You should see the ans B to see if thats ticking you by having the same equation
Inequality
1.) is x-y>10, is x-y>x+y
a. x=8
b. y =-20
taking A alone.. -y>2 => y<-2 hence y is -ve hence x-y>x+y suff
taking B... y=-20, => x>-10
suppose x = -9, x-y = 11, x+y = -29
suppose x = 9, x-y = -29, x+y = -11
Insuff
ans A
2.) Remainder when x is divided by 6..
A - x/2 - rem is 1 => x is odd
x/3 - rem is 0 => x is multiple of 3
hence x can be odd nos multiples of 3 - 3,9,15,21 ...whenever divided by 6 ..rem is 3 => A is suff
B - x/12 - rem is 3 => x = 12 * k +3 => x = 2 * (6k) + 3 => rem always 3
=> B is suff
ans D
3.) is m+z>0
neither statement alone is sufficient as nature of m and z is not known
combining we get z>0 => m>3z hence m+z is always +ve
ans C
4.) I missed this one big time...
earlier i thought it to be C..but agreed to one of the explanations provided above...
taking A alone - q=11 to 16 -> insuff
taking B alone - q = 2, 4, 8, 16 -> rem is always 1
ans B
5.) has been very well explained by ravindra_iit..it was a question i had posted day before for clarification...
is x-y/x+y>1
=> is (x-y/x+y)-1>0
=> is -2y/x+y>0.
=>is y/(x+y)<0
it is true when y<0 & x+y>0 or y>0 & x+y<0
a. x>0 insuff
b. y<0 insuff
both together insuff
ans is E...OA is E (question from OG11)
6.) is 1/p > r / (r^2 +2)
taking A - p=r => 1/r > r/(r^2 + 2) -> nature or r is not known..insuff
taking B - nor relation given -> insuff
together as r>0
it can be cross multiplied to get r^2+2 > r^2 always true
hence ans is C ....OA is also C...its a OG11 question
a. x=8
b. y =-20
taking A alone.. -y>2 => y<-2 hence y is -ve hence x-y>x+y suff
taking B... y=-20, => x>-10
suppose x = -9, x-y = 11, x+y = -29
suppose x = 9, x-y = -29, x+y = -11
Insuff
ans A
2.) Remainder when x is divided by 6..
A - x/2 - rem is 1 => x is odd
x/3 - rem is 0 => x is multiple of 3
hence x can be odd nos multiples of 3 - 3,9,15,21 ...whenever divided by 6 ..rem is 3 => A is suff
B - x/12 - rem is 3 => x = 12 * k +3 => x = 2 * (6k) + 3 => rem always 3
=> B is suff
ans D
3.) is m+z>0
neither statement alone is sufficient as nature of m and z is not known
combining we get z>0 => m>3z hence m+z is always +ve
ans C
4.) I missed this one big time...
earlier i thought it to be C..but agreed to one of the explanations provided above...
taking A alone - q=11 to 16 -> insuff
taking B alone - q = 2, 4, 8, 16 -> rem is always 1
ans B
5.) has been very well explained by ravindra_iit..it was a question i had posted day before for clarification...
is x-y/x+y>1
=> is (x-y/x+y)-1>0
=> is -2y/x+y>0.
=>is y/(x+y)<0
it is true when y<0 & x+y>0 or y>0 & x+y<0
a. x>0 insuff
b. y<0 insuff
both together insuff
ans is E...OA is E (question from OG11)
6.) is 1/p > r / (r^2 +2)
taking A - p=r => 1/r > r/(r^2 + 2) -> nature or r is not known..insuff
taking B - nor relation given -> insuff
together as r>0
it can be cross multiplied to get r^2+2 > r^2 always true
hence ans is C ....OA is also C...its a OG11 question
GCD LCM
GCD AND LCM
---------------
The greatest common divisor (GCD), of two or more non-zero integers, is the largest positive integer that divides the numbers without a remainder
The lowest common multiple (LCM), of two integers and is the smallest positive integer that is a multiple both of and of .
