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DS Question 1 - Dec 13

If ab ≠ 0, does a=b?

(1) x^a = x^b
(2) x|a| = x|b|

Source: Manhattan GMAT | Difficulty: Medium

PS Question 1 - Dec 13

In how many different ways can a group of six students be equally distributed into three classes: physics, mathematics and history ?

(A) 6
(B) 15
(C) 20
(D) 45
(E) 90

Source: GMAT Club Tests | Difficulty: Medium

statement 1:
if x is 1, no matter what value a or b contains,
they may or may not be equal, not sufficient

statement 2:
essentially cuts to |a|=|b|
a=3
b=-3
mods of these are equal but a!=b
so not sufficient

combine(1)+(2):
we can have 1^(-2) = 1^(2)
which is same as 1^(3)=1^(99)

together also they are not sufficient to conclude if a=b
Option E


3 classes 6 students

each class can have 2 student each to be considered as equal distribution
each student is identified as one unique item and each class is unique as well

ways of selecting:-
6C2 * 4C2 * 2C2 = 15*6*1 = 90 ways .... option E ?

E.
1) X is 1 or 0, can take any galue of a and b
2) similar situation - x is 0 or 1 any value of A and B is okay.
3) we still cant conclude if x is not 0 or 1.


be careful about dividing by x - we don’t know if it is 0 or not.

aah! you are correct, thanks for pointing it out, I completely ignored it!!

it’s 20, so answer c. 6! / 3! * 3! = 20

the groups are unique here

oh yes, my bad. Then it’s 6! / (2!)^3 = 90 so answer E

DS Question 2 - Dec 13

Three consecutive integers are selected from the integers 1 to 50, inclusive. What is the sum of the remainders that result when each of the three integers is divided by x ?

(1) When the greatest of the consecutive integers is divided by x, the remainder is 0.

(2) When the least of the consecutive integers is divided by x, the remainder is 1.

Source: Manhattan GMAT | Difficulty: Medium

Should be C.
1) If x is 2, then we have a remainder of 1
if x is 1, then we have no remainders.
2) we can exclude x is 1, but we have many options.
3) combining the two,
the max number has R0, the min number has R1.
X must be 3. Therefore we can conclude what the sum of remainders is.

PS Question 2 - Dec 13

At a certain party hall, it is observed that in every party at least 75% of the people take coffee, at least 30% of the people take tea and between 20% to 50% people take cold drink. Each person in the party takes at least one of the beverages but no single person takes all three of them. What is the sum of the minimum possible percentage and maximum possible percentage of people who take both coffee and tea in a party?

A. 85
B. 95
C. 100
D. 105
E. 120

Source: GMATWhiz | Difficulty: Hard

Can anybody please help me out with this one

+1 I need help on this one

Let’s assume there are 100 people in the party.

Given:

Number of people who take coffee = at least 75
Number of people who take tea = at least 30
Number of people who take cold drinks = between 20 and 50

Minimum possible percentage of people who take both coffee and tea

To minimize the overlap, lets "assume" that no people drink coffee and tea (i.e. let’s assume that there is no overlap)

If that were the case, the number of people who drink coffee + number of people who drink tea = 105

The number exceeds the number of attendees (number of attendees are 100). Hence, the extra 5 are the minimum possible people who drink both tea and coffee.

Maximum possible percentage of people who take both coffee and tea

The maximum possible number of people who drink tea and coffee will be in a scenario when all people who drink tea also drink coffee.

Before we proceed to find this number, note that its also given that "no single person takes all three" (of the beverages), so we will first exclude the minimum number of people who can take cold drink.

From the information given its 20 (as between 20 and 50 people take cold drink)

Now, the remaining 80 people can drink both tea and coffee.

This is the maximum possible number.

Sum

Minimum = 5
Maximum = 80
Total = 80 + 5 = 85

Hope this makes sense !

DS Question 1 - Dec 14

For a certain city’s library, the average cost of purchasing each new book is $28. The library receives $15,000 from the city each year,the library also receives a bonus of $2,000 if the total number of items checked out over the course of the year exceeds 5,000.Did the library receive the bonus last year ?

(1) The library purchased an average of 50 new books each month last year and received enough money from the city to cover this cost.

(2) The lowest number of items checked out in one month was 459 .

Source: Official Guide | Difficulty: Medium

DS Question 2 - Dec 14

If n is an integer, is (n-1)(n+1) a multiple of 24?

1) n is odd.
2) n is not divisible by 3.

Source: Math Revolution | Difficulty: Hard

Should be D.

1) We can deduce that the total amount spent last year is (28*50*12)=16800. Therefore, the library spent over its 15000 income from the govt, and must have recieved the bonus to cover the extra 1800.
2) If the lowest number of items checked out in one month is 459, we can calculate the worst case scenario (all 12 months have 459 books checked out)
460*12=5520-12=5508.
So we can see, even if the library has the worst case scenario, it is still well over the 5,000 books required for the bonus.

both are suff, D.