Given that:
Horizontal radius (a) = 8 cm
Vertical radius (b) = 5 cm
A = π a b
A =( π) (8)( 5)
A = 125.6 cm2
Area =$\dfrac{6m+4m}{2}\times 3m$
$5 m \times 3 m = 15 m^{2}$
let x be Carlo’s age. Twice his age is 2x. One less than twice his age is 2x - 1 which is Adrian’s age. Since the sum of their ages is 50,
we have
x + (2x - 1) = 50.
Simplifying, we have 3x - 1 = 50. This gives us 3x = 51 and x = 17.
Therefore, Carlo is 17 years old and Adrian is 2(17) - 1 = 33 years old.
Let their present ages be 4x, 7x and 9x years respectively.
Then, (4x - 8) + (7x - 8) + (9x - 8) = 56
20x = 80
x = 4.
Their present ages are 4x = 16 years, 7x = 28 years and 9x = 36 years respectively.
Muffins will be contained in boxes- 6, 12, 18, 24, 36….
Chocolates will be contained in boxes- 8,16,24,32….
Soft-toys will be contained in boxes- 9,18,27,36….
The first box which contains all three items will have to be a multiple of 6, 8 and 9.
Being the first box to contain all three items, it will be the lowest multiple of all 3, which is the LCM.
LCM of 6, 8, 9 is 72.
LCM of 7/3 and 13/4 = LCM of numerators/ HCF of denominators
= 91/1= 91.
Given,
Distance =20 km
speed=4 kmph
Time=?
Using formula,
Time = Distance / speed
= 20/4
= 5 hours.
As we know that, Distance = Speed × Time
So, in 1 hour, distance covered = 50 × 1 = 50 miles
In next 3 hours, distance covered = 60 × 3 = 180 miles
Total distance covered = 50 + 180 = 230 miles
Total Time = 1 + 3 = 4 hrs
We know that,Average speed=$\dfrac{Total distance}{Total time}$
⇒ Avg. Speed = 230/4 = 57.5 mph
Average speed=$\dfrac{Total distance}{Total time}$
Determine the total distance : 70+30=100
Determine the total time :2+1=3
Average speed=$\dfrac{100}{3}$.
Average speed=33.33mph
B covers 100m in 25 seconds B take time =(4000*25)/100=1000 sec=16 min 40 sec.
A takes time =1000 sec-25sec=975 sec= 16 min 15 sec.
The difference in the timing of A and B is 5 seconds. Hence, A beats B by 5 seconds.
The distance covered by B in 5 seconds = (80 * 5) / 25 = 16m
Hence, A beats B by 16m.
Required time =$\dfrac{12×24}{12+24}$
=$\dfrac{12×24}{36}$
=$\dfrac{24}{3}$
=8 minutes.
Let the leak can drain all the water of the tank in y hours
Part of the tank filled by the pipe in 1 hr = 1/2
Part of the tank emptied by the leak in 1 hr = 1/y
Part of the tank filled by the pipe with leak in 1 hr=$\dfrac{1}{2\dfrac{2}{3}}=\dfrac{1}{\dfrac{8}{3}}=\dfrac{3}{8}$
$\dfrac{1}{2}+(−\dfrac{1}{y})=\dfrac{3}{8}$
$\dfrac{1}{2}−\dfrac{1}{y}=\dfrac{3}{8}$
⇒$\dfrac{1}{y}=\dfrac{1}{2}−\dfrac{3}{8}=\dfrac{1}{8}$
⇒y=8
i.e., the leak can drain all the water of the tank in 8 hours
A tank is filled in 10 hours by three pipes A, B and C. The pipe C is twice as fast as B and B is twice as fast as A. How much time will pipe A alone take to fill the tank?
Suppose pipe A alone takes x hours to fill the tank.
Then, pipes B and C will take $\dfrac{x}{2}$ and $\dfrac{x}{4}$ hours respectively to fill the tank.
$\therefore \dfrac{1}{x}+\dfrac{2}{x}+\dfrac{3}{x}=\dfrac{1}{10}$
$\dfrac{7}{x}=\dfrac{1}{10}$
x=70
Part of the cistern filled by pipe P in 1 min = $\dfrac{1}{12}$
Part of the cistern filled by pipe Q in 1 min = $\dfrac{1}{16}$
Suppose Q should be closed after x minutes
i.e., Pipe P and Q will be open for initial x minutes then P will be open for another (9-x) min
x$\left(\dfrac{1}{12}+\dfrac{1}{16}\right)+(9-x)\dfrac{1}{12}$=1
$\dfrac{7x}{48}+\dfrac{9}{12}-\dfrac{x}{12}$=1
7x+36−4x=48
3x=12
x=4
The word 'DELHI' has 5 letters and all these letters are different.
Total number of words (with or without meaning) that can be formed using all these 5 letters using each letter exactly once
= Number of arrangements of 5 letters taken all at a time
= $^5P_{5}$
=5!
=5×4×3×2×1
=120