General rule
To find the HCF of 24 of 36, express both using prime factorization:
24 --> 2 * 2 * 2 * 3
36 --> 2 * 2 * 3 * 3
The HCF is thus 2 * 2* 3 = 12 (pick out the common terms)
To find the LCM, again express both using prime factorization:
The LCM is 2 * 2 * 2 * 3 * 3 = 72(pick out the common terms, then multiply in the remaining terms)
Properties of GCD and LCM
For any two consecutive integers the GCM can only be 1.
GCD X LCM = Product of two numbers
LC
-----------------------
The least common multiplier of A and B is 120, the ratio of A and B is 3:4, what is the largest common divisor?
Soln
Let the nos be 3x and 4x . Hence their GCD will be x
GCD X LCM = Product of two numbers
120 X x = 3x X 4x
x=10
-------------------
1. What is the smallest possible common multiple of two integers, which are both greater than 250.
1) 251
2) 252
3) 502
4) 750
5) 884
2. What is the greatest possible common divisor of two different positive integers which are less than 144?
a) 143
b) 142
c) 72
d) 71
e) 12
1. Smallest possible number greater than 250 is 251.
Now we have to look for smallest multiple of 251. Why ? Because then 251 will be a factor of that number .
Now smallest multiple of 251 = 251*2 = 502
So the two numbers are 251 and 502 respectively and their LCM is 502. Hence C.
2. Greatest number less than 144 is 143 = 11*13
So maximum possible value of GCD of 143 and any number less that 143 will be = 13
Now consider the largest number less than 143 which is = 142 = 2*71
Now maximum possible value of GCD of 142 and any number less that 142 will be = 71
The number which is less than 144 as well as GCD of which and of 142 is 71 will be =71
Hence D.
--------
What is the greatest common divisor of positive integers m and n?
1) m is prime number
2)m and n are consecutive integers
1) m is prime number
then it is divisible by 1 and itself
now let n = multiple of m, therefor GCD is m
let is not a multiple of m then GCD = 1
so we can not determine
2) m and n are consecutive integers
let n = (m+1) GCD will always be 1
Hence B
---------------
The greatest common divisor (GCD), of two or more non-zero integers, is the largest positive integer that divides the numbers without a remainder
The lowest common multiple (LCM), of two integers and is the smallest positive integer that is a multiple both of and of .
General rule
To find the HCF of 24 of 36, express both using prime factorization:
24 --> 2 * 2 * 2 * 3
36 --> 2 * 2 * 3 * 3
The HCF is thus 2 * 2* 3 = 12 (pick out the common terms)
To find the LCM, again express both using prime factorization:
The LCM is 2 * 2 * 2 * 3 * 3 = 72(pick out the common terms, then multiply in the remaining terms)
Properties of GCD and LCM
For any two consecutive integers the GCM can only be 1.
GCD X LCM = Product of two numbers
LC
-----------------------
The least common multiplier of A and B is 120, the ratio of A and B is 3:4, what is the largest common divisor?
Soln
Let the nos be 3x and 4x . Hence their GCD will be x
GCD X LCM = Product of two numbers
120 X x = 3x X 4x
x=10
-------------------
1. What is the smallest possible common multiple of two integers, which are both greater than 250.
1) 251
2) 252
3) 502
4) 750
5) 884
2. What is the greatest possible common divisor of two different positive integers which are less than 144?
a) 143
b) 142
c) 72
d) 71
e) 12
1. Smallest possible number greater than 250 is 251.
Now we have to look for smallest multiple of 251. Why ? Because then 251 will be a factor of that number .
Now smallest multiple of 251 = 251*2 = 502
So the two numbers are 251 and 502 respectively and their LCM is 502. Hence C.
