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Instructions

There are six spherical balls, B1, B2, B3, B4, B5, and B6, and four circular hoops H1, H2, H3, and H4.

Each ball was tested on each hoop once, by attempting to pass the ball through the hoop. If the diameter of a ball is not larger than the diameter of the hoop, the ball passes through the hoop and makes a “ping”. Any ball having a diameter larger than that of the hoop gets stuck on that hoop and does not make a ping.

The following additional information is known:
1. B1 and B6 each made a ping on H4, but B5 did not.
2. B4 made a ping on H3, but B1 did not.
3. All balls, except B3, made pings on H1.
4. None of the balls, except B2, made a ping on H2.

Question 34

What was the total number of pings made by B1, B2, and B3?


Correct Answer: 6

Let us try to arrange the hoops and the spheres in increasing order of their diameter according to the information given,

In clue 1, we are given that B1 and B6 made a ping on H4, and B5 did not.

So, we can definitely say that,

(B1, B6) < H4 < B5  --(1)

In clue 2, we are given that B4 made a ping on H3, but B1 did not.

So, we can definitely say that,

B4 < H3 < B1 --(2)

In clue 3, we are given that all balls, except B3, made pings on H1.

So, we can definitely say that,

(B1, B2, B4, B5, B6) < H1 < B3 --(3)

In clue 4, we are given that none of the balls, except B2, made a ping on H2.

So, we can definitely say that,

B2 < H2 < (B1, B3, B4, B5, B6) --(4)

Combining (1) and (2), we can definitely say that

B4 < H3 < B1 < H4 < B5 --(5)

The only ball that we do not have enough information is about B6, as it can be either less than H3 or greater than H3 and less than H4.

Combining (3), (4) and (5), we get,

B2 < H2 < B4 < H3 < B1 < H4 < B5 < H1 < B3

The sphere B6 can be placed in 2 positions, giving us two possible inequalities, which are

B2 < H2 < (B4, B6) < H3 < B1 < H4 < B5 < H1 < B3

B2 < H2 < B4 < H3 < (B1,B6) < H4 < B5 < H1 < B3

The position of B6 is not clear, and except for that, all the positions are fixed.

Pings made by B1 = H4, H1 = 2

Pings made by B2 = H2, H3, H4, H1 = 4

Pings made by B3 = 0

Total number of pings by B1, B2 and B3 = 2 + 4 + 0 = 6.

Hence, the correct answer is 6.

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