Choose the appropriate alternative :
Q75 – 77 : A,B,C,D collected one rupee coins following the given pattern.
Together they collected 100 coins.
Each one of them collected even no. of coins.
Each one of them collected at least 10 coins.
No two of them collected the same no. of coins.
75. The maximum number of coins collected by any one of them cannot exceed
(a) 64
(b) 36
(c) 54
(d) None of these.
76. If A collected 54 coins, then the difference in the number of coins between the one who collected maximum number of coins and the one who collected the second highest number of coins must be at least.
(a) 12
(b) 24
(c) 30
(d) None of these.
77. If A collected 54 coins and B collected two more coins than the twice of the number of coins collected by C. Then the number of coins collected by B could be,
(a) 28
(b) 20
(c) 26
(d) 22
Answers:
| 75 | (b) |
| 76 | (c) |
| 77 | (d) |
75. For any one of them to collect maximum number of coins the remaining three should collect minimum number of coins. And from the conditions given this has to be 10, 12 and 14. So if the three of them collect (10+12+14) = 36 coins, the fourth one has to collect (100 – 64) = 36 coins which has to be the maximum by any one person.
76. Since A has collected 54 coins out of 100, he should obviously be. the person who collected the maximum number of coins. For the difference between him and the second highest person to be minimum, the second highest person should collect maximum number of coins possible under given conditions. And for this to happen the remaining two should collect minimum number of coins. So if the two of them collect 10 & 12 coins i.e. 22 coins between themselves, the third person would have to collect (100 – 54 – 22) = 24 coins. Hence the difference between him and the highest person should at least be (54 – 34) = 30.
77. If A has collected 54 coins, the remaining three of them should collect (100 – 54) = 46 coins between themselves. Let us assume that C has collected 10 coins. So B will collect (2 x 10) + 2 = 22. So A will collect (46 – 10 – 22) = 14 coins, which is a possible combination. Let us now assume that C picks up 12 coins, so B should pick up (2 x 12) + 2 = 26. So A will have to collect (46 – 12 – 26) = 8 coins. This combination is not possible. It can be hence concluded that C cannot pick up more than 10 coins and hence B has to pick up 22 coins to satisfy the given condition.
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