TOUGH & TRICKY SET Of PROBLEMS

SOLUTION:
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

This is tough:
First note that min distance from the circle to the line would be: length of perpendicular from the origin to the line (as the circle is centered at the origin) - the radius of a circle (which is 1).

Now we can do this by finding the equation of a line perpendicular to given line \(y=\frac{3}{4}*x-3\) (we know it should cross origin and cross given line, so we can write the formula of it), then find the croos point of these lines and then the distance between the origin and this point. But it's very lengthy way.

There is another, shorter one. Though I've never seen any GMAT question requiring the formula used in it.

We know the formula to calculate the distance between two points \((x_1,y_1)\) and \((x_2,y_2)\): \(d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\) BUT there is a formula to calculate the distance between the point (in our case origin) and the line:

DISTANCE BETWEEN THE LINE AND POINT:
Line: \(ay+bx+c=0\), point \((x_1,y_1)\)

\(d=\frac{|ay_1+bx_1+c|}{\sqrt{a^2+b^2}}\)

DISTANCE BETWEEN THE LINE AND ORIGIN:
As origin is \((0,0)\) -->

\(d=\frac{|c|}{\sqrt{a^2+b^2}}\)

So in our case it would be: \(d=\frac{|3|}{\sqrt{1^2+(\frac{3}{4})^2}}=\frac{12}{5}=2.4\)

So the shortest distance would be: \(2.4-1(radius)=1.4\)

Answer: A.

OR ANOTHER APPROACH:

In an x-y Cartesian coordinate system, the circle with center (a, b) and radius r is the set of all points (x, y) such that:
\((x-a)^2+(y-b)^2=r^2\)

If the circle is centered at the origin (0, 0), then the equation simplifies to:
\(x^2+y^2=r^2\)

So, the circle represented by the equation \(x^2+y^2 = 1\) is centered at the origin and has the radius of \(r=\sqrt{1}=1\).

Then note that min distance from the circle to the line would be: length of perpendicular from the origin to the line (as the circle is centered at the origin) - the radius of a circle (which is 1).

So we should find the length of perpendicular, or the height of the right triangle formed by the X and Y axis and the line \(y = \frac{{3}}{{4}}x-3\).

The legs would be the value of x for y=0 (x intercept) --> y=0, x=4 --> \(leg_1=4\).
and the value of y for x=0 (y intercept) --> x=0, y=-3 --> \(leg_2=3\).

So we have the right triangle with legs 4 and 3 and hypotenuse 5. What is the height of this triangle (perpendicular from right angle to the hypotenuse)? As perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle: \(\frac{height}{leg_1}=\frac{leg_2}{hypotenuse}\) --> \(\frac{height}{3}=\frac{4}{5}\) --> \(height=2.4\).

\(Distance=height-radius=2.4-1=1.4\)

Answer: A.

You can check the link of Coordinate Geometry below for more.

bro bunuel,
do u think we would need to knw these formulas and such a question can ever appear on gmat?
you mentioned that u have never seen any gmat question that requires us to know formula.

SOLUTION:
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

This is with replacement case (and was solved incorrectly by some of you):

\(P=2*\frac{3}{9}*\frac{2}{9}=\frac{4}{27}\)

We are multiplying by 2 as there are two possible wining scenarios RW and WR.

Answer: D.

Hi Bunuel,

I always have problem in understand the probablity problems.
If this is asked without replacement then how we can solve this problem:????

Can we solve the original question with nCr method???

Please clear my doubts.

Thanks in advance,
Rrsnathan.

SOLUTION:
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

Probability of a randomly selected person NOT to be born in a leap year=3/4
Among 2 people, probability that none of them was born in a leap = 3/4*3/4=9/16. The probability at least one born in leap = 1- 9/16=7/16<1/2
So, we are looking for such n (# of people), when 1-(3/4)^n>1/2
n=3 --> 1-27/64=37/64>1/2

Thus min 3 people are needed.

Answer: C.

