Entrance Exam for College Math

Introduction to entrance exam for college math:

Entrance exam for college evaluates high school students, general educational improvement and their ability to complete college-level work. SAT exam is one of the entrance exams for college math. It is used to find out skills in math, reading comprehension and vocabulary of the students planning to attend college. Now, we are going to see some of the problems on entrance exam for college math.

Entrance Exam for College Math Solved Problems:

Example problem 1:

Solve the quadratic equation by factoring method: x2 - 143x + 142 = 0.

a) 1, 142

b) 1, 132

c) 1, 122

d) 1, 112

Solution:

The given quadratic equation is x^2 - 143x + 142 = 0

Here a = coefficient of x2 = 1

b = coefficient of x = -143

c = constant term = 142

We find a × c = 1 × 142 = 142 = -1 * -142, a + c = (-1) + (-142) = -143 = b.

By splitting the middle term, we get

x2 - 143x + 142 = 0

x2 - 1x - 142x + 142 = 0

x(x - 1) - 142(x - 1) = 0

(x - 1) (x - 142) = 0

x = 1, x = 142

So, the option a is the correct answer.

So, the answer is x = 1, 142. Having problem with word problems in algebra keep reading my upcoming posts, i will try to help you.

Additional Solved Problems-entrance Exam for College Math:

Example problem 2:

Solve the following simultaneous equations using substitution method:

9x – y = 18 ----------Equation (1)

1x + y = 22--------------Equation (2)

a) (2, 25)

b) (3, 27)

c) (5, 27)

d) (4, 18)

Solution:

Step 1: Let us consider the Equation (1)

9x – y = 18

Subtract 9x on both sides of the equation

9x – y –9x = 18 – 9x

-y = -9x + 18

y = 9x - 18---------Equation (3)

Step 2: Substitute the value of y in Equation (2). We get

1x + y = 22

1x + (9x – 18) = 22

10x – 18 = 22

Add 18 on both sides of the equation

10x – 18 + 18 = 22 + 18

10x = 40

Divide by 10 on both sides of the equation

`(10x) / 10 = 40 / 10`

x = 4

Step 3: Substituting this value of x in Equation (3), we get

y = 9x – 18

y = 36 – 18

y = 18

So, the option d is the correct answer.

So, the answer is (4, 18).

Thoughts on Distance Learning

Introduction to thoughts on distance learning:

In this article, we shall discuss about thoughts on distance learning. In mathematics, distance is defined as an object moves from one particular point to another point which is calculated by speed and time. That is, when we multiply the time and speed we will get the distance. Please express your views of this topic Define Permutations by commenting on blog.

Formula for the calculating distance is

Distance = speed `xx` time

Now we shall solve some example problems regarding thoughts on distance learning.

Examples to Thoughts on Distance Learning:

Example 1:

A train is driven at average speed of 3 kilometers per hour to the station 6 hours. If the train is droved at the average speed of 2, how many hour would need to reach the station?

Solution:

3 km/h needs 6 hours

2 km/h needs x hours

This can be written as,

3 = 6 and

2 = x

Also we can written as

3 `xx` 6 = 2x

18 = 2x Also we can written as

2x  = 18 now we have to divide both sided by 2

`(2x)/2`  = `18/2`

X = 9

therefore, if the train is droved at 2 kilometers per hour, it would need 9 hours
Example 2:

Bus and car are leaving from the same place in opposite direction. Bus goes at 3mph and the car goes at 2 mph. how many hours will they need to 22 miles apart?
Solution:

Distance = speed time

Speed of  bus = 3mph

Time of bus = t

Therefore, distance of the bust = 3t

Speed of  car = 2mph

Time of bus = t

Therefore, distance of the bust = 2t

Therefore,

3t + 2t  = 22 miles

5t  =  22   now we have to divide both sides by 5. so we get

`(5t)/5` = `22/5`

t =4.4 hours

Bus and car will reach 22 miles apart in 4.4 hours


Is this topic Formula for Calculating Interest hard for you? Watch out for my coming posts.

