Rhombus

Properties of a Rhombus

• Opposite angles of the rhombus have the equal measure.
• The diagonals of the rhombus intersect each other at right angles.
• Rhombus has two diagonals that connects the opposite pair of vertices.
• Rhombus is symmetric along the diagonals.
• Every rhombus is a parallelogram.
• All the parallelograms are not a Rhombus.
• Dual polygon of a rhombus is called as a Rectangle.
• Sum of the adjacent sides of any rhombus is equal to 180 degree. That is the adjacent angles of a rhombus are supplementary.

Area of a Rhombus

Area of a Rhombus is same as that of the area of a parallelogram. Area is the product of the base and the height.
The Area of a Rhombus Formula is,
A=b * h
The base of the rhombus is the length of one of its sides and the height is the perpendicular distance between the opposite sides.
Perimeter of a Rhombus

The perimeter of a rhombus is the total sum of all the side lengths. All the sides are having equal length in a Rhombus. And so, Perimeter of a Rhombus Formula is,
Perimeter = 4s, where s is length of the sides.

Factorial upto 10

n	n!
0	1
1	1
2	2
3	6
4	24
5	120
6	720
7	5,040
8	40,320
9	362,880
10	3,628,800

DATA SUFFICIENCY questions

Mark A) If the question can be answered with statement I alone but not statement II alone, or can be answered with statement II alone but not statement I alone
Mark B) If the question cannot be answered with statement I alone or with statement II alone, but can be answered if both statements are used together
Mark C) If the question can be answered with either statement alone
Mark D) If the question cannot be answered with the information provided 

1. A point P is identified as P(m,n). What is the ratio AP:BP given that the points A and B are identified as A(5,-4) and B(1,6)?

Stmt 1. m = 3. Not sufficient.
Stmt 2. n = 2.5 Not sufficient.

Either statement alone is not sufficient. Both put together, we can answer the questions. Ans B

2. A point P is identified as P(m,n). What is the ratio AP:BP given that the points A and B are identified as A(5,-4) and B(5,6)?

Stmt 1. m = 5. Not sufficient. It tells us that all points lie on the line x = 5, but this is not enough
Stmt 2. n = 3 Not sufficient.

Both put together, this is enough.

In the above question, would n = 1 have been sufficient? Think about that.

3. A survey of 100 people tried to find the number of people who can write with both their left and right hands. What is the maximum number of people who could write left-handed and right-handed?

Stmt 1. 50 people can write only with their left hand. 40 people can write only with their right hand. Sufficient: Maximum of 10 people could write left-handed and right-handed.
Stmt 2. 50 people can write with their left hand. 40 people can write with their right hand. Sufficient: Maximum of 40 people could write left-handed and right-handed.

Answer Choice C

4. What is the slope of a line?

Stmt 1: The line makes 135 degrees with the negative direction of x - axis. Sufficient: The line makes 45 degrees with positive x axis. This should be enough.
Stmt 2: The line makes an isosceles right triangle with the coodinate axes and the product of the intercepts is negative. Sufficient: Either both intercepts are positive and equal or negative and equal. Slope = -1
Answer Choice C 

DATA SUFFICIENCY

“The ultimate goal of mathematics is to eliminate any need for intelligent thought.”-A. N. Whitehead
Well, for starters try to think what’s the meaning of the above quote, it has a nice and beautiful meaning. To make things so simple that an average mind is able to see it. So lets put this into practice, with another concept lesson.
Today, we will discuss DATA SUFFICIENCY, one of the very scoring problems in cat and other mba entrance exams. The good thing about DS is we get DS both in quant and DI, and it makes for 6-8 problems in almost every paper. As there are no theorems in DS, we will take things to note
Things To Note:
1) DS problems, we need to answer if it is sufficient information to answer, means, there should be one conclusive answer.  We do not need to find the answer, just if it can be found or not?
2) Some questions ask , is this true? So if we can find that the information available is enough to prove that it is not, we are still able to answer the question, that it is not true. Hence we are able to answer the question.
3) Check for all possibilties, that is using one statement, using second, then only combine the two.

