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.
3
30,40
Ans is 25
125, 106, 88, 76, 65, 58, 53
88
435
answer is 28
>2+8=10
>reverse of 28 is 82
>subtract 28 from 82
>82-28=54
the number is 28
Capacity of the tank =(12 x 13.5) liters =162 liters.
Capacity of each bucket =9 liters
Number of buckets needed = 162/9 =18.
10C3 = 120
1/32(x)L=21h
10L=1h
X=21x10x32
=6720 Liters
Thus:
6720L=Yh
10L=1h
Yh=6720/10
Y=672h
let x be sum.
SI of 18% for 1 year= x*18/100=0.18x;
SI of 18% for 2 years is 0.36x;
SI of 12% for 1 year= x*12/100=0.12x;
SI of 12% for 2 years is 0.24x;
Given, 0.36x-0.24x=840
0.12x=840
x=840/0.12=7000.
B, 8mm x 5 mm = 40 mm^2
1000mm x 1000mm = 1,000,000mm^2
= 1,000,000 mm^2/40mm^2
= 25,000
17
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.
120
512