# Aptitude - Simple Interest - Discussion

### Discussion :: Simple Interest - General Questions (Q.No.2)

2.

Mr. Thomas invested an amount of Rs. 13,900 divided in two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be Rs. 3508, what was the amount invested in Scheme B?

 [A]. Rs. 6400 [B]. Rs. 6500 [C]. Rs. 7200 [D]. Rs. 7500 [E]. None of these

Explanation:

Let the sum invested in Scheme A be Rs. x and that in Scheme B be Rs. (13900 - x).

 Then, x x 14 x 2 + (13900 - x) x 11 x 2 = 3508 100 100 28x - 22x = 350800 - (13900 x 22) 6x = 45000 x = 7500.

So, sum invested in Scheme B = Rs. (13900 - 7500) = Rs. 6400.

Video Explanation: https://youtu.be/Xi4kU9y6ppk

 Venu said: (Sep 23, 2010) Can any one solve this in simple way?

 Rajesh said: (Nov 7, 2010) Use the method of alligation.

 Suchita said: (Nov 29, 2010) @Rajesh. Can you please tell me that method.

 Vinod said: (Jan 20, 2011) Suppose total simple interest of A & B is 13900*25%(14%+11%) =3475, but in problem total interest for 2 years is 3508, difference amount is Rs.33/-,can some one explain me, why this differnce has come ?

 Pp Depu said: (Feb 17, 2011) Can some one plese explain me the 2nd step 28X-22X = 350800 - (13900x22) how it came?

 Anky said: (Feb 28, 2011) Can some one explain why we took x and 13900 - x.

 Insaan said: (Mar 7, 2011) @ Pp depu See, 28x/100 + (13900 - x)* 22 / 100 = 3508 = 28x/100 + 305800 - 22x /100 = 3508 take 100 lcm 28x + 305800 - 22x/100 = 3508 or 28x + 305800 - 22x = 350800 or 28x + 305800 - 22x - 350800 = 0 0r 28x - 45000 - 22x = 0 or 6x - 45000 = 0 or 6x = 45000 x = 7500

 Heetu said: (May 22, 2011) Please give me any simple way I cant understand it.

 Srini said: (Jun 2, 2011) @Vinod You cannot take 25% directly from Principal since money is divided into two parts, so that diff is there in your calculation Now see, 7500 x 14% + 6400 x 11% in 2 years will give you 3508.

 Pathum said: (Jun 4, 2011) Why it has not taken X for B's principal? Then it is easy to solve without subtract again.......

 Viji said: (Jul 9, 2011) Please explain how do you get 7500 and 6400.

 Anu said: (Sep 24, 2011) In this explanation how - 22x will come can any one explain me.

 Aditi said: (Nov 12, 2011) C let us assume that principle invested in a is x & b be y total amt invested is 13900 so x+y=13900 or y=13900-x rit-----1 foorming quad eq. of simple int. x*14*2/100 + y*11*2/100 = 3508 i=prt/100 or 28x/100+22y/100=3508 28x+22y=350800 substitution of y frm 1 28x+22(13900-x)=350800 now solve.

 Zakir said: (Nov 19, 2011) Good thinking.

 Ismail Khan said: (Aug 17, 2012) pin/100 + (13900-p)in/100=3508 where p is principle i is interest rate and n is no.of years p(28/100)+(13900-p)22/100=3508 28p+13900*22-22p=3508*100 28p-22p+305800=350800 6p=45ooo p=7500 13900-7500=6400

 Renu said: (Dec 2, 2012) Formula is simple interest(S.I)=P*T*R/100; P=Principal amount or sum T=Time R=Interest rate By using this formula we can solve this.

 Swati Alone said: (Jul 16, 2013) Can anyone tell me how to do this answer easily and fast without using pen ?

