Problems 151 and 226 in the OG 13

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Problems 151 and 226 in the OG 13

by didieravoaka » Mon Jun 01, 2015 4:49 pm
Hi guys,

Does anyone can help me for the problems 151 and 226 in the official guide 13th edition?

Problem 151: I don't understand why we multiply 3/4 by 4/5 and then why we add 1 to 3/5. Another way to solve this problem will help me.
Problem 226: I don't understand why we choose answer D.

Thank you.

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by [email protected] » Mon Jun 01, 2015 5:03 pm
Problem 151
At a loading dock, each worker on the night crew loaded 3/4 as many boxes as each worker on the day crew. If the night crew has 4/5 as many workers as the day crew, what fraction of all the boxes loaded by the two crews did the day crew load?


A. 1/2
B. 2/5
C. 3/5
D. 4/5
E. 5/8
I think the fast way is to plug in some nice numbers.
Since the two pieces of information regarding the night shift are related to the information regarding the day shift, let's assign some nice values to the DAY shift.

Number of workers

Day shift: 5 workers (this is an easy number to find 4/5 of)
Night shift: 4 workers (4/5 of 5 = 4)

Boxes loaded per worker
Day shift: 4 boxes per worker
Night shift: 3 boxes per worker (3/4 of 4 = 3)

Total boxes loaded
Day shift: 5 workers times 4 boxes per worker = 20 boxes
Night shift: 4 workers times 3 boxes per worker = 12 boxes

Combined total boxes for both shifts = 20 + 12 = 32

Of the 32 boxes, the day shift loaded 20 of them.
20/32 = [spoiler]5/8[/spoiler]

Answer: E

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by GMATGuruNY » Mon Jun 01, 2015 6:54 pm
A straight pipe 1 yard in length was marked off in
fourths and also in thirds. If the pipe was then cut into
separate pieces at each of these markings, which of
the following gives all the different lengths of the
pieces, in fractions of a yard?

(A) 1/6 and 1/4 only
(B) 1/4 and 1/3 only
(C) 1/6, 1/4 and 1/3
(D) 1/12, 1/6, and 1/4
(E) 1/12, 1/6, and 1/3
Ignore the given length of 1 yard. The problem can be solved using any length.
Let the pipe = 12 yards.
Dividing 12 into 3rds will yield markings at 4 and 8.
Dividing 12 into 4ths will yield markings at 3, 6, and 9.

Listing the markings in order, we get:
0......3..4....6....8..9......12

There are only 3 distinct differences between markings: 1 yards, 2 yards, and 3 yards.
Phrasing these differences as fractions, we get:
1/12.
2/12 = 1/6.
3/12 = 1/4.

The correct answer is D.

Check here for a similar problem:

https://www.beatthegmat.com/700-question-t87924.html
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by nikhilgmat31 » Mon Jun 01, 2015 11:58 pm
yes put the good numbers to solve it fast.

day workers load = 24 box
night worker loads = 3/4 * 24 = 18

number of day workers = 10
number of night workers = 4/5 * 10 = 8

Day work done = 24 * 10 = 240
nigh work done = 18 * 8 = 144

total = 240 + 144 = 384

day ratio = 240/384 = 5/8 Enjoy

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by nikhilgmat31 » Tue Jun 02, 2015 12:03 am
selecting 12 is good way of solving problem

1/4 cuts at = 0,3, 6, 9,12
1/3 cuts at = 0,4,8,12

so 0,3,4,6,8,9,12

different lengths -> 3,1,2 i.e. 3/12,1/12, 2/12 = 1/4, 1/12,1/6

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by GMATinsight » Tue Jun 02, 2015 2:37 am
A straight pipe 1 yard in length was marked off in
fourths and also in thirds. If the pipe was then cut into
separate pieces at each of these markings, which of
the following gives all the different lengths of the
pieces, in fractions of a yard?

(A) 1/6 and 1/4 only
(B) 1/4 and 1/3 only
(C) 1/6, 1/4 and 1/3
(D) 1/12, 1/6, and 1/4
(E) 1/12, 1/6, and 1/3
Lets take the 1/4 = 0.25 and 1/3 = 0.33

The pipe is cut at

----(0.25 mark)---(0.33 mark)-----(0.50 mark)-----(0.66 mark)---(0.75 mark)---


Now the first piece is of length = 0.25 = 1/4
Now the Second piece is of length = 0.33-0.25 = (1/3) - (1/4) = 1/12
Now the Second piece is of length = 0.50-0.33 = (1/2) - (1/3) = 1/6

Hence 1/4, 1/12 and 1/6

Answer: Option D
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by [email protected] » Tue Jun 02, 2015 7:22 am
At a loading dock, each worker on the night crew loaded 3/4 as many boxes as each worker on the day crew. If the night crew has 4/5 as many workers as the day crew, what fraction of all the boxes loaded by the two crews did the day crew load?


A. 1/2
B. 2/5
C. 3/5
D. 4/5
E. 5/8
Here's an algebraic solution:

Number of workers
Day shift: Let W = the # of day shift workers
Night shift: There are 4W/5 workers (since 4/5 of W = 4W/5)

Boxes loaded per worker
Day shift: Let B = boxes per worker on day shift
Night shift: So, 3B/4 = boxes per worker on night shift

Total boxes loaded
Day shift: W workers times B boxes per worker = WB boxes
Night shift: 4W/5 workers times 3B/4 boxes per worker = 12WB/20 boxes. Simplify to get a total of 3WB/5 boxes.

