Six web surfers M, N, O, P, X, and Y each had 30 stars which they distributed among four bloggers A, B, C, and D. The number of stars received by A and B from the six web surfers is shown in the figure below.
The following additional facts are known regarding the number of stars received by the bloggers from the surfers.
- The numbers of stars received by the bloggers from the surfers were all multiples of 5 (including 0).
- The total numbers of stars received by the bloggers were the same.
- Each blogger received a different number of stars from M.
- Two surfers gave all their stars to a single blogger. 5. D received more stars than C from Y.
Solution
| A | B | C | D | Total | |
|---|---|---|---|---|---|
| M | |||||
| N | |||||
| O | |||||
| P | |||||
| X | |||||
| Y | |||||
| Total |
From the set following table will be made, in the first column we have listed all 6 web surfers (M,N,O,P,X,Y)
And in the first row we have list down all 4 bloggers (A,B,C,D)
Converting bar chart into table will help us to manipulate data & work with the total constraints.
| A | B | C | D | Total | |
|---|---|---|---|---|---|
| M | 10 | 0 | 30 | ||
| N | 25 | 0 | 30 | ||
| O | 0 | 0 | 30 | ||
| P | 5 | 25 | 30 | ||
| X | 0 | 0 | 30 | ||
| Y | 5 | 20 | 30 | ||
| Total | 45 | 45 | 45 | 45 | 180 |
Following data can be deducted from the figure and additional facts given in the set.
Data of stars received by A & B is given in the chart so we can directly plot them into the table.
As the question mentions each 6 web surfers have received 30 stars each. therefore, we can assign 30 to each surfer in the total column.
And so we get the sum of all stars as 30+30+30+30+30+30 = 180
Now using the information give in clue 2,
We know that each blogger has received same number of stars
which gives us 180/4 = 45 stars being received by each blogger.
| A | B | C | D | Total | |
|---|---|---|---|---|---|
| M | 10 | 0 | 30 | ||
| N | 25 | 0 | 30 | ||
| O | 0 | 0 | 30 | ||
| P | 5 | 25 | 0 | 0 | 30 |
| X | 0 | 0 | 30 | ||
| Y | 5 | 20 | 30 | ||
| Total | 45 | 45 | 45 | 45 | 180 |
For P, we already know by chant that it has given 5 stars to A and 25 stars to B, Which sums up to 5 +25 = 30
As we already know the total number of stars it can give is 30.
Therefore, P will give 0 stars to both C & D.
| A | B | C | D | Total | |
|---|---|---|---|---|---|
| M | 10 | 0 | 30 | ||
| N | 25 | 0 | 30 | ||
| O | 0 | 0 | 30 | ||
| P | 5 | 25 | 0 | 0 | 30 |
| X | 0 | 0 | 30 | ||
| Y | 5 | 20 | 0 | 5 | 30 |
| Total | 45 | 45 | 45 | 45 | 180 |
According to clue 5,
We are told that D receives more stars than C from Y. Considering Y has already given 25 stars, it will give 0 stars to C and 5 stars to D.
(As according to clue 1, number of stars received by each blogger from each surfer is a multiple of 5)
| A | B | C | D | Total | |
|---|---|---|---|---|---|
| M | 10 | 0 | 30 | ||
| N | 25 | 0 | 30 | ||
| O | 0 | 0 | 30/0 | 0/30 | 30 |
| P | 5 | 25 | 0 | 0 | 30 |
| X | 0 | 0 | 0/30 | 30/0 | 30 |
| Y | 5 | 20 | 0 | 5 | 30 |
| Total | 45 | 45 | 45 | 45 | 180 |
As per clue 5,
two surfers gave all stars to a single blogger
which implies there will be two surfers with a combination like (0.0.0.30)
which is only possible for O & X.
Therefore O & X will give 30 to C & D in some order.
| A | B | C | D | Total | |
|---|---|---|---|---|---|
| M | 10 | 0 | 15 | 5 | 30 |
| N | 25 | 0 | 30 | ||
| O | 0 | 0 | 30/0 | 0/30 | 30 |
| P | 5 | 25 | 0 | 0 | 30 |
| X | 0 | 0 | 0/30 | 30/0 | 30 |
| Y | 5 | 20 | 0 | 5 | 30 |
| Total | 45 | 45 | 45 | 45 | 180 |
By clue 3,
M has given different stars to each blogger,
Since he has already given 0 & 10, remaining stars should add up to 20.
The only possibility is 5 & 15.
We know that X & O both give one of C or D 30 stars.
Now M could not give 15 stars to D - as Y has given D 5 stars already which if done will make sum more than 45.
Therefore, the only possibility left for M is giving 5 stars to D and 15 stars C.
| A | B | C | D | Total | |
|---|---|---|---|---|---|
| M | 10 | 0 | 15 | 5 | 30 |
| N | 25 | 0 | 0 | 5 | 30 |
| O | 0 | 0 | 30/0 | 0/30 | 30 |
| P | 5 | 25 | 0 | 0 | 30 |
| X | 0 | 0 | 0/30 | 30/0 | 30 |
| Y | 5 | 20 | 0 | 5 | 30 |
| Total | 45 | 45 | 45 | 45 | 180 |
Hence, the stars given to C & D by N which is 0 & 5 respectively. We cannot give C 5 stars as it would make it total of C exceed 45.
In DILR sets, often the final solution can be more than one at which point you should not give up thinking that you haven't got it right but work on making cases. These are the two final cases that we get:
Case 1
| A | B | C | D | Total | |
|---|---|---|---|---|---|
| M | 10 | 0 | 15 | 5 | 30 |
| N | 25 | 0 | 0 | 5 | 30 |
| O | 0 | 0 | 30 | 0 | 30 |
| P | 5 | 25 | 0 | 0 | 30 |
| X | 0 | 0 | 0 | 30 | 30 |
| Y | 5 | 20 | 0 | 5 | 30 |
| Total | 45 | 45 | 45 | 45 | 180 |
Case 2
| A | B | C | D | Total | |
|---|---|---|---|---|---|
| M | 10 | 0 | 15 | 5 | 30 |
| N | 25 | 0 | 0 | 5 | 30 |
| O | 0 | 0 | 0 | 30 | 30 |
| P | 5 | 25 | 0 | 0 | 30 |
| X | 0 | 0 | 30 | 0 | 30 |
| Y | 5 | 20 | 0 | 5 | 30 |
| Total | 45 | 45 | 45 | 45 | 180 |
| A | B | C | D | Total | |
|---|---|---|---|---|---|
| M | 10 | 0 | 15 | 5 | 30 |
| N | 25 | 0 | 0 | 5 | 30 |
| O | 0 | 0 | 30 / 0 | 0 / 30 | 30 |
| P | 5 | 25 | 0 | 0 | 30 |
| X | 0 | 0 | 0 / 30 | 30 / 0 | 30 |
| Y | 5 | 20 | 0 | 5 | 30 |
| Total | 45 | 45 | 45 | 45 | 180 |
N distributed 25 and 0 for A & D.
Similarly, P distributed 5 & 25 for A & B.
Therefore, there are two surfers distributed their stars among exactly 2 bloggers.