1. Good post? |

## set 1 #19

Q19:

Each of the 30 boxes in a certain shipment weighs either 10 pounds or 20 pounds, and average (arithmetic mean) weight of the boxes in the shipment is 18 pounds.* If the average weight of the boxes in the shipment is to be reduced to 14 pounds by removing some of the 20-pound boxes, how many 20-pound boxes must be removed?

*

A.**** 4

B.**** 6

C.**** 10

D.*** 20

E.**** 24

SPOILER: *My ans is 6 , Official Answer is diff. Wt is the correct ans ?

2. Good post? |

Formal explaination:

Let say # of 10 pnd box - A
# of 20 pnd boxes - B
Now assume X boxes to be removed
From question we can say A*10 + B*20 = 30*18 = 540
Also A*10 + (B-X)20 = (30-X) * 14 = 420 - 14X
Further simplfy these 2 eqns , we will get 6X = 120 => X = 20 {ANS}

3. Good post? |
I also agree with D (20 boxes)

n = # of 10 lb boxes
m= # of 20 lb boxes

(10n*20m)/(n+m) =18;
n+m=30
therefore:
m=24
n=6

Remove some of the 20 lb bixes to average 14 lb
(6*10+20m/(6+m)=14
m=4
m=number of 20 lb boxes needed to make 14 lb mean

Therefore remove 20 from he original 24.

4. Good post? |
Assume No. of 10 pnd boxes=x
no. of 20 pnd boxes=30-x
[10x+20(30-x)]/30 =18(avg. wt)
x=6(10pnd boxes)& 24(20 pnd boxes)

[10*6+20(24-x)]/(30-x) =14(new avg. wt)

x=20 Ans(d)

5. Good post? |
Let x be the 10 pound boxes and y be the 20 pound boxes and A be the 20 pound boxes removed.

so we get

10x+20y=540 (30*18).........eq1

and

10x+20(y-A)/30-A=14

ie. 10x+20y-20A=420-14A

sub. value from equ 1,we get

540-20A=420-14A

so 6A=120

A=20

6. Good post? |
agree with D, its 20 boxes you need to remove

There are currently 1 users browsing this thread. (0 members and 1 guests)

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•

SEO by vBSEO ©2010, Crawlability, Inc.