Linear Races and Circular races

Linear Races

Linear Races :- when we have straight tracks for races.


Terms used to define linear races & their actual meanings:-
A gives B a start of 10 meters : B starts a race 10 meters ahead of A.

A gives B a start of 10 seconds : B starts 10 seconds before A.

A beats B by 10 meters : When A reaches finishing line, B is 10 meters behind.

A beats B by 10 seconds : B takes 10 seconds more than A to finish race.

A beats B by 10 meters or 10 seconds: B takes 10 seconds to cover 10 meters. Speed of B is 1 m/s.

Beat time : difference between time take by loser & winner.

Winner's distance: Length of the race track.

Time take by winner =  time taken by loser - beat time.

Dead Heat : Tie


Q1. X beats Y by 60 meters or by 12 seconds in a 2 Km race.Find Speed of X, Speed of Y, Time taken by X & time taken by Y.

Solution:-

Speed of Y = 60 / 12 = 5 m/s

Time taken by Y = 2000 / 5 = 400 minutes

Time taken by X = 400 - 60 = 340 minutes
Speed of X = 2000 / 340 = 5.88 m/s


Q2. In a 1000 m race, Neeta beats Geeta by 50 m & Seeta by 100 m. By what distance will Geeta beat Seeta ?
(1) 48.36 m (2) 50 m(3) 52.64 m (4) 51.28 m (5) 52.36 m


Solution :-
Distance covered by Neeta : Geeta in winning time by Neeta is = 1000 : 950 = 20:19.

Since time taken is same, ratio of their speed is also 20:19
Distance covered by Neeta : Seeta in winning time by Neeta is = 1000 : 900 = 20:18.
Since time taken is same, ratio of their speed is also 20:18
Hence Ratio of speeds of Geeta to Seeta would be 19:18.

Distance covered by Seeta when Geeta covers 1000 m = 1000 * 18/19 = 947.36
So Geeta beats Seeta by 1000 - 947.36 = 52.64 m. Hence answer option 3.


Q3. M beats N by 30 m or 5 seconds. Which of the following statements is/are true ?
I. Speed of N can not be found.
II. Speed of M can be found.
III. Distance covered by N can be found.
IV. N takes 6 more seconds to meet M.

1. Statement I is true.
2. Statement I, II & III are true.
3. Statement I & III are true.
4. All statements are true.
5. All statements are false.

Solution:-
By given data only speed of N can be found. To find other parameters we need length of track. Hence no statement is true. So answer option 5. 

Q4. A gives B a start of 15 seconds. A can run at speed of 80 mps & B can run at speed of 40 mps. In how much time will A  meet B on straight track after B starts race?
(1) 10 s (2) 12 s(3) 15 s  (4) 20 s (5) 30 s

Solution:-
Distance covered by B in early 15 seconds = 40 * 15 = 600 m
Relative speed of A & B = 80 - 40 = 40 mps
Time taken after A begin = 600 / 40 = 15 seconds.
Total time after B begins race = 15 + 15 = 30 seconds.Hence answer option 5. 

Q5. In above question, what would have been speed of A so that it would meet B after 2 mins after B begins his race ? (All other data remains same)
(1) 42.52 m/s(2) 45 m/s (3) 45.15 m/s (4) 45.71 m/s (5) 50 m/s

Solution:-
Relative Distance = 600 meters.
Time = 120 - 15 = 105 seconds
Relative Speed = 600 / 105 = 5.71 m/s
Speed of A = 40 + 5.71 = 45.71 m/s

Circular races

Circular Races : Circular races are on circular tracks where one can meet other person more than once.

When two persons A & B starts from same point at same time on a circular track then we can find

I. after how much time they meet for first time :- they meet for first time when one covers one more lap than other person. Relative distance would be length of track & using relative speed, time taken can be found.


II. After how much time they will meet for first time at starting point : this can be find out by taking LCM of time taken  by individual to cover one lap.


Q1. Two person X & Y start from the same point and move along a circular track of 60 m. Speed of X is 5 m/s & speed of Y is 7 m/s. After how much time will they meet for the first time ?

