Ayush Kalia
1 year, 5 months ago

I want to know how to calculate term coefficients

Om Anand
1 year, 5 months ago

The maximum distance illuminated by a candle decreases uniformly from 9 metres at the start to 1 metre at the end of four hours. A new candle is lighted at the centre of a large room, what is the area of the portion initially lighted which goes dark after 2 hours?

Pv Aaditya
1 Likes Posted 1 year, 5 months ago

Since the decrease is uniform, the distance illuminated after 2 hours = 5 meters

Area initially illuminated = $$\pi * 9^2$$

Area illuminated after 2 hours = $$\pi * 5^2$$

Required area = $$81 \pi - 25 \pi = 56*22/7 = 176 m^2$$

Om Anand
1 year, 5 months ago

In a triangle XYZ, the external angle bisector of angle X and the internal bisector of angle Y meet at W. If angle XZY is 50 degrees, what is angle XWY?

Pv Aaditya
3 Likes Posted 1 year, 5 months ago

Let the external angle X be 2a. So, angle XWY = a, angle YXZ = 180 - 2a

Let angle XYZ = 2b, so angle XYW = b.

Angle XZY = 180-(180-2a+2b) = 2(a-b) = 50 => a-b = 25

Angle XWY = 180-(180-2a+a+b) = a - b = 25 degrees

Required angle = 25 degrees

Abhipraay Bajpai
0 Likes Posted 1 year, 5 months ago

Can you explain the Angle XWY = 180-(180-2a+a+b) part?

Shouldn't it be equal to 180-(90-a+b) as external angle bisector of angle X should also bisect the internal angle X?

Sagnik Sarkar
1 year, 5 months ago

7 different balls in 7 diff boxes (each with max 2 balls). No. of ways is a. 7 similar boxes with 7 diff balls (each with max 1 ball). No. of ways is b. What is a/b?

Pv Aaditya
2 Likes Posted 1 year, 5 months ago

Finding a:

Case 1: All the boxes have 1 ball each => 7! ways

Case 2: One box has 0 balls and one box has 2 balls, all the other boxes have 1 ball each => 7C1 (for selecting the box with 0) * 6C1 (selecting the box with 1)* 7C2 (selecting 2 balls to be placed in the box) * 5! (distribution of the remaining balls in the 5 boxes) ways

Case 3: Case of (2 2 1 1 1 0 0) => 7C2 (for selecting 2 boxes that have 0 balls) * 5C2 (for selecting 2 boxes that will have 2 balls each)* 7C2 (balls to be placed in the first box)* 5C2 (balls to be placed in the second box) * 3! (distribution of the remaining 3 balls) ways

Case 4: Case of (2 2 2 1 0 0 0) => 7C3 (selecting 3 boxes that will have 0 balls each) * 4C3 (selecting the box with 1 ball) * 7C2 (selecting 2 balls) * 5C2 (selecting the next set of 2 balls) * 3C2 (selecting the third set of 2 balls) ways

Sum of these = 463680

Finding b: There in only 1 way in which 7 distinct balls can be put in 7 similar boxes such that each box has only 1 ball

So, a/b = 463680

SUDHIR PAL
1 year, 5 months ago

if x+y+z=6 , and x>0 , y>0 , z>0 then find the maximum value of x*y^3*z^2

Pv Aaditya
0 Likes Posted 1 year, 5 months ago

x + y/3 + y/3 + y/3 + z/2 + z/2 = 6

AM >= GM

(x + y/3 + y/3 + y/3 + z/2 + z/2)/6 >= $$(x*y/3*y/3*y/3*z/2*z/2)^{1/6}$$

=> 1 >= $$(x*y/3*y/3*y/3*z/2*z/2)^{1/6}$$

=> 1 >= $$x*y^3*z^2 / (3^3*2^2)$$

=> $$x*y^3*z^2$$ <= 27*4 = 108

Max value = 108

Divakar Bogapurapu
1 year, 5 months ago

In the chess world championship, 15 matches take place between the top two ranked players in the world. How many ways can a person predict exactly 12 results (win, draw and loss) right?

Pv Aaditya
1 Likes Posted 1 year, 5 months ago

There are three results - win, draw, loss

Select the 12 matches in $$^{15}C_{12}$$ ways. For these, the correct prediction is possible in only 1 way.

For each of the other three matches, a wrong prediction can be done in 2 ways. So, total number of ways of predicting wrong = $$2^3$$

Total number of ways = $$^{15}C_{12}$$*$$2^3$$ = 3640 ways

Monika Jha
1 year, 5 months ago

How many different scalene triangles are possible with a perimeter of 99 units given that the lengths of all the sides are integers? A 175 B 168 C 192 D 196

Pv Aaditya
0 Likes Posted 1 year, 5 months ago

The maximum length of any side can be 49. (If one side is 50, then the sum of the other two sides, 49, will be less than this side)

Let the lengths of the three sides be 49-x, 49-y, 49-z, where x, y, z can vary from 0 to 48.

Now, since the perimeter is 99, 49 - x + 49 - y + 49 - z = 99 => x+y+z = 48 and 0 <= x, y, z <= 48.

This can be found out by using the number of integral solutions formula. $$^{n+r-1}C_{r-1}$$ = $$^{48+3-1}C_{3-1}$$ = $$^{50}C_2$$ = 49*25 =1225

This is the total number of triangles.

To get scalene triangles, we have to get the number of integral solutions for x+y+z = 48, where 0 <= x < y < z <= 48

If x = y = 0, there is one solution; similarly for x = y = 1 and so on till x = y = 24 -> 25 solutions in all (This also takes care of one equilateral triangle case).

Again, two variables can be equal in three ways - x,y ; y,z or z,x. So, the number of isosceles triangles is 24*3 = 72

Number of equilateral triangles = 1

So, number of scalene triangles = 1225 - 73 = 1152

But, there are for ordered x, y and z. Since we want the general case of unordered variables, we have to divide this by 3! = 6

So, number of scalene triangles =1152/6 = 192

Faraz Ahmad
1 year, 5 months ago

A boy and his dog are climbing a mountain. When the boy is 100m from the summit, his dog starts to run to the summit of the mountain. When it reaches the summit, it runs down the slope again. When the dog reaches the boy this time, he starts to run to the summit the second time, and this process continues until both have reached the top of the mountain. What is the length of the journey his dog has run?

Pv Aaditya
0 Likes Posted 1 year, 5 months ago

The ratio of speeds of the boy and the dog are needed to answer this question

ank patel
1 year, 5 months ago

If f(x)=16^x/ 16^x+4, find the value of f(1/852)+f(2/852)+f(3/852)+ ……..+f(851/852).

Pv Aaditya
1 Likes Posted 1 year, 5 months ago

f(x) = $$16^x/(16^x + 4)$$

f(1-x) = $$16^{1-x}/(16^{1-x}+4)$$ = $$16*16^x/[16^x*(16+4*16^x)]$$ = $$4/(16^x+4)$$

f(x) + f(1-x) = 1

f(1/852)+f(851/852) = 1

...

So, required sum = 1 + 1 + ... (425 times) + f(426/852) = 425 + f(1/2) = 425 + 0.5 = 425.5

135 points

46 points

45 points

26 points

25 points

Hi, do you have any specific example in mind?