original rate of x houses per week?
(A) 0.8(80 – y)
(B) 0.8 + 0.0025y
(C) 80/y – 1.25
(D) 80/1.25y
(E) 80 – 0.25y
(A) 0.8(80 – y)
(B) 0.8 + 0.0025y
(C) 80/y – 1.25
(D) 80/1.25y
(E) 80 – 0.25y
One smurf and one elf can build a treehouse together in two hours, but the smurf would need the help of two fairies in order to complete the same job in the same amount of time. If one elf and one fairy worked together, it would take them four hours to build the treehouse. Assuming that work rates for smurfs, elves, and fairies remain constant, how many hours would it take one smurf, one elf, and one fairy, working together, to build the treehouse?
(A) 5/7
(B) 1
(C) 10/7
(D) 12/7
(E) 22/7
The combined rate of individuals working together is equal to the sum of all the individual working rates.
Let s = rate of a smurf, e = rate of an elf, and f = rate of a fairy. A rate is expressed in terms of treehouses/hour. So for instance, the first equation below says that a smurf and an elf working together can build 1 treehouse per 2 hours, for a rate of 1/2 treehouse per hour.
1) s + e = 1/2
2) s + 2 f = 1/2
3) e + f = 1/4
The three equations can be combined by solving the first one for s in terms of e, and the third equation for f in terms of e, and then by substituting both new equations into the middle equation.
1) s = 1/2 – e
2) (1/2 – e) + 2 (1/4 – e) = 1/2
3) f = 1/4 – e
Now, we simply solve equation 2 for e:
(1/2 – e) + 2 (1/4 – e) = 1/2
2/4 – e + 2/4 – 2 e = 2/4
4/4 – 3e = 2/4
-3e = -2/4
e = 2/12
e = 1/6
Once we know e, we can solve for s and f:
s = 1/2 – e
s = 1/2 – 1/6
s = 3/6 – 1/6
s = 2/6s = 1/3
f = 1/4 – e
f = 1/4 – 1/6
f = 3/12 – 2/12
f = 1/12
We add up their individual rates to get a combined rate:
e + s + f
=1/6 + 1/3 + 1/12
=2/12 + 4/12 + 1/12
= 7/12
Remembering that a rate is expressed in terms of treehouses/hour, this indicates that a smurf, an elf, and a fairy, working together, can produce 7 treehouses per 12 hours. Since we want to know the number of hours per treehouse, we must take the reciprocal of the rate. Therefore we conclude that it takes them 12 hours per 7 treehouses, which is equivalent to 12/7 of an hour per treehouse.
The correct answer is D.
(A) 5/7
(B) 1
(C) 10/7
(D) 12/7
(E) 22/7
The combined rate of individuals working together is equal to the sum of all the individual working rates.
Let s = rate of a smurf, e = rate of an elf, and f = rate of a fairy. A rate is expressed in terms of treehouses/hour. So for instance, the first equation below says that a smurf and an elf working together can build 1 treehouse per 2 hours, for a rate of 1/2 treehouse per hour.
1) s + e = 1/2
2) s + 2 f = 1/2
3) e + f = 1/4
The three equations can be combined by solving the first one for s in terms of e, and the third equation for f in terms of e, and then by substituting both new equations into the middle equation.
1) s = 1/2 – e
2) (1/2 – e) + 2 (1/4 – e) = 1/2
3) f = 1/4 – e
Now, we simply solve equation 2 for e:
(1/2 – e) + 2 (1/4 – e) = 1/2
2/4 – e + 2/4 – 2 e = 2/4
4/4 – 3e = 2/4
-3e = -2/4
e = 2/12
e = 1/6
Once we know e, we can solve for s and f:
s = 1/2 – e
s = 1/2 – 1/6
s = 3/6 – 1/6
s = 2/6s = 1/3
f = 1/4 – e
f = 1/4 – 1/6
f = 3/12 – 2/12
f = 1/12
We add up their individual rates to get a combined rate:
e + s + f
=1/6 + 1/3 + 1/12
=2/12 + 4/12 + 1/12
= 7/12
Remembering that a rate is expressed in terms of treehouses/hour, this indicates that a smurf, an elf, and a fairy, working together, can produce 7 treehouses per 12 hours. Since we want to know the number of hours per treehouse, we must take the reciprocal of the rate. Therefore we conclude that it takes them 12 hours per 7 treehouses, which is equivalent to 12/7 of an hour per treehouse.
The correct answer is D.
6 machines, each working at the same constant rate, together can complete a certain job in 12 days, How many additional machines, each working at the same constant rate, will be needed to complete the job in 8 days?
A. 2
B. 3
C. 4
D. 6
E. 8
6 machines take = 12 days.
Therefore 1 machine = 12*6 days.
8 days will take = (12*6)/8 = 9 machines.
No of additional machines required = 9-6 = 3
A. 2
B. 3
C. 4
D. 6
E. 8
6 machines take = 12 days.
Therefore 1 machine = 12*6 days.
8 days will take = (12*6)/8 = 9 machines.
No of additional machines required = 9-6 = 3
Pumps A, B, and C operate at their respective constant rates. Pumps A and B, operating simultaneously, can fill a certain tank in 6/5 hours, pumps A and C, operating simultaneously, can fill the tank in 3/2 hours; and pumps B and C, operating simultaneously, can fill the tank in 2 hours. How many hours does it take pumps A, B, and C, operating simultaneously, to fill the tank?
