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Tricky problem sets [#permalink ]
04 Oct 2013, 21:47
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Three men, Alpha,Beta, and Gamma, working together, do a job in 6 hours less time than Alpha alone, in 1 hour less time than Beta alone, and in one-half the time needed by Gamma when working alone. let h be the number of hours needed by Alpha and Beta, working together to do the job. Then h equals:

A)

\frac{5}{2} B)

\frac{3}{2} C)

\frac{4}{3} D)

\frac{5}{4} E)

\frac{3}{4}
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If a,b,c are positive real numbers such that a(b+c) = 152, b [#permalink ]
04 Oct 2013, 21:55
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Both roots of the quadratic equation x^2 - 63x + k = 0 are p [#permalink ]
04 Oct 2013, 21:58
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Both roots of the quadratic equation

x^2 - 63x + k = 0 are prime numbers. The number of possible values of k is

A) 0

B) 1

C) 2

D) 4

E) more than four

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If x,y, and z are positive numbers satisfying [#permalink ]
04 Oct 2013, 22:07
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If x,y, and z are positive numbers satisfying

x + \frac{1}{y} = 4 ,

y + \frac{1}{z} = 1 and

z + \frac{1}{x} = \frac{7}{3} then xyz =

A)

\frac{2}{3} B)

1 C)

\frac{4}{3} D)

2 E)

\frac{7}{3}
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Two non-zero real numbers, a and b, satisfy ab = a-b. which [#permalink ]
04 Oct 2013, 22:13
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Two non-zero real numbers, a and b, satisfy ab = a-b. which of the following is a possible value of

\frac{a}{b} + \frac{b}{a} - ab ?

a)

-2 b)

\frac{-1}{2} c)

\frac{1}{3} d)

\frac{1}{2} e)

2
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Re: Two non-zero real numbers, a and b, satisfy ab = a-b. which [#permalink ]
04 Oct 2013, 22:28
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fozzzy wrote:

Two non-zero real numbers, a and b, satisfy ab = a-b. which of the following is a possible value of \frac{a}{b} + \frac{b}{a} - ab ? a) -2 b) \frac{-1}{2} c) \frac{1}{3} d) \frac{1}{2} e) 2

Hi Fozzzy,

Given : ab = a-b

Solution :

a/b + b/a - ab

(a^2+ b^2)/ab - ab

[(a-b)^2 + 2ab]/ab - ab (We know a^2 + b^2 can be written as (a-b)^2 + 2ab)

Now,

[(ab)^2 + 2ab]/ab - ab

ab + 2 - ab

Answer = 2

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Re: Both roots of the quadratic equation x^2 - 63x + k = 0 are p [#permalink ]
04 Oct 2013, 22:46
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fozzzy wrote:

Both roots of the quadratic equation x^2 - 63x + k = 0 are prime numbers. The number of possible values of k is A) 0 B) 1 C) 2 D) 4 E) more than four

Given : Both the roots are prime.

Solution : From the equation we know that the sum of the roots = 63 ( which is an odd number )

We know except 2, rest of the prime numbers are odd. Also adding 2 odd number will give even number.

To get the sum as odd number (here 63), we need an even prime & an odd prime. Only even prime is 2 and the odd prime is 61.

Hence there is only one pair of roots (2,61) and there can be only one solution for K.

Cheers

Qoofi

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Re: If a,b,c are positive real numbers such that a(b+c) = 152, b [#permalink ]
04 Oct 2013, 22:51
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fozzzy wrote:

If a,b,c are positive real numbers such that a(b+c) = 152, b(c+a) = 162, and c(a+b) = 170, then abc is A) 672 B) 688 C) 704 D) 720 E) 750

From Given, we can see

ab+ac = 152 (1), bc + ab = 162 (2), ac + bc = 170 (3)

Solving (1) & (2), we get ac - bc = -10 (4) & Solving (2) & (3), we get ab - ac = -8 (5)

Now solving (4) & (5), we get ab - ac = -18 (6). Solve equation (6) with (1), we get

ab = 72 Hence ac = 80 and bc = 90.

Multing all 3 values, (ab)x(bc)x(ca) = 72 x 90 x 80

(abc) ^2 = 72 x 90 x 80

Which gives

abc = 720 This method is tedious & time consuming. I am sure there has to be a simpler way to find abc. I will work on it.

Cheers

Qoofi

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Qoofi on 05 Oct 2013, 01:13, edited 1 time in total.

