0.15% of
(A) ₹ 5
(B) ₹ 150
(C) ₹ 0.05
(D) ₹ 105
If x% of
(A) 1000
(B) 1200
(C) 1400
(D) 1500
if 50% of (x-y) =30% of (x+y), then what percentage of x is y?
(A) 25%
(B)
(C) 40%
(D) 400%
What percent of 3.6kg is 72 gms ?
(A) 32%
(B) 22%
(C) 12%
(D) 2%
If 40% of (A+B) = 60% of (A-B) then
(A)
(B)
(C)
(D)
A travels 25% faster than B. They started their journey from a point P to another point Q and reach at the same time on point Q. Distance between P and Q is 85 km. On the way, however, A lost about 20 minutes while stopping for petrol. What was the speed of B?
(A) 51 km/hr
(B) 50 km/hr
(C) 45 km/hr
(D) 75 km/hr
(E) None of these
Speed of boat in still water is 9 km/hr. Stream speed initially is 2 km/hr but it increases by 3 km/hr after every hour. Find the time after which boat will come back to the position where it started.(in hour)
(A) 5(5/8)
(B) 4(7/8)
(C) 5(3/8)
(D) 4(3/4)
(E) None of these
Akhil takes 30 minutes extra to cover a distance of 150 km if he drives 10 km/h slower than his usual speed. How much time will he take to drive 90 km if he drives 15 km per hours slower than his usual speed?
(A) 2h 45m
(B) 2h 30m
(C) 2h
(D) 2h 15m
Let's use the information given to calculate Akhil's usual speed first.
We know that Akhil takes 30 minutes (0.5 hours) extra to cover a distance of 150 km when he drives 10 km/h slower than his usual speed.
Let "S" be Akhil's usual speed in km/h. So, his slower speed would be (S - 10) km/h.
The time taken to cover a distance is equal to the distance divided by the speed: Time = Distance / Speed
At his usual speed, it takes him: Time at usual speed = 150 km / S hours
At the slower speed, it takes him: Time at slower speed = 150 km / (S - 10) hours
The difference in time between these two scenarios is 0.5 hours (30 minutes): Time at slower speed - Time at usual speed = 0.5 hours
Now, we can set up the equation and solve for S:
(150 km / (S - 10)) - (150 km / S) = 0.5
To solve this equation, we'll first get a common denominator: [150S - 150(S - 10)] / [S(S - 10)] = 0.5
Now, simplify and solve for S: [150S - 150S + 1500] / [S(S - 10)] = 0.5
1500 / [S(S - 10)] = 0.5
Now, cross-multiply: 2 * 1500 = S(S - 10)
3000 = S^2 - 10S
S^2 - 10S - 3000 = 0
Now, we can solve this quadratic equation for S using the quadratic formula:
S = [-(-10) ± √((-10)^2 - 4(1)(-3000))] / (2(1))
S = [10 ± √(100 + 12000)] / 2
S = [10 ± √12100] / 2
S = [10 ± 110] / 2
Now, we have two possible values for S, but we'll take the positive one because speed cannot be negative:
S = (10 + 110) / 2 = 120/2 = 60 km/h
So, Akhil's usual speed is 60 km/h.
Now, we want to find out how much time he will take to drive 90 km when he drives 15 km/h slower than his usual speed, which would be (60 - 15) = 45 km/h.
Time = Distance / Speed Time = 90 km / 45 km/h = 2 hours
Akhil will take 2 hours to drive 90 km when he drives 15 km/h slower than his usual speed.
Running at the speed of 3 km/hr., a person reaches his destination 10 minutes later than usual time. if he increases his speed by 1 km/hr., he reaches his destination 15 min earlier. and the distance to his destination.
(A) 10 km
(B) 12 km
(C) 5 km
(D) 4 km
(E) None of these
A man can row 52 km upstream and 42 km downstream in 10 hours. It is also known that he can row 66 km downstream and 60 km upstream in 13 hours. Find the speed of the man in still water.
(A) 8 km/hr
(B) 12 km/hr
(C) 6 km/hr
(D) 10 km/h
(E) None of these
Get the Examsbook Prep App Today