GATE Questions

---

We park car temporarily in garrage, similarly, aeroplane is temporarily parked in hangar.

Let, Hema attempted 'k' sum correctly, then

k x 3 -2 x(35-k) = 60
5k = 130
k = 26

so 26 correct sums.

We can initially arrange the six boys in 6! ways.
Having done this, now three are seven places and six girls to be arranged. This can be done in ⁷P₆ ways.

Hence required number of ways = 6! x ⁷P₆

Cost of each cake P1 & P2 = Rs. 12

Offer decrease in % = 20%

Increase in sale per day (Q2) = 250 - 18%

Total quantity per day (Q1) = 250

Hence, required change total revenue

TR = P2Q2 - P1Q1 = (12 x 0.8 x 250 x 1.18) -  (12 x 250) = - Rs.168.

Suppose they meet after 'h' hours

Then

3h + 4h = 17.5

7h = 17.5

h = 2.5 hours

So they meet at => 4 + 2.5 = 6:30 pm

Perimeter = 36

⇒ 2 ( l + w ) = 36

⇒ l + w = 18

Since length and width are positive integers in centimeters.

Therefore, possible dimensions are :

( 1, 17 )cm , ( 2, 16 ) cm , ( 3, 15 ) cm, (4 , 14) cm, ( 5,13) cm, ( 6, 12) cm , ( 7, 11) cm, ( 8, 10 ) cm, ( 9, 9) cm.

Hence, there are 9 rectangles.

The cow can graze an area of 2826 sq.m i.e it is able to cover area of circle of 2826 sq.m

Therefore, πr2 =2826
r2 =(2826×7/22) = 899.18 = 900
r = 30 m.

Therefore, the radius of the circle implies the length of the rope = 30 mts.

In the given puzzle the logic is

2 x 1 + 1 = 2 + 1 = 3

3 x 2 + 2 = 6 + 2 = 8

4 x 3 + 3 = 12 + 3 = 15

5 x 4 + 4 = 20 + 4 = 24

6 x 5 + 5 = 30 + 5 = 35

Similarly,

7 x 6 + 6 = 42 + 6 = 48

7 = 48.