This is a sequence of English letters, in numbers.
⇒ ten = 3 letters
⇒ nine = 4 letters
⇒ sixty = 5 letters
⇒ ninety = 6 letters
⇒ seventy = 7 letters
⇒ Sixty six =8 letters
⭕️This logical question goes with the no. of letters in the number!!!!!!!
⭕️∴ the biggest number with 9 letters will be ninety-six (96).
let d distance and s be speed.
d/7 = x and d/5 = x+12
solving we get d=210.
It’s choice A because you take the last to letters and move them to the front then the previous two letters go after them and so on.
In order to enhance my career, I asked my colleagues about my strength and weaknesses, the areas where I need to improve more, and acquired more knowledge from my superiors. I worked on my weaknesses to convert into a positive response.
11
Dear friends i ofcourse confused like you when i come
across to solve this problem. Really friends its very
simple if you understand the question clearly. First thing
is what is mean by “as many as” means its called ‘idiom and
phrase’ in english and it means “the same number of”. now
read the question “how many pairs of letters in STAINLESS
which has same number of letters between them in the word
as they have in english alphabet”.
In the alphabetical order, A-Z can be numbered as 1-26.
In A(INL)E which is same as in the alphabetical order A
(BCD)E. In both the cases E is in the Fourth position. so
we got one pair.
And in ST, there are no letters between them in the word
stainless. In alphabetical orer from A-Z also there is no
letters between them..so we got the second pair…
In STAINLESS it has two pairs ST and AE
Distance covered by B to meet A=Total distance – Distance covered by A hrs
[using : distance = speed x time]
Putting value of from equation (1),
hrs
Therefore, time at which both A and B will meet is = 7 a.m. + 3 hrs =10 am
33
8 games They played
A own 3 games = 18 Rs
B loss 3 Rs and own 1 games i,e 9(A own 3 Games)-6(B own
1 game)=3
c own 12 Rs and loss 4 games(A own 3+b own 1) and own 4
games i.e 24(own)-12(loss)
so totally A own 3 games
B own 1 game
c own 4 games
=8
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
FASHION= FOIHSAN
It is F-ASHIO-N to F-OIHSA-N.
So PROBLEM will be P-ROBLE-M,
the answer is PELBORM.