A) 1/3
B) 1/2
C) 2/3
D) 5/6
E) 1
1/A + 1/B = 5/6
1/B +1/C = 2/3
1/C + 1/A =1/2
Adding all three equations above we get:
1/A + 1/B + 1/B +1/C + 1/C + 1/A = 5/6 + 2/3 + 1/2
2(1/A + 1/B + 1/C) = (5 + 4 + 3)/ 6 = 12/6
1/A + 1/B + 1/C = 12/6*2 = 1
=> Pumps A, B, and C, operating simultaneously will fill the tank in 1 hr
A) 1/3
B) 1/2
C) 2/3
D) 5/6
E) 1
1/A + 1/B = 5/6
1/B +1/C = 2/3
1/C + 1/A =1/2
Adding all three equations above we get:
1/A + 1/B + 1/B +1/C + 1/C + 1/A = 5/6 + 2/3 + 1/2
2(1/A + 1/B + 1/C) = (5 + 4 + 3)/ 6 = 12/6
1/A + 1/B + 1/C = 12/6*2 = 1
=> Pumps A, B, and C, operating simultaneously will fill the tank in 1 hr
Each of the 10 machines works at the same constant rate of doing certain job. The amount of time needed by 10 machines, working together to complete the job is 16 hrs. How many hours are needed if only 8 machines working together were to complete the job?
A. 18
B. 20
C. 22
D. 24
E. 26
10 machines, working at same constant rate, take time to complete a job = 16 hrs
thus 1 machine takes time to complete a job = 10 * 16 = 160 hrs
=> 8 machines will take = 160/8 = 20 hrs to complete the job.
A. 18
B. 20
C. 22
D. 24
E. 26
10 machines, working at same constant rate, take time to complete a job = 16 hrs
thus 1 machine takes time to complete a job = 10 * 16 = 160 hrs
=> 8 machines will take = 160/8 = 20 hrs to complete the job.
If x, y, and z are positive integers such that x is less than y and y is less than z, is x a factor of the odd integer z?
(1) x and y are prime numbers, whose sum is a factor of 57
(2) z is a factor of 57
(1) states x+y is a factor of 57.The factors of 57 are 1,3 and 19. Since x and y are positive integers and both are prime, x+y cannot be 1, 3 because if its 1 or 3, one of x or y has to be 0 or 1 which is not a prime number.Thus x+y must be 19. Number 2 cannot be the factor of z as z is an odd number.Now x cannot be 17 as it is less than y, still even if it is, it does not tell us whether it is a factor of z or not.Thus option 1 is insufficient to answer the ques.
(2) states z is a factor of 57. Thus z is either 3 or 19. Since x is less than z, and z has factors z itself and 1, so x cannot be the factor of z, but if x = 1 it can be the factor of z, thus statement 2 alone is not sufficient to answer the question.
Now combine both 1 and 2.In this case numbers x and y are 2 and 17 as x is less than y and both x and y are prime numbers, now according to option 2, z is either 1,3, 19 or 57.
Nowhere it is stated that x+y=z, since z is greater than y it can be either 19 or 57, in both the cases x is not the factor of z.
Thus (1) and (2) together are sufficient to answer the question.
Hence answer is C.
Note:- You will notice that 1 and the number itself are always factors of a given number.
Mohit Gupta, one of the Gmat aspirants corrected the above explanation so correct answer is A.
But no where it has been told that X has to be odd number or cannot be even number. As X has to be a PRIME NUMBER, we can assume it to be 2. So X can be 2 and Y can be 17 and thus statement 1 if true, proves that X is not the factor of Z.Now Statement 2 in itself is not sufficient as it tells us about the possible values of Z which can be 1, 3, 19, 57 but does not tell us about the possible values of X and Y.So answer should be A.
Answer is A.
(1) x and y are prime numbers, whose sum is a factor of 57
(2) z is a factor of 57
(1) states x+y is a factor of 57.The factors of 57 are 1,3 and 19. Since x and y are positive integers and both are prime, x+y cannot be 1, 3 because if its 1 or 3, one of x or y has to be 0 or 1 which is not a prime number.Thus x+y must be 19. Number 2 cannot be the factor of z as z is an odd number.Now x cannot be 17 as it is less than y, still even if it is, it does not tell us whether it is a factor of z or not.Thus option 1 is insufficient to answer the ques.
(2) states z is a factor of 57. Thus z is either 3 or 19. Since x is less than z, and z has factors z itself and 1, so x cannot be the factor of z, but if x = 1 it can be the factor of z, thus statement 2 alone is not sufficient to answer the question.
Now combine both 1 and 2.In this case numbers x and y are 2 and 17 as x is less than y and both x and y are prime numbers, now according to option 2, z is either 1,3, 19 or 57.
Nowhere it is stated that x+y=z, since z is greater than y it can be either 19 or 57, in both the cases x is not the factor of z.
Thus (1) and (2) together are sufficient to answer the question.
Hence answer is C.
Note:- You will notice that 1 and the number itself are always factors of a given number.
Mohit Gupta, one of the Gmat aspirants corrected the above explanation so correct answer is A.
But no where it has been told that X has to be odd number or cannot be even number. As X has to be a PRIME NUMBER, we can assume it to be 2. So X can be 2 and Y can be 17 and thus statement 1 if true, proves that X is not the factor of Z.Now Statement 2 in itself is not sufficient as it tells us about the possible values of Z which can be 1, 3, 19, 57 but does not tell us about the possible values of X and Y.So answer should be A.
Answer is A.
If d represents the hundredths digit and e represents the thousandths digit in the decimal .4de, what is the value of this decimal rounded to the nearest tenth?
(1) d – e is equal to a positive perfect square.
(2) sqrt (d) > e*e
From statement (1), we know that d – e must equal a positive perfect square. This means that d is greater than e. In addition, since any single digit minus any other single digit can yield a maximum of 9, d – e could only result in the perfect squares 9, 4, or 1.
However, this leaves numerous possibilities for the values of d and e respectively. For example, two possibilities are as follows:
d = 7, e = 3 (d – e = the perfect square 4)
d = 3, e = 2 (d – e = the perfect square 1)
In the first case, the decimal .4de would be .473, which, when rounded to the nearest tenth, is equal to .5. In the second case, the decimal would be .432, which, when rounded to the nearest tenth, is .4.
