Madan Singh runs a rental car company. He charges ₹ 250 per day plus ₹ 15 for every kilometre the car is driven. Professor Dayal rents a car for 1 day, while his own car is being repaired. He assures Madan Singh that he will pay him more than ₹ 500 as rent for the day.

(1) The inequality for the rent paid by Dayal for 1 day is :

1. 3x < 100
2. x > 25
3. 3x > 50
4. x < 75

(2) The solution set for the inequality obtained above is given by :

1. {16, 17, 18, ……………}
2. {17, 18, 19, ……………}
3. {19, 20, 21, ……………}
4. {20, 21, 22, ……………}

(3) Dayal estimated that the rent for 1 day would be less than ₹ 600 as he calculated the distance he has to drive the car. The inequality for the rent in this case would be :

1. y > 30
2. 2y < 35
3. 3y < 70
4. 4y > 45

(4) The solution set for the above inequality is given by :

1. {……………, 20, 21, 22, 23}
2. {……………, 17, 18, 19, 20}
3. {……………, 18, 19, 20, 21}
4. {……………, 15, 16, 17, 18}

(1)

Total Rent = (Fixed Daily Charge) + (Charge per Kilometre x distance)

Let the distance driven be x km.

Total Rent = ₹ 250 + 15x

The total rent is more than 500:

250 + 15x > 500

15x > 500 - 250 $\quad$ [Subtracting 250 from both sides]

15x > 250

Divide both sides by 5:

3x > 50

Hence, option 3 is the correct option.

(2)

Let's solve 3x > 50:

3x > 50

x > $\dfrac{50}{3}$ $\quad$ [Dividing 3 from both sides]

x > 16.66..

The solution must be greater than 16.66..

Solution set = {17, 18, 19, ….}

Hence, option 2 is the correct option.

(3)

Let the distance driven be y km. The total rent is less than 600:

250 + 15y < 600

15y < 600 - 250 $\quad$ [Subtracting 250 from both sides]

15y < 350

Divide both sides by 5:

3y < 70

Hence, option 3 is the correct option.

(4)

Let's solve 3y < 70:

3y < 70

y < $\dfrac{70}{3}$ $\quad$ [Dividing 3 from both sides]

y < 23.33..

The distance must be 23 km or less.

Solution set = {……………, 20, 21, 22, 23}

Hence, option 1 is the correct option.