2. Greatest number less than 144 is 143 = 11*13
So maximum possible value of GCD of 143 and any number less that 143 will be = 13
Now consider the largest number less than 143 which is = 142 = 2*71
Now maximum possible value of GCD of 142 and any number less that 142 will be = 71
The number which is less than 144 as well as GCD of which and of 142 is 71 will be =71
Hence D.
--------
What is the greatest common divisor of positive integers m and n?
1) m is prime number
2)m and n are consecutive integers
1) m is prime number
then it is divisible by 1 and itself
now let n = multiple of m, therefor GCD is m
let is not a multiple of m then GCD = 1
so we can not determine
2) m and n are consecutive integers
let n = (m+1) GCD will always be 1
Hence B
Coordinate geometery
Coordinate Geometry
1. Distance formula
If two points P(x1, y1), Q(x2, y2) then the distance between the points P and Q,
D=sqrt [(x1-x2)^2 +(y1-y2)^2]
2. Section Formula
If any point (x, y) divides the line segment joining the points (x1, y1) and (x2,y2) in the ratio (m : n) internally,
X = (mx2+nx1) / (m + n)
Y = (my2+ny1) / (m + n)
If externally,
X = (mx2-nx1) / (m-n)
Y = (my2-ny1) / (m-n)
3. Area of Triangle
The area of triangle whose vertices are A(x1, y1), B(x2,y2) & C(x3,y3) is given by
[x1 (y2-y3) +x2 (y3-y1) +x3 (y1-y2)] /2
Note: Since the area cannot be negative, we have to take the modulus value given by the above equation.
a) If one of the vertices of triangle is at the origin and the two vertices are A(x1, y1), B(x2, y2)
Area= [x1y2-x2y1]/2
4. Centroid of a Triangle
It is the point of intersection of its medians. Centroid divides the medians in the ratio 2:1.
If A(x1, y1), B(x2, y2) & C(x3, y3) are the coordinates of the vertices of a triangle then the coordinates of the centroid G of that triangle are
X=(x1+x2+x3)/3 & y= (y1+y2+y3)/3
5. In-centre of a Triangle
It is the centre of a circle that touches the side of a triangle is called its In-centre. In other words, if the three sides of the triangle are tangential to the circle then the centre of that circle represents the in-centre of the triangle.
If A(x1, y1), B(x2,y2) & C(x3,y3) are the coordinates of the vertices of a triangle then the coordinates of its in-centre are
X= (ax1+bx2+cx3)/ (a +b +c) Y= (ay1+by2+cy3)/ (a +b +c)
Where BC=a, AB=c & AC=b
6. Circum centre of a Triangle
The point of intersection of the perpendicular bisector of the sides of a triangle is called its circum-centre. It is equidistant from the vertices of the triangle. It is also known as the centre of the circle that circumscribes the triangle.
Let ABC be a triangle. If O is the circum-centre of the triangle ABC, then OA=OB=OC and each of these three represent the circum-radius.
7. Collinearity of Three Points
Three given points A, B & C are said to be collinear, that is lie on the same straight line, if any of the following condition occur:
a) Area of triangle formed by these three points is Zero.
b) Slope of AB=slope of AC
8. Slope of a Line
The slope of a line joining two points A(x1, y1) and B(x2, y2) is denoted by m and is given by,
M= (y2-y1)/(x2-x1) = tan z where z is the angle that the line makes with the positive direction of x-axis.
9. Different forms of the equation of a straight line
a) General form
The general form of the equation of a straight line is
ax + by+ c=0
Where a, b and c are real constants.
Slope of line= -a/b
The general form is also given by
Y=mx+ c where m is the slope & c is the intercept on y-axis.
b) Line Parallel to the X-axis
The equation of a straight line parallel to the x-axis and at a distance b from it is given by y=b.
Equation of the x-axis is y=0
c) Line Parallel to Y-axis
The equation of a straight line parallel to the y-axis and at a distance a from it is given by x=a.
