Maths Teachers in Nicholson Lines, Delhi

Since, 5.4 cm = 405 units, 

1 cm = 405/5.4 units = 75 units.

Ans. k = 75 uniis

Answer

The denominator of the given fraction is 6 which has factors 2&3. We know that any rational number whose denominator is not in the form of" 2^n*5^m "will have a recurring decimal expansion.As the denominator of the given fracrion 5/6 contains a factor "3" which is different from "2" and "5" ,the decimal expansion  is recurring and non terminating.

Derivative of f(×)=x^x(1+logx) For max or min derivative of f(×)=0 then 1+logx=0 X=1/e and second derivative of f is positive at 1/e f(×) is min at x=1/e Min value is f(1/e)=1/e^1/e

Answer:

Two things are  to be noted while solving the problem.

1. As  Virat has to keep the strike through out the over, he can score only even runs like 2,4,6 or no runs with each ball.

2. As he has to play all the 6 balls of the over ,  some runs have  to be scored from the last ball. He can score the 12 runs from 6 balls in various sequences and the no. of ways of selecting the balls depends accordingly.

1. Only 2 balls in sequence of 6+6 =12:

  One ball is already pre selected as last ball. The remaining 1 ball can be selected out of 5 balls in 5 ways.( That is 12 runs will be made from ball 1&6  or 2&6 or 3&6 or 4&6 or 5&6.).So no.of ways for this option =  5

2. Only 3 balls in sequence of 4+4+4= 12.:

   Again as one ball is reserved as last ball, we have to select remaining 2 balls   out of 5 balls . As per laws  of  permutations, the selection of " r" objects out   of "n" objects can be done in nPr ways= n ! / (n-r)!. If out of the " n" objects , " m" objects are identical then the number of ways of selecting "r" objects is  nPr/ m! = n!/ { ( n-r)! * m!}.             As Virat has to score  identical runs of 4 each from these 2 balls they both constitute 2 identical objects.                         So number of  ways of selecting 2 identical  balls out 5 balls is  5P2/ 2!=  5!/{ ( 5-2)! * 2!} =  5!/( 3!*2!). = 10 ways.

3. Only 3 balls in combination of 6+4+2 = 12

    Here the last can be hit either for 6, 4, 2 runs. So it can be selected in 3 ways. After that we can select 2 balls out of  5 balls in 5P2 ways. As both the balls will be hit for different amount of runs they are not identical. Therefore the number of ways of selection for the 2 balls is 5P2= 5! /3!= 5*4= 20'ways. As these 20 ways are possible for each of the 3 ways of    selecting the last ball , the total number of selection for this option is 3*20 = 60 ways.

4. Only 4 balls in combination of 6+2+2+2  =12:

    Here the last ball can be hit either for  2 runs or 6 runs.

    Case1: The last ball is hit for 6 runs. Then we have to select 3 balls out of 5 balls in which all the 3 balls will be hit for 2 runs each and hence all 3 are identical. Therefore number of ways of selecting the 3 balls is 5P3 / 3! = 5*4*3/ 6= 10

    Case 2: The last ball is hit for 2 runs. Then we have to select 3 balls out of 5 balls in which 2 balls will be hit for 2 runs each and hence they 2 are identical. Therefore the number of ways of doing this is     5P3/2!  = 5*4*3/2  = 30.         So total no. of ways for this option = 10+30  =40

5. Only 4 number of balls in combination of 4+4+2+2='12.                   

 Here last ball can be selected in 2 ways.It  can be hit for 4 or 2 runs.     

 Case1: Last ball will be hit for 2 runs.From the remaining 5 balls we can select  3 balls out of which 2 balls would be hit for 4 runs each and hence identical..Therefore the number of ways of doing this is 5 P 3/2!=5*4**3/2. = 30

Case 2:  Last ball will be hit for 4 runs.

From the remaining 5 balls we can select 3 balls out of which 2 balls would be hit for 2 runs each and hence identical. This situation is same case 1 and so the number of ways of doing this is 5P3/2!= 30. So total number of ways of selection for this option would be 30+ 30  = 60 ways.

6. Only 5 balls in combination of 4+2+2+2+2  = 12.  

 Here last ball can be hit either for 4 or 2 runs.

Case 1: Last ball will be hit for 4 runs.

From the remaining 5 balls we can select 4 balls in 5 P4 ways and as all the 4 balls will be hit for 2 runs each and so there will be 4 identical balls. Therefore number of ways of selecting 4 balls = 5P4/ 4! = 5!/ ( 1!*4!)  = 5 ways.

Case2:  Last ball will be hit for 2 runs.

From the remaining 5 balls we can select 4 balls in 5P4 ways.But in this 3 balls will be hit for 2 runs each and so there will be 3 identical balls. Therefore number of ways of selecting 4 balls  = 5P4/ 3! = 5!/ ( 1!*3!)  = 20 ways. So total number of ways of selection for this option is  5+ 20  = 25 ways

7. All the 6 balls in combination of 2+2+2+2+2+2 =12.

  In this option all the 6 balls will be selected at a time and so the number of ways for this option =1

So total number of ways of hitting exactly 12 runs= 5+10+60+40+60+ 25+ 1 = 201

The ratio of milk to water is 4:1 Initially 60 litres of milk Take away 6 litres =54 litres Take away 6 litres = 48 litres Now 6+6 litres of water is added=12 Therefore 48:12 4:1

Left principal = x Then Amount= 2x So interest = 2 x -x =x We know that rate = int ×100/principle × time Rate= x *100/x * 5 Rate = 100/5 Rate =20%per annum

1. (7*3) - (4-2) = 21 - 2 =19

2. (9*3) - (7-4) = 27-3 = 24

3. (5*4) - (9-1) = 20-8=12

Let the man's age be x ans son's age be y

Man's age 18 years ago = (x-18)

Son's age 18 years ago = (y-18)

x= 2y -----------------(1)

(x-18) = 3(y-18) ----------(2)

Substituting value of x from (1) in (2)

2y-18 = 3y - 54

y = 54-18 = 36

y = 36

x= 72

Man' age = 72 

Son's age = 36

Sum of age = 72+36 = 108

We know, W = m.g 

So here, the weight of the student on earth = 78x9.8 N = 764.4 N     (Ans.)

Answer:;

A counter is an instance where a  rule/theorem/law, which is generally valid otherwise in most of the cases, is not valid in the particular case. Counter of any rule is an example of an instance, where the rule is not applicable or an exception. to the rule.To illustrate ,let us consider the rule for finding Leap Year. 

We know that there are 366 days in a leap year, that is the month of February has 29 days in a leap year. The general rule is, if the year is divisible by 4,then it is a leap year. That is if the numerical value of any year is a multiple of 4,then the year is a leap year.Thus,1980,1992,2012 ,2020 etc. are leap years. But this rule is not valid ,if the year is a century year, that is the last 2 digits of the year are 0,0.Thus the century years 1800,1900,2100,2200,2300  etc. are not leap years ,though they are multiples of 4. These are counter examples for the general rule of leap year.

Again this rule is not valid if the  century years are multiples of 400 ,because century years which are multiples of 400 are leap years .Thus century years 1600,2000,2400 etc. are century leap years having 366 days. These  are again counter examples for the general rule of  leap years for century years.