Australian Department of Defense Interview Questions, Process, and Tips

d

625

A wet stone

( b ) 216.90

9

Father’s sister

D. 17 : 3

let present age of ANAND AND BALA BE( A AND B ) RESPECTIVELY

A.T.Q

(A-10) = 1/3[B-10]. ……… – {1}

given

B = A+ 12 PUTTING IN 1

THEREFORE A-10 = 1/3 (A+2) = A+2 = 3A – 30

2A =32

A=16

The batsman on 98 is on strike. He hits the ball and they run 3. UNFORTUNATELY one of the batsmen doesn`t turn correctly for one of the runs and the umpire calls ONE SHORT and awards only two runs. Therefore the first batsman has his century. There is now 1 ball remaining and one run is required to win. The batsman on strike, however is now the one on 97 runs. He now either hits a 4 or a 6. They win the game and both batsmen scored centuries.

Read more: 3 runs required in 3 balls to win with only a wicket left. The batsmen is on 98 and the runner is on 97. How will both the batsmen score centuries … – 3 runs required in 3 balls to win with only a wicket left. The batsmen is on 98 and the runner is on 97. How will both the batsmen score centuries as well win the match ?

A = (B*H)/2
H = 2B
formula becomes A = (B*2*B)/2
this can be rewritten as A = (2*B^2)/2
the 2 in the numerator and denominator cancel out and you get:
A = B^2

40 km avrege speed

2/3 * 15 miles = 10 miles then

time for 10 miles travel is t1 = 10/40 = 1/4

then remaning distance is 5 miles

time for 5 miles travel is t2 = 5/60 = 1/12

t1+ t2= 1/3 in seconds * 60 = 20 min

4×1 = 4
4×2=8
8×3=24
24×4= 96 (not 92)
96×5 =480
480×6= 2880

So answer is 92