Question

class 8 chapter 2 exercise 2.2 question 6 and 7 ​

Answer 1

Answer:

Hope it helps!! Mark this answer as brainliest if u found it useful and follow me for quick and accurate answers...

Step-by-step explanation:

6) Let the three consecutive integers be x, x + 1 and x + 2.

As per the condition, we get

x + (x + 1) + (x + 2) = 51

⇒ x + x + 1 + x + 2 = 51

⇒ 3x + 3 = 51

⇒ 3x = 51 – 3 [transposing 3 to RHS]

⇒ 3x = 48

⇒ x = 48 ÷ 3 [transposing 3 to RHS]

⇒ x = 16

Thus, the required integers are 16, 16 + 1 = 17 and 16 + 2 = 18, i.e., 16, 17 and 18.

7) Let the three consecutive multiples of 8 be 8x, 8x + 8 and 8x + 16.

As per the conditions, we get

8x + (8x + 8) + (8x + 16) = 888

⇒ 8x + 8x + 8 + 8x + 16 = 888

⇒ 24x + 24 = 888

⇒ 24x = 888 – 24 (transposing 24 to RHS)

⇒ 24x = 864

⇒ x = 864 ÷ 24 (transposing 24 to RHS)

⇒ x = 36

Thus, the required multiples are

36 × 8 = 288, 36 × 8 + 8 = 296 and 36 × 8 + 16 = 304,

i.e., 288, 296 and 304.

Answer 2

Answer:

6. let the three consecutive integers be x,x+1 and x+2

sum of integers=51

x+x+1+x+2=51

3x+3=51

3x=48

x=16

Therefore the numbers are 16,17 and 18

7.  let the three consecutive integers be 8x, 8(x+1) and 8(x+2)

sum of multiples=888

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

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

24x+24=888

24x=864

x= 36

The three multiples are 36, 292 and 304

Hope it helps<3