Children's problem about coins that will attract cash flow

Children's tasks Sometimes it can be confusing for any adult. And it’s funny to watch a person with higher education and work experience look at a simple problem and do not understand from which side you can approach it. It is obvious that he did not teach mathematics in school!

And today's edition. "Site" It invites readers to test their knowledge and ingenuity. In what way? And very simply – try to solve the problems that we have found for you. At the same time, you will conduct a warm-up for the mind, which is quite useful and necessary at any age. Especially if you are no longer a student or a student.



Children's tasks

1. In the conditions of the first problem, it is indicated that some travelers were lucky to find 6 bags of coins at once. But when they started counting the coins, they found that each bag of coins has a different number. The first bag contained 60 coins, the second 30, the third 20, the fourth 15, and the fifth 12. Can you figure out how many coins are in the sixth bag?
   
   
   
   
2. In the second problem, you need to determine the weight of watermelon, which was recently brought from the market (but, it seems, forgot to weigh). It is only known that the father ate 1/4 of the watermelon, the mother ate 1/5 of the watermelon, and the child was able to master only 1/6 of the watermelon. The rest of the watermelon weighed 2.3 kg. Can you get your initial weight?
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3. And finally, we offer a problem where you need to find out the weight of animals. And here you can either alternately find the weight of a hare, a cat and a dog, or find the total weight of animals in a faster way. Can you find all the solutions without any clues?
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Tips to challenges

1. If you think a little about the conditions of the problem, you understand that it can be depicted differently. And if in the first bag of coins the most (60), then in the second only 1/2 of this number (30), in the third - 1/3 (20), in the fourth - 1/4 (15), and in the fifth - 1/5 (12). In this case, the sixth bag will be 1/6 of 60, which equals 10 coins. Although someone will say that the sixth bag is larger than the others, and 5 bags of coins are in it. Then you need to add all the coins 60 + 30 + 20 + 15 + 12 = 137 coins. But the first option seems better, doesn’t it?
   
   
   
   
   
   
2. First of all, find out what part of the watermelon was eaten. This requires 1/4 + 1/5 + 1/6. If we mentally return to about 6th grade, we remember that for such a mathematical operation all fractions must be brought to a common denominator (the number below the fraction). Then our example will be as follows: 15/60 + 12/60 + 10/60 = 37/60. So, the remaining part of the watermelon belongs to 60/60 – 37/60 = 23/60 = 2.3 kg. From this we understand that the eaten part of the watermelon (37/60) = 3.7 kg. Then the total weight of the watermelon was originally 3.7 + 2.3 = 6 kg.
   
   
3. If a dog with a hare weighs 20 kg, and a cat with a hare weighs 10 kg, then it is easy to understand that a dog is 10 kg heavier than a cat. Then, from the total weight of the dog and cat, we find that the cat weighs (24-10) / 2 = 7 kg. The weight of the dog is 7 + 10 = 17 kg. Then the weight of the hare is 10 - 7 = 3 kg. The total weight of the animals is 17 + 7 + 3 = 27 kg. There is another way, though, because we know the weight of each pair of animals. And 10 + 20 + 24 = 54 kg is the weight of 2 dogs, 2 cats and 2 hares together. Then 54/2 = 27 kg, which will equal the total weight of the animals.
   
   
   
   
   
   

Did you manage to solve these childhood problems? Don’t forget to share your results in the comments.