Practices Problems to Thoughts on Distance Learning :

Problem 1:

A train is driven at average speed of 6 kilometers per hour to the station 6 hours. If the train is driven at the average speed of 4, how many hour would need to reach the station?

Answer: 9 hours

Problem 2:

Bus and car are leaving from the same place in opposite direction. Bus goes at 10 mph and the car runs at 6mph. how many hours will they need to 39 miles apart?

Answer: 2.4 hours

Exact Distribution

Introduction to exact distribution:

The exact distribution refers the probability distribution the probability theory. The probability distribution is used to determine the number of possibility for the occurrence of an event. The most commonly used probability distributions are the binomial distribution, geometric distribution, normal distribution and the gamma distribution. These above mentioned distributions are included in the discrete and continuous probability distribution. The major type of the probability distribution is the discrete probability distribution and the continuous probability distribution. This article has the study about exact distribution.

Types of Exact Distribution:

The major types of the probability distribution are

Discrete probability distribution
Continuous probability distribution
Discrete probability distribution:

The probability for a countable number of occurrences for the event is calculated in the discrete probability distribution.
Continuous probability distribution:

The probability values in this are the continuous ranged value it is calculated in the continuous probability distribution.

Examples for Exact Distribution:

Example 1 to exact distribution:

If X is normally distributed the mean value is 1 and its standard deviation is 2. Determine the value of P (0 ≤ X ≤ 7).

Solution:

The given mu value is 1 and the standard deviation is 2.

Z = `(X- mu)/ sigma`

When X = 0, Z = `(0- 1)/ 2`

= -`1/2`

= -0.5

When X = 7, Z = `(7- 1)/ 2`

= `6/2`

= 3

Therefore,

P (0 ≤ X ≤ 7) = P (-0.5 < Z < 3)

P (0 ≤ X ≤ 7) = P (0 < Z < 0.5) + P (0 < Z < 3) (due to symmetry property)

P (0 ≤ X ≤ 7) = (0. 6915- 0.5) + (0.9987 - 0.5)

P (0 ≤ X ≤ 7) = 0.1915 + 0.4987

P (0 ≤ X ≤ 7) = 0.4987

The value for P (0 ≤ X ≤ 7) is 0.4987.

Between, if you have problem on these topics, please browse expert math related websites for more help on cbse 10th sample papers.

Example 2 to exact distribution:

The probability for destroying the target in only one time is 0.37. Compute the probability that it would be destroyed on the third attempt itself.

Solution:

The probability of destroying the target in one trial is p = 0.37.

The value of the q is calculated by q = 1-p

q = 1- 0.37

q = 0.63

By the geometric distribution, the probability for the success is calculated by using the formula

P(X =x) = q x p, the value of x is 0, 1, 2. . .

The target is destroyed at the third attempt, so x = 3.

P(X = 3) = (0. 63) 3 (0.37)

P(X = 3) = (0.25) (0.37)

P(X = 3) = 0.0925

The probability for destroying the target at the third trial is 0.0925.

Easy Way to Learn Statistics

Introduction to Easy Way to Learn Statistics

Statistics is the proper science of creating successful use of mathematical relating to groups of individuals or experiments. It deals with all features of this including not only the collection, analysis and interpretation of such data, but also the planning of the collection of data, in terms of the design of surveys and experiments. Now we will learn the statistics in easy way.

Examples for Easy Way to Learn Statistics

Example 1

What the mean, median,mode and range of the following group of numbers?

11,12,15,17,19.

Solution

The given numbers are 11,12,15,17,19.

Mean

We can find a mean in easy way. Mean is the average of the given number. So find the sum of the given numbers.

Sum of the given numbers are = 11+12+15+17+19

Now divided by 5 (Because 5 is the total given numbers) =74/5

=14.8.

Median

A middle value of the given number series is the median.

