Maths

Option (b) and (c) both satisfies the given condition, but 997918 is the larger of the two.

Since x > 0 then at x = 1, we get y = 2 and for the rest values we get y > 2

Also for 0 < x < 1 ; x+ $\frac{1}{x}$ > 2

and for x > 1 ; x+ $\frac{1}{x}$ > 2

Hence, y ≥ 2

We know that,

$a^2+a^2={\left(4\sqrt[]{2}\right)}^2,a^2=16,a=4$

Now, 2 (area of square) = (square of diagonal)

So, for new area 32 (16 * 2)

2 * 32 = d²

8 = d

Final ratio of wages = 5 * 4 : 7 * 3 : 9 * 2 = 20 : 21 : 18

Total wages received by P = $\frac{20}{59}*5900=2000$

Daily wage of P = $\frac{2000}{5}=400$

Let h = height of the container,

Given, h = 2r

$\therefore$ Volume v = πr²h = πr² (2r) = 2πr³

$\therefore$ Absolute error = (1.02)³ * 2πr³ − 2πr³ * 13

= 2πr³ [ (1.02)² − 1] = 2πr³ [1.0612 − 1]

= 0.0612 * 2πr³

$\therefore$ % error $=\frac{0.0612*2\pi r^3}{2\pi r^3}*100\mathrm{\%}=6.12\mathrm{\%}$

Put a = 2, b = 12 in $\sqrt[]{a^b}=5b+a^2$

$\therefore \sqrt[]{2^{12}}=5*12+4=64$

$\Rightarrow$ 26 = 64, which is true.

For every day's work, P1 can afford to miss 3 days.

Hence, to break even it has to be 7 days' work in 28 days.

Let Ab = X km and speed of bus be Y km/hr and time be T hrs original equation becomes $\frac{X}{Y}=T.....(1)$

By increasing speed by 7 km/hr equation becomes

$\frac{X}{y+7}=T-1.....(2)$

By decreasing speed by 5 km/hr equation becomes

$\frac{X}{y-5}=T+2.....(3)$

On solving equation 1, 2 & 3 we get the value of y as and

$\frac{35}{3}$ and $T=\frac{8}{3}$

$\therefore$ AB = X = $\frac{280}{9}$

≈ 31 km

A + 4 = B + 5 = C + 6 = D + 7 = P + B + C + D + 8

A + 4 = B + 5 A – 1 = B

A + 4 = C + 6 A – 2 = C

A + 4 = D + 7 A – 3 = D

Now, A + 4 = A + A – 1 + A – 2 + A – 3 + 8

A $=\frac{2}{3}$

$B=\frac{-1}{3},C=\frac{-4}{3},D=\frac{-7}{3}$

So, A + B + C + D =