Signifyd Interview Questions, Process, and Tips

trick
(r/10)*(r/10)
(20/10)*(20/10)
4% decrese

(0! + 0! + 0! + 0! + 0!)! = 120

Here ‘!’ symbol is used for factorial.
And 0! = 1
= (0! + 0! + 0! + 0! + 0!)!
= (1 + 1 + 1 + 1 + 1)!
= 5!
= 120

=12.05×5.4/0.6
=12.05×9.0
=108.45

D. 760

Hint: Assume that the speed of the stream is x and the speed of the boat in still water is x. From the statement of the question form two equations in two variables x and y. This system is reducible to linear equations in two variables. Reduce the system to a system of linear equations in two variables by proper substitutions. Solve the system of equations using any one of the methods like Substitution method, elimination method, graphical method or using matrices. Hence find the value of x satisfying both the equation. The value of x will be the speed of the stream.
Complete step-by-step answer:
Let the speed of the stream be x, and the speed of the boat in still water be y.
We have the speed of the boat upstream = y-x.
Speed of the boat downstream = y+ x.
Now since it takes 14 hours to reach a place at a distance of 48 km and come back, we have the sum of the times taken to reach the place downstream and time taken to return back upstream is equal to 14.
Now, we know that time =Distance speed
Using, we get
Time taken to reach the place =48y+x and the time taken to return back =48y−x.
Hence, we have
48y+x+48y−x=14
Dividing both sides by 2, we get
24y+x+24y−x=7 —–(i)
Also, the time taken to cover 4km downstream is equal to the time taken to cover 3km upstream.
Hence, we have 4y+x=3y−x
Transposing the term on RHS to LHS, we get
4y+x−3y−x=0 ——– (ii)
Put 1y+x=t and 1y−x=u, we have
24t+24u=7 ——-(iii)4t−3u=0 ——–(iv)
Multiplying equation (iv) by 6 and subtracting from equation (iii), we get
24t−24t+24u+18u=7⇒42u=7
Dividing both sides by 42, we get
u=742=16
Substituting the value of u in equation (iv), we get
4t−3(16)=0⇒4t−12=0
Adding 12 on both sides, we get
4t=12
Dividing both sides by 4, we get
t=18
Reverting to original variables, we have
1y+x=18 and 1y−x=16
Taking reciprocals on both sides in both equations, we have
y+ x=8 ——- (v)y−x=6 ——–(vi)
Adding equation (v) and equation (vi), we get
2y=14
Dividing both sides by 2, we get
y=7.
Substituting the value of y in equation (v), we get
7+x=8
Subtracting 7 from both sides we get
x = 8-7 =1
Hence the speed of the stream is 1 km/hr.

Greatest number of 5 digits=99, 999
Smallest number of 5 digits=10, 000
And their sum=99, 999+10, 000=109, 999

1) time =20mins,speed=15kmph=15*(5/18)=25/6 m/sec
distance=20*(25/6)=250/3 m
then 2) given as time =15min
speed=250/3*15=(50/9)*(18/5)kmph=20kmph

Soldier have to move 0.75 miles to West and then 0.375 miles to South to reach the camp.
How:-
Firstly soldier moving 1 mile to East from camp and then 1/2 miles= 0.50 miles to North .

Then it is moving 1/4 miles =0.25 miles to west
And then 1/8 miles =0.125 miles to South

Now we just have to count the difference and minus the value of camp to East with the value of North to west = 1mile – 0.25miles(1/4) = 0.75 miles

Same case with north and south = 0.50 miles(1/2) – 0.125 miles(1/8) = 0.375 miles

Hence proves to return camp the soldier have to move 0.75 miles to west and 0.375 miles to south

I have been teaching some students for 1 year.

1^1,2^2,3^3,4:^4,5^5,6^6
1,4,27,256,3125,46656

s=a+(a+10)+(a+10+10)+(a+10+10+10)
s=4a+60
a=(s-60)/4

Rs. 1500

“RAW”:
RAW IS AN INDIAN INTILLIGENCE WING.
“RESEARCH AND ANALYSIS WING”

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