Class 12th

Let the number of men be x

$\frac{4x}{3(95-x)}=\frac{8}{9}$

9 x = 570 – 6 x

x = 38, so number of men is 38

$\frac{T1speed}{T2speed}=\sqrt[]{\frac{TimetakenbyT2toreachChennai}{TimetakenbyT1toreachYeshvantpur}}$

$or\frac{45}{T2speed}=\sqrt[]{\frac{200}{242}}=\frac{10}{11}$

T2 speed = $\frac{99}{2}$ $\frac{}{}$ =49.5 km/hr.

As per the problem, we have CP of 18? SP of 18 = CP of 3

CP of 15 = SP of 18

But CP of 15 toffees = Rs. 90

So SP of 18 toffees = Rs. 90

SP per toffee $=\frac{90}{18}=Rs.\text{?}5$

Let 100 units be B's income and X units be B's expenditure A's income = 60 units

A's expenditure =   $\frac{50X}{100}unitsBut60=\left(\frac{80}{100}\right)*X\Rightarrow X=75$ $$

B's saving = (100 – 75) units = 25 units

Hence, A's saving = 

60 − $\left(\frac{50}{100}\right)*75units=22.5units$ $$

A's saving : B's saving = 22.5 : 25 = 9 : 10

f CP = 100, MP = (100 + 44) * $\frac{100}{90}$ $\frac{}{}$ = 160

If MP = 23200, CP = 23200 * $\frac{100}{160}$ $\frac{}{}$ = 14500

Speed become $\frac{2}{3}$ $\frac{}{}$

So, time also become $\frac{2}{3}*60$ $\frac{}{}$

= 40 minutes

Hence, bus will stop for 20 minutes per hour.

Y alone take 25 * 4 = 100 days

Total work LCM of (25&100) = 100

Y alone = $\frac{}{}\frac{}{}$ $\frac{100}{\left(4-1\right)}=\frac{100}{3}$

At 2:40 a.m., the hour hand is between 2 and 3 while the minute hand is at 8. Therefore, angle between the two hands at 2:40 a.m. = $\frac{}{}$ $\left(180-40*\frac{1}{2}\right)$  = 160°.

et the present ages be X, Y, and Z years (X – 5): (Y – 5): (Z – 5) = 2 : 4 : 5 and taking pairwise values. Substitute these values in the equation

2a + 4a + 5a = x + y + z – 15

11a = 165

a = 15

Five years ago ages were 30, 60, 75

Present ages are 35, 65, 80.