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2. What is the probability that a 3-digit positive integer picked at random will have one or more "7" in its digits?
(A) 271/900
(B) 27/100
(C) 7/25
(D) 1/9
(E) 1/10

Total 3 digit numbers 900, 3 digit number with no 7 =8*9*9=648, P(at least one 7)=1-P(no 7)=1-648/900=252/900=7/25

Answer: C.
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How do you derive 3 digit number with no '7 =8*9*9' ???

Time, rate and job in work problems are in the same relationship as time, speed (rate) and distance.

Time*Rate=Distance <--> Time*Rate=Job \ done.

We know that 10 people need 55 days to complete the job --> let the combined rate of 10 people be x --> Time*Rate=x*55=Job \ done;
Now, if combined rate of 10 people is x, then combined rate of 16 man will be 1.6x (1.6 times more than x as 16 is 1.6 times more than 10), so new \ time*new \ rate=same \ job --> t_2*1.6x=x*55 --> t_2=\frac{55}{1.6}\approx{34.8} (as rate increased 1.6 times then time needed to do the same job will decrease 1.6 times).

Some work problems with solutions:
time-n-work-problem-82718.html?hilit=reciprocal%20rate
facing-problem-with-this-question-91187.html?highlight=rate+reciprocal
what-am-i-doing-wrong-to-bunuel-91124.html?highlight=rate+reciprocal
gmat-prep-ps-93365.html?hilit=reciprocal%20rate
questions-from-gmat-prep-practice-exam-please-help-93632.html?hilit=reciprocal%20rate
a-good-one-98479.html?hilit=rate

Hope it helps.
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I found question 4 interesting, and I usually attempt it by plugging in. A little more time consuming but I find it helpful. Lets assume the job is to make 1100 widgets, that means that each man is expected to make 1 widget per day, and a ten man crew can make 10 widgets. So after 55 days, 550 widgets have been made, thus 550 are left. Since they hire 6 more guys, it will take a little over 34 days to compete the job at a rate of 16 widgets per day. Since they do not work on any day that it rains this means that they worked for 35 out of the 40 days it took to finish the job. 40-35=5.

Answer is B.

9. Of the applicants passes a certain test, 15 applied to both college X and Y. If 20 % of the applicants who applied college X and 25% of the applicants who applied college Y applied both college X and Y, how many applicants applied only college X or college Y?
(A) 135
(B) 120
(C) 115
(D) 105
(E) 90

Question means the following:
15 people applied to both college X and Y, these 15 people represent 20% of all applicants who applied to X and 25% of all applicants who applied to Y.


0.2*x=15 --> x=75 --> # of people who applied only to X is: 75-15=60;
0.25*y=15 --> y=60 --> # of people who applied only to Y is: 60-15=45;

# of people who applied only to X OR Y is: 60+45=105.

Answer: D.

Hope it's clear.
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Total # of arrangements:
Drivers seat: 2 (either mother or father);
Front seat: 4 (any of 4 family members left);
Back seat: 3! (arranging other 3 family members on the back seat);
So. total # of arrangements is 2*4*3!=48.

# of arrangements with sisters sitting together:
Sisters can sit together only on the back seat either by the left window or by the right window - 2, and either {S1,S2} or {S2,S1} - 2 --> 2*2=4;
Drivers seat: 2 (either mother or father);
Front seat: 2 (5 - 2 sisters on back seat - 1 driver = 2);
Back seat with sisters: 1 (the last family member left);
So, # of arrangements with sisters sitting together is 4*2*2*1=16.

48-16=32.

Hope it's clear.
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my approach was like this -

Case 1a- Father Drives - Co-driver is Daughter1/Daughter2 So it is possible in 3!*2 ways.
Case 1b -Father Drives Co-driver is occupied by Son or Mother + both daughters are at the end of back seat - this is possible in 4 ways...So totat to case 1 is 16 that is going to be symetrical when mother drives.Hence, total ways = 32

Perfect Answer

can someone plz try and show me question 2 in the C way?

1 mean 1C9.
thanks.
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I found one more solution to this problem by Bunnel but the answer is 5... that is B..
But here its mentioned C...


4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain?
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8

This one was solved incorrectly:
Days to finish the job for 10 people 110 days.
On the 61-st day, after 5 days of rain --> 5 days was rain, 55 days they worked, thus completed 1/2 of the job, 1/2 is left (55 days of work for 10 people).
Then 6 more people was hired --> speed of construction increased by 1.6, days needed to finish 55/1.6=34.375, BUT after they were hired job was done in 100-60=40 days --> so 5 days rained. They needed MORE than 34 days to finish the job, so if it rained for 6 days they wouldn't be able to finish the job in 100(40) days.

Answer: B.

4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain?
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8

This one was solved incorrectly:
Days to finish the job for 10 people 110 days.
On the 61-st day, after 5 days of rain --> 5 days was rain, 55 days they worked, thus completed 1/2 of the job, 1/2 is left (55 days of work for 10 people).
Then 6 more people was hired --> speed of construction increased by 1.6, days needed to finish 55/1.6=34.375, BUT after they were hired job was done in 100-60=40 days --> so 5 days rained. They needed MORE than 34 days to finish the job, so if it rained for 6 days they wouldn't be able to finish the job in 100(40) days.

Answer: B.[/quote]

Solution provided by Economist is not correct (the one with answer C). OA's and solutions are given in my post on the 2nd page. OA for the quoted question is indeed B:

Given: 10-man crew needs 110 days to complete the construction.

"On the 61-st day, after 5 days of rain ..." --> as it was raining for 5 days then they must have bee working for 55 days thus completed 1/2 of the job, 1/2 is left (55 days of work for 10 men).

Then contractor "hired 6 more people" --> speed of construction increased 1.6 times, so the new 16-man crew needed 55/1.6=~34.4 days to complete the construction, but after they were hired job was done in 100-60=40 days --> so 5 days rained. (They needed MORE than 34 days to finish the job, so if it rained for 6 days they wouldn't be able to finish the job in 100(40) days.)

Answer: B.[/quote]


Okay..
Thanks for clarifying...