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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
3 years, 5 months ago

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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 - $$a+b+c=1$$, $$ab+bc+ca = 1$$, $$abc=2$$

We know that $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca) = (a+b+c)((a+b+c)^2 -3(ab+bc+ca))$$

Plugging the values, we get

$$a^3+b^3+c^3-6=(1)(1^2-3(1))=-2 \Rightarrow a^3+b^3+c^3 = 4$$

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

$$\frac{AB}{DC}=\frac{BC}{AC} \Rightarrow DC = 12$$

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

Subtracting both the equations, we get -

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

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

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

4.) $$a+b = 1$$, $$a^2+b^2 = 2$$

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

Now -

$$a^5 + b^5 = a^5 + b^5 + a^3b^2 + a^2b^3 - a^3b^2 - a^2b^3$$

$$= a^3(a^2+b^2)+b^3(a^2+b^2) - a^3b^2 - a^2b^3 = (a^2+b^2)(a^3+b^3) - a^2b^2(a+b)$$

So, $$a^5+b^5 = (a^2+b^2)(a^3+b^3) - (ab)^2(a+b) = (a^2+b^2)(a+b)(a^2+b^2-ab) - (ab)^2(a+b)$$

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

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

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

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

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

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

from triangle $$BAD \rightarrow sin \theta = \frac{x}{\sqrt{36+x^2}}$$ and $$cos \theta = \frac{6}{\sqrt{36+x^2}}$$

We know that $$sin 2\theta = 2sin\theta cos\theta$$

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

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

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

Next time when you post a question in discussion, please makes sure that they are of proper difficulty level. · 3 years, 5 months ago

All these are Class X FTRE Sample Paper Questions. · 3 years, 5 months ago

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. · 3 years, 5 months ago

that was easy. Now, help me solve this... $$x^{2}+5x+5=32$$ Don't use trial and error method and the $$x$$ value should be a single one. · 3 years, 5 months ago

Are you f'ing kidding us? This is as easy as the ones posted above. Sorry. · 3 years, 5 months ago

Not kidding.Not f'ing. but making use of situations. · 3 years, 5 months ago

That's the question which the brilliant community should solve · 3 years, 5 months ago

$$x=\frac{\pm \sqrt{133}-5}{2}$$ · 3 years, 5 months ago

thank you. that's all i need. please write the steps also. · 3 years, 5 months ago