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Re: If a,b,c are positive real numbers such that a(b+c) = 152, b [#permalink ]
04 Oct 2013, 23:54

Qoofi wrote:

fozzzy wrote:

If a,b,c are positive real numbers such that a(b+c) = 152, b(c+a) = 162, and c(a+b) = 170, then abc is A) 672 B) 688 C) 704 D) 720 E) 750

From Given, we can see

ab+ac = 152 (1), bc + ab = 162 (2), ac + bc = 170 (3)

Solving (1) & (2), we get

ac - bc = -8 (4) & Solving (2) & (3), we get ab - ac = -8 (5)

Now solving (4) & (5), we get

ab - ac = -18 (6). Solve equation (6) with (1), we get

ab = 72 Hence ac = 80 and bc = 90 .

Multing all 3 values, (ab)x(bc)x(ca) = 72 x 90 x 80

(abc) ^2 = 72 x 90 x 80

Which gives

abc = 720 This method is tedious & time consuming. I am sure there has to be a simpler way to find abc. I will work on it.

Cheers

Qoofi

I believe there are some typos That should be ac - bc = -10, ab - bc = -18 ab = 62 and bc and ac ( 90,80) will switch values! kudos

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Re: Both roots of the quadratic equation x^2 - 63x + k = 0 are p [#permalink ]
05 Oct 2013, 00:08

so the only value for k is 61 * 2 = 122

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Re: Two non-zero real numbers, a and b, satisfy ab = a-b. which [#permalink ]
05 Oct 2013, 00:18

Qoofi wrote:

fozzzy wrote:

Two non-zero real numbers, a and b, satisfy ab = a-b. which of the following is a possible value of \frac{a}{b} + \frac{b}{a} - ab ? a) -2 b) \frac{-1}{2} c) \frac{1}{3} d) \frac{1}{2} e) 2

Hi Fozzzy,

Given : ab = a-b

Solution :

a/b + b/a - ab

(a^2+ b^2)/ab - ab

[(a-b)^2 + 2ab]/ab - ab (We know a^2 + b^2 can be written as (a-b)^2 + 2ab) Now,

[(ab)^2 + 2ab]/ab - ab

ab + 2 - ab

Answer = 2

Can you explain that step are you adding 2ab? if you are then you have subtract that variable too isn't it? or this is something else...

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Re: Two non-zero real numbers, a and b, satisfy ab = a-b. which [#permalink ]
05 Oct 2013, 00:41
1

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fozzzy wrote:

Qoofi wrote:

fozzzy wrote:

Two non-zero real numbers, a and b, satisfy ab = a-b. which of the following is a possible value of \frac{a}{b} + \frac{b}{a} - ab ? a) -2 b) \frac{-1}{2} c) \frac{1}{3} d) \frac{1}{2} e) 2

Hi Fozzzy,

Given : ab = a-b

Solution :

a/b + b/a - ab

(a^2+ b^2)/ab - ab

[(a-b)^2 + 2ab]/ab - ab (We know a^2 + b^2 can be written as (a-b)^2 + 2ab) Now,

[(ab)^2 + 2ab]/ab - ab

ab + 2 - ab

Answer = 2

Can you explain that step are you adding 2ab? if you are then you have subtract that variable too isn't it? or this is something else...

We know

(a - b ) ^2 = a^2 + b^2 - 2ab Now we are substituting

a^2 + b ^2 = (a-b)^2 + 2ab Expanding

(a-b)^2 + 2ab = a^2 + b^2 - 2ab + 2ab = a^2 + b^2
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Re: Two non-zero real numbers, a and b, satisfy ab = a-b. which [#permalink ]
05 Oct 2013, 00:48

@qoofi thanks makes perfect sense now kudos...

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Re: If a,b,c are positive real numbers such that a(b+c) = 152, b [#permalink ]
05 Oct 2013, 01:11

fozzzy wrote:

Qoofi wrote:

fozzzy wrote:

If a,b,c are positive real numbers such that a(b+c) = 152, b(c+a) = 162, and c(a+b) = 170, then abc is A) 672 B) 688 C) 704 D) 720 E) 750

From Given, we can see

ab+ac = 152 (1), bc + ab = 162 (2), ac + bc = 170 (3)

Solving (1) & (2), we get

ac - bc = -8 (4) & Solving (2) & (3), we get ab - ac = -8 (5)

Now solving (4) & (5), we get ab - ac = -18 (6). Solve equation (6) with (1), we get

ab = 72 Hence ac = 80 and bc = 90 .

Multing all 3 values, (ab)x(bc)x(ca) = 72 x 90 x 80

(abc) ^2 = 72 x 90 x 80

Which gives

abc = 720 This method is tedious & time consuming. I am sure there has to be a simpler way to find abc. I will work on it.

