Maths

Total volume = 2800 cubic feet

Area of ceiling = 2800/28 = 100sqfeet

Total time taken = $\frac{100}{0.125}$ = 800min

= 13hr 20min

Time taken to paint four walls= 40 – 13* $\frac{1}{3}$ = 26hr 40min

Let us find the pattern

10⁴ – 21 = 9979

10⁶ – 21 = 999979

10⁸ – 21 = 99999979

.         .         .

.         .         .

.         .         .

10ⁿ – 21 = 999 (n–2times) 79

When, n = 78

10⁷⁸ – 21 = 999 (76times) 79

Sun of all digit = 77 * 9 + 7

= 693 + 7 = 700

Let the CP of apple be p

Then, px + (p + 4)y – 8 = p * 3x + (p+4) (y–6)

px +py +4y = 3px + py – 6p + 4y – 24 + 8

2px – 6p = 16

p (x–3) = 8

The maximum value of p is 8 at x = 4

Then, CP of an apple and mango is 8 + 12 = 20 Rs.

Let fare for business class and economy class be Rs. 3x and Rs. 2x respectively. Let the passenger in business class and economy class be 12y and 25y respectively.

Total fare = 3x * 12y + 2x * 25y

=86xy = 430000

xy = 5000

Absolute difference = (50 – 36) * 5000

= Rs.70, 000

Let X = Total Amount of Gold

& Y = Total Amount of Platinum

$\frac{}{}$ $\frac{X}{Y}=\frac{2}{5}$  . (1)

Given U1 + P1 = U2 + P2. (2)

U1 + U1 = X & P1 + P2 = Y. (3)

Also, given U1 = 3U2

From (2) P2 − P1 = 2U2

Also, X = 4U2 & 2P1 = Y − 2U2

$\frac{X}{Y}=\frac{2}{5}=\frac{4U_2}{2P_1+2U_2}$

4P1 + 4U2 = 20U2

4P1 = 16U2 $=\frac{16}{3}{U}_1$

$\frac{U_1}{P_1}=\frac{4*3}{16}=\frac{3}{4}$ $\frac{}{}{}_{}$

U1 : P1 = 3 : 4

Let Gopal's share be G, Abhishek's be A, Sadiq's be S and Pawan's be P.

Given that $G+3=A+\frac{A}{3}=\frac{80S}{100}=P-4$ $\frac{}{}\frac{}{}$

You get

G = P – 7. (1)

$$ $A=\frac{3}{4}(P-4)$  . (2)

$$ $S=\frac{5}{4}(P-4)$  …. (3)

Also given that G + A + P + S = 80…. (4)

Substituting the values from (1), (2) and (3) in (4), we get

$\frac{}{}\frac{}{}$  $\left[P+\frac{3P}{4}+\frac{5P}{4}+P\right]-7-3-5=80$ or P = 23.75

For n = 14, 39 (1 + 33 + 36 + 35)

= 39 (1 + 27 + 729 + 243)

= 39 * 103.

$ForMedian=L+\frac{\frac{N}{2}-c,}{f}xh$

$14=12+\frac{\frac{a+b}{2}+13-\left(a+b\right)}{12}*6$              

->a + b = 18                                   -(ii)

Solving (i) & (ii) a = 8 & b = 10

->(a – b)² = 4

Volume of tube $=\frac{2}{3}\pi a^3+\pi a^{2}b$ $\frac{}{}{}^{}{}^{}$

When it is half-full $=\frac{1}{3}\pi a^3+\frac{1}{2}\pi a^{2}b$ $\frac{}{}{}^{}\frac{}{}{}^{}$

Volume of Cylinder

$=\frac{1}{3}\pi a^3+\frac{1}{2}\pi a^{2}b-\frac{2}{3}\pi a^3=\frac{1}{2}\pi a^{2}b-\frac{1}{3}\pi a^3$

Length in the cylinder

$=\left(\frac{1}{2}\pi a^{2}b-\frac{1}{3}\pi a^3\right)÷\pi a^2=\frac{1}{2}b-\frac{1}{3}a$

The height above the lowest point

$\frac{1}{2}b-\frac{1}{3}a+a=\frac{2}{3}a+\frac{1}{2}b$

When it is $\frac{1}{4}$ $\frac{}{}$ full, volume $=\frac{1}{6}\pi a^3+\frac{1}{4}{\pi }^{a}b$ $\frac{}{}{}^{}\frac{}{}{}^{}$

Volume of cylinder

$=\frac{1}{6}\pi a^3+\frac{1}{4}\pi a^{2}b-\frac{2}{3}\pi a^3=\frac{1}{4}\pi a^{2}b-\frac{1}{2}\pi a^3$

Length in the cylinder

$=\left(\frac{1}{4}\pi a^{2}b-\frac{1}{2}\pi a^3\right)÷\pi a^2=\frac{1}{4}b-\frac{1}{2}a$

The height above the lowest point

$=\frac{1}{4}b-\frac{1}{2}a+a=\frac{1}{2}a+\frac{1}{4}b$

When it is full, volume 

$=\frac{3}{4}\left(\frac{2}{3}\pi a^3+\pi a^{2}b\right)=\frac{1}{2}\pi a^3+\frac{3}{4}\pi a^{2}b$

Volume in cylinder

$=\frac{1}{2}\pi a^2+\frac{3}{4}\pi a^{2}b-\frac{2}{3}\pi a^3=\frac{3}{4}\pi a^{2}b-\frac{1}{6}\pi a^3$

Length in the cylinder

$=\left(\frac{3}{4}\pi a^{2}b-\frac{1}{6}\pi a^3\right)÷\pi a^2=\frac{3}{4}b-\frac{1}{6}a$

The height above the lowest point

$=\frac{3}{4}b-\frac{1}{6}a+a=\frac{5}{6}a+\frac{3}{4}b$

${\left(2x^r+\frac{1}{x^2}\right)}^{10}$

Let the constant term is (k + 1)^th term.

${\therefore }^{10}{C}_k{\left(2x^r\right)}^{10-k}.x^{-2k}$              

${=}^{10}{C}_k{2}^{10-k}.x^{10r-rk-2k}$              

              For constant term 10r – rk – 2k = 0.

$\therefore k=\frac{10r}{r+2}k\in I$              

$\therefore r=3,k=6andr=8,k=8$               

For k = 6

For k = 8  

$\therefore r=8$