Math, asked by simranbhinder286, 4 months ago

Indian Army is the third biggest military content and in the world next to US and China however there are many first that make Indian Army stand out in the world making us all Indians proud knowing them will help you celebrate Republic day vigour and gratitude. on 71 Republic Day parade in Delhi captain Rs meel is planning for trade of following two groups:
(a)first group of Army contingent of 624 members behind an army band of 32 members .
(b) 2nd of CRPF troops with 468 soldiers behind the the 228 members of bikers .
These two groups are too much in the same number of columns the sequence of a soldier is followed by different states janki which are showing the culture of respective States.
(i) what is the maximum number of a columns in which the CRPF troop can March
(a) 4 (b) 8 (c) 12 (d) 16
(ii) what is maximum number of columns in which total army troop amd crpf troop together can march past ?
(a) 2 (b) 6 ( c) 4(d) 8 ​

Answers

Answered by amitnrw
2

Given :  

Army contingent of 624 members behind an army band of 32 members.

CRPF troops with 468 soldiers behind the 228 members of bikers.

To Find :  maximum number of columns in which the army troop can march?

maximum number of columns in which the CRPF troop can march?

maximum number of columns in which total army troop and CRPF troop together can march past?

(iv) What should be subtracted with the numbers of CRPF soldiers and the number of bikers so that  their maximum number of column is equal to the maximum number of column of army troop?

(v) What should be added with the numbers of CRPF soldiers and the number of bikers so that their  maximum number of column is equal to the maximum number of column of army troop?

Solution:

army troop

624

32

HCF of 624 and 32

624 = 32 x 19  + 16

32 = 16 x 2  + 0

16 is the HCF

maximum number of columns in which the army troop can march = 16

CRPF troop

468

228

HCF

468 = 228 x 2 + 12

228 = 19 x 12   +  0

12 is the HCF

maximum number of columns in which the CRPF troop can march =  4

HCF of 12 and 16

16 = 12 x 1 + 4

12 = 4 x 3 + 0

4 is HCF

maximum number of columns in which total army troop and CRPF troop together can march  = 4

468 = 16 x 29 + 4

228  = 16 x 14  + 4

Hence  4 numbers of CRPF soldiers and the number of bikers should be subtracted so that their maximum number of column is equal to the maximum number of column of army troop

(a) 4 Soldiers and 4 Bikers

468 = 16 x 30 -  12

228  = 16 x 15  - 12

Hence  12 numbers of CRPF soldiers and the number of bikers should be added so that their maximum number of column is equal to the maximum number of column of army troop

(b) 12 Soldiers and 12 Bikers

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