Small scooters cost £80 and large scooters cost £130. A school buys few scooters for the playground for £1000. How many of the smaller and larger scooter did they buy?

Small scooters = 6

Large scooters = 4

Step-by-step explanation:

Let S represents small scooter and L represents large scooter

The cost of a small scooter is £80

Mathematically,  the cost of S number of small scooters is

80S

The cost of a large scooter is £130

Mathematically,  the cost of L number of large scooters is

130L

The total amount spent on both scooters is £1000

Mathematically,

80S + 130L = 1000

So we have 2 unknowns and only 1 equation. Therefore, we have use trial and error method .

Trial and Error Method:

80S + 130L = 1000

80S = 1000 - 130L

S = (1000 - 130L)/80

Let’s suppose they bought 1 large scooter

S = (1000 - 130(1))/80

S = 10.875

Number of scooters cannot be in fraction so 1 doesn’t work

Let’s suppose they bought 2 large scooter s

S = (1000 - 130(2))/80

S = 9.25

Number of scooters cannot be in fraction so 2 doesn’t work

Let’s suppose they bought 3 large scooter s

S = (1000 - 130(3))/80

S = 7.625

Number of scooters cannot be in fraction so 3 doesn’t work

Let’s suppose they bought 4 large scooters

S = (1000 - 130(4))/80

S = 6

So that means they bought 6 small scooters and 4 large scooters.

Similarly, other combinations for scooters don’t work since they yield either fractional value or negative value which cannot be correct.

Therefore, the only possible combination of small and large scooters is

6 Small scooters

4 Large scooters

Verification:

80S + 130L = 1000

80(6) + 130(4) = 1000

480 + 520 = 1000

1000 = 1000 (satisfied)

1/3 + 1/3 ≠ 2/6 because 1/3 alone = 2/6  

step by step explanation:

1/3 + 1/3 = 2/3       2/3=4/6

ratio is 2 : 9

the answer is three

13.5 : 3