Question

the sum of two consecutive multiples of 11 is 99..the smaller one is???​

Answer 1

Answer:

Let the three consecutive multiples of 11 be 11x , 11(x + 1) and 11(x + 2)

Total Sum of these multiples=99

So, 11x+11(x+1)+11(x+2)=99

11x+11x+11+11x+22=99

11x+11x+11x+11+22=99

33x+33=99

33x=99-33

33x=66

x=66/33=2

x=2

therefore, value of x=2

now, as we know x=2

11(x+1)=11(2+1)=11*3=33

11(x+2)=11(2+2)=11*4=44

Therefore, the smallest number is 2

Step-by-step explanation:

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Answer 2

Answer:

let the first consecutive multiple be= x

let the second consecutive multiple be=x+11

sum of these multiple=99

x+x+11=99

2x+11=99

2x=99-11

2x=88

$x = \frac{88}{2} \\ x= 44$

first multiple=x=44

second multiple=x+11=44+11=55

$\red {\underline \bold{proove: }}$

x+x+11=99

44+44+11=99

88+11=99

99=99

$\green {\underline \bold{r.h.s. = l.h.s.}}$

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