Mastering Simultaneous Equations

Let's explore some practical problems that will help you master simultaneous equations. These questions will use real-life scenarios to make learning engaging and relevant.

Question 1:
Two friends, Aditi and Rahul, are planning a trip. Aditi will travel by car at a speed of 60 km/h, while Rahul will travel by bus at a speed of 40 km/h. If they both start at the same time and Aditi has a head start of 30 km, how long will it take for Rahul to catch up to Aditi?

Answer:
Let t be the time in hours that Rahul travels. During this time, Aditi will have traveled a distance of 60(t) + 30 (the head start). Rahul will have traveled 40t. Setting these distances equal gives us the equation:

60t + 30 = 40t

To solve for t, subtract 40t from both sides:

20t + 30 = 0

Then subtract 30 from both sides:

20t = -30

Finally, divide by 20:

t = -1.5

Since time cannot be negative, this means that Rahul will never catch up to Aditi under these conditions. This question encourages you to think critically about real-life situations involving speed and distance.

Question 2:
A bakery sells two types of cakes: chocolate and vanilla. The price of a chocolate cake is Rs. 300, and the price of a vanilla cake is Rs. 200. If a customer buys a total of 10 cakes for Rs. 2500, how many of each type did they buy?

Answer:
Let x be the number of chocolate cakes and y be the number of vanilla cakes. We can set up the following equations based on the information provided:

x + y = 10 (total number of cakes)
300x + 200y = 2500 (total cost of cakes)

Now we can use substitution or elimination. Let's use substitution. From the first equation, we can express y in terms of x:

y = 10 - x

Now substitute this into the second equation:

300x + 200(10 - x) = 2500

Distributing the 200 gives us:

300x + 2000 - 200x = 2500

Combining like terms:

100x + 2000 = 2500

Subtracting 2000 from both sides:

100x = 500

Now divide by 100:

x = 5

Substituting back to find y:

y = 10 - 5 = 5

The customer bought 5 chocolate cakes and 5 vanilla cakes.

Question 3:
In a market, the price of apples is Rs. 30 per kg, and the price of oranges is Rs. 20 per kg. A shopper wants to buy a total of 15 kg of fruit and spends Rs. 400. How many kilograms of apples and oranges did the shopper buy?

Answer:
Let a be the kilograms of apples and o be the kilograms of oranges. We can set up the following equations:

a + o = 15 (total weight)
30a + 20o = 400 (total cost)

From the first equation, express o in terms of a:

o = 15 - a

Now substitute into the second equation:

30a + 20(15 - a) = 400

Distributing the 20 gives us:

30a + 300 - 20a = 400

Combining like terms:

10a + 300 = 400

Subtracting 300 from both sides:

10a = 100

Now divide by 10:

a = 10

Substituting back to find o:

o = 15 - 10 = 5

The shopper bought 10 kg of apples and 5 kg of oranges.

These problems illustrate how simultaneous equations can be used to solve real-life scenarios. As you practice, you will enhance your understanding and ability to apply these concepts in various situations. Keep up the great work!