### Discussion :: Simple Interest - General Questions (Q.No.1)

V.Malarvizhi said: (Jul 20, 2010) | |

Why not applied principal formula si*100/r*t ? |

Jayesh said: (Aug 21, 2010) | |

Can the formula be SI*100/R*T ? |

Name said: (Sep 6, 2010) | |

3y--815, 4y--854, ---------- 1y--39, So, 3y ---39 x 3 = 117 Principal is = (815-117) = 698 |

Kannan said: (Nov 25, 2010) | |

Why should not subsract with 854? Please clear my doubt.... |

Priya said: (Nov 29, 2010) | |

@Lavanya and kanna To get 1 year difference we have to subtract 4year amount-3year amount this difference amount is the 1 year simple interest.. that's why.. Rs. (854 - 815) = Rs. 39 39 is 1year S.I. hope it helps you.. |

Nag said: (Feb 8, 2011) | |

Thank you. This is explanation is very helpful for me. |

Appu said: (Feb 21, 2011) | |

Hai friends, Principle is the amount borrowed. Here, explain about SI of 3 years and 4 years. Assume P is the principal then, P+854 for 4 years and p+815 for 3yrs. So here it is necessary to find the SI of one year and subtract it from the 3 year or 4 year. That is, (815-117) = 698 and 854-156=698. |

Vishnu said: (Mar 8, 2011) | |

To get a si of one year or it might be two years if it is two year we have to divide it by two si=3yrs -2yrs willl give u one year si why should we have to go to risk. |

Ramesh said: (Mar 18, 2011) | |

PTR/100 = 815 P*3*R/100 = 815 R/100=854- 815 p * 3 * 39 = 815 P * 117 = 815 p = 815-117 = 698 |

Hari said: (Mar 26, 2011) | |

I like this type of solutions thank you I think this answer also 1year 854-815=39 3years 39*4=156 854-156=698 |

Vijay said: (May 27, 2011) | |

What is formula of simpe interest ? |

Rajini Kumar said: (Jun 3, 2011) | |

S.Ib = PNR/100 |

Narasimha Reddy said: (Jun 4, 2011) | |

S.I=PTR/100 |

Gayathri said: (Jun 22, 2011) | |

Hi Friends! Principle = the amt u borrowed Interest = the small additional amt u pay for the amt u borrowed First let me explain u d question In the question, A sum of money at simple interest amounts to Rs. 815 in 3 years means the Principle amt together wit the interest paid for 3 yrs is Rs. 815 and similarly Principle amt together wit the interest paid for 4 yrs is Rs. 854 Therefore, if we subtract (854-815) we get the simple interest for 1 year as Rs. 39 Then, we calculate simple interest for 3 yrs as (39*3)= 117 Next step, we subtract (815 - 117) to get Principle amt for a yr as Rs. 698 Brief explanation for last step: 815 = Principle amt + int for 3 yrs ......equation 1 117 = int for 3 yrs ......equation 2 now subtract both equations, we get the principle amt I hope tat this explanation will be helpful to u |

Reshmi said: (Jul 6, 2011) | |

Thank you gayathri. You explained well. |

Prathap said: (Jul 6, 2011) | |

Simple and clear explanation by gayatri. |

Raghu Karthic said: (Jul 28, 2011) | |

Superb explanation gayathri. Thanks a lot. |

Ashik said: (Jul 30, 2011) | |

S.I for one yr=854-815=39 for 4 years=4*39=156 hence,854-156=698 i think it will be simple.... |

Vikram said: (Aug 2, 2011) | |

According to the question, SI + P = 815 in three years and SI + P = 854 in four years so 815 - P = SI and 854 - P = SI Therefore, 815 - P = (P * 3 * R)/100 and ---- eq 1 854 - P = (P * 4 * R)/100 ---- eq 2 Now, eq 1 / eq 2 gives (815 - P)/(854 - P) = 3/4 Solving this eq, we get P = 698 |

Uma said: (Aug 16, 2011) | |

@gayathri your explination is very helpfull. Thank you. |

Harsh said: (Sep 13, 2011) | |

I am not deciding which is easiest way to solve of the given question so please suggest me. |

