Keystone Homeschool: Problems

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Tutoring Service Helps Many Students in Solving Problems

Interesting Probability Problems (p Nitin)

Probability of an event: If there are n elementary events associated with a random experiments and m of them are favourableto an event A, then the probability of happening or occurence of A denoted by P(A) and is defined as the ratio m/n

Thus, P(A) = m/n

Let us use this formula and solve some interesting probability problems.

Pro 1:Two dice are thrown simultaneously. Find the probability of getting :

a)an even number as the sum

b)the sum of prime number

c)a total of at least 10

d)a doublet of even number

e) a multiple of 2 on one dice and a multiple of 3 on the other

f)same number on both dice

g)a multiple of 3 as the sum

Solution: When two dice are thrown together the sample space S associated with the random experiment is given by

S= { (1,1),(1,2),(1,3),(1,4),(1,5),(1,6),

(2,1),(2,2),(2,3),(2,4),(2,5),(2,6), ,

(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),

(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),

(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),

(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

Clearly total number of events is 36

a) Let A be the event " getting an even number as the sum" i.e 2,4,6,8,10,12as the sum. Then,

A= { (1,1),(1,3),(3,1), (2,2),(1,5),(5,1),(3,3),(2,4),(4,2),(3,5),(5,3),(4,4),(6,2)(2,6),(5,5),(6,4),(4,6),(6,6)}

Favourable number of elementary events = 18

So, required probability = 18/36 =1/2

b) Let A be the event " getting the sum as a prime number" i.e 2,3,5,7,11 as the sum.Then,

A ={(1,1),(1,2),(2,1),(1,4),(4,1),(2,3),(3,2),(1,6),(6,1),(2,5),(5,2),(3,4),(4,3),(6,5),(5,6)}

Favourable number of elementary events = 15

So, required probability = 15/36 = 5/12

c)Let A be the event of "getting a total of at least 10" i.e 10,11,12.Then,

A={(6,4),(4,6),(5,5),(6,5),(5,6),(6,6)}

Favourable number of elementary events= 6

So, required probability=6/36=1/6

d)Let A be the event of getting a double of an even number.Then ,

A={(2,2),(4,4),(6,6)}

Favourable number of elementary events= 3

So, required probability=3/36=1/12

e) Let A be the event of "getting a multiple of 2 on one dice and a multiple of 3 on the other dice".Then,

A={(2,3),(2,6),((4,3),(4,6),(6,3),(6,6),(3,2),((3,4),(3,6),(6,2),(6,4)}

Favourable number of elementary events= 11

So, required probability=

f) Let A be the event of "getting the same number on both the dice."Then,

A = {(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)}

Favourable number of elementary events= 6

So, required probability=6/36 = 1/6

g) Let A be the event " getting a multiple of 3 as the sum" i.e 3,6,9,12 as the sum.Then,

A = {(1,2),(2,1),(1,5),(5,1),(2,4),(4,2),(3,3),(3,6),(6,3),(5,4),(4,5),(6,6)}

Favourable number of elementary events= 12

So, required probability=12/36 = 1/3

Some interesting probability problems:

Pro 2: Find the probability that a leap year will contain 53 sundays.

Sol: In a leap year there are 366 days.

366 days = 52 weeks and 2 days

Thus, a leap year has always 52 Sundays.The remaining 2 days can be:

(i) Sunday and Monday, (ii) Monday and Tuesday,

(iii)Tuesday and Wednesday , (iv) Wednesday and Thursday,

(v)Thursday and Friday,(vi)Friday and Saturday,

(vii)Saturday and Sunday.

If S is the sample space associated with this problem , then S consists of the above seven points.

The total number of elementary events =7

Let A be the event that a leap year has 53 Sundays.In order that a leap year, selected at random, should have 53 Sundays, one of the "over" days must be a Sunday.This is can be in any one of the following two ways

(i) Sunday and Monday or (ii)Saturday and Sunday

Favourable number of elementary events =2

Hence, required probability = 2/7

Some more interesting probability problems:

Pro 3: The number lock of a suitcase has 4 wheels, each labelled with ten digits i.e from 0 to 9.The lock opens with a sequence of four digits with no repeats.What is the probability of a person getting the right sequence to open the suitcase.

