Last visit was: 12 Sep 2024, 01:30 It is currently 12 Sep 2024, 01:30
Toolkit
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

# The amount of water in a salt water sugar solution is increased by 2/5

SORT BY:
Tags:
Show Tags
Hide Tags
Admitted - Which School Forum Moderator
Joined: 25 Oct 2020
Posts: 1124
Own Kudos [?]: 1124 [16]
Given Kudos: 629
Schools: Ross '25 (M\$)
GMAT 1: 740 Q49 V42 (Online)
Senior Moderator - Masters Forum
Joined: 19 Jan 2020
Posts: 3127
Own Kudos [?]: 2842 [0]
Given Kudos: 1511
Location: India
GPA: 4
WE:Analyst (Internet and New Media)
Dream Score Representative
Joined: 07 Oct 2021
Posts: 381
Own Kudos [?]: 327 [1]
Given Kudos: 2
Senior Moderator - Masters Forum
Joined: 19 Jan 2020
Posts: 3127
Own Kudos [?]: 2842 [0]
Given Kudos: 1511
Location: India
GPA: 4
WE:Analyst (Internet and New Media)
Re: The amount of water in a salt water sugar solution is increased by 2/5 [#permalink]
JerryAtDreamScore
PyjamaScientist
The amount of water in a saltwater sugar solution is increased by $$\frac{2}{5 }$$. Before the increase, what was the ratio of the weight of water to the weight of the solution?
(1) Due to the addition of water, the ratio of weight of the sugar in the solution to the weight of the solution decreases from $$\frac{1}{4}$$ to $$\frac{1}{5}$$.
(2) After the addition of water, the ratio of the weight of water in the solution to the weight of the solution becomes $$\frac{7}{10}$$.

Important Info:

Increasing the amount of water by 40% will decrease the percent (weight) of the solute (sugar) by $$1 - \frac{1}{1.4}$$.

Statement 1 Alone:

The math here doesn't quite work out, but we are given [Sugar : Solution = 1 : 4], so it must be true that [Water : Solution = 3 : 4]. Then statement 1 alone is sufficient.

Statement 2 Alone:

Again, the math here is weird and statements 1 + 2 would contradict, but if we set the original percent weight of water as X, we would have $$\frac{1.4X}{ 1 + 0.4X} = \frac{7}{10}$$, so we are able to solve for the original percent of water.

Don't we need salt proportion in statement 1?

Posted from my mobile device
Intern
Joined: 18 Nov 2021
Posts: 14
Own Kudos [?]: 18 [2]
Given Kudos: 285
Location: Germany
Concentration: Finance, Strategy
GMAT 1: 620 Q50 V25
GMAT 2: 700 Q50 V36
Re: The amount of water in a salt water sugar solution is increased by 2/5 [#permalink]
1
Kudos
1
Bookmarks
yashikaaggarwal
JerryAtDreamScore
PyjamaScientist
The amount of water in a saltwater sugar solution is increased by $$\frac{2}{5 }$$. Before the increase, what was the ratio of the weight of water to the weight of the solution?
(1) Due to the addition of water, the ratio of weight of the sugar in the solution to the weight of the solution decreases from $$\frac{1}{4}$$ to $$\frac{1}{5}$$.
(2) After the addition of water, the ratio of the weight of water in the solution to the weight of the solution becomes $$\frac{7}{10}$$.

Important Info:

Increasing the amount of water by 40% will decrease the percent (weight) of the solute (sugar) by $$1 - \frac{1}{1.4}$$.

Statement 1 Alone:

The math here doesn't quite work out, but we are given [Sugar : Solution = 1 : 4], so it must be true that [Water : Solution = 3 : 4]. Then statement 1 alone is sufficient.

Statement 2 Alone:

Again, the math here is weird and statements 1 + 2 would contradict, but if we set the original percent weight of water as X, we would have $$\frac{1.4X}{ 1 + 0.4X} = \frac{7}{10}$$, so we are able to solve for the original percent of water.

Don't we need salt proportion in statement 1?

Posted from my mobile device

You are totally right.

