The following answers may not be correct. If you find some discrepancy in the solution, please PM me or post it under the same topic. I took a cursory look into the OA's but am yet to match all my answers and compare the logic.
Bunuel wrote:
1. THE SUM OF EVEN INTEGERS:
The sum of the even numbers between 1 and k is 79*80, where k is an odd number, then k=?
(A) 79
(B) 80
(C) 81
(D) 157
(E) 159
Sol:
Evenly spaced sequence: 2,4,6,....(k-1)
Sum of elements, \(S_n = \frac{n}{2}(2a+(n-1)d)\)
\(a=2\)
\(d=2\)
\(S_n= \frac{n}{2}(2*2+n*2-2)\)
\(=\frac{n}{2}(2+n*2)\)
\(=n+n^2\)
\(=n(n+1)\)
\(n * (n+1) = 79 * 80\)
\(n = 79\)
\(n = \frac{k-1-2}{2}+1\)
\(k = 2n+1\)
\(k = 2*79+1= 159\)
Ans: E
Bunuel wrote:
2. THE PRICE OF BUSHEL:
The price of a bushel of corn is currently $3.20, and the price of a peck of wheat is $5.80. The price of corn is increasing at a constant rate of
5x cents per day while the price of wheat is decreasing at a constant rate of 2^1/2*x-x cents per day. What is the approximate price when a
bushel of corn costs the same amount as a peck of wheat?
(A) $4.50
(B) $5.10
(C) $5.30
(D) $5.50
(E) $5.60
Sol:
Let's do the calculations in cents
\(320 + 5x = 580 - (sqrt(2)x- x)\)
\(320 + 5x = 580 - 1.4x + x\) =>\(sqrt(2)\approx1.4\)
\(320 + 5x = 580 - 0.4x\)
\(5.4x = 580 - 320\)
\(x = \frac{260}{5.4}\)
\(x \approx \frac{27*100}{54}\)
\(x \approx 49\) =>27/54>26/54
\(320 + 5x = 320+5*49 = 320+245 = 565\)
This question requires precision in calculation as answers are quite close to one another and the approximation may result in bad result.
Closest: $5.60.
Ans: "E"
Bunuel wrote:
3. LEAP YEAR:
How many randomly assembled people are needed to have a better than 50% probability that at least 1 of them was born in a leap year?
A. 1
B. 2
C. 3
D. 4
E. 5
Sol:
For every 4 year period,
Leap Years = 1
Non-Leap = 3
Total = 4
1 Person: Probability of at least 1 person born in leap year is = 1/4 < 1/2. Not sufficient yet.
2 People: Probability of at least 1 person born in leap year is = 1/4*3/4+3/4*1/4+1/4*1/4 = 3/17 < 1/2. Not sufficient.
3 People: Probability of at least 1 person born in leap year is = 3*(1/4*3/4*3/4)+3*(3/4*1/4*1/4)+1/4*1/4*1/4 = 27/64+9/64+1/64= 37/64>1/2.
Sufficient.
Ans: "C"
Bunuel wrote:
4. ADDITION PROBLEM:
AB + CD = AAA, where AB and CD are two-digit numbers and AAA is a three digit number; A, B, C, and D are distinct positive integers. In the
addition problem above, what is the value of C?
(A) 1
(B) 3
(C) 7
(D) 9
(E) Cannot be determined
\(\quad\quad AB\)
\(+ CD\)
\(----\)
\(AAA\)
AAA can't be more than 198.
So, the hundreds place = 1= A
\(\quad \quad 1B\)
\(+ CD\)
\(----\)
\(\quad 1\quad1\quad1\)
Last two digits of the result are 11
1+C=11(if nothing is carried over from units place of the result; B+D<10)
C=11-1=10. Not possible. Single digit can't be 10.
Or, 1+C+1=11 (if 1 is carried over from units place of the result; B+D>10)
C = 11-2=9.
Sole possibility for C is 9.
Ans: "D"
Bunuel wrote:
5. RACE:
A and B ran, at their respective constant rates, a race of 480 m. In the first heat, A gives B a head start of 48 m and beats him by 1/10th of a
minute. In the second heat, A gives B a head start of 144 m and is beaten by 1/30th of a minute. What is B’s speed in m/s?
(A) 12
(B) 14
(C) 16
(D) 18
(E) 20
Let A's speed: A m/s
Let B's speed: B m/s
1/10th of a minute = 6sec
1/30th of a minute = 2sec
Total distance of the race track = 480m
First heat;
A's distance = 480m
A's speed = A m/s
A's time = t secs
B's distance = 480-48=432m
B's speed = B m/s
B's time = t+6 secs
A's time = B's time - 6 (if A took 1000 seconds; B took 1006 seconds)
\(\frac{480}{A}=\frac{432}{B}-6\) ------ 1st (\(rate*time=distance \quad or \quad time,t=D/r\))
Second heat;
A's distance = 480m
A's speed = A m/s
A's time = t secs
B's distance = 480-144=336m
B's speed = B m/s
B's time = t-2 secs
A's time = B's time + 2
\(\frac{480}{A}=\frac{336}{B}+2\) ------ 2nd
Using 1st and 2nd,
\(\frac{432}{B}-6=\frac{336}{B}+2\)
\(\frac{432}{B}-\frac{336}{B}=8\)
\(\frac{96}{B}=8\)
\(B=12\)
B's speed : 12m/s
Ans: "A"
Bunuel wrote:
6. PROBABILITY OF DRAWING:
A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red and a white ball in two successive draws, each ball
being put back after it is drawn?