The number series is 11,12,15,17,19.

Here the center value is 15.

Therefore 15 is the median.

Mode

It is also very easy way to find. Mode is a duplicate value of the given number series. Here no duplicate value.

Therefore mode is empty or null.

Range

It can find in easy way. Range is the difference between maximum value and the minimum value of the given number series.

Range=19-11

=8.

Example 2

What is the mean, median of the following group in statistics?

14,16,18,20, 22.

Solution

The given numbers are 14,16,18,20 and 22.

Mean

We learn.mean is the average of the given number. We find the sum of the given numbers.

Sum of the given numbers are = 14+16+18+20+22

= 90.

Now divided by 5 (Because 5 is the total given numbers) = 90/5

= 18.

Median

We learn,a middle value of the given number series is the median.

The number series is 14,16,18,20, 22.

The middle value of the above series is 18.

Therefore 18 is the median.

These are the examples and find the solution in easy way.

Student Directed Learning

Introduction for student directed learning:

In this article, we will discuss about the student directed learning. In online, we will get direct help with clear explanation for mathematics subjects. The students who have doubts in the topics, so they access the online for help in math. Its also useful to workout the homework problems and doubt clarification. The online topic is sequences. It has two types

1. Arithmetic sequence

2. Geometric sequence.

The student directed learning formulas and example problems are given below.

Student Directed Learning Sequences for Formulas:

Definitions:

Arithmetic sequence means that, the sequence of a numbers such that the difference between two consecutive members of the sequence is a constant. Geometric sequence means that, the sequence of a numbers such that the ratio between two consecutive members of the sequence is a constant.

Formula for arithmetic sequence:

nth term of the sequence : `a_n` = `a_1` + (n - 1)d

Series of the sequence: `S_n` = `(n(a_1 + a_n))/2 `

Formula for geometric sequence:

nth term of the sequence:` a_n` = a1 * rn-1

Series of the sequence: `S_n` = `(a_1(1-r^n))/(1 - r)`

Student Directed Learning Sequences for Example Problems:

Example problem 1:

Find the 13th term of the given series 24, 27, 30, 33, 36,...

Solution:

First term of the series, a1 = 24

Difference of two consecutive terms, d = 27 - 24

n = 13

The formula to find the nth term of an arithmetic series, `a_n = a_1 + (n-1)d`

So, the 13th term of the series 24, 27, 30, 33, 36,... = 24 + (13 - 1) 2

= 24 + 12 * 2

= 24 + 24

After simplify this, we get

= 48

So, the 13th term of the sequence 24, 27, 30, 33, 36,... is 48.

Example problem 2:

Find out the 5th term of a geometric sequence if a1 = 90 and the common ratio (C.R) r = 2

Solution:

Use the formula `a_n = a_1 * r^(n-1)` that gives the nth term to find `a_5` as follows

`a_5 = a_1 * r^(5-1)`

= 90 * (2)4

= 90 * 16

After simplify this, we get

= 1440

The 5th term of a geometric sequence is 1440

The above examples are helpful to study of Student directed learning sequences.

Simple Solutions Math Answers

Introduction for math:

Mathematics is the study of quantity, structure, space, and change. Mathematicians seek out patterns, formulate new conjectures, and establish truth by rigorous deduction from appropriately chosen axioms and definitions. There is debate over whether mathematical objects such as numbers and points exist naturally or are human creations. Albert Einstein, on the other hand, stated that "as far as the laws of mathematics refer to reality, they are not certain; and as far as they are certain, they do not refer to reality.                                                                                                                                               source Wikipedia

Problems for Simple Solutions Math Answers

Problem 1:

Find the simple solutions.

3x-3=3

Solution:

the given function is 3x-3=3

add 3 on both sides of this equation.

3x+3-3=3+3

We get

3x=6

Divide by 3 on both sides of this equation

`(3/3)x` =`6/3`

We get

x=2

answers of this simple solution is x=2

Problem 2:

Find the simple solutions

5x-3=7

Solution:

the given function is 5x-3=7

add 3 on both sides of this equation.

