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)

Question number 37 olympiad 2013 section B

Solution :
Income on the nth day is =n^2+2..........(1)
Expenditure on nth day=2n+1..............(2)
Therefore ,the saving on nth day=n^-2n+1.....(3)
The savings on day1 to nth day when summed  up =1240......(Given)
The saving series from day-1  to nth day as follows like 0,1,4,9,16,----------n^2
Sum of this type of series= n(n+1)(2n+1)/6....................................Not A.P or G.P
Here the value of n=n-1 as first day saving is 0
Threfore ,(n-1)(n)(2n-2+1)/6=1240,=>n=16(Answer)

The given figure is created by using the arcs of quadrants with radii 1cm,2cm and 3cm.Find the total area of the shaded region.Take pi is equal to 3.14

Solution:To find
1.Find the area of all the smallest petals =8xarea of the smallest arc
2.Find the area of all the smallest petals =8xarea of the intermediate  arc
3.Find the area of all the smallest petals =8xarea of the biggest arc
4.The area of shaded region =8xarea of the smallest arc+8xarea of the biggest arc-8xarea of the intermediate  arc
5.The area of an arc=area of sector-area of the triangle inside =theta/360xx(radiu)^2-1/2x(radius)^2
6.Therefore ,
         1.The area of all the smallest petals=2.28cm^2
         2.The area of all the medium size  petals=9.12cm^2
         3.The area of all the largest petals=20.25cm^2
7.The entire shaded area in the figure=2.28cm^2+(20.25cm^2-9.12cm^2)=13.41cm^2(Answer)

The given figure shows sector OAB with centre O and radius 54cm. Another circle XYZ with centre P, is enclosed by the sector OAB.the angle AOB=60 degree . Find the area of OXPY.

Solⁿ: Given: sector OAB with centre O and radius 54cm.

                      <AOB=60⁰

                        OZ=OB=OA=54cm

Construction: join O and P.

Let XP=r, then OP=54-r

In ΔOXP  right angled at X,

        <XOP=30⁰

          Sin30⁰=XP/OP=r/54-r=1/2     then r=18cm-----------(1)

          OP=54-18=36cm

          So,XP²+OX²=OP²

          18²+OX²=36² => OX=18√3cm---------------(2)

Area of OXPY quadrilateral=2xarea of ΔOXP

                                              =2x ½ x 18√3 x 18 = 561.2cm(ans.)

Comment: The radius of a circle drawn inside a sector is 1/3 of the mother circle

       

If the roots of the quadratic equation x^2+Px+q=0 are tan30 and tan15 respectively ,then the value of 2+q-p is --------------------------------

Solution :tan(A+B)=tanA+tanB/1-tanA.tanB
then ,tan45 degree=tan30+tan15/1-tan30.tan15
then tan30+tan15+tan30.tan15=1
Then  2+q-p=2+tan30+tan15+tan30.tan15=3(Answer)

Three containers have their volume in ratio 3:4:5. They are full of mixtures of milk and water. The mixtures contain milk and water in the ratio of (4:1), (3:1) and (5:2) respectively. The contents of all these three containers are poured into a fourth container. The ratio of milk and water in the fourth container is

                                                                                                                                                                                                                                                                                                                                        Three containers have their volume in ratio 3:4:5. They are full of mixtures of milk and water. The     mixtures contain milk and water in the ratio of (4:1), (3:1) and (5:2) respectively. The contents of all these three containers are poured into a fourth container. The ratio of milk and water in the fourth container is?

A. 4:1

B. 151:48

C. 157:53

D. 5:2

                                                                                                                                                                                                                                                                                                                                        Ans: Let the ratio of milk and water in the first container be,

                                   M1/W1=4/1

        The ratio of milk and water in the second container be,              

                                   M2/W2=3/1

        The ratio of milk and water in the third container be,

                                   M3/W3=5/2

        Since, the three containers are in the ratio 3:4:5,

        

         In the first container,

                                   M1+W1=3x---------(1)

                                   M1/W1=4/1

                               => M1=4W1-------------------(2)

              Substituting (2) in (1),

                                   4W1+W1=3x

                               => 5W1=3x

                               => W1=3x/5-----------(3)

               Substituting (3) in (1) we get,

                                   M1=12x/5

         In the second container,

                                 M2+W2=4x---------(4)

                                   M2/W2=3/1

                               => M2=3W2------------------(5)

               Substituting (5) in (4) we get,

                                   3W2+W2=4x

                                   4W2=4x

                                   W2=4x/4=x---------(6)

                Substituting (6) in (4) we get,

                                   M2=3x

       In the third container,

                                   M3+W3=5x---------(7)

                                   M3/W3=5/2

                               => 2M3=5W3

                               => M3=5W3/2---------(8)

                Substituting (8) in (7) we get,

                                   5W3/2+W3=5x

                               => 5W3 +2W3/7=5x

                               => 7W3=10x

                               => W3=10x/7----------(9)

                Substituting (9) in (7),

                                   M3=25x/7

       Since all three mixtures are poured into the fourth container,

                            M4=M1+M2+M3 and

                            W4=W1+W2+W3

.: M4=12x/5 + 3x + 25x/7

       =84x+105x+125x/35

       =314x/35

.: W4=3x/5 + x +10x/5

        =21x + 35x + 50x/35

        =106x/35

M4/W4=314x/35÷106x/35

            =157/53

           =157:53(answer)

(3x-1)^7, then addition of all cofactors is =?

