80. Number of students who have opted the subjects A, B, C are 60, 84, 108 respectively. The examination is to be conducted for these students such that only the students of the same subject are allowed in one room. Also the number of students in each room must be same. What is the minimum number of rooms that should be arranged to meet all these conditions?
(a) 28
(b) 60
(c) 12
(d) 21
81. How many five digit numbers can be formed using the digits 2,3,8,7,5 exactly once such that the number is divisible by 125?
(a) 0
(b) 1
(c) 4
(d) 3
82. You can collect Rubies and Emeralds as many as you can. Each Ruby is worth of Rs.4 crores and each Emerald is worth of Rs.5 crore. Each Ruby weighs 0.3 kg. And each Emerald weighs 0.4 kg. Your bag can carry at the most 12 kg. What you should collect to get the maximum wealth?
(a) 20 Rubies and 15 Emeralds
(b) 40 Rubies
(c) 28 Rubies and 9 Emeralds
(d) None of these.
83. I have one rupee coins, fifty paise coins and twenty five paise coins. The number of coins are in the ratio 2.5 : 3 : 4. If the total amount with me is Rs.210. Find the number of one rupee coins.
(a) 90
(b) 85
(c) 100
(d) 105
84. My son adores chocolates. He likes biscuits. But he hates apples. I told him that he can buy as many chocolates he wishes. But then he must have biscuits twice the number of chocolates and apples more than biscuits and chocolates together. Each chocolate cost Re.1. The cost of apple is twice of the chocolate and four biscuits are worth one apple. Then which of the following can be the amount that I spent on that evening on my son if number of chocolates, biscuits and apples brought were all integers?
(a) Rs. 34
(b) Rs.33
(c) Rs.8
(d) None of these.
Answers:
| 80 | (d) |
| 81 | (c) |
| 82 | (b) |
| 83 | (d) |
| 84 | (a) |
80. HCF of 60, 84 and 108 is 12. Hence 12 students should be seated in each room. So for subject A we wuld require (60/12) = 5 rooms, for subject B we would require (84/12) = 7 rooms and for subject C we would require (108/12) = 9 rooms. Hence minimum number of rooms required to satisfy our condition = (5+7+9) = 21 rooms.
81. Let us find some of the smaller multiples of 125. They are 125, 250, 375, 500, 625, 750, 875, 1000 …..So we find that multiples of 125 end in either 0 or 5. So from the given digits we can only have 5 at the end. The tens place digit can either be 2 or 7, which can be had in 2! = 2 ways. And the remaining two digits have to occupy the first two places. Which can be again had in 2! = 2 ways. The required number of such numbers = (2 x 2) = 4. HINT : We can even go ahead and find these 4 numbers : 38725, 38275, 83725 and 83275.
82. To maximise the value of the wealth we must carry more of the one whose value per kg. is more. Value per kg. of Ruby = (4/0.3) = Rs.13.33 crores and value per rupee of each Emerald = (5/0.4) = Rs.12.5 crore. It is obvious that we should carry entire 12 kg. of Ruby. This would amount to (12/0.3) = 40 Rubies.
83. Since the number of coins are in the ration 2.5 : 3 : 4, the value of the coins will be in the ratio (1 x 2.5) : (0.5 x 3) : (0.25 x 4) = 2.5 : 1.5 : 1 or 5 : 3 : 2. Since, they totally amount to Rs.210, if the value of each type of coins are assumed to be 5x, 3x and 2x, the average value per coin will be 210/10x. So total value of 1 Re. Coin will be 5x (210/10x) = Rs.105. So total number of 1 Re. coin will be 105.
84. The cost of each chocolate is Re.1. So the cost of apple should be Rs.2 and that of 1 biscuit should be Rs.0.5. Thus if eats x chocolates, he has to eat 2x biscuits. Hence total value of chocolates will be Rs.x and that of biscuits will be (0.5)(2x) = Rs.x. Hence we see that the value of chocolates is to the value of biscuits will always be 1:1. As per our assumption he will have to eat more than (x + 2x) = 3x apples and hence the total value of the apples will be more than (2)(3x) = 6x. In other words, the ratio of value chocolates to apples or biscuits to apples will be more than 1 : 6. In other words, if value of chocolates and biscuits is Rs.1 each, then the value of apples has to be more than Rs.6, or the combined value will be more than Rs.8. This means that the value of apples will always constitute more than 6/8 or ¾ of the entire bill. It can further be observed that the total value of chocolates and biscuits together will always be an even integer and so will be the value of apples. This means that the combined value of all three of them has to be even and not odd. So (b) Rs.33 cannot be the answer. Also (c) Rs.8 cannot be the answer as, if we take the value of chocolates and biscuits to be minimum i.e.Re.1 each, then the value of apples can be a minimum of Rs.8. Hence the total value will always be Rs.10 or higher. The only option possible is Rs.34. To verify this let us find two even numbers (one of them higher than 3/4th of 34) which add up to 34. We can find many such numbers eg. 32+2, 30+4, 28+6 and 26+8. All of these could be a possible combination. Hence the answer is (a).
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