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# Mani, Sani and Karna have x, y and z number of chocolates respectively

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Retired Moderator
Joined: 22 Aug 2013
Posts: 1443
Location: India
Mani, Sani and Karna have x, y and z number of chocolates respectively  [#permalink]

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28 Jun 2018, 11:04
1
3
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Difficulty:

45% (medium)

Question Stats:

64% (01:54) correct 36% (02:09) wrong based on 136 sessions

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Mani, Sani and Karna have x, y and z number of chocolates respectively. Does Sani have the average of the total number of chocolates among all three?

(1) y is 20% more than z and 25% less than x.

(2) x = 32.
Math Expert
Joined: 02 Aug 2009
Posts: 7684
Re: Mani, Sani and Karna have x, y and z number of chocolates respectively  [#permalink]

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28 Jun 2018, 19:27
1
Mani, Sani and Karna have x, y and z number of chocolates respectively. Does Sani have the average of the total number of chocolates among all three?

Is x+z=2y??

(1) y is 20% more than z and 25% less than x.
Since it gives us a ratio between 3, should be sufficient..
Let z=100, y=120% of 100=120
Also y=75% of X.....120=75% of X
X=120*4/3=160
So 100,120,160
Y is not the average
Sufficient

(2) x = 32.
Insufficient

A
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Joined: 13 Oct 2013
Posts: 136
Concentration: Strategy, Entrepreneurship
Mani, Sani and Karna have x, y and z number of chocolates respectively  [#permalink]

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07 Oct 2018, 01:55
Hi Chetan2u
I did not understand this part "x+z=2y??"
should not it be, y= x+y+z/3?

chetan2u wrote:
Mani, Sani and Karna have x, y and z number of chocolates respectively. Does Sani have the average of the total number of chocolates among all three?

Is x+z=2y??

(1) y is 20% more than z and 25% less than x.
Since it gives us a ratio between 3, should be sufficient..
Let z=100, y=120% of 100=120
Also y=75% of X.....120=75% of X
X=120*4/3=160
So 100,120,160
Y is not the average
Sufficient

(2) x = 32.
Insufficient

A

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Kindly press +1 Kudos if my post helped you in any way
Math Expert
Joined: 02 Aug 2009
Posts: 7684
Re: Mani, Sani and Karna have x, y and z number of chocolates respectively  [#permalink]

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09 Oct 2018, 01:01
sunita123 wrote:
Hi Chetan2u
I did not understand this part "x+z=2y??"
should not it be, y= x+y+z/3?

chetan2u wrote:
Mani, Sani and Karna have x, y and z number of chocolates respectively. Does Sani have the average of the total number of chocolates among all three?

Is x+z=2y??

(1) y is 20% more than z and 25% less than x.
Since it gives us a ratio between 3, should be sufficient..
Let z=100, y=120% of 100=120
Also y=75% of X.....120=75% of X
X=120*4/3=160
So 100,120,160
Y is not the average
Sufficient

(2) x = 32.
Insufficient

A

hi..

it is same..
$$y=\frac{x+y+z}{3}.........3y=x+y+z........2y=x+z$$
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Re: Mani, Sani and Karna have x, y and z number of chocolates respectively   [#permalink] 09 Oct 2018, 01:01
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