835, 734, 642, 751, 853, 981, 532
835,sum(digits)= 16
734 = 14
642=12
751= 13,
853=16,
981=18,
532=10
answer(c)751
C
GIVEN: 2A(B+C)+AC-2C(A-B)
THEREFORE 2AB+2AC+AC-2AC+2BC
2AB+AC+2BC
2(AB+BC)+AC
LET b1=AB b2=BC b3=AC
STEP1: b1 = b1+b2
so b1 = AB+BC
THEREFORE
NOW: b1 = AB+BC b2 = BC b3 = AC
STEP 2: b3 = b1+b3
so b3 = AB+BC+AC
THEREFORE
NOW: b1 = AB+BC b2 = BC b3 = AB+BC+AC
STEP3:
NOW: b1 = b1+b3
so b1 = AB+BC+AB+BC+AC
=2(AB+BC)+ AC
AB BC AC
STEP1 AB+BC BC AC
STEP2 AC BC AB+BC+AC
STEP3 AB+BC+AB+BC+AC BC AB+BC+AC
i.e 2(AB+BC)+AC BC AB+BC+AC
(x**2 – 6* x + 5) = (x-1)*(x-5)
(x**2 + 2 * x + 1) = (x + 1) * (x+1) = (x+1)**2
For what x is (x-1)*(x-5)/( (x+1)**2) a minimum?
One way to answer this question is by using calculus.
Take the derivative, and set to zero.
Since this is a fraction of polynomials, and a fraction is
zero only if it’s numerator is zero, we need calculate only
the numerator of the derivative and set it to zero.
The numerator of the
Derivative of (x-1)*(x-5)/( (x+1)**2) is
( (x-1) + (x-5) ) ( x+1)**2 – (x-1)(x-5)( 2 (x+1) )
= (2 x – 6) (x+1)**2 – (2) (x-1)(x-5) (x+1)
= 0
Divide through by 2 (x+1)
(x-3)(x+1) – (x-1)(x-5) = 0
(x**2 – 2 x – 3 ) – (x**2 – 6 x + 5) = 0
x**2 – x**2 – 2 x + 6 x – 3 – 5 = 0
4 x – 8 = 0
x = 2
Plugging in x = 2 into the original
(x**2-6*x+5)/(x**2+2*x+1)
gives us (2**2 – 6 * 2 + 5)/(2**2 + 2*2 + 1)
= (4 – 12 + 5) / (4 + 4 + 1) = -3/9 = -1/3
Least value is -1/3
A
$57.30
125, 106, 88, 76, 65, 58, 53
88
Initially (2*33 + 1*24)/(2+1) = 30p a gm
So, cost for 100g = 100*30 = 3000p
Later, (1*33 + 2*24)/(1+2) = 71/3 p a gm.
So, cost for 100g = 100*(71/3) = 2366.67p
He saves 3000 – 2366.67 = 633.33p
Concept of weighted average.
451 times.
Explanation: There are 60 minutes in an hour.
In ¾ of an hour there are (60 * ¾) minutes = 45 minutes.
In ¾ of an hour there are (60 * 45) seconds = 2700 seconds.
Light flashed for every 6 seconds.
In 2700 seconds 2700/6 = 450 times.
The count start after the first flash, the light will
flashes 451 times in ¾ of an hour.
3, 4, 9, 22.5, 67.5, 202.5, 810
A. 4
Answer: 66.67 km approx.
Solution:
Let the first train A move at u km/h.
Let the second train B move at v km/h.
Let the distance between two trains be d km
Let the speed of bee be b km/h
Therefore, the time taken by trains to collide = d/(u+v)
Now putting all the known values into the above equation, we get,
u = 50 km/hr
v = 70 km/hr
d = 100 km
b = 80 km/hr
Therfore, the total distance travelled by bee
= b*d/(u+v)
= 80 * 100/(50+70)
= 66.67 km (approx)