Sixt R&D Interview Questions, Process, and Tips

4 boy 3 sis

500

Slice the cake

Product of two numbers = 1320
HCF = 6
LCM = x
Formula:
Product of two numbers =(HCF *LCM)
1320=(6*x)
x=1320/6
x=220
LCM Of the numbers is220

obey

2
6*2=12
12*6=72
72*12=864

9 am

4×1 = 4
4×2=8
8×3=24
24×4= 96 (not 92)
96×5 =480
480×6= 2880

So answer is 92

56.25 % is the right answer.
area of square = side ^ 2
at first, it was 100% , means 1
then, a side was enlarged to 25%, means now, 125% means new side is 1.25
taking square will give 1.5625
1.5625 – 1 = 0.5625
so, 56.25% increment in area.

let A goes up and B comes down.
assume A takes 1 step in 1sec.
hence,B takes 5 steps in 1sec.
50 steps ll be moved by A in 50/1=50secs
125 steps ll be moved by B in 125/5=25 secs
hence speed of escalator=(difference of no of
steps)/(difference of time)
speed=(125-50)/(50-25)= 3 steps/sec
hence during upward journey of 50secs by A,total steps=50 by
A and 150 by escalator..total=150+50=200
during downward journey of 25 secs by B,total steps=125 by B
and 75 by escalator..total=125+75=200steps.
ANS:200 steps

11 times in 12hours

Ans. 175
Relative speed in m/s=40-22=18X5/18=5m/s
Total distance=125+x m
T=1min=60 s
D=sXt
125+x=5X60
x=175

put 1 red marble in one jar and all the rest (99) in the
other.
This gives you 50% from the first jar (if they pick that
jar they will get red 100% of the time). For the other jar
the chances are 49/99 or 49.494949%. Divide that by 2 and
its 24.7474%. Total odds are 50% plus 24.7474% = 74.7474%

The batsman on 98 is on strike. He hits the ball and they run 3. UNFORTUNATELY one of the batsmen doesn`t turn correctly for one of the runs and the umpire calls ONE SHORT and awards only two runs. Therefore the first batsman has his century. There is now 1 ball remaining and one run is required to win. The batsman on strike, however is now the one on 97 runs. He now either hits a 4 or a 6. They win the game and both batsmen scored centuries.

Read more: 3 runs required in 3 balls to win with only a wicket left. The batsmen is on 98 and the runner is on 97. How will both the batsmen score centuries … – 3 runs required in 3 balls to win with only a wicket left. The batsmen is on 98 and the runner is on 97. How will both the batsmen score centuries as well win the match ?

23

CPNCBZ

C. 39

3*2+1 = 7
15*2+1 = *31*
31*2+1 = 63
63*2+1 = 127
127*2+1 = 255
255*2+1 = 511

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

190.85

Circular Track sis of 11 km

Speed of Mens are

4 , 5.5 , 8 km/hr

Time at Which they are at starting Points again

11/4 , 11/5 , 11/8

11/4 , 2.2 , 11/8

We need to find LCM of these

to find at what time they Meet again at starting point

LCM 11 * 2 = 22

After 22 Hrs they will meet at starting point