Sol: There are 10C4 x 4! = 5040 sequence of 4 distinct digits out of which there is only one sequence in which the lock opens

Therefore, required probability = 1/5040

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Interesting Probability Problems (p Nitin)

Thus, P(A) = m/n

Let us use this formula and solve some interesting probability problems.

Pro 1:Two dice are thrown simultaneously. Find the probability of getting :

a)an even number as the sum

b)the sum of prime number

c)a total of at least 10

d)a doublet of even number

e) a multiple of 2 on one dice and a multiple of 3 on the other

f)same number on both dice

g)a multiple of 3 as the sum

Solution: When two dice are thrown together the sample space S associated with the random experiment is given by

S= { (1,1),(1,2),(1,3),(1,4),(1,5),(1,6),

(2,1),(2,2),(2,3),(2,4),(2,5),(2,6), ,

(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),

(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),

(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),

(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

Clearly total number of events is 36

a) Let A be the event " getting an even number as the sum" i.e 2,4,6,8,10,12as the sum. Then,

A= { (1,1),(1,3),(3,1), (2,2),(1,5),(5,1),(3,3),(2,4),(4,2),(3,5),(5,3),(4,4),(6,2)(2,6),(5,5),(6,4),(4,6),(6,6)}

Favourable number of elementary events = 18

So, required probability = 18/36 =1/2

b) Let A be the event " getting the sum as a prime number" i.e 2,3,5,7,11 as the sum.Then,

A ={(1,1),(1,2),(2,1),(1,4),(4,1),(2,3),(3,2),(1,6),(6,1),(2,5),(5,2),(3,4),(4,3),(6,5),(5,6)}

Favourable number of elementary events = 15

So, required probability = 15/36 = 5/12

c)Let A be the event of "getting a total of at least 10" i.e 10,11,12.Then,

A={(6,4),(4,6),(5,5),(6,5),(5,6),(6,6)}

Favourable number of elementary events= 6

So, required probability=6/36=1/6

d)Let A be the event of getting a double of an even number.Then ,

A={(2,2),(4,4),(6,6)}

Favourable number of elementary events= 3

So, required probability=3/36=1/12

e) Let A be the event of "getting a multiple of 2 on one dice and a multiple of 3 on the other dice".Then,

A={(2,3),(2,6),((4,3),(4,6),(6,3),(6,6),(3,2),((3,4),(3,6),(6,2),(6,4)}

Favourable number of elementary events= 11

So, required probability=

f) Let A be the event of "getting the same number on both the dice."Then,

A = {(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)}

Favourable number of elementary events= 6

So, required probability=6/36 = 1/6

g) Let A be the event " getting a multiple of 3 as the sum" i.e 3,6,9,12 as the sum.Then,

A = {(1,2),(2,1),(1,5),(5,1),(2,4),(4,2),(3,3),(3,6),(6,3),(5,4),(4,5),(6,6)}

Favourable number of elementary events= 12

So, required probability=12/36 = 1/3

Some interesting probability problems:

Pro 2: Find the probability that a leap year will contain 53 sundays.

Sol: In a leap year there are 366 days.

366 days = 52 weeks and 2 days

Thus, a leap year has always 52 Sundays.The remaining 2 days can be:

(i) Sunday and Monday, (ii) Monday and Tuesday,

(iii)Tuesday and Wednesday , (iv) Wednesday and Thursday,

(v)Thursday and Friday,(vi)Friday and Saturday,

(vii)Saturday and Sunday.

If S is the sample space associated with this problem , then S consists of the above seven points.

