Good set of PS 2

Bunuel,

For Q1, Can u please elaborate more on this >>
Another way: Total number of arrangements-arrangements with sisters sitting together=2*4*3!-2*2(sisters together)*2*2*1(arrangement of others)=48-16=32

I was able to get answer by first method as u have mentioned. But, I am failing to get the computation mentioned in another method.

Thanks!

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.

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

There are total 900 3 digit numbers;

There are 8*9*9=648 (you can write this as 8C1*9C1*9C1 if you like) 3-digit numbers without 7 in its digits (first digit can take 8 values from 1 to 9 excluding 7; second and third digits can take 9 values from 0 to 9 excluding 7),

P(at least one 7)=1-P(no 7)=1-648/900=252/900=7/25.

Answer: C.

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.

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.

OA-s and solutions are given in my post on 2nd page. Solution for this question:

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.

[quote="Bunuel"]Please find below new set of PS problems:



5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t?
(A) 2
(B) 4
(C) 8
(D) 20
(E) 45


R= 0,12, then:
\(\frac{R}{S}= \frac{12}{100}\)
\(\frac{R}{S}= \frac{3}{25}\)
\(25R = 3S\)
Since 45 is the only number which has 3 and 5 as his primes (E) should be the answer.
Please provide OA.

Please read the thread.

OA-s and solutions are given in my post on 2nd page.


5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t?
(A) 2
(B) 4
(C) 8
(D) 20
(E) 45

\(s\) divided by \(t\) yields the remainder of \(r\) can always be expressed as: \(\frac{s}{t}=q+\frac{r}{t}\) (which is the same as \(s=qt+r\)), where \(q\) is the quotient and \(r\) is the remainder.

Given that \(\frac{s}{t}=64.12=64\frac{12}{100}=64\frac{3}{25}=64+\frac{3}{25}\), so according to the above \(\frac{r}{t}=\frac{3}{25}\), which means that \(r\) must be a multiple of 3. Only option E offers answer which is a multiple of 3

Answer: E.

For those who like visualizing, here is a solution for Question 7 using graphs.

Please, refer to the attached drawing, in which the three graphs \(y=1/x,\) \(y=2x,\) and \(y=x^2\) are depicted for \(x>0\).
The exact values for A, B, and C can be worked out, but they are not important to establish the order of the three algebraic expressions.

So, the correct orderings are:
If \(x\) between 0 and A: \(x^2<2x<1/x\)
If \(x\) between A and B: \(x^2<1/x<2x\)
If \(x\) between B and C: \(1/x<x^2<2x\)
If \(x\) greater than C: \(1/x<2x<x^2\)

We can see that only the first two of the above options are listed as answers (I and II).

Answer: D.

Good challenging question.

Driving spot - only two variations. 1) Mother drives 2) Father drives

The permutations come up in the spots for the other 4 seats.

So, 2 default values * 4P4 = total 2*4! = 48 permutations

Total # ways - # ways together = # ways can't sit together

The two daughters can only sit together when they are together in the back seat.
So that's:
12B
21B
B12
B21

...where 1 and 2 represent daughter #1 and #2. "B" represents the boy child.

so that's 4 ways we counted.

4 ways * variations on other 3 seats
4 ways * (default mother/father drives and variations on other 2 seats)
4 ways * (2 * 2P2)
4 ways * 2 * 2! = 16

Total ways = 48
Variations together = 16
Variations not together = 48 - 16 = 32

Looks like the above analysis is flawed.

Ways to select 2 men out of 8 is \(8C2\)
Ways to select 3 women out of 5 is \(5C3\)
Ways to select the remaining 1 person out of the remaining 8 = 8

Total number of ways of forming a committee as above = 28 * 10 *8 = 2240 - (1)

Now for the restriction that 2 men cannot serve together consider the ways they can serve together= 1*10*8=80 - (2)

(1) - (2) = 2160 ways looks like the right answer.

OA for this question is E.

2240 has duplications in it.

I don't think they do. Why don't you try on a smaller example?

Actually there is no need, but just to illustrate.

Two sets {A, B, C} and {a, b, c}. How many ways to select a committee of 3 so that there is at least one out of each set.

My way (correct one): \(C^1_3*C^2_3+C^2_3*C^1_3=3*3+3*3=18\). We could do this in another way: \(C^3_6-2=18\).

Your way (wrong one): \(C^1_3*C^1_3*4=36\).

Let's check:
Aab;
Aac;
Abc;
Bab;
Bac;
Bbc;
Cab;
Cac;
Cbc;

ABa;
ABb;
ABc;
ACa;
ACb;
ACc;
BCa;
BCb;
BCc.

Total of 18 ways.

Hope it's clear.

Dear Bunuel,

Thanks for taking the trouble to list the set of possibilities for an example. I really appreciate it. But I think there is a basic point that is being missed here. In the above example let us say two of the possibilities are AaB and BaA given my approach. The point is that we are implicitly assuming importance to A in the first possibility and to B in the second possibility as we first satisfy the basic requirement of 1 man and 1 woman. "A" being picked to satisfy the basic requirement tells that A needs to be there and is more important than B. And vice versa. In my thinking that implicit importance and ordering makes the two possibilities different. So my reading would be in such questions the uniqueness of the committee is not only with respect to who constitute it but also the way it is selected.

An example to illustrate my point would be : In how ways a class of 10 can be formed from a pool of 100 applicants so that 8 are selected on merit.

Not sure I understand what you are trying to say here. Anyway, {AaB} and {BaA} are the same two groups.

So, the correct answer is E.

Dear Bunuel,

Consider this:

In how ways 3 scholars be selected from a pool of 10 scholars so that 2 are selected on merit.

Let us say one possible set of scholars is: A B C with A and B being selected on merit and C on other considerations. We have the same set which is : C B A but Now C and B being selected on merit and A on other considerations.

Don't you think the two groups are unique even though constituted by the same people?

Is not your question,

"How many 3 member committees be formed from 6 people so that each has at least one man and one woman?"

asking the same?

Here "one man and one woman" is analogous to "merit" in the above example. The point is you are not going to select "one man and one woman" randomly. This is not an abstract problem and so the solution should be based on the practical facts. Since whether ordering is important or not for a particular problem is crucial to using the right formula we need to be right in making that decision.

The way one would think, the selection of a committee is done in the above case would be not by the roll of dice but by some practical considerations. So to first satisfy the essential requirement that I select a man and a woman I go by some practical considerations. The fact that I first select "A" over "B" says that I prefer "A" in some way over "B" or "A" is in some way more important than "B". So to me the selection AaB is different from the selection BaA since order, though not physical, is implied. It is like in the first case "A" is selected on merit and "B" on merit in the second case.