1/x
The last person covered 120.71 meters.
It is given that the platoon and the last person moved with
uniform speed. Also, they both moved for the identical
amount of time. Hence, the ratio of the distance they
covered – while person moving forward and backword – are
equal.
Let’s assume that when the last person reached the first
person, the platoon moved X meters forward.
Thus, while moving forward the last person moved (50+X)
meters whereas the platoon moved X meters.
Similarly, while moving back the last person moved [50-(50-
X)] X meters whereas the platoon moved (50-X) meters.
Now, as the ratios are equal,
(50+X)/X = X/(50-X)
(50+X)*(50-X) = X*X
Solving, X=35.355 meters
Thus, total distance covered by the last person
= (50+X) + X
= 2*X + 50
= 2*(35.355) + 50
= 120.71 meters
Note that at first glance, one might think that the total
distance covered by the last person is 100 meters, as he
ran the total lenght of the platoon (50 meters) twice.
TRUE, but that’s the relative distance covered by the last
person i.e. assuming that the platoon is stationary.
Bus started at 8:00
it travelled with 18mph…
distance of destination is 27 miles…
we know velocity=(distance)/time
i.e., time=27/18hours=3/2hours=90 min…
i.e., the bus reached destination at 9:30
the bus stayed for 30 min…
so the bus started return journey at 10:00
now the bus returned with velocity 18 + (18/2)=27mph
the taken to the bus to travel = 27/27=1 hour = 60 min…
so bus would be returned on 11:00…
choosing my career path
i will drop a mail saying that can you pleaae meal me the the originals of the stregic plan
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
50 each
3x -y = 100
4x-2y = 100
A & B one day work= 1/18 + 1/30= 8/90
As we know that number of work is 1
1÷8/90= 1*90/8= 90/8.
Since they say “ twice the amount of work
Then ,2*90/8= 22.5 days
33
CITOXE
Initially, potatoes have 99% water by weight ; which means they have 1% solid non-water content.
1% of 100 kg = 1 kg
Now even when they dehydrate,this 1kg solid mass remains constant.
It is given that finally, 98% is water by weight , which implies that 2% is non-water solid.
This means 2% of total weight = 1 kg
Total weight *2/100 = 1 kg
Therefore, total weight finally = 50 kg.
a)0
b)206
c)0
d)250
e)39
f)92
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
If a blue stone is thrown into a red sea, several things could happen depending on the context and the properties of the stone and the sea:
Symbolically: Since blue and red are contrasting colors, the interaction of a blue stone in a red sea could be seen as a visual or metaphorical contrast. It could represent a stark difference or an unexpected element introduced into an existing situation.
Scientifically: In reality, the color of the stone and the sea would not have a direct physical impact on each other. The stone would sink or float based on its density and the water’s buoyancy. The color of the water, whether red or any other color, does not change the fundamental principles of objects interacting with liquids.
It’s important to note that red seas, in the context of bodies of water, typically do not exist naturally. The phrase “red sea” is often used metaphorically or symbolically rather than referring to an actual body of water with a red color.