D
speed = distence / time
60km/hr=distance / 9 sec
km / hr to m / s==> 60*1000 / 1* 3600 = 16.667
16.667*9 = distance
150 m is the answer
Time taken to fill repaired cistern = 12 mins
In 1 min, pipe will fill = 1/12
Time taken to fill cistern with leak = 12 + 18 = 30 mins
In 1 min, it will fill = 1/30
Consider the leak as an outlet pipe, so
1/30 = 1/12 – [ in 1 min how much the outlet pipe leaks ]
= 1/12 – 1/30
= 5/60 – 2/60 = 3/60 = 1/20
Therefore, it will take 20 mins for the leak to empty the cistern
Answer is 274 .
Let the required term be x.
Pattern :- 1) Find the difference between consecutive terms as :
2 – 1 = 1
4 – 2 = 2
13 – 4 = 9
31 – 13 = 18
112 – 31 = 81
2) Now write the above numbers obtained by differences in series as :
1, 2, 9, 18, 81, y
(Suppose y be next term after 81.)
3) Look the difference series carefully and it follows the below pattern as :
(9 * 2)/1 = 18
(18 * 9)/2 = 81
(81 * 18)/9 = 162 = y (the next term after 81)
Thus the required answer is :
x = 112 + y
x = 112 +162 =274
Take 5 pills from jar 1, 4 pills from jar2,…. and 1 pill
from jar 5 and put altogether in the scale. the ideal
weight should be (1+2+3+4+5)x10= 150gms. for eg if the jar2
is contaminated then the weight will be 146 gms. so
depending on the amount of weight loss we can identify the
contaminated jar.
pages # figures
1000 1000 4
from 999 to 100 (999-99)*3 2700
from 99 to 10 (99-9)*2 180
from 9 to 1 (9-0)*1 9
total sum= 2893
(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
time (dist covrd by husband) (dist covrd by wife)
0.5hr 20 miles 0 miles
1 hr 40 miles 25 miles
1.5 hr 60 miles 50 miles
2 hrs 80 miles 75 miles
2.5 hrs 100 miles 100 miles
Therefore wife catches up after 2.5 hrs..
8
Average = (1+2+3+4+5+6+7+8+9)/3
= 45/3 = 15
Every side will contain sum of 15.
1st side contains 8 and 7.
2nd side contains 9 and 6.
3rd side contains 1,2,3,4 and 5.
350m
on the earth
8/15 left.
1/15 * 4 + 1/20 * 4 = 7/15
1 – 7/15 = 8/15
Answer is 45
First we need to subtract those reminders from the respective numbers, then we have to find the hcf of two numbers(numbers got from the subtraction) then you will get the answer.
So,
After subtraction you will get
3026-11 = 3015
5053-13 = 5040
HCF of these two numbers
5 | 3015 5040
3 | 603 1008
3 | 201 336
| 67 112
We can’t find a common diviser since 67 is a prime number
So the HCF = 5 * 3 * 3
= 45
5*3*3 = 45
55967