35 hours
30
Let the firt number be n,
The largest number will be n+1
Their sum
n+n+1=55
2n+1=55
2n=54
n=27
Largest number =27+1=28
20 minutes
CPNCBZ
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1×2×…100=100!
Number of zeros in product of n numbers =[5n]+[52n]+[53n]+…
Number of zeros in product of 100 numbers =[5100]+[52100]+[53100]
where [.] is greatest integer function
=[20]+[4]+[0.8]=20+4=24
11 times in 12hours
Day = Night
White = Black
12
(b) 16.66%
Earlier for ₹x we could purchase y gm of sugar.
Now we pay ₹1.2x for y gm of sugar
(As there was an increase in price so, x + 20%x = 1.2x)
At current rates for ₹x you can purchase y/1.2 gm of sugar
So the reduced consumption is y-(y/1.2)
Percentage change = (reduced consumption/ original consumption ) *100
That is (0.2/1.2) *100 = 16.66% (approx)
here the 30% of 100 is 30
and 10% of the 30 = 3
so 20% = 6%
so ans = 6%
(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
4
B
11