1. The number of circles that can be drawn out of 10 points of which
7 are collinear is
a) 130
b) 85
c) 45
d) 72
e) CBD
my take option b...please confirm
my take option b...please confirm
Thanks a lot..I was really pissed off with this. I was thinking that what is it that I missed out.
3. The number of permutation of the letters a, b, c, d, e, f, g such that
niether the pattern 'beg' nor pattern 'acd' occurs is
a) 4806
b) 420
c) 2408
d) 140
e) None of these
option (e)
My answer is 4798, below is solution how i got it:
(7! -2) - (5!+5!)
please help if wrong
Hello can you please help me how to find the answe to this one. Its Number system LOD 2, Arun Sharma page: 42, Question n: 4
Q) The unit digit in the expression 36^234 X 33^512 X 39*180 - 54^29 X 25^123 X 31^512 will be?
If i use the cyclicity property to find out the unit digit of 36, 33 and 39 its coming out to be 1 in each case which is wrong i know. E.g 6 has a cyclicity of 1, and when i divide 234 by 1 there is no remainder = 0. 6^0 = 1. but I know it should be 6. please help me what am I doing wrong. Thanks
Hello can you please help me how to find the answe to this one. Its Number system LOD 2, Arun Sharma page: 42, Question n: 4
Q) The unit digit in the expression 36^234 X 33^512 X 39*180 - 54^29 X 25^123 X 31^512 will be?
If i use the cyclicity property to find out the unit digit of 36, 33 and 39 its coming out to be 1 in each case which is wrong i know. E.g 6 has a cyclicity of 1, and when i divide 234 by 1 there is no remainder = 0. 6^0 = 1. but I know it should be 6. please help me what am I doing wrong. Thanks
How can 36^(any number) end in 1?? It'll always end in 6. 33 and 39 would both end in 1.
So the first part's unit digit is 6*1*1 = 6
Second part: 54^29 = 4
25^123 = 5
31^512 = 1
Thus, 4*5*1 = 20 = Unit's digit is 0
Thus, answer should be 6-0 = 6
How can 36^(any number) end in 1?? It'll always end in 6. 33 and 39 would both end in 1.
So the first part's unit digit is 6*1*1 = 6
Second part: 54^29 = 4
25^123 = 5
31^512 = 1
Thus, 4*5*1 = 20 = Unit's digit is 0
Thus, answer should be 6-0 = 6
See I know, 36^243 or any number cant end in one. But my confusion is that when I divide 243 by its cylicity which is 1, it leaves no remainder. Then you raise the last digit of 36 i.e. 6 to the remainder which is 0. 6^0 = 1. hence the confusion
take for example i am sure youalready know:
3^57 {cyclicity of 3 is 4, when you divide 57 by 4 you get 1 as a remainder. now you raise 3 to the power of 1 and will get 3^1 = 3 which is the last digit of 3^57). I am applying the same logic above but it doesnt work.
See I know, 36^243 or any number cant end in one. But my confusion is that when I divide 243 by its cylicity which is 1, it leaves no remainder. Then you raise the last digit of 36 i.e. 6 to the remainder which is 0. 6^0 = 1. hence the confusion
take for example i am sure youalready know:
3^57 {cyclicity of 3 is 4, when you divide 57 by 4 you get 1 as a remainder. now you raise 3 to the power of 1 and will get 3^1 = 3 which is the last digit of 3^57). I am applying the same logic above but it doesnt work.
You probably didn't get the funda of cyclicity properly. I'll try to explain using some other example, should be clear to you now:
The cyclicity of 2 is 4. Why??
2x1 = 2 --------------------------- (1)
2x2 = 4 --------------------------- (2)
4x2 = 8 --------------------------- (3)
8x2 = 6 (unit's place of 16) -------- (4)
6x2 = 2 again (unit's place of 12)
Thus, the cycle repeats after every 4 steps. Therefore we say, the cyclicity of 2 is 4.
