Updated On : August 7, 2023
Summary: This article clears the basics of IPM Aptitude Questions Based on Remainders for your IPMAT preparations. Solve the questions listed in this article!
Taking the IPMAT is difficult in itself, and having less preparation time adds to the difficulty. You can manage your time, though, if you know the right strategy to prepare for each section of the paper.
The IPM Aptitude Questions Based on Remainders assess your mathematical reasoning skills.
Practising more number of questions from this concept will help you fetch more marks and outrun your competitors.
Students with non-math backgrounds have previously passed the exam with the proper preparation methods.
To ease your preparation, we have compiled the important IPM aptitude questions and answers based on remainders problems in this post.
In arithmetic, the remainder is defined as the integer left over after dividing one integer by another to produce an integer quotient. (You are advised to watch the above video to have a better understanding of IPM Aptitude Questions Based on Remainders).
Many questions based on the remainder concept are generally asked in competitive examinations, including IPMAT.
Therefore, following IPMAT Maths Preparation Tips would definitely help you solve these questions in a matter of seconds.
We have listed important tips and theorems related to the remainders along with previous year's questions with solutions.
Starting from the basics, the general equation for finding remainders is.
Dividend(n) = Divisor(d) x Quotient(q) + Remainder(r)
Let’s understand this concept with the help of an example; if we divide 100/3, we will have:-
Divisor(d) = 3 Dividend(n) = 100
Quotient(q) = 33
Remainder(r) = 1
100 = 3 x 33 + 1
If x, y, and z be three random numbers, we know that
(i) remainder of (x + y) z
(ii) remainder of (x -y) z
(iii) remainder of (x * y) z
The Trick to solve this question is to replace the unknown number by their respective remainder i.e., (Rx + Ry) / z. The remainder of this term will be the answer now.
Three cases that are possible using this approach are explained below.
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To help you get an idea about the type of questions asked in the exam, we have provided few sample questions that are curated from the previous year's Question Papers for IPMAT.
In a division problem, the product of quotients and the remainder is 24 while their sum is 10. If the divisor is 5 then the dividend is __________.
In this problem, we have given that the product of quotient(q) and the remainder(r) is 24.
i.e q x r = 24 -( i )
And it is also given that their sum is 10
i.e q + r = 10 -( ii )
Divisor(d) = 5.
( i ) and ( ii ) represent the sum and product of a quadratic equation; from ( ii ) putting r = q - 10, in ( i ), we will get q x (10 - q) = 24,
i.e q2-10q + 24 = 0,
Which will give q = 6 or 4.
Therefore, one of them will be 6, and the other one will be 4. We do not know yet whether Quotient is six and Remainder is four or vice versa.
We know that
Dividend(n) = divisor(d) x quotient(q) + remainder(r) .
From the given data, n = 5 x q + r. - (iii),
As the divisor is 5, we infer that the remainder obviously will be smaller than 5; therefore the remainder will be 4, and the quotient will be 6. Putting the above-decided value in (iii).
n = 5(6) + 4 , therefore n = 34.
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What is the remainder when 123 x 124 x 125 is divided by 9?
When 123 is divided by 9, the remainder will be -3.
While it is -2 when 124 is divided by 9.
Similarly, when 123 is divided by 9, the remainder will be -1
Therefore, final remainder = (-3)(-2)(-1)
The required positive remainder = 9-6
What is the remainder when 2^50 is divided by 49?
Using Euler's theorem, the remainder when 2^50 is divided by 49 is 1.
Explanation: Euler’s theorem states that if a and n are relatively prime (i.e., the greatest common divisor of a and n is 1), then a^(ϕ(n)) ≡ 1 (mod n). Here, ϕ(49) is 42 since 49 is a prime number. Therefore, 2^42 gives a remainder of 1 when divided by 49. We know that 50 > 42, and the next multiple is 84, which is too high, so the highest multiple of 42 under 50 is 42. So, we can rewrite 2^50 as 2^(42+8) = 2^42 * 2^8. As per Euler's theorem, 2^42 will give us 1, and so the equation simplifies to 2^8, which is 256. The remainder, when 256 is divided by 49, is 18.
What is the remainder when 7^200 is divided by 6?
The remainder is 1.
Explanation: 7^n leaves a remainder of 1 when divided by 6, for all natural numbers n. Therefore, 7^200 will also leave a remainder of 1 when divided by 6.
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If the divisor is 14 and the dividend is 214, what is the remainder?
The remainder is 2.
Explanation: When you divide 214 by 14, the quotient is 15, and the remainder is 2.
What is the remainder when 5^99 is divided by 13?
The remainder is 8.
Explanation: If you notice the pattern, the remainder of powers of 5 is divided by 13 cycles every four numbers: 5^1 (remainder 5), 5^2 (remainder 12), 5^3 (remainder 8), 5^4 (remainder 1). Thus, when you find the remainder of 5^n divided by 13, if n is divisible by 4 (like 5^96), the remainder is 1. Since 99 = 96 + 3, the remainder of 5^99 divided by 13 will be the same as the remainder of 5^3 divided by 13, which is 8.
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