Felipe's school is hosting a math competition against other schools in the same district. Each school can only allow 10 students to compete. Felipe and his classmates are taking tests to determine the 10 best math students to send. Felipe did well, but tied with John Roy for the last spot. His teacher decided to set-up a one-problem challenge, whoever got it right the fastest would win.

Knowing that "H" is equal to 10, and T is half of M, how could MATH be 42, TEAM be 40, and MEET be 37?

• If H=10, T=7,M=14,A=11 and E=8, it could result MATH as 42,TEAM as 40 and MEET as 37.
  Explanation:
  Let us take the word "MATH" first: M+A+T+H=2T+A+T+10=42=>3T+A=32
  Similarly, for "TEAM"=> 3T+A+E=40=>E=8(since 3T+A=32)
  upon further substitution in the word "MEET", we get T=7 and M=14.
  from this, A=11 can be obtained.
  
• 13 years agoHelpfull: Yes(3) No(0)
• Let T be 'x' then M will be '2x', A be 'y' and E be 'z' and given H is 10.so, 3x+y=32 3x+y+z=40 and 3x+2z=37.By solving,x=7,y=11 and z=8.So values of E is 8,A is 11,T is 7 and M is 14.
• 13 years agoHelpfull: Yes(2) No(0)
• a=2.21, h=10, m=1.948, e=9.5, t=0.974
  explanation:
  by substituting these values and on multiplication v get that concerned answers
• 13 years agoHelpfull: Yes(0) No(0)