CAT 2025 Slot 3 DILR Set 5 Q18–Q22

A DILR caselet from the Data Interpretation & Logical Reasoning (DILR) section of CAT 2025 Slot 3, with 5 questions (Charts & Tables). Answers are CAT's official key.

The set

Three countries — Pumpland (P), Xiland (X) and Cheeseland (C) — trade among themselves and with the (other countries in) Rest of World (ROW). All trade volumes are given in IC (international currency). The following terminology is used:
• Trade balance = Exports – Imports
• Total trade = Exports + Imports
• Normalized trade balance = Trade balance / Total trade, expressed in percentage terms

The following information is known.
1. The normalized trade balances of P, X and C are 0%, 10%, and –20%, respectively.
2. 40% of exports of X are to P. 22% of imports of P are from X.
3. 90% of exports of C are to P; 4% are to ROW.
4. 12% of exports of ROW are to X, 40% are to P.
5. The export volumes of P, in IC, to X and C are 600 and 1200, respectively. P is the only country that exports to C.

Question 18

How much is exported from C to X, in IC?

TITA: type in the answer. There are no options and no negative marking.

Answer: 48 Charts & Tables TITA

Solution

  1. Only P exports to C, so C's imports \(= 1200\). Normalised balance \(-20\%\): \(E_C - 1200 = -0.2(E_C + 1200) \Rightarrow E_C = 800\): 720 to P, 32 to ROW, 48 to X.
  2. C to X \(= 800 - 720 - 32 = 48\) (6% of C's exports).

Question 19

How much is exported from P to ROW, in IC?

TITA: type in the answer. There are no options and no negative marking.

Answer: 200 Charts & Tables TITA

Solution

  1. Only P exports to C, so C's imports \(= 1200\). Normalised balance \(-20\%\): \(E_C - 1200 = -0.2(E_C + 1200) \Rightarrow E_C = 800\): 720 to P, 32 to ROW, 48 to X.
  2. Let ROW's exports be \(R\). X's imports \(= 600 + 48 + 0.12R\), and \(E_X = \frac{11}{9}I_X\) (10% balance).
  3. P's imports \(= 0.4E_X + 720 + 0.4R\), with \(0.4E_X = 0.22 I_P\). Solving gives \(R = 2100\), \(E_X = 1100\), \(I_X = 900\), \(I_P = 2000\).
  4. P is balanced, so \(E_P = 2000 = 600 + 1200 + \) (P to ROW), giving P to ROW \(= 200\).

Question 20

How much is exported from ROW to ROW, in IC?

TITA: type in the answer. There are no options and no negative marking.

Answer: 1008 Charts & Tables TITA

Solution

  1. Only P exports to C, so C's imports \(= 1200\). Normalised balance \(-20\%\): \(E_C - 1200 = -0.2(E_C + 1200) \Rightarrow E_C = 800\): 720 to P, 32 to ROW, 48 to X.
  2. Let ROW's exports be \(R\). X's imports \(= 600 + 48 + 0.12R\), and \(E_X = \frac{11}{9}I_X\) (10% balance).
  3. P's imports \(= 0.4E_X + 720 + 0.4R\), with \(0.4E_X = 0.22 I_P\). Solving gives \(R = 2100\), \(E_X = 1100\), \(I_X = 900\), \(I_P = 2000\).
  4. ROW sends 12% to X, 40% to P and nothing to C, so 48% goes to ROW: \(0.48 \times 2100 = 1008\).

Question 21

What is the trade balance of ROW?
  1. A
    200
  2. B
    100
  3. C
    0
  4. D
    -200

Answer: (A) 200 Charts & Tables MCQ

Solution

  1. Only P exports to C, so C's imports \(= 1200\). Normalised balance \(-20\%\): \(E_C - 1200 = -0.2(E_C + 1200) \Rightarrow E_C = 800\): 720 to P, 32 to ROW, 48 to X.
  2. Let ROW's exports be \(R\). X's imports \(= 600 + 48 + 0.12R\), and \(E_X = \frac{11}{9}I_X\) (10% balance).
  3. P's imports \(= 0.4E_X + 720 + 0.4R\), with \(0.4E_X = 0.22 I_P\). Solving gives \(R = 2100\), \(E_X = 1100\), \(I_X = 900\), \(I_P = 2000\).
  4. P is balanced, so \(E_P = 2000 = 600 + 1200 + \) (P to ROW), giving P to ROW \(= 200\).
  5. ROW's exports to P, X, C: \(840 + 252 + 0 = 1092\). Imports from P, X, C: \(200 + 0.6 \times 1100 + 32 = 892\).
  6. Trade balance \(= 1092 - 892 = 200\) (the internal ROW-to-ROW trade cancels either way).

Question 22

Which among the countries P, X, and C has/have the least total trade?
  1. A
    Both X and C
  2. B
    Only X
  3. C
    Only C
  4. D
    Only P

Answer: (A) Both X and C Charts & Tables MCQ

Solution

  1. Only P exports to C, so C's imports \(= 1200\). Normalised balance \(-20\%\): \(E_C - 1200 = -0.2(E_C + 1200) \Rightarrow E_C = 800\): 720 to P, 32 to ROW, 48 to X.
  2. Let ROW's exports be \(R\). X's imports \(= 600 + 48 + 0.12R\), and \(E_X = \frac{11}{9}I_X\) (10% balance).
  3. P's imports \(= 0.4E_X + 720 + 0.4R\), with \(0.4E_X = 0.22 I_P\). Solving gives \(R = 2100\), \(E_X = 1100\), \(I_X = 900\), \(I_P = 2000\).
  4. P is balanced, so \(E_P = 2000 = 600 + 1200 + \) (P to ROW), giving P to ROW \(= 200\).
  5. Total trade: P \(= 2000 + 2000 = 4000\), X \(= 1100 + 900 = 2000\), C \(= 800 + 1200 = 2000\).
  6. X and C tie for the least.

Same concept in other years

See every question on Trade flows between countries: imports, exports, tariffs, balances.

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All 130 Charts & Tables questions asked since 2017.