At a dinner party, every two guests used a bowl of rice between them, every three guests used a bowl of dal among them and every four guests used a bowl of curd among them. There are altogether 65 bowls. What is the number of guests present at the party ?

Correct — D, (d) 60.

THE STEP THAT DECIDES THIS QUESTION is turning each sharing arrangement into a FRACTION OF A BOWL PER GUEST, so that the three kinds of bowl can be added.

Let there be n guests.

Every TWO guests share a bowl of rice, so the number of rice bowls is n/2. Every THREE guests share a bowl of dal, so the number of dal bowls is n/3. Every FOUR guests share a bowl of curd, so the number of curd bowls is n/4.

The total number of bowls is the sum of the three:

n/2 + n/3 + n/4 = 65

Take the lowest common denominator, which is 12:

6n/12 + 4n/12 + 3n/12 = 13n/12 = 65 n = 65 × 12 / 13 = 60

Sixty guests, which is option (d). Check it against the stem: 60 guests use 30 bowls of rice, 20 of dal and 15 of curd, and 30 + 20 + 15 = 65 exactly.

THE DIVISIBILITY CHECK, which answers this item in about a second and is the most useful thing on this card. For all three counts to be whole numbers of bowls, n must be divisible by 2, by 3 and by 4 — that is, by their lowest common multiple, which is 12. Now look at what is offered: 90, 80, 70 and 60. Ninety divided by twelve is seven and a half; eighty gives six and two-thirds; seventy gives five and five-sixths. Only SIXTY is a multiple of twelve, so only sixty can be the number of guests at all, whatever the total number of bowls happens to be. The arithmetic above then confirms it.

BACK-SUBSTITUTION SAYS THE SAME THING with the totals visible. The bowls used by n guests are 13n/12, that is one and one-twelfth bowls for each guest:

n = 90 → 97·5 bowls n = 80 → 86·67 bowls n = 70 → 75·83 bowls n = 60 → 65 bowls

Only one of the four both is a whole number and matches the 65 the stem gives.

A POINT ABOUT WHAT IS BEING COUNTED, because it is where the question can confuse a careful reader. The three groupings are three DIFFERENT partitions of the SAME sixty people, not three separate sets of guests. Every guest belongs simultaneously to a rice pair, a dal trio and a curd quartet; nobody is left out of any of them and nobody is counted twice among the guests. What is being added is bowls, not people, and the bowls are of three distinct kinds, so there is no double counting to correct for. A candidate who tries to allocate guests to one dish or another has misread the arrangement.

THE SHORTCUT WORTH KEEPING is the per-guest rate. Each guest accounts for 1/2 + 1/3 + 1/4 = 13/12 of a bowl, so bowls and guests stand in the fixed ratio 13 : 12. Sixty-five bowls therefore means 65 × 12/13 = 60 guests, and the whole problem is one multiplication. Notice also that this ratio makes the number of bowls slightly LARGER than the number of guests, which is a useful sanity check: the answer must be a little below 65, and only one option is.

The words "dal" and "curd" are printed in roman without italics or gloss, and the options are bare numerals in descending order, as the booklet sets them.

Why the others are wrong

• (a)90 — Ninety guests is the largest option and it fails two tests at once. First, ninety is not divisible by four, so ninety guests could not have been grouped into quartets for the curd at all — twenty-two bowls would serve eighty-eight of them and two guests would be left over. Since the stem describes a clean arrangement in which every four guests share a bowl, the number of guests must be a multiple of twelve, and ninety is not. Second, the totals are far apart: ninety guests would use forty-five bowls of rice, thirty of dal and twenty-two and a half of curd, ninety-seven and a half in all, against the sixty-five the stem states. The general check that disposes of it in a moment is the per-guest rate. Each guest accounts for thirteen-twelfths of a bowl, so the bowls always slightly OUTNUMBER the guests — and with sixty-five bowls the guests must be fewer than sixty-five, which rules out all three of the larger options together.
• (b)80 — Eighty guests is not divisible by three, so the dal could not have been shared three to a bowl without a remainder — twenty-six bowls would serve seventy-eight guests and two would go without. The requirement that n be a multiple of twelve is not a technicality here; it is the arithmetic expression of the stem's own description, which has every two guests, every three guests and every four guests neatly grouped. On the totals, eighty guests would call for forty bowls of rice, twenty-six and two-thirds of dal and twenty of curd, some eighty-seven bowls in all, which is more than a third above the stated sixty-five. It is worth noticing that the four options descend evenly in steps of ten, so no option is marked out by its position; the only structural feature that distinguishes them is divisibility, and it distinguishes them completely.
• (c)70 — Seventy is the nearest wrong option and it has a definite arithmetical origin, which makes it the most instructive of the three. A candidate who forms the per-guest rate correctly as 13/12 of a bowl and then MULTIPLIES the sixty-five bowls by 13/12 instead of dividing gets 70·4, which rounds to seventy. The direction matters: there are more bowls than guests, so converting bowls into guests must make the number smaller, not larger. Seventy also fails the divisibility test — it is not divisible by three or by four — and it would call for seventy-five and five-sixths bowls, which is not a whole number of anything. The safeguard is to state the ratio in words before using it: thirteen bowls for every twelve guests. Written that way, sixty-five bowls is five lots of thirteen, so the guests are five lots of twelve, which is sixty, and no inversion is possible.

