School-Safe Puzzle Games

## Heavy Fish

Big Thanks go out to Shawn for submitting this neat challenge. As many of our regulars know, Shawn owns a ridiculous amount of puzzle solving ability, consistently getting not only Smart-Kit’s hardest word puzzles & riddles, but also the math & pictorial challenges too. He’s been visiting since the very early days of the site, and easily recognizable in the comments section by the helpful answers & explanations that beginners (and sometimes experts too!) can rely upon.

My uncle went fishing and had a very good day, catching several fish.  The heaviest three fish together made up 35% of the total weight of the catch, and he quickly sold them.  The three lightest fish together made up 5/13 of the total weight of the remaining fish.  How many fish did my uncle catch?

Answers to this challenge can be entered into the section below; submissions will automatically be revealed when time is up!

### 12 Comments to “Heavy Fish”

1. Bobo The Bear | PUZZLE MASTER | Profile

I’m coming up with 10 total fish, using the following reasoning:

Heaviest 3 = 35% of total.
Lightest 3 = (5/13)*65% = 25% of total.
Remaining N = 40% of total.
N>3, otherwise those three would be the heaviest (40% > 35%)
N<5, otherwise they would average at most 8% apiece (40/5 = 8), and the lightest three would be at most 24%, making them the lightest (24% < 25%).
Therefore N=4, and there are a total of 10 fish.

Nicely done, Shawn. It's not immediately obvious that there's enough information to solve the problem, but the parameters neatly confine the possibilities to only one outcome. Good puzzlesmithing.

2. Falwan | Profile

total wt.=t
heavy ones=0.35t
light ones=(5/13)*(t-0.35t)=0.25t
medium ones=t-(0.35t+0.25t)=0.4t
weight of each medium one=(0.35t+0.25t)/6=0.1t
no. of medium ones= 0.4t/0.1t=4

total no. of fish = 3+4+3 = 10 fish

3. engjs1960 | Profile

10 fish all up

4. Vik | Profile

if n is avg. wt, x total wt & y nos of fish over 6, then x/n=6+y. Also 6n=.35x+5/13(x-.35x) i.e x/n=10. Therefore 6+y=10 fish?

5. suineg | PUZZLE MASTER | Profile

Ok, let me give you my approach to this problem:
There are 3 heavier fish and 3 lighter fish:
3 heavier fish= 35% of weight
remaining= 65% of weight
Of this 65%, three lighter fish represent 5/13 or 25% of the total weight
the new remainder is 40% for the unknown quantity of fish
For logic the number of remaining fish have to be more than 3 because 40% is greater than 35% that represent the three heavier fish.

You could tackle this problem by average so you could say that in “average” each heavier fish represent 35/3 % and each lighter fish represent 25/3 %, and you know that the middle fish represent 40%.

Now you try with 4 fishes 4 x average percent = 40% — Average percent is 10 that is 30/3 that is between 40\$ and 25%, now could the number of fish be higher?
you try with 5 fish: 40/ 5 is 8% that is lower than 25/3 percent so it is impossible that the average of the middle group was lower that the one of the lighter group therefore the number of fish is

3 heavier + 4 middle + 3 lighter = 10 fish

Cool man!

6. bizarette18 | PUZZLE MASTER | Profile

10

7. Shawn | PUZZLE GRANDMASTER | Profile

Very good, the total number of fish must be 10.

I ran into this puzzle some time ago, and I thought it would be a good challenge for this group. As Bobo commented, I like a puzzle in which the stated facts seem to be inadequate, and you have to search for the missing link.

8. shaks | Profile

I have 12 fish as my total – the 3 lightest fish make up (5/13)*(13/20)of the total weight of the catch – that give me a quarter – so if 3 fish is a quarter of the total, then 12 is the number of fish caught

9. shaks | Profile

I see my mistake, thanks!

10. soumen023 | Profile

3 heaviest = 35/100
3 lightest = 25/100 (5/13 of remaining => 25/65 of remaining)
So,
1 – (35+25)/100 = 40/100 weight remaining
On Average Each fish should weigh less than 1/3 of 35/100 or 12
and also weigh more than 1/3 of 25/100 or 8
Hence 4 fishes should b remaining
So a total of 10 fishes was caught.

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