GATE Questions

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Second generation computers were developed during 1956 - 1963.

In the given puzzle,

The logic is in the format of lets say, a/b

In that case b doesn't matter, because

2/2 = 1,

6/5 = 6/2 = 3,

10/11 = 10/2 = 5

This means that equation is a/2

So 16/19 is equal 16/2 = 8.

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

Given letters are k, l, m, n, o = 5

number of letters to be in the words = 3

Total number of words that can be formed from these 5 letters taken 3 at a time without repetation of letters = 5P3 ways.

⇒ 5P3 = 5 x 4 x 3 = 60 words.

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

Let the cost price of the article be 'P'

Then from given data,

$$\begin{gathered} P + 28P100 = 4544 \\ => 128P = 454400 \\ => P = 3550 \end{gathered}$$

Hence, the cost price of the article = P = Rs. 3550

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