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# TOUGH & TRICKY SET Of PROBLEMS

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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]

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03 Nov 2010, 21:24
sap wrote:
Bunuel wrote:
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: $$2.4-1(radius)=1.4$$

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: $$\frac{height}{leg_1}=\frac{leg_2}{hypotenuse}$$ --> $$\frac{height}{3}=\frac{4}{5}$$ --> $$height=2.4$$.

$$Distance=height-radius=2.4-1=1.4$$

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

co-ordinates of line are (1,0) and (0,-3/4). This line will be a secant to the circle, the least distance then should le less than 1. Please correct me if I m wrong somewhere.

I think you mean not the coordinates but X and Y intercepts of the line. If the X and Y intercepts of the line $$y=\frac{3}{4}*x-3$$ indeed were (1,0) and (0,-3/4), you would be right: the line would intersect the circle and the least possible distance between a point on the circle and a point on the line would be zero.

But the the X and Y intercepts of the line $$y=\frac{3}{4}*x-3$$ are (4, 0) and (0, -3) respectively.

Hope it's clear.
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]

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03 Nov 2010, 21:36
I beg to ask again ( not too good at maths ) the equation of the line is x/1 - y/(-3/4) = 1. What is the difference between co-ordinates and intercepts ? I know I m missing something but not able to identify what . I m still in a position that it is a secant to the circle. Can you please explain a little more, if possible with another example ?
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]

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03 Nov 2010, 21:48
sap wrote:
I beg to ask again ( not too good at maths ) the equation of the line is x/1 - y/(-3/4) = 1. What is the difference between co-ordinates and intercepts ? I know I m missing something but not able to identify what . I m still in a position that it is a secant to the circle. Can you please explain a little more, if possible with another example ?

There is no such thing as "co-ordinate of a line".

As for X and Y intercepts: X-intercept is the point of intersection of a curve (line) and X-axis, so the point (x, 0), similarly Y-intercept is the point of intersection of a curve (line) and Y-axis, so the point (0, y).

So, X and Y intercepts of the line $$y=\frac{3}{4}*x-3$$ are (4, 0) and (0, -3) respectively and the line does not intersect the circle $$x^2+y^2=1$$.

For more on this issues please check Coordinate Geometry chapter of Math Book: math-coordinate-geometry-87652.html

Hope it helps.
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]

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02 Dec 2010, 10:00
Bunuel wrote:
SOLUTION:
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 $$\sqrt{2}*x-x$$ 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 Note that we are not asked in how many days prices will cost the same. Let $$y$$ be the # of days when these two bushels will have the same price. First let's simplify the formula given for the rate of decrease of the price of wheat: $$\sqrt{2}*x-x=1.41x-x=0.41x$$, this means that the price of wheat decreases by $$0.41x$$ cents per day, in $$y$$ days it'll decrease by $$0.41xy$$ cents; As price of corn increases $$5x$$ cents per day, in $$y$$ days it'll will increase by $$5xy$$ cents; Set the equation: $$320+5xy=580-0.41xy$$, solve for $$xy$$ --> $$xy=48$$; The cost of a bushel of corn in $$y$$ days (the # of days when these two bushels will have the same price) will be $$320+5xy=320+5*48=560$$ or$5.6.

In the equation above- can someone explain why cents is not being converted to $? The two sides are being added without /100 so shouldn't the equations read : 3.2 + 0.05xy --- LHS and 5.8 - 0.0041xy -- RHS? Apples to apples? Math Expert Joined: 02 Sep 2009 Posts: 36548 Followers: 7076 Kudos [?]: 93101 [0], given: 10552 Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] ### Show Tags 02 Dec 2010, 10:23 sixsigma1978 wrote: Bunuel wrote: SOLUTION: 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 $$\sqrt{2}*x-x$$ 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

Note that we are not asked in how many days prices will cost the same.

Let $$y$$ be the # of days when these two bushels will have the same price.

