Maths

Speed of A with respect to B

AB = 36 – 24 = 12 km/hr

Time taken by A to catch B

AB = 120/12 = 10 hours

(20*18*10)/1500 = (45*16*4)/Q

Q = 1200 tons

Let total work is 60 (LCM of 20, 12 & 15)

Per day work 

(Raj + Mohan) = 3 units

(Mohan + Sohan) = 5 units

(Raj + Shohan) = 4 units

(Raj + Mohan + Sohan) = (3 + 5 + 4)/2 = 6 units

Raj's per day work = 6 – 5 = 1 unit

Total time taken by Raj = 60 days

Total days taken by Mahesh alone = 45*4 = 180 days

Let the total work = 180 units (LCM of 180 and 60)

Mahesh per day work = 1 units

Mahesh and Anand per day work = 180/60 = 3 units

Anand per day work = 3 – 1 = 2

Time taken = 180/2 = 90 days

Food for 80 person = 150 days

Food for 1 person = 150 * 80

Food for 150 person = (150 * 80/150)

= 80 days

Let the total work is 300 units (LCM of 150,100 and 60)

A's per day work = 2 units

B's per day work = 3 units

C's per day work = 5 units

(A+ B + C)'s per day work = 2 + 3 + 5 = 10 units

Time taken = 300/10 = 30 Days

Let y = 2^111x then, $\frac{1}{4}{y}^3+4y=2y^2+1$  

y³ – 8y² +16y – 4 = 0

Let three real roots be a, b and c then abc = and corresponding values of x be x₁,   x₂, and x_3\

$2^{111 (x_1 + x_2 + x_3)}= 4$

111 (x₁ + x₂ + x₃) = 2

x₁ + x₂ + x₃ = 2/111

m + n = 2 + 111 = 113

p and q are the roots of the equation x² – 4ax + 3 = 0

pq = 3

Similarly the constant term of x² – 2bx + c = 0

Product of $\left(\frac{2}{p}-q\right)and\left(\frac{2}{q}-p\right)$

Hence, c = $\frac{4}{pq}-2-2+pq=\frac{1}{3}$

f (x) = $\frac{x}{4x^2+7x+9}$ ………. (1)

Divide by x in numerator and denominator equation (1) can be written as f (x) = $\frac{1}{4x+7x+\frac{9}{x}}$

f (x) is maximum, when the denominator is minimum we have to find the minimum value of $4x+\frac{9}{x}$ , so, denomination will also minimum

we know that, AM ≥ GM

$\frac{4x+\frac{9}{x}}{2}\ge \sqrt[]{4x*\frac{9}{x}}$

$\frac{4x+\frac{9}{x}}{2}=6$

$4x+\frac{9}{x}=12$

The minimum value of $4x+\frac{9}{x}$ is 12

Therefore, the maximum value of f (x) = $\frac{1}{12+7}=\frac{1}{19}$

For Aman,

SP of 1 book = 14Rs.

CP of 1 book = $\frac{100}{(100+x)}*14$

For Baman,

SP of 1 book = 12Rs.

CP of 1 book = $\frac{100}{(100+y)}*12$

Price of all notebooks is same, therefore.

$\frac{100}{(100+x)}*14$ = $\frac{100*12}{100+y}$

100 + 7y = 6a

100 = 12y – 7y; [x=2y]

5y = 100

y = 20

a = 40

CP each notebook = $\frac{100*12}{100+20}$ = 10Rs.

In his sales with Daman, he sold 10 books to her for Rs. 130

CP of 10books = Rs. 100

Profit percent = $\frac{130-100}{30}*100$

=30percent