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Ans = 8
Use simple box method
[1][2][2][2]= 1x2x2x2 =8
Logic >
once place have 2 number (2,6)
Tens place have also 2 number (7,3)
So that number is divisible by 4
Now, 2 numbers fixed at once and tens place left 2 numbers which is choice at hundredth place
Now 3 number fix left 1 number which is placed at thousands place
Total number form is 4×4 = 16
But divisible by 4 is 1x2x2x2 =8
E non of those
9
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 ?
15, 16, 34, 105, 424, 2124, 12576
The series is 15, 16, 34, 105, 424, 2124, 12576
1) (15*1) + 1 = 16
2) (16*1) + 2 = 34
3) (34*3) + 3 = 105
4) (105*4) + 4 = 424
5) (424*5) + 5 = 2125
Clearly, they gave 2124 instead of 2125, so it is the wrong number in the series.
(c) 25%
20*55*65=y*65*75
y=(20*55*65)/(65*75)
y=14.667 =15
y=15
% reduction=(20-15)/20=25%
ans: none
let it may be any number the square cant end in 8
A 1×1 cube in the middle of an edge of the 3×3 cube will
have two faces painted. A cube has 12 edges, so the answer
is 12.
9, 12, 11, 14, 13, (…..), 15
16
let square be x (squares are 4 sides)
i.e., X+X+X+X=4X
4X+3=1460
4X=1460-3
4X=1457
X=364.25