(1) 30 s (2) 15 s (3) 12 s (4) 16 s (5) Can not be determined


Solution:-

Since we don't know the directions of X & Y we can not determined answer. It is possible that they are running in same direction or they might be running in opposite direction. Hence answer option 5.



Q2. Two friends Raj & Rahul start a race on circular track of 240 m from same point in same direction at same time. Speed of the Raj is 20 m/s & that of Rahul is 25 m/s. After how much time will they meet for first time ?

(1) 10 s (2) 12 s (3) 24 s (4) 36 s (5) 48 s




Solution:-

Since same direction is same their related speed is : 25 - 20 = 5 m/s

Related distance to meet for first time is one lap of track = 240 m

Time taken : 240 / 5 = 48 seconds.




Q3. In above question, what would be time taken if they are running in opposite direction ?

(1) 3.33 s (2) 5.33 s (3) 8.33 s (4) 10 s (5) 12 s





Solution:-

Since opposite direction is same their related speed is : 25 + 20 = 45 m/s

Related distance to meet for first time is one lap of track = 240 m

Time taken : 240 / 45 = 5.33 seconds.



Q4. Two friends Raj & Rahul start a race on circular track of 500 m from same point in same direction at same time. Speed of the Raj is 20 m/s & that of Rahul is 25 m/s. After how much time will they meet for first time at starting point?

(1) 20 s (2) 25 s (3) 100 s (4) 200 s (5) 500 s


Solution:- 

Time taken to meet at starting point would be when both complete laps at same time. That is LCM of their time taken to complete track.

Time taken by Raj to complete track = 500/20 = 25 s

Time taken by Rahul to complete track = 500/25 = 20 s

Time taken to meet for first time at starting point = LCM(20,25) = 100 s.


Q5. In above question, what would be time taken to meet for first time if they are moving along circular track in opposite direction ?

(1) 20 s (2) 25 s (3) 100 s (4) 200 s (5) 500 s


Solution:- 

Time taken to meet at starting point would be when both complete laps at same time. That is LCM of their time taken to complete track.

Time taken by Raj to complete track = 500/20 = 25 s

Time taken by Rahul to complete track = 500/25 = 20 s

Time taken to meet for first time at starting point = LCM(20,25) = 100 s.


Meeting at starting point in circular races is independent of direction.



when more than 2 people are running in circular track. For e.g. 3 persons X, Y & Z.

I. after how much time they meet for first time :- It can be found by determining the time taken between two people & then between three.
II. After how much time they will meet for first time at starting point : LCM of time taken by X, Y & Z.

Q6. If X, Y & Z are starts their race by moving along a circular track of length 120 m from same point at same time in same direction. Find the time taken for them to meet for first time if speed of X is 2m/s, Y is 3m/s & that of Z is 5 m/s.
(1) 40s (2) 60s (3) 100 s (4) 120s (5) 240s

Solution:-
All three will meet only when X meets Y.
relative distance = 120 m
relative speed of X & Y = 3 - 2 = 1 m/s
time taken for them to meet for first time = 120 s


All three will meet only when X meets Z.
relative distance = 120 m
relative speed of X & Z = 5 - 2 = 3 m/s
time taken for them to meet for first time = 120 /3 =  40 s

All three will meet for first time when X meets Y & Z together for first time : LCM (120,40)= 120 s


Q7. in above question, what time they will meet for first time at starting point ?
(1) 40s (2) 60s (3) 100 s (4) 120s (5) 240s


Solution:- 

Time taken to meet at starting point would be when all complete laps at same time. That is LCM of their time taken to complete track.

Time taken by X to complete track = 120/2 = 60 s

Time taken by Y to complete track = 120/3 = 40 s
Time taken by Z to complete track = 120/5 = 24 s

Time taken to meet for first time at starting point = LCM(60,40,24) = 120 s.
It is independent of directions. Let them run in any direction. You don't worry whenever we are finding their first meet at starting point.