Permutation and Combination

Conversion : Distance and weight

1 mile = 1760 yards
1 yard = 3 feet
1 mile² = 640 acres
I gallon = 4 quarts
1 quart = 2 pints
1 pint = 2 cups
1 cup = 8 ounces
1 pound = 16 ounces
1 ounce = 16 drams
1 kg = 2.2 pounds

Time and Work - Examples

Q
If A can do a work in 10 days, B can do it in 20 days and C in 30 days in how many days will the three together do it?

Soln:
The efficiencies are A = 1/10, B = 1/20 and C = 1/30
So work done per day by the three = 1/10 + 1/20 + 1/30 = 11/60 => No of days = 60/11 = 5.45 days.


Q
If A and B can do a work in 10 days , B and C can do it in 20 days and C and A can do it in 40 days in what time all the three can do it?

Soln:

A+B = 1/10
B+C = 1/20
C+A = 1/40
Adding all the three we get 2(A+B+C) = 7/40 => A+B+C = 7/80 => No of days = 80/7 days.

If A can do a work in 12 days, B can do it in 18 days and C in 24 days. All the three started the work. A left after two days and C left three days before the completion of the work. How many days are required to complete the work?

Soln:

Let the total no of days be x.

A worked only for 2 days, B worked for x days and C worked for x-3 days.

So, mA + nB + oC = 1
ð      2(1/12) + x(1/18) + (x-3)(1/24) = 1
ð      12 + 4x + 3(x-3) = 72
ð      x = 69 / 7 days.

Note:

The ratio of dividing wages = ratio of efficiencies = ratio of parts of work done

Q:

A can do a work in 10 days and B can do it in 30 days and C in 60 days. If the total wages for the work is Rs. 1800 what is the share of A?

Soln:

Ratio of wages = 1/10 : 1/30 : 1/60 = 6 : 2 : 1  (Multiplying each term by LCM 60)

So total 9 equal parts in Rs. 1800 => each part = Rs. 200 => share of A = 6 parts = Rs. 1200.
Applying the same logics to pipes and cistern

Q:

A pipe can fill a tank in 5 hrs but because of a leak a the bottom it takes 1 hr extra. In what time can the leak alone empty the tank?

Soln:

Let the filling pipe be A.
A = 1 / 5.

But with the leak L,  A – L = 1 / 6   ( A-L because leak is outlet)

So, 1/L = 1 / 5 – 1/ 6 = 1/30 => Leak can empty the tank in 30 hrs.

Q:

A pipe A can fill the tank in 10 hrs, B can fill it in 20 hrs and C can empty in 40 hrs. All are opened at the same time. After how many hours shall the pipe B be closed such that the tank can be filled in 10 hrs?

Soln:

Let the pipe B be closed after x hrs.

Then A worked for 10 hrs, B worked for x hrs and C worked for 10 hrs.

mA + nB – oC = 1    (since C is outlet)

10(1/10) + x(1/20) – 10(1/40) = 1

x = 5 hrs.

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.
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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.


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 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.
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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.
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5. Product of three consecutive numbers is always divisible by 6.
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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.
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 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.

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 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.
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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.
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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.
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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.
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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.)
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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.
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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.
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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.

Numbers : Converting a recurring decimal to vulgar fraction

A decimal with recurring value is called recurring decimal.
E.g: 2/9 will give 0.22222222...... where 2 is recurring number.

Method:
a) Separate the recurring number from the decimal fraction.
b) Annex denominator with "9" as many times as the length of the recurring number.
c) Reduce the fraction to its lowest terms.

Example: Consider 0.2323232323

Step a: The recurring number is 23
 Step b: 23/99 [the number 23 is of length 2 so we have added two nines] 
Step c: Reducing it to lowest terms : 23/99 [it can not be reduced further].

How to Convert a mixed-recurring decimal to vulgar fraction ? A decimal with both recurring and non-recurring value is called mixed recurring decimal.
E.g: 28/25 will give 1.1199999999...... where 11 is non-recurring number and 9 is recurring number.

Method:
a) Separate the recurring number, non recurring number from the decimal fraction.
b) Round the decimal after point to the first recurring value.
c) Result of step b - non recurring number.
d) Annex as many "0" as non-recurring number length and as many "9" as recurring number length.
e) Step c / Step d
f) Add the fraction with the number before decimal point.
Example: Consider 1.11999999...

