Since getting the basics right is the most important thing in CAT prep, We can discuss and revise them through this thread.
Different formula's for calculating area of a triangle??
An odd function is symmetrical about?
X-Axis
Y-Axis
Origin
when p happens, q happens..valid logical deduction is??
Regarded to be
considered as.
Correct or not?
Could The English verb "could" is used as both an auxiliary verb and a modal verb. When using as a model verb or an auxiliary verb the verb "could" stays in its base form. Could doesn't change form according to the subject. Could is followed by a main verb. The main verb also stays in its base form. (The main verb that is used with could doesn't change form according to the subject.) Generally, we use could to:
- talk about past possibility or ability
- make requests
'Could' can be used to talk about the past, the present or the future. Most modal verbs behave quite irregularly in the past and the future. Study the chart below to learn how "could" behaves in different contexts. Modal Use Positive Forms
1. = Present 2. = Past 3. = Future Negative Forms
1. = Present 2. = Past 3. = Future You can also use: could
possibility 1. John could be the one who stole the money. 2. John could have been the one who stole the money. 3. John could go to jail for stealing the money. 1. Mary couldn't be the one who stole the money. 2. Mary couldn't have been the one who stole the money. 3. Mary couldn't possibly go to jail for the crime. might,
may could
conditional
of can 1. If I had more time, I could travel around the world. 2. If I had had more time, I could have traveled around the world. 3. If I had more time this winter, I could travel around the world. 1. Even if I had more time, I couldn't travel around the world. 2. Even if I had had more time, I couldn't have traveled around the world. 3. Even if I had more time this winter, I couldn't travel around the world. could
suggestion 1. NO PRESENT FORM 2. You could have spent your vacation in Hawaii. 3. You could spend your vacation in Hawaii. NO NEGATIVE FORMS could
past ability I could run ten miles in my twenties. I could speak Chinese when I was a kid. "Could" cannot be used in positive sentences in which you describe a momentary or one-time ability. Yesterday, I could lift the couch by myself. Not Correct I couldn't run more than a mile in my twenties. I couldn't speak Swahili. "Could" can be used in negative sentences in which you describe a momentary or one-time ability. Yesterday, I couldn't lift the couch by myself. Correct be able to could
polite request Could I have something to drink? Could I borrow your stapler? Requests usually refer to the near future. Couldn't he come with us? Couldn't you help me with this for just a second? Requests usually refer to the near future. can,
may,
might REMEMBER: "Could not" vs. "Might not"
"Could not" suggests that it is impossible for something to happen. "Might not" suggests you do not know if something happens. Examples: · Jack might not have the key. Maybe he does not have the key. · Jack could not have the key. It is impossible that he has the key.
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Divisibilty Methods For checking divisibility by any 'prime/odd' number except for factors of 5', you have the concept of base number.
Number Add Base Number Subtract Base Number
---------------------------------
3: 1 2
7: 5 2
9: 1 8
11: 10 1
13: 4 9
17: 12 5
19: 2 17
21: 19 2
23: 7 ?
27: ? 8
29: 3 ?
...
...
And so on....(i'll describe a method to get these below)
Now for checking divisibility either add the last digit*add base number to the number "formed by removing last digit" or you can subract last digit*subtract base number from the "formed by removing last digit"
for e.g. check for 51/17 either 5- 1*5 =0 or 5 + 1*12 =17 hence divisible.
to check 312/13 we can 31+2*4 =39 hence dvisible or we can 31-2*9 = 13 ence divisible.
to check 61731/19 = 6173 + 1*2 = 6175 = 617 + 5*2 = 627 = 62 + 7*2 = 76 hence divisible.
to check 357976/29 = 35797 + 6*3 = 35815 = 3581 +5*3 = 3596 = 359 + 6*3 = 377 = 37 + 7*3 = 58 hence divisible..
to check 382294/11 = 38229-4*1 =38225 = 3822-5 = 3817 = 381 - 7 = 374 = 37 -4 =33 Hence divisible..
The Subract base Number for a number can be obtained as the {(samllest multiple of number which ends in one)-1}/10
i.e. for 3 or 7 it is (21-1)/10 =2
for 13 it is 91-1/10 = 9.
The AddbaseNumber for a number can be obtained as the {(samllest multiple of number which ends in nine)+1}/10
i.e. for 13 it is (39+1)/10 =4.
for 7 it is 49+1/10 = 5
Proof:
For SubtactBaseNumber say the number abcde...
I want to check divisibility by 17 where subtractbasenumber is 5
I can always write abcde... as 10X+Y (where Y is last digit and X is number formed by removing last digit)
Now X-Y*5 = (10X -50Y)/10 = (10X + Y -51Y)/10 = (OriginalNumber - 51 Y ) / 10
The number '51 Y' is a multiple of 17 so if "OriginalNumber" is divisible by 17 then "OriginalNumber - 51*Y" got to be.. i.e. "10X - 50Y"
as 10 and 17 are co-prime if "10X- 50Y" is divisible the "X-5Y" got to be.....
same theory hold's for addbasenumbers too....
This also defines why it is so easy to check divisibility by 3 or 9 just keep on adding the digits...
And you can check divisibility by 11 just by keeping on subrating digits form previous number.. (which is same as taking sum of even/odd location separately..) Divisibility by 7
Only for those interested in Number theory (Not a Cat short-cut)
say the number is :
38,391,787
Separate into pairs of digits
38 39 17 87
Consider the difference between each pair of digits and the nearest multiple of seven, beginning for the first pair at right, lower (upper) for the first, upper (lower) for the second and so on, alternating for each new pair.
4 -----4 (21-17)
38 39 17 87
---4 ------3 (87-84)
The resulting digits, read from right are 3444 (which is also a number multiple of 7).
Proceed in the same way with 3,444
1
34 44
----2
The final pair 21 is a multiple of seven, so is the original number 38,391,787.
ANOTHER EXAMPLE
Look how fast this method is.
Consider the 15-digit number 531,898,839,909,822
2 ----2--- 3 ----0
5 31 89 88 39 90 98 22
---3 ---4 ----6 ----1
Now we have 10,634,232
4 -----0
10 63 42 32
----0 ----4
And now 4,004
2
40 04
---4
Which gives 42, a multiple of 7.
We only need three steps for a 15-digit number. This is called TOJA's method of divisibility. Incidentally this also works for 11 and 13. Just a little manupulation is required, (in case you get a remainder of more than 9)
Let A = 5,962
7
59 62 77 which is a multiple of 1
--7
EXAMPLE 2
Let A = 5, 971,845
6---- 4
5 97 18 45
--9 ----1
8
14 96 -> 88 ->divisible
----8
EXAMPLE 3
Let A = 80,714,546
8 ----10
80 71 45 46
----5-------2
The resulting numbers ( 2 10 5 8 ) don’t form a decimal number, so proceed in this way: Put the exceeding number 1 from 10, below the 2 and sum.
2 0 5 8 -> 3 0 5 8
1
3
30 58 33
---3
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