Leadsquared Interview Questions, Process, and Tips

10000

one-legged =5% of 10000=500

remaining=10000-500=9500

barefooted=9500/2=4750

remaining people= 9500-4750=4750

hence required number of shoes= 4750*2+500*1=100005% of 10000 = 500 one legged

9500 / 2 = 4750 bare foot

minium no of shoes = 4750*2 + 500*1 = 10000

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(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

Given y/x = 1/3, x+2y =10
3y=x
Then substitute x=3y in x+2y=10

3y+2y=10
5y=10
y=2

Then substitute y=2 in x=3y
x=3*2
x=6
: x=6, y=2

Three man’s on day work = 1/6+1/7+1/8=73/168
three man’s can complete the work in 1/73/168 days
= 168/73 days
Now if they works together fro the alternet days they will
complete the worksin 2*168/73 days
(If three mans working for the alternate days then work
completion time will be doubbled)

6+8+10+10= 36
6²+8²+10²+12²=344

2x+x+2x+x = 300m
6x = 300
x = 300/6 = 50m
50m x 100m = 5,000 m2

Given:

In a group of 15 students,

7 have studied Latin,

8 have studied Greek,

3 have not studied either.

To find:

The number of students who studied both Latin and Greek.

Solution:

In a group of 15 students, have studied Latin, 8 have studied Greek, 3 have not studied either.

Therefore,

n(A∪B) = 15 – 3

n(A∪B) = 12

7 have studied Latin,

n(A) = 7

8 have studied Greek,

n(B) = 8

n(A∩B) is the number of students who studied both Latin and Greek.

n(A∩B) = n(A) + n(B) – n(A∪B)

n(A∩B) = 7 + 8 – 12

n(A∩B) = 15 – 12

n(A∩B) = 3

The number of students who studied both Latin and Greek is 3

Final answer:

3 of them studied both Latin and Greek.

Thus, the correct answer .3

Collective performance
Postive attitude towards working
Shaping team player
Turning individuals into team players

The queen gave two rings to two daughter’s and she kept the
third ring in her jewel box and she presented the ring to
her third daughter along with that jewel box.

427 is the answer
Because
385 means (3+5=8) or (8-3=5) ,
462 means (4+2=6) or (6-2=4) ,
572 means (5+2=7) or (7-5=2) ,
396 means (6+3=9) or (9-3=6) ,
671 means (6+1=7) or (7-6=1) ,
264 means (4+2=6) or (6-2=4) ,
but only 427 is the number in which if we do addition of any of its two digits (e.g 4+2 or 4+7) then its answer doesn’t come 3rd digit
like if we do 4+2 then answer will be 6 so the number should be 426 but it is not
or if we do 7-4 then answer will be 3 (not 2)
so the right answer is 427

Let the numbers be x and x + 2.
Then, (x + 2)2 – x2 = 84
⇒ 4x + 4 = 84
⇒ 4x = 80
⇒ x = 20.
∴ The required sum
= x + (x + 2)
= 2x + 2
= 42

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

24 times

Ans is 25