Logical Math Riddles for Class 9: There’s no better workout for the brain than an intricate question. Have you ever noticed yourself spending hours scratching your head on a riddle, brain teaser or a fun puzzle? Well, if yes, then you’re not alone.
There are times when even the most impulsive people get laser-focused on solving a puzzle or completing a task. In this day and age, overstimulation of the mind has become a plague. You are constantly bombarded with memes, news and unnecessary information.
This can dull your mind and make you feel tired and gloomy. But there are plenty of ways to cheer yourself up and unlock the potential of your brain. The best way to stimulate the mind and also have a little fun is to solve riddles, puzzles and brain teasers.
Today, we bring you one such mathematical riddle that’s sure to melt your mind. Hopefully, you didn’t skimp on math in school because you’ll need to know polynomials to crack this riddle. So, sit back and relax your mind. Grab a pen and paper if possible.
Ready?
Let’s Go!
Dive in to solve this polynomial math riddle in 45 seconds or less
Polynomial Math Riddles for Students: Find the Value of P, Q, R, S & T in 45 Seconds
Given above is a viral math riddle that has been confusing the netizens for quite a while. Test your mathematical knowledge with this polynomial riddle.
Here’s a polynomial equation for you. You have to solve it within 45 seconds.
P2 + Q2 + R2 = S2 + T2
The conditions are as follows:
Find the value of P, Q, R, S and T.
P, Q, R, S & T are consecutive and positive integers
Don’t exceed the time limit or cheat.
Your time starts now!
GO!
Related: Math Riddles For Students: Can You Solve this BODMAS Question? 99% of People Fail!
Polynomial Math Riddle Solution
Hopefully, you had fun solving this polynomial riddle and found the answer on time. Now, it’s time to verify your answers.
The value of P, Q, R, S and T is 10, 11, 12, 13 and 14 respectively.
Here’s how to solve the riddle:-
Assume P = any positive integer x then
Q = x+1
R = x+2
S = x+3
T = x+4
The equation becomes
P2 + Q2 + R2 = S2 + T2 => x2 + (x+1)2 + (x+2)2 = (x+3)2 + (x+4)2
Using the identity for (a+b)2 = a2 + 2ab +b2, we get
=> x2 + x2 + 2x + 1 + x2 + 4x + 4 = x2 + 6x + 9 + x2 + 8x + 16
=> 3x2 + 6x + 5 = 2x2 + 14x + 25
=> x2 – 8x – 20 = 0
Upon solving for zeroes, we get
X = -2, 10
Since the solution is a positive integer, x = P = 10
And Q = 11, R = 12, S = 13, T = 14
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