U is the answer as
T is sitting opposite to R and W is sitting opposite to U
let t = total no of students.. then
students who passed one or both subjects,
n(e U h) = n(e) + n(h) – n(e intersection h)
=> t = 0.8t + 0.7t – 144
=> t = 1.5t – 144
students who failed both subjects is 10% i.e. 0.1t
=>t-n(e U h) = 0.1t,
=>t -(1.5t – 144) = 0.1t
=>t- 1.5t- 0.1t = -144
=> -0.6t = -144
=>t = 240
1 – (-1) = 1 +1 = 2
The ans is -20.
Solution:
A-1
B-2
C-3 D-4 …like wise till Z-26.
ELECTRICITY-5+12+5+3+20+18+9+3+9+20+25==129
GAS-7+1+9==27
ELECTRICITY-GAS=129-27-(Minus 2)=100
so
JACK-JILL=(10+1+3+11-(10+9+12+12)-(minus(2))==(-20)
4
Servlets are the Java programs that run on the Java-enabled web server or application server. They are used to handle the request obtained from the webserver, process the request, produce the response, then send the response back to the webserver.
Niece
40
x – 30 = 1/4 x || *4
4x – 120 = x || -x + 120
3x = 120 || /3
x = 40
(c) 25%
20*55*65=y*65*75
y=(20*55*65)/(65*75)
y=14.667 =15
y=15
% reduction=(20-15)/20=25%
Its 2 times faster than the other train
v1*t=v2
v2*t=4*v1
solving these two,we get
v2/v1=2
G.c.d=num1 *num2/lcm
Num1=20
Gcd=5
Lcm=60
5=(20*num2)/60
Num2= 5*60/20
Num2=15
Lets call the 5 litre jug as jug A and 3 litre jug as jug B. Now, follow the steps:
Fill jug A completely. Now it contains 5 litres.
Slowly pour the water from jug A to jug B until jug B is completely filled. Now, jug A contains 2 litres and jug B contains 3 litres.
Throw away the water in jug B so that it is completely empty. Now, jug A contains 2 litres and jug B is empty.
Transfer the water from jug A to jug B. Now, jug A is empty and jug B contains 2 litres.
Fill jug A completely. Now, jug A contains 5 litres and jug B contains 2 litres.
Transfer water from jug A to jug B until jug B is completely filled. Now, jug A contains 4 litres and jug B contains 3 litres.
Now you have 4 litres of water in jug A.
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