During a certain week, a seal ate 50 percent of the first 80 smelt it

The number of smelt is some particular number you cannot assume that it's 100.

Hi,

I am getting a different answer. My approach:
1. First Part - 80 smelts - 50% eaten = 40 ....... Statement 1
2. 2nd Part - x smelts - 30% eaten = 0.3x ....... Statement 2

Total Smelts = 1st part + 2nd part = 80+x
Seal ate 40% of total smelts => 0.4 (80+x) ....... Statement 3

Therefore , combining Statement 1,2,3
40 + 0.3x = 0.4(80+x) . This is giving x = 80 and 32 as the answer.

Kindly assist where I have gone wrong. Thanks

We are told that the seal ate 30% of the remaining smelt it came across. Thus x in your solution must be the number of the remaining smelt. Which makes the total number of smelt equal to 80 + x = 160. 40% of 160 is 64.

Does this make sense?

Yes it clears out my confusion
Thanks Bunuel...

Responding to a pm:

The seal ate 50% of first portion and 30% of second portion. Overall (average), it ate 40% of the entire portion.

Using scale method, Second portion/First portion = (50 - 40)/(40 - 30) = 1/1

So both portions need to be equal i.e. of 80 smelts each. So both portions together had 160 smelts and the seal ate 40% of it i.e. (40/100)*160 = 64 smelts

Let total no of smelts be X.
eaten smalts:- 50% of first 80==40 && 30 % of remaining (X-80)==30(X-80)/100
total 40% of X
40+ 30(X-80)/100=40X/100
X=160.
so eaten smelts= 40*160/100==64
Ans D.

First the seal ate 50% of 80 i.e 40 smelt.

Suppose the number of remaining smelts is x. Seal then ate .30x of these remaining smelts.

Total smelts are 80+ x and seal ate 40% of 80+x

.4(80+x)= 40+.3x
x= 80

Total smelts eaten are .4(80+80)= 64

D is the answer

Hi,

I did the same thing and I think what you have to do is assume the remaining smelts as x. Thus, x will be 80 and 30% of the remaining smelts will be 24 which will give total as 64. Hope its clear.

(.5) 80 + .3(x-80) =.4x

x= 160

.5(80) + .3(80)= 64

Thus
"D"

A seal ate 50 percent of the first 80 smelt it came across => \(\frac{50}{100} * 80\)

Let x be the remaining number of smelt the seal came across

Total number is smelt the seal ate is \(\frac{50}{100} * 80 + \frac{30}{100} * x\)

=> \(\frac{50}{100} * 80 + \frac{30}{100} * x = \frac{40}{100} * (80 + x)\)

=> \((50 * 80) + 30x = 40 * (80 + x)\)

=> \(4000 + 30x = 40x + 3200\)

=> \(10x = 4000 - 3200\)

=> \(x = 80\)

UPDATE :

At first I solved only for 'x' and chose option E but the question actually asks for the number of smelt the seal ate which is

\(\frac{50}{100}* 80 + \frac{30}{100}* 80 = 64\)

Hence option D Thanks EgmatQuantExpert for the correct solution


Important learning for me

After solving the variables, do explicitly check what the question has asked for. Don't just stop after calculating the variables you have assumed and pick an answer.

let x+40 be the total no of smelts
50% of the first 80 smelt = 40
remaining = x
30% of the remaining =0.3x
it ate 40% of the total
0.4(x+40)=40+0.3x
0.4x+16=40+0.3x
0.1x=24
x=240
total no of smelts = 280
it ate 40% of 280
=112

help me out here !!


Corrected above !!

let x+40 be the total no of smelts
50% of the first 80 smelt = 40

remaining would be (x+40)-80=x-40
so the equation becomes
0.4(x+40)=40+0.3(x-40)
0.4x+16=40+0.3x-12
x=120

total no of smelts =120+40=160
it ate 40% of 160 = 64

D

CounterSniper

it ate 40% of the total, so the equation becomes

0.4(x+80)=40+0.3x

Also, here the total number of smelts is x+80 not x+40

Solution

Given:

• In a certain week, a seal ate 50 percent of the first 80 smelt it came across.
• The seal ate 30% of remaining seal it came across.
• In total, seal ate 40 percent of the smelt it came across

To find:

• How many smelts did seal eat in the week?

Approach and Working:
Let us assume that seal came across x smelts after eating 50% of first 80 smelts it came across.
• 50% of 80+ 30% of x= 40% of (x+80)
• 40+0.3x= 0.4x+32
• x=80

Hence, seal ate 40+ 24= 64

Hence, the correct answer is option D.

Answer: D

Hey workout,

I think you found x but did not calculate the number of smelts the seat ate.

Please check once.

It looks like I solved for the wrong thing. The question asks for the number of smelts the seal ate which is

\(\frac{50}{100}* 80 + \frac{30}{100}* 80 = 64\)

Hence option D Thanks EgmatQuantExpert for the correct solution

let assume the no. of smelt be x.

Case 1:

80*50% = 40

Case 2:

Remaining smelt :

(x -80)

So, the no. of smelt ate at second phase :

(x - 80)*30%

3(x - 80) / 10

Case 3: we can form an equation based on the information that seal ate a total of 40% smelt.

40 + 3(x -80) / 10 = x * 40%

(400 + 3x - 240) / 10 = 2x / 5

x = 160.

No of smelt ate : 40 + ( 160- 80) 30% = 40 + 24 = 64.

Thus the best answer is D.