Math, asked by suhanirathi62, 1 year ago

1)karan is 3 times older to teja now. After 5 year their ages will be in the ratio 11:10.Find their present ages?

2)the distance between two towns on a national highway is 760 km. A and B start simultaneously from these cities on their bikes towards each other. They meet after 4 hrs and if the speed of B is 10kmph more than that of A. Find their speeds?


suhanirathi62: please do fast

Answers

Answered by Anonymous
2
hey mate

here is your answer

The first question needs some corrections
Attachments:

suhanirathi62: mean
suhanirathi62: ur 1 question is right or not
Anonymous: mine answer is right according to the question
Anonymous: u just check the first question
Anonymous: it's wrong i guess
suhanirathi62: ol
suhanirathi62: ok
suhanirathi62: and the second one
Answered by nilesh102
5

Q.[1]

Given data :-

  • Karan is 3 times older to Teja
  • After 5 year their ages will be in the ratio 11:10

Solution :-

Let, age of Teja be x

→ age of Teja = x years ..... ( 1 )

Hence, according to given

→ age of Karan = 3x years ..... ( 2 )

According to given after 5 years

→ age of Karan : age of Teja

= 3x + 5 : x + 5 ..... ( 3 )

According to given ratio of their ages after 5 year

→ age of Karan : age of Teja

= 11:10 ..... ( 4 )

Now, from ( 3 ) & ( 4 )

→ 3x + 5 : x + 5 = 11 : 10

We can write it as

→ {3x + 5}/{x+5} = 11/10

→ 10 {3x + 5} = 11 {x + 5}

→ 30x + 50 = 11x + 55

→ 30x - 11x = 55 - 50

→ 19x = 5

→ x = 5/19

→ age of Teja = x years

→ age of Teja = 5/19 years

Put value of x in eq. ( 2 )

→ age of Karan = 3x years

→ age of Karan = 3 × 5/19 years

→ age of Karan = 15/19 years

Put value of x in eq. ( 3 )

→ age of Karan : age of Teja

= 3x + 5 : x + 5

→ age of Karan : age of Teja

= 3 × 5/19 + 5 : 5/19 + 5

→ age of Karan : age of Teja

= 15/19 + 5 : 100/19

→ age of Karan : age of Teja

= 110/19 : 100/19

Hence, before 5 year age of Karan and age of Teja is 15/19 and 5/19 respectively. and after five year age of Karan and age of Teja is 110/19 and 100/19 respectively.

Q.[2]

Given data :-

  • The distance between two towns on a national highway is 760 km.
  • A and B start simultaneously from these cities on their bikes towards each other.
  • They meet after 4 hrs and if the speed of B is 10kmph more than that of A.

Solution :-

Let, speed of A be ' x '

→ Speed of A = x

Hence, according to given

→ Speed of B = x + 10 .....( 1 )

→ Total time taken by A and B = 4 hour

→ Total distance travel by A and B = 760km

Now, we use formula of relative speed

→ Relative speed of A and B

= {Total speed}/{Total time}

→ x + x + 10 = 760/4

→ 2x + 10 = 190

→ 2x = 190 - 10

→ 2x = 180

→ x = 180/2

→ x = 90 km/hr

Hence, speed of A is 90 km/hr.

Put value of x in eq. ( 1 )

→ Speed of B = x + 10

→ Speed of B = 90 + 10

→ Speed of B = 100 km/hr

Hence, speed of B is 100 km/hr.

Hence, speed of A and B is 90 km/hr and 100 km/hr respectively.

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