@Torque024 bhai tum haamesha coordinate kyun use karte ho... koi khas reason...
@Improver said:@Torque024 bhai tum haamesha coordinate kyun use karte ho... koi khas reason...
No specific reason, I'm more comfortable through coordinates.
@Zedai said:ABCD is a square. P is the mid point of AB. The line passing through A and perpendicular to DP intersects the diagonal at Q and BC at R. If AB=2 then PR=_______?A) 1/2B) sqrt(3)/2C) sqrt(2)D) 1E) None of these
the triangle ABR and APD will be similar.
so applying the similarity condition, AP/BR=AD/AB
therefore BR=1
in triangle PBR pr=sqrt(1^2+1^2)
therefore PR=sqrt2
f(x,y)=yth digit to the right of the decimal point of decimal representation of x.
then
f((1 + sqrt(2))^3000,1000) is
a)8
b)9
c)6
d)none
then
f((1 + sqrt(2))^3000,1000) is
a)8
b)9
c)6
d)none
@Zedai said:ABCD is a rectangle with AD =10. P is a point on BC such that Ang(APD)=90. If DP=8 then the length of BP is________??(1) 6.4(2)5.2(3)4.8(4)3.6(5) none of the above
triangles APD, ABP and PDC are all similar to each other.
from triangles ABP and PDC,
BP*PC= AB^2-------(1)
also BP+PC=10------(2)
from triangle APD and ABP,
BP/6=AB/8,
substituting this in (1) and using (2)
we get BP= 3.6
@Torque024 said:f(x,y)=yth digit to the right of the decimal point of decimal representation of x.thenf((1 + sqrt(2))^3000,1000) isa)8b)9c)6d)none
It will be 9.
(1 + √2)^3000 + (√2 - 1)^3000 = Integer, say I
and (√2 - 1) = 0.41 (approx)
Now, (√2 - 1)^3
=> (√2 - 1)^3000
So, we can say that (1 + √2)^3000 > I - 10^(-2000) and (1 + √2)^3000
Hence, 1000th digit to the right of the decimal will be 9
@Torque024 said:x^2 - |5x - 3| - x OA is -5
x
x^2+5x-3-x=>x^2+4x-5=>x^2+5x-x-5=>(x+5)(x-1)
-5
x>3/5
x^2-5x+3-x=>x^2-6x+1=>3-2root2 3+2root2
==> -5
x^2+5x-3-x=>x^2+4x-5=>x^2+5x-x-5=>(x+5)(x-1)
-5
x>3/5
x^2-5x+3-x=>x^2-6x+1=>3-2root2
==> -5
@Torque024 said:x^2 - |5x - 3| - x OA is -5
When x
x^2 + 5x - 3 - x
x^2 + 4x - 5
-5
So, -5
When x ≥ 3/5
x^2 - 5x + 3 - x
x^2 - 6x + 1
3 - 2√2
Now, 3 - 2√2
So, 3/5 ≤ x
Combining (1) and (2) we will get
-5
@krum said:xx^2+5x-3-x=>x^2+4x-5=>x^2+5x-x-5=>(x+5)(x-1)-5x>3/5x^2-5x+3-x=>x^2-6x+1=>3-2root2==> -5
Ya got it, I was taking the lower limits of the common solution. :/
Rajiv is a student in a business school. After every test he calculates his cumulative average. QT and OB were his last two tests. 83 marks in QT increased his avg by 2. 75 marks in OB further increased his avg by 1. Reasoning is the next test, if he gets 51 in reasoning, his avg will be_____?
@Zedai said:Rajiv is a student in a business school. After every test he calculates his cumulative average. QT and OB were his last two tests. 83 marks in QT increased his avg by 2. 75 marks in OB further increased his avg by 1. Reasoning is the next test, if he gets 51 in reasoning, his avg will be_____?
if i understood the question right..i dont think it can be determined..
we need the no.of exams before QT
we need the no.of exams before QT
@Zedai said:Rajiv is a student in a business school. After every test he calculates his cumulative average. QT and OB were his last two tests. 83 marks in QT increased his avg by 2. 75 marks in OB further increased his avg by 1. Reasoning is the next test, if he gets 51 in reasoning, his avg will be_____?
a - avg.
