We are asked to find the ratio in which two varieties of sugar, costing Rs. 18 per kg and Rs. 24 per kg, should be mixed to get a mixture that costs Rs. 20 per kg. Here's the step-by-step approach:
x
kg.y
kg.18x
and the cost of the sugar at Rs. 24 per kg is 24y
.x + y
kg, and we want the cost of the mixture to be Rs. 20 per kg.20(x + y)
rupees.Thus, the cost equation becomes:
18x + 24y = 20(x + y)
18x + 24y = 20x + 20y
18x - 20x = 20y - 24y
-2x = -4y
x = 2y
x
to y
is x/y = 2/1
.The two varieties of sugar should be mixed in the ratio 2:1.
According to question ,we can say that
From equation ?. x4 = 398 + 227 = 625
? x4 = 54
From equation ?. y2 = 346 - 321 = 25
? y2 = 52
But, Feb 2004 not included because we are calculating from March 2004 to March 2005. So it has 1 odd day only.
∴ The day on 6th March, 2005 will be 1 day beyond the day on 6th March, 2004.
Given that, 6th March, 2005 is Monday.
∴ 6th March, 2004 is Sunday (1 day before to 6th March, 2005).
Couldn't be determined, since the total amount of money is not given in either of the case
Given that a /b = b/c
? b2 = ac
? a4 : b4 = a4 : a2 c2 = a2 : c2
Given, A = 6 km/h, B = 3 km/ h
According to the formula, Average speed = 2AB/(A + B)
? Required average speed = 2 x 6 x 3/(6 + 3)
= 36/9 = 4 km/h
Given that, W + D = 6 ...(i)
[ w = Time taken while walking and
D = Time taken while driving ]
From Eq. (i)
5 +D = 6
? D = 1
2D = 2 x 1 = 2
? He will take 2 h to drive both ways.
175760 = 24 x 5 x 13 3
= 23 x 2 x 5 x 133
To make perfect cube, it must be divided by 2 x 5 = 10
On comparing it with ax2 + bx + c = 0 , we get a = 1, b = p, c = q
The roots of the equation x2 + px + q = 0 are equal if
b2 - 4ac = 0
? p2 - 4q = 0 ? p2 = 4q.
Thus , required answer is option B .
Let the volume be x3 and 27x3
? Their edges are x and 3x
Now Ratio of their surface area = 6x2 : 54x2 = 1 : 9
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