1. Good post? |

2 Tricky DS

1. A certain jar contains only b black marbles, w white marbles, and r red marbles. If one marble is to be chosen at random from the jar, is the probability that the marble chosen will be red greater than the probability that the marble chosen will be white?

a) r/(b+w) > w/(b+r)
b) b-w > r

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2. The integers m and p are such that 2<m<p and m is not a factor of p. If r is the remainder when p is divided by m, is r>1?

a) The greatest common factor of m and p is 2
b) The least common multiple of m and p is 30

2. Good post? |
I think the answers should be B, D

Note if HCF is 2, remainder canot be less than 2.
Also for LCM as 30 , so in this case remainder will always be equal to 1 or 0( but this is out of scope as m is not the factor of p)

3. Good post? |
Originally Posted by hitzs
I think the answers should be B, D

4. Good post? |

Not sure about 2...might be E

5. Good post? |

1. r/(b+w)>w/(b+r) .... since all numbers are positive ... cross multiply
==> rb +r^2 > wb + w^2
==> rb + r^2 + rw > wb + w^2 + rw
==> r(r+b+w) > w(b+w +r)
==> r/(b+w+r) > 2/(b+w +r)

statement 2. this statment does not lead to any conclusion.

Q2. stat 1. is insufficient. we know that 2 is common factor in both of the numbers.
stat 2. is insufficient on its own. we know that LCM = 30 = 2*3*5

We know that p > m; the only possible answer for p =10 and m =6
Combine the two statments and we can answer the question.

I would choose C.

6. Good post? |
Sorry, in first Quest I mis-read probability of black and white ...it should be A only

7. Good post? |
I think the answer to number 2 is A.

Statement 1 tells us that m and p are both even. Since m is not a factor of p, the remainder cannot be 0. And if m and p are both even, the remainder cannot be 1. Therefore the remainder is greater than 1. Statement 1 is sufficient.

With statement 2, it is possible that m = 3 and p = 10, in which case the remainder is 1 and the answer to the question is no. But it is also possible that m = 6 and p = 10, in which case the remainder is 4 and the answer to the question is yes. Two different answers... statement 2 is insufficient.

8. Good post? |
Yep. The answers are A, A.

In problem two statement a says that both integers are even. Hence the remainder will always be => 2
Statement b - gives two diff. answers. so insuff.

9. Good post? |

we need to find whether r/[b+w+r] > w/[b+r+w]
take r/b+w > w/b+r : w,r,b all are integers >0
r/[b+w]+1 > w/[b+r]+1

r+b+w/b+w > w+b+r/ b+r so 1/b+w>1/b+r that comes to b+r > b+w

or in other words r>w

10. Good post? |
the answer to the secound question is A

The HCF of m and P is 2 That is m and P are even integers and p > m so the other factor(s) of p must be greater than the other factor(s) of m so when p/m the remainder r should be >1

The LCM of m and p = 30 , take 5 and 6 r> 1 but take 30 and 6 LCM = 30 but r = 0

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