Q8. If X, Y & Z are starts their race by moving along a circular track of length 120 m from same point at same time. Y & Z are running in same direction while X is running in opposite direction. Find the time taken for them to meet for first time if speed of X is 2m/s, Y is 3m/s & that of Z is 5 m/s.
(1) 40s (2) 60s (3) 100 s (4) 120s (5) 240s

Solution:-
All three will meet only when X meets Y.
relative distance = 120 m
relative speed of X & Y = 3 + 2 = 5 m/s
time taken for them to meet for first time = 120/5 = 24 s


All three will meet only when Y meets Z.
relative distance = 120 m
relative speed of X & Z = 5 - 3 = 2 m/s
time taken for them to meet for first time = 120 /2 =  60 s

All three will meet for first time when Y meets X & Z together for first time : LCM (24,60)= 120 s.

Q9. If A overtakes B for the first time in the middle of 6th lap. Find ratio of speed of A to B. We know they started their race from same point at same time.
(1) 6:5 (2) 11:9 (3) 5:6 (4) 9:11 (5) Can not be determined.

Solution:-
Overtakes means same direction.
A overtakes B for first time when he covers 5.5 laps. Same time A would cover 4.5 laps.
Ratio of speeds = ratio of distance covered = 5.5 : 4.5 = 55:45 = 11:9

Number System questions 1

1. When x is a prime number                      a^x - a is always divisible by x.


Q1. What is remainder when 5¹⁰¹ - 5 is divided by 101.

(1) 0   (2)  100   (3) 5 (4)  50  (5) 20 

Solution :- As concept states, it is divisible by 101. Hence remainder is zero.



Q2. 91¹⁰¹ -91 is divisible by numbers except ?



(1) 91   (2) 13   (3) 7   (4) 101   (5) Divisible by all above


Solution :- As concept states, it is divisible by 101 hence option 5 is eliminated.

91¹⁰¹ -91 = 91(91¹⁰⁰ -1), hence divisible by 91

91 = 13 * 7 hence option 5.
------------------------------------------------------------------------------------------------------

2. When a base of a number increases, the number decreases and vice-versa.
Q3. (2345)₆ is equal to

(1) (569)₁₀ (2) (3458)₁₀  (3)  (4286)₁₀  (4) (569)₅ 

Solution :- Only option 1 follows concept. No need to solve.


------------------------------------------------------------------------------------------------------
 3. Relation between a two digit number and number formed using its digits.

    ab = 10a +b                     &
    ba = 10b + a

hence ab - ba = (10a +b) - (10b + a) = 9(a-b)
  &     ab + ba = (10a +b) + (10b + a) = 11(a+b)


Q4. The difference between a two digit number and number formed by reversing its digits is 45. What is difference between the digits of its number ?

(1)  9  (2)  5  (3)  3  (4)  4  (5) None of these


Solution :- According to concept 9(a-b) = 45 hence (a-b)  5. Hence option 2.
------------------------------------------------------------------------------------------------------
4.IMP - Divisible count in a range.
Steps to follow.
I. Subtract extreme numbers & divide by number (x)
II. Check remainder (n)
III. Check first n digits of range, if one of the number is divisible then add 1 to x.


Q5. How many three digit numbers are divisible by 7 ?

(1)  128  (2)  129  (3)  127  (4)  142  (5) 140

Solution :-
Extreme three digit numbers are 100 - 999
I. 999 - 100 = 899 the 899 / 7 = 128 * 7 + 3
II. Remainder = 3
III. None of the first three digits (100,101 & 102) are divisible by 7 hence answer is option 1.


Q6. How many four digit numbers are divisible by 7 ?

(1)  1284  (2)  1285  (3)  1286 (4)  1428  (5) 1287

Solution :-
Extreme four digit numbers are 1000 - 9999
I. 9999 - 1000 = 8999 the 8999 / 7 = 1285 * 7 + 4
II. Remainder =4
III.Out of the first four digits (1000,1001,1002 & 1003) number 1001 is divisible by 7 hence answer = 1285 + 1 = 1286. Hence option 3.
------------------------------------------------------------------------------------------------------
5. Product of three consecutive numbers is always divisible by 6.
------------------------------------------------------------------------------------------------------
6. To find highest power of a number in factorial .
I. Factorize the number.
II. Find the power of Individual factor in factorial by simply dividing number continuously with ignoring remainder. Add all the numbers.


Q7. Find highest power of 15 in 30!

