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VP
Joined: 19 Oct 2018
Posts: 1289
Location: India
Re: Find 0.42-0.35, 0.ab=0.ababababa......  [#permalink]

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16 Sep 2019, 22:24
Thanks for letting me know. Edited!!
Bunuel wrote:
nick1816 wrote:
Find 0.42-0.35

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?
Math Expert
Joined: 02 Sep 2009
Posts: 60460
Re: Find 0.42-0.35, 0.ab=0.ababababa......  [#permalink]

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16 Sep 2019, 22:30
nick1816 wrote:
Thanks for letting me know. Edited!!
Bunuel wrote:
nick1816 wrote:
Find 0.42-0.35

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.
_________________
Manager
Joined: 10 May 2018
Posts: 68
Re: Find 0.42-0.35, 0.ab=0.ababababa......  [#permalink]

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16 Sep 2019, 22:37
1
Find 0.42-0.35

.42 nonterminating repeating will be equivalent to 42/99

Similarly .35 (NTR)= 35/99

42/99-35/99
=7/99

E

Posted from my mobile device
VP
Joined: 19 Oct 2018
Posts: 1289
Location: India
Seven balls of different weights are randomly painted red, orange, yel  [#permalink]

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16 Sep 2019, 22:51
1
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 co-prime positive integers, find a + b.

A. 3
B. 5
C. 7
D. 9
E. 11
Director
Joined: 16 Jan 2019
Posts: 534
Location: India
Concentration: General Management
WE: Sales (Other)
Re: Find 0.42-0.35, 0.ab=0.ababababa......  [#permalink]

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16 Sep 2019, 23:02
1
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)
Director
Joined: 30 Sep 2017
Posts: 570
GMAT 1: 720 Q49 V40
GPA: 3.8
Find 0.42-0.35, 0.ab=0.ababababa......  [#permalink]

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16 Sep 2019, 23:57
1
x=0.4242.. and y=0.3535.., so Z = x-y = 0.0707...

If Z= 0.0707.. and 100Z= 7.0707.., then:
--> 100Z-Z = 7.0707.. - 0.0707..
--> 99z = 7
--> Z = 7/99

VP
Joined: 19 Oct 2018
Posts: 1289
Location: India
n is a positive integer. Is n^3+2n is divisible by 12. 1) n is a prim  [#permalink]

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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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Joined: 18 Aug 2017
Posts: 5692
Location: India
Concentration: Sustainability, Marketing
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Re: n is a positive integer. Is n^3+2n is divisible by 12. 1) n is a prim  [#permalink]

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17 Sep 2019, 10:48
1
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
VP
Joined: 19 Oct 2018
Posts: 1289
Location: India
The figure shows a cube of side 10 cm. If A and C are vertices of the  [#permalink]

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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 mid-points 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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VP
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Posts: 1231
Location: India
Concentration: Entrepreneurship, Marketing
WE: Education (Education)
The figure shows a cube of side 10 cm. If A and C are vertices of the  [#permalink]

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04 Oct 2019, 00:53
3
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 mid-points 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
VP
Joined: 20 Jul 2017
Posts: 1231
Location: India
Concentration: Entrepreneurship, Marketing
WE: Education (Education)
Re: X, Y and Z are three positive integers such that X+Y+Z = 36  [#permalink]

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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 (n-1)C(r-1)

= (36-1)C(3-1) = 35c2 = 595

Posted from my mobile device
VP
Joined: 19 Oct 2018
Posts: 1289
Location: India
Re: X, Y and Z are three positive integers such that X+Y+Z = 36  [#permalink]

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06 Oct 2019, 01:09
1
X+Y+Z=36
X,Y,Z>0

X=A+1
Y=B+1
Z=C+1
A,B,C are non-negative integers

then
A+B+C=33

A|B|C
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
Senior Manager
Status: PhD-trained. Education, Research, Teaching, Training, Consulting and Advisory Services
Joined: 23 Apr 2019
Posts: 438
What is the maximum value of 'N' such that the product 570 X 60  [#permalink]

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06 Oct 2019, 09:53
5
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
VP
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Posts: 1289
Location: India
Re: What is the maximum value of 'N' such that the product 570 X 60  [#permalink]

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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
VP
Joined: 20 Jul 2017
Posts: 1231
Location: India
Concentration: Entrepreneurship, Marketing
WE: Education (Education)
Re: X, Y and Z are three positive integers such that X+Y+Z = 36  [#permalink]

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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 Hovkial

Can you post the solution ?
I think it’s 595

Posted from my mobile device
Senior Manager
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Joined: 23 Apr 2019
Posts: 438
Re: X, Y and Z are three positive integers such that X+Y+Z = 36  [#permalink]

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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+3-1)C(3-1) = 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.
Intern
Joined: 09 Apr 2018
Posts: 6
Re: X, Y and Z are three positive integers such that X+Y+Z = 36  [#permalink]

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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+3-1)C(3-1) = 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 3-1 for combination?
Intern
Joined: 19 Sep 2019
Posts: 2
Re: What is the maximum value of 'N' such that the product 570 X 60  [#permalink]

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08 Oct 2019, 06:31
nick1816

Can you elaborate your explanation?

Thank you
VP
Joined: 19 Oct 2018
Posts: 1289
Location: India
Re: X, Y and Z are three positive integers such that X+Y+Z = 36  [#permalink]

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08 Oct 2019, 06:31
X|Y|Z

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 non-negative integers

Total integral solutions= (N+r-1) C (r-1)

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+3-1)C(3-1) = 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 3-1 for combination?
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Status: PhD-trained. Education, Research, Teaching, Training, Consulting and Advisory Services
Joined: 23 Apr 2019
Posts: 438
Two springs are separated by a certain fixed distance between them  [#permalink]

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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   [#permalink] 08 Oct 2019, 15:26

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