The problem describes a scenario where Nimol needs to buy two types of medicine, A and B, for his mother. He needs to determine the price of each type of medicine. The problem gives two pieces of information: (1) A single day's dosage of medicine A (1 pill twice a day) and medicine B (2 pills three times a day) costs $156. (2) A 7-day supply of medicine A and a 3-day supply of medicine B costs $660. The problem asks us to set up two simultaneous equations and solve them to find the price of a single pill of medicine A and medicine B.

1. Problem Description

The problem describes a scenario where Nimol needs to buy two types of medicine, A and B, for his mother. He needs to determine the price of each type of medicine. The problem gives two pieces of information:

(1) A single day's dosage of medicine A (1 pill twice a day) and medicine B (2 pills three times a day) costs $

1

5

6. (2) A 7-day supply of medicine A and a 3-day supply of medicine B costs $

6

6

0. The problem asks us to set up two simultaneous equations and solve them to find the price of a single pill of medicine A and medicine B.

2. Solution Steps

Let $a$ be the price of one pill of medicine A and $b$ be the price of one pill of medicine B.

From the first piece of information, one day's dosage of medicine A is 2 pills (1 pill twice). One day's dosage of medicine B is 6 pills (2 pills three times). The total cost of one day's dosage is $

1

5

6. So the first equation is:

$2a + 6b = 156$

From the second piece of information, the cost of a 7-day supply of medicine A is the price of $7 \times 2 = 14$ pills. The cost of a 3-day supply of medicine B is the price of $3 \times 6 = 18$ pills. The total cost is $

6

6

0. So the second equation is:

$14a + 18b = 660$

Now we have a system of two linear equations:

$2a + 6b = 156$

$14a + 18b = 660$

We can simplify the first equation by dividing by 2:

$a + 3b = 78$

$14a + 18b = 660$

From the simplified first equation, we can express $a$ in terms of $b$:

$a = 78 - 3b$

Substitute this expression for $a$ into the second equation:

$14(78 - 3b) + 18b = 660$

$1092 - 42b + 18b = 660$

$1092 - 24b = 660$

$24b = 1092 - 660$

$24b = 432$

$b = \frac{432}{24}$

$b = 18$

Now substitute the value of $b$ back into the expression for $a$:

$a = 78 - 3(18)$

$a = 78 - 54$

$a = 24$

3. Final Answer

The price of one pill of medicine A is 24 Rupees and the price of one pill of medicine B is 18 Rupees.