Angle of 360° is covered by hour hand in 12 h .
So, angle of 135° is covered by hour hand in 12/360° x 135° = 4.5 h
Thus, required time = (3 + 4.5)h = 7.5 h = 7 : 30
Let x, y and z be the length breadth and deapth of a cuboid.
? x + y + z = 19
x2 + y2 + z2 = (5?5)2 = 125
Surface area of the cuboid = 2(xy + yz + zx)
= (x + y + z)2) - (x2 + y2 + z2)
= 361 - 125 = 236 cm3
In first situation.
Radius = r1, height = h1 and volume = v1
In second situation,
Radius =2r1, height = h2 and volume = v2
In the volume is fixed, then
v1 = v2
? (1/3)?r21h1 = (1/3)?(2r1)2h2
? h1 = 4h2
? h2 = h1/4
Therefore, height of the the cone will be one-fourth of the previous height.
Required percentage
= 100 ? (40 + 25 + 15)
= 100 ? 80 = 20%
6 x 12 men = 8 x 18 women = 18 x 10 children
? 12 men = 24 women = 30 children
? 2 men = 4 women = 5 children
Now, 4 men + 12 women + 20 children = 4 men + 6 men + 8 men = 18 men
Time to do remaining work in 1 man = (6 x 12 - 18 x 2) = 36 days
? Required number of men to finish the work in 1 days = 36
Solution 1.
Let the question solved correctly by Mohan = X ;
and number of question solved wrongly by Mohan = 30 - X ;
According to the question,
3X - (30 - X) x 2 = 40
? 3x - 60 + 2x = 40
? 5x - 60 = 40
? 5x = 40 + 60 ? x = 100/5 = 20
? Mohan attempted 20 questions correctly .
Solution 2.
Let the question solved correctly by Mohan = X ;
and number of question solved wrongly by Mohan = Y ;
According to given question;
X + Y = 30 ; -----------i
3X - 2Y = 40 ;-------ii
Now multiply the first equation by 2;
we will get
2X + 2Y = 60 ;--------iii
After Add the equation ii and iii , we will get
3X - 2Y + 2X + 2Y = 40 + 60 ;
? 5X = 100 ;
? X = 20;
? Mohan attempted 20 questions correctly .
Concentration of glucose are in the ratio =
Quantity of glucose taken from A = 1 liter out of 2
Quantity of glucose taken from B = 3/5 x 3 = 1.5 lit
Quantity of glucose taken from C = 0.8 lit
So, total quantity of glucose taken from A,B and C = 3.6 lit
So, quantity of alcohol = (2 + 3 + 1) - 3.6 = 2.4 lit
Ratio of glucose to alcohol = 3.6/2.4 = 3:2
Let quantity of mixture be x liters.
Suppose a container contains x units of liquid from which y units are taken out and replaced by Water. After operations , the quantity of pure liquid = units, Where n = no of operations .
So, Quantity of Milk =
Given that, Milk : Water = 9 : 16
=> Milk : (Milk + Water) = 9 : (9+16)
=> Milk : Mixture = 9 : 25
Therefore,
=> x = 15 liters
Let the tin contain 5x litres of liquids
=> 5(4x - 36) = 2(x + 36)
=> 20x - 180 = 2x + 72
=> x = 14 litres
Hence, the initial quantity of mixture = 70l
Quantity of liquid B
=
= 50 litres.
Copper in 4 kg = 4/5 kg and Zinc in 4 kg = 4 x (4/5) kg
Copper in 5 kg = 5/6 kg and Zinc in 5 kg = 5 x (5/6) kg
Therefore, Copper in mixture = kg
and Zinc in the mixture = kg
Therefore the required ratio = 49 : 221
As water costs free, water sold at cost price of milk gives the profit.
Required profit % = 5/20 x 100 = 5 x 5 = 25%.
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