Quant Boosters  Rajesh Balasubramanian  Set 3

For this type of question, we need to consider only the internal arrangement within the M and 2As.
M and 2As can be rearranged as AMA, AAM, or MAA.
So, the probability that M will feature between the 2As is 1/3.
Now, let us think why we need to consider only the M and 2As.
Let us start by considering a set of words where the M and 2 As are placed at positions 2, 3 and 5.
The other three letters have to be in slots 1, 4 and 6
Three letters can be placed in three different slots in 3! = 6 ways.
Now with __ M A __ A __ there are 6 different words.
With __ A M __ A __ there are 6 different words.
With __ A A __ M __ there are 6 different words.
For each selection of the positions for A,A and M, exactly onethird of words will have M between the two A’s.
This is why only the internal arrangement between A, A and M matters.
So, probability of M being between 2 As is 1/3.

Q27) N is a 3digit number that is a multiple of 7; what is the probability that it will be a multiple of 5?
a) 1/5
b) 11/54
c) 13/64
d) 13/66

N is a three digit multiple of 7.
N could be 105, 112, 119, 126…..994.
Or, 15 × 7, 16 × 7…..142 × 7.
Or there are 142–14 = 128 numbers.
Within these we need to locate the multiples of 5.
Or, we need to isolate multiples of 35.
Or, we need to see how many numbers there are in the list 105, 140, 175…..980.
35 × 3, 35 × 4…35 × 28…
Or, there are 26 such numbers.
Probability 26/128 = 13/64
Answer choice (c).

Q28) A boss decides to distribute Rs. 2000 between 2 employees. He knows X deserves more that Y, but does not know how much more. So he decides to arbitrarily break Rs. 2000 into two parts and give X the bigger part. What is the chance that X gets twice as much as Y or more?
a) 2/5
b) 1/2
c) 1/3
d) 2/3

The bigger part could be any number from 1000 to 2000.
Now, if the bigger part is to be at least twice as much as the smaller part, we have
X ≥ 2Y or X ≥ 2(2000 – X)
Or X ≥ 4000/3
Given that X lies between 1000 and 2000, what is the probability that X lies between 4000/3 and 2000?
This probability is equal to (2000 − 4000/3)/(2000 − 1000) = 2/3
Answer choice (d).

Q29) What is sum of all rearrangements of the 4digit number 3214?
a) 66660
b) 55554
c) 60048
d) 65024

Total number of rearrangements = 4! Ways = 24.
Each of the four digits is likely to appear in the units place , therefore  th of 24 = 6 numbers will have 2 in the units place, 6 will have 1 in the units place and so on..
Therefore, sum of units place = 6*(2+3+1+4) = 60
Sum of tens place = 6*(20+30+10+40) = 600.
Sum of hundreds place = 6*1000 =6000
Sum of thousands place = 6*10000 =60000. Therefore, sum of all rearrangements of 3214 = 66660.

Q30) A bag contains 4 red and 3 black balls. A second bag contains 2 red and 3 black balls. One bag is selected at random. If from the selected bag one ball is drawn, then what is the probability that the ball drawn is red?
a) 39/70
b) 41/70
c) 29/70
d) 17/35

Let Red balls be 'r' and brown balls be 'b'
P(r) = p(b1) * p(r) + p(b2) * p(r)
P(R) = 1/2 ∗ 4/7 + 1/2 ∗ 2/5
P(R) = 17/35

@rajesh_balasubramanian Sir, if we take x=0 and y = 1050, then HCF(0,1050) = 1050. Or am I missing something?