n - no. of subjects
x - total marks
x/n=a
(x+83)/(n+1)=a+2
(x+83+75)/(n+2)=a+3
=> xn+83n=(n+1)(x+2n)
=> 83n=2n^2+x+2n
=> 2n^2-81n+x=0
=> xn+158n=(n+2)(x+3n)
=> 158n=3n^2+2x+6n
=> 3n^2-152n+2x=0
-243n+3x+304n-4x=0
=> 61n=x
2n^2=20n
=>n=10, x=610
average after reasoning = (610+83+75+51)/13 = 63
n - no. of subjects
x - total marks
x/n=a
(x+83)/(n+1)=a+2
(x+83+75)/(n+2)=a+3
=> xn+83n=(n+1)(x+2n)
=> 83n=2n^2+x+2n
=> 2n^2-81n+x=0
=> xn+158n=(n+2)(x+3n)
=> 158n=3n^2+2x+6n
=> 3n^2-152n+2x=0
-243n+3x+304n-4x=0
=> 61n=x
2n^2=20n
=>n=10, x=610
average after reasoning = (610+83+75+51)/13 = 63
@wovfactorAPS said:if i understood the question right..i dont think it can be determined..we need the no.of exams before QT
@krum said:a - avg.n - no. of subjectsx - total marksx/n=a(x+83)/(n+1)=a+2(x+83+75)/(n+2)=a+3=> xn+83n=(n+1)(x+2n)=> 83n=2n^2+x+2n=> 2n^2-81n+x=0=> xn+158n=(n+2)(x+3n)=> 158n=3n^2+2x+6n=> 3n^2-152n+2x=0-243n+3x+304n-4x=0=> 61n=x2n^2=20n=>n=10, x=610average after reasoning = (610+83+75+51)/13 = 63
for each p>1, a sequence {An} is defined by Ao=1 and An=(p*n)+[(-1)^n *(An-1)] for each n>=1. For how many integer values of p, 1000 is a term of the sequence?
8
7
5
4
P.S: An-1 means (n-1)th term
@Zedai said:Rajiv is a student in a business school. After every test he calculates his cumulative average. QT and OB were his last two tests. 83 marks in QT increased his avg by 2. 75 marks in OB further increased his avg by 1. Reasoning is the next test, if he gets 51 in reasoning, his avg will be_____?
63?
Tests before QT and OB=x
Avg before QT and OB=y
xy+83=xy+2x+y+2;
xy+83+5=xy+3x+2y+6;
x=10,y=61;
After OB avg=(xy+83+5+51)/y+3= 63?
Tests before QT and OB=x
Avg before QT and OB=y
xy+83=xy+2x+y+2;
xy+83+5=xy+3x+2y+6;
x=10,y=61;
After OB avg=(xy+83+5+51)/y+3= 63?
@Zedai said:Rajiv is a student in a business school. After every test he calculates his cumulative average. QT and OB were his last two tests. 83 marks in QT increased his avg by 2. 75 marks in OB further increased his avg by 1. Reasoning is the next test, if he gets 51 in reasoning, his avg will be_____?
n tests and 'a' average
an = t
t + 83 = (n + 1)(a + 2)
2n + a = 81
t + 83 + 75 = (n + 2)(a + 3)
3n + 2a = 152
n = 10, a = 61
Average = (610 + 83 + 75 + 51)/13 = 63
@Zedai said:Rajiv is a student in a business school. After every test he calculates his cumulative average. QT and OB were his last two tests. 83 marks in QT increased his avg by 2. 75 marks in OB further increased his avg by 1. Reasoning is the next test, if he gets 51 in reasoning, his avg will be_____?
let his average be = x
number of subjects = y
(xy + 83)/(y + 1) = x + 2
(xy + 83 + 75)/(y + 2) = x + 3
4y + 2x = 162
2x + 3y = 152
y = 10 and x = 61
(610 + 83 + 75 + 51)/13 = 63
@Zedai said:for each p>1, a sequence {An} is defined by Ao=1 and An=(p*n)+[(-1)^n *(An-1)] for each n>=1. For how many integer values of p, 1000 is a term of the sequence?8754P.S: An-1 means (n-1)th term
0--1
1--p-1
2--3p-1
3--1
4--4p+1 and so on with 2 as 3p-1 and 6 as 7p-1 and so on
So for p=1001 it is possible
and factors of 1001=1,7,11,13,77,91,143,1001;
factor+1 should be of 4k type
possible factors=7,11,91,143
Total 5 ways?
1--p-1
2--3p-1
3--1
4--4p+1 and so on with 2 as 3p-1 and 6 as 7p-1 and so on
So for p=1001 it is possible
and factors of 1001=1,7,11,13,77,91,143,1001;
factor+1 should be of 4k type
possible factors=7,11,91,143
Total 5 ways?
@Zedai said:for each p>1, a sequence {An} is defined by Ao=1 and An=(p*n)+[(-1)^n *(An-1)] for each n>=1. For how many integer values of p, 1000 is a term of the sequence?8754P.S: An-1 means (n-1)th term
A(0) = 1
A(1) = p - 1
A(2) = 2p + (p - 1) = 3p - 1
A(3) = 3p - (3p - 1) = 1
A(4) = 4p + 1
A(5) = 5p - (4p + 1) = p - 1
A(6) = 7p - 1
A(7) = 1
.
.
and so on
p - 1 = 1000
=> p = 1001
(4k)p + 1 = 1000 (not possible)
(4k - 1)p - 1 = 1000
(4k - 1)p = 1001 = 7*11*13
(4k - 1) can be 7, 11, 91 or 143
So, p can take 4 values (7, 11, 91, 143)
Total = 5 values