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# TOUGH & TRICKY SET Of PROBLEMS

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Re: A and B ran, at their respective constant rates [#permalink]

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01 Nov 2012, 03:07
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jordanshl wrote:
SOLUTION:
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

I really don't understand Bunuel's explanation, can someone explain this answer as they would to a child. Thanks!

Let x be the speed of B.
Write the equation:

(480-48)/x (time of B for first heat) - 6 (seconds, time B lost to A first heat) = TIME OF A (in both heats A runs with constant rate, so the time for first and second heats are the same)=(480-144)/x (time of B for second heat) + 2 (seconds, time B won to A second heat)

(480-48)/x-6=(480-144)/x+2
x=12

Let A's speed = a, B's speed = b

First Heat:

Let time taken by B to run the distance (480-48)m be t seconds.
A travels the distance of 480m in t-6 seconds

$$\frac{480-48}{b} = t$$

$$\frac{480}{a} = t-6$$

So,

$$\frac{480}{a} = \frac{480-48}{b}-6$$

Similarly

$$\frac{480}{a} = \frac{480-144}{b}+2$$

So,

$$\frac{480-48}{b}-6 = \frac{480-144}{b}+2$$

$$\frac{144-48}{b} = 8$$

So,
b=12

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Re: A and B ran, at their respective constant rates [#permalink]

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01 Nov 2012, 06:44
jordanshl wrote:
SOLUTION:
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

I really don't understand Bunuel's explanation, can someone explain this answer as they would to a child. Thanks!

Merging topics.

Check alternate solutions given by Karishma:

tough-tricky-set-of-problms-85211-80.html#p970679
tough-tricky-set-of-problms-85211-80.html#p970681

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

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13 Nov 2012, 13:35
Bunuel wrote:
SOLUTION OF 8-10
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

do you have more these kind of questions?

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

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16 Dec 2012, 22:30
Bunuel wrote:
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.

Hi Bunuel

I guess the question should be re-worded to restrict the number of draws to 2.

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

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16 Dec 2012, 23:08
geneticsgene wrote:
Bunuel wrote:
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.

Hi Bunuel

I guess the question should be re-worded to restrict the number of draws to 2.

I think we are given that: "What is the probability of drawing a red and a white ball in two successive draws..."
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]

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20 Dec 2012, 16:20
Bunuel wrote:
SOLUTION OF 8-10

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

Hi Bunuel
amazing collection, but my answer to Q 9 is 25% and why i say so is shown below, Correct me if i am wrong:-

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

N to be divisible by 8 it must be a multiple be a 4 since we require three 2's; 4 has two 2's (n=2) i.e (4+2) brings another one.................
So to find the probability we must find total multiple of 4 from 1 to 96 and that is 96/4 = 24
total outcomes = 96
hence Probability = 24 /96 = 1/4 = 25%
i.e. option APls correct me if i am wrong!!!!!!!!!!!!!!!!!!!!!!!!!

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

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20 Dec 2012, 16:26
I think i made mistake since 6 can be also a probable number but is not a multiple of 4..............Bunuel can u pls help me solve this one using multiple principle

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

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21 Dec 2012, 12:57
First of all +1 for your explanation.................

LAst part i converted it into OR statement and solved...Pls confirm whether I am correct.

Prob = Prob of even number or Prob of multiple of 8

= 48/96 + 12/96
= 1/2 + 1/8
= 5/8 Ans

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

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17 Jan 2013, 22:44
Bunuel wrote:
SOLUTION OF 8-10

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

The "AND" should really be OR, since you mean either of the two cases will work.

Anyway, thx for the questions!

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

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25 Jan 2013, 01:22
Bunuel wrote:
angel2009 wrote:
Bunuel wrote:
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*3/9*2/9=4/27

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

I've some problem to understand this ; as the two events are mutually exclusive , independent.
So why not the answer is 3/9*2/9 = 2/27

Winning scenario consists of TWO cases RW and WR. Probability of each case is: 3/9*2/9, so 3/9*2/9+3/9*2/9=2*3/9*2/9.

Only 3/9*2/9 would be correct answer if we were asked to determine the probability of FIRST ball being red and SECOND white, OR FIRST white and SECOND red. These are the cases in which order of drawing matters.

