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Pls shed some light on these few official problems!

Expert replies
by larrybird » Mon Sep 07, 2009 6:17 am
Experts, here are 3 problems where I need help. Thanks a lot for the support!

1. Plz someone show how do this in a quick way (not just adding and finding the mean). Thanks!

70,75,80,85,90,105,105,130,130,130
The numbers above are the seconds it took each of 10 students to run 400 mts. If the standard deviation is 22.4, how many students are 1 standard deviation below the mean ?
OA after some discussion

2. A certain manufacturer of cakes, muffins and bread mixes has 100 buyers, of whom 50 purchase cake, 40 purchase muffin and 20 purchase muffin both cake and muffin mix. If a buyer is selected at random from the 100 buyers, what is the probability that the buyer selected will be one who purchases neither cake nor muffins?
OA after some discussion

3. A car averages 25 miles per galon (mpg) in the city and 40 mpg in the highway. Which of the following is the closest to the avergae mpg when car is driven 10 miles in the city and 50 miles in the highway?

a.28
b.30
c.33
d.36
e.38

OA after some discussion
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Source: — Problem Solving |

by bharathh » Mon Sep 07, 2009 7:00 am
Not really sure how to do the 1st one faster than the method below

I divided all the numbers by 10 (as you would the sum) and got the mean by adding them up. In this case it's a 100.

You can figure out that 5 numbers are below the std dev

2. 100 = 50 + 40 - 20 + n

where n is the number that don't order either M or F

n = 30

30 out of 100 ppl don't order either. So probability of picking one out of this lot is 3/7

3. I solved this quickly on considering this to be a mixture problem

Setting up the eqn

25 mpg 40 mpg

gives X mpg on average if I drive at a ratio of 10:50 miles

So 40-X:X-25 = 1:5
5(40-X) = X-25
X = 27.5 ~ 28
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by larrybird » Mon Sep 07, 2009 8:28 am
1. OK, tks
2. OA is 3/10. I guess you wanted to divide by 100 , right?
A question on this one: can you assume that all purchases regarding bread miexs are zero?? If not, there is no way to solve this.
3. OA is 36
Initially I did a weighted average (1/6)* 25 + (5/6)*40 = 37.5 which is approx = 38
The right way to do this is calculating total number of miles divided by total number of gallon.
Total miles: 60
Total gallons: 10/25 + 50/40 = 1.65
therefore: 60/1.65 = 36.36 which is approx = 36
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by tom4lax » Mon Sep 07, 2009 11:17 am
#3 is interesting. I guess I wasnt the only person to do the weighted average. The way to approach I think is to consider it a "rate" problem, and we know we cant just take the average of those. Need total/total.
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