(1)  2  (2)  6  (3)  21  (4)  7  (5) 14

Solution: Prime Factors of 15 = 3 * 5.
To find highest power of 15 just find highest power of 5.
Highest power of 5 = 30/5 + 6/5  (ignore remainders for next calculation)
= 6 + 1 = 7.
As proved highest power of 15 in 30! is 7.Hence answer option 4.


Q8. Find highest power of 12 in 30!.

(1)  3  (2)  26  (3)  14  (4)  13  (5) 12

Solution: Prime Factors of 12 = 2² * 3.

Highest power of 2 = 30/2 + 15/2 + 7/2 + 3/2  (ignore remainders for next calculation)
=15 + 7 + 3 + 1 = 26 => 13 pairs of 2.
Highest power of 2² = 13.


Highest power of 3 = 30/3 + 9/3 + 3/3   (ignore remainders for next calculation)
= 10 + 3 + 1 = 14.

As proved highest power of 12 in 30! is 13.Hence answer option 4.
------------------------------------------------------------------------------------------------------
 7. Sum of numbers of numbers formed using n digits.

Sum = Sum of digits * (n-1)! * 1111.... n times.
                    provided no digit is zero & digits are not repeated. 


Q9. Find sum of all digits formed using 2,3,4 & 6 such then digits are not repeated.

(1)  99990  (2)  62525  (3)  99900  (4)  98900  (5) None of these.

Solution :-
Since zero is not one of digits & digits are not repeated, we can use formula

Sum = Sum of digits * (n-1)! * 1111.... n times.

sum =  15 * 3!* 1111 = 99990. Hence option 1.

------------------------------------------------------------------------------------------------------
 8. All numbers which are perfect squares of odd numbers are of the form 8k+1 but vice-versa not true.
OR a perfect square of a odd number leaves a remainder of 1 when divided by 8.
------------------------------------------------------------------------------------------------------
9. A four digit number formed with its 2 digits repeating (abab) = ab * 101
     A six digit number formed with its 3 digits repeating (abcabc) = abc * 1001 = abc * 11 * 13 * 7


(Very frequently asked question pattern in any exams. Concept saves a lot of time)


Q10. Number 2828 is divisible by except

(1)  2  (2)  7  (3)  14  (4) 101   (5) 11


Solution :- I know your reaction would be "Wow i can answer this without calculations, answer is option 5.



Q11. Number 123123 is divisible by except

I. 123
II. 11
III. 77
IV. 143
V. 91


(1) Statement I, II but not III, IV.
(2) Statement II, III but not I.
(3) Statement I, II, IV but not III.
(4) Statement I, II, III, IV but not V.
(5) Divisible by all



Solution :
123123 = 123 * 1001 = 123 * 7 * 11 * 13 = 123 * 7 * 143 = 123 * 77 * 13 = 123 * 91 * 11
hence answer option 5.
------------------------------------------------------------------------------------------------------
10.Number of Factors of a number is determined by power of its prime factors.


Number form in prime factors = a^x * b^y * c^z  ....
where a, b , c ... are prime numbers. then

Number of factors = (x+1)(y+1)(z+1)....


Q12. Find the number of factors of number 96 ?

Solution:  N = 96 = 2⁵ *3¹   
 Number of factors = (5+1)(1+1) = 6 * 2 = 12


Note:
1. In a perfect square power of prime factors are always even and they always have odd number of factors.
2. In a perfect cube power of prime factors is always multiple of three.


Q13. How three many numbers with odd number of factors are divisible by 3 ?

(1)  7  (2)  21  (3)  149  (4) 150   (5) 299

Solution : Only perfect squares have odd number of factors. And three digit perfect squares divisible by 3 are 144, 225, 324, 441, 576, 729 & 900. Hence option 1.

Q14. Which of the following can not be number of factors of a number which is a perfect cube ?
(1)  4  (2)  16  (3)  7 (4) 28   (5)9

Solution : Power of prime factors of a perfect cube are always multiple of 3.



let a & b any prime factors. then



a³ then number of factors = 3 + 1 = 4

a⁶ then number of factors = 6 + 1 = 7
a³ * b³ then number of factors = 4 * 4 = 16
a⁶ * b³ then number of factors = 7 * 4 = 28
hence option 5.

Or We know that only perfect square can have odd number of factors so now we are only left with option 3 & 5. We can easily find the number a⁶ is a perfect cube as well as perfect square. Hence option 5.


