150
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
The correct Answer is 69
let the average of 25 person be x
total age of a person=average×no.of person
total age of 25 person=X×25
=25X
when a new person of 46kg come average decrease by 5kg(X-5)
Total age of 26 person =25x+46
Average age of 26 person=X-5
A/Q
.25X+46=26(X-5)
25X+46=26X-130
26X-25X=130+46
X =176
hence the av. of 25 person be 176kg
ANSWER is ==> 1
1st step : 0.5
2nd step : 0.5+0.05 = 0.55
3rd step : 0.55+0.10 = 0.65
4th step : 0.65+0.15 = 0.8
5th step : 0.80+0.20 = 1.00
twenty members
127.179(app)
given distance of the train along the wind is 695
and againt the wind is 498
and time = distance/speed
as we know that time is equal in both the cases hence equate
695/s1=498/s2———–(1);
where s1=speed of the plane + speed of the wind
and s2=speed of the plane -speed of the wind
given that speed of the wind is 21k/h
s1=sp+21
s2=sp-21
substitu in eq 1
we get the answer as 27.17(app)
Let ‘N’ is the smallest number which divided by 13 and 16 leaves respective remainders of 2 and 5.
Required number = (LCM of 13 and 16) – (common difference of divisors and remainders)
= (208) – (11) = 197.
If both agree stating the same fact, either both of them speak truth of both speak false.
∴ Probability
=3/5×4/7+2/5×3/7=12/35+6/35=18/35
Speed = (45×518)m/sec=252m/sec
Total distance covered = (360+140) m = 500 m.
∴Required time= 500×225sec = 40 sec.
Rs. 1500