Thus, statement (1) is not sufficient on its own to answer the question.
Statement (2) tells us that sqrt d = e2. Since d is a single digit, the maximum value for d is 9, which means the maximum square root of d is 3. This means that e2 must be less than 3. Thus the digit e can only be 0 or 1.
However, this leaves numerous possibilities for the values of d and e respectively. For example, two possibilities are as follows:
d = 9, e = 1
d = 2, e = 0
In the first case, the decimal .4de would be .491, which, when rounded to the nearest tenth, is equal to .5. In the second case, the decimal would be .420, which, when rounded to the nearest tenth, is .4.
Thus, statement (2) is not sufficient on its own to answer the question.
Taking both statements together, we know that e must be 0 or 1 and that d – e is equal to 9, 4 or 1.
This leaves the following 4 possibilities:
d = 9, e = 0
d = 5, e = 1
d = 4, e = 0
d = 1, e = 0
These possibilities yield the following four decimals: .490, .451, .440, and .410 respectively. The first two of these decimals yield .5 when rounded to the nearest tenth, while the second two decimals yield .4 when rounded to the nearest tenth.
Thus, both statements taken together are not sufficient to answer the question.
The correct answer is E: Statements (1) and (2) TOGETHER are NOT sufficient.
(1) d – e is equal to a positive perfect square.
(2) sqrt (d) > e*e
From statement (1), we know that d – e must equal a positive perfect square. This means that d is greater than e. In addition, since any single digit minus any other single digit can yield a maximum of 9, d – e could only result in the perfect squares 9, 4, or 1.
However, this leaves numerous possibilities for the values of d and e respectively. For example, two possibilities are as follows:
d = 7, e = 3 (d – e = the perfect square 4)
d = 3, e = 2 (d – e = the perfect square 1)
In the first case, the decimal .4de would be .473, which, when rounded to the nearest tenth, is equal to .5. In the second case, the decimal would be .432, which, when rounded to the nearest tenth, is .4.
Thus, statement (1) is not sufficient on its own to answer the question.
Statement (2) tells us that sqrt d = e2. Since d is a single digit, the maximum value for d is 9, which means the maximum square root of d is 3. This means that e2 must be less than 3. Thus the digit e can only be 0 or 1.
However, this leaves numerous possibilities for the values of d and e respectively. For example, two possibilities are as follows:
d = 9, e = 1
d = 2, e = 0
In the first case, the decimal .4de would be .491, which, when rounded to the nearest tenth, is equal to .5. In the second case, the decimal would be .420, which, when rounded to the nearest tenth, is .4.
Thus, statement (2) is not sufficient on its own to answer the question.
Taking both statements together, we know that e must be 0 or 1 and that d – e is equal to 9, 4 or 1.
This leaves the following 4 possibilities:
d = 9, e = 0
d = 5, e = 1
d = 4, e = 0
d = 1, e = 0
These possibilities yield the following four decimals: .490, .451, .440, and .410 respectively. The first two of these decimals yield .5 when rounded to the nearest tenth, while the second two decimals yield .4 when rounded to the nearest tenth.
Thus, both statements taken together are not sufficient to answer the question.
The correct answer is E: Statements (1) and (2) TOGETHER are NOT sufficient.
Is x^2 + y^2 > 4a?
(1) (x + y)^2 = 9a
(2) (x – y)^2 = a
Official Answer to the above problem.
(1) INSUFFICIENT: If we multiply this equation out, we get:
x2 + 2xy + y2 = 9a
If we try to solve this expression for
x2 + y2, we getx2 + y2 = 9a – 2xy
Since the value of this expression depends on the value of x and y, we don't have enough information.
(2) INSUFFICIENT: If we multiply this equation out, we get:
x2 – 2xy + y2 = a
If we try to solve this expression for x2 + y2,
we getx2 + y2 = a + 2xy
Since the value of this expression depends on the value of x and y, we don't have enough information.
(1) AND (2) INSUFFICIENT: We can combine the two expanded forms of the equations from the two statements by adding them:
x2 + 2xy + y2 = 9ax2 – 2xy + y2 = a----- 2x2 + 2y2 = 10ax2 + y2 = 5a
If we substitute this back into the original question, the question becomes: "Is 5a > 4a?"If a > 0, the answer is yes.We know from the question stem that a is nonnegative.However, if a = 0 the answer is no.
The correct answer is E.
(1) (x + y)^2 = 9a
(2) (x – y)^2 = a
Official Answer to the above problem.
(1) INSUFFICIENT: If we multiply this equation out, we get:
x2 + 2xy + y2 = 9a
If we try to solve this expression for
x2 + y2, we getx2 + y2 = 9a – 2xy
Since the value of this expression depends on the value of x and y, we don't have enough information.
(2) INSUFFICIENT: If we multiply this equation out, we get:
x2 – 2xy + y2 = a
If we try to solve this expression for x2 + y2,
we getx2 + y2 = a + 2xy
Since the value of this expression depends on the value of x and y, we don't have enough information.
(1) AND (2) INSUFFICIENT: We can combine the two expanded forms of the equations from the two statements by adding them:
x2 + 2xy + y2 = 9ax2 – 2xy + y2 = a----- 2x2 + 2y2 = 10ax2 + y2 = 5a
If we substitute this back into the original question, the question becomes: "Is 5a > 4a?"If a > 0, the answer is yes.We know from the question stem that a is nonnegative.However, if a = 0 the answer is no.
The correct answer is E.
Is the integer x divisible by 3?
1) The last digit in x is 3.
2) x+5 is divisible by 6.
Statement (1) is not sufficient.
If x is 33, then x is divisible by 3, and (1) holds, but if x=43, then (1) is true but x is not divisible by 3.
Statement (2) is sufficient. According to it , there exists an integer k such that x+5 = 6k => x = 6k-5 => x/3 = (6k-5)/3 = x/3 = (2k) - 5/3 => x is not divisible by 3.
