4*A + 7/B + 2*C + 5/D + 6*E =47.2506
=> 4 * A + 7/B + 2 *C + 5/D + 6*E = 40 + 7 + 0.2 + 0.05 + 0.0006
Comparing the terms on both sides, we get :
4*A =40, 7/B = 7, 2*C = 0.2, 5/D = 0.05, 6*E = 0.0006
A = 10,
B = 1,
C = 0.1,
D = 100,
E = 0.0001.
5*A + 3*B + 6*C + D + 3*E =(5*10) + (3*1) + (6*0.1) +100 +(3*0.0001)
=50 + 3 + 0.6 + 100 + 0.003 = 153.6003.
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
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