IPM Cross Multiplication Questions & Answers [Solving Steps]

As an IPMAT aspirant, you must already know that understanding all concepts thoroughly is not sufficient. Mastering techniques like IPM Cross Multiplication Questions & Answers can significantly enhance your ability to solve mathematical problems quickly and accurately.

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This can save you a large amount of time to smartly cover as many questions to obtain good scores in these entrance exams. 

With a time duration of just 1.5 hours to finish the IPMAT exams, solving mathematical equations using conventional methods may not be a smart move. Saving time for your entire paper is your ultimate goal, so that you can attempt maximum questions in a go.

If you do not know how to solve lengthy calculations in a matter of seconds? Here is a simple technique named cross multiplication that will help you solve complex multiplication problems in less time.

Read through the post that will give you a complete insight on IPM Cross Multiplication Questions & Answers.

Did you know? Vedic Maths is taught to students from the UK, the US, Australia, Canada and West Asia across the globe?

What is Cross Multiplication?

Let's begin to understand the basics of cross multiplication - 

Cross Multiplication or Vedic Multiplication is a concept of Vedic maths that enables faster calculation than the usual conventional method. It is based on mental calculation.

As said above, the QA section in IPMAT is a bit difficult, lengthy and time-consuming. Therefore following maths tips for IPMAT will help maximize your scores. Implementing these tips offered by our experts have been found extremely useful at solving Multiplication Aptitude questions. 

Let us understand this concept using an example, say 24 X 35.

Step-1: Multiply the one’s digits of both the numbers and then note the ones digit a number of the product in the answer.

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Here we need to multiply 4 x 5, which would give us 20 and then note 0(one-digit number of the product) in the answer and take the tens digit to carry forward.

24

X

35

0

Step-2: Multiply the diagonals and add them together. In this problem, we have 2 x 5 = 10 and 3 x 4 = 12 as the diagonals. Add the results together to obtain 22+2(a carryforward), write its one-digit value down next to 0 and take the tens digit to carry forward.

24

X

35

= ( 2 X 5 + 3 X 4 + 2(carry forward) = 24 )

40

Read More: Short Tricks to Prepare for Maths Section in IPMAT Rohtak

Step 3: Multiply the digits in the ten’s column together and write the number to the left of the previous value.

24

X

35

= (2 X 3+2 (carry forward) = 8)

 Answer: 840

IPM Cross Multiplication Questions & Answers

Learning simple tricks will not only help manage your time in the final exam but also help you solve the questions with accuracy. You can increase your speed by more than 100% by learning a few simple tricks.

As per experts, solving previous year's Question Papers of IPMAT will help you get an idea about the type of questions asked and the difficulty level of questions.

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Let us understand how to multiply more significant numbers by using the cross multiplication technique.

Question: Multiply 235 X 275

Step-1: multiply the unit figures, that is, multiply 5 of 235 by 5 of 275 (5 x 5 = 25), put five at the answer figure area, and carry over 2 for the next step.

Step-2: do cross multiplication, that is, multiply 3 and 5 of 235 by 5 and 7 of 275 respectively and add the two products (3 x 5 + 5 x 7 = 15 + 35= 50).

Now, add the number 2 carried from step1 with 50 (50 + 2 = 52), put two on the left side of 5 in the answer figure, and take over 5 for the next step.

Step-3: do three way multiplications. You need to multiply 2, 3 and 5 of 235 by 5, 7 and 2 of 275 respectively and add all the products (2 x 5 + 3 x 7 + 5 x 2 = 10 + 21 + 10 = 41).

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Now, add 5 (carried from Step2) with 41 (41 + 5 = 46), put 6 at the left side of 25 in the answer figure, and carry over 4 to the next step.

Step-4: Again do cross multiplication. But this time it should be the tenth and hundredth digits of both the numbers. Multiply 2 and 3 of 235 by 7 and 2 of 275 respectively and add the products (2 x 7 + 3 x 2 = 14 + 6 = 20).