Equation of the y-axis is x=0
d) Slope Intercept Form
The equation of a straight line passing through the point A(x1, y1) and having a slope m is given by,
(y- y1)=m(x-x1)
e) Two points form
The equation of a straight line passing through two points A(x1, y1) and B(x2, y2) is given by
(y- y1)= [(y2-y1) (x-x1)]/(x2-x1)
Slope= (y2-y1)/(x2-x1)
f) Intercept Form
The equation of a straight line making intercepts a and b on the axes of x & y respectively is given by,
x /a + y/b =1
10) Condition for Two lines to be parallel
Two lines are said to be parallel if their slopes are equal. For this to happen, the ratio of coefficients of x and y in both the lines should be equal.
In a general form, this can be stated as:
Line parallel to ax +by +c=0 is ax +by +k=0
Or dx+ ey+ k=0 if a/d = b/e where k is a constant.
11) Condition for two lines to be perpendicular
Two lines are said to be perpendicular if product of the slopes of the lines is equal to -1.
12) Length of Perpendicular of a Point from a Line.
The length of perpendicular from a given point (x1,y1) to a line ax +by +c=0 is
|ax1+by1+c|/sqrt (a^2+b^2)
a) Distance b/w two parallel lines will always be the same.
When two straight lines are parallel whose equation are ax +by +c=0 & ax+by+c1=0
Distance b/w them = |c-c1|/sqrt (a^2+b^2)
probability
A five-member committee is to be formed from a group of five military officers and nine civilians. If the committee must include at least two officers and two civilians, in how many different ways can the committee be chosen?
119
1,200
3,240
3,600
14,400
119
1,200
3,240
3,600
14,400
This is a combination problem, so you will be using the formula:
nCr = (n!)/(r!*(n-r)!)
You must break the problem up into two Cases. Find the combination of each part and then add them.
Case 1 is selecting exactly two officers and three civilians.
For the officers,
5C2 = (5!)/(2!*(5-2)!) = 10
For the civilians,
9C3 = (9!)/(3!*(9-3)!) = 84
So, 10*84 combinations for Case 1 = 840.
Case 2 is selecting exactly three officers and two civilians.
For the officers,
5C3 = (5!)/(3!*(5-3)!) = 10
For the civilians,
9C2 = (9!)/(2!*(9-2)!) = 36
So, 10*36 combinations for Case 2 = 360.
Adding Case 1 and Case 2 = 1200.
The answer is B.
nCr = (n!)/(r!*(n-r)!)
You must break the problem up into two Cases. Find the combination of each part and then add them.
Case 1 is selecting exactly two officers and three civilians.
For the officers,
5C2 = (5!)/(2!*(5-2)!) = 10
For the civilians,
9C3 = (9!)/(3!*(9-3)!) = 84
So, 10*84 combinations for Case 1 = 840.
Case 2 is selecting exactly three officers and two civilians.
For the officers,
5C3 = (5!)/(3!*(5-3)!) = 10
For the civilians,
9C2 = (9!)/(2!*(9-2)!) = 36
So, 10*36 combinations for Case 2 = 360.
Adding Case 1 and Case 2 = 1200.
The answer is B.
Arithmetic series Problem
In a certain sculpture, coils of wire are arranged in rows. The second row has two more coils than the first, the third two more than the second, and so on, to the tenth and final row. If there is a total of 120 coils of wire in the sculpture, how many coils are in the final row?
let first row = x
second row = x+2
third row = x+4
...
tenth row = x+18
total = 10x + 2(1+2+3+...+9)
total = 10x + 2*45 = 120
x = 3
thus, tenth row = x+18 = 21
second row = x+2
third row = x+4
...
tenth row = x+18
total = 10x + 2(1+2+3+...+9)
total = 10x + 2*45 = 120
x = 3
thus, tenth row = x+18 = 21
you can you use n(n+1)/ 2 for 1 to 9
Alternatively you can use formulas for arithmetic progression for this