5x+3-3=7+3

We get

5x=10

Divide by 5 on both sides of this equation

`(5/5)x` =`10/5`

We get

x=`10/5`

answers of this simple solution is x=2

Problem 3:

Find the simple solutions

2x-3=3

Solution:

the given function is 2x-3=3

add 4 on both sides of this equation.

2x+3-3=3+3

We get

2x=6

Divide by 2 on both sides of this equation

`(2/2)x` =`6/2`

We get

x=3

answers of this simple solution is x=3

Problem 4:

Find the simple solutions

2x+7=-21

Solution:

the given function is 2x+7=-21

subtract 7 on both sides of this equation.

2x+7-7=-21-7

We get

2x=28

Divide by 2 on both sides of this equation

`(2/2) x` =`28/2`

We get

x=14

answers of this simple solution is x=14

Some more Problems are Simple Solutions Math Answers:

Problem 5:

Find the area and perimeter of the rectangle,

Length of the rectangle=10 mm

Height of the rectangle = 12 mm

Solution:

Formula for area of the rectangle = length*height

Area of the rectangle = 12*10

=120 mm^2

Formula for perimeter of the rectangle =2*length + 2* height

=2*12+2*10

=24+20

=42 mm

answers

Area of the rectangle = 120mm^2

Perimeter of the rectangle = 42 mm

Problem 6:

Find the area and circumference of the triangle,

Radius of the circle =6 mm and `pi` = 3.14

Solution:

Formula for area of the circle = `pi` r2

Area of the circle = 3.14*62

=3.14*36

=113.04 mm^2

Formula for circumference of the circle =2 `pi` r

=2*3.14*6

=37.68 mm

answers

Area of the circle = 113.04 mm^2

Circumference of the circle = 37.68 mm

Very Small Numbers Learning

Introduction to very small numbers learning:

Let us study about very small numbers learning. The mathematics world is seemed to be fully composed of various kinds of numbers.Understanding List of Prime Numbers to 100 is always challenging for me but thanks to all math help websites to help me out.
The two basic categories that numbers will get split into are named as small numbers and small numbers.
The small numbers are the numbers that are seemed to be found to out by comparing each number with others. Some of the examples for very small numbers learning are given.
Very Small Numbers Learning:

Very small numbers learning – example 1:

Do solving the given number series (5 * 4), (4 - 2), (3 + 2.5), (2 * 0.2), (0.4 * 2.5), (46 - 10), `(0.8/0.2)` and find out the small number that are calculated in the series?

Solution:

The number series provided with the various operations for calculations is as follows: (5 * 4), (4 - 2), (3 + 2.5), (2 * 0.2), (0.4 * 2.5), (46 - 10), `(0.8/0.2)` .
The operations in the given number series are calculated as below:
(5 * 4), (4 - 2), (3 + 2.5), (2 * 0.2), (0.4 * 2.5), (46 - 10), `(0.8/0.2)`
(20), (2), (5.5), (0.4), (1), (36), (4)
20, 2, 5.5, 0.4, 1, 36, 4
To find the very small value in the given number series arrange the calculated values of the number series in the order smallest to the greatest values as follows:
0.4 < 1 < 2 < 4 < 5.5 < 20 < 36
Therefore in the given number series after calculation the very small number is found be as ‘0.4’.

Please express your views of this topic Number Place Value by commenting on blog.

Prepare for small numbers – example 2:

Do solving the given number series `(10/4)` , `(4/4)` , (3 * 6), (2 * 0.16), (-0.4 - 2.5), (5 + 10), (0.8 + (-0.8)) and find out the small number that are calculated in the series?