 Jyoti .K.Khanchandani said: (Aug 18, 2013) Amount invested is nothing but the principal. Now principal is divided into two parts. Therefore P1+P2 = 13900 Rs-------equation no.1. Now, S.I with rate 14%=S.I1=(P1*2*14)/100 -----as for two years N=2. S.I with rate 11%=S.I2=(P2*2*11)/100 -----as for two years N=2. Therefore total S.I=S.I1+S.I2 ---------equation 2. But S.I = 3508 Rs. Therefore Equation 2 becomes, 3508 = [ (P1*2*14)/100 ] + [(P2*2*11)/100 ]. 3508 = [ P1*14 +P2*11 ] [2/100]---TAKE 2& 100 common. (3508*100)/2=[ P1*14 +P2*11 ] -----equation 3. Solve equation 1 & 3 simultaneously, Hence P1 = 7500 ----for A scheme at rate 14%. P2 = 6400 ----for B scheme at rate 11%.

 Pranali said: (Aug 27, 2013) Can someone please explain me the 2nd step 28X-22X = 350800 - (13900x22) how it came?

 Rajesh said: (Oct 15, 2013) Assume the amount invested in scheme B be 6400. Then the amount invested in scheme A is 13900-6400 = 7500. 14% in 7500 = 1050. For two years (1050*2) = 2100. 11% in 6400 = 704. For two years (704*2) = 1408. Adding these two gives (2100 + 1408) = 3508. Therefore we assumed correctly and the answer is 6400.

 Bugit said: (Dec 2, 2013) I see. the key issue here is they did not say that they are substituting (total money in - "? scheme-A money in") for "? scheme-B money in". A part that may be confusing (was for me at first) is the use of x: x being the variable and the multiplicative mechanism. the variable should be noted differently than the multiplication - capitalizing X at least or using * to denote multiplication. Let me draw this out in simple terms. A = money put into scheme A. Simple interest for scheme A is 14% B = money put into scheme B. Simple interest for scheme B is 11% 3508 is the total interest returned from scheme A and scheme B. 13900 is the total money invested into A and B. (A + B) = 13900 Because the interest accumulates over 2 years, the interest gained is doubled. Therefore the interests in both schemes are doubled. (A * 14 * 2)/100 +(B * 11 *2)/100 = total interest gained. note the 100 in the denominator is because the 14 and 11 are percentages. To simplify things the .14 is represented as 14/100 and .11 to 11/100. The numbers workout in the end without any converting or anything. since we cannot do anything with the 2 variables A and B, we substitute B for ("total money invested" - A). With this we can solve for just A. (A * 14 * 2)/100 +((13900- A)* 11 * 2)/100 = 3508 Getting rid of the common denominators yields: (A * 14 * 2)+((13900- A) * 11 * 2) = 350800 Simplify: (28 * A)+((13900- A) * 22) = 350800 distribute the 22 into the parenthesis (28 * A) + (13900 *22) -(22 * A) = 350800 Simplify: 28A - 22A + (305800) = 350800 Add like terms: 6A + (305800) = 350800 6A = 350800 - 305800 A = 7500. Since A + B = 13900: 7500 + B = 13900 B = 6400 \$7500 was initially invested into Scheme A and \$6400 was initially invested into Scheme B. At the interest 14% and 10%, Scheme A and Scheme B yielded \$3508. Sorry if it is too simple, i just wanted to make sure there was a little confusion as possible.

 Chaitanya said: (Dec 19, 2013) Is we use the formula ptr/100?

 Bikash Kumar said: (Apr 10, 2014) Let the sum invested in Scheme A be Rs. x and that in Scheme B be Rs. (13900 - x). Then, (X*14*2/100)+{(13900-X)*11*2/100} = 3508. 28x/100 + (13900*22)-(x*22)/100 = 3508. 28x/100 + 305800-22x/100 = 3508. 28x+305800-22x/100 = 3508. Now, 28x+305800-22x=3508*100. 28x-22x=350800-305800. 6x=45000. x=45000/6. x=7500. So, sum invested in Scheme B = Rs. (13900 - 7500) = Rs. 6400.

 Sameth said: (Jun 5, 2014) What is Rs. Stand for?

 Mani Smart said: (Jul 4, 2014) The formula is (pnr/100). p=inicial amount, n=no of years, r=rate of interest

 Teja said: (Sep 6, 2014) See, it is given that both a&b together combinedly having principal amount of 13,900 for two years having si of 3508. Now by adding both schemes si's with the total si given for two years by using formula si = p*t*r/100. By considering scheme b = 13,900-a, We can find investment on b scheme.