Venkatesh said: (Sep 19, 2011) | |

Gayathri thank you so much before that I want to say myself I am dull even though I understood you are explantion it is very clear keep it up. |

Jaya said: (Sep 26, 2011) | |

Gayathri thanks a lot its vry clr & simple |

Kichu said: (Oct 24, 2011) | |

Thanks gayathri. |

Muskan said: (Nov 20, 2011) | |

Please give me the solution of this question. A certain sum given on simple interest became double in 20 years. In how many years will it be 4 times. |

Ranjitha said: (Dec 27, 2011) | |

Its same friends.. you can do it in either way (39*4)=156 854-159=698 only |

Sonia said: (Dec 29, 2011) | |

Thanks Gayathri. Really your explanation is helpful. |

Venkatesh Kumar said: (Jan 17, 2012) | |

Thanks Gayathti for giving detailed explanation. |

Saurabh said: (May 1, 2012) | |

Thanks gayathri. |

Tejashree said: (Jun 8, 2012) | |

Thanks gayathri |

Radhika said: (Jun 23, 2012) | |

Thank you so much Gayathri.... very good explaination |

Moksh said: (Jul 21, 2012) | |

@Gayathri. Thanks a lot dear. Very useful answer :). |

Ismail Khan said: (Aug 17, 2012) | |

Amount for three years is 815 so AMOUNT= p + SI A = p + pin 815= p + 3pi.......... (1) similerly for 4 years 854 = p +4pi..........(2) subtracting equation (1) from(2) we get 39 = pi or pi =39 put this value of pi in equation (1) we get 815= p+ 3*39 815= p+ 117 815-117=p p=698 |

Satish said: (Sep 3, 2012) | |

@Gayahthri explained good . |

Sudesna said: (Sep 4, 2013) | |

When we apply the formula SI = PRT/100. And when we apply the formula Amount = P+SI. |

Sumit Singh said: (Dec 23, 2013) | |

815 ------ 854 (This value 39 increase in 1 year ). So in four year 39*4/156. This 156 is interest in four year. So principle is equal to / 854-156 equal to 698. |

Varunraj said: (Jan 4, 2014) | |

If we take for four years. 39*4 = 156. 854-156 = 698 is the answer. If we take three years. 39*3 = 117. 815-117 = 698 the answer is same ! We can do both the way. |

Ravi said: (Mar 28, 2014) | |

If in this question asked rate also then how to solve this question? |

Nishan said: (Jul 10, 2014) | |

In my view the interest between the 3rd & 4th year of Rs. 39/- cannot determine the same interest component for the first three years as the interest will be paid on top of the capital? As such the interest component should reduce proportionately over the previous years? |

Narendra said: (Aug 13, 2014) | |

SI1 = 815. SI2 = 854. DIFFERENCE = SI2-SI1. D = 854-815. D = 39. FOR 3 YEARS = 3*39. = 117. SO NOW P = 815-117. P = 698. |

Rafee And Guru said: (Aug 19, 2014) | |

Hi Friends! Principal = the amount you borrowed. Interest = the small additional amount you pay for the amount you borrowed. First let me explain you the question. In the question, A sum of money at simple interest amounts to Rs. 815 in 3 years means the Principal amount together with the interest paid for 3 yrs is Rs. 815 and similarly Principal amount together with the interest paid for 4 yrs is Rs. 854 Therefore, if we subtract (854-815) we get the simple interest for 1 year as Rs. 39. Then, we calculate simple interest for 3 yrs as (39*3) = 117. Next step, we subtract (815 - 117) to get Principal amt for a yr as Rs. 698. Brief explanation for last step: 815 = Principal amt + int for 3 yrs ......equation 1. 117 = int for 3 yrs ......equation 2. Now subtract both equations, we get the principal amount I hope that this explanation will be helpful to you. |

Priyanka said: (Nov 16, 2014) | |

Hi friends. Here goes the explanation, SI for 4 years = Rs 854. SI for 3 years = Rs 815. {4-3} = 1. So SI for 1 year can be {854-815}. = Rs 39. So the SI for 3 years can be {39*3}. = 117. So principal is 815-117. = RS 698{ANS}. |

