GATE Questions

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Given that to find 15 of 80

=> What % of 80 is 15

Let it be 'p'

=> p% of 80 = 15

=> $$\begin{gathered} p x 80100 = 15 \\ ⇒p = 15 x 10080 \\ ⇒p = 150080 = 18.75 \end{gathered}$$

Therefore, 15 is 18.75% of 80 => 18.75% of 80 = 15

Example::

Now, in the similar way we can find 15% of 80 =?

=> 15x80/100 = p

=> p = 4 x 3 = 12

Therefore, 15% of 80 = 12

Let number of boys = B

Number of girls = G

From the given data,

1B/3 + 1G/2 = 300

But given B/3 = 100

=> B = 300

G/2 = 300 - 100

G = 200 x 2

=> G = 400

Therefore, total number of students in the college = B + G = 700.

Time taken to cover 600 km = 600/100 = 6 hrs.

Number of stoppages = 600/75 - 1 = 7

Total time of stoppages = 4 x 7 = 28 min

Hence, total time taken = 6 hrs 28 min.

Let the present ages of the mother and daughter be 2x and x years respectively.
Then, (2x - 18) = 3(x - 18) => x = 36
Required sum = (2x + x) = 108 years.

Given G + B= 41 and B = G-1
Hence, G = 21 and B = 20
Now we have 2 options,
1G and 3M
(or)
3G and 1M
(2G and 2M or 0G and 4M or 4G and oM are not allowed),

Total : C121×C320+C321×C120= 50540 ways.

An anti-lock braking system (ABS) is a safety anti-skid braking system used on aircraft and on land vehicles , such as cars, motorcycles, trucks and buses.

ABS is only on the front wheels and it can improve your ability and the stability of your car by keeping the front wheels from locking under severe braking it will allow the wheels to continue to roll so the car can stop straight and stop as safe and as soon as possible

Class can have a public method for specific data type conversions.
for example:
class B
{
double value;
public  B(int i )
operator double()
{
return value;
}
};
B BObject;
double i = BObject; // assigning object to variable i of type double.
now conversion operator gets called to assign the value.

Draw the two chords as shown in the figure. Let O be the center of the circle. Draw OC
perpendicular to both chords. That divides the two chords in half.
So CD = 10 and AB = 14. Draw radii OA and OD, both equal to radius r.
We are given that BC = 5, the distance between the two chords. Let
OB = x.

We use the Pythagorean theorem on right triangle ABO

AO² = AB² + OB²
r² = 14² + x²

We use the Pythagorean theorem on right triangle DCO

DO² = CD² + OC²

We see that OC = OB+BC = x+5, so

r² = 10² + (x+5)²

So we have a system of two equations:

r² = 14² + x²
r² = 10² + (x+5)²

Since both left sides equal r², set the right sides
equal to each other.

14² + x² = 10² + (x+5)²
196 + x² = 100 + x² + 10x + 25
196 = 125 + 10x
71 = 10x
7.1 = x

r² = 14² + x²
r² = 196 + (7.1)²
r² = 196 + 50.41
r² = 246.41
r = √246.41
r = 15.69745202 cm