Corporate and Other LawsMCQQuestion 6032 of 221
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11. For availing foreign exchange for studying abroad, which of the following option is correct:

Select Your Answer (Keyboard: A, B, C, D)
For any discrepancies in this question, email contact@cadada.in

Correct Answer

✅ Option A — The transaction of withdrawal of foreign exchange of USD 2,25,000 for studying abroad is prohibited.

All Options:

  • AThe transaction of withdrawal of foreign exchange of USD 2,25,000 for studying abroad is prohibited. ✓
  • BThe transaction of withdrawal of foreign exchange of USD 2,25,000 for studying abroad requires prior approval of Government of India.
  • CThe transaction of withdrawal of foreign exchange of USD 2,25,000 for studying abroad requires prior approval of RBI.
  • DThe transaction of withdrawal of foreign exchange of USD 2,25,000 for studying abroad do not require prior approval of RBI.

Detailed Solution & Explanation

Correct Answer: Option **A** Explanation: As per the official suggested answer, option **A** is correct. Hence, **Option A** is the correct answer.

More Questions from Corporate and Other Laws

If 3x+y+2x−y=−1\displaystyle 3x+y+2x−y\displaystyle \frac{3}{x+y} + \frac{2}{x-y} = -1 and 1x+y−1x−y=43\displaystyle 1x+y−1x−y=43\displaystyle \frac{1}{x+y} - \frac{1}{x-y} = \frac{4}{3} then (x,y)\displaystyle (x, y) is:

2x+510+3x+1015=5\displaystyle \frac{2x+5}{10} + \frac{3x+10}{15} = 5, find x\displaystyle x

Find value of x2−10x+1\displaystyle x^2 - 10x + 1 if x=15−26\displaystyle x = 15−26\displaystyle \frac{1}{5-2\sqrt{6}}

The cost of 2\displaystyle 2 oranges and 3\displaystyle 3 apples is $28$. If the cost of an apple is doubled then the cost of $3orangesand\displaystyle oranges and5$ apples is $75$. The original cost of 7\displaystyle 7 oranges and 4\displaystyle 4 apples (in Rs) is:

In a multiple choice question paper consisting of $100questionsof\displaystyle questions of1\displaystyle mark each, a candidate gets $60\% marks. If the candidate attempted all questions and there was a penalty of $0.25$ marks for wrong answers is:

The values of x\displaystyle x and y\displaystyle y satisfying the equations 3x+y+2x−y=3\displaystyle 3x+y+2x−y=3\displaystyle \frac{3}{x+y} + \frac{2}{x-y} = 3 and 2x+y+3x−y=233\displaystyle 2x+y+3x−y=233\displaystyle \frac{2}{x+y} + \frac{3}{x-y} = \frac{23}{3} given by

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