
Chandana counted 32 wheels of some buses and cars. If a bus has 6 wheels and a car has 4 wheels. How many buses and cars were there?
Answer
615k+ views
Hint: Here, first take the number of buses as $x$ and the number of cars as $y$. Now, we can express 32 in the form, $32=6x+4y$. Now, consider the possible values of $x$ and $y$ where we will get the sum as 32, which will be the required number of buses and cars.
Complete step-by-step solution -
We are given that Chandana counted 32 wheels of some buses and cars. The number of wheels of buses are 6 and the number of wheels of cars are 4.
Now, we have to find the number of buses and cars.
Total number of wheels counted by Chandana = 32
Number of wheels of buses = 6
Number of wheels of cars = 4
Let $x$ be the number of buses and $y$ be the number of cars.
Now, we can express 32 as the sum of two numbers in such a way that one is divisibe by 6 and the other is divisible by 4.
Hence, 32 can be expressed as:
$32=6x+4y$
Here, 32 can be expressed in two ways.
For $x=2$ and $y=5$, we will get:
$\begin{align}
& 32=6\times 2+4\times 5 \\
& 32=12+20 \\
\end{align}$
Similarly, for $x=4$and $y=2$, we will obtain:
$\begin{align}
& 32=6\times 4+4\times 2 \\
& 32=24+8 \\
\end{align}$
Here, there are two possibilities:
One of the possibilities is that there would be 2 buses and 5 cars.
The other possibility is that there would be 4 buses and 2 cars.
Note: Here, without checking randomly we can also find the solution by considering the HCF of 4 and 6. The HCF is 2, now for $x=2$, find the value of $y$. Similarly, for $y=2$, find the value of $x$.
Complete step-by-step solution -
We are given that Chandana counted 32 wheels of some buses and cars. The number of wheels of buses are 6 and the number of wheels of cars are 4.
Now, we have to find the number of buses and cars.
Total number of wheels counted by Chandana = 32
Number of wheels of buses = 6
Number of wheels of cars = 4
Let $x$ be the number of buses and $y$ be the number of cars.
Now, we can express 32 as the sum of two numbers in such a way that one is divisibe by 6 and the other is divisible by 4.
Hence, 32 can be expressed as:
$32=6x+4y$
Here, 32 can be expressed in two ways.
For $x=2$ and $y=5$, we will get:
$\begin{align}
& 32=6\times 2+4\times 5 \\
& 32=12+20 \\
\end{align}$
Similarly, for $x=4$and $y=2$, we will obtain:
$\begin{align}
& 32=6\times 4+4\times 2 \\
& 32=24+8 \\
\end{align}$
Here, there are two possibilities:
One of the possibilities is that there would be 2 buses and 5 cars.
The other possibility is that there would be 4 buses and 2 cars.
Note: Here, without checking randomly we can also find the solution by considering the HCF of 4 and 6. The HCF is 2, now for $x=2$, find the value of $y$. Similarly, for $y=2$, find the value of $x$.
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