Step b and c: 119-11 [rounded value of number after decimal point - non recurring value]Step d: 900
Step e: 108/900 [c/d]
Step f: 1+¹⁰⁸/₉₀₀ [adding with number before decimal point ]

Reducing it to lowest terms : 900+108 / 900 = 1008/900 = 28/25.Step a: The recurring number is 11, non-recurring number is 9

Special Number - Ramanujam’s Number

It is the smallest number which can be written as the sum of two cubes in two different ways. It is a very famous number, so if you dare to proclaim maths as your hobby, you must know it.
Number is 1729=10^3+9^3=1^3+12^3.

Ramanujam Number is 1729.

But allowing negative perfect cubes (the cube of a negative integer) gives the smallest solution as 91 (which is a factor of 1729):

91 = 6^3 + (−5)^3 = 4^3 + 3^3

There is one interesting story on this. Do google search if you have enough time.

Clocks funda for cat

I. 1 minute space = 6° degrees.

II. 60 minute space = 360°

III. In one hour, minute hand moves 60 minutes spaces where hour hand moves 5. Hence In an hour, minute hand gain 55 minute space over hour hand.  

IV. In 1 minute, hour hand moves 0.5° and minute hand moves 6°. Hence in 1 minute, the minute hand gains 5.5° or 11/2 degrees over the hour hand.

V. The hour & the minute hand coincide every 720/11 minutes. 

Incorrect Clock :-

I. It is said to gain time (fast) if hour hand & minute hand coincide in less than 720/11 minutes.

II. It is said to lose time (slow) if hour hand & minute hand coincide in more than 720/11 minutes.

Q1. After 5 O' clock, at what time will the hour & minute hand of the clock coincide for the first time ? (Time in HH:MM:SS format)

(1) 05:22:16 (2) 05:24:27 (3) 05:27:16 (4) 05:27:27 (5) 05:28:27

Solution :- 

At five of clock distance between hour and & minute hand is 150°
Speed of hour hand is 0.5° per minute.
Speed of minute hand is 6° per minute. 
Relative speed of minute hand & hour hand - 6°- 0.5° = 5.5° per minute
Time taken to meet = 150 / 5.5 = 27.27 minutes = 27 minutes 16 seconds

Q2. What is smallest angle between the minute hand & hour hand at 10.30 ?
(1) 225° (2) 120° (3) 132° (4) 315° (5) 135°

Solution:-
At 10 O' clock distance will be 300°.
In next 30 minutes, minute hand will move 180°(6° per minute) & hour hand will move 15°(0.5° per minute) in clockwise direction. Angular distance between them will be 135° or 225°. Hence answer is 135°. Answer option 5.

Q3. Minute hand & hour hand coincided after 60 minutes. Which of the following options is true ?
(1) Clock is gaining time.
(2) Clock is losing time.
(3) Clock is neither losing nor gaining time.
(4) It will again coincide after 61 minutes.


Solution :- We know that clock is said to gain time (fast) if hour hand & minute hand coincide in less than 720/11 minutes. Hence option 1.


Q4. A watch was running 8 minutes behind time on Sunday noon on date X. It was running 13 minutes ahead of time on next Sunday noon. After how much time after noon on X the clock was showing correct time ?
(1) 24 hours (2)  32 hours (3)  64 hours (4)  76 hours (5)  108 hours

Solution:-
First Sunday Noon : - 8 minutes
Second Sunday Noon : + 13 minutes
Time to cover in week time(168 Hours) = 21 minutes.
Time to gain from first Sunday noon to show correct time = 8 minutes.
21 minutes gain takes 168 hours.
8 minute gain will take = 168 * 8 / 21 = 64 hours from first Sunday noon. Hence answer option 3.

Time and distance

TIME & DISTANCE :

• Distance = Speed * Time
• 1 km/hr = 5/18 m/sec
• 1 m/sec = 18/5 km/hr
• Suppose a man covers a certain distance at x kmph and an equal distance at y kmph. Then, the average speed during the whole journey is 2xy/(x+y) kmph
• Likewise, travelling 3 equal distance with speeds x,y and z, average speed = 2xyz / (xy+yz+zx)
• Time taken after meeting (same distance):-
  if 2 persons starts journey from A & B and move towards each other and meet at C, after meeting they reach opposite points in x & y hrs respectively.
  
  
  then speed before they meet is  √y : √x
  Q. Seeta start her journey from Mumbai towards Pune at speed of  20 kmph & her sister Geeta starts her journey from Pune towards Mumbai at speed of 30 kmph. They meet at Lonavala. What is ratio of time taken by them to reach their destinations after they meet ?
  (1) 9:4 (2) 2:3 (3) 3:2 (4) 1:1 (5) Data insufficient
  Solution :-
  Speed before they meet is x:y i.e. 20:30
  time after they meet will be y²:x² = 900 : 400 = 9:4. Hence answer option 1.
   