Hence B is the answer i.e the (2) statement alone is sufficient to answer the question but (1) statement alone is not sufficient to answer the question.
1) The last digit in x is 3.
2) x+5 is divisible by 6.
Statement (1) is not sufficient.
If x is 33, then x is divisible by 3, and (1) holds, but if x=43, then (1) is true but x is not divisible by 3.
Statement (2) is sufficient. According to it , there exists an integer k such that x+5 = 6k => x = 6k-5 => x/3 = (6k-5)/3 = x/3 = (2k) - 5/3 => x is not divisible by 3.
Hence B is the answer i.e the (2) statement alone is sufficient to answer the question but (1) statement alone is not sufficient to answer the question.
What is the first term of an arithmetic progression of positive integers ?
a)Sum of the squares of the first and second term is 116.
b)The seventh term is divisible by 10.
Explanation - let x be the first and y be the second term.
Hence 0 less than x less than y
From Statement 1: x^2 + y^2 = 116 only when x=2 and y=10.
Hence the first term is 4. This is the only combination that works - hence sufficient
Statement 2: let 10z be the 7th term (z is an integer)
Thus x+6(y-x)=10z => y=(10z+5x)/6. Hence the last digit of the numerator can be 5 or 0. Now y is also an integer => x must be even => least possible value of y =10.
Assuming different values of z we get
z=1; x=12 This is impossible (x should be less than y)
z=2; x=8; 8 10 12 14 16 18 20
z=3; x=6; 6 10 14 18 22 26 30 and so on -- hence insufficient
a)Sum of the squares of the first and second term is 116.
b)The seventh term is divisible by 10.
Explanation - let x be the first and y be the second term.
Hence 0 less than x less than y
From Statement 1: x^2 + y^2 = 116 only when x=2 and y=10.
Hence the first term is 4. This is the only combination that works - hence sufficient
Statement 2: let 10z be the 7th term (z is an integer)
Thus x+6(y-x)=10z => y=(10z+5x)/6. Hence the last digit of the numerator can be 5 or 0. Now y is also an integer => x must be even => least possible value of y =10.
Assuming different values of z we get
z=1; x=12 This is impossible (x should be less than y)
z=2; x=8; 8 10 12 14 16 18 20
z=3; x=6; 6 10 14 18 22 26 30 and so on -- hence insufficient
If x is a non-zero integer, what is the value of x ^ y?
(1) x = 2
(2) (128 ^ x)[6 ^ (x + y)] = (48 ^ 2x)(3 ^ -x)
Official Answer and Explanation to the above problem
One of the most effective ways to begin solving problems involving exponential equations is to break down bases of the exponents into prime factors and combine exponents with the same base. Following this approach, be sure to simplify each statement as much as possible before arriving at the conclusion, since difficult problems with exponents often result in unobvious outcomes.
(1) INSUFFICIENT: While this statement gives us the value of x, we know nothing about y and cannot determine the value of x^y.
(2) SUFFICIENT: (128^x)[6^(x + y)] = (48^2x)(3^-x)
[(2^7)^x][(2 × 3)^(x + y)] = {(2^4 ) 3]^2x}(3^-x)
[(2^7)^x][2^(x + y)][3^(x + y)] = (2^8x)(3^2x)(3^-x)
[2^(8x + y)][3^(x + y)] = (2^8x)[3^(2x - x)]
(2^8x)( 2^y)(3^x)(3^y) = (2^8x)(3^x)
( 2^y)(3^y) = 1
(2 × 3)^y = 1
6^y = 1
y = 0
Since y = 0 and x is not equal to zero (as stated in the problem stem), this information is sufficient to conclude that x^y = x^0 = 1.
The correct answer is B.
(1) x = 2
(2) (128 ^ x)[6 ^ (x + y)] = (48 ^ 2x)(3 ^ -x)
Official Answer and Explanation to the above problem
One of the most effective ways to begin solving problems involving exponential equations is to break down bases of the exponents into prime factors and combine exponents with the same base. Following this approach, be sure to simplify each statement as much as possible before arriving at the conclusion, since difficult problems with exponents often result in unobvious outcomes.
(1) INSUFFICIENT: While this statement gives us the value of x, we know nothing about y and cannot determine the value of x^y.
(2) SUFFICIENT: (128^x)[6^(x + y)] = (48^2x)(3^-x)
[(2^7)^x][(2 × 3)^(x + y)] = {(2^4 ) 3]^2x}(3^-x)
[(2^7)^x][2^(x + y)][3^(x + y)] = (2^8x)(3^2x)(3^-x)
[2^(8x + y)][3^(x + y)] = (2^8x)[3^(2x - x)]
(2^8x)( 2^y)(3^x)(3^y) = (2^8x)(3^x)
( 2^y)(3^y) = 1
(2 × 3)^y = 1
6^y = 1
y = 0
Since y = 0 and x is not equal to zero (as stated in the problem stem), this information is sufficient to conclude that x^y = x^0 = 1.
The correct answer is B.
Give that n is an integer, is n-1 divisible by 3?
1. n^2 +n is not divisible by 3
2. 3n+5>=k+8, where k is a positive multiple of 3
Sttement (1) Sufficient -- It can be simplified as n(n+1) is not divisible by 3
i.e to satisfy the above condition n can take a value immediately following 3 or multiples of 3.
Such as n = 4, 7, 10 .....
Thus case n-1 is always divisible by 3.
Statement (2) Insufficient -- 3n >= k + 3
This implies n >= (k+3)/3
the possible values of k includes 3,6,9,12 .....
So this can be either multiple of 3 or any other number..
1. n^2 +n is not divisible by 3
2. 3n+5>=k+8, where k is a positive multiple of 3
Sttement (1) Sufficient -- It can be simplified as n(n+1) is not divisible by 3
i.e to satisfy the above condition n can take a value immediately following 3 or multiples of 3.
