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# Tina and Ethan collect only red and black marbles. Together, they have

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Math Expert
Joined: 02 Sep 2009
Posts: 65831
Tina and Ethan collect only red and black marbles. Together, they have  [#permalink]

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13 Jul 2020, 01:14
00:00

Difficulty:

85% (hard)

Question Stats:

54% (02:59) correct 46% (02:08) wrong based on 83 sessions

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Competition Mode Question

Tina and Ethan collect only red and black marbles. Together, they have a total of 100 (red and black) marbles. How many marbles does Tina currently have?

(1) If Tina gives half of her red marbles to Ethan and Ethan gives half of his black marbles to Tina, then Tina would have a total of 20 more marbles than Ethan.

(2) If Tina gives half of her black marbles as well as half of her red marbles to Ethan, then Ethan would have a total of 60 more marbles than Tina.

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Re: Tina and Ethan collect only red and black marbles. Together, they have  [#permalink]

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13 Jul 2020, 01:48
1
IMO B

Let T(R) + T(B) + E(R) + E(B) = 100

Statement 1
After exchange , Tina = 1/2 T(R) + 1/2 E(B) + T(B)
Ethan = 1/2 T(R) + 1/2 E(B) + E(R)
A/C ques, Tina-Ethan =20
T(B) -E(R) = 20
Not sufficient

Statement 2
After exchange , Tina = 1/2 T(R) + 1/2 T(B)
Ethan = 1/2 T(R) + 1/2 T(B) + E(R)+ E(R)
Ethan -Tina = 60
So, E(R) + E(B) = 60
So T(B) + T(R) = 40
Sufficient

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Re: Tina and Ethan collect only red and black marbles. Together, they have  [#permalink]

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13 Jul 2020, 01:51
1
Let Tina have x red marbles and y black marbles.
Ethan have p red marbles and q black marbles.
Therefore
T = X(r) + Y(b)
E = P(r) + Q(b)

Statement 1: If Tina gives half of her red marbles to Ethan and Ethan gives half of his black marbles to Tina, then Tina would have a total of 20 more marbles than Ethan.
T = {X(r)}/2 + Y(b) + {Q(b)}/2
E = P(r) + {X(r)}/2 + {Q(b)}/2

=> {X(r)}/2 + Y(b) + {Q(b)}/2 - [P(r) + {X(r)}/2 + {Q(b)}/2] = 20
=> {X(r)}/2 + Y(b) + {Q(b)}/2 - P(r) - {X(r)}/2 - {Q(b)}/2 = 20
=> Y(b) +P(r) = 20
Black marble of Tina + Red marbles of Ethan = 20
(Insufficient)

Statement 2: If Tina gives half of her black marbles as well as half of her red marbles to Ethan, then Ethan would have a total of 60 more marbles than Tina.
T = {X(r)}/2 + {Y(b)}/2
E = P(r) + {X(r)}/2 + Q(b) + {Y(b)}/2

=> P(r) + {X(r)}/2 + Q(b) + {Y(b)}/2 - [{X(r)}/2 + {Y(b)}/2] = 60
=> P(r) + {X(r)}/2 + Q(b) + {Y(b)}/2 - {X(r)}/2 - {Y(b)}/2 = 60
=> P(r) + Q(b) = 60
Therefore Ethan have 60 in starting, so Tina will have 100-60 = 40 marble (sufficient)

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Re: Tina and Ethan collect only red and black marbles. Together, they have  [#permalink]

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13 Jul 2020, 02:14
1
Quote:
Tina and Ethan collect only red and black marbles. Together, they have a total of 100 (red and black) marbles. How many marbles does Tina currently have?

(1) If Tina gives half of her red marbles to Ethan and Ethan gives half of his black marbles to Tina, then Tina would have a total of 20 more marbles than Ethan.

(2) If Tina gives half of her black marbles as well as half of her red marbles to Ethan, then Ethan would have a total of 60 more marbles than Tina.

Attachment:

pic1.png [ 8.32 KiB | Viewed 595 times ]

Given:
let the red and black marbles with Tine be a & c.
let the red and black marbles with Ethan be b & d.
a + b + c + d = 100
statement 1:
a/2 + c + d/2 -(b + a/2 + d/2) = 20
c - d = 20
we can not find a+c.
not sufficient

statement 2:
b + a/2 + d + c/2 - (a/2 + c/2) = 60
b + d = 60
a +c = 100 - 60 = 40

Ans: B
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Re: Tina and Ethan collect only red and black marbles. Together, they have  [#permalink]

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13 Jul 2020, 20:26
1
Ans: B
(1) If Tina gives half of her red marbles to Ethan and Ethan gives half of his black marbles to Tina, then Tina would have a total of 20 more marbles than Ethan.
from here we get TB-ER=20
Not sufficient

(2) If Tina gives half of her black marbles as well as half of her red marbles to Ethan, then Ethan would have a total of 60 more marbles than Tina.
we get, ER+EB=60
From here we can get TR+TB
So sufficient
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Re: Tina and Ethan collect only red and black marbles. Together, they have  [#permalink]

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14 Jul 2020, 20:16
Solution:

Let Tina's marbles be RT and BT
And Ethan's marbles be RE and BE.

NOW,
(RT+BT) + (RE+BE) = 100 ---- (X)

We need to find Tina's current marbles. Therefore, we need RT+BT=?

STATEMENTS:

(1) Forming an equation, we get :

(RT + BT) - (1/2)RT + (1/2)BE = 20 + (RE + BE) + (1/2)RT -(1/2)BE

On solving further, we have :

BT = 20 + RE ----- (A)

We just get a relation between BT and RE. Can't find Tina's marbles.

Insufficient

(2) Again, forming equations, we get :

(RT + BT) -(1/2)(RT + BT) + 60 = (RE + BE) +(1/2)(RT + BT)

On solving, we get

RE + BE = 60 ------ (B)

Also. Combining (X) and (B), we get:

RT + BT = 40

We get Tina's marbles.

Sufficient

Hence, the answer is Option (B).
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Re: Tina and Ethan collect only red and black marbles. Together, they have  [#permalink]

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25 Jul 2020, 15:13
Bunuel wrote:

Competition Mode Question

Tina and Ethan collect only red and black marbles. Together, they have a total of 100 (red and black) marbles. How many marbles does Tina currently have?

(1) If Tina gives half of her red marbles to Ethan and Ethan gives half of his black marbles to Tina, then Tina would have a total of 20 more marbles than Ethan.

(2) If Tina gives half of her black marbles as well as half of her red marbles to Ethan, then Ethan would have a total of 60 more marbles than Tina.

My thought process was not be lost in calculations as it is a DS question.

(1) Taking the variables for each of the Red and Black balls that each person had and we also know the total of them, the statement does not help much in creating a proper equation for us.
(2) The same variables of Tina is used for Ethan and we understand that we can get rid of the variables using the two equations, also the total will help us get the marbles Tina had.

Therefore, B

Might not be the ideal way but when time is short and we need to mark some answer choice, the strategy works best

B
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Re: Tina and Ethan collect only red and black marbles. Together, they have   [#permalink] 25 Jul 2020, 15:13