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
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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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.
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
Ans. No. For all the numbers between 0 and 1 the square root is greater than the number itself.
Ans. Here is some of the ways to get the desired result.
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Ans. We need to remember the basic formulae
Speed = distance / time
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.
On solving the both equations
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.
Ans. Answer is 218.
The difference between
20-2 = 18
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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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
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
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.
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.
7200/20=360 revolution for the 20-back wheel
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.
*The article might have information for the previous academic years, please refer the official website of the exam.