(x+2)^2 -x^2 = 84
X=20
So (20,22)
Sum= 42
40
x – 30 = 1/4 x || *4
4x – 120 = x || -x + 120
3x = 120 || /3
x = 40
we need to take half tabulate among the 4 tablets. den it ll be like 1 tabulate is of fever and one tabulate is of cough
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 ?
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
How will you know the odd is in lighter one or heavier one from only one weighing. It will require 2 weighing to find the odd set and one weighing for odd coin in that set i.e total 3 weighings.
35 km/h = 10 m/s
Speed = Distance/Time
Distance= (100+150)m = 250m
Speed = 10 m/s
Time= 250/10 = 25s
Answer: z and u
Explanation:
Y is to the right of U and exactly in front of V. Therefore,
U Y
V
Z is behind W and W and X are at extreme ends. So, W has to be to the right of Y. The final arrangement is as follows.
U Y W
X V Z Therefore, Z and U are at extreme ends is true.
The greatest number that will divide 187, 233 and 279 leaving the same remainder in each case.
To find : The number ?
Solution :
First we find the difference between these numbers.
The required numbers are
233-187=46
279-233=46
279-187=92
Now, We find the HCF of 46 and 92.
Therefore, The required largest number is 46.
A. Rs 140
7+8+11=26
B: 7/26*520= 140
When A runs 1000 meters, B runs 900 meters and when B runs 800 meters, C runs 700 meters.
Therefore, when B runs 900 meters, the distance that C runs = (900 x 700)/800 = 6300/8 = 787.5 meters.
So, in a race of 1000 meters, A beats C by (1000 – 787.5) = 212.5 meters to C.
So, in a race of 600 meters, the number of meters by Which A beats C = (600 x 212.5)/1000 = 127.5 meters.
let MP = 100x
25% discount = 75x -> SP
Profit% = (Profit/CP)x100
125 = (75x/CP)x100
.: CP = 60x
Now if discount = 10%
SP = 90x
.: Profit% = ((90x-60x)/60x)x100 = 50%
East
( a ) Treatment
$57.30
A
Son in law