the answer is At 9:48 PM
At 1:00 pm the difference between A & B = 8 km
after 2:00 pm ………………. = 11 km (as B’s speed
is 1 and A’s 4 km, then eqv speed=(4-1)=3 km)
After 3:00………………….. = 13 km (as B’s speed 2 km)
After 4:00………………….. = 14 km
after 5:00………………….. = 14 km (A’s speed= B’s
speed)
after 6:00………………….. = 13 km
after 7:00………………….. = 11 km
after 8:00………………….. = 8 km
after 9:00………………….. = 4 km
and now the eqv speed is= (9-4) =5 km/hr;
and the renaming distance is 4 km;
then, time=(60*4)/5=48 min;
then the meeting time is=9:00+48 min=9:48 pm;
Very great things your doing
Alice — 3844
Liu —30976
TOTAL NUMBER OF BOOKS 40+40=80
ISSUED BOOKS 30
REMAINING BOOKS 80-30=50
RATE=50/80=5/8
In a cube all the diagonal and sides are equal, we can go diagonally.
5c1 / 9c1
5 + 6 + 7 + 8 + 9 = 35 so two primes 5 and 7.
let no. of boys not participating be x
then the no. of girls not participating = x+5
no. of boys : girls participating = 3:2
given no.of boys participating = 15
therefore, the ratio is now 15:y(say)
then 3:2 = 15 : x
on solving 3/2 =15/y ie.., 3y =30 we get y =10
hence no. of girls participating =10
therefore total no of students paricipating = 15+10=25
total no of students in class =60 given
hence no. of students not participating = 60-25=35
therefore x+(x+5)=35
2x=30
x=15
therfore no of girls not participating =15+5=20
therefore total no of girls in class = no of girls
participating + no of girls not participating
=10+20
=30 is the answer
suppose
pipe:
A -30 hours A’s effeciency (60/30) =2
60( lcm of 30 and 20)
B- 20 hours B’s effeciency (60/20)=3
time taken by both to fill = 60/5 =12 as given in question (effeciencies of both a+b =2+3=5)
time taken by faster pipe i.e b = 60/3 =20
X-340 = X x 4 x 8/ 100
100X – 34000 = 32X
68X = 34000
X = 34000/68
X= 500.
Take 5 pills from jar 1, 4 pills from jar2,…. and 1 pill
from jar 5 and put altogether in the scale. the ideal
weight should be (1+2+3+4+5)x10= 150gms. for eg if the jar2
is contaminated then the weight will be 146 gms. so
depending on the amount of weight loss we can identify the
contaminated jar.
400%
e.g:-
l = 5 b = 2
Area= l*b =10
New after 100% increament
l=10 b = 4
Area = 10*4
D = 180m
S = 42 – 6 = 36 * 5/18 = 10
T = D/S = 180/10
T = 18s