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40 km avrege speed
1920
30
1st gets 15 RS less than average
4x+60=s
x=(s-60)/4 = s/4-15
s/4 is average
1st gets 15 RS less than average
The number is greater than the number obtained in reversing the digits and the ten’s digit is greater than the unit’s digit
Let ten’s and unit’s digit be 2x and x respectively
Then, (10×2x+x)−(10x+2x)=36
9x=36
x=4
Required difference=(2x+x)−(2x−x)=2x=8
3500*10/100
3500*11.5/100 = 402.5 per year
402.5×3 = R1207.5
160600
12
6.25
(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
Example 1:
Assign, A=20, B=10, C=5, D=5(Because C is equal to D as
given), E=1.
A/B = 20/10 = 2. So A/B = 2
A/C = 20/5 = 4. So A/C = 4
A/E = 20/1 = 20. So A/E = 20
Therefore “A/E is Greatest”
Example 2:
Assign, A=100, B=50, C=20, D=20(Because C is equal to D as
given), E=10.
A/B = 100/50 = 2. So A/B = 2
A/C = 100/20 = 5. So A/C = 5
A/E = 100/10 = 10. So A/E = 10
Therefore “A/E is Greatest”
12.9
2+3+5+7+11+13+17+19+23+29=129
129/10=12.9
500