Q15. Find the number of factors in 6⁶ - 5⁶
 (1)  4  (2)  8  (3)  16 (4) 36   (5) 49


Solution:
6⁶ - 5⁶  =  (6³)²-  (5³)²
 =  (216)²-  (125)²   =   91 * 341 = 13 * 7 * 31 * 11

Number of factors = 2 * 2 * 2 * 2 = 16. Hence answer option 3.
------------------------------------------------------------------------------------------------------
11. Number of ways in which a number can be written as a product of two co-prime numbers = 2^M-1. where M is number  of prime factors.

Q16. In how many number of ways 540 can be written as a product of two co-prime numbers ?
(1)  1  (2)  2  (3)  3 (4) 4   (5) 6


Solution:- 540 = 2² * 3³ * 5¹
 Number of ways in which a number can be written as a product of two co-prime numbers
= 2^{M-1 =} 2^M-1  = 2^3-1 = 4. Hence option 4.
------------------------------------------------------------------------------------------------------
12. A perfect square will never end in 2,3,7 &8.

CET & SNAP Pattern
 Q17. Which of the following is not a perfect square ?
 (1)  2025  (2)  3844  (3)  6568 (4)  7056  (5) 3025


Solution: - As concept states answer is option 3. ( Don't try to calculate in exams.)
 ------------------------------------------------------------------------------------------------------
13. A square of n digit number will have either 2n or 2n-1 digits.

Q18. Which of the following number , with which some of digits expressed as 'x' can be a perfect square of 5 digit number ?

(1) 30xxxxxxxxx8 
(2) 30xxxx25
(3) 4xxxxxx6
(4) 1xxxxxxx0
(5) 4xxxxx25


Sol:- Square of a 5 digit number will have 9 digits or 10 digits. Hence option 2, 3 & 5 are eliminated.
A square can not end with 8 so option 1 is eliminated. Hence option 4 is answer.
 ------------------------------------------------------------------------------------------------------
14. Series  type.

I. Sum of first 'n' natural numbers = n(n+1) / 2
II. Sum of first 'n' odd numbers = n²
III. Sum of first 'n' even numbers = n² + n
IV. Sum of the square of first 'n' natural numbers = n(n+1)(2n+1)/6
V. Sum of the cube of first 'n' natural numbers = (n(n+1) / 2)²

Q19. What is sum of all three digit numbers lesser than 251 ?


(1)  4950  (2)  14825  (3)  26250 (4)  26425  (5) 31375

Sol:-
Sum of all digits from 1 - 250(Series formula I) = 250 * 251 / 2 = 31375
Sum of all digits from 1-99 = 99 * 100 / 2  = 4950
Sum of all three digits numbers less than 251 i.e. 100-250 = 31375 - 4950 = 26425
Hence answer option 4.


Or
Total number between 100 - 250 = 151
Avg of numbers = 175
sum of numbers = 151 * 175 = 26425


Q20. What is sum of all numbers which are perfect cube and less than 8001  ?
(1) 21000  (2)  32600  (3)  36800 (4) 42100   (5) 44100
Solution :-
8000 is cube of 20. Hence question is asking of sum of all first 20 cubes. Which can be calculated using 5th formula.
sum = (20(20+1) / 2)²
= (20(21) / 2)²
= 210 * 210
= 44100. Hence answer option 5.
 ------------------------------------------------------------------------------------------------------

Indices and Surds

Indices

1. If x^N is given then x is the base and N is the index or power or exponent
2. A³ means A multiplied with itself 3 times i.e. A x A x A
3. A ^p X a ^q = a ^(p+q)
4. A ^p / A ^q = A ^(p-q)
5. (A ^p)^q = A ^pq
6. A ^–p = 1 / A^p
7. p√a = a ^1/p i.e. pth root of a
8. (ab)^p=a^p x b^p
9. A⁰ = 1 (provided a ≠0)
10.  A¹ = a
11. _A mⁿ = a ^p, where p = mⁿ ie. A raised to the base m raised to the power
12. If a ^p = b ^p, then if p is ≠0, then a = b, if p is odd and [a = b or a = (-b) if p is even]
13. If a ^p = a ^q and a ≠ 0 or -1, then p = q
14. A ^-1 = 1/A
15. (A/B)^-1 = B/A
16.  (A)^m/n = (n√A)^m
17.  √A X √B = √(AB)