First let's simplify the formula given for the rate of decrease of the price of wheat: $$\sqrt{2}*x-x=1.41x-x=0.41x$$, this means that the price of wheat decreases by $$0.41x$$ cents per day, in $$y$$ days it'll decrease by $$0.41xy$$ cents;

As price of corn increases $$5x$$ cents per day, in $$y$$ days it'll will increase by $$5xy$$ cents;

Set the equation: $$320+5xy=580-0.41xy$$, solve for $$xy$$ --> $$xy=48$$;

The cost of a bushel of corn in $$y$$ days (the # of days when these two bushels will have the same price) will be $$320+5xy=320+5*48=560$$ or $5.6. Answer: E. In the equation above- can someone explain why cents is not being converted to$?
The two sides are being added without /100
so shouldn't the equations read :

3.2 + 0.05xy --- LHS
and 5.8 - 0.0041xy -- RHS?
Apples to apples?

Even though it didn't affect the final answer I've still changed it to: $$320+5xy=580-0.41xy$$, so now everything is in cents.
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Re: TOUGH & TRICKY SET Of PROBLMS Question 9 [#permalink]

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07 Dec 2010, 02:17
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%

N=n*(n+1)*(n+2)

N is divisible by 8 in two cases:
When n is even:
No of even numbers (between 1 and 96)=48
AND
When n+1 is divisible by 8. -->n=8p-1 --> 8p-1<=96 --> p=12.3 --> 12 such nembers

Total=48+12=60

Probability=60/96=0.62

so if I were to write out n(n+1)(n+2) = n^3 + 3n^2 + 2n meaning that every even number will be divisible by 8 because every even number will have at least three 2's as factors, also taking care of the n+2 because it would be even as well, then add the 12 numbers divisible by 12 when you add 1. Is the way I am thinking correct?
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Re: TOUGH & TRICKY SET Of PROBLMS Question 9 [#permalink]

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07 Dec 2010, 02:32
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Expert's post
mmcooley33 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%

N=n*(n+1)*(n+2)

N is divisible by 8 in two cases:
When n is even:
No of even numbers (between 1 and 96)=48
AND
When n+1 is divisible by 8. -->n=8p-1 --> 8p-1<=96 --> p=12.3 --> 12 such nembers

Total=48+12=60

Probability=60/96=0.62

so if I were to write out n(n+1)(n+2) = n^3 + 3n^2 + 2n meaning that every even number will be divisible by 8 because every even number will have at least three 2's as factors, also taking care of the n+2 because it would be even as well, then add the 12 numbers divisible by 12 when you add 1. Is the way I am thinking correct?

Expanding is not a good idea. Below is a solution for this problem:

If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8?
A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4

$$n(n+1)(n+2)$$ is divisible by 8 in two cases:

1. When $$n$$ is even:
$$n=2k$$ --> $$n(n+1)(n+2)=2k(2k+1)(2k+2)=4k(2k+1)(k+1)$$ --> either $$k$$ or $$k+1$$ is even so 8 is a multiple of $$n(n+1)(n+2)$$.

# of even numbers between 1 and 96, inclusive is $$\frac{96-2}{2}+1=48$$ (check this: totally-basic-94862.html?hilit=last%20first%20range%20multiple)

AND

2. When $$n+1$$ is divisible by 8. --> $$n+1=8p$$ ($$p\geq{1}$$), $$n=8p-1$$ --> $$8p-1\leq{96}$$ --> $$p\leq{12.1}$$ --> 12 such numbers.

Also note that these two sets have no overlaps, as when $$n$$ and $$n+2$$ are even then $$n+1$$ is odd and when $$n+1$$ is divisible by 8 (so even) then $$n$$ and $$n+2$$ are odd.