Such as n = 4, 7, 10 .....
Thus case n-1 is always divisible by 3.
Statement (2) Insufficient -- 3n >= k + 3
This implies n >= (k+3)/3
the possible values of k includes 3,6,9,12 .....
So this can be either multiple of 3 or any other number..
If set S = {7, y, 12, 8, x, 9}, is x + y less than 18?
(1) The range of set S is less than 9.
(2) The average of x and y is less than the average of set S.
(1) INSUFFICIENT: Statement (1) tells us that the range of S is less than 9. The range of a set is the positive difference between the smallest term and the largest term of the set. In this case, knowing that the range of set S is less than 9, we can answer only MAYBE to the question "Is (x + y) <>
Consider the following two examples:Let x = 7 and y = 7. The range of S is less than 9 and x + y <>
Let x = 10 and y = 10. The range of S is less than 9 and x + y > 18, so we conclude NO.
Because this statement does not allow us to answer definitively Yes or No, it is insufficient.
(2) SUFFICIENT: Statement (2) tells us that the average of x and y is less than the average of the set S. Writing this as an inequality:
(x + y)/2 < (7 + 8 + 9 + 12 + x + y)/6 (x + y)/2 < (36 + x + y)/6 3(x + y) <>
2(x + y) <>
x + y <>
Therefore, statement (2) is SUFFICIENT to determine whether x + y <>
(1) The range of set S is less than 9.
(2) The average of x and y is less than the average of set S.
(1) INSUFFICIENT: Statement (1) tells us that the range of S is less than 9. The range of a set is the positive difference between the smallest term and the largest term of the set. In this case, knowing that the range of set S is less than 9, we can answer only MAYBE to the question "Is (x + y) <>
Consider the following two examples:Let x = 7 and y = 7. The range of S is less than 9 and x + y <>
Let x = 10 and y = 10. The range of S is less than 9 and x + y > 18, so we conclude NO.
Because this statement does not allow us to answer definitively Yes or No, it is insufficient.
(2) SUFFICIENT: Statement (2) tells us that the average of x and y is less than the average of the set S. Writing this as an inequality:
(x + y)/2 < (7 + 8 + 9 + 12 + x + y)/6 (x + y)/2 < (36 + x + y)/6 3(x + y) <>
2(x + y) <>
x + y <>
Therefore, statement (2) is SUFFICIENT to determine whether x + y <>
A building has two types of apartments, big and small. 65 percent of the apartments are small. The number of occupied big apartments is twice the number of unoccupied small apartments. What percent of the apartments in the building are occupied ?
(1) The number of occupied big apartments is six times the number of unoccupied big apartments.
(2) The building has a total of 160 apartments
Statement (1 ) -- Sufficient.
Percentage occupied = (big & occupied + small & occupied) / total (say, n)
(35/100)n * (6/7) = big & occupied
(small ) - (small & unoccupied) = (small & occupied)
(65/100)n - [(1/2) big & occupied] = small & occupied
(65/100)n - [(1/2) (35/100)n * (6/7)] = small & occupied
Total = [(35/100)n * (6/7) + (65/100)n - (35/100)n * (3/7) ] / n
(multiplied (6/7) by (1/2) here ).
n is removed from the numerator and divisor
(35/100) * (6/7) + (65/100) - (35/100) * (3/7) --- This is approx equal to 0.8
so whatever n is, it is the answer.
Statement (2) -- Insufficient -- 160 total apartments does not give any information regarding any number of occupied apartments -- there could be 2 big and 1 small, or 40 big and 20 small.
(1) The number of occupied big apartments is six times the number of unoccupied big apartments.
(2) The building has a total of 160 apartments
Statement (1 ) -- Sufficient.
Percentage occupied = (big & occupied + small & occupied) / total (say, n)
(35/100)n * (6/7) = big & occupied
(small ) - (small & unoccupied) = (small & occupied)
(65/100)n - [(1/2) big & occupied] = small & occupied
(65/100)n - [(1/2) (35/100)n * (6/7)] = small & occupied
Total = [(35/100)n * (6/7) + (65/100)n - (35/100)n * (3/7) ] / n
(multiplied (6/7) by (1/2) here ).
n is removed from the numerator and divisor
(35/100) * (6/7) + (65/100) - (35/100) * (3/7) --- This is approx equal to 0.8
so whatever n is, it is the answer.
Statement (2) -- Insufficient -- 160 total apartments does not give any information regarding any number of occupied apartments -- there could be 2 big and 1 small, or 40 big and 20 small.
Find an integer S.
(1) S=3
(2) S^3=3^S
Statement (1) - insufficient ---- S=3 or –3
Statement (2) - sufficient ---- S = 3
X^Y = Y^X only if X and Y are positive integers of equal value
=> X and Y have to be positive, otherwise one side will give a negative value when X or Y is odd while other side will give a decimal value
Given that Y is equal to 3, X has to be equal to 3.
Hence B.
(1) S=3
(2) S^3=3^S
Statement (1) - insufficient ---- S=3 or –3
Statement (2) - sufficient ---- S = 3
X^Y = Y^X only if X and Y are positive integers of equal value
=> X and Y have to be positive, otherwise one side will give a negative value when X or Y is odd while other side will give a decimal value
Given that Y is equal to 3, X has to be equal to 3.
Hence B.
When positive integer n is divided by 3, the remainder is 2; and when positive integer t is divided by 5, the remainder is 3. What is the remainder when the product nt is divided by 15 ?
(1) n- 2 is divisible by 5.
(2) t is divisible by 3.
Statement (1) alone - no sufficient information about t
Statement (2)alone - no sufficient information about n
From the ques --- n = a*3 + 2
t = b*5 + 3
Using Statement (1) --- n = p*5 + 2
Thus n = 15*k + 2
Using Statement (2) --- t = q*3 + 3
Thus t = 15*j + 3
nt = 15*k(15*j + 3) + 30*j + 6
=> always leaves a remainder of 6 when divided by 15.
