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Hello all. Welcome to the 11th of our Gifted World challenge series. Below the question, you will also find the solution and featured responses we received for Challenge number 11.
This week's challenge is a series of fascinating questions, that you can even experiment on at home and interestingly, ties up with our workshop on Conceptual Learning happening on Nov 9th.
Challenge #12 - The "Think More Deeply" Challenge
This challenge consists of 5 compound questions. You can answer as many as you like. The answers of each question will be published as separate blogs on our website.
Post your answer in the form embedded below, or directly on the - https://forms.gle/c1CabVXLsuwpCqzi7
Last day to submit answers is Saturday, 9 November. The 3 best answers (as judged by the Gifted World team) will be discussed in our workshop on Conceptual Learning on November 9. Looking forward to your answers
Post your answer here - https://forms.gle/c1CabVXLsuwpCqzi7
This challenge has been given by Vishnu Agnihotri, one of our amazing mentors and co-founders.
Solution to Challenge #11
Challenge number 11 required you to do a little bit of algebra. 10 of you got the answer right!
Here are our favourite answers -
Jivaana Luthra, Grade 11, Fravashi International Academy, Nashik
"The possible sums for 13 are: 1+9+3, 2+7+4, 3+6+4, 8+4+1, 8+3+2, 5+6+2, 5+7+1. Now assuming, P,Q,R,S,T,U,V,W,X to be different numbers to satisfy the two conditions: all are different digits and 3 add up to 13, the only possibility is P=3, Q=9, R=1, S=8, T=4, U=7, V=2, W=6, X=5. Hence, T=4"
Sammit Basu, 7th Grade, Vidyashilp Academy, Bengaluru, Karnataka
"By using the process of elimination, I found that instead of ruling a failed combination as failed i simply re-arranged the numbers until I found the correct combination. Also by using the formula, P+Q = S+T = 13 - R and simply substituting the letters for the others relevant in the question. The complete answer is as follows:
P+Q+R = 13 = 9+3+1
R+S+T = 13 = 1+8+4
T+U+V = 13 = 4+7+2
V+W+X = 13 = 2+5+6"
Lakshmi Murthy, teacher at BASE Educational Services, gave us a great strategy for finding the answer
"I did by my maternal granny's method. Trial and error. I decided to give a lesser digit to the repeated alphabet (to allow reuse). So, I started with P + Q + R as 9 + 3 + 1. So, R + S + T had to be 1 + 8 + 4 (check for yourself - can't be anything else). So, T + U + V had to be 4 + 7 + 2, and nothing else. V + W + X is 2 + 5 + 6. If the value of P, Q W, X is any one of 9, 6, 5 or 3; the value of R should be 2, of T should be 4, of V should be 1."
Others who got it correct were -
Abhinav Krishna, Woodlem Park School Ajman, Grade 7
IZAAN HUSSAIN
Sachin Bhaveshkumar Amlani Grade 7. School DPS Ajman, U.A.E.
Dinithi ,7 , Gems our own high school, Dubai
Nithilan Ashok Kumar | Grade 7 | Delhi Private School Sharjah
Tvesha Mishra, 7C, Greenwood high Bannerghatta Bangalore
Sanjiv Sachin Sethuraman 7 PSBBSSS
Thank you all for your enthusiastic participation!
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