Question :(3x-1)⁷, then addition of all cofactors is =? if cofactors are A7,A6,A5,A4,A3,A2,A1,A0

Solution

1.       3x-1=3x-1

2.       (3x-1)²=9x²−6x+1

3.       (3x-1)³=27x³−27x²+9x−1

4.       (3x-1)⁴=81x⁴−108x³+54x²−12x+1

5.       (3x-1)⁵ =243x⁵−405x⁴+270x³−90x²+15x−1

6.       (3x-1)⁶ =729x⁶−1458x⁵+1215x⁴−540x³+135x²−18x+1

7.       (3x-1)⁷ =2187x⁷−5103x⁶+5103x⁵−2835x⁴+945x³−189x²+21x−1
  Key to answer :Addition of An,A(n-1),A(n-2).............A0=2^n

Algebra	Cofactors for the algebraic expression (3x-1)7
	A7		A6	A5	A4	A3		A2	A1	A0	Total
3x-1									3	-1	2
(3x-1)2								9	-6	1	4
(3x-1)3							27	-27	9	-1	8
(3x-1)4					81		-108	54	-12	1	16
(3x-1)5				243	-405		270	-90	15	-1	32
(3x-1)6		729		-1458	1215		-540	135	-18	1	64
(3x-1)7	2187	-5103		5103	-2835		945	-189	21	-1	128

If the polynomial 16X4—24X3+41X2-mx+16 is a perfect square, then find the value of m (A)12 (B) -12 (C) 24 (D) -24

If the polynomial 16X⁴—24X³+41X²-mx+16 is a perfect square, then find the value of m

(A)12 (B) -12 (C) 24 (D) -24

Solution :Let find it using division method to find square root of an algebraic expression

A does half as much work as B in three-fourth of the time. If together they take 18 days to complete the work, how much time shall B take to do it?

A does half as much work as B in three-fourth of the time. If together they take 18 days to complete the work, how much time shall B take to do it?

(A) 30 days

(B) 35 days

(C) 40 days

(D) None of these

Solution:

1.  Let B does the work in B days, then B in 1 day will do 1/B part of work

2.  A takes three-fourth of working days of B to do half of work done by B, therefore A in 3/4XB days will do 1/2 work, then in 1 day A would do 1/2÷3/4XB=2/3B part of work

3.  A and B together in 1 day will do =1/B+2/3B=5/3B part of work

4.  As given, A and B do 1 work in 18days, then A and B together in 1 day will do 1/18 part of the work

5.  Therefore, looking at step 3 and 4 ,5/3B=1/18, THEN B=30DAYS (Answer)

The ratio of the incomes of P and Q is 5 : 4 and the ratio of their expenditures is 3 : 2. If at the end of the year, each saves 1600, then the income of P is

The ratio of the incomes of P and Q is 5 : 4 and the ratio of their expenditures is 3 : 2. If at the end of the year, each saves 1600, then the income of P is=?

Solution :

1.       Let income of P and Q are PI and QI,then PI/QI=5/4,THEN QI=4/5.PI

2.       Let expenditure of P and Q are PE and QE,then PE/QE=3/2 ,THEN PE=3/2.QE

3.       Saving =Income -expenditure

4.       Saving of P is PI-PE=1600 AND QI-QE=1600, then QE=QI-1600

5.       We need income of P=PI

   From above (4) we know PI-PE=1600, Then replacing PE value from above (2), PI-3/2XQE=1600, then replacing QE from above (4), PI-3/2(QI-1600) =1600, then replacing QI from above (1), PI-3/2(4PI/5-1600) =1600, then PI-12PI/10+2400=1600, then -2PI/10=-800, then 2PI=8000, then PI=4000(ANSWER). 

If x/(b -c)(b+c—2a)= y/(c -a)(c+a-2b)= z/(a -b)(a+b-2c) then the value of (x + y * z) is?

If x/(b -c)(b+c—2a)= y/(c -a)(c+a-2b)= z/(a -b)(a+b-2c)

then the value of (x + y * z) is?

Solution :Let all three ratios are equal to a constant value of K
Then it goes like 

1. X=K{(b -c)(b+c—2a)}
2. Y=K{(c -a)(c+a-2b)}
3. C=K{(a -b)(a+b-2c)}

Then adding X,Y and Z would be =K{(b -c)(b+c—2a)+(c -a)(c+a-2b)+(a -b)(a+b-2c)}=KX0=0(Answer).