Cheers

Qoofi

I believe there are some typos That should be ac - bc = -10, ab = 62 and bc and ac ( 90,80) will switch values! kudos

Sorry I made a typo mistake.

ac - bc = -10
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Re: Three men, Alpha,Beta, and Gamma, working together, do a job [#permalink ]
05 Oct 2013, 02:46

fozzzy wrote:

Three men, Alpha,Beta, and Gamma, working together, do a job in 6 hours less time than Alpha alone, in 1 hour less time than Beta alone, and in one-half the time needed by Gamma when working alone. let h be the number of hours needed by Alpha and Beta, working together to do the job. Then h equals: A) \frac{5}{2} B) \frac{3}{2} C) \frac{4}{3} D) \frac{5}{4} E) \frac{3}{4}

This one is bit of a toughie. Or I can say it required computation. Lets begin.

Let x be the time required for (A, B, G) together to complete the work. Their rate =

1/x ( Work done per time )

now Time for A to complete work = x + 6, its rate =

1 / (x+6) Similarly Time for B = x + 1 and rate =

1/(x+1) & Time for G = 2x and rate =

1/2x - Eqn (1)

Work done by Alpha A & Beta B per hour =

1/(x+6) + 1/(x+1) Now, we can say work done by Gamma G by subtracting the work done per time of all 3 together with work done per time by Alpha A & Beta B

Which is,

(1/x) - ( 1/(x+6) + 1/(x+1) ) - Eqn (2) Now equate Eqn (1) & (2)

1/2x = 1/x - ( 1/(x+6) + 1/(x+1) ) After cross multiplying, the above equation will come down to

(3x+2)(x+3) x= 2/3 or -3 Leaving the negative solution, substitute x = 2/3 in

1/(x+6) + 1/(x+1) We get rate of work done by A & B together = 3/4.

Time req = Work / rate = 1/ (3/4) =

4/3 Tough one. I hope it is clear and really looking forward for a simpler solution.

Cheers

Qoofi

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Re: Tricky problem sets [#permalink ]
05 Oct 2013, 05:05

A teacher gave a test to a class in which 10% of the students are juniors and 90% are seniors. the average score on the test was 84. The juniors all received the same score, and the average score of the seniors was 83. what score did each of the juniors receive on the test?

A) 85

B) 88

C) 93

D) 94

E) 98

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Re: Tricky problem sets [#permalink ]
05 Oct 2013, 07:38
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fozzzy wrote:

A teacher gave a test to a class in which 10% of the students are juniors and 90% are seniors. the average score on the test was 84. The juniors all received the same score, and the average score of the seniors was 83. what score did each of the juniors receive on the test? A) 85 B) 88 C) 93 D) 94 E) 98

Lets say there are 100 students and there are 10 Juniors & 90 Seniors.

X is the score of Juniors.

(10 X + 83*90) / 100 = 84

Solving you get X = 93

Hope it is clear.

Cheers

Qoofi

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Re: Tricky problem sets [#permalink ]
05 Oct 2013, 08:31

New problem added! Kudos for correct solutions

call a number "prime-looking" if it is composite but not divisible by 2,3, or 5. The three smallest prime-looking numbers are 49,77 and 91. There are 168 prime numbers less than 1000. How many prime-looking numbers are there less than 1000?

A) 100

B) 102

C) 104

D) 106

E) 108

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Re: If x,y, and z are positive numbers satisfying [#permalink ]
05 Oct 2013, 12:36
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fozzzy wrote:

If x,y, and z are positive numbers satisfying

x + \frac{1}{y} = 4 ,

y + \frac{1}{z} = 1 and

z + \frac{1}{x} = \frac{7}{3} then xyz =

A)

\frac{2}{3} B)

1 C)

\frac{4}{3} D)

2 E)

\frac{7}{3} This one is indeed a Tricky one. Took me sometime.

Multiply all 3 eqns.

(x + \frac{1}{y})(y + \frac{1}{z})(z + \frac{1}{x}) = 4*1*\frac{7}{3} xyz + x + \frac{1}{y} + y + \frac{1}{z} + z + \frac{1}{x} + \frac{1}{xyz} = \frac{28}{3} xyz + \frac{1}{xyz} + 4 + 1 + \frac{7}{3} = \frac{28}{3} xyz + \frac{1}{xyz} = \frac{28}{3} - \frac{22}{3} xyz + \frac{1}{xyz} = 2 Now multiply xyz on both sides, we get

(xyz)^2 - 2xyz + 1 = 0 This is

(xyz - 1)^2) = 0 Hence

xyz = 1 Option B. Hope it helped.

Cheers

Qoofi

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Re: Tricky problem sets [#permalink ]
05 Oct 2013, 22:02

New problem added! Kudos for correct solutions

If

60^a = 3 and

60^b = 5, then

12^ {\frac{{ 1-a-b}}{2(1-b)}} A)

\sqrt{3} B) 2

C)

\sqrt{5} D) 3

E)

\sqrt{12}
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Re: Tricky problem sets
[#permalink ]
05 Oct 2013, 22:02