Fuel to go = x + (x/4)=5x/4
Fuel to come = x
now,
x+(5x/4) = 4.5
9x/4 = 4.5
9x = 18
x = 2 (Fuel to comeup)
Fuel to go will be:
2+(2/4) = 2 + (0.5) = 2.5
Hint: Assume that the speed of the stream is x and the speed of the boat in still water is x. From the statement of the question form two equations in two variables x and y. This system is reducible to linear equations in two variables. Reduce the system to a system of linear equations in two variables by proper substitutions. Solve the system of equations using any one of the methods like Substitution method, elimination method, graphical method or using matrices. Hence find the value of x satisfying both the equation. The value of x will be the speed of the stream.
Complete step-by-step answer:
Let the speed of the stream be x, and the speed of the boat in still water be y.
We have the speed of the boat upstream = y-x.
Speed of the boat downstream = y+ x.
Now since it takes 14 hours to reach a place at a distance of 48 km and come back, we have the sum of the times taken to reach the place downstream and time taken to return back upstream is equal to 14.
Now, we know that time =Distance speed
Using, we get
Time taken to reach the place =48y+x and the time taken to return back =48y−x.
Hence, we have
48y+x+48y−x=14
Dividing both sides by 2, we get
24y+x+24y−x=7 —–(i)
Also, the time taken to cover 4km downstream is equal to the time taken to cover 3km upstream.
Hence, we have 4y+x=3y−x
Transposing the term on RHS to LHS, we get
4y+x−3y−x=0 ——– (ii)
Put 1y+x=t and 1y−x=u, we have
24t+24u=7 ——-(iii)4t−3u=0 ——–(iv)
Multiplying equation (iv) by 6 and subtracting from equation (iii), we get
24t−24t+24u+18u=7⇒42u=7
Dividing both sides by 42, we get
u=742=16
Substituting the value of u in equation (iv), we get
4t−3(16)=0⇒4t−12=0
Adding 12 on both sides, we get
4t=12
Dividing both sides by 4, we get
t=18
Reverting to original variables, we have
1y+x=18 and 1y−x=16
Taking reciprocals on both sides in both equations, we have
y+ x=8 ——- (v)y−x=6 ——–(vi)
Adding equation (v) and equation (vi), we get
2y=14
Dividing both sides by 2, we get
y=7.
Substituting the value of y in equation (v), we get
7+x=8
Subtracting 7 from both sides we get
x = 8-7 =1
Hence the speed of the stream is 1 km/hr.
b
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 ?
24 times
The answer for this question is very simple: it is called Ceaser Cipher.
It consists of replacing a character by another located in (current character position + (key-1)), where key = how many positions you want to skip.
For Example: VICTORY -> YLFWRUB with the key = 3.
Then SUCCESS -> QXFFHQQ
Note: it should work like a circle(Z+1 = A).
Correct option is C)
Let be students are consider as child
Let the age of child added later be x years.
average age of 12 children =20 years
∴ Total age of 12 children =20×12=240 years
After one more child is added-
Average of 13 children =20−1=19 years
Average age of 13 children =
13
sum of 12 children+x
19=
13
240+x
⇒240+x=247
⇒x=247−240=7 years
Hence the age of child added later is 7 years.
10000*8:10000*12
2:3
5x=25000
x=5000
p:q
10000:15000
Sunday
Monday
Tuesday
55 + 70 = 125
125/14hrs = 60 mph
14 hours