A person travels a total distance of 360 km by a car. He covers the first half at a speed of 50 km * h ^ - 1 and the other half at 5 km * h ^ - 1 . Find:a) the total time taken to complete the entire journey.5) average speed of the car.
Answers
Answer:
G I V E N :
A person invested a sum of money for 5years at the rate of 6% per annum simple interest. after 5 years, he re invested the same principal for 6 years at the rate of 8% per annum simple interest. If the total interest earned from both these investments is Rs.12480, find the sum invested by him.
S O L U T I O N :
Case I
According to the question we are given with several datas. They are
Let the principal be P
Time (T) = 5 years
Rate (R) = 6 % p.a
Now, to get the simple interest we need to apply
\sf{S.I. = \frac{P × R × T}{100}}S.I.=100P×R×T
\sf{S.I. = \frac{P × 6 × 5}{100}}S.I.=100P×6×5
\sf{S.I. = \frac{30P}{100}}S.I.=10030P
\star \: \underline{ \boxed{\sf{ \pink{S.I. = \frac{3P}{10}} }}}⋆S.I.=103P
For case 1 we got the simple interest as 3P/10
Case II
Now, here also we are loaded with the new datas. Noting them down
The pricipal is P1
Time (T1) = 6 years
Rate (R1) = 8 % p.a
Now, getting the simple interest for case II
\sf{S.I. = \frac{P_{1} × R_{1} × T_{1}}{100}}S.I.=100P1×R1×T1
\sf{S.I. = \frac{P× 8×6}{100}}S.I.=100P×8×6
\sf{S.I. = \frac{48P}{100}}S.I.=10048P
\star \: \underline{ \boxed{ \sf{ \red{S.I. = \frac{12P}{25}} }}}⋆S.I.=2512P
For case II the interest is 12P/25
Now, adding both these cases which equals the total interest earned
Case I + Case II = Total Interest earned
\rightharpoonup \sf{ \frac{3P}{10} + \frac{12P}{25} = 12480 }⇀103P+2512P=12480
Now, getting the L.C.M. The L.C.M of 10, 25 is 50
\rightharpoondown\sf{ \frac{3P ( 5)}{10 ( 5)} + \frac{12P(2)}{25(2)} = 12480 }⇁10(5)3P(5)+25(2)12P(2)=12480
\rightharpoonup \sf \frac{15P}{50} + \frac{24P}{50} = 12480⇀5015P+5024P=12480
\rightharpoondown \sf \frac{39P}{50} = 12480⇁5039P=12480
\rightharpoonup \sf39P = 12480(50)⇀39P=12480(50)
\rightharpoondown \sf39P = 624000⇁39P=624000
\rightharpoonup \sf P = \frac{624000}{39}⇀P=39624000
\star \: \: \boxed{ \green{\frak{ {\sf P} = 16000}}}⋆P=16000
Hence, the sum invested is ₹ 16000
Answer:
Given distance=360 km.
Let the speed of the train be x km/hr.
Speed when increased by 5 km/hr =(x+5) km/hr
x
360
−
(x+5)
360
=1
x(x+5)
[360x+1800−360x]
=1
x
2
+5x−1800=0
x
2
+45x−40x−1800=0
x(x+45)−40(x+45)=0
(x−40)(x+45)=0
x=40,−45
The speed of the train is 40 km/hr.