P(1R1W) = 3C1 x 2C1 / (9C2) = 1/6 WHY? what am i doing wrong?

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

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25 Jan 2013, 04:55
MegW wrote:
Bunuel wrote:
angel2009 wrote:
I've some problem to understand this ; as the two events are mutually exclusive , independent.
So why not the answer is 3/9*2/9 = 2/27

Winning scenario consists of TWO cases RW and WR. Probability of each case is: 3/9*2/9, so 3/9*2/9+3/9*2/9=2*3/9*2/9.

Only 3/9*2/9 would be correct answer if we were asked to determine the probability of FIRST ball being red and SECOND white, OR FIRST white and SECOND red. These are the cases in which order of drawing matters.

P(1R1W) = 3C1 x 2C1 / (9C2) = 1/6 WHY? what am i doing wrong?

Hint: notice that each ball is put back after it is drawn, so each time we choose 1 ball out of 9.
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]

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02 Feb 2013, 11:33
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.

I did it the following way

Line passes thr' points (0,-3) & (4,0) - The Area of triangle formed by the line = 6 (A = 1/2 (4*3))
Hypotenuse of this triangle = 5 therefore A=1/2 b*h i.e. 6 = (1/2) * 5 * h therefore h=2.4 distance = 2.4-1 = 1.4

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

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07 Feb 2013, 00:09
Bunuel wrote:
SOLUTION:
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 x be the speed of B.
Write the equation:

(480-48)/x (time of B for first heat) - 6 (seconds, time B lost to A first heat) = TIME OF A (in both heats A runs with constant rate, so the time for first and second heats are the same)=(480-144)/x (time of B for second heat) + 2 (seconds, time B won to A second heat)

(480-48)/x-6=(480-144)/x+2
x=12

how come
(480-48)/(x-6)=(480-144)/(x+2) is equal to x=12?
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]

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07 Feb 2013, 00:19
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.

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

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07 Feb 2013, 00:20
Bunuel wrote:
SOLUTION OF 8-10
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

Average=50, Sum of temperatures=50*5=250
As the min temperature is 45, max would be 250-4*45=70 --> The range=70(max)-45(min)=25

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

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

Sum=A, next 5 consecutive will gain additional 5*5=25, so sum of the next five consecutive integers in terms of A is: A+25

I don't understand the following:
>n=8p-1 --> 8p-1<=96 --> p=12.3 --> 12 such nembers

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

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13 Apr 2013, 16:10
Hello,

Can you help me understand "Probability of a randomly selected person NOT to be born in a leap year=3/4"?

-Thanks,

Bunuel wrote:
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.

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

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13 Apr 2013, 16:27
NBA1234 wrote:
Hello,

Can you help me understand "Probability of a randomly selected person NOT to be born in a leap year=3/4"?

-Thanks,

Bunuel wrote:
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.

Check here: tough-tricky-set-of-problems-85211-100.html#p1069456
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Re: TOUGH & TRICKY SET Of PROBLEMS [#permalink]

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27 Jun 2013, 12:17
In GMAT Club Math Book v3 - Jan-2-2013.pdf , Page 9

Sum of n first positive even numbers:n(n+1)

Since sum is: 79x80 so 79 positive integers = 79x2=158.
Since K is odd, K = 159

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

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24 Aug 2013, 08:01
Hi Brunel,

Thank you for the questions and the explanations as well.

I have a question. I was only able to solve 3 out of 10. (Q8/9/10) - I found the other ones really hard. It has demotivated me. I am aiming for 49-50 in Maths. Please suggest.

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

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24 Aug 2013, 16:25
Bunuel wrote:
As asked I'm combining all the problems from Tough & tricky problems in single thread. Here are the first ten questions. Next set and solutions to these ten will follow in couple of hours.

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

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

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

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

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

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

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

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

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%

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

THE OA WITH SOLUTIONS WILL BE PROVIDED.

upload these sorts of questions every week.... So amazing to have all in one post............ great work Boss
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Re: TOUGH & TRICKY SET Of PROBLEMS   [#permalink] 24 Aug 2013, 16:25

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