Now add 4 (carried from Step3) with 20 (20 + 4 = 24), put 4 at the left side of 625 in the answer figure, and carry over 2 to the next step.

Step-5: It is the final step. Multiply 2 of 235 by 2 of 275 (2 x 2 = 4), add 2 (carried from Step4) with 4, and put it at the left side of 4625 in the answer figure and you will get the final answer 64625.

Question: Multiply 1235 X 275

Step-1: multiply 5 of 1235 with 5 of 275 (5 x 5 = 25), put 5 at the answer figure area and carry over 2 for the next step.

Step-2: do cross multiplication, that is, multiply 3 and 5 of 1235 with 5 and 7 of 275 respectively and add the products (3 x 5 + 5 x 7 = 15 + 35 = 50).

Now, add 2 (carried from Step1) with 50 (50 + 2 = 52), put 2 on the left side of 5 in the answer figure and carry over 5 for the next step.

Step-3: Do three way multiplications, that is, you need to multiply 2, 3 and 5 of 1235 with 5, 7 and 2 of 275 respectively and add the products (2 x 5 + 3 x 7 + 5 x 2 = 10 + 21 + 10 = 41).

Now add 5 (carried from Step2) with 41 (41 + 5 = 46), put 6 at the left side of 25 in the answer figure and carry over 4 to the next step.

Read More: Short Tricks to solve multiplication and division questions in IPMAT

Step-4: Again do three way multiplications. But this time you need to multiply 1, 2 and 3 of 1235 with 5, 7 and 2 of 275 respectively and add all the products (1 x 5 + 2 x 7 + 3 x 2 = 5 + 14 + 6 = 25).

Now add 4 (carried from Step3) with 25 (25 + 4 = 29), put 9 at the left side of 625 in the answer figure and carry over 2 for the next step.

Step-5: Now do cross multiplication that consisting 1 and 2 of 1235 with 7 and 2 of 275 respectively and add the products (1 x 7 + 2 x 2 = 7 + 4 = 11).

Now, add 2 (carried from Step4) with 11 (11 + 2 = 13), put 3 at the left side of 9625 in the answer figure and carry over 1 to the next step.

Step-6: this is the final step. Multiply 1 of 1235 with 2 of 275 (1 x 2 = 2) and add 1 (carried from Step5) with 2 (2 + 1 = 3), put 3 at the left side of 39625 and the final answer of the question is 339625.

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Question: How many positive integer divisors are less than but do not divide?

Solution:

First we prime factorize n, n = apbqcr.....

No. of factors = (1+p)(1+q)(1+r)....

For example factors of 12 are = 22 x 31, no. of divisors = (2+1)(1+1) = 6.

Number of factors of n = (29 + 1)( 17 +1 ) = 30 x 18

n2 = 258 x 334

Number of factors of n2 = (58 + 1 )(34 +1) = 59 x 35.

No. of factors of n2 that are less than n = (59 x 35 - 1) / 2

(solving 59 x 35 using cross multiplication to save time, 59x 35 = 2065)

Ans: 1032

How to Find Square of any number using Cross Multiplication?

Learning important Maths Formulas will help you solve the questions quickly in the upcoming exam. You can follow the simple technique given below to find the square of any number.

Example:

Finding square of 72 using cross multiplication

72 X 72

Step1: Multiply the one’s digits of both the numbers and then note the ones digit-number of the product in the answer.

Here we need to multiply 2 x 2 which would give us 4 and then note 4 in the answer.

72

X

72

=4

Step 2: Multiply the diagonals and add them together. In this problem, we have 7 x 2 = 14 and 7 x 2 = 14 as the diagonals. Add the results together to obtain 28 and write its one-digit value down next to 4 and take the tens digit to carry forward.

72

X

72

= ( 2X 7 + 2X 7 = 28 )

84

Step 3: Multiply the digits in the ten’s column together and write the number to the left of the previous value.

72 

X

72

=(7 X7+2(carry forward) = 51)

Answer: 5184