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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