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problem solving q's

by hareesh860 » Tue Jan 17, 2012 8:31 pm
1) There are 11 books on a shelf. How many ways are there to choose 4 of them so that no two of the chosen books stand next to each other?

(a) 70 (b) 165 (c) 495 (d) none of these


2) Two brothers, each aged between 10 and 90, "combined" their ages by writing them down one after the other to create a four digit number, and discovered this number to be the square of an integer. Nine years later they repeated this process (combining their ages in the same order) and found that the combination was again a square of another integer. What was the sum of their original ages?

(a) 37 (b) 55 (c) 63 (d) 71
Last edited by hareesh860 on Sun Jan 22, 2012 8:28 am, edited 2 times in total.
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by [email protected] » Tue Jan 17, 2012 9:01 pm
Q1.
we can consider the books that are chosen as 1, 2, 3 and 4. Now the place on the left of 2, 3, and 4 are reserved so that no two books will be adjacent. Hence only 8 ( 11 - 3 reserved places) are left. Out of these 4 are chosen. So number of ways = 8C4 = 70
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by [email protected] » Tue Jan 17, 2012 9:41 pm
Let their original ages be x and y.

Combining their original ages yields:
100x+y = n^2

Their ages in 9 years are x+9 and y+9 respectively.
So, combining their ages 9 years later yields:
100(x+9)+(y+9) = (n+9)^2
100x+y+909 = n^2+18n+81

Substitute 100x+y = n^2, from the original combining of ages, and solve for n:
n^2+909 = n^2+18n+81
909 = 18n+81
18n = 828
n = 46

So, 100x+y= 46^2 = 2116
Thus, x = 21, y = 16, and x+y = 37
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by GMATGuruNY » Tue Jan 17, 2012 10:34 pm
hareesh860 wrote:1) There are 11 books on a shelf. How many ways are there to choose 4 of them so that no two of the chosen books stand next to each other?

(a) 70 (b) 165 (c) 495 (d) none of these
This question is no different from the following:

How many ways can 11 books be PLACED on a shelf so that the last 4 books placed are all separated from each other?

The first 7 books can be placed anywhere on the shelf.
The last 4 books can be placed to the left or right of any of the first 7 books, yielding 8 options for the last 4 books.
Number of combinations of 4 that can be formed from 8 choices = 8C4 = 70.

The correct answer is A.
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by GMATGuruNY » Tue Jan 17, 2012 11:43 pm
hareesh860 wrote: Two brothers, each aged between 10 and 90, "combined" their ages by writing them down one after the other to create a four digit number, and discovered this number to be the square of an integer. Nine years later they repeated this process (combining their ages in the same order) and found that the combination was again a square of another integer. What was the sum of their original ages?

(a) 37 (b) 55 (c) 63 (d) 71
Let the greater perfect square formed by the brothers' ages = x².
Let the smaller perfect square formed by the brothers' ages = y².

The age of one of the brothers forms the LAST two digits of y².
9 years later, this brother's age forms the last two digits of x².
This implies a difference of 9 between x² and y².
To illustrate, XX29 - XX20 = 9.

The age of the other brother forms the FIRST two digits of y².
9 years later, this brother's age forms the first two digits of x².
This implies a difference of 900 between x² and y².
To illustrate, 29XX - 20XX = 900.

Thus, the difference between x² and y² = 9+900 = 909:
x² - y² - 909.
(x+y)(x-y) = 909.

909 can be divided into the following factor pairs: 101*9, 303*3, and 909*1.
Since x and y are two-digit integers, the only viable option is (x+y)(x-y) = 101*9, implying that x+y = 101 and x-y = 9.
The other options -- x+y = 303 and x+y = 909 -- will not allow x and y to be two-digit integers.
Adding together x+y = 101 and x-y = 9, we get:
(x+y) + (x-y) = 101 + 9
2x = 110
x = 55.

Since x+y = 101:
55+y = 101
y = 46.

Thus, y² = 46² = 2116.

Thus, the two ages are 21 and 16, and their sum = 21+16 = 37.

The correct answer is A.
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by hareesh860 » Wed Jan 18, 2012 12:37 am
solutions

1) Its same as asking we have 7 books in a shelf and we have to choose place
for 4 more books so that no two of these 4 will be together.
Answer 8C4 = 70.


2) Let x, y be the ages.
Then 100x + y = a^2,
also 100(x + 9) + b + 9 = b^2,
subtracting we get b^2 - a^2 = 909 = 101*9 = 303*3 = 909*1
thus, (b,a) are (55, 46) (156, 150), (455, 454)
rejecting the last 2 pairs as a <= 99, we get (b, a)
as (55, 46) a^2 = 2116,
thus the sum of the ages is 21 + 16 = 37
Hence, choice (a) is the right answer.
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by hareesh860 » Wed Jan 18, 2012 1:24 am
3) every town-station and for 3 minutes in every village-station started from A with a speed
of 60km/h towards B and at the same time a train T2 with a speed of 80km/h which does not
stop in any intermediate station started from B towards A. They met at C which is 560 km
away from B. If the number of town-stations between A and C is less than the number of
village-stations, then at least how many stations - town or village - are there
between A and C? Assume T1 stops only at town or village-stations.

(A) 12 (B) 13 (C) 14 (D) 16 (E) 20


4) The students lockers at a high school are numbered consecutively beginning with locker number 1. The plastic digits used to number the lockers cost 2 cents apiece. thus, it costs 2 cents to label locker number 9 and four cents to label locker number 10. If it costs $137.94 to label all the lockers, how many lockers are there at the school?

