This one is today's Quantitative Question of the Day (QQAD) . Let's see how many of you are up for this challenge.
Let A be a subset of {1, 2, 3, ... , 27} such that no two subsets of A have the same sum. What is the largest possible sum for A?
(1) 101 (2) 121 (3)129 (4) 135 (5) none of these
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Team BV -Vineet
scrabbler
March 8, 2013, 5:00am
25130
@bodhi_vriksha said: This one is today's
I get 135.
nramachandran
March 8, 2013, 5:14am
25131
@scrabbler said: I get 135.
scrabbler
March 8, 2013, 5:25am
25132
@nramachandran said: Sir, Is there a method to solve it without trial and error?
Not sure. What I did was, I started with 27 and 26, and just noted possible totals. 1 at a time, 2 at a time and 3 at a time (since not more than 6 numbers came this was sufficient). With this combo I could not find any overlap.
dexian
March 8, 2013, 6:03am
25133
@scrabbler said: I get 135.
27 26 25 24 23 10 .......... this can give 135 na????
Min value of (x+y^2+z^3) (1/x+1/y^2+1/z^3) for +ve x,y,z
gsathe89
March 8, 2013, 6:16am
25135
find the sum of all two-digit positive integers which exceed the product of their digits by 12??
dexian
March 8, 2013, 6:24am
25139
@Gsathe89 said: find the sum of all two-digit positive integers which exceed the product of their digits by 12??
two numbers are 28 and 39
scrabbler
March 8, 2013, 6:27am
25140
@Dexian said: 27 26 25 24 23 10 .......... this can give 135 na????
no two subsets of A have the same sum
Consider ordered quadruple of integers (a,b,c,d) as interesting if 1 a+d > b+c . How many interesting ordered quadruples are there?
@Gsathe89 said: find the sum of all two-digit positive integers which exceed the product of their digits by 12??
The number can be written as ab or (10a + b)
10a + b - a*b = 12
a*(10-b) + b = 12
a*(10-b) - (10-b) = 2
(a - 1)*(10 - b) = 2
LHS can be written as (1*2), ( -1 * -2) , (2 *1), ( -2 * -1)
Case 1)
a -1 = 1 => a =2
10 - b = 2 => b = 8 , So number is 28
Case 2)
a -1 = -1 => a = 0 (rejected)
Case 3)
a - 1 = 2 => a = 3
10 - b = 1 => b = 9, So number is 39
Case 4)
a - 1 = -2 (rejected)
So, sum of the numbers = 28 + 39 = 67 ?
@Gsathe89 said: thanks ..
Don't post random replies here 😃 .. This thread is only for asking and replying to QA questions :)
Welcome :)
@amresh_maverick said: Min value of (x+y^2+z^3) (1/x+1/y^2+1/z^3) for +ve x,y,z
AM>=HM
so we get the relation : (a1+a2+a3+....a(n))(1/a1+1/a2+......1/a(n))>=n^2
in this ques ,there are three terms which are product of temrs and their reciprocals...so product >=3^2=9 ..hence answer would be 9 :)
Team BV-- Pratik Gauri
@Dexian said: ok q samajh hi nhi paya tha main......
Now that you have, go for it!
Team BV - Vineet
