CSAT Solved Papers/ 2024/Q50

2024 CSAT — Q50

Quant Statement validity 2.5 marks Medium

The total cost of 44 oranges, 66 mangoes and 88 apples is equal to twice the total cost of 11 orange, 22 mangoes and 55 apples. Consider the following statements:

  1. The total cost of 33 oranges, 55 mangoes and 99 apples is equal to the total cost of 44 oranges, 66 mangoes and 88 apples.

  2. The total cost of one orange and one mango is equal to the cost of one apple.

Which of the statements given above is/are correct?

  1. A 1 only
  2. B 2 only
  3. C Both 1 and 2 Answer
  4. D Neither 1 nor 2

Worked rationale

Let unit costs be o, m, ao,\ m,\ a (orange, mango, apple). The given relation:

4o+6m+8a=2 (o+2m+5a)=2o+4m+10a.4o + 6m + 8a = 2\,(o + 2m + 5a) = 2o + 4m + 10a.

Collect terms:

4o+6m+8a−2o−4m−10a=0  ⇒  2o+2m−2a=0  ⇒  o+m=a.(⋆)4o + 6m + 8a - 2o - 4m - 10a = 0 \;\Rightarrow\; 2o + 2m - 2a = 0 \;\Rightarrow\; o + m = a. \tag{$\star$}

The single relation o+m=ao+m=a is the engine for both statements.

Statement 2 is exactly (⋆)(\star): one orange ++ one mango == one apple. True.

Statement 1 compares 3o+5m+9a3o+5m+9a with 4o+6m+8a4o+6m+8a. Their difference is

(4o+6m+8a)−(3o+5m+9a)=o+m−a=0by (⋆).(4o+6m+8a) - (3o+5m+9a) = o + m - a = 0 \quad\text{by } (\star).

So the two totals are equal. True.

Both statements follow from (⋆)(\star).

Answer: (c) Both 1 and 2.

Why the other options miss

  • A
    doesn’t recognise its own derived identity: verifies statement 1 (the difference o+m−ao+m-a) but fails to recognise that (⋆)(\star) is statement 2 verbatim.
  • B
    reads off o+m=ao+m=a for statement 2 but does not test statement 1 by differencing the two baskets, so dismisses it.
  • D
    mishandles the “twice” relation (e.g. doubles the wrong side or forgets to distribute the 22), never reaching o+m=ao+m=a, and rejects both.

Specialist insight

The whole problem collapses to the one derived identity o+m=ao+m=a. Don’t try to solve for three unknowns (you can’t, and you don’t need to) — extract the single linear relation the data forces, then test each statement against it by differencing baskets. Statement 1 is true not by computing totals but because its basket differs from the reference basket by exactly (o+m−a)(o+m-a), which is 00. This “difference of baskets” move is the time-saver: it turns a scary three-variable system into one subtraction.

The trap, in one line

Reduce the "twice" relation to o+m=ao+m=a, then difference baskets — statement 1's gap is o+m−a=0o+m-a=0 and statement 2 *is* o+m=ao+m=a, so both hold: (c)(c).

← All 2024 CSAT questions