Aptitude · Ratio and Proportion · Model 5

Twenty coins and five rupees are the same pile

Coin questions look hard because they quietly mix two different units. The ratio counts coins while the total counts rupees, or the other way round. Convert one into the other with a single number per denomination and the question becomes an ordinary split.

Convert a coin pile in either direction →

Coins to rupees: divide. Rupees to coins: multiply. The divisor is how many of that coin make one rupee — 1 for ₹1, 2 for 50p, 4 for 25p.

01 The idea

Two units in one question

Twenty 25-paise coins are worth ₹5. That sentence contains everything this model needs. Twenty is a count of coins; five is an amount of money; and the number that converts between them is 4, because four 25-paise coins make a rupee. Call that the magic number of the denomination. Rupees to coins, multiply by it. Coins to rupees, divide by it.

The magic numbers are the easy part: 1 for a rupee coin, 2 for 50 paise, 4 for 25 paise, 5 for 20 paise, 10 for 10 paise, 20 for 5 paise. In each case it is just 100 divided by the coin’s value in paise, and you can rebuild the whole table in your head from that.

The difficulty is that a paper deliberately mismatches the units. It gives you the ratio of the numbers of coins and then the total value in rupees. Or the ratio of the values and then the total number of coins. Either way you cannot equate the ratio with the total until one of them has been converted, and every wrong answer in this model comes from equating them anyway.

So there are exactly two setups. If the ratio is of coin counts and the total is money, divide each ratio term by its magic number to turn the ratio into a value ratio. If the ratio is of values and the total is a coin count, multiply each term by its magic number to turn it into a coin ratio. After that, it is Model 2: add the parts, divide the total, multiply out.

Make the ratio the same kind of thing as the total before you divide. Coins to rupees divides by the magic number; rupees to coins multiplies by it.

Magic numberHow many coins of that denomination make ₹1. For 50 paise it is 2; for 25 paise, 4; for 20 paise, 5; for 10 paise, 10. It is always 100 ÷ value in paise.

Coin ratio and value ratioThe same pile described two ways. Counts of 8 : 8 : 12 in ₹1, 50p and 25p coins is a value ratio of 8 : 4 : 3, because the second and third get divided by 2 and 4.

The two setupsRatio of counts with a money total → divide. Ratio of values with a coin total → multiply. Reading which one you have is the whole decision the model asks you to make.

02 Worked example

3 : 8 : 20 coins worth ₹372

One bag runs the whole lesson. A bag holds ₹1, 50-paise and 25-paise coins in the ratio 3 : 8 : 20. The total value is ₹372. Find the total number of coins.

Name the units in the questionThe ratio 3 : 8 : 20 counts coins. The total, ₹372, is money. Those cannot be equated, and noticing that is the whole first step.ratio = coins total = rupees ⇒ convert the ratio

Divide each term by its magic numberCoins to rupees is a division. Three rupee coins are worth ₹3; eight 50-paise coins are worth ₹4; twenty 25-paise coins are worth ₹5.3/1 + 8/2 + 20/4 = 3 + 4 + 5 = 12 value parts

Divide the money total by the value partsBoth sides are now in rupees, so one division gives the scale. Twelve parts of money make up the ₹372.₹372 ÷ 12 = 31 ⇒ x = 31

Multiply the original coin ratio by that scaleThe counts come from the coin ratio, not the value ratio. Three, eight and twenty parts, each worth 31 coins.coins = 3(31), 8(31), 20(31) = 93, 248, 620

Total the coins, and check the money both waysThe counts add to the answer. The values must add back to ₹372, which is the check that catches a multiply-instead-of-divide slip.93 + 248 + 620 = 961 coins ₹93 + ₹124 + ₹155 = ₹372 ✓

Look at the two totals side by side: 961 coins and ₹372. They are nowhere near each other, and that gap is the entire content of the model. A student who writes 3x + 8x + 20x = 372 gets x = 12 and an answer of 372 coins, which is wrong for a reason worth naming out loud — they added coins and set the sum equal to rupees.

03 The method

The conversion table, and the two directions

Six denominations and two directions. That is the whole model, and the right-hand column is where the marks are.

Magic number = 100 ÷ the coin’s value in paise. Coins → value: divide. Value → coins: multiply. Then it is an ordinary split: add the converted parts, divide the given total by that, and multiply back out.

Decide the direction by looking at the total, not the ratio. If the total is in rupees, the ratio must be turned into rupees, so divide. If the total is a number of coins, the ratio must be turned into coins, so multiply. Two further checks worth building in: the counts must all come out as whole numbers, because coins do not come in fractions; and the values must add back to the money total. If the counts are fractional, the question is broken rather than difficult — and published practice sets do contain such questions.

Coin	Magic number	Coins → value	Value → coins
₹1	`1`	÷ 1	× 1
50 paise	`2`	÷ 2	× 2
25 paise	`4`	÷ 4	× 4
20 paise	`5`	÷ 5	× 5
10 paise	`10`	÷ 10	× 10
5 paise	`20`	÷ 20	× 20
Ratio of coins set equal to a rupee total	`the error the model prevents`	3+8+20 = 372 gives 372 coins	the answer is 961

05 Cheat sheet

Model 5 on one page

Two directions, one table of magic numbers, and the two checks that tell you whether your answer is a bag that could exist.

