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# Hard Math Problems

Online math help isn’t reserved for students who find the subject tough. If you are the kind of person who relishes the thought of putting your numerical skills to test, online math solvers can meet that need. Log onto math websites to find really hard math problems which will help you improve your problem solving skills and get ahead in the subject. Math help websites have all levels of math problems starting from the simple ones to extremely hard math problems. Starting from middle school math, students will find hard 5th grade, 6th grade and 7th grade math problems. The problems are easily available online and are accessible any time students may need help.

## Hard Math Problems with Answers

Online math solvers are a great help to students wondering if they’ll eve get the answer to super hard math problems. Students are given a step-by-step break down of solutions and explanations to clear any doubts they may have. Hard math problems for 8th graders, 9th graders and 10th graders cover all types of questions and range from questions that are slightly difficult to very hard math problems. As any math veteran knows, math is all about the practice so getting started early definitely gives students the edge they need to get ahead in the subject.

### Solved Examples

Question 1: Car A began a journey from a point at 9 am, traveling at 30 mph. At 10 am car B started traveling from the same point at 40 mph in the same direction as car A. At what time will car B pass car A?
Solution:
Let after t hours the distances D1 traveled by car A

=> D1 = 30 t

Car B starts at 10 am and will therefore have spent one hour less than car A when it passes it.

After (t - 1) hours, distance D2 traveled by car B

=> D2 = 40 (t - 1)

When car B passes car A, they are at the same distance from the starting point and therefore D1 = D2

=> 30 t = 40 (t - 1)

Solve the equation for t,

=> 30 t = 40t - 40

=> 10 t = 40

=> t = 4

=> Car B passes car A at = 9 + 4 = 13 pm.

Question 2: Four times the first of three consecutive even integers is six more than the product of two and the third integer. Find the integers.

Solution:
First integer = x

Second integer = x + 2

Third integer = x + 4

Since four times the first integer equals six more than the product of two and the third integer.

=> 4x = 6 + 2(x + 4)

=> 4x = 6 + 2x + 8

=> 2x = 14

=> x = 7.

Hence,

First integer = x = 7

Second integer = x + 2 = 7 + 2 = 9

Third integer = x + 4 = 7 + 4 = 11.

## Hard Math Problems for College

College students who want to gain an edge on their math courses will find hard math problems for college students a useful and convenient way to prepare for tests or exams. The hard college math problems also have answers and are sometimes fully solved so that students can correct any mistakes and learn the right methods. The online math helpers also have the answers and in some cases, worked out solutions so that students can correct any mistakes and make sure they are on the right track.

### Solved Examples

Question 1: Find the general solution for differential equation

(D4 - 5D3 + 5D2 + 5D - 6)y = 0

Solution:
Given differential equation, (D4 - 5D3 + 5D2 + 5D - 6)y = 0

=> For general solution of equation,

Solve D4 - 5D3 + 5D2 + 5D - 6 = 0

=> D4 - 5D3 + 6D2 - D2 + 5D - 6 = 0

=> D2 (D2 - 5D + 6) - (D2 - 5D + 6) = 0

=> (D2 - 5D + 6)(D2 - 1) = 0           ................................(1)

Now

D- 1 = (D - 1)(D + 1)  and

Factors of D2 - 5D + 6

D2 - 5D + 6 = D2 - 2D - 3D + 6

= D(D - 2) - 3(D - 2)

= (D - 3)(D - 2)

Therefore, equation (1) implies

(D2 - 5D + 6)(D2 - 1) = (D - 3)(D - 2)(D - 1)(D + 1) = 0

=> D = 3, 2, 1, -1 or D = -1, 1,, 2, 3

=> General solution of differential equation is,

=> y = C1 e-x + C2 ex + C3 e2x + C4 e3x .

Question 2: Find the 20th term and the sum of the first 14 terms of this progression, 2, 5, 8, 11,.........?
Solution:
Given progression, 2, 5, 8, 11,.........  is a arithmetic progression

Here First term (a) = 2

and Common difference (d) = 5 - 2 = 3

We know, nth term can be represented as, an = a + (n - 1)d

and sum of n terms = $\frac{n}{2}${2a + (n - 1)d}

To find 20th term of the series

=> a20 = a + 19 d

= 2 + 19 * 3

= 2 + 57

= 59

=> 20th term = 59

Sum of first 14 terms = $\frac{14}{2}${2 * 2 + (14 - 1) * 3}

= 7{4 + 39}

= 7 * 43

= 301

=> Sum of first 14 terms = 301.