Set theory for cat

27) n(A U B U C) = n(A) + n (B)+ n(C) – n(A n B) – n(A n C) – n(B n C) + n(A n B n C)
28) n(A_only) = n(A) – n(A n C) – n(A n B) + n(A U B U C)
To refresh, the union of sets is all elements from all sets. The intersection of sets is only those elements common to all sets. Let’s call our sets A, B, and C. If n = intersection and u = union. The need-to-know formulas:
P(A u B u C) = P(A) + P(B) + P(C) – P(A n B) – P(A n C) – P(B n C) + P(A n B n C)
To find the number of people in exactly one set:
P(A) + P(B) + P(C) – 2P(A n B) – 2P(A n C) – 2P(B n C) + 3P(A n B n C)
To find the number of people in exactly two sets:
P(A n B) + P(A n C) + P(B n C) – 3P(A n B n C)
To find the number of people in exactly three sets:
P(A n B n C)
To find the number of people in two or more sets:
P(A n B) + P(A n C) + P(B n C) – 2P(A n B n C)
To find the number of people in at least one set:

P(A) + P(B) + P(C) – P(A n B) – P(A n C) – P(B n C) + 2 P(A n B n C)
For questions involving set theory, it may be helpful to make a Venn diagram to visualize the solution.
To find the union of all set: (A + B + C + X + Y + Z + O)
Number of people in exactly one set: (A + B + C)
Number of people in exactly two of the sets: (X + Y + Z)
Number of people in exactly three of the sets: O
Number of people in two or more sets: (X + Y + Z + O)

Conversion : Distance and weight

Distance
1 mile = 1760 yards
1 yard = 3 feet
1 mile = 1.6 km (nearly)

Area
1 mile² = 640 acres

Speed
1 km/hr = 5/18 m/sec
1 m/sec = 18/5 km/hr

Volume
I gallon = 4 quarts
1 quart = 2 pints
1 pint = 2 cups
1 cup = 8 ounces
1 pound = 16 ounces
1 ounce = 16 drams
1 kg = 2.2 pounds

Conversion : Money

Dollar
1 Nickel = 5 cents
1 dime = 10 cents
1 quarter = 25 cents
1 half = 50 cents
1 dollar = 100 cents

Quadratic Equation

Quadratic equations looks like: ax² + bx + c = 0

where a,b,c are real numbers, and a ≠ 0. Every quadratic equation can have 0, 1 or 2 real decidions derived by the formula:

The number D = b² - 4ac is called discriminant.

If D < 0 then the quadratic equation have no decidions. If D = 0 then the quadratic equation have 1 decidion x = - ^b/_2a. If D > 0 then the quadratic equation have 2 decidions.

Example:

If we have equation: x² + 3x - 4 = 0

a = 1, b = 3, c = -4

Parabola

The graph of a quadratic equatin is called a parabola.

If a > 0 then graph horns pointing down:

if a < 0 then graph horns pointing up:

The midpoint of any parabola is the point x = -b/2a.

Sign of Quadratic Equation

Let f(x) = ax^2 + bx + c , where a,b,c &# 949; R and a ≠ 0

Vieta's formulas

If x₁ and x₂ are the roots of the quadratic equation ax² + bx + c = 0 then:

These formulas are called Vieta's formulas.

We can find the roots x₁ and x₂ of a quadratic equation by solving the system above.

Maxima and minima

Maxima a>0  x= - b/2a  Max = -(b^2 -4ac)/4a

Minima a<0  -----------------------------------------(same)

AP GP and HP

An arithmetic progression is a sequence of numbers such that the difference of any two successive members of the sequence is a constant.
For example, the sequence 3, 5, 7, 9, 11,... is an arithmetic progression with common difference 2.

