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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

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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
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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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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

Answer is (E)
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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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nick1816
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 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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nick1816
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
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Hovkial
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

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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
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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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
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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Hovkial
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
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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.
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Hovkial
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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nick1816

Can you elaborate your explanation?

Thank you
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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
Hovkial
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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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%
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