Hey guys..
Chapter : Time Speed n Distance
LOD II
Quests : 35, 38, 46
In question 46 can some1 please explain the concepts of gaining and losing time?
CAT_Aspirant101 SaysI don't know the answer mate! Are you sure? Anyone else?
I just remember seeing a similar question in time material which was solved in this way ... I'll confirm with my friends and let u knw 😃 :
Can any one help me out with this problem..
Arun sharma quant, Coordinate geo ,pg 598
The equations of two equal sides AB and A of an isosceles triangle ABC are x+y =5 and 7x-y = 3 respectively.What will be the equation of the side BC if area of triangle ABC is 5 sq units.
A) x+3y-1 = 0 B) x-3y+1 = 0
C)2x -y =5 D) x+2y =5
Thanks in advance
Can any one help me out with this problem..
Arun sharma quant, Coordinate geo ,pg 598
The equations of two equal sides AB and A of an isosceles triangle ABC are x+y =5 and 7x-y = 3 respectively.What will be the equation of the side BC if area of triangle ABC is 5 sq units.
A) x+3y-1 = 0 B) x-3y+1 = 0
C)2x -y =5 D) x+2y =5
Thanks in advance
Am getting B) x-3y+1=0
I drew the co-ordinate system, and line segments for all the given options. The given two lines intersect at (1,4). Let 'A' be that point
Thus, after drawing the 3 lines, i.e. the two given lines and the one given option, we get a triangle. Only triangle formed in option B resembles an isosceles triangle.
So, verifying for option B,
the line segments x+y=5 and x-3y+1=0 intersects at B(7/2, 3/2) and
the line segments 7x-y=3 and x-3y+1=0 intersects at C( 1/2,1/2)
Distance BC=sqrt((7/2-1/2)+(3/2-1/2))
=sqrt(10)
Also, distance of perpendicular from A(1,4) to line BC is
((1)-3(4)+1)/sqrt(10)) = sqrt(10)
So, area= 1/2 (sqrt10)(sqrt10) = 5. Thus verified.
Am getting B) x-3y+1=0
I drew the co-ordinate system, and line segments for all the given options. The given two lines intersect at (1,4). Let 'A' be that point
Thus, after drawing the 3 lines, i.e. the two given lines and the one given option, we get a triangle. Only triangle formed in option B resembles an isosceles triangle.
So, verifying for option B,
the line segments x+y=5 and x-3y+1=0 intersects at B(7/2, 3/2) and
the line segments 7x-y=3 and x-3y+1=0 intersects at C( 1/2,1/2)
Distance BC=sqrt((7/2-1/2)+(3/2-1/2))
=sqrt(10)
Also, distance of perpendicular from A(1,4) to line BC is
((1)-3(4)+1)/sqrt(10)) = sqrt(10)
So, area= 1/2 (sqrt10)(sqrt10) = 5. Thus verified.
It would be great if anyone could suggest any other approach without using options...
Can any one help me out with this problem..
In a quadratic equation, (whose coefficients are not necessarily real) the constant term is not 0. The cube of the sum of the squares of its roots is equal to the square of the sum of the cubes of its roots.Which of the following is true?
a)Both roots are real
b)neither if the roots is real
c)at least one root is non-real
d)at least one root is real
e) exactly one root is non-real
Can any one help me out with this problem..
In a quadratic equation, (whose coefficients are not necessarily real) the constant term is not 0. The cube of the sum of the squares of its roots is equal to the square of the sum of the cubes of its roots.Which of the following is true?
a)Both roots are real
b)neither if the roots is real
c)at least one root is non-real
d)at least one root is real
e) exactly one root is non-real
I wrote the equations for the roots and got an equation : (a-b) = sqrt (2 ( a^2 + b^2 )) .. If i substitute an integer for one of the roots , I'm getting the other as imaginary and vice versa ... So i suppose (e) is the answer
Well the ans is C
i took the eq as ax^2 + bx + c = 0...after solving i got the condition as
3(b^2) = 8ac
it means b^2 - 4ac => the eq doesn't have real roots
so i chose B...but OA is given as C...it would be great if someone could give justification to C and why B is wrong?..
missshararat SaysI wrote the equations for the roots and got an equation : (a-b) = sqrt (2 ( a^2 + b^2 )) .. If i substitute an integer for one of the roots , I'm getting the other as imaginary and vice versa ... So i suppose (e) is the answer
the question is from Permutation & combination,
There are V lines parallel to x axis and W lines parallel to y axis. So how many rectangles can be formed with the intersection of these lines..??
Say I is the number of employees working in at least one department = 100%
II is the number of employees working in at least two department
III is the number of employees working in all three department
=> I II III
Exactly 1 = I - II
Exactly 2 = II - III
Exactly 3 = III
To find the maximum value of III, put II = III, i.e, exactly 2 = 0
Also, I + II + III = 69 + 78 + 87 = 234
=> 100 + III + III = 234
=> III(max) = 67
To get the least value of III, out I = II = 100
=> III(min) = 234 - 200 = 34
Thanks for the reply. However I could not get the II part i.e.
min (III)... ??
please explain a bit more..
Thanks
HI all
Im new to PG and need ur help on a few Q's from Arun Sharma
1) a+b+c+d=90 and a,b,c,d=0. This one is a question from P&C.;
Thanks
Im new to PG and need ur help on a few Q's from Arun Sharma
1) a+b+c+d=90 and a,b,c,d=0. This one is a question from P&C.;
Thanks
the question is from Permutation & combination,
There are V lines parallel to x axis and W lines parallel to y axis. So how many rectangles can be formed with the intersection of these lines..??
HI
the solution is vC2 * wC2 , where C is for combination.
Q) what is the remainder when 128^1000 is divided by 153?
Ans a)103 b) 145 c) 118 d)52
Q) what is the remainder when 128^1000 is divided by 153?
Ans a)103 b) 145 c) 118 d)52
is it 145?
bbCAT005 Saysis it 145?
no..the ans is 52
Q) what is the remainder when 128^1000 is divided by 153?
Ans a)103 b) 145 c) 118 d)52
153=17*9....128^1000=2^7000.....
Use Chinese remainder theorem........
Rem=Rem=2*-1=-2=7....
Rem=Rem=1.....
9x+18y=1.....x=2 and y=-1 satisfies.....
So Rem=9*2*1+(-1)*17*7=18-119=-101=52.....
153=17*9....128^1000=2^7000.....
Use Chinese remainder theorem........
Rem=Rem=2*-1=-2=7....
Rem=Rem=1.....
9x+18y=1.....x=2 and y=-1 satisfies.....
So Rem=9*2*1+(-1)*17*7=18-119=-101=52.....
how 9x+18y=1 has come??
Find the last two digit in 122*123*125*127*129?
Ans 1)20 2)50 3)30 4)40 5)60
rsgage Sayshow 9x+18y=1 has come??
Thats something that has to be satisfied.....go through Chinese Remainder Theorem and you'll understand....
Find the last two digit in 122*123*125*127*129?
Ans 1)20 2)50 3)30 4)40 5)60
The product is divisible by 50....so from options last two digits have to be 50.....