 R.Srinivasulu said: (Nov 5, 2014) Shortcut method tell me sir.

 Sou said: (Jan 29, 2015) Its a heavy. Is there any shortcut?

 Phani Kumar said: (Mar 13, 2015) Here the shortcut for this kind of problems. P1 = ((100*I) - (P*T*R2))/(difference of R1 and R2). And then find P2 by: P2 = P-P1. Here, P = 13900. P1 = Part1; P2 = Part2; I = 3508; T = 2; R1 = 14; R2 = 11;

 Ganesh said: (Mar 30, 2015) Someone's tell me how did you get 3508? Please answer me.

 Ashwini said: (May 24, 2015) How to take rate percentages? Please explain me

 Vinoth Kumar said: (Jun 1, 2015) Here 3508 is a total interest. Say 3500. 3500 is approximately 25% of 14000 (13900). So for one year it slightly more than 12.5%. So invest in A (14%) must be more. So the answer must be less than 6950(13900/2). Two opinion A are B. Now pair up. (6400/7500) or (6500/7400). Take any one pair find out their simple interest. And then add up both SI which result in 3508 is the answer.

 Bala said: (Jun 4, 2015) Any one can solve it two or three line?

 King Kartik (The King Of Dewas said: (Jun 18, 2015) Can you answer me at what rate of simple interest will some amount be doubled in 5 years?

 Mahesh said: (Jun 29, 2015) @King Kartik. Formula : RT = (n-1)100. Here: N = 2 (amount is double). T = 5yr. R? Then 5R = 100. R = 100/5. R = 20.

 Priya said: (Jul 28, 2015) Can anyone solve by taking x as a value of B? Am getting different answer if I take x for B.

 Eeswar Kumar said: (Aug 19, 2015) By using simple formula [(I*100-P*R2)/R1-R2] = P2. Here p = Total amount shared between A & B as P = P1+P2. I = Total interest; R1 = P1's rate of interest, R2 = P2's rate of interest.

 Jay said: (Aug 27, 2015) We can solve this problem this way - I = Pti when i=interest, p= actual amount, i=rate of interest, t=time.

 Lakshman said: (Sep 13, 2015) Anyone please can explain how it came 45000?

 Vinithagopal said: (Oct 1, 2015) How this 28x-22x = 350800-(13900 x 22) came ? Please somebody explain!

 Manikant Thakur said: (Oct 25, 2015) In the question it is not mention that A having x or B having x so I am confuse here I got these two answer 7500 & 6400 so which should be the right one?

 Guru said: (Mar 14, 2016) Any other easy way to solve immediately.

 Lavanya said: (Mar 21, 2016) Any other easy way to solve immediately. And wish formula use.

 Suvas said: (Apr 4, 2016) How you got 3508?

 Nagaraju said: (May 10, 2016) Here why we had taken (13900 - x)? Please, can anyone explain it in detail?

 Ashok said: (May 15, 2016) It can simply solve like this using allegation concept as; Total amount can be given in 2 types of interests as one A is @14% & B is @11%. He gets average in totally 2 years is given 3508/2 = 1754. Its mean % is 1754/13900 = 12.6%. Apply Allegation principle on interests. Then, we get; -> A : B. -> 8 : 7. Totally we have 15 parts = 13900 We require the amount invested in Scheme: B = 7/15 * 13900. => 6486 (approx).

 Chinnaribunga said: (May 24, 2016) How it came 45000? I didn't understand this, Please tell me.

 Ramesh K said: (Jun 5, 2016) Can any one one solve this in simple method?

 Sithara Prasanth said: (Jul 2, 2016) How it can do through allegation.

 Jayesh said: (Jul 3, 2016) Anyone know the Shortcut method?

 Narender said: (Jul 20, 2016) How to use it with alligation?

 Nethi Pradeep said: (Jul 21, 2016) Please explain me in simple method.

 Saurabh Chavan said: (Jul 23, 2016) Please, anybody explain it in a simple way.

 Ans Seekers said: (Jul 28, 2016) Sir, please explain in a simple way.

 Chiran said: (Aug 11, 2016) Tell me how it became (13900 - x)?

 Jayesh said: (Aug 20, 2016) Can't we solve it using alligation and mixture? Because I tried and could not get the answer.