Surds

1. They are irrational numbers
2. When an irrational number is simplified, the remainder which cannot be simplified and is normally expressed in the form of square root is called a surd.
3. Normally for exam questions, number whose square root cannot be further found out as a perfect rational number are surds. 4 is not a surd, as square-root of 4 is 2, where 2 is a surd as square root of 2 is 1.414… which is not a rational number.
4. To solve simplification problems regarding surds, square the numbers
5. For 1 / (p + √q) or 1 / (p + √q + √r) kind of problems, to simplify, multiply by the conjugate, which is (p – √q) or (- p + √q) for 1^st case and (p + √q – √r) or (p – √q + √r) for the 2^nd case
6.  = .555555 hence, whenever in a decimal form there is a repeated number; a dot is mentioned over it.
7. Rationalising Surds:
   When you have a fraction where both the nominator and denominator are surds, rationalising the surd is the process of getting rid of the surd on the denominator. To rationalise a surd you multiply top and bottom by fraction that equals one. Take the example shown below
   1/√2
   To rationalise this multiply by effectively 1
   1/√2 * √2 /√2
   Can you see why √2 /√2 was chosen? This is because √2 * √2 = 2 so the denominator becomes surd free.
   
   For a more complex term
   
   
   
   
   Rationalizing the surd now
   
   
   
   

Practice-Divisibility Rules 1

 Practice-Divisibility Rules (10 Qs)

__________________________________________________________________________________

q1. A number when divided by 13 gives a natural number, but when divided by 23 gives a real number. The smallest such number is

(1) 0         (2)   1     (3)   437    (4)  13   (5)   23

Sol:-Option 4. Very easy one. To get a natural number it has to be least 11. Real number can be a fraction hence answer option D.

__________________________________________________________________________________

q2. The  number 276A621B is divisible by 9. How many possible values can A+B take ?

(1) 0         (2)  1     (3)   2    (4)  3   (5)  Infinite

Sol:- Option 3.  To be divisible by 9, sum of digits must be divisible by 9.

sum of digit = 2+7+6+A+6+2+1+B = 24+B

hence A+B can be 3 or 12. Can't be more as A & B can max be 9.

__________________________________________________________________________________

q3. A number 2524232221..........109876543210 is divided by 8. What is remainder ?

(1) 0         (2)  2     (3)   4    (4)  6   (5) 1

Sol:- Option 2. To check divisibility by 8, we are only concerned about last 3 digits. Remainder of 210 divided by 8 is 2.

__________________________________________________________________________________

q4.  A number is divisible by 12 and 10.Which of the following is/are false ?

 I. Number is also divisible by 15.

II. Number is also divisible by 30

III. Smallest such number is 120.

IV. Number is always co-prime with 7.

(1) Statement I & II

(2)  Statement III

(3) Statement IV

(4)  Statement III & IV

(5) None of the statement.

Sol:-Option 4.

Number is divisible by 12 & 10 hence divisible by 2,3,4,5,6 and their product.

Statement I - True as it is divisible by 3*5.

Statement II - True as it is divisible by 2*3*5. (Also If Statement 1 is true, this one has to be true)

Statement III - False, smallest such number is 60.

Statement IV - False, 7 is not factor of 12 or 10 but still a number can be formed eg. 840

 __________________________________________________________________________________

q5. 14MN27 is divisible by 9. What can be maximum value of M if (M+N) is at its minimum value ?

(1) 0         (2)  2     (3)  4    (4)  8   (5) 9

Sol:- Option 3. Number is divisible by 9 hence M+N can be 4 or 13. As M+N is at its minimum,  M can maximum be 4 when N is zero. __________________________________________________________________________________

q6. How many four digit numbers can be formed using  the digits 2,3,5,7 exactly once, such that number is divisible by 25 ?

 (1) 0         (2)  1     (3) 2    (4)  3   (5) 4

Sol :- Option 5. Number should have last two digits as 00, 25, 50 or 75 to be divisible by 25. Such possible numbers are 3725, 7325, 2375 & 3275.