Here is my solution; I hope it helps.
We want to know $$\frac{wa}{wa+sa+su} = ?$$

Statement 1:
(A):
$$\frac{su}{wa+sa+su} = \frac{1}{4} <=> \frac{wa+sa}{wa+sa+su} = \frac{3}{4}$$
$$<=> wa+sa = \frac{3}{4} * (wa+sa+su)$$

(B)
(note: I will use sometimes decimal numbers instead of fractions because I do not know how to make a fraction in an other fraction in this forum; However, I recommend using only fractions)
$$\frac{su}{(1.4)*wa+sa+su} = \frac{1}{5} <=> \frac{(1.4)*wa+sa}{(1.4)*wa+sa+su} = \frac{4}{5}$$
$$<=> (1.4)*wa+sa = \frac{4}{5} * ((1.4)*wa+sa+su) <=> \frac{7}{5}wa+sa = \frac{4}{5} * (\frac{7}{5}wa+sa+su)$$
$$<=> wa+sa = \frac{4}{5} * (\frac{7}{5}wa+sa+su) - \frac{2}{5}wa = \frac{28}{25}wa+\frac{4}{5}sa+\frac{4}{5}su - \frac{10}{25}wa = \frac{18}{25}wa+\frac{4}{5}sa+\frac{4}{5}su$$

(A) & (B)
$$=> \frac{3}{4}wa+\frac{3}{4}sa+\frac{3}{4}su = \frac{18}{25}wa+\frac{4}{5}sa+\frac{4}{5}su$$
$$<=> \frac{75}{100}wa+\frac{15}{20}sa+\frac{15}{20}su = \frac{72}{100}wa+\frac{16}{20}sa+\frac{16}{20}su$$
$$<=> \frac{3}{100}wa = \frac{1}{20}sa+\frac{1}{20}su$$
$$<=> \frac{3}{5}wa = sa+su$$
$$<=> sa+su = (0.6)*wa$$

$$=>$$
$$\frac{wa}{wa+sa+su} = \frac{wa}{wa+(0.6)*wa} = \frac{wa}{(1.6)*wa} = \frac{5}{8}$$

=> Sufficient

Statement 2:
$$\frac{(1.4)*wa}{(1.4)*wa+sa+su} = \frac{7}{10}$$
$$<=> \frac{10}{7}*\frac{7}{5}wa = \frac{7}{5}wa+sa+su$$
$$<=> \frac{10}{5}wa = \frac{7}{5}wa+sa+su$$
$$<=> \frac{3}{5}wa = sa+su$$
$$<=> sa+su = (0.6)*wa$$

$$=>$$
$$\frac{wa}{wa+sa+su} = \frac{wa}{wa+(0.6)*wa} = \frac{wa}{(1.6)*wa} = \frac{5}{8}$$

=> Sufficient

=> D
Intern
Joined: 14 Nov 2022
Posts: 2
Own Kudos [?]: 2 [1]
Given Kudos: 6
Re: The amount of water in a salt water sugar solution is increased by 2/5 [#permalink]
1
Kudos
S1 -
Assume,
salt + water + sugar = 40
sugar = 10, salt +water = 30
sugar : soln increases to 1:5 from 1:4
therefore, salt +water becomes 40
now, the water is increased by 10 units which is 2/5 of water content earlier lets call it (x).
2/5*x = 10
x=25
so earlier soln contents were, water = 25, salt =5, sugar 10
therefore, s1 is sufficient
Intern
Joined: 24 Aug 2021
Posts: 2
Own Kudos [?]: 0 [0]
Given Kudos: 34
Location: India
Re: The amount of water in a salt water sugar solution is increased by 2/5 [#permalink]
Someone can please suggest some other method to solve the question or elaborate the solutions step by step. I didn't understand either of the solutions provided in the previous threads.
Intern
Joined: 19 Jun 2022
Posts: 20
Own Kudos [?]: 4 [1]
Given Kudos: 19
Location: India
Concentration: Technology, Marketing
Schools: Kellogg
GPA: 3.85
WE:Management Consulting (Computer Software)
Re: The amount of water in a salt water sugar solution is increased by 2/5 [#permalink]
1
Kudos
MAJJOY
Someone can please suggest some other method to solve the question or elaborate the solutions step by step. I didn't understand either of the solutions provided in the previous threads.

