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Director
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Re: Find 0.420.35, 0.ab=0.ababababa......
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16 Sep 2019, 22:24
Thanks for letting me know. Edited!! Bunuel wrote: nick1816 wrote: Find 0.420.35
0.ab=0.ababababa......
A. \(0.7\)
B. 0.7
C. \(\frac{7}{9}\)
D. \(\frac{4}{13}\)
E. \(\frac{7}{99}\) Options B and C are the same. What is the source of this question?



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Re: Find 0.420.35, 0.ab=0.ababababa......
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16 Sep 2019, 22:30
nick1816 wrote: Thanks for letting me know. Edited!! Bunuel wrote: nick1816 wrote: Find 0.420.35
0.ab=0.ababababa......
A. \(0.7\)
B. 0.7
C. \(\frac{7}{9}\)
D. \(\frac{4}{13}\)
E. \(\frac{7}{99}\) Options B and C are the same. What is the source of this question? Could you please tell me what is the source of this question? Thank you.
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Re: Find 0.420.35, 0.ab=0.ababababa......
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16 Sep 2019, 22:37
Find 0.420.35
.42 nonterminating repeating will be equivalent to 42/99
Similarly .35 (NTR)= 35/99
42/9935/99 =7/99
E
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Seven balls of different weights are randomly painted red, orange, yel
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16 Sep 2019, 22:51
Seven balls of different weights are randomly painted red, orange, yellow, green, blue, indigo and violet, each ball being painted a distinct color. The green ball is found to be heavier than the blue ball, and the red ball is found to be heavier than the yellow ball. Given just this information, if the probability that the red ball is heavier than the blue ball is a/b , where a and b are coprime positive integers, find a + b.
A. 3 B. 5 C. 7 D. 9 E. 11



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Re: Find 0.420.35, 0.ab=0.ababababa......
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16 Sep 2019, 23:02
Difference would be 0.0707070.....
Obviously (A) and (B) are wrong and we can easily see that (C) and (D) are greater than 0.1
So Answer must be (E)



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Find 0.420.35, 0.ab=0.ababababa......
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16 Sep 2019, 23:57
x=0.4242.. and y=0.3535.., so Z = xy = 0.0707...
If Z= 0.0707.. and 100Z= 7.0707.., then: > 100ZZ = 7.0707..  0.0707.. > 99z = 7 > Z = 7/99
Answer is (E)



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n is a positive integer. Is n^3+2n is divisible by 12. 1) n is a prim
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17 Sep 2019, 10:39
n is a positive integer. Is \(n^3+2n\) is divisible by 12?
1) n is a prime number. 2) n>10



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Re: n is a positive integer. Is n^3+2n is divisible by 12. 1) n is a prim
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17 Sep 2019, 10:48
nick1816 wrote: n is a positive integer. Is \(n^3+2n\) is divisible by 12?
1) n is a prime number. 2) n>10 given n is an +ve integer and check whether n(n^2+2) is divisible by 12 #1 n is prime ; for n = 2 we get yes and for any other value no ; insufficient #2 n>10 again yes for n=12 and no for n=11 insufficient from 1 &2 n is prime and >10 , n(n^2+2) is not divisible by 12 IMO C



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The figure shows a cube of side 10 cm. If A and C are vertices of the
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04 Oct 2019, 00:01
The figure shows a cube of side 10 cm. If A and C are vertices of the cube and B and D are midpoints of the sides shown, what is the area of Quadrilateral ABCD? A. \(20\sqrt{5}\) B. \(50\sqrt{6}\) C. \(100\sqrt{3}\) D. 125 E. \(125\sqrt{2}\)
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The figure shows a cube of side 10 cm. If A and C are vertices of the
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04 Oct 2019, 00:53
nick1816 wrote: The figure shows a cube of side 10 cm. If A and C are vertices of the cube and B and D are midpoints of the sides shown, what is the area of Quadrilateral ABCD?
A. \(20\sqrt{5}\) B. \(50\sqrt{6}\) C. \(100\sqrt{3}\) D. 125 E. \(125\sqrt{2}\) Nice Question! We can see that all the sides AB, BC, CD & DA are equal in length and forms a right angled triangle of sides 10 & 5 and each of AB, BC, CD & DA as hypotenuse > ABCD is a Rhombus as length of diagonal AC = \(10\sqrt{3}\) & BD = \(10\sqrt{2}\) are different > Area of Rhombus = 1/2*product of diagonlas = 1/2*\(10\sqrt{3}\)*\(10\sqrt{2}\) = \(50\sqrt{6}\) IMO Option B



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Re: X, Y and Z are three positive integers such that X+Y+Z = 36
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05 Oct 2019, 20:10
Hovkial wrote: X, Y and Z are three positive integers such that X+Y+Z = 36. How many solutions are possible for this equation?
(A) 319
(B) 341
(C) 361
(D) 385
(E) 405 number of positive integral solutions of equation x1+x2+⋯+xr=n is equal to (n1)C(r1) = (361)C(31) = 35c2 = 595 Posted from my mobile device