Arithmetic progression property:

a₁ + a_n = a₂ + a_n-1 = ... = a_k+a_n-k+1

Formulae for the n-th term can be defined as:

a_n = ¹/₂(a_n-1 + a_n+1)

If the initial term of an arithmetic progression is a₁ and the common difference of successive members is d, then the n-th term of the sequence is given by

a_n = a₁ + (n - 1)d, n = 1, 2, ...

The sum S of the first n values of a finite sequence is given by the formula:

S = ¹/₂(a₁ + a_n)n, where a₁ is the first term and a_n the last.

or

S = ¹/₂(2a₁ + d(n-1))n

 

GP

 

S = a + a.r + a.r² + a.r³ + ... + a.r^(n-1) [Equation 1]

Now multiply each side of the equation by r:

S.r = a.r + a.r² + a.r³ +... + a.r^(n) [Equation 2]

Now subtract Equation 1 from Equation 2: [Note that the terms from a.r to a.r^(n-1) are in both series and cancel out]

S.r - S = a.r^n - a

S.(r-1) = a.r^n - a

= a(r^n - 1)

So,

S = a.(r^n - 1) ÷ (r - 1)

In cases like Example 2, where r < 1, it is often written as:

S = a.(1 - r^n) ÷ (1 - r)

Convergence of GP

If |r| < 1 then a_n -> 0, when n -> ∞ So the sum S of such a infinite geometric progression is:

S =  1 / (1-r)

which is valid only for |r| < 1

 

Formulae

ü The sum of first n natural numbers = n(n+1)/2

ü The sum of squares of first n natural numbers is n(n+1)(2n+1)/6

ü The sum of cubes of first n natural numbers is (n(n+1)/2)²/4

ü The sum of first n even numbers= n (n+1)

ü The sum of first n odd numbers= n²

CALENDAR notes

• Calendar repeats after every 400 years.
• Leap year- it is always divisible by 4, but century years are not leap years unless they are divisible by 400.
• Century has 5 odd days and leap century has 6 odd days.
• In a normal year 1st January and 2nd July and 1st October fall on the same day. In a leap year 1st January 1st July and 30th September fall on the same day.
• January 1, 1901 was a Tuesday.
  

Cram misc

 

Certain numbers which didn't occurred in previous cramming posts are here:

 

• 210 = 45 = 322 = 1024
• 38 = 94 = 812 = 6561
• 7 * 11 * 13 = 1001
• 11 * 13 * 17 = 2431
• 13 * 17 * 19 = 4199
• 19 * 21 * 23 = 9177
• 19 * 23 * 29 = 12673
  

Number of solutions of linear equations

Consider the two equations

a₁x+b₁y=c₁

a₂x+b₂y=c₂

Then,

ü If a₁/a₂ = b₁/b₂ = c₁/c₂, then we have infinite solutions for these equations.

ü If a₁/a₂ = b₁/b₂ <> c₁/c₂, then we have no solution.

ü If a₁/a₂ <> b₁/b₂, then we have a unique solution.

Finding the cube roots

Finding Cube Roots requires some background

Background	Last digit
13 =	1	1
23 =	8	8
33 =	27	7
43 =	64	4
53 =	125	5
63 =	216	6
73 =	343	3
83 =	512	2
93 =	729	9

From the above illustration we can take out that last digit of 2³ is 8, 3³ is 7 and vice-versa. All other repeats itself.

Procedure of finding a cube: -

• Start from right and put a comma when three digits are over

Examples: -

9,261

1,728

32,768

175,616

• After putting the comma see the last digit of the number; compare that with table provided above. You get the last digit.
• Now see the first group of numbers and ascertain cube of which number is less than the group. That number is your first digit.
• You have thus found first digit and last digit.

Let us take an example: -

Steps: -

• Counting from last we put comma after 9.
• By seeing the last digit we ascertain that last digit of cube root will be 1.
• Now we see 9 and ascertain that 2³ = 8, is less than 9 and 3³ = 27 is more.
• Our first digit thus comes to 2, and the answer is 21.

Another Example: -

• By seeing last digit we find last digit of cube root is equal to 2.
• By seeing 32 we put 3, as our first digit as 3³ = 27 is less than 32 and 4³ = 64 is more.
• Our answer is 32.