Simultaneous (Linear) Equations — Case-Study Based Questions

Case study:
Jasjit goes from Connaught Place to Qutub Minar, a distance of 13 km, by a taxi and after spending some time there, he returned back to Connaught Place by a route which is 16 km long in some other taxi.

Case 1: If the taxi takes a fixed charge of ₹ x and per kilometer charges as ₹ y, calculate the total fare for going from Connaught Place to Qutub Minar and then returning back.

Case 2: If the taxi takes a fixed charge of ₹ x for the first kilometer and an additional charge of for ₹ y per km for subsequent kilometers, calculate the total fare for going from Connaught Place to Qutub Minar and then returning back.

Answer

Given,

Distance (Connaught Place to Qutub Minar while going) = 13 km

Distance (Connaught Place to Qutub Minar while returning) = 16 km

Case 1:

Charges applied:

Fixed charge = ₹ x

Per km charge = ₹ y

Total fare = Fixed charge + (Distance × per km charge)

For trip while going (13 km):

Total fare = x + (13 × y) = x + 13y

For trip while returning (16 km):

Total fare = x + (16 × y) = x + 16y

Total fare of whole journey = (x + 13y) + (x + 16y)

= 2x + 29y.

Hence, total fare = ₹ (2x + 29y).

Case 2:

Charges :

Fixed charge = ₹ x for first kilometer

Per km = ₹ y for remaining distance

Going (13 km):

Charge for first km = ₹ x

Remaining distance = 12 km

Charge for remaining distance = 12y

Total fare = x + 12y

Returning (16 km):

First km = ₹ x

Remaining distance = 15 km

Charge for remaining distance = 15y

Total fare = x + 15y

Total fare of whole journey = (x + 12y) + (x + 15y)

= 2x + 27y.

Hence, total fare = ₹ (2x + 27y).

Case study:
The school administration planned to renovate the common room and asked Rohan to calculate its dimensions so that the new furniture could be arranged properly. The common room is cuboidal in shape.

After collecting the necessary data, Rohan found that :

The volume of the common room is 144 cubic units.

The total surface area of the common room is 192 sq. units.

The height of the room is 3 units.

Based on the above information answer the following:

(a) Form equations for volume and total surface area assuming x and y as the length and breadth.

(b) Solve the equations for finding the dimensions of the common room.

(c) While construction is going on, what length of the biggest iron rod can be put in the room?

Answer

Given,

Volume of the common room = 144 cubic units.

Total surface area of common room = 192 sq. units

Height of the room = 3 units.

(a) Let, length = x units and breadth = y units.

Volume of cuboid = l × b × h

144 = x × y × 3

3xy = 144

xy = $\dfrac{144}{3}$

xy = 48 .........(1)

By formula,

Total surface area of cuboid = 2(lb + bh + hl)

192 = 2(x × y + y × 3 + 3 × x)

192 = 2(xy + 3y + 3x)

xy + 3y + 3x = $\dfrac{192}{2}$

xy + 3y + 3x = 96........(2)

Hence, equation for volume : 3xy = 144 and for TSA = 2(xy + 3y + 3x) = 192.

(b) Substituting the value of equation (1) in equation (2), we get :

48 + 3y + 3x = 96

3x + 3y = 96 - 48

3x + 3y = 48

3(x + y) = 48

x + y = $\dfrac{48}{3}$

x + y = 16

x = 16 - y .........(3)

Substituting the value of x from equation (3) in equation (1) :

(16 - y) × y = 48

16y - y² = 48

y² - 16y + 48 = 0

y² - 12y - 4y + 48 =0

y(y - 12) - 4(y - 12) = 0

(y - 12) (y - 4) = 0

y = 12 or y = 4

If y = 12, then

x = 16 - y = 16 - 12 = 4.

If y = 4, then

x = 16 - y = 16 - 4 = 12.

Since length is longer than breadth,

∴ x = 12 and y = 4

Hence, x = 12 and y = 4.

(c) Length of the biggest rod that can be put in a room = Diagonal of room

Diagonal of room (cuboid) = $\sqrt{l^2 + b^2 + h^2}$

$\Rightarrow d = \sqrt{12^2 + 4^2 + 3^2} \\[1em] \Rightarrow d = \sqrt{144 + 16 + 9} \\[1em] \Rightarrow d = \sqrt{169} \\[1em] \Rightarrow d = 13 \text{ units}.$

Hence, length of the biggest iron rod can be put in the room = 13 units.