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.
### (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.