(1) n- 2 is divisible by 5.
(2) t is divisible by 3.
Statement (1) alone - no sufficient information about t
Statement (2)alone - no sufficient information about n
From the ques --- n = a*3 + 2
t = b*5 + 3
Using Statement (1) --- n = p*5 + 2
Thus n = 15*k + 2
Using Statement (2) --- t = q*3 + 3
Thus t = 15*j + 3
nt = 15*k(15*j + 3) + 30*j + 6
=> always leaves a remainder of 6 when divided by 15.
a, b, c all are positive, is a/b > (a+c)/(b+c) ?
1. a > c
2. a > b
statement (1). - insufficient as we do not know the relation between a and b
statement (2). - a > b gives answer irrespective of whether ( c <> b) that, a / b > (a + c) / (b + c)
a/b > [a+c]/[b+c]
= a(b+c) > b(a+c)
= ab+ac > ab+bc
= ac > bc
=a > b
Hence B
1. a > c
2. a > b
statement (1). - insufficient as we do not know the relation between a and b
statement (2). - a > b gives answer irrespective of whether ( c <> b) that, a / b > (a + c) / (b + c)
a/b > [a+c]/[b+c]
= a(b+c) > b(a+c)
= ab+ac > ab+bc
= ac > bc
=a > b
Hence B
Is the integer n odd?
(1) n is divisible by 3.
(2) 2n is divisible by twice as many positive integers as n.
Statement (1) - insufficient as n can be both even and odd.
Statement (2) - sufficient - Here n will have a unique prime factorisation as a product of
p1 ^ q1 * p2 ^ q2 * ....... pk ^qk
where p1 is less than p2 , p2 is less than p3 and so on.
Now p1 is either equal to 2 or it is an odd integer greater than 2.
Thus if n is odd then 2n has a unique prime factorisation of
2 * p1^q1 * p2^q2 * .... * pk ^ qk
but if n is even then 2n can be factorised as
p1^(q1+1) * p2^q2 * ... where p1=2, q1 >= 1
Hence , now the total number of factors is based on choice of prime powers for each prime in term.
Thus if n is odd, then 2n must have twice as many factors as n.
Alternatively in simpler language
If n is even --- 2n will not have twice as many divisors as n
e.g . For n=2; 2n or 4 has 1,2,4 as divisors ,
For n = 4 divisors are 1,2,4; 2n=8 has 1,2,4,8
But if n is odd
then if n=3 divisors are 1,3 and 2n=6 divisors are 1,2,3,6
n=5 divisors are 1,5 and 2n=10 divisors are 1,2,5,10
Thus if n is odd the 2nd condition is satisfied. Hence ans is B
(1) n is divisible by 3.
(2) 2n is divisible by twice as many positive integers as n.
Statement (1) - insufficient as n can be both even and odd.
Statement (2) - sufficient - Here n will have a unique prime factorisation as a product of
p1 ^ q1 * p2 ^ q2 * ....... pk ^qk
where p1 is less than p2 , p2 is less than p3 and so on.
Now p1 is either equal to 2 or it is an odd integer greater than 2.
Thus if n is odd then 2n has a unique prime factorisation of
2 * p1^q1 * p2^q2 * .... * pk ^ qk
but if n is even then 2n can be factorised as
p1^(q1+1) * p2^q2 * ... where p1=2, q1 >= 1
Hence , now the total number of factors is based on choice of prime powers for each prime in term.
Thus if n is odd, then 2n must have twice as many factors as n.
Alternatively in simpler language
If n is even --- 2n will not have twice as many divisors as n
e.g . For n=2; 2n or 4 has 1,2,4 as divisors ,
For n = 4 divisors are 1,2,4; 2n=8 has 1,2,4,8
But if n is odd
then if n=3 divisors are 1,3 and 2n=6 divisors are 1,2,3,6
n=5 divisors are 1,5 and 2n=10 divisors are 1,2,5,10
Thus if n is odd the 2nd condition is satisfied. Hence ans is B
If j and k are positive integers where k > j, what is the value of the remainder
If j and k are positive integers where k > j, what is the value of the remainder when k is divided by j?
The reason is as follows:
k/j = (jm + 5)/j = m + 5/j
If 5 < j, then the remainder will be 5.
But when 5 > j, the remainder can't be 5.
The examples of gokul showed this (Thank you, gokul :D)
So, (1) is insuff.
(1) and (2) is suff.
(1) There exists a positive integer m such that k = jm + 5.
(2) j > 5
The reason is as follows:
k/j = (jm + 5)/j = m + 5/j
If 5 < j, then the remainder will be 5.
But when 5 > j, the remainder can't be 5.
The examples of gokul showed this (Thank you, gokul :D)
So, (1) is insuff.
(1) and (2) is suff.
If a and b are integers and a is not equal to b, is ab > 0?
If a and b are integers and a is not equal to b, is ab > 0?
Is
? Question basically asks are 
and 
both positive or both negative.
(1)
--> both and could be positive as well as could be positive and negative. Not sufficient.
(2) a^b is a non-zero integer --> if
and 
answer would be NO but if and 
answer would be YES. Not sufficient.
(1)+(2) Example from (2) is still valid. Not sufficient.
Answer: E.
(1) a^b > 0
(2) a^b is a non-zero integer
Is
? Question basically asks are and both positive or both negative.(1)
--> both and could be positive as well as could be positive and negative. Not sufficient.(2) a^b is a non-zero integer --> if
and answer would be NO but if and answer would be YES. Not sufficient.(1)+(2) Example from (2) is still valid. Not sufficient.
Answer: E.
Thursday, September 09, 2010
If the probability of rain on any given day in Chicago during the summer is 50%, independent of what happens on any
other day, what is the probability of having exactly 3 rainy days from July 4 through July 8, inclusive?
