Advertisements
Advertisements
प्रश्न
90% and 97% pure acid solutions are mixed to obtain 21 litres of 95% pure acid solution. Find the quantity of each type of acid to be mixed to form the mixture.
उत्तर
Let x litres and y litres be respectively the amount of 90% and 97% pure acid solutions.
As per the given condition
0.90x + 0.97y = 21 × 0.95
⇒ 0.90x + 0.97y = 21 × 0.95 ……….(i)
And
x + y = 21
From (ii), substitute y = 21 – x in (i) to get
0.90x + 0.97(21 – x) = 21 × 0.95
⇒ 0.90x + 0.97× 21 – 0.97x = 21 × 0.95
⇒ 0.07x = 0.97× 21 – 21 × 0.95
`⇒ x =( 21 × 0.02)/ 0.07= 6`
Now, putting x = 6 in (ii), we have
6 + y = 21 ⇒ y = 15
Hence, the request quantities are 6 litres and 15 litres.
APPEARS IN
संबंधित प्रश्न
Find the value of k for which each of the following system of equations have no solution :
2x + ky = 11
5x − 7y = 5
For what value of k, the following system of equations will represent the coincident lines?
x + 2y + 7 = 0
2x + ky + 14 = 0
Solve for x and y:
`2x - 3/y = 9, 3x + 7/y = 2`
Solve for x and y:
23x - 29y = 98, 29x - 23y = 110
Solve for x and y:
6(ax + by) = 3a + 2b,
6(bx – ay) = 3b – 2a
Find the value of k for which the system of linear equations has an infinite number of solutions:
5x + 2y = 2k,
2(k + 1)x + ky = (3k + 4).
Find the values of a and b for which the system of linear equations has an infinite number of solutions:
2x + 3y = 7, (a + b + 1)x - (a + 2b + 2)y = 4(a + b) + 1.
The sum of the numerator and denominator of a fraction is 8. If 3 is added to both of the numerator and the denominator, the fraction becomes `3/4`. Find the fraction.
The length of a room exceeds its breadth by 3 meters. If the length is increased by 3 meters and the breadth is decreased by 2 meters, the area remains the same. Find the length and the breadth of the room.
The sum of two numbers is 80. The larger number exceeds four times the smaller one by 5. Find the numbers.