Answer: 8:10, 7:10
Explanation:
The bus b1, which started at P, reached S at 10:40, passing through the intermediary cities Q and R.
The time taken to travel from P to S
= 3 * 40 + 2 * 15 = 150minute
(journey)+(stoppage) = 2 hrs 30 minutes.
Hence, b1 started at 10:40 – 2:30 = 8:10 at P.
b2 reached Q, starting at U, through the city T, S and R.
The time taken by it to reach S = 4 * 40 + 3 * 15 = 205 minutes = 3 hr 25 minutes.
Hence, b2 started at, 10:35 – 3:25 = 7:10, at U.
21
50
the answer is 34
let the for digit number be = pqrs
p=q/3 ==>q = 3p
r=p+q=p+3p =4p
s=3q = 3(3P)=9p
number:
p 3p 4p 9p
let p=1 answer is 1349
if it 2 answer 2 6 8 18
so it becomes five digit number so correct answer is 1349
There are 18 numbers between 100 and 300 that are divisible by 11: 110, 121, 132, 143, 154, 165, 176, 187, 198, 209, 220, 231, 242, 253, 264, 275, 286, 297.
put 1 red marble in one jar and all the rest (99) in the
other.
This gives you 50% from the first jar (if they pick that
jar they will get red 100% of the time). For the other jar
the chances are 49/99 or 49.494949%. Divide that by 2 and
its 24.7474%. Total odds are 50% plus 24.7474% = 74.7474%
Let the person ate x banana on first day, so below is the equation for five days
X+x+6+x+12+x+18+x+24=100
So, he ate 8 banana first day
If n = no. of revolutions for back wheel
9n = 7n + 10*7
n=315
Let 10’splace digit is x and unit’s place digit y
First milestone : 10x+y
Second milestone : 10y+x
Third milestone: 100x+y
Since the speed is uniform so
Distance covered in first Hr = Distance covered in Second Hr
(10y+x)-(10x+y) = (100x+y)-(10y+x)
After solving, we get —-> y=6x but since x and y are digits so only possible combination is x=1 and y=6,
So average speed = 45 KM/HR
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 ?
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