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

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During a certain week, a seal ate 50 percent of the first 80 smelt it [#permalink]

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New post 12 Jul 2018, 01:14
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During a certain week, a seal ate 50 percent of the first 80 smelt it came across, and 30 percent of the remaining smelt it came across. If the seal ate 40 percent of the smelt it came across during the entire week, how many smelt did it eat?


A. 24

B. 32

C. 55

D. 64

E. 80

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During a certain week, a seal ate 50 percent of the first 80 smelt it [#permalink]

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New post Updated on: 12 Jul 2018, 06:07
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.
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Originally posted by workout on 12 Jul 2018, 02:32.
Last edited by workout on 12 Jul 2018, 06:07, edited 4 times in total.
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During a certain week, a seal ate 50 percent of the first 80 smelt it [#permalink]

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New post Updated on: 12 Jul 2018, 06:55
Bunuel wrote:
During a certain week, a seal ate 50 percent of the first 80 smelt it came across, and 30 percent of the remaining smelt it came across. If the seal ate 40 percent of the smelt it came across during the entire week, how many smelt did it eat?


A. 24

B. 32

C. 55

D. 64

E. 80



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

Originally posted by CounterSniper on 12 Jul 2018, 05:04.
Last edited by CounterSniper on 12 Jul 2018, 06:55, edited 2 times in total.
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During a certain week, a seal ate 50 percent of the first 80 smelt it [#permalink]

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New post 12 Jul 2018, 05:28
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CounterSniper wrote:
it ate 40% of the total
0.4(x+40)=40+0.3x



CounterSniper

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

0.4(x+80)=40+0.3x
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Re: During a certain week, a seal ate 50 percent of the first 80 smelt it [#permalink]

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New post 12 Jul 2018, 05:32
CounterSniper wrote:
let x+40 be the total no of smelts


Also, here the total number of smelts is x+80 not x+40
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Re: During a certain week, a seal ate 50 percent of the first 80 smelt it [#permalink]

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New post 12 Jul 2018, 05:48
1

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
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Re: During a certain week, a seal ate 50 percent of the first 80 smelt it [#permalink]

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New post 12 Jul 2018, 05:50
Hey workout,

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

Please check once.
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Re: During a certain week, a seal ate 50 percent of the first 80 smelt it [#permalink]

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New post 12 Jul 2018, 05:54
workout wrote:
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 \(\frac{50}{100} * 80 + \frac{30}{100} * x\)

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

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

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

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

=> \(x = 80\)

Hence option E


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
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During a certain week, a seal ate 50 percent of the first 80 smelt it [#permalink]

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New post 12 Jul 2018, 06:36
Bunuel wrote:
During a certain week, a seal ate 50 percent of the first 80 smelt it came across, and 30 percent of the remaining smelt it came across. If the seal ate 40 percent of the smelt it came across during the entire week, how many smelt did it eat?


A. 24

B. 32

C. 55

D. 64

E. 80


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.
During a certain week, a seal ate 50 percent of the first 80 smelt it   [#permalink] 12 Jul 2018, 06:36
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