Mr. Bundda travels in this car and covers 1/4 part of his journey with 8 km/h, 3/5 part with 6 km/h and remaining 3/20 part with a speed of 10 km/h. Find out his average speed during the whole journey.

Let the two consecutive numbers are N and N+2

According to the question
N (N + 2) = 19043
? N² + 2N - 19043 = 0
? N² + 139N - 137N - 19043= 0
? N( N + 139) - 137 ( N - 137 ) = 0
? N( N + 139) ( N - 137) = 0
? N = 137 and N = - 139
? N = 137

A will reach at starting point in 5 * 2 / 5 = 2 hours ;
B will reach at starting point in 5 / 3 hours ;
C will reach at starting point in 5 / 2 hours ;
Then, on the starting point all three will meet after the L.C.M. of 2, 5 / 3, 5 / 2, 10 / 1 = 10 hours.

Circumference = 2?r
= 2 x (22 / 7) x 70 cm
= 440 cm

Distance travelled in 10 revolutions = 440 x 10 cm
= 4400 cm
= 44 m

? Speed = distance / time
= 44 / 5 m/sec
= (44 / 5) x (18 / 5) km/hr
= 31.68 km/hr

Regarding all copies of the same book as one book, we have only 5 books. These 5 books can be arranged in 5! ways. But all copies of the same book being identical can be arranged in only one way.

? Required number = 5! x 1! x 1! x 1! x 1! = 120

We known that,
LCM of fractions = (LCM of numerators) / (HCF of denominators)

? Required LCM = (LCM of 1, 2, 5, and 4) / (HCF of 3, 9, 6 and 27)
? = 20/3

The speed of A and B are in the ratio 11 : 8.
Let, speeds be 11s and 8s (in m/sec).
Let, race be of P meter.
Then, time taken by A to run P meter is same as that of B to run (P - 120) meter.
? P / 11 = (P - 120) / 8
? P / 11 = (P - 120 / 8
? 8P = 11 x (P - 120 )
? 8P = 11P - 120 x 11
? 11P - 8P = 120 x 11
? 3P = 11 × 120
? P = 440.

? = 124.35% of 8096
= (8096 x 124.35)/100 = 1006737.6/100
= 10067.376 = 10000

log₁₀5 = log₁₀(10/2)
= log₁₀10 - log₁₀2
= 1-0.3010
= 0.6990

Friday will fall on 3, 10, 17, 24, 31
So, it will be 5th Friday on 31 st