Aman has certain number of notes of denomination ₹20 and ₹10 which amount to ₹390. If the numbers of notes of each kind are interchanged, then the new amount is ₹90 less than before. Find the number of notes of ₹20 denomination.

Correct — A. Let x be the number of ₹20 notes and y the number of ₹10 notes.

Value now: 20x + 10y = 390 → 2x + y = 39
After the swap the total falls by ₹90: 20y + 10x = 300 → x + 2y = 30

Substitute y = 39 − 2x: x + 2(39 − 2x) = 30 → 78 − 3x = 30 → x = 16
Then y = 39 − 32 = 7

Check: 20(16) + 10(7) = ₹390, and swapped 20(7) + 10(16) = ₹300, exactly ₹90 less. So 16 notes of ₹20 → option (a)

Why the others are wrong

• (b)
  14 — 14 twenties leaves ₹110 in tens, so 11 notes. Swapping those gives 20(11) + 10(14) = ₹360, a fall of only ₹30.
• (c)
  15 — 15 twenties leaves ₹90 in tens, so 9 notes. The swap gives 20(9) + 10(15) = ₹330, a fall of ₹60 rather than ₹90.
• (d)
  12 — 12 twenties leaves 15 tens, and swapping those raises the total to 20(15) + 10(12) = ₹420. The stem needs the total to drop.

Two sentences, two equations: one fixes the value of the collection, the other fixes what happens when the counts are exchanged.

The swap has a shortcut worth learning. Exchanging the counts changes the total by (20 − 10) × (x − y), so a ₹90 fall means x − y = 9 straight away.

Pair that with 2x + y = 39 and adding the two equations gives 3x = 48, so x = 16 in a single line.

Key facts

• Aman holds 16 notes of ₹20 and 7 notes of ₹10, which is ₹390 in 23 notes.
• Swapping the two counts changes the total by 10 × (x − y), so the ₹90 drop fixes x − y = 9.
• The swapped collection is worth ₹300, and 390 − 300 = 90 as the stem requires.

Common traps

• Reading ₹90 as the new total rather than the size of the fall
• Solving correctly and then marking the ₹10 count, which is 7
• Setting up the swapped equation as 10y + 20x, which just repeats the first equation

No directly related past PYQ was found.