Error is at 46. Correct one is 29+1=30
group wise work is use full and effectively
1/2:1/3:1/4 = 6:4:3
Ram = 6/13 * 3250 = 1500,
Shyam = 4/13 * 3250 = 1000,
Mohan = 3/13 * 3250 = 750
2* pi* r ( h + r )
r=3cm
h=7cm
2 * 22/7 *7 ( 10 ) = 440 sq cm
total sum = 205 Rs.
Assume there are 52 weeks in one year.
Since he supposed to have a new order for every two weeks, he
needs 52/2 = 26 orders to break the office record.
Now after 28 weeks,he has got only 28/2 -6 = 8 orders.
Hence,he needs 26-8 = 18 new orders in the remaining 24= 52-28 weeks to break the office record.
Compute 24 orders/18 weeks = 4/3 orders/week ,
we see that averagely he has a new order for
every 4/3 weeks in the remaining weeks to break the office record.
b
u c 5 numbers…..
we add the combination of digits…..
a.17->1+7=8(even)
b.23->2+3=5(odd)!!!!!
c.37->3+7=10(even)
d.13->1+3=4(even)
e.73->7+3=10(even)
all combinations are even xept 23
ans:23
2:1
(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
1/2 Hr -> 80 Cheques
Per Hour -> 80 * 2 = 160
For 7 Hours -> 160 * 7 = 1120
For 7 1/2 Hour -> 1120 + 80 = 1200
clerk can process 1200 cheques in Sever & one half an hour day.
130
520 = 26 * 20 = 2 * 13 * 22 * 5 = 23 * 13 * 5
Required smallest number = 2 * 13 * 5 = 130
130 is the smallest number which should be multiplied with 520 to make it a perfect square.
It is assumed to be constant. Now, to cross past the pole, the train should cover a distance of x meters. Now, the time taken by the train to cross a platform of length 100 m is 25 seconds. Hence, the length of the train is 150 m.