 Loja said: (Aug 20, 2016) Can anyone explain in a simple way?

 Maheswaran said: (Aug 21, 2016) Thanks for the solution and explanations.

 Azly said: (Sep 4, 2016) We can solve this using simultaneous equations. Here we go! Total interest for 2 years - 3508. Then for a year - 3508/2 = 1754. Assume investment in A scheme as x and B as Y. 14/100 * x + 11/100 * y = 1754 --------> 1st eqn x + y = 13900 -------> 2nd eqn. Solve this. You will get the answer.

 Sumit said: (Sep 11, 2016) Please explain me. I can't understand this.

 Venu Sabbani said: (Sep 21, 2016) 13900 is then two parts 14 * 2 and 11 * 2. Then 28 and 22(scheme A and scheme B difference is 6%). 28% of 13900 = 3892. 3892 -> given amount, then3892 - 3508 = 384. Two schemes difference is 6%(28 - 22). 6% = 384. Sum 100% =? Then cross multiplication 384 * 100/6 = 6400 is the answer.

 Naveen Be Cool said: (Sep 23, 2016) A = 14 * 2 yrs. B = 11 * 2 yrs. Now you'll get 28, & 22. => 28 + 22 = 50 * 3508 * 2 = 350800. Now you need to find the amount of B = 22 * 13900 = 305800. So. 350800 - 305800 = 45000 . 28 - 22 difference is = 6. Now 45000/6 = 7500. => 13900 - 7500 = 6400. Then the value of A is 7500 and B is 6400.

 Soumya said: (Nov 18, 2016) Can anyone explain it in Alligation method?

 Bushra said: (Nov 19, 2016) What is the answer for the question? A farmer has a rectangular garden plot surrounded by 200ft of the fence. Find the length and width of the garden if its area is 2400ft2.

 Pranav said: (Nov 27, 2016) @Bushra. Area of rectangle = l * w Perimeter of rectangle = 2 * (l + w) -------------------------------------------- 2 * (l + w) = 200 l + w = 100 ---> equa(1) l * w = 2400 so, w = 2400/l ---> equa(2) Putting equa(2) in equa(1). We get, l + (2400/l) = 100 l^2 + 2400 = 100 * l l^2 - 100 * l -2400 = 0 l^2 - 60 * l - 40 * l -2400 = 0 l (l - 60) - 40 (l - 60) = 0 (l - 60) (l - 40) = 0 Hence, if l = 60 w = 40 (from equa(1)) Else, l = 40 then w = 60.

 Harsha said: (Dec 9, 2016) Can anyone explain easily?

 Paply said: (Dec 29, 2016) Anyone can solve it in easy way? Please.

 Rekha said: (Jan 2, 2017) Thank you all for giving the solution.

 Chandrasekar.S said: (Jan 8, 2017) Anyone can solve it in easy way?

 Loke said: (Jan 23, 2017) Let A amount be "x" rupees so B must be "13900-x "(because 13900 divided two members we don't know the exact amount so we consider "x" rs by A remaining amount 13900-x by B). Apply formula. PTR/100. P = PRINCIPAL T = TIME R = RATE OF INTEREST X * 14 * 2/100 + (13,900-X) * 11 * 2/100 = 3508. (x is the principal amount we don't know), (14 is rate of interest A), (2 is the year), After calculation we get 6400.

 Krishna Kant Sharma said: (Feb 3, 2017) Thanks for giving the solution. I understand it now.

 Sri said: (Mar 16, 2017) @Depu. Thanks for giving the explanation of answer.

 Madhu said: (Mar 19, 2017) @Phani Kumar Can you please solve using the formula you stated? i.e P1= {(100*1)-(PTR2)} / (R1-R2), P2=P-P1. Please anyone help me with this solution.

 Manish said: (Mar 28, 2017) Let x be the amount of B and 13900-x be the amount of A then without subtract we can find the value o B.

 Tiama said: (Apr 11, 2017) I am confused with the following stem. 28x + 305800 - 22x = 3808 * 100. Where did the other denomination (ie 100) gone?

 Karthik said: (May 3, 2017) Why the amount invested in the scheme a should be taken as x and in scheme b should be taken as 13900-x, as both amounts are part of 13900?

