Hey, Here's the search for the most intelligent guy!!

Here are some of the most trickling Maths and Science problems, If you are able to solve it. Write Detailed solutions below, with your email id. The best performer may be highly rewarded!- Q1.)If a,b,c are distinct zeroes of polynomial x^3 - x^2 + x - 2 then a^3 + b^3 + c^3 = ? Q2.)In a Triangle ABC, AB=10, BC=5, AC=6. BC is extended to D such that Trngle ABC is similar to Trngle DCA then lenght of DC = ? Q3.) If 1/x + 2/y + 3/z = 0 and 1/x - 6/y - 5/z = 0 then find x/y + y/z + z/x? Q4.)If a+b= 1 and a^2+b^2= 2 then find a^5+b^5? Q5.)ABCD is a square. AB is extended to P and DP intersects AC and BC at Q and R respectively such that DQ= 3cm and QR=2cm , then find PR? Q6.)In Trngle ABC, B=90 degree, AB=6 and AC = 10 then the length of angle bisector AD from Angle A is?

Note by Aßhĩмanyu Singh
5 years, 10 months ago

No vote yet
2 votes

</code>...<code></code> ... <code>.">   Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in </span>...<span></span> ... <span> or </span>...<span></span> ... <span> to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

First of all , I want to tell that they are no tricky problems. Tooooooo easy problems.

The complete solutions are -

1.) Clearly, by Vieta's Formula - a+b+c=1a+b+c=1, ab+bc+ca=1ab+bc+ca = 1, abc=2abc=2

We know that a3+b3+c33abc=(a+b+c)(a2+b2+c2abbcca)=(a+b+c)((a+b+c)23(ab+bc+ca))a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca) = (a+b+c)((a+b+c)^2 -3(ab+bc+ca))

Plugging the values, we get

a3+b3+c36=(1)(123(1))=2a3+b3+c3=4a^3+b^3+c^3-6=(1)(1^2-3(1))=-2 \Rightarrow a^3+b^3+c^3 = 4

2.) Since ABC is similar to DCA, the ratio of their corresponding sides must be equal. So,

ABDC=BCACDC=12\frac{AB}{DC}=\frac{BC}{AC} \Rightarrow DC = 12

3.) The given equations can be simplified to yz+2xz+3xy=0yz + 2xz + 3xy = 0 and yz6xz5xy=0yz - 6xz - 5xy = 0

Subtracting both the equations, we get -

z+y=0z=yz + y = 0 \Rightarrow z=-y

Substituting z=yz=-y in first equation, we get - x2y2=0x^2-y^2 = 0

Now, put z=yz=-y in x/y+y/z+z/xx/y + y/z + z/x. So, what you will get is x/y+y/z+z/x=1x/y + y/z + z/x = -1

4.) a+b=1a+b = 1, a2+b2=2a^2+b^2 = 2

We can easily get ab=12ab = \frac{-1}{2}

Now -

a5+b5=a5+b5+a3b2+a2b3a3b2a2b3a^5 + b^5 = a^5 + b^5 + a^3b^2 + a^2b^3 - a^3b^2 - a^2b^3

=a3(a2+b2)+b3(a2+b2)a3b2a2b3=(a2+b2)(a3+b3)a2b2(a+b) = a^3(a^2+b^2)+b^3(a^2+b^2) - a^3b^2 - a^2b^3 = (a^2+b^2)(a^3+b^3) - a^2b^2(a+b)

So, a5+b5=(a2+b2)(a3+b3)(ab)2(a+b)=(a2+b2)(a+b)(a2+b2ab)(ab)2(a+b) a^5+b^5 = (a^2+b^2)(a^3+b^3) - (ab)^2(a+b) = (a^2+b^2)(a+b)(a^2+b^2-ab) - (ab)^2(a+b)

Now plugging the values, we get - a5+b5=194a^5+b^5 = \frac{19}{4}

5.) First, let DRC=BRp=θ\angle DRC = \angle BRp = \theta and RP=xRP = x

Then AD=(5+x)cosθAD = (5+x)cos \theta and RC=5cosθRC = 5cos \theta

Now observe that - triangle QAD is similar to QCR. So, on solving for PR gives PR=52PR = \frac{5}{2}

6.) Let BD=xBD = x, BAD=DAC=θ\angle BAD = \angle DAC = \theta

Then, from triangle ABCsin2θ=810=45ABC \rightarrow sin 2\theta = \frac{8}{10} = \frac{4}{5}

from triangle BADsinθ=x36+x2BAD \rightarrow sin \theta = \frac{x}{\sqrt{36+x^2}} and cosθ=636+x2cos \theta = \frac{6}{\sqrt{36+x^2}}

We know that sin2θ=2sinθcosθsin 2\theta = 2sin\theta cos\theta

Plugging their values, we get the following quadratic - s215x+36x=12,3s^2-15x+36 \Rightarrow x = 12, 3

Since, BC=8BC = 8, so we can't take x=12x=12

Hence, AD=32+62=45AD = \sqrt{3^2 + 6^2} = \sqrt{45}

Next time when you post a question in discussion, please makes sure that they are of proper difficulty level.

Kishlaya Jaiswal - 5 years, 10 months ago

Log in to reply

All these are Class X FTRE Sample Paper Questions.

Bhargav Das - 5 years, 10 months ago

Log in to reply

I'm not trying to be rude at all but these questions were really easy and they aren't science problems. Though if you are having trouble with anyone of them just let me know because I will help you. I can give you some hints if you like too: 1. this involves complex numbers 2. I used the cosine rule 3. this is just algebra 4. this just involves a quadratic 5. just a bit of trigonometry 6. this just involves Pythagoras/tan. Hope this helps.

Joel Jablonski - 5 years, 10 months ago

Log in to reply

These are very easier problems and way below than the level of Brilliant problems. You must not post these types of problems to this discussion room.

Ram Prakash Patel Patel - 5 years, 10 months ago

Log in to reply

that was easy. Now, help me solve this... x2+5x+5=32x^{2}+5x+5=32 Don't use trial and error method and the xx value should be a single one.

Fahad Shihab - 5 years, 10 months ago

Log in to reply

That's the question which the brilliant community should solve

Fahad Shihab - 5 years, 10 months ago

Log in to reply

x=±13352x=\frac{\pm \sqrt{133}-5}{2}

Joel Jablonski - 5 years, 10 months ago

Log in to reply

@Joel Jablonski thank you. that's all i need. please write the steps also.

Fahad Shihab - 5 years, 10 months ago

Log in to reply

Are you f'ing kidding us? This is as easy as the ones posted above. Sorry.

Guilherme Dela Corte - 5 years, 10 months ago

Log in to reply

Not kidding.Not f'ing. but making use of situations.

Fahad Shihab - 5 years, 10 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...