Yeah =/
i mean i'm looking for a way to figure this out mentally/without calculator.
Math help needed
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Yeah =/
i mean i'm looking for a way to figure this out mentally/without calculator.
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if it was me, i would just round up and down wildly and hope to get it in the ball park for the answer
like 11x11x11/9x9x9 to 12x10x11/10x8x9, essentially making one 11 a 12 and one 9 a 10 and the other way making one 11 a 10 and one 9 an 8... and cancel them out to 11/6 which is little less than 2...
with that said, I usually look at the answer choices to see if they are apart enough, then I would even go for canceling two 11 from the top with two nine from bottom and getting 11/9= 1.2ish (vs. the actual 1.8ish) and I take notice of that and pick something close to 1.5!
like 11x11x11/9x9x9 to 12x10x11/10x8x9, essentially making one 11 a 12 and one 9 a 10 and the other way making one 11 a 10 and one 9 an 8... and cancel them out to 11/6 which is little less than 2...
with that said, I usually look at the answer choices to see if they are apart enough, then I would even go for canceling two 11 from the top with two nine from bottom and getting 11/9= 1.2ish (vs. the actual 1.8ish) and I take notice of that and pick something close to 1.5!
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Yeah =/
i mean i'm looking for a way to figure this out mentally/without calculator.
What are the numbers? I can't see the image on this computer.
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1.22 is about 1.2. 1.2 cubed is 1.728 (you should be able to do 144 * 12 mentally). Add on a bit to compensate for the 0.02 you dropped. 1.8 wouldn't be a bad guess.
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never seen anything like this in my MCAT prep. I think this is it, and assuming its a question with multichoice answers, you are working with palindrome numbers here. These are numbers which are the same forward as they are backward but the rules are: 11 ia palindromic prime number whilst 9 is a palindromic even number. All multiples of 11 end in 1 and produce palindromic numbers, all multiples of 9 end in either 1 or 9 but not necessarily produce another palindromic number. The gist of this is that an odd number cubed gives an odd number. Odd numbers divided by other odd numbers give you even ending numbers. Okay lets look at your problem.
11^3/9^3. Firstly we know the top and bottom numbers will be odd-does it help? Not yet
Do we know that 11 is a prime number? yes,it is only divisible by 11 or 1 without leaving a remainder. The even 9 is going to produce a left over number as its not 11 or 1. So far we know then the answer will include a remainder and that the answer will have an even digit at the end. Seek it out in the choices you have.
11^3/9^3. Firstly we know the top and bottom numbers will be odd-does it help? Not yet
Do we know that 11 is a prime number? yes,it is only divisible by 11 or 1 without leaving a remainder. The even 9 is going to produce a left over number as its not 11 or 1. So far we know then the answer will include a remainder and that the answer will have an even digit at the end. Seek it out in the choices you have.
Choose something you know to set a lower and upper bound for the closest estimate. 11/9 is close to what? 9/9 and 12/9. These bounds are somewhat large especially if all the answer choices are between 1 and 3. So we'll eventually have to narrow this down, but this is a good starting point.
12/9 = 4/3. Now (4/3)^3 is (4/3) multiplied three times. Can you actually do this in your head? 64/27? If 64/27 is difficult, then break this down also to something easier. 64/32 = 2, so the answer's a little larger than 2. A good guess would be about 2.2-2.3 Now put everything together. You know 1.33333 (4/3) < (11/9)^3 < 2.2-2.3((4/3)^3). You can guess that it'll be 1.7 or 1.8 by averaging.
12/9 = 4/3. Now (4/3)^3 is (4/3) multiplied three times. Can you actually do this in your head? 64/27? If 64/27 is difficult, then break this down also to something easier. 64/32 = 2, so the answer's a little larger than 2. A good guess would be about 2.2-2.3 Now put everything together. You know 1.33333 (4/3) < (11/9)^3 < 2.2-2.3((4/3)^3). You can guess that it'll be 1.7 or 1.8 by averaging.