Now if you have to find the units place of 92^331, all you have to do is to find the remainder of 331/4 (where 4 is the cyclicity). Remainder comes out to be 3, so we take the unit's digit of 2^3, which, as we found out earlier, was 8. So, 92^331 would end in 8.
Now taking the previous example of 36^243, since the cyclicity of 6 is 1, no matter what power you raise it to, the unit's digit would always come out to be 6. For checking this, you don't even need to find out any remainder etc., just putting in 6 would do.
Connection problem. Repeat post
n ya thr is one more... (pg 342 prob 4)
The sides of a triangle are 21cm, 20cm and 13 cm.find the area of the larger triangle into which the given triangle is divided by the perpendicular upon the longest side from the opposite vertex.
(a)72 sq. cm
(b)96 sq cm
(c)168 sq cm
(d)144 sq cm
(e)150 sq cm
Ans: 96.
Use triplets 5,12,13. 13 is already known , now u know height as 12 .
guys i am not able to solve this simple question. Guess I am doing something wrong: find the last two numbers of the product
65X29X37X63X71X87X85
i divided 85 and 100 (denominator) by 5 so i got:
(65X29X37X63X71X87X17)/20
next step:
(5X9X17X3X11X7X17)/20
next:
45X51X77X17
next
5X11X17X17
next
55X289
next
15X9
135/20 therefore remainder = 15
now multiply 15X5 because I divide 100 by 5 earlier. Arun Sir says the answer is 85. but how? please help
please if anyone knows answer for this question please post it with explanation.
Vijay and Pallavi went for 100, mts. ski race. Initially Pallavis speed was 1m/smore so Pallavi gave Vijay some lead in terms of time, when Pallavi caught up with Vijay, then Vijay increased his speed by 2m/s and he was the winner by 7 minutes and 8 seconds. Had the race been 500mts longer, he would have won by 25 more seconds. A) at what point Pallavi caught up with Vijay b.) what was the lead given to Vijay. C)speed of Pallavi, speed of Vijay.
Hi,
If someone has done this question then please confirm
pg 202 LOD 3 Question 1
Can someone please confirm that is the answer given by the book correct/incorrect?
guys i am not able to solve this simple question. Guess I am doing something wrong: find the last two numbers of the product
65X29X37X63X71X87X85
i divided 85 and 100 (denominator) by 5 so i got:
(65X29X37X63X71X87X17)/20
next step:
(5X9X17X3X11X7X17)/20
next:
45X51X77X17
next
5X11X17X17
next
55X289
next
15X9
135/20 therefore remainder = 15
now multiply 15X5 because I divide 100 by 5 earlier. Arun Sir says the answer is 85. but how? please help
it has to be 75, 65x85 will give 25 as last two digits, and 25 multiplied by anything is either 00,25,50,75 as last 2 dig. now multiply d last digits of d remainin nos to get the single last dig no, which mulitiplied by 25 will give 75 as the ans.
We have 40kg of iron.We have to divide it into four parts such that we can measure any weight upto 40 kg..
manukathuria SaysWe have 40kg of iron.We have to divide it into four parts such that we can measure any weight upto 40 kg..
i think we can split it up to 6 parts...1,2,4,7,15,30....:wow::wow:
The question is to divide them into four parts not 6...i know the answer but want to check u guys
manukathuria SaysWe have 40kg of iron.We have to divide it into four parts such that we can measure any weight upto 40 kg..
it should be 1,3,9,27...you can measure all the weights with these....
rahicecream Saysit should be 1,3,9,27...you can measure all the weights with these....
How can you give the following weights 2,5,6,7..
manukathuria SaysWe have 40kg of iron.We have to divide it into four parts such that we can measure any weight upto 40 kg..
is the answer as follows: 3^0(1), 3^1(3), 3^2(9), 3^3(27)
Ganu02 SaysHow can you give the following weights 2,5,6,7..
look bro,
suppose you have to weight 2kg..then just put 1kg on one side and 3 kg on another..
Now the difference between the two is equal to 2 kg
suppose u have to weigh 2 kg rice..then keep on adding the rice to 1 kg side,till they become balanced.
Similarly for others,you have to either add or subtract..