Originally Posted by K.C.SATHEESH:.

I am a student of SSLC Karnataka Board and now searching for Solved Maths Problems of SSLC Karnataka Board so tell me where I can get its answers?

Pls contact me for any maths questions in class 10. Email:vamsi22@rediff.com
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Krishna

The HCF h ( y ) = 2y ( y + 2 ) and LCM m ( y ) = 24y ( y + 2 ) ( y – 2 ) for two polynomials p ( y ) and q ( y ) .
If p ( y ) = 8y + 32y + 32y & if h ( y ) m ( y ) = - p ( y ) . q ( y ) . Find q ( y )

Originally Posted by Unregistered

if a cyclist had gone 3 km/hr faster he would have taken i hour and 20 minutes less to ride 80 km. What time did he take?

Let the time taken be x Hrs.
Then Time taken to ride 80Kms.=80/x Hrs
If he had gone 3Kms/hr faster, He would have taken 80/(x+3) Hrs. This is 1 Hr 20 Mins less than 80/x Hrs.
Therefore, we get the equation 80/(x+3) = 80/x-4/3 (1Hr 20Min converted to fraction=4/3)

Bringing -4/3 to LHS and sending 80/(x+3) to RHS,
we get 4/3=80/x - 80/(x+3)

Solving for x, we get x=12 Hrs.

[kulkarni102]

Originally Posted by Unregistered

if a cyclist had gone 3 km/hr faster he would have taken i hour and 20 minutes less to ride 80 km. What ti me did he take?

Let the time taken by the cyclist be x Hrs.
Then time taken by him to cover 80 Kms=80/x Hrs.
If he rides 3Kms /Hr faster, the time taken by him to cover the same 80 Kms would be=80/(x+3), which is 1 Hr and 20M less than 80/x Hrs.

Therefore,we can set up the equation as,

80/(x+3)=80/x - 4/3 ( 1Hr 20 M converted into fraction will be 4/3 )

Transposing 80/(x+3) to RHS and -4/3 to RHS,

We get 4/3=80/x - (80 x+3)

Solving for x, we get x=12 Hrs.,which is the required answer.

I as 10 std student want to inform you friends that nothing more comes from this papers but in sense its confusion so refer only needed self study is 90 percent and 10 is what we are referring