Question

plz answer the this question with step

Answer 1

hey mate,
here is the answer
ask if any queries
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Total expenditure = Rs. 6,000
let no. of students going for the trip previously be 'x' and money per student be 'y'

A/C - I
$x.y = 6000 \: \: \: \: \: \: - (i)$

A/C - II

$(x+10)(y-20)= 6000$

$xy - 20x + 10y - 200 = 6000$

$6000 - 20x + 10( \frac{6000}{x} ) \\ - 200 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 6000$
(: by eq. (i)

$- 20 {x}^{2} + 60000 - 200x = 0$

$20 {x }^{2} + 200x - 60000 = 0$

${x }^{2} + 10x - 3000 = 0$

${x}^{2} + 60x - 50x - 3000 = 0$

$x(x + 60) - 50(x + 60) = 0$
$(x + 60)(x - 50) = 0$
$x = 50 \: \: \: or \: \: \: - 60$
negative no. neglected
x = 50

so number of students going for trip is
$x + 10$
$50 + 10 = 60$
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