Here, is how I solved the problem:

Let's assume,
Salt- SA, Sugar- SU, Water-WA.
TOT=SA+SU+WA
WAnew ->new portion of water(after adding 2/5 of water)
WAold -> old potion of water(before any addition)
TOTold -> old total
Given:
WAnew-WAold+2/5WAold.

Statement 1:
change in ratio is from 1/4 to 1/5.

when ratio was 1/4:
[SU][/SU+WA+SA]=1/4

4SU=SU+WA+SA-- 1

now when new ratio is 1/5:
SU/(SU+(WA+WA*2/5)+SA)=1/5-- 2

now taking values from eqn 1 and putting in 2
SU/(4SU+(2/5)WA)=1/5-- 3
Solving eqn 3
we get SU=(2/5)WA

Putting this value in the inital eqn 1:

(2/5)WA/(SU+WA+SA)=1/4

WA/(SU+WA+SA)= 5/8 is the asked ratio. Sufficient

Statement2:
After the addition of water, the ratio of the weight of water in the solution to the weight of the solution becomes 7/10
WAnew/TOTnew=7/10
equivallent to:
[WAold+(2/5)WAold]/[TOTold+(2/5)WAold]=7/10
10WAold+(20/5)WAold=7TOTold+(14/5)WAold
10WAold+(6/5)WAold=7TOTold
56WAold=35TOTold
WAold/TOTold=35/56=>5/8 Sufficient

Manager
Joined: 19 Aug 2023
Posts: 68
Own Kudos [?]: 36 [0]
Given Kudos: 82
Location: Indonesia
GMAT Focus 1:
695 Q85 V83 DI85
GMAT 1: 640 Q49 V27
GMAT 2: 640 Q50 V26
Re: The amount of water in a salt water sugar solution is increased by 2/5 [#permalink]
Tough question

First of all we need to understand that there are 3 components

Salt - Water - Sugar

(I) Sugar concentration drop from 25% to 20%

Here we got the concentration of the leftover (Salt and Water) are 75% and 80% (before and after water addition)

Hence we got 2 equations

1... Salt + Water = 75% *O
2... Salt + Water*(1+2/5) = 80% *(O + 2/5W)

O is a new variable which indicates Original Solution of Salt Water Sugar

Solving those 2 equations we will obtain W = 5/8 O

(SUFFICIENT)

II) The new solution has Water concentration of 70%
Again new concentration means there is an addition of 2/5 W (~0.4W)
Data II we can translate to the following equation

1.4W / (O+0.4W) = 70%

it will give us that W = 0.7/1.12 O = 62.5% (~5/8)

(SUFFICIENT)

VP
Joined: 29 Oct 2015
Posts: 1204
Own Kudos [?]: 542 [0]
Given Kudos: 737
GMAT 1: 570 Q42 V28
Re: The amount of water in a salt water sugar solution is increased by 2/5 [#permalink]
egmat KarishmaB MartyMurray Would you like to help on this question ?­
Tutor
Joined: 11 Aug 2023
Posts: 1122
Own Kudos [?]: 2592 [1]
Given Kudos: 93
GMAT 1: 800 Q51 V51
The amount of water in a salt water sugar solution is increased by 2/5 [#permalink]
1
Kudos
­The amount of water in a salt water sugar solution is increased by 2/5. Before the increase, what was the ratio of the weight of water to the weight of the solution?

Let

W = Orignal Water Weight

S = Sugar Weight

T = Original Total Weight

Information given:

W is increased to 7/5 * W.

S1) Due to the addition of water, the ratio of weight of the sugar in the solution to the weight of the solution decreases from 1/4 to 1/5.

This information tells us that the increase in the amount of water in the solution increased the total amount of solution by 1/4.

After all, if

S = 1/4 * T

and

S = 1/5 * (T + 2/5 * W)

then,

1/4 * T = 1/5 * (T + 2/5 * W)

5/4 * T = T + 2/5 * W

So,

1/4 * T = 2/5 * W

We could solve that to find the relationship between W and T.

Sufficient.

S2) After the addition of water, the ratio of the weight of water in the solution to the weight of the solution becomes 7/10.

Here we have 7/5 * W = 7/10 * (T + 2/5 * W).

