A has some coins. He gives half of the coins and 2 more to B. B gives half of the coins and 2 more to C. C gives half of the coins and 2 more ta D. The number of coins D has now’, is the smallest two-digit number. How many coins does A have in the beginning?

To determine the initial number of coins A had, we can set up equations based on the sequential distribution of coins.

Correct Option

Let the initial number of coins A possessed be $x$.

• A gives half of his coins and 2 more to B. Therefore, the number of coins B receives is:

  $C_B=\frac{x}{2}+2$

• B then gives half of the coins he received and 2 more to C. The number of coins C receives is:

  $C_C=\frac{C_B}{2}+2=\frac{\frac{x}{2}+2}{2}+2=\frac{x}{4}+1+2=\frac{x}{4}+3$

• C subsequently gives half of the coins he received and 2 more to D. The number of coins D receives is:

  $C_D=\frac{C_C}{2}+2=\frac{\frac{x}{4}+3}{2}+2=\frac{x}{8}+\frac{3}{2}+2=\frac{x}{8}+\frac{7}{2}$

The problem states that the number of coins D has is the smallest two-digit number, which is 10.

Therefore, we set $C_D$ equal to 10:

$\frac{x}{8}+\frac{7}{2}=10$

To solve for $x$:

$\frac{x}{8}=10-\frac{7}{2}$

$\frac{x}{8}=\frac{20}{2}-\frac{7}{2}$

$\frac{x}{8}=\frac{13}{2}$

$x=\frac{13}{2}\times 8$

$x=13\times 4$

$x=52$

Thus, A had 52 coins in the beginning.

Incorrect Options

Options 1 (76), 2 (68), and 3 (60) are incorrect because substituting these values for $x$ into the derived equation $\frac{x}{8}+\frac{7}{2}=10$ does not satisfy the condition. For instance, if $x=76$, then $\frac{76}{8}+\frac{7}{2}=9.5+3.5=13\ne 10$. Similarly, for $x=68$, $\frac{68}{8}+\frac{7}{2}=8.5+3.5=12\ne 10$, and for $x=60$, $\frac{60}{8}+\frac{7}{2}=7.5+3.5=11\ne 10$. Only $x=52$ correctly results in D receiving 10 coins.