A man visits four relatives on a day. He purchases some sweets and gives a part of it to the first relative. Next, he purchases the same number of sweets he is left with after he visited his first relative. He repeats this process while visiting his other relatives. Finally, he gives all the sweets to his fourth relative. It is known that he gives equal number of sweets to all the relatives and is left with no sweets after the fourth relative. Assuming that he initially purchased more than 20 sweets, the minimum number of sweets he gave to each relative is _______________. (Answer in integer)
Distributing Sweets Logic Puzzle Solution
This problem involves a sequence of actions: giving sweets, then purchasing more sweets based on the remaining amount. We need to find the minimum number of sweets given to each relative ($x$) under specific conditions.
Understanding the Process
Let $A_0$ be the initial number of sweets purchased. Let $x$ be the equal number of sweets given to each of the 4 relatives. Let $S_i$ be the number of sweets remaining *after* visiting the $i$-th relative. The process involves giving $x$ sweets and then purchasing an additional amount equal to the remaining sweets.
Working Backwards to Find Relationships
We work backward from the final state:
- After Relative 4: The man is left with 0 sweets. He gave $x$ sweets to the 4th relative. Therefore, just before giving sweets to the 4th relative, he must have had $x$ sweets. Let this amount be $A_3^*$. $ A_3^* = x $ The sweets remaining after the 4th relative is $S_4 = A_3^* - x = x - x = 0$.
- Before Relative 4 (Purchase Step): He purchased sweets equal to the amount he had *after* visiting the 3rd relative ($S_3$). So, the amount he had just before giving sweets to the 4th relative ($A_3^*$) is twice the amount he had after the 3rd relative ($S_3$). $ A_3^* = S_3 + S_3 = 2 S_3 $ Substituting $A_3^* = x$, we get: $ x = 2 S_3 \implies S_3 = \frac{x}{2} $
- After Relative 3: He has $S_3 = x/2$ sweets left. He gave $x$ sweets to the 3rd relative. So, before giving sweets to the 3rd relative, he must have had $S_3 + x$ sweets. Let this amount be $A_2^*$. $ A_2^* = S_3 + x = \frac{x}{2} + x = \frac{3x}{2} $
- Before Relative 3 (Purchase Step): He purchased sweets equal to the amount he had *after* visiting the 2nd relative ($S_2$). So, the amount he had just before giving sweets to the 3rd relative ($A_2^*$) is twice the amount he had after the 2nd relative ($S_2$). $ A_2^* = S_2 + S_2 = 2 S_2 $ Substituting $A_2^* = 3x/2$, we get: $ \frac{3x}{2} = 2 S_2 \implies S_2 = \frac{3x}{4} $
- After Relative 2: He has $S_2 = 3x/4$ sweets left. He gave $x$ sweets to the 2nd relative. So, before giving sweets to the 2nd relative, he must have had $S_2 + x$ sweets. Let this amount be $A_1^*$. $ A_1^* = S_2 + x = \frac{3x}{4} + x = \frac{7x}{4} $
- Before Relative 2 (Purchase Step): He purchased sweets equal to the amount he had *after* visiting the 1st relative ($S_1$). So, the amount he had just before giving sweets to the 2nd relative ($A_1^*$) is twice the amount he had after the 1st relative ($S_1$). $ A_1^* = S_1 + S_1 = 2 S_1 $ Substituting $A_1^* = 7x/4$, we get: $ \frac{7x}{4} = 2 S_1 \implies S_1 = \frac{7x}{8} $
- After Relative 1: He has $S_1 = 7x/8$ sweets left. He gave $x$ sweets to the 1st relative. The initial number of sweets purchased ($A_0$) is the amount he had after the 1st relative plus the $x$ sweets given away. $ A_0 = S_1 + x = \frac{7x}{8} + x = \frac{15x}{8} $
Finding the Minimum Number of Sweets
We have the relationship $A_0 = \frac{15x}{8}$. We are given two conditions:
- $x$ (number of sweets given to each relative) must be an integer.
- $A_0$ (initial number of sweets purchased) must be an integer.
- $A_0 > 20$.
For $A_0$ to be an integer, $x$ must be a multiple of 8. Let $x = 8k$, where $k$ is a positive integer.
Substituting $x = 8k$ into the equation for $A_0$:
$ A_0 = \frac{15(8k)}{8} = 15k $Now, apply the condition $A_0 > 20$:
$ 15k > 20 $ $ k > \frac{20}{15} $ $ k > \frac{4}{3} $ $ k > 1.333... $Since $k$ must be an integer, the smallest possible integer value for $k$ is 2.
We need the minimum number of sweets given to each relative, which is $x$. Using the minimum value of $k$:
$ x = 8k = 8 \times 2 = 16 $The minimum number of sweets given to each relative is 16. The initial purchase would be $A_0 = 15 \times 2 = 30$, which satisfies $A_0 > 20$. Let's verify:
- Start: 30 sweets. Give 16 to Rel 1. Left: 14.
- Buy 14. Have 28. Give 16 to Rel 2. Left: 12.
- Buy 12. Have 24. Give 16 to Rel 3. Left: 8.
- Buy 8. Have 16. Give 16 to Rel 4. Left: 0.
The process is consistent.
Important Questions from Numerical Reasoning
- Let $p_1$ and $p_2$ denote two arbitrary prime numbers. Which one of the following statements is correct for all values of $p_1$ and $p_2$?
- A 'frabjous' number is defined as a 3 digit number with all digits odd, and no two adjacent digits being the same. For example, 137 is a frabjous number, while 133 is not. How many such frabjous numbers exist?
Ankita has to climb 5 stairs starting at the ground, while respecting the following rules:
1. At any stage, Ankita can move either one or two stairs up.
2. At any stage, Ankita cannot move to a lower step.
Let $F(N)$ denote the number of possible ways in which Ankita can reach the $N^{th}$ stair. For example, $F(1) = 1$, $F(2) = 2$, $F(3) = 3$. The value of $F(5)$ is ________.- In a zoo, three lions and four tigers eat 390 kg of food every week. In another zoo, four lions and five tigers eat 500 kg of food every week. Lions and tigers eat different amounts of food, but all individuals of the same species eat the same amount. The amount of food a single lion eats per week is ________ kg.
(Answer in integer) Consider a spherical globe rotating about an axis passing through its poles. There are three points P, Q, and R situated respectively on the equator, the north pole, and midway between the equator and the north pole in the northern hemisphere. Let P, Q, and R move with speeds $v_P$, $v_Q$, and $v_R$, respectively.
Which one of the following options is CORRECT?