We could solve that to find the exact relationship between W and T.

Sufficient.

Originally posted by MartyMurray on 27 Apr 2024, 11:01.
Last edited by MartyMurray on 29 Apr 2024, 02:34, edited 1 time in total.
Tutor
Joined: 16 Oct 2010
Posts: 15296
Own Kudos [?]: 67986 [1]
Given Kudos: 442
Location: Pune, India
Re: The amount of water in a salt water sugar solution is increased by 2/5 [#permalink]
1
Kudos

PyjamaScientist
The amount of water in a salt water sugar solution is increased by $$\frac{2}{5 }$$. Before the increase, what was the ratio of the weight of water to the weight of the solution?

S1) Due to the addition of water, the ratio of weight of the sugar in the solution to the weight of the solution decreases from $$\frac{1}{4}$$ to $$\frac{1}{5}$$.

S2) After the addition of water, the ratio of the weight of water in the solution to the weight of the solution becomes $$\frac{7}{10}$$.
­Given: In a water salt sugar solution, quantity of water is increased by 40%.
Question: Concentration of water in the solution before the increase?

Statement 1: Due to the addition of water, the ratio of weight of the sugar in the solution to the weight of the solution decreases from 1/4 to 1/5

Sugar was 25% of previous solution and is 20% of new solution though its amount has not changed. Why? Because water was added.

$$\frac{25}{100} * Total_{Original} = \frac{20}{100} * Total_{New}$$ (Recall C1V1 = C2V2 discussed in replacement)

$$\frac{25}{100} * Total_{Original} = \frac{20}{100} * [Total_{Original} + \frac{40}{100} * Water_{Original}]$$

Here we can easily find $$\frac{Water_{Original}}{Total_{Original}}$$ and that is what we need. Sufficient alone.

Statement 2: After the addition of water, the ratio of the weight of water in the solution to the weight of the solution becomes 7/10

$$\frac{Water_{Original} + \frac{40}{100} * Water_{Original}}{Total_{Original} + \frac{40}{100} * Water_{Original}} = \frac{7}{10}$$­

By cross multiplication, we can easily find $$\frac{Water_{Original}}{Total_{Original}}$$ and that is what we need. Sufficient alone.

VP
Joined: 29 Oct 2015
Posts: 1204
Own Kudos [?]: 542 [0]
Given Kudos: 737
GMAT 1: 570 Q42 V28
Re: The amount of water in a salt water sugar solution is increased by 2/5 [#permalink]
Thank you KarishmaB maa'm
KarishmaB

PyjamaScientist
The amount of water in a salt water sugar solution is increased by $$\frac{2}{5 }$$. Before the increase, what was the ratio of the weight of water to the weight of the solution?

S1) Due to the addition of water, the ratio of weight of the sugar in the solution to the weight of the solution decreases from $$\frac{1}{4}$$ to $$\frac{1}{5}$$.

S2) After the addition of water, the ratio of the weight of water in the solution to the weight of the solution becomes $$\frac{7}{10}$$.
­Given: In a water salt sugar solution, quantity of water is increased by 40%.
Question: Concentration of water in the solution before the increase?

Statement 1: Due to the addition of water, the ratio of weight of the sugar in the solution to the weight of the solution decreases from 1/4 to 1/5

Sugar was 25% of previous solution and is 20% of new solution though its amount has not changed. Why? Because water was added.

$$\frac{25}{100} * Total_{Original} = \frac{20}{100} * Total_{New}$$ (Recall C1V1 = C2V2 discussed in replacement)

$$\frac{25}{100} * Total_{Original} = \frac{20}{100} * [Total_{Original} + \frac{40}{100} * Water_{Original}]$$

Here we can easily find $$\frac{Water_{Original}}{Total_{Original}}$$ and that is what we need. Sufficient alone.

Statement 2: After the addition of water, the ratio of the weight of water in the solution to the weight of the solution becomes 7/10

$$\frac{Water_{Original} + \frac{40}{100} * Water_{Original}}{Total_{Original} + \frac{40}{100} * Water_{Original}} = \frac{7}{10}$$­

By cross multiplication, we can easily find $$\frac{Water_{Original}}{Total_{Original}}$$ and that is what we need. Sufficient alone.