hafgola wrote:
Hi I struggled with this one, I am wondering
Is there any good shortcut in how to take the root of 1089 ? how do you do that without spending too much time on it ?
many thanks
Let me take your question first:
It is a good idea to remember squares of numbers till 20. They come in handy sometimes.
To find the square root of bigger numbers (which are perfect squares), observe the following:
1. A square of a number only ends in 0/1/4/5/6/9.
-If it ends in 0, it will have even number of 0s at the end and the square root will end in 0 (I know 1340 cannot be a perfect square.)
-If it ends in 1, the square root will end in 1/9 (e.g. root 361 = 1
9, root 441 = 2
1)
-If it ends in 4, the square root will end in 2/8
-If it ends in 5, there will be a 2 right before it and its square root will end in 5 (e.g. root 6
25 = 25, root 6
75 is not an integer)
-If it ends in 6, the square root will end in 4/6.
-If it ends in 9, the square root will end in 3/7.
These are all derived by just observing the last digits when we square a number.. You don't need to learn this.
Since 1089 ends in a 9, its square root will end in 3/7.
Also \(30^2 = 900\) so root of 1089 will be greater than 30.
But \(40^2 = 1600\) so root of 1089 will be much less than 40. The number that can be the square root of 1089 is 33 (if 1089 is a perfect square). You can square 33 to check.
Also, an equation like \(\frac{90}{{v -3}} - \frac{90}{{v + 3}} = \frac{1}{2}\), doesnt need to be solved.
99% chance is that v will be an integer.
The options give me values for \(\frac{90}{{v + 3}}\)
Let me say if it is 2.5, then 90/2.5 = 36 and v = 33
Then \(\frac{90}{{v - 3}}\) = 3. I get the difference of 1/2 an hour. 2.5 satisfies my condition.
Had 2.5 not been the first option, you might say that it would take me a long time to check each option.
No. 90/2.4 = 900/24 which is not an integer since 900 is not divisible by 8 so it will not be divisible by 24
90/2.3. No ways 900 is divisible by 23
90/2.2. No ways 900 is divisible by 11
90/2.1. No ways 900 is divisible by 7
In fact, when I look at the options, the first one I try is 2.5. With practice, you will gain the instinct.
I have been able to solve most of the questions from work, speeed distane & mixtures. However i always solve them using the algebraic approach which eats up a lot of time. I have seen you solving questions by a logical approach that saves a considerable bit of time. I have tried solving that way but I just cant get going. Could you please advice me on how to develop this skill??