(A) 1/32
(B) 2/25
(C) 5/16
(D) 8/25
(E) 3/4
Answer: Correct ans is C. For OE click on the link below.
other day, what is the probability of having exactly 3 rainy days from July 4 through July 8, inclusive?
(A) 1/32
(B) 2/25
(C) 5/16
(D) 8/25
(E) 3/4
Answer: Correct ans is C. For OE click on the link below.
A number is selected at random from first 30 natural numbers. What is the probability that the number is a multiple
of either 3 or 13?
(A) 17/30
(B) 2/5
(C) 7/15
(D) 4/15
(E) 11/30
The first 30 natural nos are 1,2,3.....28,29,30.
There are 10 multiples of 3 in the above range and there are 2 multiples of 13 in the same range.
Hence there are total 12 i.e (10+2) nos which can be either a multiple of 3 or 13.
Total numbers = 30.
So the probability is 12/30 = 2/5.
of either 3 or 13?
(A) 17/30
(B) 2/5
(C) 7/15
(D) 4/15
(E) 11/30
The first 30 natural nos are 1,2,3.....28,29,30.
There are 10 multiples of 3 in the above range and there are 2 multiples of 13 in the same range.
Hence there are total 12 i.e (10+2) nos which can be either a multiple of 3 or 13.
Total numbers = 30.
So the probability is 12/30 = 2/5.
If n is an integer from 1 to 96, what is the probability for n*(n+1)*(n+2) being divisible by 8?
A) 25%
B) 50%
C) 62.5%
D) 72.5%
E) 75%
n is even - anytime n is even, it is divisible by 8
total nos using sequence theorm 96 = 2 + (#-1) 2,
hence # = 48
n is odd - again 48 no
but in 1-8, only one combination is divisible by 8, when n is 7,15, 22... hence 12 cases
probalility = possible outcomes / total outcomes = 48+12 / 96 = 60/96 = .625 = 62.5%
OR
we need to check for what values of n, n(n+1)(n+2) is divisible 8....
n(n+2) is divisible by 8 for all values of even numbers and there are 48 even nos in 1 to 96....
now the remaining part (n+1) is divisible by 8 for 12 odd numbers....such as 7, 15, 23, 31, 39.... to find this u
can divide 96 by 8....
so totally 48 + 12 = 60 numbers are there between 1 to 96 for which n(n+1)(n+2) is divisible 8....
now when calculate the % it would be ( 60 / 96 ) * 100 = 62.5 %....
A) 25%
B) 50%
C) 62.5%
D) 72.5%
E) 75%
n is even - anytime n is even, it is divisible by 8
total nos using sequence theorm 96 = 2 + (#-1) 2,
hence # = 48
n is odd - again 48 no
but in 1-8, only one combination is divisible by 8, when n is 7,15, 22... hence 12 cases
probalility = possible outcomes / total outcomes = 48+12 / 96 = 60/96 = .625 = 62.5%
OR
we need to check for what values of n, n(n+1)(n+2) is divisible 8....
n(n+2) is divisible by 8 for all values of even numbers and there are 48 even nos in 1 to 96....
now the remaining part (n+1) is divisible by 8 for 12 odd numbers....such as 7, 15, 23, 31, 39.... to find this u
can divide 96 by 8....
so totally 48 + 12 = 60 numbers are there between 1 to 96 for which n(n+1)(n+2) is divisible 8....
now when calculate the % it would be ( 60 / 96 ) * 100 = 62.5 %....
During a behavioral experiment in a psychology class, each student is asked to compute his or her lucky number by
raising 7 to the power of the student's favorite day of the week (numbered 1 through 7 for Monday through Sunday
respectively), multiplying the result by 3, and adding this to the doubled age of the student in years, rounded to
the nearest year. If a class consists of 28 students, what is the probability that the median lucky number in the
class will be a non-integer?
(A) 0%
(B) 10%
(C) 20%
(D) 30%
(E) 40%
Since any power of 7 is odd, the product of this power and 3 will always be odd. Adding this odd number to the
doubled age of the student (an even number, since it is the product of 2 and some integer) will always yield an odd
integer. Therefore, all lucky numbers in the class will be odd.
The results of the experiment will yield a set of 28 odd integers, whose median will be the average of the 14th and
15th greatest integers in the set. Since both of these integers will be odd, their sum will always be even and
their average will always be an integer. Therefore, the probability that the median lucky number will be a non-
integer is 0%.
raising 7 to the power of the student's favorite day of the week (numbered 1 through 7 for Monday through Sunday
respectively), multiplying the result by 3, and adding this to the doubled age of the student in years, rounded to
the nearest year. If a class consists of 28 students, what is the probability that the median lucky number in the
class will be a non-integer?
(A) 0%
(B) 10%
(C) 20%
(D) 30%
(E) 40%
Since any power of 7 is odd, the product of this power and 3 will always be odd. Adding this odd number to the
doubled age of the student (an even number, since it is the product of 2 and some integer) will always yield an odd
integer. Therefore, all lucky numbers in the class will be odd.
The results of the experiment will yield a set of 28 odd integers, whose median will be the average of the 14th and
15th greatest integers in the set. Since both of these integers will be odd, their sum will always be even and
their average will always be an integer. Therefore, the probability that the median lucky number will be a non-
integer is 0%.
According to the directions on a can of frozen orange juice concentrate, 1 can of concentrate is to be mixed with 3 cans of water to make orange juice. How many 12-ounce cans of the concentrate are required to prepare 200 6-ounce servings of orange juice?
A. 25
B. 34
C. 60
D. 67
E. 100
4 cans of orange juice need 1 can of concentrate.