Magesh said: (Nov 23, 2014) | |

For four years the interest is, 854-815 = 39. 39*4 = 156. So 854-156 = 698. It may simple one no need to apply any formulas here. Time waste one. |

Stuti said: (Mar 21, 2015) | |

Hi. There is a simple shortcut for solving this question. P/Sum = A1n2-A2n1/n2-n1. = 815*4-854*3/4-3. = 3260-2562/1 = 698. |

Shaminzu said: (Jul 20, 2015) | |

3y---815. 4y---854. ------------ 1y---39. 3y---39*3 = 117 or 4y---39*4 = 156. 815-117 = 698 or 854-156 = 698. So we could subtract by 854 or 815. All The best. Think smart. |

Monika said: (Oct 16, 2015) | |

Can explain me logic behind it? |

Prasad said: (Oct 29, 2015) | |

Will you elaborate it in understanding way? |

Md Sahil said: (Jan 9, 2016) | |

Sum amounts to 815 in 3 years. Sum amounts to 854 in 4 years. Find sum? Let the sum be Rs. x. x = principal amounts in 3 years = Rs. 815. x = principal amounts in 4 years = Rs. 854. Principal amount in 1 year = 39 I. Interest of 1 year = Rs. 39. Interest of 3 year = 3x39= Rs. 117. Principal = x. Rate = ? Time = 3 years. For 3 years. Interest = Rs. 117. Amount = Rs. 815. Time = 3 years. Let rate = R%. A = p+I. P = A-I. = R(815-117). Rate 698 answer. |

Naren said: (Feb 9, 2016) | |

815-p = 3*x.1 -----> equation 1. 854-p = 4*x.2 -----> equation 2. Divide equation 1 with equation 2. |

Sumair Mudliar said: (Jun 3, 2016) | |

@Md Sahil. Thank you for explaining the solution. |

Imran Syed said: (Jun 18, 2016) | |

For simple interest, this formula is very helpful. That is I = Prt. |

Salomi said: (Jun 27, 2016) | |

@Gayathri. Thank you for explaining the solution clearly. |

Kabali said: (Jul 6, 2016) | |

@Gayathri nice explanation. |

Mick Sudama said: (Jul 10, 2016) | |

Here you can do it by another S. I 1 yr = 39, in 4 years =156, & principle=854 - 156 = 698. |

Aparna said: (Jul 22, 2016) | |

Crystal clear explanation, Thanks @Gayathri. |

Kitty said: (Jul 24, 2016) | |

@Appu. Your explanation is good, Thank you. |

Sachin said: (Jul 27, 2016) | |

Thanks @Vikram and @Gayatri. |

Sahid Hussain said: (Aug 5, 2016) | |

Let the sum is Rs. X and rate of interest is y%. We know A = P + SI where SI = (P * R * T) /100 Therefore, A = P + (P * R * T) /100. Case 1 : A = 815, P = x, T = 3 years, R = y%. A = x + (x * y * 3) /100->1 Case 2 : A = 854, P = x, T = 4 years, R = y%. A = x + (x * y * 4) /100 ->2 Subtracting equation (1) from equation (2), we get x * y = 3900. Now by putting this value of x * y = 3900 in any one of above two equation, we'll get sum = x = 698 Answer. |

Rakesh.H said: (Aug 11, 2016) | |

I will explain in a simple way. We have A = P + I. 815 = P + I -----> Eq 1. 854 = P + I -----> Eq 2. Subtract Eq 1 and 2. Simple interest I = 815 - P - 854 + P = 39 for 1year. Then take for 3 years or 4 years. You will get the same answer. For 3 years: 39 * 3 = 117. Or For 4 years: 39 * 4 = 156. Then use formula A = P + I. For 3 years: 815 = P + 117, P = 815 - 117, = 698. For 4 years: 854 = P + 156, P = 854 - 155, = 698. Take for 3 or 4 years answer will be same. I think the explanation is very useful. |

Aryan said: (Aug 15, 2016) | |

Let principal = x, Principal + simple interest = total sum (after any number of years), For 3 years, X + x * r * 3 = 815 ----> Eqution 1, For 4 years, X + x * r * 4 = 854 ----> Equation 2, Now there are 2 unknown and 2 equations solve for x and we get. X = 698. |

