Shortcut Trick
Given the cost of 15 apples and 12 oranges is ₹447, we divide by 3 to get the cost of 5 apples and 4 oranges as ₹149.
An inconsistent statement will have the same ratio of apples to oranges (5:4) but an equivalent cost that does not match ₹149.
In the fourth statement, 35 apples and 28 oranges are in a 5:4 ratio.
However, dividing its cost ₹1,029 by 7 gives ₹147 for 5 apples and 4 oranges.
Since ₹147 ≠ ₹149, it contradicts the given data, leading to no possible solutions.
∴ The correct answer is Purchased 35 apples and 28 oranges and paid a sum of ₹1,029.
Alternate Method
Given:
Cost of 15 apples and 12 oranges = ₹447
Formula Used:
For a system of linear equations a1x + b1y = c1 and a2x + b2y = c2:
If a1/a2 = b1/b2 ≠ c1/c2, the system is inconsistent and yields no possible solution.
Calculation:
⇒ Let the price of 1 apple be a and 1 orange be o.
⇒ 15a + 12o = 447
⇒ Checking Statement 1: 25a + 20o = 745
⇒ 15/25 = 12/20 = 447/745 (ratio simplifies to 3/5 for all). Consistent (Infinite solutions).
⇒ Checking Statement 2: 10a + 13o = 353
⇒ 15/10 ≠ 12/13. Consistent (Unique solution).
⇒ Checking Statement 3: 12a + 8o = 340
⇒ 15/12 ≠ 12/8. Consistent (Unique solution).
⇒ Checking Statement 4: 35a + 28o = 1029
⇒ a1/a2 = 15/35 = 3/7
⇒ b1/b2 = 12/28 = 3/7
⇒ c1/c2 = 447/1029 = 149/343
⇒ Here, a1/a2 = b1/b2 ≠ c1/c2
⇒ Because the ratios of coefficients are equal but differ from the constant ratio, the system is inconsistent.
∴ The correct answer is Purchased 35 apples and 28 oranges and paid a sum of ₹1,029.
Additional Information
Unique Solution
A system of equations has a unique solution if the lines intersect at exactly one point. Condition: a1/a2 ≠ b1/b2.
No Solution (Inconsistent System)
A system has no solution if the lines are perfectly parallel and never intersect. Condition: a1/a2 = b1/b2 ≠ c1/c2.
Infinite Solutions (Dependent System)
A system has infinitely many solutions if the equations represent the exact same line. Condition: a1/a2 = b1/b2 = c1/c2.