A and B have coins of Rs.1, Rs. 2, Rs. 5 and Rs. 10, in the ratio 4 ∶ 3 ∶ 6 ∶ 2 and 3 ∶ 5 ∶ 7 ∶ 3, respectively. A has Rs.6/‐ more than B. Which of the following can be the number of coins with A and B, respectively?
60, 54
Understanding the Coin Ratios and Values
The problem provides information about the coins held by two individuals, A and B, in different denominations (Rs.1, Rs.2, Rs.5, and Rs.10). The number of coins of each denomination is given in a ratio for both A and B.
Let's break down the ratios for A and B:
- For A: The ratio of coins of Rs.1, Rs.2, Rs.5, and Rs.10 is 4 ∶ 3 ∶ 6 ∶ 2.
- For B: The ratio of coins of Rs.1, Rs.2, Rs.5, and Rs.10 is 3 ∶ 5 ∶ 7 ∶ 3.
Total Parts in the Ratios
To find the total number of coins for a person based on the ratio, we first sum the parts in their respective ratio:
- Total parts for A's coins: \(4 + 3 + 6 + 2 = 15\) parts.
- Total parts for B's coins: \(3 + 5 + 7 + 3 = 18\) parts.
This means that the total number of coins A has must be a multiple of 15, and the total number of coins B has must be a multiple of 18.
Calculating the Total Value
Let's calculate the total value of coins for A and B based on their ratios. Suppose the number of coins for A is \(k_A\) times the ratio parts, and for B is \(k_B\) times the ratio parts, where \(k_A\) and \(k_B\) are positive integers.
Value for A:
- Number of Rs.1 coins: \(4k_A\)
- Number of Rs.2 coins: \(3k_A\)
- Number of Rs.5 coins: \(6k_A\)
- Number of Rs.10 coins: \(2k_A\)
Total value for A (\(V_A\)) = \((4k_A \times 1) + (3k_A \times 2) + (6k_A \times 5) + (2k_A \times 10)\)
\(V_A = 4k_A + 6k_A + 30k_A + 20k_A = 60k_A\)
The total number of coins for A is \(4k_A + 3k_A + 6k_A + 2k_A = 15k_A\).
Value for B:
- Number of Rs.1 coins: \(3k_B\)
- Number of Rs.2 coins: \(5k_B\)
- Number of Rs.5 coins: \(7k_B\)
- Number of Rs.10 coins: \(3k_B\)
Total value for B (\(V_B\)) = \((3k_B \times 1) + (5k_B \times 2) + (7k_B \times 5) + (3k_B \times 10)\)
\(V_B = 3k_B + 10k_B + 35k_B + 30k_B = 78k_B\)
The total number of coins for B is \(3k_B + 5k_B + 7k_B + 3k_B = 18k_B\).
Using the Value Difference Condition
We are given that A has Rs. 6 more than B. So, \(V_A - V_B = 6\).
Substituting the values of \(V_A\) and \(V_B\):
\(60k_A - 78k_B = 6\)
We can divide the entire equation by 6:
\(10k_A - 13k_B = 1\)
We need to find integer values for \(k_A\) and \(k_B\) that satisfy this equation. The possible number of coins for A is \(15k_A\) and for B is \(18k_B\). We will check the given options to see which pair satisfies both conditions: the total number of coins are multiples of 15 and 18 respectively, and the corresponding \(k_A\) and \(k_B\) values satisfy \(10k_A - 13k_B = 1\).
Evaluating the Options
| Option | A's Coins | B's Coins | Is A's coins multiple of 15? (\(15k_A\)) | Is B's coins multiple of 18? (\(18k_B\)) | Value of \(k_A\) (A's coins/15) | Value of \(k_B\) (B's coins/18) | Check \(10k_A - 13k_B = 1\) | Result |
|---|---|---|---|---|---|---|---|---|
| 1 | 42 | 36 | No (42 / 15 is not an integer) | Yes (36 / 18 = 2) | - | \(k_B = 2\) | Cannot evaluate \(k_A\) | Incorrect |
| 2 | 45 | 54 | Yes (45 / 15 = 3) | Yes (54 / 18 = 3) | \(k_A = 3\) | \(k_B = 3\) | \(10(3) - 13(3) = 30 - 39 = -9\) (\(-9 \neq 1\)) | Incorrect |
| 3 | 60 | 54 | Yes (60 / 15 = 4) | Yes (54 / 18 = 3) | \(k_A = 4\) | \(k_B = 3\) | \(10(4) - 13(3) = 40 - 39 = 1\) (\(1 = 1\)) | Correct |
| 4 | 60 | 72 | Yes (60 / 15 = 4) | Yes (72 / 18 = 4) | \(k_A = 4\) | \(k_B = 4\) | \(10(4) - 13(4) = 40 - 52 = -12\) (\(-12 \neq 1\)) | Incorrect |
From the evaluation, only Option 3 satisfies both conditions: the total number of coins for A (60) is a multiple of 15 (\(60 = 15 \times 4\)), the total number of coins for B (54) is a multiple of 18 (\(54 = 18 \times 3\)), and the corresponding values \(k_A = 4\) and \(k_B = 3\) satisfy the equation \(10k_A - 13k_B = 1\).
Therefore, the possible number of coins with A and B are 60 and 54, respectively.
Important Questions from Miscellaneous
A stone is thrown horizontally from the top of a 20 m high building with a speed of 12 m/s. It hits the ground at a distance R from the building. Taking g = 10 m/s2 and neglecting air resistance will give :
A sphere of volume V is made of a material with lower density than water. While on Earth, it floats on water with its volume f1V (f1 < 1) submerged. On the other hand, on a spaceship accelerating with acceleration a < g (g is the acceleration due to gravity on Earth) in outer space, its submerged volume in water is f2V. Then:
A railway wagon (open at the top) of mass M1 is moving with speed v1 along a straight track. As a result of rain, after some time it gets partially filled with water so that the mass of the wagon becomes M2 and speed becomes v2. Taking the rain to be falling vertically and the water stationery inside the wagon, the relation between the two speeds v1 and v2 is :
Consider the following statements:
1. Distance between the longitudes becomes zero on North Pole and South Pole.
2. Distance between the longitudes is maximum on the Equator.
3. Number of longitudes is more than number of latitudes.
Which of the statements given above is/are correct?
One block of 2⋅0 kg mass is placed on top of another block of 3⋅0 kg mass. The coefficient of static friction between the two blocks is 0⋅2. The bottom block is pulled with a horizontal force F such that both the blocks move together without slipping. Taking acceleration due to gravity as 10 m/s2, the maximum value of the frictional force is :