Rahul Dev Singh said: (Sep 20, 2016) | |

If certain principal amounts to A1 in T1 year and to A2 in T2 year ,then the sum is given by = A2 .T1--A1.T2 _____________ T2--T1 Now use the formula = 854 * 3 -- 815 * 4 ________________ 4--3 = 2562--3260 ____________ 1 = 698. |

Karthikeyan said: (Oct 3, 2016) | |

@Gayathri. Superb explanation, You are genius Thank you. |

Chirag said: (Oct 19, 2016) | |

@Rahul. How? Please explain your solution clearly. |

Tanuja said: (Oct 22, 2016) | |

Yeah, this is the very easy solution. But can you explain this method with some other examples? Like 9 years SI is 657 and 5 years SI is 555 so then? |

Ryan said: (Nov 17, 2016) | |

@Tanuja Use Vikram's explanation as reference In case you didn't understand still, just logically understand that the: Resultant sum = Principal + (SI * No. of years). In the case of your example which is 9 years SI is 657 and 5 years SI is 555 so: P + 9SI = 657 -------------(1) P + 5SI = 555 -------------(2) Subtracting (1) with (2) you get: 4SI= 102, SI = 102/4, => SI = 25.5. Substituting SI in (1) equation you get: P + 9 (25.5) = 657, P = 657 - 9 (25.5), => P = 427.5. |

Chikuuuu said: (Dec 27, 2016) | |

Your explanation helped me a lot, Thanks @Gayathri. |

Jitendra said: (Jan 7, 2017) | |

@Gayathri. Very good explanation, easily understand. Thanks. |

Sachin said: (Feb 2, 2017) | |

Thanks for all the solution and explanation. |

Deepti said: (Feb 7, 2017) | |

It is given in sum that SI of 3yrs is 815 so why we need to multiply 3 with 39 to get SI for 3 years? |

Suraj.D said: (Feb 12, 2017) | |

Please tell me if we subtract 854- 815 we get principal+ s.I for 1 year. Then can we consider it only s.I for 1 year? Please explain me. |

Pradeep said: (Feb 19, 2017) | |

Let 1 yr int .....x. 3yr int.......3x. 4yr int........4x. (P+4x) - (P+3x) = 854 - 815, x = 39. |

Nirmala said: (Mar 10, 2017) | |

Why we are not substructing 854? Please explain. |

Manya said: (Mar 13, 2017) | |

Thanks a lot @Gayatri. |

Anusha said: (Mar 17, 2017) | |

Very helpful to understand the solution. Thanks for all your explanation. |

Surjeet said: (Mar 17, 2017) | |

Because we calculate with 3 years that is why we subtract from 815. |

Deepthy said: (Apr 5, 2017) | |

Thanks @Pradeep. |

Sunil said: (Apr 24, 2017) | |

Thanks, I like this explanation. This is helpful to understand the simple interest. |

Pawan said: (May 27, 2017) | |

How can we apply formula 100*SI/RT? Here R is not given. |

Marikkannan said: (Jun 2, 2017) | |

Thanks for all the explanation. It is a very helpful solution. |

Ashwini said: (Jun 2, 2017) | |

A2T1-A1T2/T1-T2. The above formula is applies here, 854 * 3-815 * 4/3-4, 2562-3260/-1, = 698. |

Aman said: (Jun 12, 2017) | |

Hi, I cannot understand one thing that how could be the amount increasing every year is constant? I mean if suppose 100 is the principle and rate is 10 percent then for the first year the amount will be 110, 2nd year it will be 121, 3rd year it will be 133.10. Now 133.1 -121 is not equal to 121-110 also not equal to 110 -100 so the increasing amount is not constant. In this problem, how can we take the increasing amount constant and directly subtract it and divide it? I mean how is it possible that the increasing amount remains 39 over three years? |

Mayaank said: (Jun 19, 2017) | |

Is this the sum of money for 3 years? |

K Mahendra Naidu said: (Jul 21, 2017) | |

Hi, guys this problem can be solved in 2 ways i.e, Sum of 3 years ------------ 815 Sum of 4 years------------- 854 So, the difference is 1 year. Amount 39. So, solution1: 39*3=117(difference*years). 815-117=698. Sol2: 39 * 4 = 156(difference*years), 854 - 156 = 698. |

