A conical vessel of radius 6 cm and height 8 cm is filled with water. A sphere is lowered into the water and its size is such that when it touches the sides of the conical vessel, it is just immersed. How much water will remain in the cone after the overflow?

A conical vessel of radius 6 cm and height 8 cm is filled with

water. A sphere is lowered into the water and its size is such

that when it touches the sides of the conical vessel, it is just

immersed. How much water will remain in the cone after the 

overflow? 

Solution :Let the radius of the spear=r and slant height of cone =h
Then r=A/S(The radius of a circle which is drawn inside a triangle) where A=area of trianle and S=semiperimeter of the triangle 
Here slant height=10cm
Area of trianle=48cm^2
Semiperimeter of the triangle=16cm
So radius=48/16=3cm
The left water content in the cone after submerging the spear=1/3x6^2x8-4/3x3^3=188.57cm^2(Answer)

Mohit went from Delhi to Shimla via Chandigarh by car. The distance from Delhi to Chandigarh is 3/4times the distance from Chandigarh to Shimla. The average speed from Delhi to Chandigarh was 1 and 1/2 times that from Chandigarh to Shimla. If the average speed for the entire journey was 49km/hr, what was the average speed from Chandigarh to Shimla?

Solution :
           

                    Delhi________________Chandigarh______________Shimla
Let the distance from Chandigarh to Shimla be X.km=d2
So, the distance from Delhi to Chandigarh be 3X/4.km=d1

Let the speed from Chandigarh to Shimla be y km/hr=S2
So,the speed from Delhi to Chandigarh =3/2y=S1

Let T1 be the time taken from Delhi to Chandigarh=(3x/4)/(3y/2)=1/2.x/y hrs
Let T2 be the time taken from Chandigarh to Shimla=x/y hrs
So total time of journey from Delhi to Simla=3/2xy

The total distance from Delhi to Simla=7x/4
The average speed from Delhi to Simla=(7x/4)/(3/2.x/y)
                                                               =>7y/6=49
                                                               =>y=42 km/hr(Answer)

There are three pipes fitted in a tank. First two pipes when operated simultaneously, fill the tank in same time as the third pipe alone. the second pipe fills the tank 7 hours faster than the first pipe and 3hours slower than the third pipe. the approximate time required by each pipe to fill the tank simultaneously is

There are three pipes fitted in a tank. First two pipes when operated simultaneously, fill the tank in same time as the third pipe alone. The second pipe fills the tank 7 hours faster than the first pipe and 3hours slower than the third pipe. The approximate time required by each pipe to fill the tank simultaneously is
Solution :
Let  the third pipe fills the tank in x hours
So, second pipe will fill the tank in x+3 hours
       first pipe will fill the tank in x+3+7=x+10 hours
The third pipe in 1hour will fill 1/x part of the tank
The second pipe in 1hour will fill 1/x+3 part of the tank
The first pipe in 1hour will fill 1/x+10 part of the tank

First two pipes when operated simultaneously, fill the tank in same time as the third pipe alone ,so
(1/x+3) + (1/x+10)=1/x
=>x=(30)^1/2=5.4...=5 and 1/2 hours=Filling hour of third pipe
:. filling hour of first pipe=x+10=15 and 1/2 hrs
    filling hour of second pipe=x+3=8 and 1/2hrs(Answer)

___?____ times the sum of the squares of sides of a triangle is equal to four times the sum of the square of the medians of the same triangle?

____?____ times the sum of the squares of sides of a triangle is equal to four times the sum of the square of the medians of the same triangle?

Solution: Let, the be triangle be a equilateral triangle ABC as follows

AO, BM and CN are the medians of the triangle.
Here, they are all equal.
Three times of the  sum of the squares of sides of a triangle=3(3a^2)=9a^2=a^2-(a/2)^2........(1)
Four times the sum of the square of the medians of the same triangle=4[3{a^2-(a/2)^2}]=9a^2....(2)
Therefore, (1)=(2)......solved.

THE RATIO OF THE SUM OF n terms of two A.P. is(7n+1):(4n+27).find the ratio of 11th terms

Given ratio of sum of n terms of two AP’s = (7n+1):(4n+27)͢   (1)
Let’s consider the ratio these two AP’s mth terms as a_m : a’_m →(2)
Recall the nth term of AP formula, a_n = a + (n – 1)d
Hence equation (2) becomes,
a_m : a’_m = a + (m – 1)d : a’ + (m – 1)d’
On multiplying by 2, we get
a_m : a’_m = [2a + 2(m – 1)d] : [2a’ + 2(m – 1)d’]
= [2a + {(2m – 1) – 1}d] : [2a’ + {(2m – 1) – 1}d’]
= S_{2m – 1} : S’_{2m – 1} 
= [7(2m – 1) + 1] : [4(2m – 1) +27] [from (1)]
= [14m – 7 +1] : [8m – 4 + 27]
= [14m – 6] : [8m + 23]
Thus the ratio of mth terms of two AP’s is [14m – 6] : [8m + 23]
Therefore ,the 11th term ratio=14x11-6/8x11+23=148/111(Answ).

f Cosec-Sin=a3 and Sec-Cos=b3, then a2b2(a2+b2) =?

If Cosec-Sin=a³ and Sec-Cos=b^3, then a²b²(a²+b²) =?

Answer: Cosec-Sin=a³ and Sec-Cos=b^3, then a²b²(a²+b²)

=> Cosec-Sin=a³=> a³=Cos²/sin and Sec-Cos=b³=>b³= Sin²/cos

=>a²b²(a²+b²) =a⁴b²+a²b⁴= (a²b+ab²)²-2a³b³=[(a⁶b³)^1/3+(a³b⁶)^1/3]²-2a³b³

=>Applying the values of a³=Cos²/sin and b³= Sin²/cos in the expression [(a⁶b³)^1/3+(a³b⁶)^1/3]²-2a³b³, [(a⁶b³)^1/3+(a³b⁶)^1/3]²-2a³b³=(Sin+Cos)²- 2Sin.Cos=1(Answer)