GATE Questions

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Let us assume that the lotion has 50% alcohol and 50% water.
ratio = 1:1
As the total solution is 9ml
alcohol = water = 4.5ml
Now if we want the quantity of alcohol = 30%
The quantity of water = 70%
The new ratio = 3:7
Let x ml of water be added
We get,
4.54.5+x = 37
=> x=6
Hence 6ml of water is added.

Let C1 be the cost price of the first article and C2 be the cost price of the second article.
Let the first article be sold at a profit of 22%, while the second one be sold at a loss of 8%.

We know, C1 + C2 = 600.
The first article was sold at a profit of 22%. Therefore, the selling price of the first article = C1 + (22/100)C1 = 1.22C1
The second article was sold at a loss of 8%. Therefore, the selling price of the second article = C2 - (8/100)C2 = 0.92C2.

The total selling price of the first and second article = 1.22C1 + 0.92C2.

As the merchant did not make any profit or loss in the entire transaction, his combined selling price of article 1 and 2 is the same as the cost price of article 1 and 2.

Therefore, 1.22C1 + 0.92C2 = C1+C2 = 600
As C1 + C2 = 600, C2 = 600 - C1. Substituting this in 1.22C1 + 0.92C2 = 600, we get
1.22C1 + 0.92(600 - C1) = 600
or 1.22C1 - 0.92C1 = 600 - 0.92*600
or 0.3C1 = 0.08*600 = 48
or C1 = 48/(0.3) = 160.
If C1 = 160, then C2 = 600 - 160 = 440.
The item that is sold at loss is article 2. The selling price of article 2 = 0.92*C2 = 0.92*440 = 404.80.

Let last digit is 2
when second last digit is 4 remaining 4 digits can be filled in 120 ways, similarly second last digit is 6 remained 4 digits can be filled in 120 ways.
so for last digit = 2, total numbers=240

Similarly for 4 and 6
When last digit = 4, total no. of ways =240
and last digit = 6, total no. of ways =240
so total of 720 even numbers are possible.

$$\begin{gathered} 3 \\ 32 + 3 = 12 \\ 5 \\ 52 + 5 = 30 \end{gathered}$$

Here the given numbers are related in such a manner that,

$$\begin{gathered} 52 = 25 \\ 5+12 + 1 = 37 \\ Similarly, \\ 72 = 49 \\ 7+12 + 1 = 64 + 1 = 65 \end{gathered}$$

Hence, 25 : 37 :: 49 : 65.

When the switch is ON, in a tube light choke is nothing but the coil/ballast ( inductor) which is used to induce the high voltage across it. as we know that inductor has property to induce high voltage for the brief period of time, this high voltage is required to ionize the gases in the starter.

Then gas inside the starter gets ionized due to this full voltage and heats the bimetallic strip that is caused to be bent to connect to the fixed contact. Current starts flowing through the starter. Although the ionization potential of the neon is the little bit more than that of the argon but still due to small electrode gap high voltage gradient is appeared in the neon bulb and hence glow discharge is started first in a starter.

As voltage gets reduced due to the current causes a voltage drop across the inductor, the strip cools and breaks away from the fixed contact. At that moment a large L di/dt voltage surge comes across the inductor at the time of breaking.

This high valued surge comes across the tube light electrodes and strikes penning mixture (mixture argon gas and mercury vapor).

Gas discharge process continues and current gets the path to flow through the tube light gas only due to low resistance as compared to the resistance of starter.

Number of liters of water in 150 liters of the mixture = 20% of 150 = 20/100 x 150 = 30 liters.
P liters of water added to the mixture to make water 25% of the new mixture.
Total amount of water becomes (30 + P) and total volume of mixture is (150 + P).
(30 + P) = 25/100 x (150 + P)
120 + 4P = 150 + P => P = 10 liters.

Let the number be x. Then,

3x2 - 4x = x + 50

3x2- 5x - 50 = 0

(3x + 10)(x - 5) = 0

x = 5