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05 Mar 2021, 06:15
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
59% (02:18) correct
41%
(02:20)
wrong
based on 98
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How much water should be added to 50 L of a mixture at 3/2 L for $20 so as to have a new mixture worth $8 1/3 per litre?
(A) 30 L
(B) 28 L
(C) 20 L
(D) 18 L
(E) 16 L
(A) 30 L
(B) 28 L
(C) 20 L
(D) 18 L
(E) 16 L
XSatishX
Joined: 05 Jul 2020
Last visit: 17 Nov 2022
Posts: 100
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Given Kudos: 36
Location: India
Concentration: Leadership, General Management
Schools: LBS '24 (D) ISB '23 (II) Tepper '24 (S)
GMAT 1: 650 Q49 V30
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05 Mar 2021, 06:47
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So basically we need to bring down the cost from (2*20)/3 to 25/3 per liter.
Now the ratio of water and mixture will be= (40/3-25/3):(25/3-0) = 15/3:25/3 = 3:5 Assuming cost of water =0
Now we have 50 L of the mixture so we will add 5x = 50 so x =10 and 3x = 30 Ltrs
Answer A
Now the ratio of water and mixture will be= (40/3-25/3):(25/3-0) = 15/3:25/3 = 3:5 Assuming cost of water =0
Now we have 50 L of the mixture so we will add 5x = 50 so x =10 and 3x = 30 Ltrs
Answer A
14 Mar 2021, 21:51
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Bunuel
1.5L cost =20
1L costs =\(\frac{40}{3}\)
50L cost= \(\frac{40}{3}*50\)
We want cost per L to be \(\frac{25}{3}\)
So we can form an equation:
\(\frac{\frac{40}{3}*50}{50+x}=\frac{25}{3}\)
Where \(x\) is the quantity of water added
\(x=30\)
Hope it's clear.
