Math, asked by sujay1161, 5 months ago

What should be subtracted from the number 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?​

Answers

Answered by meenakshipmurali
13

Answer:

4 soldiers and 4 bikers

Step-by-step explanation:

1 . find the HCF of army troop soliders which is 16.

2. find the HCF of number of CRPF soldiers and bikers which is 12.

Now we have to find the number of CRPF soldiers and bikers that has to be subtracted so that they can March in 16 coloumns i.e same as the number of columns of army troop soldiers.

So the new number has to be divisible by 16

when we divide 468 ÷ 16

remainder is 4

when we divide 228 ÷16

remainder is 4

so we have to subtract 4 from CRPF soldiers and bikers so that they can March in 16 columns.

verification: to check if HCF of 466 and 224 is 16

464 = 2×2×2×2×29

224= 2×2×2×2×2×7

HCF = 2×2×2×2

= 16

hence verified

Answered by amitnrw
10

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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