__________________________________________________________________________________

q7.  If a positive integer n is divided by 7, the remainder is 4. Which of the numbers below yields a remainder of 0 when it is divided by 7?

(1) n + 3         (2)  n + 2     (3) n - 1    (4)  n - 2   (5) n + 1

Sol:-  n divided by 7 yields a remainder equal to 4 is written as follows
n = 7 k + 4 , where k is an integer.
add 3 to both sides of the above equation to obtain
n + 3 = 7 k + 7 = 7(k + 1)
The above suggests that n + 3 divided by 7 yields a remainder equal to zero. The answer is Option 1.

__________________________________________________________________________________

q8.If n is an integer, when (2n + 2)² is divided by 4 the remainder is

  (1) 0         (2)  1     (3) 2    (4)  3   (5) 4

Sol:-
We first expand (2n + 2)²
(2n + 2)² = 4n ² + 8 n + 4
Factor 4 out.
= 4(n ² + 2n + 1)
(2n + 2)² is divisible by 4 and the remainder is equal to 0. The answer is Option 1.

 __________________________________________________________________________________

q9.Which of these numbers is not divisible by 3?

 (1) 339      (2)  342     (3) 552    (4)  1111   (5) 672

Sol:-   One may answer this question using a calculator and test for divisibility by 3. However we can also test for divisibility by adding the digits and if the result is divisible by3 then the number is divisible by 3.

3 + 3 + 9 = 15 , divisible by 3.

3 + 4 + 2 = 9 , divisible by 3.

5 + 5 + 2 = 12 , divisible by 3.

1 + 1 + 1 + 1 = 4 , not divisible by 3.

The number 1111 is not divisible by 3 the answer is Option 4.

__________________________________________________________________________________

q10.  How many 4 digit numbers can be formed using 5, 8, 3, 4 exactly once, such that the number is divisible by 11 ?

 (1) 16     (2) 5     (3) 1     (4) 22     (5) None of these

Sol:- 

Dividing by 11
The difference between the sum of the odd numbered digits (1st, 3rd, 5th...) and the sum of the even numbered digits (2nd, 4th...) is divisible by 11.

No such number can be formed. Hence Option 5.

Cyclicity & Remainders (for CAT)

Number Cyclicity:-

Number cyclicity is very important for every chapter of the quant. Any exams will definitely need the basics of number cyclicity. Questions can directly be asked on this topic and you will need this to solve many questions on other topics.

Don’t forget time is biggest constrain for any entrance exam and cyclicity will have one to save time.

Cyclicity is basically use to find the unit digit or tens digit of the number.

Unit Digit Cyclicity:-

Q1. Find the unit digit of 2^2548.

Sol: - You will need more than 6 hours to solve this problem if you don’t use cyclicity theorem.

We notice that

2^1 end with 2

2^2 end with 4

2^3 end with 8

2^4 end with 6

2^5 end with 2

2^6 end with 4

2^7 end with 8

2^8 end with 6

We notice that 5^th power end in 2 and number repeats after 4 powers. Hence cyclicity for 2 is 4. It will always end with 2, 4, 6 and 8.

So Divide 2548 by 4 and we get remainder = 2

Hence unit digit of 2^2548 with be 4.

Remember:-

When exponent is 1 number ends with 2

When exponent is 2 number ends with 4

When exponent is 3 number ends with 8

When exponent is 0 or 4 number ends with 6.

Similarly we can find of all other numbers.

Number	Unit digit of Cyclicity
0,1,5,6	1
4,9	2
2,3,7,8	4

Tens Digit Cyclicity:-

Similarly we can arrive for tens digit cyclicity.

Number Tens Digit Cyclicity

Number	Unit digit of Cyclicity
0,1,5,6	1
7	4
6	5
4,9	10
2,3,8	20

Base System

Numbers : Basic Operations on Numbers

Let a and b any two numbers then find the results of basic operations below.

Points to remember:-
1. Sum and Difference of any two odd or two even numbers is always even.
2. a^b always depends on a.
ie Even raise to any number is even and Odd raise to any number is odd.
eg 2^ 3 = 8 and 3^ 2 = 9.
3. If atleast one number is even - Product is always even.