A group of students decided to collect as many paise from each member of group as is the number of members. If the total collection amounts to Rs. 59.29, the number of the member is the group is:

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