KVPY 2017 NEWS

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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.

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**.

**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 10^{21} cubic meters. For the sphere of the size of earth the volume shall remain same. Hence it can 1.097 X 10^{21} 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.

- 3
^{3 }+ 3 + (3/3) - 3
^{3 }+ 3! –(3!/3) - (3! X3!) – 3! + (3/3)

If digit 3 is considered then it can also be done: - 33 – (3+3)/3
- 33 – 3 +3/3

**Check KVPY Result**

**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.

**Ans.** Answer is 218.

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.

**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

6^{th} 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.

30X240=7200 distance

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.

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