The total number of elementary events =7

(i) Sunday and Monday or (ii)Saturday and Sunday

Favourable number of elementary events =2

Hence, required probability = 2/7

Some more interesting probability problems:

Sol: There are 10C4 x 4! = 5040 sequence of 4 distinct digits out of which there is only one sequence in which the lock opens

Therefore, required probability = 1/5040

Processing ...

avogadro's law problems with solutions (dave hook)

Relationships between properties of gases can be measured by changing one property while holding the others constant. Avogadro's Law relates the properties of gases to the amount of gas present. Avogadro's law states that the volume of a gas at a given temperature and pressure is directly proportional to the amount of gas, in moles. Thus, V is proportional to n, the number of moles of the gas, AT CONSTANT TEMPERATURE and PRESSURE.

Avogadro said that quantity and volume are directly related:

V 'alpha' n 'rArr' V/n is a constant

Or for the same sample of gas at constant temperature and pressure,

V1/ n1 = V2 / n2

Let us take an example of avogadro's law problems with solutions:

If we have 8 molecules of H2 and 4 molecules of O2, we get 12 molecules and a combined volume. If we then react the mixture, we end up in 8 molecules of gaseous water molecules, which occupy the same volume as the 8 molecules of any other gas.

8 H2 (g) + 4 O2 (g) ? 8 H2O (g)

Avogadro Law Concept Based Questions: avogadro's law problems with solutions

1. Which sample represents the smallest number of moles?
a) 1L H2 at STP
b) 1L Argon at STP
c) 1L of H2 at 27 ? C and 760mm of Hg

Solution:

Let us consider each situation.
At STP, 22.4 L volume will have 1 mole of the gas.
So, in a) 1L of H2, there will be (1L) (1 mole) / 22.4L = 0.0446 moles
˜ 0.045 moles

Similarly in b) 22.4 L of Ar would correspond to 1 mole.

So, 1L would be = (1L) (1 mole) /22.4L = 0.045 moles

In c) we should use the ideal gas equation: PV = nRT
P = 760 mm of Hg V = 1 L T = 27 ? C = (27 +273) K
= 1 atm = 300 K

Therefore, n = PV/ RT
R is the gas constant that has a value of 0.0821 atm L mol -1 K-1
n = [(1 atm) (1 L)] / [(0.0821 atm L mol -1 K-1) (300 K)]

= 0.0406 moles
So the correct option is c) 1L H2 at 27 ? C and 760 mm of Hg.

2. Which of the samples would have the same number of particles?

a) 1L He at STP and 1L O2 at STP
b) 2L He at STP and 1L of He at STP
c) 1L He at 27 ? C and 760 mm of Hg; and 2L He at STP

Solution:

a) In 22.4 L of gas, there will be 1 mole of the gas

So, in 1L of He there will be (1L) (1 mole) /22.4L = 0.0446 moles
˜ 0.045 moles

In 1L of O2, the number of moles will be the same, 0.045 moles.

Now, we know that in every mole, there will be Avogadro number of particles = 6.023 x 10 ^23 particles.

So in 0.045 moles, there will be
(0.045 moles) x (6.023 x 10 ^23 particles)/ 1 mole
= 2.7 x 10 22 number of particles

So, this is the required answer: a) 1L He at STP and 1L O2 at STP

Application based questions on Avogadro's Law: avogadro's law problems with solutions

1. Suppose we have 12.2 L sample containing 0.5 moles oxygen gas at a pressure of 1 atm and temperature of 25 ? C. If all this oxygen were converted into ozone, at the same temperature and pressure, what would be the volume of ozone?

Solution:
The equation for this reaction can be given as follows-

3O2 ? 2 O3
(Oxygen) (Ozone)

In this balanced equation 3 moles of oxygen produce 2 moles of ozone.
In the question given, 0.5 moles of oxygen are reacting. So the number of moles of ozone that it would result

= (0.5 ? 2) / 3
= 0.33 moles of ozone

So, V1 = 12.2 L n1 = 0.5
V2 =? n2 = 0.33

So from Avogadro's Law equation, V1/n1 = V2 / n2,
V2 = (V1 n2) / n1
V2 = (12.2 L) (0.33 moles) / (0.5 moles)
= 8.133 L

2. 11.2 L sample of gas is determined to contain 0.5 moles of nitrogen. At the same temperature and pressure, how many moles of gas would there be in a 20 L sample?
Solution:

V1 = 11.2 L n1 = 0.5 moles
V2 = 20 L n2 =?