Visithra said: (Jul 25, 2017) | |

Very useful discussion. Thank you all. |

Rajeswari said: (Aug 9, 2017) | |

3years = 815. 4years = 854. Diff = 39. 39*3 = 117. 815-117 = 698. 854-117 = 698. |

Jayraj said: (Aug 29, 2017) | |

Good method, Thanks @Vikram. |

Lalit said: (Sep 1, 2017) | |

Hi, Please explain me why SI calculated on 3 years why not 4 year? |

Koishik Mandal said: (Sep 5, 2017) | |

S.I. for 1 year = Rs. (854 - 815) = Rs. 39. S.I. for 3 years = Rs.(39 x 3) = Rs. 117. Principal = Rs. (815 - 117) = Rs. 698. |

Pavithrasai said: (Sep 26, 2017) | |

Here, si=p*r*t/100. So for 3 yrs, 815=p*r*3/100 => r=81500/3p and for 4yrs, 854=p*r*4/100 => r=85400/4p. Verify options by placing each option as p values and option which gives same values of r in both the equations is the answer. Hence, option c. |

Monisha Lodhi said: (Sep 30, 2017) | |

Why we find out only for three years. Is principle is found on only for the first year money? |

Divya said: (Oct 13, 2017) | |

Is 698 is principal amount? |

Sase M said: (Oct 20, 2017) | |

Why should we apply like this; s.i for 1 year = Rs 854-815=39 1yr = 39 then 4y = 156 then 854-156=698. |

Sase M said: (Oct 20, 2017) | |

@Divya. Its a initial amount. |

Nipan said: (Oct 20, 2017) | |

Thanks for explaining. |

Madhu said: (Oct 22, 2017) | |

Basic formula of to find r% and p%. When give two amounts and two years to find p formula, (A1*T2-A2*T1)/T2-T1. R% FORMULA. (A2-A1)/(A1*T2-A2*T1). |

Aquarius said: (Nov 16, 2017) | |

Can't this question be solved using formula p=100*si/r*t. Total amt = p+si. 815 = p+si. si = 815-p. 854 = p+si. si = 854-p. Substituting in formula, p = 100*815-si/r*3. Can anyone please help how to solve using this method of substitution. |

Pramod Nepali said: (Nov 30, 2017) | |

Let p=x nd after 3year A1=815 nd4year A2=854... A( is total amount)=p+i we no P=(A*100)/(100+TR) .......p nd R is equal for both A1 nd A2 . so P1 =p2-------(eq1) we have formula P1=(A1*100)/(100+T1R)--------(2) And P2=(A2*100)/(100+T2R)---------(3) put eq2 and eq3 on eq 1 and putting all value we go Rate (R)=5.58 value of R input on eq 2 or eq 3 we got the value of p. |

Pragna said: (Dec 1, 2017) | |

Amount(A) = Principle(p) + Simple Interest(SI). The sum of money i.e Principle(p) invested or borrowed results an Amount of 815 in 3 years and 854 in 4 years. Since Rate of interest(R) is not mentioned assume it as same for both. A1 = P + SI1 815 = P + SI1 P = 815 - SI1--- eq1 A2 = P + SI2 854 = P + SI2 P = 854 - SI2 ----- eq2 eq1= eq2 therefore -----> SI2 - SI1 = 854 - 815 =39 Now, SI1 = P*T1*R/100 =P*3*R/100 SI2 = P*T2*R/100 = P*4*R/100 SI2 - SI1 = P*R/100 39 = P*R/100 -------> Here we can say T=1year, so this is SI for 1 year now calculate SI for 3 or 4 years SI for 1 year = 39 SI for 3 years = 39*3 = 117 Which is SI1 or SI for 4 years = 39*4 = 156 which is SI2. A1 = P + 117 815 = P + 117 P = 815 - 117 = 698 A2 = P + 156 854 = P+156 P = 854 -156 = 698 So sum of money i.e Principle P = 698. |

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