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## 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
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## 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 D

=> D

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 D

=> D

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

=> 30 t = 40 (t - 1)

Solve the equation for t,

=> 30 t = 40t - 40

=> 10 t = 40

=> t = 4

=>

_{1}traveled by car A=> D

_{1}= 30 tCar 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 D

_{2}traveled by car B=> D

_{2}= 40 (t - 1)When car B passes car A, they are at the same distance from the starting point and therefore

**D**_{1}= D_{2}=> 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.

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

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## Solved Examples

**Question 1:**Find the general solution for differential equation

(D

^{4}- 5D

^{3}+ 5D

^{2}+ 5D - 6)y = 0

**Solution:**

Given differential equation, (D

=> For general solution of equation,

Solve D

=> D

=> D

=> (D

Now

D

Factors of D

D

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

= (D - 3)(D - 2)

Therefore, equation (1) implies

(D

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

=> General solution of differential equation is,

^{4 }- 5D^{3}+ 5D^{2}+ 5D - 6)y = 0=> For general solution of equation,

Solve D

^{4 }- 5D^{3}+ 5D^{2}+ 5D - 6 = 0=> D

^{4 }- 5D^{3}+ 6D^{2}- D^{2 }+ 5D - 6 = 0=> D

^{2}(D^{2}- 5D + 6) - (D^{2}- 5D + 6) = 0=> (D

^{2}- 5D + 6)(D^{2}- 1) = 0 ................................(1)Now

D

^{2 }- 1 = (D - 1)(D + 1) andFactors of D

^{2}- 5D + 6D

^{2}- 5D + 6 = D^{2}- 2D - 3D + 6= D(D - 2) - 3(D - 2)

= (D - 3)(D - 2)

Therefore, equation (1) implies

(D

^{2}- 5D + 6)(D^{2}- 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 = C**._{1}e^{-x}+ C_{2}e^{x}+ C_{3}e^{2x}+ C_{4}e^{3x}**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, n

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

To find 20th term of the series

=> a

= 2 + 19 * 3

= 2 + 57

= 59

=>

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

= 7{4 + 39}

= 7 * 43

= 301

=>

Here First term (a) = 2

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

We know, n

^{th}term can be represented as, a_{n}= a + (n - 1)dand sum of n terms = $\frac{n}{2}${2a + (n - 1)d}

To find 20th term of the series

=> a

_{20}= 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.**