Hi Bunuel,

If I shall encounter this question or one like it on the GMAT, will I be given the information as to what the ratio of regular year to leap year is?
If not, is there a list of little things like this that are important to know?

SOLUTION:
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

The solution given in the file was 79, which is not correct. Also the problem was solved with AP formula thus was long and used the theory rarely tested in GMAT. Here is my solution and my notes about AP. Some may be useful:

The number of terms in this set would be: n=(k-1)/2 (as k is odd)
Last term: k-1
Average would be first term+last term/2=(2+k-1)/2=(k+1)/2
Also average: sum/number of terms=79*80/((k-1)/2)=158*80/(k-1)
(k+1)/2=158*80/(k-1) --> (k-1)(k+1)=158*160 --> k=159

Answer E.

MY NOTES ABOUT AP:
ARITHMETIC PROGRESSION
Sequence a1, a2,…an, so that a(n)=a(n-1)+d (constant)
nth term an = a1 + d ( n – 1 )
Sn=n*(a1+an)/2 or Sn=n*(2a1+d(n-1))/2
Special cases:
I. 1+2+…+n=n(1+n)/2 (Sum of n first integers)
II. 1+3+5+… (n times)=n^2 (Sum of n first odd numbers). nth term=2n-1
III. 2+4+6+… (n times)=n(n+1) (Sum of n first even numbers) nth term=2n

SOLUTION WITH THE AP FORMULA:
Sequence of even numbers
First term a=2, common difference d=2 since even number
Sum to first n numbers of AP:
Sn=n*(a1+an)/2=n(2*2+2(n-1))/2=n(n+1)=79*80
n=79 (odd)
Number of terms n=(k-1)/2=79 k=159

OR
Sum of n even numbers n(n+1)=79*80
n=79
k=2n+1=159

Dear Brunel ,

Please explain special cases from AP

MY NOTES ABOUT AP:
ARITHMETIC PROGRESSION
Sequence a1, a2,…an, so that a(n)=a(n-1)+d (constant)
nth term an = a1 + d ( n – 1 )
Sn=n*(a1+an)/2 or Sn=n*(2a1+d(n-1))/2
Special cases:
I. 1+2+…+n=n(1+n)/2 (Sum of n first integers)
II. 1+3+5+… (n times)=n^2 (Sum of n first odd numbers). nth term=2n-1
III. 2+4+6+… (n times)=n(n+1) (Sum of n first even numbers) nth term=2n

Bunuel
Hi Bunuel.. U r really a math expert ! Kudos for all your post.
My Query :-
Q4.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
AB and CD are two digit integers, their sum can give us only one three digit integer of a kind of AAA it's 111.
So, A=1. 1B+CD=111.
How did you creach to the conclusion that 111 can be the only result of addition? how is A=1

Q8. " As the min temperature is 45, max would be 250-4*45=70 --> The range=70(max)-45(min)=25 " Can you please explain the logic behind -4*45. How do we get the max value here ?

Can you please explain !
Thanks in advance

Hi Bunnel,

In question number 3 why have you taken 3/4 for no probability of any child in leap year.

SOLUTION:
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

Probability of a randomly selected person NOT to be born in a leap year=3/4
Among 2 people, probability that none of them was born in a leap = 3/4*3/4=9/16. The probability at least one born in leap = 1- 9/16=7/16<1/2
So, we are looking for such n (# of people), when 1-(3/4)^n>1/2
n=3 --> 1-27/64=37/64>1/2

Thus min 3 people are needed.

Answer: C.

If there are 3 people required then shouldn't this too be true?
case 1: when only 1 person was there, the probability would be 1/4 = 0.25
case 2: when only 2 people are there, the probability would be (1/4)*(1/4) = 0.0625
case 3: when 3 people are there, the probability would be (1/4)*(1/4)*(1/4) = 0.015625

So, shouldn't the answer be: 0.25 + 0.0625 + 0.015625 = 0.0328125?

(I know my answer is incorrect, but fail to understand why my way of calculating is wrong?)