From the Avogadro's law,
n2 = (V2 n1)/ (V1)
n2 = (20 L) (0.5 moles) / (11.2 L)
= 0.89 moles

3. Consider the following chemical equation:

2NO2 (g) ? N2O4 (g)

If 25 mL of NO2 gas is completely converted to N2O4 gas, under the same conditions, what volume will the N2O4 occupy?

Solution:
Here, from the equation given, 2 moles of NO2 (g) forms 1 mole of N2O4 (g).

So, V1 = 25 mL n1 = 2 mole
V2 =? n2 = 1mole

V2 = (V1 n2) / n1

V2 = (25 mL) (1 mole) / (2 moles)
= 12.5 mL

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Easy Division Problems (Omkar Nayak)

Introduction to easy division problems:

Let us see some content of easy division problems. Division is the operations in mathematics. This is one of the processes in arithmetic operation. The division is very easy for separate the group of events. The division is the opposite process of multiplication. The division problems are may be between integers or fractions. The integers are very easy to divide.

Definition:

Division is the process of dividing the objects together collection of larger objects. This process is used to easily separate from large number of objects. The division is very sufficient when compared to the subtraction. The solution is may be integer or floating. The division is signified as (/).

Steps for easy division:

Step 1: First change way of the given problem like should be written as division way.

Step 2: Next, we need to look at the first number of the given dividing number.

Step 3: Next, we want to see how many times that number divided by the given number.

Step 4: From that last step, we need to subtract the answer from the number. Repeat the step up to getting remainder as zero.

Examples:

Let us see some easy division problems.

Problem 1:

Find the easy value of the problems by division where the values are 34 and 6.

Solution:

The given values are 34 and 6.

Always dividend value should write outside and divisor value should write inside. The subtraction can be taken from left to right.

First write the problem like given below. Like the division format.
Look the first number, it is 3. So, it cannot be divided by the number of 6. So, take another digit with that.
Now the total number is 34. Put calculation how many time will multiply by 6 for getting 34.
But, the number 34 cannot get from the 6th table. But we can get below number of 30 at 5 times.

So subtract from 34. getting remainder as 4 and 5 as quotation. 4 cannot be divided by 6. So 4 is remainder.
____
6 ) 34 ( 5
30
_________
4
________

4 is remainder and 5 is quotation.

Problem 2:

Find the easy value of the problems by division where the values are 25 and 5.

Solution:

The given values are 25 and 5.

Always dividend value should write outside and divisor value should write inside. The subtraction can be taken from left to right.

First write the given problem like division format.
Look the first number It is 2. It cannot divide by 5.
So, take another number 25. The 25 can be divided by 5.
Put calculation how many times want to multiple for getting 25. It is 5 times in mathmetics multiplication table.
Subtract both answers, we are getting 0 as remainder and 5 is quotation.
___
5) 25 ( 5
25
________
0
_________

0 is remainder and 5 is quotation.

Problem 3:

Find the easy value of problems by division where the values are 12 and 6.

Solution:

The given values are 12 and 6.

Always dividend value should write outside and divisor value should write inside. The subtraction can be taken from left to right.

First write the problem as division format.
Look the first. It is 1. It cannot be divided by 6.
So take next number too. Now it is 12. It can be divided by the number of 6.
Put multiplication for getting 12. It is in 2 times from 6th table.
Now, subtract 12 from given number. Remainder 0 and the quotation is 2.
___
6)12 ( 2
12
______
0
_________

0 is remainder and 2 is quotation.

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