The answers below may not be correct. They're not matched with the OA's. Please comment should you find any discrepancy in the logic or solution.
Bunuel wrote:
Please find below new set of PS problems:
1. A family consisting of one mother, one father, two daughters and a son is taking a road trip in a sedan. The sedan has two front seats and three back seats. If one of the parents must drive and the two daughters refuse to sit next to each other, how many possible seating arrangements are there?
(A) 28
(B) 32
(C) 48
(D) 60
(E) 120
Sol:
***
Total ways when father is driving: 4!=24
M*
***
Total ways when Mother is driving: 4!=24
We need to subtract the arrangements when two daughters are sitting together in the back seat;
(2!*2!) for every seating arrangement when Mother, Father and Brother(once when father is driving and again when mother is driving) are sitting on the co-driver's seat which totals to 4.
So; 4*(2!*2!) = 16 arrangements when both daughters are sitting together.
Thus total number of possible seating arrangements = 48-16=32.
Ans: "B"
Bunuel wrote:
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
Sol:
Total number of 3 Digit positive integers= 999-100+1=900
Numbers without any 7 = 8*9*9=648
Probability that none will be 7 = 648/900
Probability that at least one is 7 = 1-648/900 = 252/900=7/25
Ans: "C"
Bunuel wrote:
3. A sphere is inscribed in a cube with an edge of 10. What is the shortest possible distance from one of the vertices of the cube to the surface of the sphere?
(A 10( sqrt3- 1)
(B) 5
(C) 10( sqrt2 - 1)
(D) 5( sqrt3 - 1)
(E) 5( sqrt2 - 1)
Sol:
The distance of the vertex to the center of the sphere
\(5\sqrt{2}\) (Hypotenuse for a isosceles right angled triangle 45-90-45 and side 5)
The shortest distance between vertex and point on the circumference of the circle
\(5\sqrt{2}-5\)
\(5(\sqrt{2}-1)\)
Ans: "E"
Bunuel wrote:
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
Sol:
It rained from 56th to 60th day. The crew of 10 men worked for 55 days prior to the rain.
If they finish 1 work in 110 days.
They must have finished 1/2 work in 55 days.
Thus; work left to be completed = 1/2
6 men added. Total workers: 16
1 work:
10 men -> 110 days
1 man -> 110*10 Days
16 men -> 110*10/16 Days
1/2 Work;
16 Men -> (110*10)/(16 *2) Days = 34.33 or 35 days
It is mentioned in the question that the job was done on the 100th day.
From 61 to 100; there are 40 days. However, the actual number of days required for the work is 35 days.
Thus; the crew didn't work for those 5 days it rained.
Ans: "B"
Bunuel wrote:
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
Sol:
\(\frac{s}{t}=64.12\)
\(\frac{s}{t}=64+\frac{12}{100}{\)
Remainder r=12
Divisor, d=100
\(\frac{r}{d}=\frac{12}{100}\)
\(\frac{r}{d}=\frac{3}{25}\)
\(25r=3d\)
Implies; r contains a multiple of 3 and d contains a multiple of 25.
If Remainder must be a multiple of 3; the only option is 45.
Ans: "E"
Bunuel wrote:
6. A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together?
(A) 3510
(B) 2620
(C) 1404
(D) 700
(E) 635
Sol:
Let's find the number of selections without the restriction of 2 mutually exclusive men.
3Men of 8Men and 3Women of 5Women
OR
2Men of 8Men and 4Women of 5Women
\(C^8_3*C^5_3+C^8_2*C^5_4\)
\(560+140=700\)
Find the number of possible selections with those supposedly mutually exclusive men put together.
\(C^6_1*C^5_3+C^6_0*C^5_4\)
\(60+5=65\)
Thus the number of selections where 2 particular men are not together = 700-65=635.
Ans: "E"
Bunuel wrote:
7. If x is positive, which of the following could be the correct ordering of 1/x,2x and x^2 ?
I. x^2<2x<1/x
II. x^2<1/x<2x
III. 2x<x^2<1/x
(A) None
(B) I only
(C) III only
(D) I and II only
(E) I II and III
Sol:
We'll test it with 4 numbers; x=0.01,1,2,5
x=0.01
-------
x^2=0.0001
2x=0.02
1/x=1/0.01=100
x^2<2x<1/x.
I. could be correct.
x=1
----
x^2=1
2x=2
1/x=1
x^2=1/x<2x. No such inequality given. Ignore.
x=2
----
x^2=4
2x=4
1/x=0.5
1/x<x^2=2x. Ignore
x=5
---
x^2=25
2x=10
1/x=1/5=0.2
1/x<2x<x^2. Not given.
Don't want to try any more numbers. Will go with I only.
Ans: "B"
Bunuel wrote:
8. In the xy plane, Line k has a positive slope and x-intercept 4. If the area of the triangle formed by line k and the two axes is 12, What is the y-intercept of line K ?
(A) 3
(B) 6
(C) -3
(D) -6
(E) -4
Sol:
Positive slope means that the line is making an acute angle with x-axis. From the given information; the line should intersect y axis below x axis; meaning thereby the y-intercept is negative.
The line is touching Ist, IIIrd and IVth quadrant. Say the line intersect y-axis at Y and x-axis at X. The triangle hence formed \
will be XOY where O is the origin(0,0)
\(\triangle{XOY}=\frac{1}{2}*\bar{OY}*4=12\)
\(\bar{OY}=6\)
Since y is minus; the intercept is -6.
Ans: "D"
Bunuel wrote:
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
Sol:
Say x applied for college X
And y applied for college y
\(x \cap y = 15\)
\(15=0.2x\)
\(x=75\)
\(15=0.25y\)
\(y=60\)
The candidates who applied only for x and y should be: 75+60-2(15) = 105
Ans: "D"
Bunuel wrote:
10. What is the lowest positive integer that is divisible by each of the integers 1 through 7, inclusive?
(A) 420
(B) 840
(C) 1260
(D) 2520
(E) 5040
Sol:
Find LCM(1,2,3,4,5,6,7)
2=2
3= 3
4=2^2
5= 5^1
6=2^1 * 3^1
7= 7^1
Choose Highest Powers of prime factors= 2^2*3^1*5^1*7^1 = 420
Ans: "A"
Thanks for the questions, Bunuel.
_________________
~fluke
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