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07 Feb 2026, 02:31
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C
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Difficulty:
45%
(medium)
Question Stats:
68% (02:40) correct
32%
(03:16)
wrong
based on 38
sessions
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A jar contains only 2-euro coins, each weighing 8.50 grams, and 0.5-euro coins, each weighing 7.80 grams.The jar contains x coins whose average weight is 8.00 grams. How many euros are in the jar?
(A) x/2
(B) 2x/3
(C) 13x/14
(D) x
(E) 5x/4
(A) x/2
(B) 2x/3
(C) 13x/14
(D) x
(E) 5x/4
07 Feb 2026, 05:36
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Let n be the number of 2-euro coins and m be the number of 0.5-euro coins,
where n+m=x.
The total weight equation is 8.5n+7.8m=8x.
Solution
Substitute m=x−n into the weight equation 8.5n+7.8(x−n)=8x.
Simplify:
8.5n+7.8x−7.8n=8x,
so, 0.7n=0.2x, hence
n = \(\frac{0.2}{0.7}\) x = \(\frac{2}{7}\) x.
Then
m = x − \(\frac{2}{7}\) x = \(\frac{5}{7}\) x
Total Euros Value = 2n+0.5m= 2 * \(\frac{2}{7}\) x + 0.5 * \(\frac{5}{7}\) x = \(\frac{4}{7}\) x * \(\frac{2.5}{7}\) x = \(\frac{6.5}{7}\) x = \(\frac{13}{14}\) x.
ANS = C
where n+m=x.
The total weight equation is 8.5n+7.8m=8x.
Solution
Substitute m=x−n into the weight equation 8.5n+7.8(x−n)=8x.
Simplify:
8.5n+7.8x−7.8n=8x,
so, 0.7n=0.2x, hence
n = \(\frac{0.2}{0.7}\) x = \(\frac{2}{7}\) x.
Then
m = x − \(\frac{2}{7}\) x = \(\frac{5}{7}\) x
Total Euros Value = 2n+0.5m= 2 * \(\frac{2}{7}\) x + 0.5 * \(\frac{5}{7}\) x = \(\frac{4}{7}\) x * \(\frac{2.5}{7}\) x = \(\frac{6.5}{7}\) x = \(\frac{13}{14}\) x.
ANS = C
07 Feb 2026, 08:37
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This is a clean weighted average problem disguised as a coin question. The key is recognizing that you're not actually solving for the number of coins—you're solving for the value in euros.
Here's the systematic approach:
Step 1: Set up what we know
Let's say there are:
- a = number of 2-euro coins (weight: 8.50g each)
- b = number of 0.5-euro coins (weight: 7.80g each)
- Total coins: x (so a + b = x)
- Average weight: 8.00g
Step 2: Write the weighted average equation
(8.50a + 7.80b) / x = 8.00
Since a + b = x, we can substitute:
8.50a + 7.80b = 8.00(a + b)
8.50a + 7.80b = 8.00a + 8.00b
Step 3: Simplify
8.50a - 8.00a = 8.00b - 7.80b
0.50a = 0.20b
a/b = 0.20/0.50 = 2/5
So for every 2 of the 2-euro coins, there are 5 of the 0.5-euro coins.
Step 4: Calculate the euro value
Value from 2-euro coins: 2 × 2 = 4 euros
Value from 0.5-euro coins: 5 × 0.5 = 2.5 euros
Total value: 4 + 2.5 = 6.5 euros
But wait—this is for the ratio 2:5. The question asks for the total in terms of x coins.
If we have x coins total, with ratio 2:5, that means:
- 2-euro coins: (2/7)x
- 0.5-euro coins: (5/7)x
Total euros = 2 × (2/7)x + 0.5 × (5/7)x
= (4/7)x + (2.5/7)x
= (6.5/7)x
= (13/14)x
The answer is (C) 13x/14.
Common trap: Students sometimes forget to convert from "number of coins" to "euro value." The 2-euro coins are heavier and worth more, so you need both the ratio and the value calculation.
Takeaway: In weighted average problems, always track what you're averaging (here: weight) separately from what you're solving for (here: value).
Free gamified GMAT prep: edskore.com | 725 GMAT Focus
Here's the systematic approach:
Step 1: Set up what we know
Let's say there are:
- a = number of 2-euro coins (weight: 8.50g each)
- b = number of 0.5-euro coins (weight: 7.80g each)
- Total coins: x (so a + b = x)
- Average weight: 8.00g
Step 2: Write the weighted average equation
(8.50a + 7.80b) / x = 8.00
Since a + b = x, we can substitute:
8.50a + 7.80b = 8.00(a + b)
8.50a + 7.80b = 8.00a + 8.00b
Step 3: Simplify
8.50a - 8.00a = 8.00b - 7.80b
0.50a = 0.20b
a/b = 0.20/0.50 = 2/5
So for every 2 of the 2-euro coins, there are 5 of the 0.5-euro coins.
Step 4: Calculate the euro value
Value from 2-euro coins: 2 × 2 = 4 euros
Value from 0.5-euro coins: 5 × 0.5 = 2.5 euros
Total value: 4 + 2.5 = 6.5 euros
But wait—this is for the ratio 2:5. The question asks for the total in terms of x coins.
If we have x coins total, with ratio 2:5, that means:
- 2-euro coins: (2/7)x
- 0.5-euro coins: (5/7)x
Total euros = 2 × (2/7)x + 0.5 × (5/7)x
= (4/7)x + (2.5/7)x
= (6.5/7)x
= (13/14)x
The answer is (C) 13x/14.
Common trap: Students sometimes forget to convert from "number of coins" to "euro value." The 2-euro coins are heavier and worth more, so you need both the ratio and the value calculation.
Takeaway: In weighted average problems, always track what you're averaging (here: weight) separately from what you're solving for (here: value).
Free gamified GMAT prep: edskore.com | 725 GMAT Focus
