agil ki pratiroop ki vakhya kare in hindi
Answers
Explanation:
Other Number = 60
Step-by-step explanation:
Given:
HCF of two numbers is 12.
LCM of two numbers is 180.
One number is 36.
To Find:
What is the other number ?
Solution: For two numbers a and b , we know that
(a × b) = {HCF(a , b) × {LCM(a , b)
or
Product of two numbers = HCF × LCM
Let the other number be b. Then we have
One number = 36
Another number = b
\implies{\rm }⟹ 36 × b = 12 × 180
\implies{\rm }⟹ 36b = 2160
\implies{\rm }⟹ b = 2160/36
\implies{\rm }⟹ b = 60
Hence, the other number is 60.
[ Verification ]
a × b = LCM × HCF
36 × 60 = 180 × 12
2160 = 2160
\large\bold{\texttt {Verified }}Verified
Explanation:
Given :
A boy runs for 10 min at a uniform speed of 9km/hr and 20 min at average speed of 12km/hr
To Find :
Speed the boy will take in next 20 min = ?
Solution :
We know,
\sf{}Average\ Speed=\dfrac{Total\ Distance}{Total\ Time}Average Speed=Total TimeTotal Distance
Total time => 10 + 20 = 30min
Convert minute into hour. [1h= 60min,so 1min = 1/60h]
Thus, total time in min = 30/60
\sf{}\implies 12=\dfrac{Toatl\ distance}{\frac{30}{60}}⟹12=6030Toatl distance
\sf{}\implies 12\times \dfrac{30}{60}=Total\ distance⟹12×6030=Total distance
\sf{}\therefore Total\ distance=6km∴Total distance=6km
Therefore,total distance is equal 6m
Distance covered in 10min/distance which boy walk earlier) :-
We know,
\sf{}Speed=\dfrac{Distance}{Time}Speed=TimeDistance
Covert 10 min into hr
=> 10/60hour
\sf{}\implies 9=\dfrac{Distance}{\frac{10}{60}}⟹9=6010Distance
\sf{}\implies 9\times \dfrac{10}{60}=Distance⟹9×6010=Distance
\sf{}Distance=1.5kmDistance=1.5km
Then remaining distance,
\sf{}\bf{}Remaing\ distance=Total\ distance-distance\ travelled\ earlierRemaing distance=Total distance−distance travelled earlier
\sf{}\implies 6-1.5⟹6−1.5
\sf{}\therefore 4.5km∴4.5km
Therefore,for next 20 the boy should run with a speed,
\sf{}Speed=\dfrac{Distance}{Time}Speed=TimeDistance
Covert 20min into hr
=> 20/60hr
\sf{}Speed=\dfrac{4.5}{\frac{20}{60}}Speed=60204.5
\sf{}\implies \dfrac{4.5}{\frac{2}{6}}⟹624.5
\sf{}\implies \dfrac{4.5}{\frac{1}{3}}⟹314.5
\sf{}\implies 4.5\div \dfrac{1}{3}⟹4.5÷31
\sf{}\implies 4.5\ \times 3⟹4.5 ×3
\sf{}\therefore 13.5km/h∴13.5km/h
Therefore,speed needed for next 20min is equal to 13.km/h