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) *
Answers
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