Combined total boxes for both shifts = WB + 3WB/5 = 8WB/5

Of the 8WB/5 boxes, the day shift loaded WB of them.
WB/(8WB/5) = 5/8 = E

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by j_shreyans » Tue Jun 02, 2015 8:15 am
Hi Mitch ,

Can you please explain your below step, I didn't get it.

There are only 3 distinct differences between markings: 1 yards, 2 yards, and 3 yards.
Phrasing these differences as fractions, we get:
1/12.
2/12 = 1/6.
3/12 = 1/4.



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by GMATinsight » Tue Jun 02, 2015 8:31 am
j_shreyans wrote:Hi Mitch ,

Can you please explain your below step, I didn't get it.

There are only 3 distinct differences between markings: 1 yards, 2 yards, and 3 yards.
Phrasing these differences as fractions, we get:
1/12.
2/12 = 1/6.
3/12 = 1/4.



Thanks,

The differences of 1 yards, 2 yards, and 3 yards have been taken on the base value of 12 yard (assumed total length of stick)

hence 1 yard in 12 yard represents 1/12 of 1 yard
hence 2 yard in 12 yard represents 2/12 = 1/6 of 1 yard
hence 3 yard in 12 yard represents 3/12 = 1/4 of 1 yard
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by GMATGuruNY » Tue Jun 02, 2015 8:57 am
j_shreyans wrote:Hi Mitch ,

Can you please explain your below step, I didn't get it.

There are only 3 distinct differences between markings: 1 yards, 2 yards, and 3 yards.
Phrasing these differences as fractions, we get:
1/12.
2/12 = 1/6.
3/12 = 1/4.



Thanks,
The markings are as follows:
0......3..4....6....8..9......12.

When the pipe is cut into pieces, we get the following lengths:
0......3 --> 3 yards
3..4 ----> 1 yard
4....6 ---> 2 yards
6....8 ---> 2 yards
8..9 ----> 1 yard
9.....12 -> 3 yards.

From the 12-yard pipe, 3 different lengths are yielded:
1 yard, 2 yards and 3 yards.

Question stem:
Which of the following gives all the different lengths of the pieces, in fractions of a yard?
(1 yard)/(12 yards) = 1/12.
(2 yards)/(12 yards) = 1/6.
(3 yards)/(12 yards) = 1/4.
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by didieravoaka » Tue Jun 02, 2015 3:55 pm
Thank you very much Brent. I understand much better with this method.
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by [email protected] » Tue Jul 07, 2015 3:27 am
didieravoaka wrote:Hi guys,

Does anyone can help me for the problems 151 and 226 in the official guide 13th edition?

Problem 151: I don't understand why we multiply 3/4 by 4/5 and then why we add 1 to 3/5. Another way to solve this problem will help me.
Problem 226: I don't understand why we choose answer D.

Thank you.

Marc.
At a loading dock, each worker on the night crew loaded 3/4 as many boxes as each worker on the day crew. If the night crew has 4/5 as many workers as the day crew, what fraction of all the boxes loaded by the two crews did the day crew load?

(A) 1/2
(B) 2/5
(D) 4/5
(E) 5/8

Solution:

One way to solve this question is to choose convenient values for the number of day workers and for the number of boxes loaded by each day worker.

We are given that each worker on the night crew loaded 3/4 as many boxes as each worker on the day crew and that the night crew has 4/5 as many workers as the day crew. We can set up an equation relating the number of day workers and the number of night workers:

4/5(number of day workers) = number of night workers

Let's choose the convenient value of 20 for the number of day workers and substitute it into the equation above:

4/5 x 20 = number of night workers

16 = number of night workers

Next, we set up the equation for the number of boxes loaded by each worker.

(¾) x (number of boxes loaded by each day worker) = number of boxes loaded by each night worker

Now let's choose the convenient value of 8 for the number of boxes loaded by each day worker and substitute it into the above equation:

¾ x 8 = number of boxes loaded by each night worker

6 = number of boxes loaded by each night worker

Thus, the total number of boxes loaded by each type of worker is as follows:

Day workers = 20 x 8 = 160

Night workers = 16 x 6 = 96

Total boxes loaded = 160 + 96 = 256

The question asks us to determine the fraction of all the boxes loaded by the day crew, and that is:

(Total boxes loaded by day crew)/(Total boxes loaded)

160/256

40/64

5/8

Answer: E

Note that, in this "fraction of total" question, we could have chosen any realistic values for the number of day workers and for the number of boxes loaded by each day worker. We chose the convenient values of 20 and 8, respectively, but other values would have yielded the same fractional answer.

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by nikhilgmat31 » Mon Jul 13, 2015 1:26 am
For workers problem we need not to put numbers to solve too much.

let Day's rate = D
number of workers in Day = d

let Night's rate = N = 3/4 D
number of workers in Night = n= 4/5 d

Work done in Day = d*D
Work done in night = n*N = 3/4 * 4/5 d*D = 3/5dD

Work done in Day & night = dD + 3/5 dD = 8/5 dD

Ratio = dD / (8/5 dD) = 5/8

Answer is E