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On 71th republic day Parade in Delhi Captian RS Meel is planing for parade of following two group:
(a) First group of Army contingent of 624 members behind an army band of 32 members.
(b) Second group of CRPF troops with 468 soldiers behind the 228 members of bikers.
These two groups are to march in the same number of columns. This sequence of soldiersis followed by different states Jhanki which are showing the culture of the respective states.
(i) What is the maximum number of columns in which the army troop can march?
(a) 8 (b) 16
(c) 4 (d) 32
(ii) What is the maximum number of columns in which the CRPF troop can march?
(a) 4 (b) 8
(c) 12 (d) 16
(iii) What is the maximum number of columns in which total army troop and CRPF troop
together can march past?
(a) 2 (b) 4
(c) 6 (d) 8
(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?
(a) 4 Soldiers and 4 Bikers
(b) 4 Soldiers and 2 Bikers
(c) 2 Soldiers and 4 Bikers
(d) 2 Soldiers and 2 Bikers
Answers
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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Explanation:
Here is ur answer mate..
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