MAT

Solving the given equations simultaneously, we get x = 2 and y = 3.

The relative speed of the second train with respect to the first train is 75 - 60 = 15 km/hr. Therefore, the second train will overtake the first train in 15/1 = 1 hour. Since the second train leaves at 11:00 AM, it will overtake the first train at 11:00 AM + 1 hour = 12:00 PM + 30 minutes = 1:30 PM.

Let the cost price of the article be ($x). Then, the profit is ($120 - $x). Since the profit is 20% of the cost price, we have ($120 - $x) = 0.2 ($x). Solving for ($x), we get ($x) = ($100).

In the given code, each letter is assigned a number based on its position in the alphabet. For example, 'A' is assigned 1, 'P' is assigned 5, and so on. Therefore, 'ORANGE' is written as 15027951.

The area of the path is equal to the area of the outer rectangle minus the area of the inner rectangle. The outer rectangle has a length of ($(20 + 2x)) meters and a width of ($(15 + 2x)) meters. Therefore, the area of the outer rectangle is ($(20 + 2x)(15 + 2x)) square meters. The inner rectangle has a length of 20 meters and a width of 15 meters. Therefore, the area of the inner rectangle is ($20 \times 15) square meters. Therefore, the area of the path is ($(20 + 2x)(15 + 2x) - 20 \times 15) square meters = ($2x(20 + 15)) square meters.

First, convert the speed of the train from km/hr to m/s: ($72 \times 5/18 = 20) m/s. Then, use the formula ($distance = speed \times time) to find the length of the train: ($100 = 20 \times 10). Therefore, the length of the train is 200 meters.

Let the sum of money be ($P) and the rate of interest be ($r)%. Then, the amount after 10 years is ($P + P \times r \times 10/100 = 2P). Therefore, ($r = 10)%. Now, let the number of years required for the sum to triple itself be ($n). Then, the amount after ($n) years is ($P + P \times r \times n/100 = 3P). Substituting ($r = 10)%, we get ($3P = P + P \times 10 \times n/100). Simplifying, we get ($n = 15) years.

Let the number of apples of variety A sold be ($x) and the number of apples of variety B sold be ($y). Then, we have the following system of equations: ($x + y = 100) and ($2x + 3y = 250). Solving this system of equations, we get ($x = 60) and ($y = 40).

The number of male employees is ($100 \times 60/100 = 60). Therefore, the number of female employees is ($100 - 60 = 40).

The relative speed of the second train with respect to the first train is ($75 - 60 = 15) km/hr. Therefore, the second train will be 15 km ahead of the first train in ($15/15 = 1) hour. Since the second train leaves at 11:00 AM, it will be 15 km ahead of the first train at 11:00 AM + 1 hour = 12:00 PM + 30 minutes = 12:30 PM.

Let the cost price of the article be ($x). Then, the profit is ($120 - $x). Since the profit is 20% of the cost price, we have ($120 - $x) = 0.2 ($x). Solving for ($x), we get ($x) = ($100).

The area of the path is equal to the area of the outer rectangle minus the area of the inner rectangle. The outer rectangle has a length of ($(20 + 2x)) meters and a width of ($(15 + 2x)) meters. Therefore, the area of the outer rectangle is ($(20 + 2x)(15 + 2x)) square meters. The inner rectangle has a length of 20 meters and a width of 15 meters. Therefore, the area of the inner rectangle is ($20 \times 15) square meters. Therefore, the area of the path is ($(20 + 2x)(15 + 2x) - 20 \times 15) square meters = ($2x(20 + 15)) square meters.

First, convert the speed of the train from km/hr to m/s: ($72 \times 5/18 = 20) m/s. Then, use the formula ($distance = speed \times time) to find the length of the train: ($100 = 20 \times 10). Therefore, the length of the train is 200 meters.

Let the sum of money be ($P) and the rate of interest be ($r)%. Then, the amount after 10 years is ($P + P \times r \times 10/100 = 2P). Therefore, ($r = 10)%. Now, let the number of years required for the sum to triple itself be ($n). Then, the amount after ($n) years is ($P + P \times r \times n/100 = 3P). Substituting ($r = 10)%, we get ($3P = P + P \times 10 \times n/100). Simplifying, we get ($n = 15) years.