2 questions from Number Systems, LOD-III
q44) N= 202 X 20002 X 200000002 X 20000000000000002 X 20000...2 (31 zeroes). The sum of the digits in this multiplication will be:
a)112 b)160 c)144 d)Cannot be determined
q45) 25 sets of problems on DI-one each for the DI sections of 25 CATALYST tests were prepared by the AMS research team. The DI section of each CATALYST contained 50 questions of which exactly 35 questions were unique i.e, they had not been used in the DI section of any of the other 24 CATALYSTs. What could be the maximum number of questions prepared for the DI sections of all the 25 CATALYSTs put together?
a)1100 b)975 c)1070 d)1055
(Answers: 44-b, 45-d)
This is my logic for q45 which leads to a wrong answer, please help to point out the flaw...
The number of unique questions would add up to 35X25=875
After this number of questions remaining is 15X25=375.
In order to maximise the number of these (non-unique) questions, let each question be repeated just once at most. Thus maxm. no. of such questions=374/2+1=188
....which leads to a total of 875+188=1063 😞 😞 what's wrong??
you are right!!!!
Q)A hundred and twenty digit number is formed by writing the first x natural number in front of each other as 12345678910111213......Find the remainder when this number is divided by eight.
options: 6,7,2,0
According to book last there digits will be 646. Can anyone please explain how?
See, first 9 digits are single digit nos
So, 120-9 = 111 digits remaining.
Now each next no. is 2 digit no.
So, 111/2= 55.5
Means after 55 nos from 9, you'll arrive at 119th digit.
So, check 54th no. from 9 to arrive at 117th digit.
=> 9+ 54 =63.
Further it should be 646566...and so on. where '646' will be the 118th, 119th and 120th digit.
Then by divisibility method for 8, answer is 6.
please someone explain the below question:
two trains A and B start from station AA and BB respectively at same time towards each other. After passing each other they take 12 and 3 hours to reach BB and AA respectively. If A is traveling at speed of 48 km/hr, the speed of B is
a. 24km/hr
b. 22km/hr
c. 21km/hr
d. 96km/hr
e. 120 km/hr
please someone explain the below question:
two trains A and B start from station AA and BB respectively at same time towards each other. After passing each other they take 12 and 3 hours to reach BB and AA respectively. If A is traveling at speed of 48 km/hr, the speed of B is
a. 24km/hr
b. 22km/hr
c. 21km/hr
d. 96km/hr
e. 120 km/hr
Let the distance between AA and BB be d.
then at the point of meeting of trains, the following equation holds true.
48t+vt= d -------------- (1)
, where v is the speed of train B and t is the time after which they meet when they start from either stations at t=0.
since the train a reaches BB after 12 hrs of meeting. hence the total time taken by A to cover d is 12+t hrs.
the the equation (1) becomes
48t+vt= (t+12)48
solving which gives vt= 12x48
now since train B reaches AA after 3hrs from meeting and the distance covered by both trains is d. Hence,
(t+12)48 = (t+3)v ------------ (2)
solving eq. 1&2 .. we get v=16t
putting this in (1) we get 16t^2= 48x12, therefore t= 6hrs
hence v= 16x6= 96 km/hr.
NOTE: It can be directly inferenced from the question that speed of B will be greater than A. So we can also put the options C and D and see which satifies the eqs to directly reach at the answer.
Cheers!!
You can go in this way as well.
Lets say their are 100 men out of which 45 are married. This 45 is equal to 25% of the total women population, which means their are 180 womens. Then 45 (men) + 45 (women)/ 260 (Total population)=32.14
Sources of misunderstanding!
Okay finally I got it.
But tell me also how to do this part... (I am a noob)
"Rem=2*-1=-2=7
Rem=1"
)Rem=-1^2333=-1.......
Similarly, Rem=-1^1750=-1......
hi Puys,
Please explain how to solve the following ques..
and also explain the approach for all such type of questions
Q12. Number Systems LOD 3.
Find all two-digit numbers such that the sum of the digits constituting the number is not less that 7;
the sum of the squares of the digits is not greater than 30;
the number consistingof the same digits written in the reverse order is not larger than half the given number.
(a) 52 (b) 51 (c) 49 (d) 53 (e) 50 .
Please help me solve the following question from arun sharma book (Number systems) :
Q-1) What is the remainder when 2(8!)-21(6!) divides 14(7!) + 14(13!)?
options: 1,7!,8!,9!
Q-2)How many integer values of x and y are there such that 4x+7y =3, while xoptions:144, 141,143,142
Q-3) A candidate takes a test and attempts all the questions in it. While any correct answer fetches 1 mark, wrong answers are penalized as follows: one-tenth of the questions carry 1/10 negative mark each, one-fifth of the question carry 1/5 negative marks each and the rest of the questions carry 1/2 negative mark each. Unattempted questions carry no marks. What is the difference between the maximum and the minimum marks that he can score?
options:100,120,140,none
Please give me detailed solution as my answers are not matching book's answer. thank you...
