Question

Please solve this question​

Answer 1

Question :

Nine friends had a tea party party. All the boys took only coffee and all girls took only tea. The cost per cup of coffee in rupees is numerically 2 less than the number of girls and the cost per cup of tea in rupees is numerically less than 2 than the number of boys. If the ratio of the total expenses of the boys and girls is 5 : 6, then what is the cost of each coffee ( In Rs. )

1. 2
2. 5
3. 7
4. 3

Answer :

Let the number of boys be B and the number of girls be G

Given :

Total number of friends who had party = 9

⇒ Number of girls + Number of boys = 9

⇒ G + B = 9

⇒ B = 9 - G → ( 1 )

Given : All the boys took only coffee and all the girls took only tea

∴ Total no. of coffees = No. of boys → ( 2 )

∴ Total no. of teas = No. of girls → ( 3 )

According to the question

Cost per cup of coffee = 2 less than No. of Girls = Rs. ( G - 2 )

Total cost of coffees = No. of coffees × Cost per cup of coffee =  Rs. B( G - 2 )              [ ∵ From ( 2 ) ]

Cost per cup of tea = 2 less than No. of boys = Rs. ( B - 2 )

Total cost of teas = No. of teas × Cost per cup of tea = Rs. G( B - 2 )              [ ∵ From ( 3 ) ]

Given :

Ratio of the total expenses of boys and girls = 5 : 6

$\Rightarrow \rm \dfrac{B(G-2)}{G(B-2)} =\dfrac{5}{6}$

[ Since all boys took coffees and all girls took teas ]

$\Rightarrow \rm \dfrac{BG-2B}{BG-2G} =\dfrac{5}{6}$

⇒ 6( BG - 2B ) = 5( BG - 2G )

⇒ 6BG - 12B = 5BG - 10G

⇒ 6BG - 5BG - 12B + 10G = 0

⇒ BG - 12B + 10G = 0

From ( 1 )

⇒ ( 9 - G )G - 12( 9 - G ) + 10G = 0

⇒ 9G - G² - 108 + 12G + 10G = 0

⇒ - G² + 31G - 108 = 0

⇒ G² - 31G + 108 = 0

⇒ G² - 27G - 4G + 108 = 0

⇒ G( G - 27 ) - 4( G - 27 ) = 0

⇒ ( G - 4 )( G - 27 ) = 0

⇒ G - 4 = 0    or     G - 27 = 0

⇒ G = 4         or      G = 27

As Total no. of friends are 9 no. of girls can't be more than the sum. Hence, G ≠ 27

⇒ G = 4

∴ No. of girls = 4

Cost per cup of coffee = Rs. ( G - 2 ) = 4 - 2 = Rs. 2

Thus, the cost of each coffee is Option( 1 ) Rs. 2.