(A) 2/27
(B) 1/9
(C) 1/3
(D) 4/27
(E) 2/9
Sol:
RW+WR
\(\frac{3}{9}*\frac{2}{9}+\frac{2}{9}*\frac{3}{9}\) ---(the denominator remains 9 as the balls are put back)
\(\frac{6}{81}+\frac{6}{81}\)
\(\frac{12}{81}\)
\(\frac{4}{27}\)
Ans: "D"
Bunuel wrote:
7. THE DISTANCE BETWEEN THE CIRCLE AND THE LINE:
What is the least possible distance between a point on the circle x^2 + y^2 = 1 and a point on the line y = 3/4*x - 3?
A) 1.4
B) sqrt (2)
C) 1.7
D) sqrt (3)
E) 2.0
Sol:
This solution can be best described with a geometric figure; I don't know/have proper tool to create such image(it'd be great if someone suggest
me a good tool for the same)
\(x^2+y^2=1\)
This is equation of the circle with center at origin (0,0) and radius r=1 unit.
y = 3/4*x - 3
when x=0, y=-3; y-intercept is 3
y=0, x=4.
We can join points K(0,-3) and L(4,0) and form the line.
We can imagine this line KL to be the hypotenuse of a right triangle in the IVth quadrant;
OL = Altitude: 4
OK = Base: 3
Hypotenuse= LK = \(sqrt{OL^2+OK^2} = 5\)
Let's draw a perpendicular from origin O on the hypotenuse LK to intersect LK at point D and circle at point G
For \(\triangle LOK\)
\(\frac{1}{2}*3*4=\frac{1}{2}*OD*5\)
\(OD=\frac{12}{5}\)
\(OD=2.4\)
Minimum distance from circle to line, GD = (OD - radius of the circle) (=>minimum distance from a point away from a line to the line is always
the perpendicular distance from point to the line)
GD = 2.4-1 = 1.4
Ans: "A"
Bunuel wrote:
8. THE AVERAGE TEMPERATURE:
The average of temperatures at noontime from Monday to Friday is 50; the lowest one is 45, what is the possible maximum range of the temperatures?
A. 20
B. 25
C. 40
D. 45
E. 75
Sol:
Range= Maximum value in a dataset - Minimum value in the dataset
\(\bar x = 50\)
\(S_t = \bar x * count = 50 * 5 = 250\)
Lowest temperature is: 45;
We get the maximum range if 4 out of 5 readings have the same lowest temperature as 45.
So,
\(45*4+t_5=250\)
\(t_5=250-180\)
\(t_5=70\)
\(t_5=t_{max}=70\)
\(t_{min}=45\)
\(R(t) = 70-45 = 25\)
Ans: "B"
Bunuel wrote:
9. PROBABILITY OF INTEGER BEING DIVISIBLE BY 8:
If n is an integer from 1 to 96 (inclusive), what is the probability for n*(n+1)*(n+2) being divisible by 8?
A. 25%
B 50%
C 62.5%
D. 72.5%
E. 75%
Sol:
Let f(n) = n*(n+1)*(n+2)
The product of the three consecutive number will be divisible by 8 if the first of the 3 consecutive numbers is even.
f(2); the function becomes 2*3*4 will be divisible by 8 because 2 is even.
f(52), the function becomes 52*53*54 will be divisible by 8 because 52 is even
f(61), the function becomes 61*62*63 will NOT be divisible by 8 because 61 is odd
However, there is one more case where the function will be divisible by 8 even when first number of the three consecutive number is odd;
f(7): the function becomes 7*8*9; the middle term 8 is divisible by 8 and thus the entire function becomes divisible by 8.
f(63): the function becomes 63*64*65; the middle term 64 is divisible by 8 and thus the entire function becomes divisible by 8.
Probability = favorable/total
How many functions are there;
f(1) = 1*2*3
f(2) = 2*3*4
f(3) = 3*4*5
f(4) = 4*5*6
.
.
.
f(95) = 95*96*97
f(96) = 96*97*98
Total = 96
To find the favorable cases; we need to find the count of all even numbers from 1 to 96 and the count of all numbers that are divisible by 8.
Count of even numbers = (96-2)/2+1=47+1=48
Total numbers divisible by 8 = (96-8)/8+1 = 12
\(Probability = \frac{48+12}{96} = \frac{60}{96}=0.625=62.5%\)
Ans: "C"
Bunuel wrote:
10. SUM OF INTEGERS:
If the sum of five consecutive positive integers is A, then the sum of the next five consecutive integers in terms of A is:
A. A+1 inquiry
B. A+5
C A+25
D 2A
E. 5A
Sol:
Let: m,n,o,p,q be the first set of 5 consecutive integer
\(A = m+n+o+p+q\)
Let: r,s,t,u,v be the immediate next set of 5 consecutive integer
\(A_2=r+s+t+u+v\)
r=m+5
s=n+5
t=o+5
u=p+5
v=q+5
Now; subtract 1st set from second
\(A_2=r+s+t+u+v\)
\(A_2=m+5+n+5+o+5+p+5+q+5\)
\(A_2=m+n+o+p+q+25\)
\(A_2=A+25\)
Ans: "C"
_________________
~fluke
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