A person has a total of 100 coins consisting of ₹2 and ₹5 coins. If the total value of the coins is ₹320, then the number of ₹2 coins is
- (a)40
- (b)50
- (c)60
- (d)70
Correct — C, 60. Let x be the number of two-rupee coins, so the rest, 100 − x, are five-rupee coins. Their total value is 2x + 5(100 − x) = 320, which becomes 500 − 3x = 320, so 3x = 180 and x = 60. Check it: 60 two-rupee coins are worth ₹120 and the remaining 40 five-rupee coins are worth ₹200, and ₹120 + ₹200 = ₹320.
- (a)40 — Forty two-rupee coins and sixty five-rupee coins come to ₹80 + ₹300 = ₹380, sixty rupees too much. This is the answer to the mirror question — it is the number of five-rupee coins that is 40.
- (b)50 — An even split gives ₹100 + ₹250 = ₹350, still above the required total.
- (d)70 — Seventy two-rupee coins and thirty five-rupee coins give ₹140 + ₹150 = ₹290, thirty rupees short.
Two quantities mixed to a known count and a known total form a standard pair of linear equations — one counting the items, the other valuing them. Substituting the count equation into the value equation collapses the pair to a single unknown.
The alligation view is quicker still. If all 100 coins were five-rupee coins the total would be ₹500; the actual total is ₹180 less, and every two-rupee coin substituted for a five-rupee one costs ₹3 of that shortfall. So the number of two-rupee coins is 180 ÷ 3 = 60. The same 'assume the extreme, then account for the gap' move solves hens-and-goats, tickets and marking-scheme problems.
- Count equation x + y = 100; value equation 2x + 5y = 320; substitution gives x = 60 and y = 40.
- Assume-the-extreme shortcut: (500 − 320) ÷ (5 − 2) = 60 two-rupee coins.
- Always confirm which coin the question asks for — the two counts here are 60 and 40, and both appear among the options.
- The same structure appears whenever items of two values are mixed to a known total.
Check by valuing both heaps: ₹120 + ₹200 = ₹320.
- Answering 40, which is the number of five-rupee coins.
- Setting up the value equation with the coin values swapped.
- Assuming equal numbers of the two coins because the total is a round figure.
Asked as a fixed number of items of two denominations with the total value given, and the count of one denomination wanted.
In an exam, a candidate attempts 20 questions and scores 72 marks. If 5 marks are awarded for each correct answer and 2 marks are deducted for each wrong answer, then how many questions were answered correctly by him?
- (a) 18
- (b) 17
- (c) 16
- (d) 15
Answer(c) 16
The same count-and-value pair wearing an examination hall instead of a purse — a fixed number of items, two different per-item values and one grand total. The assume-the-extreme shortcut works there too.
- practice — not a real PYQ
A box has 50 notes of ₹10 and ₹50 denominations worth ₹1,300 in all. The number of ₹50 notes is
- (a)20
- (b)25
- (c)30
- (d)35
Answer(a) 20 — if all 50 were ₹10 notes the total would be ₹500; the ₹800 gap divided by the ₹40 difference per note gives 20 fifty-rupee notes.
- practice — not a real PYQ
A farmer has hens and goats totalling 30 animals with 80 legs between them. The number of goats is
- (a)10
- (b)15
- (c)20
- (d)25
Answer(a) 10 — all hens would give 60 legs; the 20 extra legs divided by the 2 extra legs per goat gives 10 goats.