KVPY Interview: Maths Questions

Updated On -

Oct 8, 2020

Shivani Tiwari

Content Curator

Kishore Vaigyanik Protsahan Yojana (KVPY) is the ongoing initiative of Government of India, for school students, which aims to encourage students to pursue courses in basic science and develop a research aptitude in the field. The selection for this prestigious fellowship is considered after two rounds. First round is written test while the second round is interview. Get the latest KVPY news here

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Since it is aimed to enhance the scientific know-how, one can easily guess that Mathematics forms an important part of the selection procedure. Here we take a look at popular mathematics questions, and solutions that may be asked to a candidate appearing for KVPY.

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#### Ques. Explain what you mean by: Integers, rational numbers, composite numbers, prime numbers and their properties.

Ans. Integers: Numbers which can be written without any fractional denominator. There can be positive integer or negative integer. Some define integers as set of whole numbers and their opposites.

Rational Numbers: Any number which can be expressed in the form of fraction p/q where q can be any integer except zero.

Composite Numbers: The positive integers which can be composed by the multiplication of two smaller integers.

Prime Numbers: The natural numbers greater than which are either divisible by 1 or by itself.

#### Ques. What is the volume of the earth? How much water can be filled in the sphere of the size of the earth?

Ans. The volume of earth is nearly 1.097 X 1021 cubic meters. For the sphere of the size of earth the volume shall remain same. Hence it can 1.097 X 1021 cubic meters hold of water

#### Ques. Is the Square root of the number always less than the number itself?

Ans. No. For all the numbers between 0 and 1 the square root is greater than the number itself.

#### Ques. How you can get the sum of 31 from five 3's? It is necessary to use all 3's.

Ans. Here is some of the ways to get the desired result.

1. 33 + 3 + (3/3)
2. 33 + 3! –(3!/3)
3. (3! X3!) – 3! + (3/3)
If digit 3 is considered then it can also be done:
4. 33 – (3+3)/3
5. 33 – 3 +3/3

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#### Ques. A car covers a certain distance taking 7 hours for going one-way. Speed was increased by 12 kmph at the time of fro journey and the journey gets completed in 5 hours. What is total distance covered by the car?

Ans. We need to remember the basic formulae

Speed = distance / time

Assumptions:
let the distance be x
and let the speed of going one-way be y

Now, during forward journey, time is given as 7hrs .Therefore, y=x/7

Considering the return journey, since the distance travelled is same in both directions, we will consider the distance as x

Given is that now the speed is increased by 12kmph so the speed is y+12 and the time taken is 5 hrs.

Therefore, y+12=x/5
On solving the both equations
x=7y and
5y+60=x
we get x as 210 km i.e. distance of one-way.
total distance - distance in forward journey + distance in return journey
420 km is the answer for round trip distance.

#### Ques. What would be the next number in the sequence 2, 20, 74, 110?

The difference between
20-2 = 18

74-20=54

110-74=36.

See that the third difference is twice of second difference. Going by that the fourth difference must be twice of second difference, i.e. 108.

Therefore the result is 110+108=218.

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#### Ques. What number would next appear in the sequence 8, 20, 28, 44.

Ans. This series has many possible forms which makes the subsequent number differ in accordance with the method employed.

Couple of methods is:

Method A) This series can be generated by (8*n + 4) where n is the prime number. Thus the result will be

8*(2) +4=20

8*(3) +4=28

8*(5) +4=44

8*(7) +4=60

8*(11) +4=92

8*(13) +4=108

Method B) Diff between 1 & 3 no is 28-8=20; diff between 2 & 4 no is 44-20=24
That shall make

5th number to be 28+20=48
6th number shall be 44+24=68

#### Ques. Imagine that you are driving a car from Mysore to Bangalore, which lays 250 kilometers away from each other, at a speed of 75mph. Car no is KA-19 DB1955. Now on the basis of this information, find the date of birth of the driver.

Ans. This is a particularly witty question. Since you are the one driving the car as mentioned in the first sentence of question, all you have to do is to write your own birthdate and that will be the correct answer.

#### Ques. Perimeter of front wheel of a vehicle is 30 and that of rear wheel is 20. If front wheel revolves 240 times, how many revolutions will the rear wheel take?

Ans. Consider this simple logic, the distance covered by vehicle remains constant. If the front wheel of perimeter 30, revolves 240 times the back wheel of perimeter 20 must go the same distance and revolve more times since it is smaller.

30X240=7200 distance
7200/20=360 revolution for the 20-back wheel

#### Ques. Find the smallest four-digit number which is divisible from 1 to 10.

Ans. 2520. Multiply 9*8*7*5. It will give us the result. We ignore 2, 3, 4 and 6 because their factors are taken care of in other numbers that we multiply.

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*The article might have information for the previous academic years, please refer the official website of the exam.