Official Quant thread for CAT 2013

MBA CAT 2024
16243

the maximum value of function (2-x)^5 (x-6)^7 when 211/3

3
5
4

16244
@raopradeep
155
16245
@hatemonger said:
as min value of this fn is 165
answer is 155
but how to approach these type of questions ?

16246
@ishu1991
plz tell how to approach .
16247
@raopradeep
11/3
16248
@raopradeep said:
@ishu1991plz tell how to approach .
smallest postive integer is 1 by putting 1 we are getting a value 165 so 155 can never be possible
16249
@ishu1991 said:
@raopradeep11/3
buddy tell the method
16250
@raopradeep said:
buddy tell the method
just apply differentiation with product rule
16251
@raopradeep said:
answer is 155 but how to approach these type of questions ?
in denomination u have 19abc ok...
divide ( 5a^2+a+5)/a so u will have (5a+1+5/a)
[5(a+1/a)+1] min value of (a+1/a) is 2 so min value of first fn is 11
llly for second fn min value is 15 and for third it is 19

16252
@hatemonger said:
in denomination u have 19abc ok...divide ( 5a^2+a+5)/a so u will have (5a+1+5/a) [5(a+1/a)+1] min value of (a+1/a) is 2 so min value of first fn is 11 llly for second fn min value is 15 and for third it is 19
thanks

16253
@hatemonger said:
1092??
method pls !!
16254
@Calvin4ever said:
The sum of all possible factors of 500 ??
This is a formulas based question.
500 = 2^2.5^3
Sum = 1092
For factor theorems refer http://quantexpert.co.in/qenotes/quantnotes/numbers/47-factor-theories.html

16255
@Calvin4ever said:
method pls !!
500=5^3*2^2

so sum of factors =[[ 5^4-1]/(5-1)]*[[2^3-1]/(2-1)]
16256

if f(x) =min(x-3,x+2) &g;(x)=max(x-2,x-3) and if h(x) =fx +gx then find the range of x for which h(x)

(-inf ,0]
(-inf ,6]
[0,6]
[6, inf)

16257
@Zedai said:
In an exhibition, some paintings were kept for sale. On the first day, 1 painting plus 1/7 th of the remaining paintings were sold. On the second day, 2 paintings plus 1/7 th of the remaining paintings were sold. A similar pattern continued till the kth day, when 'k' paintings were sold and no painting was left after that. If the exhibition ran for exactly k days (k > 1), then what is the minimum number of paintings sold during the exhibition?

Supp there are x paintings

First day: Sells [1+ (1/7*(x-1))] and [(6/7*x) - 6/7] left
=> (x-1) has to be divisible by 7. Or x= 7k+1

Second Day: Sells [2+(1/7)*((6/7)*x - 20/7)] = [(6/49)*x - (29/49)] and [(38/49)*x - (13/49)] left
=> x has to be 49n+36

So min value itself shows to be 36
16258
@raopradeep said:
if f(x) =min(x-3,x+2) &g;(x)=max(x-2,x-3) and if h(x) =fx +gx then find the range of x for which h(x)(-inf ,0] (-inf ,6][0,6][6, inf)
(-inf, 6] ??
16259
@raopradeep said:
if f(x) =min(x-3,x+2) &g;(x)=max(x-2,x-3) and if h(x) =fx +gx then find the range of x for which h(x)(-inf ,0] (-inf ,6][0,6][6, inf)

(-inf,6)
16260
@ananyboss said:
Hi Everyone,Help me understand the concept.Q - How many numbers from 1 - 400 ( both included) are not divisible by 2,3,5 and 7.
numbers divisible by 2, 200, by 3 - 133, by 5 - 80, by 7 - 57
divisible by 6 - 66, by 10 - 40, by 14 - 28.
by 15- 26, 21 -19, 35 -11
by 30- 13, 42 - 9
by105 - 3.,by 70 - 5
by 210 - 1
Numbers not divisble= 400 - (div by 2,3,5,7) + (div by 6,10,14,15,21)- (div by 30, 42,105,70) +(div by 210)
= 400-470+190-30+1=91?
-please let there bhi no calc mistake :banghead:
16261
@Calvin4ever said:
(-inf, 6] ??
@ishu1991 said:
(-inf,6)
bhaiyo plz explain

16262
@hatemonger said:
500=5^3*2^2so sum of factors =[[ 5^4-1]/(5-1)]*[[2^3-1]/(2-1)]
Thanx. I have found another way too. (1+2+4)(1+5+25+125)=1092. perhaps u already know this but still . .