Abt the repost, since I didnt get the reply I posted again.Anyway ans is right can u give explanation
since I didnt get reply I posted again .anyway ans right can u give explanation plz
ndhadha
since I didnt get any reply on this I posted again. Anyway ans is right can u give explanation plz
ndhadha
since I didnt get any reply on this I posted again. Anyway ans is right can u give explanation plz
deepak has explained it well.. chk out post no. 3330.. :-P
Set R contains five numbers that have an average value of 55. If the median of the set is equal to the mean, and the largest number in the set is equal to 20 more than three times the smallest number, what is the largest possible range for the numbers in the set?
A)78
B)77 1/5
C)66 1/7
D)55 1/7
E)52
Set R contains five numbers that have an average value of 55. If the median of the set is equal to the mean, and the largest number in the set is equal to 20 more than three times the smallest number, what is the largest possible range for the numbers in the set?
A)78
B)77 1/5
C)66 1/7
D)55 1/7
E)52
I'm no more than a nomadic visitor to this thread but this question has piqued my curiosity. I'm nearly sure about my answer but would love to double check before I compose a detailed solution - Is the answer 78?
If m and n are positive integers such that m/n = 13.24, which of the following could NOT be the remainder when m is divided by n?
A)12
B)16
C)30
D)48
E)120
reachmonil SaysI'm no more than a nomadic visitor to this thread but this question has piqued my curiosity. I'm nearly sure about my answer but would love to double check before I compose a detailed solution - Is the answer 78?
Monil, I'll wait for a couple of responses before i post the OA..
Imo e (120)...
if m and n are positive integers such that m/n = 13.24, which of the following could not be the remainder when m is divided by n?
A) 12
b) 16
c) 30
d) 48
e) 120
ndhadha SaysMonil, I'll wait for a couple of responses before i post the OA..
reachmonil SaysI'm no more than a nomadic visitor to this thread but this question has piqued my curiosity. I'm nearly sure about my answer but would love to double check before I compose a detailed solution - Is the answer 78?
Sounds good. I'll post my solution at the risk of getting it wrong (hopefully not)
5 numbers - a, b, c, d, e (in ascending order)
55 is the average of 5 nos.
Therefore, a + b + c + d + e = 275
For range to be maximum, (e - a) should be maximum.
Therefore, 'e' should be max and 'a' should be min.
Given a particular value of 'a', to make 'e' max, 'b' & 'd' should take min values. Therefore,
c = 55
e = 3a + 20
d = 55 (since d has to take least value, the min it can take is value of c)
b = a (since b also has to be min, least possible value of b = a)
Therefore, a + a + 55 + 55 + (3a + 20) = 275
5a + 130 = 275
a = 29
e = 3(29) + 20 = 107
Range = e - a = 78
Nipunbans SaysImo e (120)...
I would urge you to reconsider your answer :)
Hint: Express as fraction to further think of a logical pattern to emerging remainders.
PS: ndhadha, nice questions! I see you are quite a regular on this thread. Keep up the enthusiasm and all the best for your GMAT
If m and n are positive integers such that m/n = 13.24, which of the following could NOT be the remainder when m is divided by n?
A) 12
B) 16
C) 30
D) 48
E) 120
Ans B ...
for fractional values, decimal is of significance.
.24 = 24 /100 = 6/25 ....
Hence, all multiples of 6 can be remainder ..
Off late... have been making horrible mistakes in a hurry ! hope am correct this time
Sounds good. I'll post my solution at the risk of getting it wrong (hopefully not)
5 numbers - a, b, c, d, e (in ascending order)
55 is the average of 5 nos.
Therefore, a + b + c + d + e = 275
For range to be maximum, (e - a) should be maximum.
Therefore, 'e' should be max and 'a' should be min.
Given a particular value of 'a', to make 'e' max, 'b' & 'd' should take min values. Therefore,
c = 55
e = 3a + 20
d = 55 (since d has to take least value, the min it can take is value of c)
b = a (since b also has to be min, least possible value of b = a)
Therefore, a + a + 55 + 55 + (3a + 20) = 275
5a + 130 = 275
a = 29
e = 3(29) + 20 = 107
Range = e - a = 78
Agree with u buddy
..Similar approach with same ans ...
brilliant bhavin :):)
for fractional values, decimal is of significance.
.24 = 24 /100 = 6/25 ....
Hence, all multiples of 6 can be remainder ..
Off late... have been making horrible mistakes in a hurry ! hope am correct this time 
guys,
am facing a problem with answer key of the 1000PS document! could anybody tell me from where i can download the answer key to the same? pls let me know....have started solving problems from the document and am totally clueless about the answers!!
thanx
Ans B ...
for fractional values, decimal is of significance.
.24 = 24 /100 = 6/25 ....
Hence, all multiples of 6 can be remainder ..
Off late... have been making horrible mistakes in a hurry ! hope am correct this time
bhavin.. why should the remainder by a multiple of 6.. i understood the problem upto that point, but then got a lil confused..... :help:
ndhadha Saysbhavin.. why should the remainder by a multiple of 6.. i understood the problem upto that point, but then got a lil confused..... :help:
Keeping the decimal value constant, ie. 0.24, if we change the denominator, the corresponding value of numerator will change.
If denominator is 25, numerator (ie. remainder) has to be 6.
If denominator is 50, numerator (ie. remainder) has to be 12.
If denominator is 100, numerator (ie. remainder) has to be 24, so on and so forth.
Hence you see, all multiples of 6 can be remainders. 😃
Ans B ...
for fractional values, decimal is of significance.
.24 = 24 /100 = 6/25 ....
Hence, all multiples of 6 can be remainder ..
ndhadha Saysbhavin.. why should the remainder by a multiple of 6.. i understood the problem upto that point, but then got a lil confused..... :help:
Thanx monil for the explanation...
Nisha , alternatively we can say m/n = (13*25 + 6) / 25
= say (k+6)/25 where k is a multiple of 25 ...
Now when we multiple or divide by the same no, value remains unchanged
If the multiplying factor is 2 then
(2k + 12) / 50 where 2k is mult of 50 hence rem is 12 ...
So if the multiplying factor is n, then
(nk+6n) / 25n ...where nk is multiple of 25 n ...so 6n is remainder ...
Hence, to generalize any multiple of 6 can be a remainder !
Hope that helps !
Sounds good. I'll post my solution at the risk of getting it wrong (hopefully not)
5 numbers - a, b, c, d, e (in ascending order)
55 is the average of 5 nos.
Therefore, a + b + c + d + e = 275
For range to be maximum, (e - a) should be maximum.
Therefore, 'e' should be max and 'a' should be min.
Given a particular value of 'a', to make 'e' max, 'b' & 'd' should take min values. Therefore,
c = 55
e = 3a + 20
d = 55 (since d has to take least value, the min it can take is value of c)
b = a (since b also has to be min, least possible value of b = a)
Therefore, a + a + 55 + 55 + (3a + 20) = 275
5a + 130 = 275
a = 29
e = 3(29) + 20 = 107
Range = e - a = 78
@Monil,
b & d shouldn't be a & 55 here since this is a set of 5 numbers. A set should technically contain all different numbers (or am I getting too technical these days??
)
@Monil,
b & d shouldn't be a & 55 here since this is a set of 5 numbers. A set should technically contain all different numbers (or am I getting too technical these days??
You are
There is no such condition that all elements in a set should be distinct. {1,2,3,3,3,4} is a valid set. 😃