Total=48+12=60

Probability: $$\frac{60}{96}=\frac{5}{8}=62.5%$$

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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]

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07 Dec 2010, 10:57
1. k*(k-1)/2 - (k-1)*(k-1)/4 = 79*80
k^2/4 - 1/4 = 6320
k^2 = 25,281
K=159
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]

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15 Jan 2011, 10:51
Bunuel wrote:
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
C can not be less than 9, because no to digit integer with first digit 1 (mean that it's<20) can be added to two digit integer less than 90 to have the sum 111 (if CD<90 meaning C<9 CD+1B<111).
C=9

Bunuel I have doubt here-- consider following AB as 25 CD 86 which sums to 111 and ||ly
if we consider AB as 15 & CD as 96 then also we end up with 111 so I am doubtfull with
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]

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15 Jan 2011, 13:13
yogesh1984 wrote:
Bunuel wrote:
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
C can not be less than 9, because no to digit integer with first digit 1 (mean that it's<20) can be added to two digit integer less than 90 to have the sum 111 (if CD<90 meaning C<9 CD+1B<111).
C=9

Bunuel I have doubt here-- consider following AB as 25 CD 86 which sums to 111 and ||ly
if we consider AB as 15 & CD as 96 then also we end up with 111 so I am doubtfull with

SOLUTION:
AB and CD are two digit integers, their sum can give us only one 3-igit integer of a kind of AAA: 111. So, A=1 --> 1B+CD=111.

C can not be less than 9, because no 2-digit integer with first digit 1 (mean that it's <20) can be added to 2-digit integer less than 90 to have the sum 111 (if CD<90 meaning C<9 CD+1B<111) --> C=9.

So, as A=1 then AB can not be 25.

Hope it's clear.
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]

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16 Jan 2011, 03:17
Bunuel wrote:
yogesh1984 wrote:
Bunuel wrote:
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
C can not be less than 9, because no to digit integer with first digit 1 (mean that it's<20) can be added to two digit integer less than 90 to have the sum 111 (if CD<90 meaning C<9 CD+1B<111).
C=9

Bunuel I have doubt here-- consider following AB as 25 CD 86 which sums to 111 and ||ly
if we consider AB as 15 & CD as 96 then also we end up with 111 so I am doubtfull with

SOLUTION:
AB and CD are two digit integers, their sum can give us only one 3-igit integer of a kind of AAA: 111. So, A=1 --> 1B+CD=111.

C can not be less than 9, because no 2-digit integer with first digit 1 (mean that it's <20) can be added to 2-digit integer less than 90 to have the sum 111 (if CD<90 meaning C<9 CD+1B<111) --> C=9.

So, as A=1 then AB can not be 25.

Hope it's clear.

Aaha my bad sir, actually I was missing the link of AB and AAA, now am clear..thanks
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]

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24 Jan 2011, 10:40
Bunuel wrote:
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

I did not understand. How do you know that the center of the circle is on (0,0)? I didn't understand the rest either.
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]

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24 Jan 2011, 10:43
mariyea wrote:
Bunuel wrote:
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

I did not understand. How do you know that the center of the circle is on (0,0)? I didn't understand the rest either.

Check this: math-coordinate-geometry-87652.html
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]

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07 Feb 2011, 13:29
Bunuel wrote:
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

there is an easier approach and faster...
we find the height of the triangle (which will be perpendicular to the line and crosses the center and then substract from it the radius...
to find the height, then we calculate the area of the triangle
4*3*0.5 = 12
5*height*0.5 = 12, so height = 2.4 (12/5==>24/10)
2.4-1 = 1.4
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]

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09 Feb 2011, 23:19
Can you put dis entire set in some pdf file as attachment
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]

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10 Feb 2011, 08:06
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:
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"
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]

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11 Feb 2011, 04:06
Bunuel wrote:
SOLUTION:
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: $$\frac{height}{leg_1}=\frac{leg_2}{hypotenuse}$$ --> $$\frac{height}{3}=\frac{4}{5}$$ --> $$height=2.4$$.