Solution:

The number series provided with the various operations for calculations is as follows: `(10/4)` , `(4/4)` , (3 * 6), (2 * 0.16), (-0.4 - 2.5), (5 + 10), (0.8 + (-0.8)).
The operations in the given number series are calculated as below:
`(10/4)` , `(4/4)` , (3 * 6), (2 * 0.16), (-0.4 - 2.5), (5 + 10), (0.8 + (-0.8))
(2.5), (1), (18), (0.32), (-2.9), (15), (0)
2.5, 1, 18, 0.32, -2.9, 15, 0
To find the very small value in the given number series arrange the calculated values of the number series in the order smallest to the greatest values as follows:
-2.9 < 0 < 0.32 < 1 < 2.5 < 18 < 15
Therefore in the given number series after calculation the very small number is found be as ‘-2.9’.


Prepare for small numbers – exercises:

Which one is very small even square number between 400 and 600? (Answer: 441)
Which is the very small cube number that lies between 1 and 600? (Answer: 8)
Mention the very small odd prime number from 600 and 800. (Answer: 601)

Solving Mental Math Problems

Solving Mental Math Problems

Mental math practice comprises arithmetical calculations using only the human brain, with no help from calculators, computers, or pen and paper. People use mental math when computing tools are not available, when it is faster than other means of calculation or in a competition context. Mental math often involves the use of specific techniques devised for specific types of problems. Source: Wikipedia.

Solving Mental Math Problems - Example Problems

Solving example problems

Example 1: Linda has $74 and Lisa has $99. How much did they has altogether?

Solution:

We have to add 74 and 99 to know the sum.

Step 1: Add the two ones' place digits: 9 + 4 = 13.

Step 2: Add the two tens' place digits: 9 + 7 = 16.

Step 3: The sum of the ones' place digits is a two-digit number so decrease the ones' place sum by 10: 13 - 10 = 3 and increase the tens' place sum by 1: 16 + 1 = 17.

Step 4: Combine the tens' and ones' place sums.

Answer is $173.

Example 2: A pair of shoes cost is $57. A pant cost is $81. Find the difference between the pair of shoes and pant.

Solution:

We have to subtract 57 from 81 to know the difference.

Step 1: Subtract the two tens' place digits: 8 - 5 = 3

Step 2: Bottom ones' digit is larger than the top ones' digit.

Decrease the answer for the tens' place by 1: 3 - 1 = 2 and increase the top ones' place value by 10: 1 + 10 = 11.

Step 3: Subtract the two ones' place values: 11 - 7 = 4

Step 4: Combine the tens' and ones' place value.

Answer is $24.

Example 3: A bucket cost is $73, oven cost is 5 times more than bucket cost. What is the oven cost?

Solution:

We have to multiply 73 by 5 to know the computer cost.

Step 1: Multiply the ones' digit by 5, 5 x 3 = 15

Step 2: Multiply the tens' digit by 5, 5 x 7 = 35

Step 3: Product of ones' place is two digits (15), so increase the product of tens' digit by 1 (15), 35 + 1 = 36

Step 4: Combine the ones' and twos' place value

Answer is $365.

Example 4: Lisa had 72 candies. She gave 6 candies to each her friend. How many friends she had?

Solution:

We have to divide 72 by 6 to know the answer.

Step 1: Divide the tens' digit by 6, 7/6 = 1 with remainder 1

Step 2: Multiply remainder 1 by 10, 1 x 10 = 10 and add it to ones' place, 10 + 2 = 12

Step 3: Divide 12 by 6, 12/6 = 2

Step 4: Combine the ones' and twos' place values

Answer is 12.

Solving Mental Math Problems - Practice Problems

Solving practice problems

Problem 1: Shan saved $73 and Sham saved $78. How much did they save altogether?

Problem 2: A shirt cost is $73, Pant cost is $99. Find the cost of difference of two items.

Problem 3: A wall clock costs 3 times as much as a watch. If the watch costs $51, what is the cost of wall clock?

Problem 4: Pond had 75 candies. He put the candies equally into 5 bags. How many candies were there in each bag?

Answer: 1) $151 2) $26 3) $153 4) 15