The cost of tickets to enter the central zoo is Rs 150 for an adult and Rs 50 for a child. A family paid Rs 700 for 6 tickets altogether.

(a) Express the above statements in the form of linear equations.

(b) How many tickets were purchased in each category?

(c) If the cost of tickets in each category is decreased by 10%, how many tickets of each category can be bought for Rs 540?



Answer :

Sure, let's solve this problem step-by-step.

### (a) Express the above statements in the form of linear equations:

Let's define:
- [tex]\( a \)[/tex] as the number of adult tickets.
- [tex]\( c \)[/tex] as the number of child tickets.

From the given information, we know:
1. The total cost of the tickets purchased by the family is Rs 700.
2. The total number of tickets purchased is 6.

From these, we can write two equations:
1. The cost equation:
[tex]\[ 150a + 50c = 700 \][/tex]
2. The number of tickets equation:
[tex]\[ a + c = 6 \][/tex]

So, the linear equations are:

[tex]\[ 150a + 50c = 700 \][/tex]

[tex]\[ a + c = 6 \][/tex]

### (b) How many tickets were purchased in each category?

We have the system of equations:

[tex]\[ 150a + 50c = 700 \][/tex]
[tex]\[ a + c = 6 \][/tex]

To solve this system, we can use substitution or elimination. Here, we will eliminate [tex]\( c \)[/tex].

1. From the second equation, express [tex]\( c \)[/tex] in terms of [tex]\( a \)[/tex]:
[tex]\[ c = 6 - a \][/tex]

2. Substitute [tex]\( c \)[/tex] in the first equation:
[tex]\[ 150a + 50(6 - a) = 700 \][/tex]

3. Distribute and simplify:
[tex]\[ 150a + 300 - 50a = 700 \][/tex]
[tex]\[ 100a + 300 = 700 \][/tex]
[tex]\[ 100a = 400 \][/tex]
[tex]\[ a = 4 \][/tex]

4. Substitute [tex]\( a = 4 \)[/tex] back into the equation [tex]\( c = 6 - a \)[/tex]:
[tex]\[ c = 6 - 4 \][/tex]
[tex]\[ c = 2 \][/tex]

Therefore, the family purchased 4 adult tickets and 2 child tickets.

### (c) If the cost of tickets of each category is decreased by 10%, how many tickets of each category can be bought for Rs 540?

First, let's find the new prices after a 10% decrease:
- New cost of an adult ticket:
[tex]\[ 150 - 0.1 \times 150 = 150 \times 0.9 = 135 \][/tex]

- New cost of a child ticket:
[tex]\[ 50 - 0.1 \times 50 = 50 \times 0.9 = 45 \][/tex]

Now, we form the new system of equations, considering the total cost is Rs 540 and the total number of tickets is still 6:

1. The new cost equation:
[tex]\[ 135a + 45c = 540 \][/tex]

2. The number of tickets equation remains the same:
[tex]\[ a + c = 6 \][/tex]

We solve this system using similar steps as before:

1. Express [tex]\( c \)[/tex] in terms of [tex]\( a \)[/tex]:
[tex]\[ c = 6 - a \][/tex]

2. Substitute [tex]\( c \)[/tex] in the first new equation:
[tex]\[ 135a + 45(6 - a) = 540 \][/tex]

3. Distribute and simplify:
[tex]\[ 135a + 270 - 45a = 540 \][/tex]
[tex]\[ 90a + 270 = 540 \][/tex]
[tex]\[ 90a = 270 \][/tex]
[tex]\[ a = 3 \][/tex]

4. Substitute [tex]\( a = 3 \)[/tex] back into the equation [tex]\( c = 6 - a \)[/tex]:
[tex]\[ c = 6 - 3 \][/tex]
[tex]\[ c = 3 \][/tex]

Therefore, if the costs are decreased by 10%, the family can buy 3 adult tickets and 3 child tickets for Rs 540.