1. 2(8!)-21(6!)=6!(2*7*8-21)=7*6!(16-3)=7*13*6!=13*7!......
In 14(7!) + 14(13!)...first term gives remainder 1 with 13*7! and second term gives remainder 0.....hence net remainder is 1.....
2. 4x+7y =3 with conditions x|
Case 1: x +ve and y -ve.....let x=500-p and y=-500+q, where p and q are +ve and less than 500.......
The equation becomes 7q-4p=1503......as 4p is always even, 7q has to be odd and so q has to be odd.....
7q has to be greater than 1503....so q has to be at least 215......but p is not an integer for this value......so q=217 is our first value and subsequent values of q occur with a difference of 4....max q can be 400.......so to find out max q we use following equation.....
217+4k
Case 2: x -ve and y +ve.....let x=-500+p and y=500-q, where p and q are +ve and less than 500.......
Equation obtained is 7q-4p=1497.......first value of q will be 215.....and using 215+4k
Therefore toal=71+72=143.....
3. None as number of questions is not given.....
please someone explain the below question:
two trains A and B start from station AA and BB respectively at same time towards each other. After passing each other they take 12 and 3 hours to reach BB and AA respectively. If A is traveling at speed of 48 km/hr, the speed of B is
a. 24km/hr
b. 22km/hr
c. 21km/hr
d. 96km/hr
e. 120 km/hr
Speed of train A = Va
speed of train B = Vb
so, Va/Vb = sqrt (Tb/Ta )
(Note : Tb & Ta are the time taken by B & A respectively to reach their destination after they pass each other)
= sqrt (3/12)
= sqrt (1/4)
= 1/2.
so, Vb = 2Va = 96 km/h
this rule can be apllied only when both the trains start at the same time in oppsite direction.
Please help me solve the following question from arun sharma book (Number systems) :
Q-1) What is the remainder when 2(8!)-21(6!) divides 14(7!) + 14(13!)?
options: 1,7!,8!,9!
Q-2)How many integer values of x and y are there such that 4x+7y =3, while xoptions:144, 141,143,142
Q-3) A candidate takes a test and attempts all the questions in it. While any correct answer fetches 1 mark, wrong answers are penalized as follows: one-tenth of the questions carry 1/10 negative mark each, one-fifth of the question carry 1/5 negative marks each and the rest of the questions carry 1/2 negative mark each. Unattempted questions carry no marks. What is the difference between the maximum and the minimum marks that he can score?
options:100,120,140,none
Please give me detailed solution as my answers are not matching book's answer. thank you...
q2. 4x + 7y = 3.
7y = 3-4x
y = (3-4x)/7
for x = -1 , y = 1
the next value of x for which y is an integer is 6.
so x = 6 , y = -3
next value of x = 13 , y = -7
so what we see here is that values of x satisfying the eqn are in A.P
where 1st term a = -1 & c.d d = 7.
therefore number of terms in a + (n-1)d = 499 (as max value of x can be 499 .)
put a = -1 and d = 7 we get n = 72.28.
therefore n = 72.
now going in -ve direction ..
when x = -8 , y= 5.
so again , take a = -8 and d = -7 in
a + (n-1)d = -499 (least value of x can be -499.)
solving we get n = 71.1
so n = 71.
so total integral values of x = 71 + 72 = 143.
same will be for y as for ever x we have a y.
hello'
I have joined career launchar for CAt prepration..
should I buy this book ,I m doing all my CL material????
hello'
I have joined career launchar for CAt prepration..
should I buy this book ,I m doing all my CL material????
yes you should buy, it is a good book with problems ranging from level-1 to level 3 and will provide enough practice 😃
hello'
I have joined career launchar for CAt prepration..
should I buy this book ,I m doing all my CL material????
No harm if u have... I would say it's a wonderful book..........
find the last two digits of (99)^45329
karthick vs Saysfind the last two digits of (99)^45329
can any1 pls explain me the solution for the above??
karthick vs Saysfind the last two digits of (99)^45329
(100 - 1)^45329
last two terms of binomial expansion are:-
C(45329, 1)*100 - 1
So, last two digits will be given by 900 - 1
So, 99 will be the last two digits
karthick vs Saysfind the last two digits of (99)^45329
write 99=(100-1)
on binomial xpansion...u will get "00" as last two digits upto second last term and "-1" as ur last term .so, last two digits would be 99.
Q. 99^45329 -- Last two digits
last two digits is the same as remainde when divided by 100.
now as 99 and 100 are co prime we can use Euler's theorem.
E(100) = 100*(1-1/2)(1-1/5) = 40. (Note : 100 = 2^2 * 5^2)
so dividing the power 45329 by 40 we get rwemainde as 9.
Since 100 shows a cyclicity of 40 i.e (99 ^ 40) % 100 = 1.
therefore, now the expression becomes
(1* 99^ 9 )%100
= ((100 -1) ^9 )% 100
= (-1)^ 9
= -1.
therefor the remainder is 100 -1 = 99.