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12 Days of Christmas GMAT Competition 2025 - Day 1: A painter mixes

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Bunuel

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Bunuel

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GMAT Club Official Explanation:

A painter mixes two batches of paint to create a custom color. Batch A has a gloss finish concentration of G% by volume. Batch B has a gloss finish concentration of H% by volume. The painter combines M liters of Batch A with N liters of Batch B. What is the gloss finish percentage, by volume, of the resulting mixture?

(1) The ratio of G to H is 7 to 5.

We know G:H = 7:5, so G = 7k and H = 5k. The resulting mixture percentage still depends on the actual value of k and also on how much of each batch is used. Not sufficient.

(2) The ratio of M to N is 3 to 4.

We know M:N = 3:4. The resulting mixture percentage would be (3G + 4H)/7, which still depends on the actual values of G and H. Not sufficient.

(1)+(2) Substitute G = 7k and H = 5k into (3G + 4H)/7. We get (3 * 7k + 4 * 5k)/7 = 41k/7. Different values of k give different mixture percentages. Not sufficient.

Answer: E.

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15 Dec 2025, 09:25

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volume will be

M/(M + N) = 3/7
N/(M + N) = 4/7
final gloss fininsh will be
3/7 * G + 4/7 * H

can't solve for this since we don;t have actual value.
hence E

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Final % = (G*M + H*N )/ (M+N) = ?

S1
G:H = 7:5
G= 7H/5
We still have H,M and N unknown
Insufficient

S2
M:N = 3/4
Let M = 3k, N = 4k
=(G*3k + H * 4k)/(3k + 4k)
= (3G+4H)/7
We still have G and H unknown
Insufficient

Combined, we have G = 7H/5 and Final % = (3G + 4H)/7
Final % = (21H/5 +4H)/7
H is still unknown
Combined as well, they are insufficient

Answer E

Bunuel

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Batch A
Volume = M liters
Gloss Finish = G % of M liters = G*M/100 liters

Batch B
Volume = N liters
Gloss Finish = H% of N liters = H*N/100 liters

Resultant Mixture
Volume = M+N liters
Gloss Finish = G*M/100 + H*N/100 liters

So, Gloss Finish by volume for resultant mixture = (G*M + H*N)/100(M+N)
So, Gloss Finish % by volume for resultant mixture = (G*M + H*N)/(M+N) ------- (a)

Considering only statement (1), we won't be able to compute the value of (a)
Considering only statement (2), we again won't be able to compute the value of (a)
Considering both statements together, we still need explicit values of one of G/H and without them we can't compute the value of (a)

Hence, both statements together are not sufficient and additional data is needed.

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Updated on: 17 Dec 2025, 01:24

Originally posted by MANASH94 on 15 Dec 2025, 09:43.
Last edited by MANASH94 on 17 Dec 2025, 01:24, edited 1 time in total.

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The Mixture % of gloss should be:

(MG+NH) / (M + N)
We will need to know all the four parameters.
Statement 1 Gives only G and H values and we don't know how much paint is used from which quantity: Insufficient.
Statement 2: Gives only M and N values and we don't know what percentage gloss concentration was available: Insufficient
Statement 1 and 2 together:
We still know only the ratios of G, H, M, and N, not their absolute values.
Substituting ratios into (MG+NH)/(M+N)(MG + NH)/(M + N)(MG+NH)/(M+N) leaves the result dependent on an unknown variable, so the mixture percentage is not uniquely determined.
Hence, E.

Bunuel

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M liters of Batch A: -
Gloss finish = G%*M = MG/100 liters

N liters of Batch B: -
Gloss finish = H%*N = HN/100 liters

Resultant mixtures (M+N) liters: -
Gloss finish = (MG+HN)/100

Gloss finish percentage = (MG+HN)/(M+N) %

(1) The ratio of G to H is 7 to 5
G = 7k ; H = 5k

Gloss finish percentage = (7Mk + 5Nk)/(M+N) %
The information is not sufficient to calculate gloss finish percentage.
Not sufficient

(2) The ratio of M to N is 3 to 4
M = 3l ; N = 4l
Gloss finish percentage = (3lG + 4lH)/(3l + 4l) % = (3G+4H)/7 %

Since G & H are unknown
Not sufficient

(1) + (2)
(1) The ratio of G to H is 7 to 5
(2) The ratio of M to N is 3 to 4
G = 7k; H = 5k
M = 3l; N = 4l

Gloss finish percentage = (MG+HN)/(M+N) % = (3l*7k + 5k*4l)/(3l+4l) = (21+20)k/7 % = 41k/7 %
Since k is unknown

Not sufficient

IMO E

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15 Dec 2025, 09:45

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Statement 1 and 2 even together are not sufficient hence the gloss finish percentage cannot be uniquely determined

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since we need to find the gloss % for the resulting mixture we need both the statement to find the solution

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Gloss finish percentage of gloss mixture = (MG+NH)/(M+N)

(1) G:H = 7:5
Only gives relative comaprison and not actual values
Insufficient

(2) M:N =3:4
It doesn’t tell us about how gloss mixture is, only info about how much paint is mixed.
Insufficient

(1)&(2)
Let, G=7a & H=5a and M=3x and N=4x
Then, (MG+NH)/(M+N) = ((3x)(7a)+(4x)(5a))/(3x+4x) = 41a/7
We still don’t know value of a
Insufficient

E

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Deconstructing the Question
We are mixing two batches of paint:
Batch A: Volume $M$, Concentration $G\%$.
Batch B: Volume $N$, Concentration $H\%$.