Let me put in my way to find out the height in this case.

we now have a right triangle with 3,4 and 5.

let me view this triangle from two different directions

1) the base is $$3$$ and the height is $$4$$==> the area = $$\frac{3*4}{2}$$
2) the base is $$5$$ (hypotenuse) and the height = $$h$$ ==> area = $$\frac{5*h}{2}$$

Now both these areas shud be equal
==> $$\frac{3*4}{2}$$ = $$\frac{5*h}{2}$$

==> $$h = \frac{12}{5}$$ = $$2.4$$

Regards,
Murali.
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]

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20 Feb 2011, 16:24
Bunuel wrote:
SOLUTION:
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 $$\sqrt{2}*x-x$$ 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 Note that we are not asked in how many days prices will cost the same. Let $$y$$ be the # of days when these two bushels will have the same price. First let's simplify the formula given for the rate of decrease of the price of wheat: $$\sqrt{2}*x-x=1.41x-x=0.41x$$, this means that the price of wheat decreases by $$0.41x$$ cents per day, in $$y$$ days it'll decrease by $$0.41xy$$ cents; As price of corn increases $$5x$$ cents per day, in $$y$$ days it'll will increase by $$5xy$$ cents; Set the equation: $$320+5xy=580-0.41xy$$, solve for $$xy$$ --> $$xy=48$$; The cost of a bushel of corn in $$y$$ days (the # of days when these two bushels will have the same price) will be $$320+5xy=320+5*48=560$$ or$5.6.

Bunuel,

While solving for xy i get xy = 56

5xy - 0.41xy = 580-320
4.59xy = 260
xy = 56.6 cents

Let me know if i missed something!
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]

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20 Feb 2011, 16:38
mbafall2011 wrote:
Bunuel wrote:
SOLUTION:
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 $$\sqrt{2}*x-x$$ 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 Note that we are not asked in how many days prices will cost the same. Let $$y$$ be the # of days when these two bushels will have the same price. First let's simplify the formula given for the rate of decrease of the price of wheat: $$\sqrt{2}*x-x=1.41x-x=0.41x$$, this means that the price of wheat decreases by $$0.41x$$ cents per day, in $$y$$ days it'll decrease by $$0.41xy$$ cents; As price of corn increases $$5x$$ cents per day, in $$y$$ days it'll will increase by $$5xy$$ cents; Set the equation: $$320+5xy=580-0.41xy$$, solve for $$xy$$ --> $$xy=48$$; The cost of a bushel of corn in $$y$$ days (the # of days when these two bushels will have the same price) will be $$320+5xy=320+5*48=560$$ or$5.6.

Bunuel,

While solving for xy i get xy = 56

5xy - 0.41xy = 580-320
4.59xy = 260
xy = 56.6 cents

Let me know if i missed something!

From 320+5xy=580-0.41xy --> 5xy+0.41xy=580-320 --> xy=48.
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]

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20 Feb 2011, 18:50
Bunuel wrote:
mbafall2011 wrote:
Bunuel wrote:
SOLUTION:
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 $$\sqrt{2}*x-x$$ 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 Note that we are not asked in how many days prices will cost the same. Let $$y$$ be the # of days when these two bushels will have the same price. First let's simplify the formula given for the rate of decrease of the price of wheat: $$\sqrt{2}*x-x=1.41x-x=0.41x$$, this means that the price of wheat decreases by $$0.41x$$ cents per day, in $$y$$ days it'll decrease by $$0.41xy$$ cents; As price of corn increases $$5x$$ cents per day, in $$y$$ days it'll will increase by $$5xy$$ cents; Set the equation: $$320+5xy=580-0.41xy$$, solve for $$xy$$ --> $$xy=48$$; The cost of a bushel of corn in $$y$$ days (the # of days when these two bushels will have the same price) will be $$320+5xy=320+5*48=560$$ or$5.6.

Bunuel,

While solving for xy i get xy = 56

5xy - 0.41xy = 580-320
4.59xy = 260
xy = 56.6 cents

Let me know if i missed something!

From 320+5xy=580-0.41xy --> 5xy+0.41xy=580-320 --> xy=48.

Sorry! my bad! that was silly of me
Re: TOUGH & TRICKY SET Of PROBLMS   [#permalink] 20 Feb 2011, 18:50

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