Question #f5a58

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Question #f5a58

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Answer 1 · 2017-09-19T06:23:20

8

Explanation:

Let friends be x and bill comes Rs. 1120. So 1 has to pay Rs. 1120/x.

As 3 have no wallets, so ( x - 3 ) friends have to pay Rs. 84 each as an extra. Means Rs (1120/x + 84) each.

Now as per question,
$(\frac{1120}{x} + 84) (x - 3) = 1120$ $(1120/x +84)(x-3) = 1120$

$\Rightarrow \frac{1120}{x} . x + 84 x - \frac{3.1120}{x} - 84.3 = 1120$ $rArr 1120/x. x+84x-3.1120/x-84.3 = 1120$

$\Rightarrow 84 x - \frac{3360}{x} = 1120 - 1120 + 252$ $rArr 84x - 3360/x = 1120-1120+252$

$\Rightarrow 84 x^{2} - 252 x - 3360 = 0$ $rArr 84x^2 -252x-3360 = 0$

$\Rightarrow 84 (x^{2} - 3 x - 40) = 0$ $rArr 84(x^2-3x-40) = 0$

$\Rightarrow x^{2} - 3 x - 40 = 0$ $rArr x^2 - 3x - 40 = 0$

$\Rightarrow x^{2} - 8 x + 5 x - 40 = 0$ $rArr x^2-8x+5x-40 = 0$

$\Rightarrow x (x - 8) + 5 (x - 8) = 0$ $rArr x(x-8)+5(x-8)=0$

$\Rightarrow (x - 8) (x + 5) = 0$ $rArr (x-8)(x+5)=0$

$\Rightarrow x - 8 = 0, x + 5 = 0$ $rArr x-8 =0, x+5=0$

x = 8 , -5 [ x cannot be -5]

hence x = 8

SO, TOTAL FRIENDS BE 8

Answer 2 · 2017-09-19T06:33:03

Eight.

Explanation:

This situation can be expressed as the equation

$T = (\frac{T}{P} + E) (P - 3)$ $T = (\frac{T}{P} + E ) (P-3)$

where $T$ $T$ is the total dinner price, $P$ $P$ is the number of people and $E$ $E$ is the extra money.

(Think about this until it makes sense to you. The point of this problem is to practice converting a situation into an equation that models it.)

Each person who is paying is paying what they would have paid if everyone was paying, plus an extra $$ 84$ $$84$ . That is, each person is paying $\frac{T}{P}$ $\frac{T}{P}$ (what they would have paid if the price was evenly divided between the friends), plus some extra amount $E$ $E$ , so $\frac{T}{P} + E$ $\frac{T}{P} + E$ each in total. To find how much that is collectively, multiply by the number of people paying that much, $P - 3$ $P-3$ . This should be equal to the total price for the dinner.

From here, simply expand the equation and solve for $P$ $P$ using the quadratic formula.

$T = (\frac{T}{P} + E) (P - 3)$ $T = (\frac{T}{P} + E ) (P-3)$
$T = T + E P - \frac{3 T}{P} - 3 E$ $T = T + EP - \frac{3T}{P} - 3E$
$0 = E P - \frac{3 T}{P} - 3 E$ $0 = EP - \frac{3T}{P} - 3E$
$0 = E P^{2} - 3 E P - 3 T$ $0 = EP^2 - 3EP - 3T$

Substitute $E = 84$ $E = 84$ and $T = 1120$ $T = 1120$ and solve using the quadratic formula for finding roots of equations of the form $a x^{2} + b x + c$ $ax^2 + bx + c$ :

$P = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 b}$ $P = \frac{-b \pm \sqrt{b^2 - 4ac}}{2b}$
$P = \frac{3 E \pm \sqrt{9 E^{2} + 12 E T}}{2 E}$ $P = \frac{3E \pm \sqrt{9E^2 + 12ET}}{2E}$
$P = \frac{1344}{168}$ $P = \frac{1344}{168}$
$P = 8$ $P = 8$

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