Question

what is the area of cross-section of a box of mass 5kg if it exerts a pressure of 100 Pa. (take 1kgf = 10 newton) *​

Answer 1

Answer:

Correct question :

I have a total of rupees 300 in coins of denominator rupees 1 rup 2 and rupees 5. The number of rupees 2 coins is 3 times the number of rupees 5 coins. The total number of coin is 160. How many coins of each denominatoion is with me?

Solution :

Let the number of Rs 5 coins be x.

Number of Rs 2 coins = 3 x Number of Rs 5 coins = 3x

Number of Re 1 coins = 160 - (Number of coins of Rs 5 and of Rs 2)

= 160 - (3x + x) 160 - 4x

Amount of Re 1 coins

= Rs [1 x (160 - 4x)]

= Rs (160 - 4x)

Amount of Rs 2 coins

= Rs (2 x 3x)

= Rs 6x

Amount of Rs 5 coins

= Rs (5 x X)

= Rs 5x

It is given that the total amount is Rs 300.

160 - 4x + 6x + 5x = 300

160 + 7x = 300

On transposing 160 to R.H.S, we obtain

7x = 300 - 160

7x = 140

On dividing both sides by 7, we obtain

7x = 140

x = 140/7

x = 20

Number of Re 1 coins

= 160 - 4x

= 160 - 4 x 20

= 160 - 80

= 80

Number of Rs 2 coins

= 3x

= 3 x 20

= 60

Number of Rs 5 coins

= x

= 20