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Home Forums Quantitative-Ability Quantitative Ability teaser

discussion topic: Quantitative Ability teaser

Showing 1 to 9 out of 9 Answers

2010-01-05 20:46:12
Message 1
  

About today's teaser,

how is |f(x)= x3+y3




Teaser Question



If f(x) = x2 + y2; g(x) = xy and h(x) = x+y, then which of the following equals |f(x)|?

 


Choose the correct answer option:


  • 1.  g(x)[{h(x)}2 – 2f(x)]  

  • 2. {f(x)}1/3 [g(x) – h(x)]  

  • 3. h(x) [{h(x)}2 – 3g(x)] 

  • 4. g(x)[f(x) – {h(x)}3]

  • 5. none of these







Solution



Correct Answer: 3

Explanation: x3+ y3= (x+y)3– 3xy(x+y) = (x+y) [(x+y)2– 3xy].

2010-02-02 16:29:38
Message 2
  



Teaser Question



E,
F, G, H and K are friends. Among them any one thinner than E is thinner
than G. Any one stouter than F is stouter than H. someone in the group
who is Leaner than E is not leaner than F. H is stouter than either G
or K ( but not both). Not two in the group are equally stout. The
stoutest is.


Choose the correct answer option:

  • 1. E


  • 2. K


  • 3. G


  • 4. H


  • 5. F












Solution



Correct Answer: 5



Explanation:
Among them any one thinner than E is thinner than G means E is thinner than G.




Any one stouter than F is stouter than H means F is stouter than H.


someone in the group who is Leaner than E is not leaner than F means E is leaner than F


So, Both G and F are stouter than E. Also F is Stouter than H

H is stouter than G or K, so K is the thinnest. ( One person is thinner than E)


So the order is K  E  G H F


Hence the stoutest is F.


---------------------------------------------------------------------------------------------------

But:
Lets say, based on the 1st and 2nd  arguments we get the follwing order in increasing order of stoutness.

H<F   and E<G

3rd:
someone in the group who is Leaner than E is not leaner than F means E is leaner than F

let the someone be x.
cond satisfied wen F<x<E            => x is leaner thn E but not leaner thn F.
From this how can the conclusion be E<F ... ???


If we continue with F<E as i assume, then the order is H<F<E<G and there is a place for K somewhere in between all 4.
And K may be between E n F coz the cond says someone must be leaner thn E but not F.
So H<F<K<E<G. 
 
4th:
H is stouter than either G
or K ( but not both).


This cond doesnt fit in.. :( :(  something is wrong.

If we continue with E<F, n follow the solution given , we arrive at K  E  G H F.
But this does not satisfy the cond : "H is stouter than either G
or K ( but not both).
"  :( :(
so somethng is wrong with the Q ...
===========================================================
is tht rite??



2010-02-02 17:11:04
Message 3
  

Quantitative Ability Teaser



Correct Answer Wins points







Direction: Mark the best answer:



M and C are the centres of 2 equal
coins. The lower coin is fixed and the upper coin is rolling without
slipping on the outside of the lower coin in clockwise direction.
Initially the point T of the upper coin is in contact with the point of
the lower coin as shown in the figure. When the straight Line CM makes
an angle of 450 with the vertical, then the radius TM with make an angle (with the vertical) of :





 
















Choose correct answer:


?/8 radians

?/4 radians  

?/2 radians  

3?/14 radians








2010-03-05 22:17:26
Message 4
  

If x + y + z = 21, then the maximum value of (x – 6)(y + 7) (z – 4) is:
in this question how do you reach at the conclusion that :

(x – 6)(y + 7)(z – 4) will be maximum only when

(x – 6) = (y + 7) = (z – 4) = k (say) for a given value of (x + y + z) = 21

i did not get it.
i mean it can also happen that ano is greater than the other

2010-12-18 20:48:54
Message 5
  

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2010-12-19 02:57:12
Message 6
  
2010-12-19 08:21:43
Message 7
  
2010-12-19 10:52:14
Message 8
2011-12-19 12:01:23
Message 9
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Showing 1 to 9 out of 9 Answers

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