His walking distance is 7 km.
We know, 1 hour 15 minutes = 1.25 hours
Let, time taken by the man to walk is x hours
Time taken by the man to walk is (1.25 -x) hours
So distance covered by walking is 7x km
and distance covered by walking is 12(1.25 -x) km
As of question, 7x + 12(1.25 -x) = 1O
=> 7x+15-12x = 10
=>12x-7x=15-10
=> 5x = 5
x = 1
He walked 1 hour to catch the bus.
His walking distance is = 7x km
= 7 x 1 = 7 km.
3ab$^{2}$
LHS = $a^{3}+2c^{3}$
=$(x+y)^{3}+2(x^{3}+y^{3})$
=$x^{3}+3x^{2}y+3xy^{2}+y^{3}+2x^{3}+2y^{3}$
=$3{x^{2}(x+y)+y^{2}(x+y)}$
=$3(x+y)(x^{2}+y^{2})$
=$3ab^{2}$
=RHS
Percentage of boys not living in the hostel out of total 8.33%
Let, the number of girls = x
the number of boys = x
Number of boys residing in the hostel = $\frac{5x}{6}$
Number of boys not residing in the hostel = (x-$\frac{5x}{6}$)
= $\frac{6x-5x}{6}$
= $\frac{x}{6}$
Total students = x + x = 2x
Percentage of boys not living in the hostel out of total = $\frac{\frac{x}{6}}{2x}\times 100\%$
= $\frac{x}{6}\times \frac{1}{2x}\times 100\%$
= 8.33%
AD 2.4cm
Here, three arms of the triangle are a = BC = 5cm
b = AC = 4cm
c = AB = 3cm
So perimeter of the triangle is a + b + c = (5 + 4 + 3) = 12cm
Half of the perimeter, s = $\frac{12}{2}$= 6
Area of the triangle is = $\sqrt{s(s-a)(s-b)(s-c)}$
= $\sqrt{6(6-5)(6-4)(6-3)}$
= $\sqrt{6\times 1\times2\times 3 }$
= $\sqrt{36}$
= $\sqrt{6^{2}}$
= 6
And again, Area of the triangle is = $\frac{1}{2}\times$ base $\times$ height
= $\frac{1}{2}\times$5$\times$AD
As of question, $\frac{1}{2}\times$5$\times$AD = 6
=>AD = $\frac{6\times 2}{5}$
=> $\frac{12}{5}$
AD = 2.4cm
36
Let, Rahim drives x km. at 72 km/h
(48 —x) km is driven at 48 kmfh
According to question,
$\frac{x}{72}+\frac{48-x}{48}=45\times\frac{1}{60}=\frac{3}{4}$
=>$\frac{2x+3(48-x)}{144}=\frac{3}{4}$
=>$\frac{144-x}{144}=\frac{3}{4}$
=>x=36
