Question

answer me guys it's urgent if anyone gives useless answer instant will reported ..​

Answer 1

Answer:

since we have to form groups of boys and girls separately ,so that no body is left behind ,the number of students in a group must divide number of boys as well as number of girls m

therefore, maximum number of students in a group

= HCF(105,140)

let a=140 and b=105 so that a>b

applying Euclid's division Lemma to 140 and 105

we get ,140=105*1+35

r=35 not equal to 0

therefore ,let a= 105 and b=35

applying Euclid's division Lemma to 105 and 35

we get 105=35*3+0

r=0

HCF=35

$no.of \: groups = \frac{total \: no.of \: students}{no.of \: students \: in \: 1 \: group}$

$\frac{105 + 140}{35} = \frac{245}{35} = 7$

hence,7 groups of 35 students should be formed..

Answer 2

$\huge\mathbb\red{Answer:-}$

since we have to form groups of boys and girls separately ,so that no body is left behind ,the number of students in a group must divide number of boys as well as number of girls m

therefore, maximum number of students in a group

= HCF(105,140)

let a=140 and b=105 so that a>b

applying Euclid's division Lemma to 140 and 105

we get ,140=105*1+35

r=35 not equal to 0

therefore ,let a= 105 and b=35

applying Euclid's division Lemma to 105 and 35

we get 105=35*3+0

r=0

HCF=35

$no.of \: groups = \frac{total \: no.of \: students}{no.of \: students \: in \: 1 \: group}$

$\frac{105 + 140}{35} = \frac{245}{35} =7$

hence,7 groups of 35 students should be formed

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