Theory: Weighted Average / Mixture Formula
The resulting concentration ($X$) is given by:
$X = \frac{\text{Total Gloss Volume}}{\text{Total Mixture Volume}} = \frac{M \cdot G + N \cdot H}{M + N}$

Target: Find the unique value of $X$.

Analyze Statement (1)
"The ratio of G to H is 7 to 5."
$\frac{G}{H} = \frac{7}{5} \implies G = 1.4H$.
Substitute into the formula:
$X = \frac{M(1.4H) + N(H)}{M + N} = H \cdot \frac{1.4M + N}{M + N}$
The result depends on the unknown values of $H$, $M$, and $N$.
If $H$ doubles, the final percentage doubles. We cannot find a specific value.
INSUFFICIENT

Analyze Statement (2)
"The ratio of M to N is 3 to 4."
$\frac{M}{N} = \frac{3}{4}$. Let $M = 3k$ and $N = 4k$.
Substitute into the formula:
$X = \frac{3k \cdot G + 4k \cdot H}{3k + 4k} = \frac{k(3G + 4H)}{7k} = \frac{3G + 4H}{7}$
This gives us the weights, but we do not know the values of the concentrations $G$ and $H$.
The mixture could be low gloss or high gloss depending on $G$ and $H$. INSUFFICIENT

Combine Statements (1) and (2)
From (1): $G = 1.4H$.
From (2): $M = 3k, N = 4k$.
Substitute both into the mixture equation:
$X = \frac{3k(1.4H) + 4k(H)}{7k}$
Cancel $k$:
$X = \frac{4.2H + 4H}{7} = \frac{8.2H}{7} \approx 1.17H$

Critical Insight: The final percentage is still a function of $H$.
Without an absolute value for at least one concentration (e.g., "H is 10%"), we cannot calculate the specific gloss percentage of the mixture.
INSUFFICIENT

Answer: E

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15 Dec 2025, 09:54

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the statements are not sufficienat

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the percentage of the resulting mixture will be 41% volume.

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In the final mixture: G% of M : H% of N => GM/HN required to be found out
clearly A alone or B alone are not sufficient but combining it's sufficient.
hence C.

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Bunuel

Volume of glossy paint = MG/100 + NH /100 = (MG+NH)/100

Percentage = { (MG + NH)/100 }/(M+N)

Statement 1

G/H = 7/5

We know the ratio of G and H so we can represent G in terms of H or vice versa. However we can't cancel the terms in numerators and the term in denominator to arrive at a final answer. Hence the statement alone is not sufficient.

Statement 2

Like statement 1, we can represent M in terms of N, or vice versa but we can't cancel G and H from the numerator. Hence not sufficient.

(1) + (2)

From 1, we can represnet G in terms of H, and from 2 we can represent M in terms of N. However we can't cancel H and N. Hence together as well the statements the statements are not sufficient.

Option E

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15 Dec 2025, 10:01

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bringing the numerator & denominator into a single common term helps us giving a value for gloss finish percentage in the resulting mixture,

the gloss finish for Batch A with M liters is G*M/100

the gloss finish for batch B with N liters is H*N/100

total gloss finish of the mixture is given by (GM+HN)/100(M+N)*100

the above algebraic equation can be solved by considering G=(7/5)H & M=(3/4)N

which gives us the percentage value that is required as asked in the question

Bunuel

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(1) alone
Let's say batch A has a gloss finish g%=7x and batch B has h%=5x.
You don't have any measure of how much of each batch is in the final mixture, so you can't calculate its gloss finish %.

(2) alone
You don't have any measure of how much gloss finish percentage is in each batch.

(1) and (2) together
You can find that Mixture% = [M(7k) + N(5k)] / (M+N).
But you still don't know the actual measures of g% and h%, so you can't find a unique value for the mixture %.

E) Neither statement alone is sufficient.

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15 Dec 2025, 10:08

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Gloss % in mixture x % = (m *G% + n*H%)/(m+n)
m/n =3/4
then dividing the eq, by n , we can replace m/n by 3/4
But with out knowing G and H, we cant calculate the x
G = 7 % and H = 5%, gives us different gloss percentage as compared to G = 14% and H = 10% even though their ratio is same.

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15 Dec 2025, 10:16

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$Final glass%=\frac{MG+NH}{M+N}$ --Eq.1

(A) G/H=7x/5x => But no info. about M & N. Thus, It's Insufficient.
(B) M/N=3Y/4Y => But no info. about G&H. Thus, It's Insufficient.

Now. Option A&B together gives, G=7x,H=5x,M=3y &N=4y.
Thus after putting all values in the Eq.1, we get a defined gloss finish percentage.

Therefore, Option C is the correct answer.

Bunuel

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15 Dec 2025, 10:26

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batch A - gloss quantity = GM/100
batch B- gloss quantity = BN/100
final mixture concentration= GM+BN/(M+N)
need G,M,B,N

statement1- G=7p, H= 5p
not sufficient. 7pm + 5pn / (m+n)
still unknown
statement 2= M=3x , N= 4x
not sufficient= 3gx + 4bx / 7

combine 1 & 2 - 21px + 20px / 7x
41p/7
nothing about p. it could be anything.

so ans = E

Bunuel