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Re: X, Y and Z are three positive integers such that X+Y+Z = 36
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06 Oct 2019, 01:09
X+Y+Z=36 X,Y,Z>0 X=A+1 Y=B+1 Z=C+1 A,B,C are nonnegative integers then A+B+C=33 ABC Total number of ways to arrange 33 objects and 2 lines=\(\frac{(33+2)!}{33!2!}\)=35*34/2 = 595 Hovkial wrote: X, Y and Z are three positive integers such that X+Y+Z = 36. How many solutions are possible for this equation?
(A) 319
(B) 341
(C) 361
(D) 385
(E) 405



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What is the maximum value of 'N' such that the product 570 X 60
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06 Oct 2019, 09:53
What is the maximum value of 'N' such that the product 570 X 60 X 30 X 90 X 100 X 500 X 700 X 343 X 720 X 81 is completely divisible by \(30^N\)?
(A) 10
(B) 11
(C) 12
(D) 13
(E) 14



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Re: What is the maximum value of 'N' such that the product 570 X 60
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06 Oct 2019, 12:08
X=570 X 60 X 30 X 90 X 100 X 500 X 700 X 343 X 720 X 81 30=2*3*5 The highest power of 2 that can divide X=a= 1+2+1+1+2+2+2+0+4+0=15 The highest power of 3 that can divide X=b= 1+1+1+2+0+0+0+0+2+4=11 The highest power of 5 that can divide X=c= 1+1+1+1+2+3+2+0+1+0=12 N= minimum(a,b,c)=11 Hovkial wrote: What is the maximum value of 'N' such that the product 570 X 60 X 30 X 90 X 100 X 500 X 700 X 343 X 720 X 81 is completely divisible by \(30^N\)?
(A) 10
(B) 11
(C) 12
(D) 13
(E) 14



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Re: X, Y and Z are three positive integers such that X+Y+Z = 36
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07 Oct 2019, 02:57
Hovkial wrote: X, Y and Z are three positive integers such that X+Y+Z = 36. How many solutions are possible for this equation?
(A) 319
(B) 341
(C) 361
(D) 385
(E) 405 Hi HovkialCan you post the solution ? I think it’s 595 Posted from my mobile device



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Re: X, Y and Z are three positive integers such that X+Y+Z = 36
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07 Oct 2019, 16:00
We are given that:
X + Y + Z = 36.
The total number of solutions to this equation will be given by:
(36+31)C(31) = 38C2 = 703.
Note: "C" denotes combination.
The number of solutions where X will be equal to Y will be 19 [ from (X, Y) = (0, 0) to (18, 18)].
So, the number of solutions where X will not be equal to Y = 703  19 = 684.
Out of these 684 solutions, half of the solutions will have X > Y and the other half will have X < Y.
Hence, the total number of solutions where X will be greater than or equal to Y will be:
19 + (684/2) = 361.



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Re: X, Y and Z are three positive integers such that X+Y+Z = 36
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08 Oct 2019, 06:10
Hovkial wrote: We are given that:
X + Y + Z = 36.
The total number of solutions to this equation will be given by:
(36+31)C(31) = 38C2 = 703.
Note: "C" denotes combination.
The number of solutions where X will be equal to Y will be 19 [ from (X, Y) = (0, 0) to (18, 18)].
So, the number of solutions where X will not be equal to Y = 703  19 = 684.
Out of these 684 solutions, half of the solutions will have X > Y and the other half will have X < Y.
Hence, the total number of solutions where X will be greater than or equal to Y will be:
19 + (684/2) = 361. Why are you doing 31 for combination?



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Re: What is the maximum value of 'N' such that the product 570 X 60
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08 Oct 2019, 06:31
nick1816Can you elaborate your explanation? Thank you



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Re: X, Y and Z are three positive integers such that X+Y+Z = 36
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08 Oct 2019, 06:31
XYZ Number of ways to arrange 36 similar objects and 2 lines= 38!/36!2! or you can remember the formula \(x_1\)+\(x_2\).....+\(x_r\)= N where \(x_1\), \(x_2\) ....... \(x_r\) are nonnegative integers Total integral solutions= (N+r1) C (r1) In the above question N=36, and r=3 Anant2410 wrote: Hovkial wrote: We are given that:
X + Y + Z = 36.
The total number of solutions to this equation will be given by:
(36+31)C(31) = 38C2 = 703.
Note: "C" denotes combination.
The number of solutions where X will be equal to Y will be 19 [ from (X, Y) = (0, 0) to (18, 18)].
So, the number of solutions where X will not be equal to Y = 703  19 = 684.
Out of these 684 solutions, half of the solutions will have X > Y and the other half will have X < Y.
Hence, the total number of solutions where X will be greater than or equal to Y will be:
19 + (684/2) = 361. Why are you doing 31 for combination?



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Two springs are separated by a certain fixed distance between them
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08 Oct 2019, 15:26
Two springs are separated by a certain fixed distance between them and attached to two opposite posts on the left and right side facing each other. They are released towards each other at the same time. They meet at a point located between themselves at a distance that is 40 per cent of the total distance between them as measured from the left side. The springs continue to move back and forth between the two attached points. What is the distance measured in percentage of the total separation distance from the left side when the fourth meeting occurs?
(A) 30%
(B) 40%
(C) 50%
(D) 60%
(E) 80%




Two springs are separated by a certain fixed distance between them
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