200 cans 6-ounce of orange juice need 50 cans 6-ounce concentrate.
so 12-ounce concentrate = 50/2 = 25 cans
OR
Let number of 12 ounce cans required = x
number of 12 ounce cans of water required = 3x
Total required servings of Orange juice = 200*6 ounce
Hence
x(12) + 3x(12) = 200*6
48x= 200*6
x=25
Hence A
A. 25
B. 34
C. 60
D. 67
E. 100
4 cans of orange juice need 1 can of concentrate.
200 cans 6-ounce of orange juice need 50 cans 6-ounce concentrate.
so 12-ounce concentrate = 50/2 = 25 cans
OR
Let number of 12 ounce cans required = x
number of 12 ounce cans of water required = 3x
Total required servings of Orange juice = 200*6 ounce
Hence
x(12) + 3x(12) = 200*6
48x= 200*6
x=25
Hence A
E is a collection of four odd integers and the greatest difference between any two integers in E is 4. The standard
deviation of E must be one of how many numbers?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
Suppose the integers are 1, 3 and 5. Therefore the four integers can be:
1, 5, 5, 5
1, 3, 5, 5
1, 3, 3, 5
1, 1, 5, 5
1, 1, 1, 5
1, 1, 3, 5
Here two pairs have the same standard deviation. thus in all we have four different standard deviations. Hence B
deviation of E must be one of how many numbers?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
Suppose the integers are 1, 3 and 5. Therefore the four integers can be:
1, 5, 5, 5
1, 3, 5, 5
1, 3, 3, 5
1, 1, 5, 5
1, 1, 1, 5
1, 1, 3, 5
Here two pairs have the same standard deviation. thus in all we have four different standard deviations. Hence B
If eleven consecutive integers are listed from least to greatest, what is the average (arithmetic mean) of the
eleven integers?
(1) The average of the first nine integers is 7.
(2) The average of the last nine integers is 9.
Let the numbers be a, b, c, d, e, f, g, h , i, j, k
i) For odd number of consecutive integers median = mean
ii)We also know that the median is the "middle" number in a group (when arranged in ascending or descending order)
consisting of an odd number of numbers
We have to find f
From statement (1): It is given that average of first nine numbers = 7
Hence this implies e = 7 ..since it is given numbers are consecutive hence f = 8
Thus sufficient
From statement (2): It is given that average of last nine numbers = 9
Hence this implies g = 9..since it is given numbers are consecutive hence f = 8
Thus sufficient
eleven integers?
(1) The average of the first nine integers is 7.
(2) The average of the last nine integers is 9.
Let the numbers be a, b, c, d, e, f, g, h , i, j, k
i) For odd number of consecutive integers median = mean
ii)We also know that the median is the "middle" number in a group (when arranged in ascending or descending order)
consisting of an odd number of numbers
We have to find f
From statement (1): It is given that average of first nine numbers = 7
Hence this implies e = 7 ..since it is given numbers are consecutive hence f = 8
Thus sufficient
From statement (2): It is given that average of last nine numbers = 9
Hence this implies g = 9..since it is given numbers are consecutive hence f = 8
Thus sufficient
List K consists of 12 consecutive integers, if -4 is the least integer in list K, what is the range of the positive
integers in the list K?
A. 5
B. 6
C. 7
D. 11
E. 12
The least number in the list is -4, thus the list is: -4,-3,-2,-1, 0, 1, 2, 3, 4, 5, 6, 7
Positive integers in the above list: 1, 2, 3, 4, 5, 6, 7
Therefore the range of the positive integers is 7-1 = 6
integers in the list K?
A. 5
B. 6
C. 7
D. 11
E. 12
The least number in the list is -4, thus the list is: -4,-3,-2,-1, 0, 1, 2, 3, 4, 5, 6, 7
Positive integers in the above list: 1, 2, 3, 4, 5, 6, 7
Therefore the range of the positive integers is 7-1 = 6
A certain characteristic in a large population has a distribution that is symmetric about the mean m.If 68 percent
of the distribution lies within one Standard Deviation d of the mean, what percent of the distribution is less than
m+d?
A. 16%
B. 32%
C. 48%
D. 84%
E. 92%
In a normal bell curved distribution, 50% are below the mean and 50% are over it
If 68% are distributed within 1 S.D of the mean then this implies that 34% are 1 S.D above the mean and 34% are 1
S.D below the mean i.e 34% between m and m+d and 34% between m-d and m
The distribution is symmetric about m also => 32/2 = 16% between 0 and m-d and 16% m+d and above.
Hence total that is less than m+d = 100-16 = 84%
OR
Distribution is symmetric around mean => 68/2 = 34% =>(Mean-S.D, Mean) = (Mean, Mean+S.D] = 34% . Thus below Mean
+S.D = 50+34 = 84%
NOTE: For a normal bell-curve distribution, the percentage is approx 34% between the mean and 1 S.D. the percentage
is approximately 13.6% between 1 SD and 2 SD, the percentage is approximately 2% between 2 S.D and on...
of the distribution lies within one Standard Deviation d of the mean, what percent of the distribution is less than
m+d?
A. 16%
B. 32%
C. 48%
D. 84%
E. 92%
In a normal bell curved distribution, 50% are below the mean and 50% are over it
If 68% are distributed within 1 S.D of the mean then this implies that 34% are 1 S.D above the mean and 34% are 1
S.D below the mean i.e 34% between m and m+d and 34% between m-d and m
The distribution is symmetric about m also => 32/2 = 16% between 0 and m-d and 16% m+d and above.
Hence total that is less than m+d = 100-16 = 84%
OR
Distribution is symmetric around mean => 68/2 = 34% =>(Mean-S.D, Mean) = (Mean, Mean+S.D] = 34% . Thus below Mean
+S.D = 50+34 = 84%
NOTE: For a normal bell-curve distribution, the percentage is approx 34% between the mean and 1 S.D. the percentage
is approximately 13.6% between 1 SD and 2 SD, the percentage is approximately 2% between 2 S.D and on...
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