Question

# A rental Company can set up 3 small tents and 1 large tent in 115 min. They can set up 2 small tents and 2 large tents in 130 min. How much time is required to set up a small tent ?

- 15 min
- 25 min
- 35 min
- 40 min

Hint:

### Frame proper equations and then solve.

## The correct answer is: 25 min

### Complete step by step solution:

Let’s take the time required to make one small tent = s minutes

And let the time required to make one large tent = l minutes

Let 3s + 1l = 115 minutes…(i)

and 2s + 2l = 130 minutes….(ii)

On multiplying (i) with 2, we get 2(3s + 1l = 115)

⇒ 6s + 2l = 230…(iii)

Now, we have the coefficients of l in (ii) and (iii) to be the same.

On subtracting (ii) from (iii),

we get LHS to be 6s + 2l - (2s + 2l) = 6s - 2s = 4s

and RHS to be 230 – 130 = 100

On equating LHS and RHS, we have 4s = 100

⇒ s = 25

On substituting the value of s in (ii), we get 2 × 25 + 2l = 130

⇒ 50 + 2l = 130

⇒ 2l = 130 - 50

⇒ 2l = 80

⇒ l = 40

Hence we get s = 25 and l = 40

Hence time to set up small tent = 25 minutes

and time to set up large tent = 40 minutes

Here, option B is the right answer.

25 minutes is time required to set up a small tent.

Note: We can also solve these system of equations by making the coefficients of s

to be the same in both the equations.

On multiplying (i) with 2, we get 2(3s + 1l = 115)

Now, we have the coefficients of l in (ii) and (iii) to be the same.

On subtracting (ii) from (iii),

we get LHS to be 6s + 2l - (2s + 2l) = 6s - 2s = 4s

and RHS to be 230 – 130 = 100

On equating LHS and RHS, we have 4s = 100

On substituting the value of s in (ii), we get 2 × 25 + 2l = 130

Hence we get s = 25 and l = 40

Hence time to set up small tent = 25 minutes

and time to set up large tent = 40 minutes

Here, option B is the right answer.

25 minutes is time required to set up a small tent.

Note: We can also solve these system of equations by making the coefficients of s

to be the same in both the equations.