Math, asked by taniket561, 10 months ago

Find the two consecutive multiples of 8 such that their sum is equal to 248.

Answers

Answered by Anonymous
0

Step-by-step explanation:

Let the first multiple of 8 be 8x.

Therefore the second consecutive multiple of 8 will be 8(x+1)

Also the third consecutive multiple of 8 will be 8(x+2).

It is given that the sum of these three consecutive multiples of 8 is 888

=> 8x + 8(x+1) + 8(x+2) = 888

=> 8x + 8x + 8 + 8x + 16 = 888

=> 24x + 24 = 888

Take 24 on the RHS

=> 24x = 888 - 24

=> x = 864/24

=> x = 36.

Therefore First multiple of 8 be 8x = 8 x 36 = 288

Second Multiple of 8 be 8(x + 1) = 8(36 + 1) = 8 x 37 = 296

Third Multiple of 8 be 8(x + 2) = 8(36 + 2) = 8 x 38 = 304

If we sum up these three multiples i.e (288 + 296 + 304) we get 888

Answered by RepalaKavyasri
2

Answer:

15 and 16 are the multiples of 8.

Step-by-step explanation:

Let take consecutive multiples be 8x, 8x+8.

So, 8x + 8x + 8 = 248

16x +8 = 248

6x = 248 - 8

16x = 240

x = 240/16

x = 15

So, x = 15

8x = 8(15)

= 120

8x + 8 = 8(15) +8 nothing but 8(16)

= 120 + 8

= 128

Verification :

8x + 8x + 8 = 248

we know the values of 8x, 8x + 8

120 + 128 = 248

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