Faster Calculation methods applied on ‘Real Exam Questions’ – Part III

Today we’ll take a problem and discuss a very important strategy that helps us to get the answers even without solving- Rejecting out options! In fact I’ll go on to say this – you should try solving every question not only by the ‘usual’ methods but also by ‘rejecting out the can’t-be options’ ; you might not be able to reject out all the options but even rejection of 2 or 3 options can increase your probability of getting the correct answer (after all most of the competitive exams test your ‘aptitude’ test and not your ‘math knowledge’ !)

Try this below problem:


If we use the ‘conventional’ method, we need to multiply them taking two at a time, get the result and again multiply the third with the result to arrive at the final answer. Even if we are good at multiplying we would take approximately 2 to 3 minutes to get the answer (plus there is chance that we might like to ‘redo’ or ‘recheck’ the multiplication due to the presence of the 5th option ‘NONE OF THESE’).

Instead of multiplying what we can do is- divide the second and third multiplicand by 10 each. So that the problem in the L.H.S becomes 8.88 * 8.88 * 8.8

Since we are dividing L.H.S. by 100, we definitely need to shift decimal point in the options by two places to the left which is essentially dividing the final result by 100. Hope you get the point! But why are we doing this?

We divided the multiplicands by 10 each to make every multiplicand look like ‘EIGHT POINT SOMETHING’ ! From here can we say that the product 8.88 * 8.88 * 8.8 is definitely less than 9 * 9 * 9 or 729! If you imagine shifting decimal two places left in the options you can directly reject out (2) and (4) as they are ‘793-point-something’ and ‘762-point-something’.

Second very important result you can use is the ‘divisibility test of 11‘. Why? There is one multiplicand 88 which is divisible by 11 so my options should also be divisible by 11. (Sum of the digits at odd places [from right] minus sum of the digits at even places must be either 0 or a multiple of 11 for the ‘reference number’ to be divisible by 11).

Divisibility test of 11 on 68301.142 [ (2 + 1 + 0 + 8) – (4 + 1 + 3 + 6) = -3, NOT divisible]

Divisibility test of 11 on 65356.824 [ (4 + 8 + 5 + 5) – (2 + 6 + 3 + 6) = 5, NOT divisible]

So, the only option left is (5) None of these. You could also use the divisibility test of 9 on options (1) and (3) as first two multiplicands are multiples of 3. That way, you should be able to reject out option (1). One more way in which options are rejected is by checking the ‘digit’ at the unit’s place. Here you could also reject out option (3) on the grounds that units digit should be 2 here ( 8 * 8 * 8 = 2; taking last digit of every multiplicand and multiplying them to get the unit’s place digit).

Use all these ‘concepts’ and formulate some of your ‘own’ to solve questions without solving or calculate smartly to save time! 🙂

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