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Number System: Level 3 Test -6
That is 17+3 = 2/3 (total number of matches)
This means total number of matches = 30
Now we know they win more than 3/4th of the total number of matches that is they win more than 22.5 matches.
Obviously you cannot win half a match, so we chose the next integral value, that is 23.
They need 6 more to win from their remaining games and since they have played 20 games, from the remaining 10 games,
they can lose a maximum of 4 games.
M wins over N
N wins over M
M does not play with N
None of these
Since none of the matches ended in a draw, the scores for each of the players have to be even (since a win gives 2 points).
So the highest score possible for a player would be 50 and the lowest would be 0.
Since all 26 of them had different scores varying between 0 and 50, the scores should indeed be all the even number between 0 and 50.
And since the ranks obtained by players are in alphabetical order,
it can be concluded that A scored 50, B scored 48, C scored 46 and so on and Z scored 0.
Now the only way A can score 50 is, if he wins all his matches, i.e. he defeats all other players. Now B has scored 48.
So he has lost only one of his matches, which incidentally is against A. He must have defeated all other players.
Similarly, C has scored 46 matches. So he must have lost two matches, (i.e. to A and B) and defeated all other players.
So we conclude that a player whose name appears alphabetically higher up in the order has defeated all the players whose name appears alphabetically lower down.
Hence, M should win over N.
2 ≤ x ≤ 6
5 ≤ x ≤ 8
9 ≤ x ≤ 12
11≤ x ≤ 14
Amount of wheat remaining = x – (x/2+1/2) = x-1/2 kgs
Amount of wheat bought by the second customer = 1/2(x-1/2)
Amount of wheat remaining = (x-1/2) - (x-1/4) =x-3/4
Amount of wheat bought by the third customer = ½ (x-3/4) +1/2=x+1/8 kgs
As per the information given in the question
Because there is no wheat left after the third customer has bought the wheat.
Therefore, the value of ‘x’ = 7 kgs.
⇒ x + 10y + 50z = 107 Now the possible values of z could be 0, 1 and 2.
For z = 0: x + 10y = 107
Number of integral pairs of values of x and y that satisfy the equation x + 10y = 107 will be 11.
These values of x and y in that order are (7, 10); (17, 9); (27, 8)… (107, 0).
For z = 1: x + 10y = 57 Number of integral pairs of values of x and y that satisfy the equation x + 10y = 57 will be 6. These values of x and y in that order are (7, 5); (17, 4); (27,3); (37, 2); (47, 1) and (57, 0).
For z = 2: x + 10y = 7
There is only one integer value of x and y that satisfies the equation x + 10y = 7 in that order is (7, 0).
Therefore total number of ways in which you can pay a bill of 107 Wons = 11 + 6 + 1 = 18
Over Rupees 13 but less than Rupees 14
Over Rupees 4 but less than Rupees 5
Over Rupees 22 but less than Rupees 23
Over Rupees 18 but less than Rupees 19
As per the question: 3 × (100X + Y) = (100Y + X) – 50 ⇒ 299X = 97Y – 50 Y= 299X+ 50/97
Now the value of Y should be an integer.
By simply checking the options, we can identify option (d) as the answer.