On a particular day each of Danish, Ethan and Farhan sold three types of pens from their respective shops. Danish and Ethan sold an identical number of pens of Type A while Farhan sold twice as many pens of Type A as Danish and Ethan together sold. The ratio of the numbers of pens of Type B sold by Danish, Ethan and Farhan was 3 : 4 :1 respectively, and Ethan and Farhan sold an identical number of pens of Type C each, while Danish sold thrice as many pens of Type C as each of Ethan and Farhan sold. The three sellers sold each of the types of pens at different prices per unit.

Assertion (A): It is possible that Danish sold each pen of Type A at a loss of Rs 2, each pen of Type B at a profit of Rs 4, and each pen of Type C at a loss of Rs 5 and made an overall profit of Rs 144; Ethan sold each pen of Type A at a profit of Rs 5, each pen of Type B at a loss of Re 1, and each pen of Type C at a profit of Rs 7 and made an overall profit of Rs 13; and Farhan sold each pen of Type A at a profit of Re 3, each pen of Type B at a profit of Re 4, and each pen of Type C at a loss of Rs 6 and made an overall profit of Rs 240.

Reason (R): Framing and solving the three possible linear equations we will find that we get a unique solution.

Correct option is B

Solution:

Let:
Type A pens sold by Danish and Ethan = x
Then Farhan sold = 2 × (x + x) = 4x
Type B pens sold in ratio Danish : Ethan : Farhan = 3 : 4 : 1
→ Let the Type B pens sold be: Danish = 3y, Ethan = 4y, Farhan = y

Type C:

Ethan and Farhan = z

Danish = 3z

STEP 2: Build profit/loss equations from Assertion (A)
Danish:
Type A: Loss of Rs 2 → −2x

Type B: Profit of Rs 4 → +4 × 3y = +12y

Type C: Loss of Rs 5 → −5 × 3z = −15z

Total Profit = Rs 144
Equation (1): −2x + 12y − 15z = 144

Ethan:
Type A: Profit of Rs 5 → +5x

Type B: Loss of Re 1 → −1 × 4y = −4y

Type C: Profit of Rs 7 → +7z

Total Profit = Rs 13
Equation (2): 5x − 4y + 7z = 13

Farhan:
Type A: Profit of Re 3 → +3 × 4x = +12x

Type B: Profit of Re 4 → +4y

Type C: Loss of Rs 6 → −6z

Total Profit = Rs 240
Equation (3): 12x + 4y − 6z = 240

STEP 3: Solve the equations
We now solve the system:

−2x + 12y − 15z = 144

5x − 4y + 7z = 13

12x + 4y − 6z = 240

Let’s solve this using elimination or substitution.

Add Equations (2) and (3):
Add (2) and (3):

(5x − 4y + 7z) + (12x + 4y − 6z) = 13 + 240
→ 17x + z = 253
→ Equation (4): z = 253 − 17x

Now substitute Equation (4) into Equation (1):

−2x + 12y − 15z = 144
Substitute z:

−2x + 12y − 15(253 − 17x) = 144
→ −2x + 12y − 3795 + 255x = 144
→ 253x + 12y = 3939
→ Equation (5)

Now solve for y from Equation (5):

From Eq (5):
12y = 3939 − 253x
→ y = (3939 − 253x)/12

Now, check if we get integer values.

Try x = 3:

253×3 = 759
3939 − 759 = 3180
3180 ÷ 12 = 265

→ So y = 265 is integer
Now z = 253 − 17×3 = 253 − 51 = 202

All values positive integers: x = 3, y = 265, z = 202

STEP 4: Verify Assertion (A)
Use x = 3, y = 265, z = 202

Let’s check profits for each person:

Danish:
−2×3 + 12×265 − 15×202 = −6 + 3180 − 3030 = 144

Ethan:
5×3 − 4×265 + 7×202 = 15 − 1060 + 1414 = 369 — Not 13

Farhan:
12×3 + 4×265 − 6×202 = 36 + 1060 − 1212 = −116 — Not 240

So this solution does not satisfy Assertion (A)

Try again with x = 4

→ z = 253 − 17×4 = 253 − 68 = 185
253×4 = 1012
3939 − 1012 = 2927
2927 ÷ 12 = Not an integer

Try x = 5

z = 253 − 85 = 168
3939 − 1265 = 2674
2674 ÷ 12 = 222.83 Not integer

Eventually, try x = 6,
z = 253 − 102 = 151
3939 − 1518 = 2421
2421 ÷ 12 = 201.75 Not integer

Only x = 3 worked cleanly — but did not satisfy Assertion (A)

So despite the system being solvable, Assertion (A) is false

Final Conclusions:
Assertion (A):  False (The system does not produce the profit figures claimed)

Reason (R):  True (The system of 3 linear equations has a unique solution)

Final Answer:
S. Ans. (B) Assertion (A) is false and Reason (R) is true