SHARING PROBLEMS ARE RATE PROBLEMS IN DISGUISE, and the translation is always the same: "every k people share one item" means the items number one k-th of the people. Once each phrase has been converted into a fraction of an item per person, the fractions simply add, because the different kinds of item are independent of one another.

THE ARITHMETIC BACKBONE is the addition of unit fractions over a common denominator. Here 1/2 + 1/3 + 1/4 becomes 6/12 + 4/12 + 3/12 = 13/12, and the whole problem collapses into the statement that bowls and guests are in the ratio 13 : 12. Expressing a compound rate as a single fraction, and then reading it as a ratio in words, is the habit that keeps the direction of the final division right.

THE DIVISIBILITY INSIGHT is what makes such items quick on a multiple-choice paper. If a population is to be divided exactly into groups of 2, of 3 and of 4, its size must be a multiple of the lowest common multiple of those numbers, which is 12. That constraint is often enough to identify the answer on its own, and it is worth applying before any algebra: compute the LCM of the group sizes, and strike out every option that is not a multiple of it. The technique carries over to any question about exact grouping — sweets distributed among children, rows of chairs, batches on an assembly line, tiles laid without cutting.

WHAT IS NOT HAPPENING HERE, and this is the conceptual trap. The three groupings are not a partition of the guests into rice-eaters, dal-eaters and curd-eaters. They are three separate and overlapping ways of grouping the SAME people, and every guest takes part in all three. Nothing is double counted, because the quantity being added is bowls, and a bowl of rice, a bowl of dal and a bowl of curd are different objects. Had the question instead asked how many guests ate at least one of three dishes, with overlaps between them, the inclusion-exclusion principle would have been needed and the arithmetic would have been entirely different. Distinguishing a problem where quantities simply add from one where sets overlap is a distinction worth being deliberate about.

A RELATED FAMILY of problems uses the same machinery: pipes filling a tank at rates of one-sixth and one-eighth of the tank an hour; men and women completing fractions of a job in a day; three machines producing at different rates. In every case the method is to express each contributor's rate as a fraction of the whole per unit of time, add the fractions, and invert once at the end. The commonest error across the whole family is that single inversion being applied in the wrong direction, which is exactly the error one of the wrong options here is built on.

Key facts

• "Every k guests share one bowl" means the bowls number one k-th of the guests, so the three bowl counts are n/2, n/3 and n/4.
• Adding them gives n/2 + n/3 + n/4 = 13n/12, so bowls and guests stand in the fixed ratio 13 : 12.
• Setting 13n/12 = 65 gives n = 65 × 12/13 = 60 guests, who use 30 bowls of rice, 20 of dal and 15 of curd — sixty-five in all.
• For every group to be exact, the number of guests must be divisible by 2, 3 and 4, that is by their lowest common multiple of 12.
• Of the four options only 60 is a multiple of 12, so the divisibility check alone identifies the answer before any equation is written.
• Because each guest accounts for thirteen-twelfths of a bowl, the bowls always outnumber the guests, so the answer must be smaller than 65.
• The three groupings are three overlapping ways of grouping the same guests, not a division of them into three separate sets, so nothing is double counted.
• What is being added is bowls of three distinct kinds, which is why the fractions add directly and the inclusion-exclusion principle is not needed.
• Back-substituting the other options gives 97·5, 86·67 and 75·83 bowls respectively, none of them a whole number and none equal to 65.

Common traps

• Inverting the final ratio and multiplying by 13/12 instead of dividing, which gives about seventy and is the nearest wrong option.
• Reading "every two guests used a bowl between them" as two bowls per guest, which reverses the whole calculation.
• Trying to divide the guests into three separate groups by dish. All the guests take part in all three arrangements.
• Ignoring divisibility. Three of the four options cannot be grouped exactly into twos, threes and fours at all.
• Forgetting that the bowls outnumber the guests here, so the answer must be smaller than the total number of bowls.

• practice — not a real PYQ

  At a party every 3 guests shared one plate of snacks and every 5 guests shared one jug of water. If 32 plates and jugs were used altogether, how many guests were present at the party ?

  • (a)45
  • (b)50
  • (c)60
  • (d)75

  Answer(c) 60 — each guest accounts for 1/3 + 1/5 = 8/15 of an item, so items and guests are in the ratio 8 : 15 and 32 items mean 32 × 15/8 = 60 guests. Checking, 60 guests use 20 plates and 12 jugs, which is 32 in all.

• practice — not a real PYQ

  At a meal every 2 diners shared a bowl of rice and every 3 diners shared a bowl of dal. If 45 bowls were used in all, how many diners were present at the meal ?

  • (a)36
  • (b)45
  • (c)54
  • (d)60

  Answer(c) 54 — each diner accounts for 1/2 + 1/3 = 5/6 of a bowl, so 5n/6 = 45 and n = 54. Checking, 54 diners use 27 bowls of rice and 18 of dal, which is 45 in all; here the bowls are fewer than the diners, so the conversion works the other way from the original item.