(A)2003 (B)1999 (C)2000 (D)2005 (e)2001
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by hareesh860 » Wed Jan 18, 2012 5:31 am
solutions for 3rd & 4th qs

3) As the train T2 covered 560km, both trains travelled for 7 hours, but T1 travelled only
360 kms.
=> Halting time = 60 minutes
Let the number of town-stations be x and number of village stations be y, x < y
=> 5x + 3y = 60 => x=3, y=15, x=6, y = 10, or x=0, y = 20
=> least is 16


4)The locker labeling require 137.94/.02 = 6897 digits. Lockers 1 through 9 require 9 digits, lockers 10 through 99 require 2x90 = 180 digits, and lockers 100 to 999 require 3x900 = 2700 digits. Hence the remaining lockers require 6897-2700-180-10 = 4008 digits. So there muxt be 4008/4 = 1002 lockers, each using 4 digits. In all there are 1002 + 999 = 2001 lockers.
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by hareesh860 » Wed Jan 18, 2012 8:27 am
5) At the Scholarly Text Printing Company,each of n printing presses can produce on the average t books every m minutes. If all presses work without interruption,how many hours will be required to produce a run of 10,000 books?

(A)10,000(60)mn/t
(B)10,000(60)tm/n
(C)10,000mn/60t
(D)10,000m/60nt
(E)10,000/60mnt


6) In a certain community, property is assessed at 60% of its appraised value and taxed annually at the rate of $4.00 per $100 of assessed value. If a taxpayer is assessed $240 per quarter in property taxes, what is the appraised value of the property?

(A) $6,000
(B) $22,500
(C) $24,000
(D) $40,000
(E) $60,000
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by hareesh860 » Wed Jan 18, 2012 4:49 pm
solutions 5&6

5)The correct answer is (D). This question asks you to express a certain relationship in algebraic notation. Each machine operates at the rate of t books per m minutes or t/m . But there are n such machines, so the overall rate of operation will be n times t/m which is nt/m . To find the time it will take to produce 10,000 books, we divide that number by the rate of operation:

10,000/(nt/m) = 10,000m/nt
Finally, we divide that by 60 since there are 60 minutes in every hour: 10,000m/60nt.


6)The correct answer is (C). First of all, (D) and (E) are impossible on logical grounds since they are greater than 1,and the proportion of something that has a characteristic cannot be greater than 1. That would be like saying, "Five out of three doctors recommend. . . ." We need the total of upper and middle management
with production line experience. The ratio 4:3 tells us that the total number of middle- and upper-Management personnel in the company can be divided into 7 equal parts, with 4 of them in upper
management and 3 in middle management.Of the 4 parts in upper management, 75%, or 3/4, have experience on the production line. Three-quarters of 4 parts amounts to 3 parts ( 3/7 of the total).
You are not told how many of the middlemanagement personnel have production line experience, but the key word "greatest" tells you that you should consider all of the middle-management personnel as
having production line experience. This means that there are 3 parts from the upper- management personnel who have production line experience and that there are 3 more parts from the middle-management personnel that are assumed to have production line experience, for a total of 6 parts out of 7, or 6/7 .
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by hareesh860 » Wed Jan 18, 2012 5:00 pm
7) if 5 geometric means are inserted between 8 & 5832, the fifth term in the g.m series is?

a)648 b)832 c)1168 d)1944 e)none


8)as the number of sides of a polygon increase from 3 to n, the sum of the exterior angles formed by extending each side in succession :

a)increase b)decrease c)remains constant d)can not be predicted e) n-3 straight angles
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by shankar.ashwin » Wed Jan 18, 2012 8:25 pm
@hareesh860

As a rule of the forum, please stick to one question per post.. Not only do multiple questions/post receive less response, it is also tiresome to see continuity in the post. Also, please stick to questions pertaining to the GMAT.. notice few of the questions posted have only 4 answer choices..

Thanks :)
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by ArunangsuSahu » Thu Jan 19, 2012 7:43 am
Q1:

We have to take out 4 books out of 11 and need place them in (6+2)=* places in 8C4 ways=70 ways

Q2:

Let their original ages be x and y.

Combining their original ages yields:
100x+y = n^2 (example: 2116=21*100+16)

Their ages in 9 years are x+9 and y+9 respectively.
So, combining their ages 9 years later yields:
100(x+9)+(y+9) = (n+9)^2
100x+y+909 = n^2+18n+81

Substitute 100x+y = n^2, from the original combining of ages, and solve for n:
n^2+909 = n^2+18n+81
909 = 18n+81
18n = 828
n = 46

So, 100x+y= 46^2 = 2116
Thus, x = 21, y = 16, and x+y = 37
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by hareesh860 » Thu Jan 19, 2012 7:12 pm
sol

7) a
denote the terms in gp by a1=8, a2=8r,..., a6= 8r^5 = 5832

=> r^5 = 729 => r= 3 => a5= 8r^4 = 648.

8) c
sum of the exterior angles of any (convex) polygon is constant
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by hareesh860 » Sun Jan 22, 2012 2:36 am
9)For any integer k from 1 to 10, inclusive, the kth of a certain sequence is given
by [(-1)^(k+1)]*(1 / 2^k). If T is the sum of the first 10 terms of the sequence,
then T is:
A. greater than 2
B. between 1 and 2
C. between 1/2 and 1
D. between 1/4 and 1/2
E. less than 1/4
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