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A group of store managers must assemble 280 displays for an upcoming

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Bunuel
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A group of store managers must assemble 280 displays for an upcoming [#permalink] New post   13 Mar 2014, 02:27
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A group of store managers must assemble 280 displays for an upcoming sale. If they assemble 25 percent of the displays during the first hour and 40 percent of the remaining displays during the second hour, how many of the displays will not have been assembled by the end of the second hour?

(A) 70
(B) 98
(C) 126
(D) 168
(E) 182
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C
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Bunuel
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Re: A group of store managers must assemble 280 displays for an upcoming [#permalink] New post   13 Mar 2014, 02:27
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SOLUTION

A group of store managers must assemble 280 displays for an upcoming sale. If they assemble 25 percent of the displays during the first hour and 40 percent of the remaining displays during the second hour, how many of the displays will not have been assembled by the end of the second hour?

(A) 70
(B) 98
(C) 126
(D) 168
(E) 182

During the first hour 0.25*280 = 70 displays were assembled.
During the second hour 0.4*(280-70) = 84 displays were assembled.

Thus, 280 - (70+84) = 126 displays had not been assembled by the end of the second hour.

Answer: C.
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Re: A group of store managers must assemble 280 displays for an upcoming [#permalink] New post   13 Mar 2014, 19:20
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Answer = (C) 126

In first hour, 25% of 280 = 70

Remaining = 210

40 percent of the remaining displays during the second hour, how many of the displays will not have been assembled by the end of the second hour?

Means 60% have not been assembled

So 60% of 210 = 126
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Re: A group of store managers must assemble 280 displays for an upcoming [#permalink] New post   20 May 2014, 13:35
Total Displays:280

Remaining after first hr: 280*75/100

Remaining after 2nd hr: 280*75/100*60/100 = 126

280*75/100*60/100 = 126
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Re: A group of store managers must assemble 280 displays for an upcoming [#permalink] New post   20 May 2014, 23:53
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Total number of displays to be assembled = 280

After one hour , 25% were assembled.

Remaining (75%) = 75% of 280 = 210

40% of remaining displays assembled in second hour. Remaining 60% = 60% of 210 = 126

Hence C.
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Re: A group of store managers must assemble 280 displays for an upcoming [#permalink] New post   17 Jul 2014, 21:17
Another way to solve this is 25% + 30% (40% of 100-25) = 55% of displays were assembled. So number of displays that were not assembled = 45% of 280 = 126.
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Re: A group of store managers must assemble 280 displays for an upcoming [#permalink] New post   29 Jul 2016, 10:51
displays assembled in the first hour => 1/4*280=70
in the second hour => 40/100*210=> 84
Displays not assembled => 280-(70+84)=> 126
Smash C
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Re: A group of store managers must assemble 280 displays for an upcoming [#permalink] New post   Updated on: 18 Dec 2019, 14:46
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The alternate method could be


In first hour they did 25 % of 280, So remaining is 75%. and
in second hour they did 40% of 75% implies the did 30 % 280
So in two hours they assembled 55% of stuff. and 45 % is yet to be done
So 45% of 280 is what
50 % 280 is 140
10 % is 28 and 5 % 14
thus 45 % would be 140-14= 126
that is 126

Alternatively,
We have in the first hour we have 25 % of 280 displayed
in second hour we have 40% of 75% of 280 displayed .
Remaining displays are 60% 75% of 280
so
\(\frac{60}{100} * \frac{75}{100} * 280\)

this calculation is 126

Originally posted by Probus on 18 Jun 2019, 12:34.
Last edited by Probus on 18 Dec 2019, 14:46, edited 1 time in total.
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Re: A group of store managers must assemble 280 displays for an upcoming [#permalink] New post   21 Jun 2019, 11:39
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Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

A group of store managers must assemble 280 displays for an upcoming sale. If they assemble 25 percent of the displays during the first hour and 40 percent of the remaining displays during the second hour, how many of the displays will not have been assembled by the end of the second hour?

(A) 70
(B) 98
(C) 126
(D) 168
(E) 182

We can create the expression:

0.25 x 280 + 0.4(0.75 x 280)

70 + 0.4(210) = 70 + 84 = 154

Thus, 280 - 154 = 126 are left to be assembled.

Answer: C
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Re: A group of store managers must assemble 280 displays for an upcoming [#permalink] New post   30 Mar 2024, 08:19
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Re: A group of store managers must assemble 280 displays for an upcoming [#permalink]
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