Although the outcome of the last quiz (4 sample practice questions from the 2015 standardized Scholastic Aptitude Test [SAT] – click on SAT Practice Quiz) was only one responder getting both math problems correct I know that people really like quizzes – about as much as I like "Man on the Street" interviews.
Below is the latest math challenge quiz from The Little Book Of Big Mind Benders.
I will post all correct answers on RTE or will send the answer to anyone who requests it if no one gets the answer right. In responding please include your reasoning or explain how you got your answer.
Challenge Quiz:
Zane is older than Mabel, & both have two-digit ages. Zane's age is the same as Mabel's age backward, & Mabel's age doubled is within a year of Zane's age. Without using trial & error - how old is Zane?
My late brother was very good on those questions, I was not; therefore I have to leave the answer to others.
ReplyDeletePatrick, here's a math quiz to be solved. Send me the answer. Have a productive week. Dad
ReplyDeleteI would have to say 66, 66 is the same backwards, two digits, and can be a year apart.
ReplyDeleteYou did not take into account the doubling of Mabel’s age. Try again.
DeleteJeez Doug. You ask a guy who got a D in Algebra. Here is the bassackward way I got to it.
ReplyDeleteZane is (XY) age. Mabel is (YX)
YX*2 +/-1 = XY.
Therefore Y must be an odd number.
Y can only be 1 or 3, so she is in her teens or 30’s. Anything higher would result in 100+.
Y as 1X (2 to 9) *2 would range from 24 to 38. Therefore the only possibilities, 21 or 31 for Zane cannot work.
Y must equal 3. X must be higher than 4 or Zane would be younger.
3X (4 to 9)* 2 would equal 68 to 78. That means Zane can only be 63 or 73. Since the minimum is 68, he must be 73.
Zane is 73. Mabel is 37. 37 * 2 = 74. +/- 1 = 73.
In common core I am sure I would get another D only because I got the answer.
You get an “A” from me. You really dove into the solution. Congrats.
DeleteI used the tens & units digits to solve the problem coupled with your observations to narrow the variables – less trial & error with this approach than yours
Let AB = Zane’s age & BA = Mabel’s age
Basic Equations: 2(BA) + 1 = AB or 2(BA) – 1 = AB, therefore B must be an odd number (2 times any whole number produces an even number & adding or subtracting 1 makes the resultant an odd number). Since the ages are both two digits we know that B has to be 1 or 3. Using 5 or higher results in a three digit number.
Breakdown to tens & units form of basic equation: 20B + 2A + 1 = 10A + B or 20B + 2A – 1 = 10A + B
Combining: 19B + 1 = 8A or 19B – 1 = 8A
We quickly see that B = 1 does not work.
For B = 3 we get 19(3) – 1 = 8A or A = 7 meaning that Zane is 73 & Mabel is 37
2(37) – 1 = 73 so it confirms
I figured there had to be a more algebraic solution. My logic course was one of my highlights in college, so I went with it.
DeleteThis morning I woke up around 4.30 and in a few minutes I got it solved while still in bed. Zane's age is 73.
ReplyDeleteExplanation follows later.
Here's my mental solution. I narrowed the double digit nos. to between mid 30's and mid 40's and started to solve using no.35 as starting pt. and with increasing nos.and luck, I found the doubling of MABEL'S AGE NARROWING for correct solution and it hit the jackpot with 37.
DeleteAlgebraically,
2(10a+b)=10b+a where a & b are the first and second nos. of the 2 digit no. and the only way I can solve is by trial and error.
Mabel 37 & Zane 73 ??
ReplyDeleteCorrect answer, but need methodology to get full credit. Remember, trial & error is not allowed so please explain how he got it.
DeleteI figured Mabel's first digit had to be an odd number or it would never be within a year of the other one. I also knew the Mabel had to be under 50 or the doubling would put the older one over 100 which you said it couldn't be. So at this point i reduced the younger one's age to the teens or the thirties and did trial and error from that point and it was easy to solve.
DeleteNot sure if this makes sense , but it worked for me.
x = units digit of Mable & y = tens digit 10 y + x = Mable's age & 10 x + y = Zane's age
ReplyDelete1. x + y = 20 - x - y or 2x + 2y = 20 sum of the digits = 20 - the difference of that sum
2. 10 x + y = 2 ( 10 y + x ) - 1 or 10 x + y = 20 y + 2 x - 1 or 8 x - 19 y = - 1
3. 8 x - 19 y = - 1
4. Multiply Eq. # 1 by - 4 - 8 x - 8 y = - 80
5. Add: - 27 y = - 81 or y = 3
6. 2 x + 2 y = 20, then 2 x + 2 ( 3 ) = 20 or x = 7
Therefore Mable's age is 37 & Zane's age is 73.
N. B. x + y = 10 would have worked only because the sum of the digits in this example happens to be 10.
Excellent solution. I started with your equation 2 & worked the problem from there. Please explain equation 1 – I don’t quite follow where you got it.
DeleteThanks for asking. Actually I lucked out because the sum of the digits was 10. I could have used x + y = 10 which would not work if the ages were 57 & 75.
DeleteGot the correct answer from Jon but no explanation about how he did it. I wrote to him - I couldn't believe he cheated (?). Jerry pulled hair out, came up with somebody in the 20s. Showed him the answers. Next time make it easier please!
ReplyDeleteI had given it some thought when you originally sent it over, so it wasn't as fast as it seemed. I just recalled my previous answer. I knew the younger age had to be roughly in the 30s in order to have the math work out, so I just tried doubling and reversing numbers until I found a pair that worked.
DeletePaul told me the answer: 73. I just tried to solve it and trial & error seems the only way.
ReplyDelete