GATE Questions

---

Let the numbers be 3x, 3x + 3 and 3x + 6.
Then,
3x + (3x + 3) + (3x + 6) = 72
9x = 63
x = 7
Largest number = 3x + 6 = 27.

=> Second largest number = 27 - 3 = 24

Let the total distance to be covered is 84 kms.
Time taken to cover the distance without stoppage = 84/42 hrs = 2 hrs
Time taken to cover the distance with stoppage = 84/28 = 3 hrs.
Thus, he takes 60 minutes to cover the same distance with stoppage.
Therefore, in 1 hour he stops for 20 minutes.

We cannot find out the answer because the number of days of the current month is not given.

C alone can finish the work in 40 days.
As given C does half as much work as A and B together
=> (A + B) can do it in 20 days
(A + B)s 1 days wok = 1/20.
A's 1 days work : B's 1 days Work = 1/2 : 1 = 1:2(given)
A's 1 day’s work = (1/20) x (1/3) = (1/60) [Divide 1/20 in the raio 1:2]
B's 1 days work = (1/20) x (2/3) = 1/30
(A+B+C)'s 1 day's work = (1/60) + (1/30) + (1/40) = 9/120 = 3/40

All the three together will finish it in 40/3 = 13 and 1/3 days.

1/8 - 1/x = 1/10

 => x = 40 hrs

Since their HCFs are 7, numbers are divisible by 7 and are of the form 7x and 7y

Difference = 14
=> 7x - 7y = 14
=> x - y = 2

product of numbers = product of their hcf and lcm
=> 7x * 7y = 441 * 7
=> x * y = 63

Now, we have
x * y = 63 , x - y = 2
=> x = 9 , y = 7

The numbers are 7x and 7y
=> 63 and 49

Let the amounts to be received by P, Q and R be p, q and r.
p + q + r = 1200
p = 1/2 (q + r) => 2p = q + r

Adding 'p' both sides, 3p = p + q + r = 1200
=> p = Rs.400

q = 1/3 (p + r) => 3q = p + r

Adding 'q' both sides, 4q = p + q + r = 1200
=> q = Rs.300

r = 1200 - (p + q) => r = Rs.500.

Observe that in each column, square of sum of digits in the first row = Respective second row elements.

In the same way, square of sum of digits in the second row = Respective third row elements.

=> 6+42=102=100

=> 4+92=132=169

=> 8+12=92=81.