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manisha patil

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1.The range of a set is 56. Its median number is 50 more than the lowest number, and mean > median, at least how many numbers in this set?

 

A 13

 

B 15

 

C 17

 

D 19

 

E 21

 

2. 2 nos divided by a certain divisor yield remainders of 431,379 resp.if sum of the two nos divided by the same divisor yields a remainder of 211,find the divisor

 

3.the least number by which 72 must be multiplied inorder to produce a multiple of 112 is

a.6

b.12

c.14

d.18

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1.The range of a set is 56. Its median number is 50 more than the lowest number, and mean > median, at least how many numbers in this set?

 

A 13

 

B 15

 

C 17

 

D 19

 

E 21

 

2. 2 nos divided by a certain divisor yield remainders of 431,379 resp.if sum of the two nos divided by the same divisor yields a remainder of 211,find the divisor

 

3.the least number by which 72 must be multiplied inorder to produce a multiple of 112 is

a.6

b.12

c.14

d.18

 

 

1-21

2-599

3-14

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let minimum num a, then median (a+50), maximum number (a+56)

 

to satisfy the above conditions we need to take x median and x maximum number

 

then we get,

 

(a+(a+50)*x+(a+56)*x)/(2x+1) > (a+50)

or, a+ 6x> a+50

or, 6x>50

so, x should be 9

 

so, total members = 9*2+1=19

 

 

yes, you are right, I had made a mistake

 

PROBLEM 2:

 

X=x1*a+431

Y=y1*a+379 a: the divisor we want to calculate

 

X+Y= (x1+y1)*a+810

X+Y= m*a+211

 

(x1+y1)*a+810=m*a+211 ===è (m-x1-y1)*a=599

 

since m-x1-y1 is an integer ,599 is a prime number, and a must be>431, then a=599

 

 

PROBLEM 3:

 

72=2^3*3^2

112=2^4*7

 

LCM/72= (2^4*3^2*7)/(2^3*3^2)=14

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I really doubt if this would appear on a GRE, but here is more detail behind Stickler's answer for the mean/median question.

 

The mean is greater than the median, and the fastest way to increase the mean is to load all the values in your dataset to be at the maximum value. We also need to be sure that our median meets the criteria. The (a+50) below is multiplied by x so that we can add many copies of the median (a+50) to the dataset, if we add the same nbr of "max values" to the dataset, we will have enough median values so that the median really is (a+50). The denominator of (2x+1) is a count of the values in our dataset, which is:

+ the minimum value

+ the copies of the median value (we made "x" copies each)

+ the copies of the max values (we made "x" copies each)

 

Using Sticklers variables, let a=the minimum value in the set, so the mean is:

 

(min_value + median_values + max values)/(number of values)> (median value)

 

( a + (a + 50)*x + (a+56)*x )/(2x+1) > (a+50)

 

I just let a=0 to make it easier, then multiply each side by (2x +1) to clear the denominator:

 

( 50*x + 56*x) > (100*x + 50)

 

and solve:

 

6x > 50

x=50/6, round up to 9,

but the number of values in the set is (2x + 1) as we used in the denominator, so the count is 2*9+1 = 19.

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  • 2 years later...

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