Question

in a school 60% of the students are above 10 years of age and of these 75% are boys and the rest are girls if there are 99 boys above 10 years then the numbers of girls above 10 years in the school is

Answer 1

ANSWER:

• Number of girls above ten years = 33.

GIVEN:

• In a school 60% of the students are above 10 years of age.

• Among those 75% are boys.

• There are 99 boys above 10 years.

TO FIND:

• The numbers of girls above 10 years.

EXPLANATION:

Let the number of students be x.

Let the number of students above ten years be y.

$\sf \dfrac{y}{x} \times 100 = 60 \%$

$\sf \dfrac{y}{x} =0.6$

$\sf y =0.6x$

Among these 75% are boys.

And 99 boys are above 10 years.

$\sf \dfrac{99}{y} \times 100= 75\%$

$\sf \dfrac{99}{y} = 0.75$

$\sf \dfrac{33}{y} = 0.25$

$\sf \dfrac{33}{0.25} = y$

$\sf We \ know \ that \ y =0.6x$

$\sf\dfrac{33}{0.25}=0.6x$

$\sf \dfrac{33}{0.15} =x$

$\sf We \ know \ that \ y =0.6x$

$\sf y = 0.6 \left(\dfrac{33}{0.15}\right)$

$\sf y = 4 \times 33$

$\sf y = 132$

Number of students above ten years = 132.

Number of boys above ten years = 99.

Number of girls above ten years = 132 - 99

Number of girls above ten years = 33

HENCE 33 GIRLS ARE ABOVE 10 YEARS.

VERIFICATION:

$\sf Substitute \ y = 132 \ in \ y=0.6x$

$\sf 132 =0.6x$

$\sf 22 =0.1x$

$\sf x = 220$

$\sf We \ know \ that \ \dfrac{y}{x} \times 100 = 60 \%$

$\sf \dfrac{132}{220} \times 100 = 60 \%$

$\sf \dfrac{132}{22} \times 10= 60 \%$

$\sf 6\times 10= 60 \%$

$\sf 60 \%= 60 \%$

$\sf We \ know \ that \ \dfrac{99}{y} \times 100= 75\%$

$\sf \dfrac{99}{132} \times 100= 75\%$

$\sf \dfrac{9}{12} \times 100= 75\%$

$\sf \dfrac{3}{4} \times 100= 75\%$

$\sf 0.75\times 100= 75\%$

$\sf 75\%= 75\%$

HENCE VERIFIED.