Two Equal Amounts of Money are lent out at 6% and 5 % simple Interest respectively at the same time. The former is recovered two years earlier than the latter and the amount so recovered in each case is RS, 2800. Determine the amount that is lent out?
A) 1950
B) 1500
C) 1800
D) 1375

• Rs 1750 in each case which is not given in the options.
  check
  Interest in each case = 1750*6*10/100 = 1750*5*12/100 = Rs 1050
• 10 years agoHelpfull: Yes(9) No(16)
• Ans: Rs.1800
  let p be the amount lent out
  interest in 1st case= p*r*t/100
  = p*6*t/100
  =6pt/100------------------(equation 1)
  interest in 2nd case= p*r*t/100
  =p*5*(t+2)/100 (since recovered 2 years later than 1st one)
  since amounts are equal(each 2880) and lent out sums are equal, intersts will be equal
  => 5p(t+2)/100=6pt/100
  =>5t+10=6t
  =>t=10yrs
  final amt=2880
  =>p + interst = 2880
  => p + 5p*12/100 = 2800
  =>160p=2800*100
  => p= Rs.1750
• 10 years agoHelpfull: Yes(9) No(3)
• Ans: Rs.1800
  let p be the amount lent out
  interest in 1st case= p*r*t/100
  = p*6*t/100
  =6pt/100------------------(equation 1)
  interest in 2nd case= p*r*t/100
  =p*5*(t+2)/100 (since recovered 2 years later than 1st one)
  since amounts are equal(each 2880) and lent out sums are equal, intersts will be equal
  => 5p(t+2)/100=6pt/100
  =>5t+10=6t
  =>t=10yrs
  final amt=2880
  =>p + interst = 2880
  => p + 5p*12/100 = 2880
  =>160p=2880*100
  => p= Rs.1800
• 10 years agoHelpfull: Yes(5) No(1)
• Ans: Rs.1800
  let p be the amount lent out
  interest in 1st case= p*r*t/100
  = p*6*t/100
  =6pt/100------------------(equation 1)
  interest in 2nd case= p*r*t/100
  =p*5*(t+2)/100 (since recovered 2 years later than 1st one)
  since amounts are equal(each 2880) and lent out sums are equal, intersts will be equal
  => 5p(t+2)/100=6pt/100
  =>5t+10=6t
  =>t=10yrs
  final amt=2880
  =>p + interst = 2880
  => p + 6p*10/100 = 2880
  =>16p=2880*10
  => p= Rs.1800
• 10 years agoHelpfull: Yes(3) No(2)
• ans will be 1750
• 10 years agoHelpfull: Yes(1) No(2)