### Discussion :: True Discount - General Questions (Q.No.5)

Murali said: (Nov 23, 2010) | |

Please provide some more explanation. Then it is easily understood by me and others. |

Karthika said: (Dec 4, 2010) | |

Please explain. I can't get the concept. |

Revathy said: (Dec 27, 2010) | |

Please help us out here. |

Meha said: (Jan 5, 2011) | |

Whar ever the formula is given at biginig that was not comparable here. Which formula you are used here? please expline the formula. |

Karthick said: (Jan 5, 2011) | |

Please explain the sum and which formula you are used here. |

Raghubabu said: (Feb 5, 2011) | |

I cant understand it please explain by another form. |

Abhishek said: (Feb 23, 2011) | |

According to question, Amount=110,True Discount=10,Present Worth=110-10=100; Now,we know,P.W.=(Amount*100)/100+(R*T); So,100=(110*100)/100+(R*T);that implies (R*T)=10; Again in question, (R*2T)=(2RT)=2*10=20;Now, P.W.=(110*100)/100+(2RT)=(110*100)/100+20=91.66; So,T.D.=Amount-P.W.=110-91.66=Rs.18.34/- (clear).... |

Shubha said: (Apr 15, 2011) | |

How you have got p.W=(Amt*100)/(100+(r*t)) |

Vinay said: (Aug 28, 2011) | |

The discount allowed on the same sum 110 is double the time means 10 becomes 20% So, 20% will come. Therefore, principal amount is 100 for both the cases (20x 110)/120= Rs. 18.33 |

Anu said: (Oct 20, 2011) | |

Thank you Abhishek. |

Kuldeep Kumar Singh said: (Jan 20, 2012) | |

@Abhishek thanks for explaation. |

Dhana said: (Jan 24, 2012) | |

Thanks abhishek. |

Aejaz said: (Feb 26, 2012) | |

This can be taken as principal as 100 and amount as 110 then from amount = p + PTR/100 then we get RT as 10 : Now 2RT =20 and Amount is 110 then find the principal from above formula we will get amount as 91.65 so the discount is 110-91.67=18.33 |

Madhukumar E said: (Nov 14, 2013) | |

It's given that, If, TD on 110 = 10 at certain time hence, then, TD on 100 =20 at double the time hence, 110 = 20. 100 = ? = 18.18. In the same way. If, TD on 120 = 20 at certain time hence, then, TD on 110 = x. 120 = 20. 110 = ? 20/120 = 1/6. 1/6*110 = 18.33. So 18.33 answer. |

Kiruthi said: (Jun 15, 2015) | |

Please explain without formula generaly. |

Anvesh said: (Nov 26, 2015) | |

Let suppose 110 should be paid after 1 year with a discount of Rs. 10. -> Present worth (PW) = 110-10 = 100. -> Rate = Discount*100/(PW*time) = 10. The question is what will be the present worth, if the amount becomes 110 after 2 years with rate 10% per year. -> Present worth = 110/(1+10*2/100) = 91.67. -> Discount = 110-91.67 = 18.33. |

Mamta said: (Dec 1, 2015) | |

PW = Amt*100/100+rt = td*100/rt. => 11000/100+rt = 1000/rt. => rt = 10. Now, we have amt 110 on t double at are % and need to find out new TD on new PW or you can say. TD = Amt - PW. = 110 - 110*100/100+r*2t. = 110(1-100/100+2*10). = 110(1-100/120) = 18.33. |

Swati said: (Jan 11, 2016) | |

What is the full form of S.I? |

Balasubramanian said: (Apr 19, 2016) | |

Good explanations for discount and time & work problem. It helps to understand easily. |

Keerthi said: (Jul 18, 2017) | |

Given. T.D = 10, AMOUNT = 110. SO, BY USING THE FORMULA. 100*AMT/(100+RT)=100 * T.D/(RT), WE KNOW THAT RT= 10, Since the question asked to find out the T.D after double the time RT=20, So, by using the same above formula T.D=18.3333. |

Arvindkumargp said: (Aug 25, 2017) | |

Thanks @Abhishek. |

Devendra said: (Jun 18, 2018) | |

Thank you for explaining the answer. |

Swetha said: (Jun 27, 2018) | |

Thank you @Abhishek. |

Manisha said: (Jul 26, 2018) | |

Thanks @Abhishek. |

Jubair said: (Jan 16, 2019) | |

Thanks @Abhishek. |

Sudip Dhara said: (Apr 11, 2019) | |

See, the true discount allowed is Rs. 10 and the amount to be paid after a certain time is Rs. 110. We don't know the time and the present worth. So lets assume a value. Let's say the present worth is Rs.100, and the time is 1 year. Thus the Rate has to be 10%. Now double the time, so what is the present worth if Rs.110 was due in 2 years at 10%? It comes out to be Rs. 91.66. Thus the true discount becomes Rs.18.33.) |

Gohil Rajnikant said: (Nov 18, 2019) | |

Thanks @Abhisek. |

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