# Answer to Question #7590 in Algebra for jadine

Question #7590

Makhayaz weight is 110% of mohammeds weight.the ratio of josephs weight to tumelos weight is 10:11,the ratio of ceciliaz weight to makhayaz weight is 6:7.Ceciliaz weight is 120% of josephs weight,who weighs the least?

Expert's answer

Let the Makhayaz's weight be M. Then Mohammed's weight is 1.1M.

Let the Joseph's weight be J and Tumelos's weight be T. Then J/T = 10/11.

Let the Ceciliaz's weight be C. Then C/M = 6/7.

Also, C = 1.2J.

So, we've got the system of equations:

C/M = 6/7 ==> C = 6M/7.

C = 1.2J ==> J = C/1.2 = 6M/7/1.2 = 5/7 M.

J/T = 10/11 ==> T = 11J/10 = 11/10*5/7 M = 55/70 M.

Finally, we've got:

Makhayaz: M

Mohammed: 1.1 M

Ceciliaz: 6/7 M (≈0.8571 M)

Joseph: 5/7 M (≈0.7143 M)

Tumelos:& 55/70 M (≈0.7857 M)

So, Joseph weighs the least.

Let the Joseph's weight be J and Tumelos's weight be T. Then J/T = 10/11.

Let the Ceciliaz's weight be C. Then C/M = 6/7.

Also, C = 1.2J.

So, we've got the system of equations:

C/M = 6/7 ==> C = 6M/7.

C = 1.2J ==> J = C/1.2 = 6M/7/1.2 = 5/7 M.

J/T = 10/11 ==> T = 11J/10 = 11/10*5/7 M = 55/70 M.

Finally, we've got:

Makhayaz: M

Mohammed: 1.1 M

Ceciliaz: 6/7 M (≈0.8571 M)

Joseph: 5/7 M (≈0.7143 M)

Tumelos:& 55/70 M (≈0.7857 M)

So, Joseph weighs the least.

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