The ans is -20.
Solution:
A-1
B-2
C-3 D-4 …like wise till Z-26.
ELECTRICITY-5+12+5+3+20+18+9+3+9+20+25==129
GAS-7+1+9==27
ELECTRICITY-GAS=129-27-(Minus 2)=100
so
JACK-JILL=(10+1+3+11-(10+9+12+12)-(minus(2))==(-20)
13. One group of each = 2+3+5 = 10 and 1 remainder for each = 1+1+1 = 3. 10+3 = 13
The answer is 42
Let the principal be P and rate of interest be R%.
Therefore Required ratio
=P×R×6/100/P×R×9/100
=6PR/9PR
=6/9
=2:3
3000
7 : 3
Its very simple..
consider the fraction of s in the mixture = 1/3
So if we add one more R the the fraction wil be = 1/4
Automaticaly S becomes 25% of the mixture
1(100-1)+ 2(100-2)+…..99(100-99)
N=100
solution: N(N-1)
100(99)= 9900
x (A to B by car) + y (B to A by cycle) is 7 hours
x (A to B by car) + x (B to A by car) is 7 minus 3 is 4 hours
2x is 4 and hence x is 2
2 + y is 7 and hence y is 5
y (A to B by cycle) + y (B to A by cycle) is 5 + 5 is 10 hours
Soldier have to move 0.75 miles to West and then 0.375 miles to South to reach the camp.
How:-
Firstly soldier moving 1 mile to East from camp and then 1/2 miles= 0.50 miles to North .
Then it is moving 1/4 miles =0.25 miles to west
And then 1/8 miles =0.125 miles to South
Now we just have to count the difference and minus the value of camp to East with the value of North to west = 1mile – 0.25miles(1/4) = 0.75 miles
Same case with north and south = 0.50 miles(1/2) – 0.125 miles(1/8) = 0.375 miles
Hence proves to return camp the soldier have to move 0.75 miles to west and 0.375 miles to south
3/7
6 cuts
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 ?
C. 22%