Without performing any tedious calculation
We know the jug is five litre and the solution is 4ltr then 1.5 litre is removed making the remain solution 2.5litre
To fill the jug full we need 2.5more litre

Adding the same quantity of water will reduce the solution concentration by half
Hence we have 7.5 as the new concentration

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-----------ASIDE------------
IMPORTANT CONCEPT: If EQUAL VOLUMES of 2 solutions are combined, the concentration of the resulting mixture will be the AVERAGE of the 2 solutions.
Example: If 3 liters of a 10% alcohol solution are combined with 3 liters of a 30% alcohol solution, the resulting solution will be 20% alcohol.
Likewise, if 11 liters of 6% alcohol solution are combined with 11 liters of a 7% alcohol solution, the resulting solution will be 6.5% alcohol.
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Once we pour out 1.5 liters, we're left with 2.5 liters of 15% salt solution
Since the jug holds 5 liters, we then add 2.5 liters of water (aka a 0% salt solution)

Since we're combining EQUAL VOLUMES of 15% and 0% solutions, the salt content of the resulting solution will equal the average of 15% and 0%
(15 + 0)/2 = 7.5

So, the resulting solution will be 7.5% salt.

Answer: A

ScottTargetTestPrep
can you pls post your solution

Nsp10

These are easy to do...

Step 1. 5 ltr jug filled with 4 ltr water with 15% salt soln
Step 2. 1.5 lts spilled out of 4 lts, remaining 2.5 lts
Step 3. Normal water added is 2.5, now we have 2.5 lts salt water + 2.5 normal water in total 5 lts.
Step 4. In 5lts we know 15% salt soln we have half normal water , so now salt concentration is half of 15= 7.5%.

OA . A

We have a 5 -liter jug that initially contains 4 liters of a saltwater solution. This solution is $15 \%$ salt. Then, 1.5 liters of this solution spills out, leaving us with some amount of the original solution in the jug. After that, we fill the jug back to its full capacity (which is 5 liters) by adding water. We need to find out what percentage of the final solution is salt.

Breaking It Down

1. Initial Solution:
- Total volume: 4 liters
- Salt concentration: 15\%
- Amount of salt: $15 \%$ of 4 liters $=0.15 * 4=0.6$ liters

2. After Spilling:
- Amount spilled: 1.5 liters
- The spilled solution has the same concentration as the original, so it's also 15\% salt.
- Amount of salt spilled: $15 \%$ of 1.5 liters $=0.15 * 1.5=0.225$ liters
- Remaining solution in the jug:
- Original volume: 4 liters
- Spilled: 1.5 liters
- Remaining: 4-1.5 = 2.5 liters
- Remaining salt:
- Original salt: 0.6 liters
- Salt spilled: 0.225 liters
- Remaining salt: 0.6-0.225 = 0.375 liters

3. Filling the Jug Back to Capacity:
- Current volume in jug: 2.5 liters
- Jug capacity: 5 liters
- Water to add: 5-2.5 = 2.5 liters
- We're adding pure water, so no additional salt is added.
- Total salt remains: 0.375 liters
- Total volume now: 5 liters

4. Final Salt Concentration:
- Salt: 0.375 liters
- Total solution: 5 liters
- Percentage of salt: (0.375 / 5) * 100%
- Calculate: 0.375 / $5=0.075$
- 0.075 * 100% = 7.5%

Comparing with the Options

The final salt concentration is $7.5 %$. Let's see how this matches with the given options:
(A) $71 / 2 %=7.5 %$
(B) $93 / 8 %=9.375 %$
(C) $101 / 2 %=10.5 %$
(D) $12 %$
(E) $15 %$

Our calculated percentage is exactly $7.5 %$, which corresponds to option (A).

great explanation