Case	Rule	On 3 : 8 : 20 worth ₹372
Coin ratio, money total	`divide by the magic numbers`	3 + 4 + 5 = 12 parts
Value ratio, coin total	`multiply by the magic numbers`	8:4:3 → 8:8:12
One part	`total ÷ converted parts`	372 ÷ 12 = 31
The counts	`coin ratio × one part`	93, 248, 620
Total coins	`add the counts`	961
Counts must be whole	`otherwise the bag cannot exist`	2:3:4 with ₹372 needs 165.33 coins
Coins added to equal a money total	`never`	gives 372 coins instead of 961

The magic number is 100 over the paise value50 paise gives 2, 25 gives 4, 20 gives 5, 10 gives 10, 5 gives 20. You never have to memorise the table; you can rebuild any row of it in two seconds.

Whole coins are a real constraintCounts of 2 : 3 : 4 give a value of 4.5x rupees, so the total value must be a multiple of ₹4.50. A question setting that ratio against ₹120 is broken, and two published practice sets contain exactly that error.

Both totals are worth writing downThe pile is 961 coins and ₹372. Papers ask for either, and sometimes for one denomination’s value rather than its count. Having both totals on the page means any of those is one line away.

06 Where & why

Where this shows up

Coin questions are a reliable two or three marks and they are heavily flagged as high-frequency in bank and railway papers.

Bank Prelims · RRB

“Find the total number of coins”

The classic form: the coin ratio and the money total, so you divide. Chosen numbers always make the counts whole, and 25-paise coins turn up more often than any other denomination.

SSC CGL

“The ratio of their values is …”

The reverse direction, so you multiply. Note the phrase carefully — “ratio of their values” rather than “ratio of the coins” is the only signal that you are in the other setup.

Denominations beyond the usual three

20p, 10p and 5p coins

Nothing changes but the magic numbers: 5, 10 and 20. A question with 25p, 10p and 5p coins and no rupee coin at all works exactly the same way.

Notes rather than coins

₹10, ₹20 and ₹50 notes

The same conversion in the other direction: the magic number is now less than one, or more usefully you multiply the count by the note’s value. Same two setups, same two directions.

One question decides every problem in this model: is the total a number of coins or an amount of money? Answer that first, convert the ratio to match, and the rest is Model 2.

07 Interview questions

What gets asked

Ten. The third and fourth are the same question in opposite directions, which is exactly how a paper tests whether you read the wording.

What is the magic number of a coin?

How many of that coin make one rupee: 1 for a rupee coin, 2 for 50 paise, 4 for 25 paise, 5 for 20 paise, 10 for 10 paise. It is 100 divided by the coin’s value in paise, so you can rebuild it rather than memorise it.

Which way do you convert — multiply or divide?

Divide to go from a count of coins to an amount of money, and multiply to go the other way. Twenty 25-paise coins are 20 ÷ 4 = ₹5; ₹5 in 25-paise coins is 5 × 4 = 20 coins.

A bag has ₹1, 50p and 25p coins in the ratio 3 : 8 : 20 and a total value of ₹372. Find the total number of coins.

961. The ratio counts coins and the total is money, so divide each term by its magic number: 3 + 4 + 5 = 12 value parts. Then 372 ÷ 12 = 31, so the counts are 93, 248 and 620, which add to 961.

A box has 378 coins of ₹1, 50p and 25p whose values are in the ratio 13 : 11 : 7. How many 50p coins are there?

132. Here the ratio is of values and the total is a coin count, so multiply: 13, 11 × 2 = 22, and 7 × 4 = 28, giving 63 coin parts. Then 378 ÷ 63 = 6, so the 50-paise coins number 22 × 6 = 132.

What happens if you set 3x + 8x + 20x equal to 372?

You get x = 12 and a total of 372 coins, which is wrong. That equation adds coin counts and sets the sum equal to an amount of rupees, which is the exact mistake the model exists to prevent. The correct total is 961 coins.

A bag has 1 rupee, 50p and 10p coins in the ratio 3 : 5 : 7 with a total value of ₹124. How many 10-paise coins are there?

140. The magic numbers are 1, 2 and 10, so the value parts are 3 + 2.5 + 0.7 = 6.2. Then 124 ÷ 6.2 = 20, so the 10-paise coins number 7 × 20 = 140. Nothing about the method changes when the denominations do.

Coins of 25p, 10p and 5p are in the ratio 1 : 2 : 3 and total ₹30. How many 5-paise coins?

150. The magic numbers are 4, 10 and 20, so the value equation is x/4 + 2x/10 + 3x/20 = 30. Multiplying through by 20 gives 5x + 4x + 3x = 600, so x = 50 and the 5-paise coins number 150. Clearing the denominators first keeps the arithmetic clean.

Can a coins question be impossible?

Yes, and published practice sets contain some. Counts in the ratio 2 : 3 : 4 for ₹1, 50p and 25p coins give a value of 4.5x rupees, so the total value must be a multiple of ₹4.50. Set against ₹120 it needs 26.67 parts, which means fractional coins. That question is broken, not hard.

What checks should you run before writing the answer down?

Hard

A bag is said to hold ₹1, 50-paise and 25-paise coins whose counts are in the ratio 2 : 3 : 4, with a total value of ₹120. (a) Show that no such bag exists. (b) Find the smallest total value in whole rupees for which counts in the ratio 2 : 3 : 4 are possible, and give the three counts. (c) State the general condition the ratio 2 : 3 : 4 imposes on the total value, and explain why ₹120 fails it.

Follow-up

This problem is taken from a published practice set that reports an answer without noticing the bag is impossible. A ratio of counts fixes the value of the pile as a multiple of one fixed amount, and checking that divisibility is a real habit worth building — it is also the fastest way to eliminate wrong options in a working question.

Show the hint

Work out the value of the smallest possible such bag, with counts exactly 2, 3 and 4, and then ask what multiples of that value are available.