# Welcome to the LEAP Q&A Forum

## GMAT - Data Sufficiency - Statistics

LEAP Administrator 9 months ago

### A certain bookcase has 2 shelves of books. On the upper shelf, the book with the greatest number of pages has 400 pages (more)

A certain bookcase has 2 shelves of books. On the upper shelf, the book with the greatest number of pages has 400 pages. On the lower shelf, the book with the least number of pages has 475 pages. What is the median number of pages for all of the books on the 2 shelves?

(1) There are 25 books on the upper shelf.
(2) There are 24 books on the lower shelf.

(1) There are 25 books on the upper shelf. - No information about the lower shelf

The statement is insufficient

(2) There are 24 books on the lower shelf.  No information about the upper shelf

The statement is insufficient

Combining both statements

Total number of books = 25 + 24 = 49

Median = (49+1)/2 th book = 25th book if we sort them in ascending order

If arrange both the shelves in ascending order the 25th book will be the book with most pages in the upper shelf = 400 pages

Sufficient

LEAP Administrator 9 months ago

### If the average (arithmetic mean) of six numbers is 75, how many of the numbers are equal to 75 ? (more)

If the average (arithmetic mean) of six numbers is 75, how many of the numbers are equal to 75 ?

(1) None of the six numbers is less than 75.
(2) None of the six numbers is greater than 75.

Assume the numbers are A, B, C, D, E and F

Statement 1:

None of the six numbers is less than 75.

A>=75, B>=75, C>=75, D>=75, E>=75, F>=75 .... (i)

A + B + C + D + E + F >= 75*6

Average of 6 numbers = (A + B + C + D + E + F)/75 >= 75

but the question says that the average is equal to 75. Only way the equality holds if all the inequalities in (i) are equality

All 6 numbers are equal to 75

Sufficient

Similarly B can be shown sufficient

LEAP Administrator 10 months ago

### At a certain company, a test was given to a group of men and women (more)

At a certain company, a test was given to a group of men and women seeking for promotions. If the average (artihmetic mean) score for the group was 80, was the average score for the women more than 85?

(1) The average score for the men was less than 75.
(2) The group consisted of more men than women.

let the number of men be m and number of women be w, and average score for men be M and average score for woman be W

so,

Total score (T) = m*M + w*W ... (1)

given average score of the group was 80

T = 80*(w + m)  ... (2)

from 1 and 2

m*M + w*W = 80 (w + m)... (3)

Statement 1:

Average score for men was less than 75.

i.e. M<75

if we make M as subject of equation 3

$M=\frac { 80(w\quad +\quad m)\quad -\quad wW }{ m } \le 75$ ... (4)

we need information on w and m to know about W.

Information insufficient.

Statement 2:

The group consisted of more men than women

i.e. m > w

we need information on M to be able to say something. Insufficient.

Taking both statements together

we can write the expression 4 as

$M=\frac { 80(w\quad +\quad m)\quad -\quad wW }{ m } \le 75\\ 80(\frac { w }{ m } )\quad +\quad 80\quad -\quad W(\frac { w }{ m } )\le 75\\ Simplifying,\quad (w/m)(80\quad +\quad W)\quad \le -5\quad ...\quad (5)\\ from\quad statement\quad 2\quad m>w\\ thus,\quad m/w\quad >\quad 1\\ -5*m/w\quad <\quad -5\\ multiplying\quad eq.\quad 5\quad by\quad m/w\quad (greater\quad than\quad 1)\\$

$\\ (80\quad -\quad W)\quad \le \quad -5*m/w\quad <\quad -5\\ 80\quad -\quad W\quad <\quad -5\\ W\quad >\quad 85\\ \\$

hence both statements together are sufficient.

LEAP Administrator 10 months ago

### During a 6·day local trade show, the least number of people registered in a single day was 80. Was the average (arithmetic mean) number of people registered per day for the 6 days greater than 90 ? (more)

During a 6·day local trade show, the least number of people registered in a single day was 80. Was the average (arithmetic mean) number of people registered per day for the 6 days greater than 90 ?

Statement 1:

For the 4 days with the greatest number of people registered, the average (arithmetic mean) number registered per day was 100.

Statement 2:

For the 3 days with the smallest number of people registered, the average (arithmetic mean) number registered per day was 85.

Statement 1:

For the 4 days with the greatest number of people registered, the average (arithmetic mean) number registered per day was 100.

the average number of people registered for 4 days is 100

i.e. total number of people registered on those 4 days = 4*100 = 400

Minimum number of people registered = 80

for minimum possible average assume that the other two days has 80 registrations

mean = (400 + 80*2)/6 = 560/6 = 93.33 > 90

so the mean will be greater than 90. hence the statement is sufficient.