We appreciate your visit to Direction Q 1 to Q 5 There are 6 dice for each question On these dice there is no black dot on the top surface. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!
Answer :
To solve these types of dice problems, we need to know that dice have opposite faces that sum up to a constant number, which is 13 according to the conditions given in this problem set.
Q.1: We are given that the number of dots on the top surfaces for even-numbered dices is odd. This means for dice II, IV, and VI, the top surfaces must have an odd number of dots. Typically, the opposite face's dots sum up to a fixed number with the top side. Therefore, if we assume that tops could be odd numbers such as 1, 3, or 5, we need to compute accordingly. Given the sum of all top faces should be odd, we ensure that the sum of the top faces matches the constraints. Calculating with the given conditions provides the total dots as 17.
Q.2: For dices I, II, and IV having odd top surfaces, the problem asks us to find the sum of these dots. Again, place odd numbers like 1, 3, or 5 in the context of opposites equaling 13. The calculation results in the total sum being 17.
Q.3: If the number of dots on each of the top surfaces of dices II, V, VI is even, similarly choose even numbers like 2, 4, or 6. The resulting computation shows that the sum of these particular top surfaces reaches 18.
Q.4: Here, the condition states that dices I, II, III have odd tops, and IV, V, VI have even bottoms. Solving for these conditions: odd tops can be 1, 3, or 5, and even bottoms make the opposite the even number. With this, we calculate to find the sum to be 0, since opposites effectively cancel each other.
Q.5: Finally, with even bottom surfaces for odd-numbered dices, we harmonize the top choices so they add up based on the opposite sum constraint being 13. This results in the total sum being 18.
The key to the solutions lies in understanding the sum constraint imposed by opposing dice faces and logically assigning odd or even numbers as required by the question to reach the given total.
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