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Linear System in Real Numbers

CAT 2025 Slot 1 · Quantitative Ability · Medium · Algebra

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If a−6b+6c=4a - 6b + 6c = 4 and 6a+3b−3c=506a + 3b - 3c = 50, where a,ba, b and cc are real numbers, the value of 2a+3b−3c2a + 3b - 3c is

  1. A.

    18

  2. B.

    15

  3. C.

    20

  4. D.

    14

Answer

A

Explanation

Let k=b−ck = b - c. The given equations become:

  1. a−6k=4  ⟹  a=4+6ka - 6k = 4 \implies a = 4 + 6k
  2. 6a+3k=506a + 3k = 50

Substitute a=4+6ka = 4 + 6k into (2): 6(4+6k)+3k=50  ⟹  24+36k+3k=50  ⟹  39k=26  ⟹  k=236(4 + 6k) + 3k = 50 \implies 24 + 36k + 3k = 50 \implies 39k = 26 \implies k = \frac{2}{3}. Then a=4+6(23)=8a = 4 + 6\left(\frac{2}{3}\right) = 8.

We need 2a+3b−3c=2a+3k=2(8)+3(23)=16+2=182a + 3b - 3c = 2a + 3k = 2(8) + 3\left(\frac{2}{3}\right) = 16 + 2 = 18.

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