Maths

Let the total work be 240 units.

Total Work = 240 Units

Pipe 1: 40 hours = 6 units/h

Pipe 1 + Leak: 48 hours = 5 units/h

Leak can empty 1 unit in 1 hour.

Pipe 2 can fill 3 units per hour (half as efficient as the first one)

Therefore, effective filling is 2 units per hour.

Therefore, 120 units (half the tank) will get filled in 60 hours.

Weighted means give us

$\frac{}{}$ $\left(\frac{4}{25}+\frac{7}{40}+\frac{1}{4}+\frac{9}{200}\right)$  max. = 63

$\frac{}{}$ $\frac{\left(32+35+50+9\right)}{200}$  max. = 63

max = $\frac{63*200}{126}$ $\frac{}{}$

= 100

Time taken by two trains will be 9 hours and 6 hours respectively. So, Let the total distance be LCM of (9 and 6) = 18.

Speed of trains will be 2 km/hr & 3 km/hr.

At 9 am distance between two trains is 14 km

Time taken =   $\frac{14}{\left(2+3\right)}$ $\frac{}{}$ = 2 hours 48 minutes

So, trains will meet at 11:48 am.

If Z earns Rs 100a, X earns 120a & Y earns 160a

160a – 100a = 60a = 7980

a = 133

X= 120 * 133 = 15960

Y = 160 * 133 = 21280

Z = 100 * 133 = 13300

L.C.M. of 5, 6, 7, 8

= 35 * 24 = 840

Therefore, the required number = 840x + 3,

Which is exactly divisible by 11.

For x = 2, it is divisible by 11.

Hence,

Required number = 840x + 3

= 840 * 2 + 3

= 1683

32% of CP = Rs. 8320 or CP = Rs. 26000

Now the Aman wants a profit of 62%.

Therefore, he has to sell it at a price of 26000 * 1.62 = 42120

The solid sphere is melted and recast into cones. The volume of material is the same before and after casting.

Volume of the sphere = $\left(\frac{4}{3}\right)\pi r^3$ . (1)

Volume of the right circular cone = $\frac{1}{3}\pi R^{2}H$ . (2)

Diameter of the base of the cone

= $\sqrt[]{2}$ slant height

= $\sqrt[]{2}$ S (say). (3)

4R = 2S = 2 (H² + R²)

2R² = 2H

R = H. (4)

Volume of the cone = $\frac{1}{3}\pi R^{2}H$

$=\frac{1}{3}\pi r^2.(r)$  (since, R = r)

$=\frac{1}{3}\pi r^3$

From equations (1) and (4), it can be seen that the melted material creates exactly 4 cones of the specified dimensions. No material is left over.

(6x² + 43x – 49) – (5x² + 2x – 7) = 0

x² + 41x – 42 = 0

x² + 42x – x – 42 = 0

x (x + 42) – 1 (x + 42) = 0

(x – 1) (x + 42) = 0

x = 1 satisfy both the equations. So, there is only one common root.

Q can be 0 or 5, but 65 is not divisible by 4.

So, Q must be 0.

Now, sum of the digit should be divisible by 9.

So, P is 8.

Let y gold coins were given to 1st son.

Number of gold coins given to 2nd son = $\frac{5}{7}y$

Number of gold coins given to 3rd son = $\frac{3}{7}y$

So,  $y-\frac{3}{7}y=60\Rightarrow \frac{4y}{7}=60$

$\Rightarrow y=\frac{60*7}{4}=15*7=105$

Hence, the required number of gold coins:

$=105+\frac{5}{7}*105+\frac{3}{7}*105$

= 105 + 75 + 45 = 225