Given: Four friends X, Y, Z, and W each have some marbles.
Let the number of marbles X has → x
Let the number of marbles Y has → y
Let the number of marbles Z has → z
Let the number of marbles W has → w
- X gives Y as many marbles as he already has, gives Z twice the amount of marbles Z already has, and gives W thrice the amount of marbles W already has.
| Marbles | X | Y | Z | W |
|---|
| Initial number of marbles | x | y | z | w |
| i) X gives Y as many marbles as he already has | -y | +y | | |
| ii) X gives Z twice the amount of marbles Z already has | -2z | | +2z | |
| iii) X gives W thrice the amount of marbles W already has | -3w | | | +3w |
| Marbles left | x - y - 2z - 3w | 2y | 3z | 4w |
- Then, W gives 1/6th of his marbles to Y. Afterward, X gives 10% of the marbles he now owns to Z and 20% to Y. Finally, all of them have 50 marbles each.
1/6th of W's marbles will be → 4w × 1/6 = 2/3w
10% of the X's marbles will be → 10% of (x - y - 2z - 3w) = 0.1(x - y - 2z - 3w)
20% of the X's marbles will be → 20% of (x - y - 2z - 3w) = 0.2(x - y - 2z - 3w)
| Marbles | X | Y | Z | W |
|---|
| Initial number of marbles | x | y | z | w |
| i) X gives Y as many marbles as he already has | -y | +y | | |
| ii) X gives Z twice the amount of marbles Z already has | -2z | | +2z | |
| iii) X gives W thrice the amount of marbles W already has | -3w | | | +3w |
| Marbles left | x - y - 2z - 3w | 2y | 3z | 4w |
| iv) W gives 1/6th of his marbles to Y | | +2/3w | | -2/3w |
| v) X gives 10% of the marbles he now owns to Z | -0.1(x - y - 2z - 3w) | | +0.1(x - y - 2z - 3w) | |
| vi) X gives 20% of the marbles he now owns to Y | -0.2(x - y - 2z - 3w) | +0.2(x - y - 2z - 3w) | | |
| Marbles left | 0.7(x - y - 2z - 3w) | 2y + 2/3w + 0.2(x - y - 2z - 3w) | 3z + 0.1(x - y - 2z - 3w) | 10/3w |
| Finally, all of them have 50 marbles each. | 0.7(x - y - 2z - 3w) = 50 | 2y + 2/3w + 0.2(x - y - 2z - 3w) = 50 | 3z + 0.1(x - y - 2z - 3w) = 50 | 10/3w = 50 |
So, solving for each of them:
For X: 0.7(x - y - 2z - 3w) = 50 ------- (1)
For Y: 2y + 2/3w + 0.2(x - y - 2z - 3w) = 50 ------- (2)
For Z: 3z + 0.1(x - y - 2z - 3w) = 50 ------- (3)
For W: 10/3w = 50 ------- (4)
In equation 4 we get:
10/3w = 50
w = 50 × 3/10
w → 15
In equation 1 we get:
0.7(x - y - 2z - 3w) = 50 → (x - y - 2z - 3w) = 50/0.7
Putting the value in equation 2 we get:
2y + 2/3w + 0.2(x - y - 2z - 3w) = 50 → 2y + (2/3 × 15) + 0.2(50/0.7) = 50
→ 2y + 10 + 100/7 = 50
→ 2y + (2/3 × 15) + 0.2(50/0.7) = 50
→ y = 90/7
Putting the value in equation 3 we get:
3z + 0.1(x - y - 2z - 3w) = 50 → 3z + 0.1(50/0.7) = 50
→ 3z + (50/7) = 50
→ z = 100/7
Putting the value of w, y and z equation 1 we get:
→ 0.7(x - y - 2z - 3w) = 50
→ x = 50/0.7 + y + 2z + 3w
→ x = (50/0.7) + (90/7) + (2 × 100/7) + (3 × 15)
→ x = 1105/7
Thus, the original number of marbles each friend had was:
X = 1105/7
Y = 90/7
Z = 100/7
W → 15
Hence, "Option 4" is the correct answer.