a1, a2, a3.....a8 are eight real numbers belonging to the interval [–8, 8]. What is the minimum
possible value of a1a2 + a2 a3 +...... a7a8 + a8a1?
a) –64
(b) 0
(c) –512
(d) –16
@akansh_1 said:a1, a2, a3.....a8 are eight real numbers belonging to the interval [–8, 8]. What is the minimum possible value of a1a2 + a2 a3 +...... a7a8 + a8a1?a) –64 (b) 0(c) –512 (d) –16
-512
@akansh_1 said:a1, a2, a3.....a8 are eight real numbers belonging to the interval [–8, 8]. What is the minimum possible value of a1a2 + a2 a3 +...... a7a8 + a8a1?a) –64 (b) 0(c) –512 (d) –16
-512
@Ashmukh said:a circle which has a diameter of 10 cm,and two diam of the same circle intersect at O,let P be a point in the first quadrant and the foot of the perp drawn to x axis be A and foot of perp drawn to y-axis be B,calculate the max value of hypotenuse so formed.
The value of hypotenuse will be maximum when the point lies on the circle in the first quadrant, which will be equal to radius of the circle, i.e 5 cm.
@akansh_1 said:a1, a2, a3.....a8 are eight real numbers belonging to the interval [–8, 8]. What is the minimum possible value of a1a2 + a2 a3 +...... a7a8 + a8a1?a) –64 (b) 0(c) –512 (d) –16
-64x8 => -512
@akansh_1 said:a1, a2, a3.....a8 are eight real numbers belonging to the interval [–8, 8]. What is the minimum
possible value of a1a2 + a2 a3 +...... a7a8 + a8a1?
a) –64
(b) 0
(c) –512
(d) –16
-512
8 and -8 seperated alternately
The value of (222)baseX in base 'X' when converted to base 10 is 'P'. The value of (222)baseY in base 'Y' when converted to base 10 is Q. If (P – Q)base10 = 28, then what is the value of (Q – X)base10?
(a) 10
(b) 92
(c) 79
(d) Cannot be determined
(a) 10
(b) 92
(c) 79
(d) Cannot be determined
@akansh_1 said:a1, a2, a3.....a8 are eight real numbers belonging to the interval [–8, 8]. What is the minimum possible value of a1a2 + a2 a3 +...... a7a8 + a8a1?a) –64 (b) 0(c) –512 (d) –16
-512??
@Ashmukh said:@akansh_1 i thot as much ,but this is not d ans,5sqrt(2)
I don't know.. may be I'm going wrong somewhere. But for 5root(2), point P will lie outside the circle in the first quadrant.
@akansh_1 said:The value of (222)baseX in base 'X' when converted to base 10 is 'P'. The value of (222)baseY in base 'Y' when converted to base 10 is Q. If (P – Q)base10 = 28, then what is the value of (Q – X)base10?(a) 10(b) 92(c) 79(d) Cannot be determined
2x^2+2x+2=P
x^2+x+1=P/2
y^2+y+1=Q/2
(x-y)(x+y+1)=(p-q)/2
solving...
x=4, y=2
Q-x=10...........option a..
@akansh_1 said:The value of (222)baseX in base 'X' when converted to base 10 is 'P'. The value of (222)baseY in base 'Y' when converted to base 10 is Q. If (P – Q)base10 = 28, then what is the value of (Q – X)base10?
(a) 10
(b) 92
(c) 79
(d) Cannot be determined
79
getting N-15 should be perfect square, possible for c)
@surajsrivastav said:2x^2+2x+2=Px^2+x+1=P/2y^2+y+1=Q/2(x-y)(x+y+1)=(p-q)/2solving...x=4, y=2Q-x=10...........option a..
y=2 is not valid in base2.
@vijay_chandola said:'X' is the largest sum of rupees which can never be paid using any number of coins of denominations Rs.4, Rs.8, Rs.13 and Rs.18. What is the sum of the digits of 'X'?(a) 9 (b) 10 (c) 11 (d) None of these
4k-------->=4
4k+1--------->=13
4k+2------------>=18
4k+3------------->=31
31-4=27 can not be formed
9
@akansh_1 said:The value of (222)baseX in base 'X' when converted to base 10 is 'P'. The value of (222)baseY in base 'Y' when converted to base 10 is Q. If (P – Q)base10 = 28, then what is the value of (Q – X)base10?(a) 10(b) 92(c) 79(d) Cannot be determined
2 +2X + 2X^2 = P (X>=3)
2 + 2Y + 2Y^2 = Q (Y >=3)
P-Q = 2*(X-Y)*[ 1 + X + Y] = 28 => (X-Y)*(1 + X + Y) = 14
14 = 1*14, 2*7 (Negative values rejected)
X-Y = 1 | (X+Y+1) = 14 => X = 7, Y = 6 => (Q - X) = 79
X-Y = 2 | (X+Y+1) = 7 => X = 4, Y = 2 (rejected)
So, 79 ?
@surajsrivastav said:How come yaar??
sry for late reply
just reached home
the numbers are 2(X^2+X+1) and 2(Y^2+Y+1)
subtract Q from P
2(X^2-Y^2+X-Y)=28
(X^2-Y^2+X-Y)=14...............(1)
also
we need to find Q-X= N
=Y^2+Y+1-X.......(2)
in the above equation (1) u substitute (2)
u get X^2+15 =N