A & B one day work= 1/18 + 1/30= 8/90
As we know that number of work is 1
1÷8/90= 1*90/8= 90/8.
Since they say “ twice the amount of work
Then ,2*90/8= 22.5 days
(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
6/5
let original radius=100
original surface area lets say it x =4 * (22/7) * 100 *100
new radius=100 + 50% of 100=150
new surface area lets say it y = 4 * (22/7)* 150* 150
increase in surface area=((y-x)/x) * 100
i.e.
((4*(22/7)*150*150 – 4*(22/7)*100 *100) / (4*(22/7)*100*100))*100
=125
so the answer is 125%.
hope this helps you.
( a ) 28
=12.05×5.4/0.6
=12.05×9.0
=108.45
Joe is fencing in a square area of 576 square feet.
Fence posts, which are needed every three feet, cost 32.00 each.
The fencing cost 4.50 per foot. What is the total cost of the fencing materials?
Solution:-
Area of Square is 576 sq ft.
so, the length of side = root(576) = 24 ft
Since the fencing would be done on the perimeter, we would need that
formula is 4*side = 96 ft.
Number of Fence post needed = 96/3 = 32
1 Fence post cost = 32
32 Fence Post cost = 32 * 32 = 1024
Fencing cost = Perimeter * fencing cost = 96 * 4.5 = 432
Total Cost = 432 + 1024 = 1456
I would design a voice integrated clock that reads out the time aloud whenever people ask for time
100
Answer:
Since the car has met the person 20 minutes before hand, it has actually saved 10 mins of journey (to and fro)
since the man has started 1.30 hrs before and car has met him 10 mins before actual time he takes to reach daily is 1hr and 20 mins
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0L -> 1 way
1L -> 3 ways
2L -> 7 ways
3L -> 4 ways
4L -> 1 way
total 16 ways
s=a+(a+10)+(a+10+10)+(a+10+10+10)
s=4a+60
a=(s-60)/4
Had is correct ! Anil went in right direction but ended up
with wrong answers
Soln: B lied ( see anil’s comment ) which implies B has not
stolen mule. So B could steal camel or sheep.
1) Lets assume B stole camel.
In that case, A lies ( refer – A says “B had stolen sheep”
), which implies A had stolen sheep ( becoz a lier cant
steal mule ). So A->sheep B->camel C->mule. But this cant
happen because C cant lie ( refer – C says ” B had stolen
mule” ).
2) Lets assume B stole sheep.
In that case, A is true ( refer – A says “B had stolen
sheep” ), which implies A had stolen mule. So A->Mule
B->Sheep C->camel. Here C lies ( refer – C says ” B had
stolen mule” )
SO ANS: answer A- mule, B-sheep, C camel
(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)