 Jitender said: (Jun 11, 2017) Simple interest for two year 3508, therefore simple interest for one year will be1754 x+y=13900 ---> Equation (1). 14x+11y=175400 ---> equation 2. To solve the equation we are multiplying the equation 1 by 14. 14x+14y=194600 ---> equation 1 after multiply by 14. 14x+11y=175400. After subtracting the equation we will get. 3y = 19200, Y = 6400.

 Ganesh Naik said: (Jul 1, 2017) @ Aditi. Nice explanation. they asked for scheme B so we shld take x=13900-y instead of y=13900-x.

 Tushar Waghchaure said: (Jul 4, 2017) Let's assume any answer between option, suppose 6400, and total amount is given 13900 so, 13900-6400=7500 so we got two values of principle(p), we know formula i,e SI= P*R*T/100. so put value...SI=7500*14*2/100=2100 AND SI=6400*11*2/100=1408 and addition of this is =3508. So, A invest 7500 AND B invest 6400.

 Lucy said: (Aug 12, 2017) 350800 - (13900 x 22) =45000 How its came?

 Anshul said: (Aug 19, 2017) The answer can be 6400 or 7500. It is not necessary that A will invest X and B will invest the difference.

 Amrutha said: (Oct 14, 2017) From the question we know 13900 is the amount he invested. Let assume, 'x' be the amount invested on A. so 13900-x be the amount B have SI=PNR/100. Case of A: for 2 years, as per the question. SI1=x*2*14/100 Similarly Case of B: for 2 years SI2=(13900-x)*2*11/100 its given that the total amount of simple interest is 3508 ie,SI1+SI2=3508 substituting the above two equations Then, x x 14 x 2 /100 + (13900 - x) x 11 x 2/100 = 3508 28x - 22x = 350800 - (13900 x 22) 6x = 45000 x = 7500. So, sum invested in Scheme B = Rs. (13900 - 7500) = Rs. 6400.

 Renu said: (Oct 19, 2017) @Phani Kumar. Can you plz tell me how to take a difference between r1 and r2 (14% - 11%) or (14*2-11*2).

 Surya said: (Nov 30, 2017) How 45000 will come? Please explain it.

 Pragna said: (Dec 1, 2017) @Renu. It has Some correction @Phani Kumar . Given -----> P = P1+P2 (Principle is divided between two) Sum of interests on P1 and P2 is given as I = SI1+ SI2 = 3508 S11---> Simple interest on P1 SI2 ---> Simple interest on P2 I = ( P1*T1*R1 /100) + (P2*T2*R2/100) 100*I = (P1*T1*R1) + (P2*T2*R2) But Given T1= T2=T so, 100*I /T = P1*R1 + P2*R2 = P1*R1 + (P-P1)*R2 = P1*R1 + P*R2 - P1*R2 (100*I/T) - P*R2 = P1*(R1-R2) ((100*I)-( P*T*R2))/ T = P1*(R1-R2) P1 = (100*I - P*T*R2) / T*(R1-R2) (R1-R2 = 14 -11 =3) Find P2 = P - P1

 Punna Vandana said: (Feb 8, 2018) Please, can anyone explain this problem in percentages?

 Dhilip said: (Feb 8, 2018) It is a PTR/100 formula total SI is given we find amount invested in B so subtract x from p here x is an amount invested and p total amount. Two different scheme so we add two amount and the SI is placed other side by the formula SI=P*R*T/100.

 Manish Kumar said: (Apr 22, 2018) In this question money is divided into two parts. On one part he got 14% interest and on other part he got 11%. The difference is 3%(14%-11%) in one year. If we got 14% on both then 14% of 13900 = 1946. But we got interest 3504 which is in two years so in one year 3504/2 =1754. The difference is 1946-1754=192. So we got 3% difference on 192. 3% =192. 1%=192/3=64. 100%=6400. So, one part is 6400 and other (13900-6400) = 7500.

 Sreelatha said: (May 10, 2018) C let us assume that principle invested in a is x. & b be y. The total amt invested is 13900, so x+y=13900 or y=13900-x, by using formula i=ptr/100, x*14*2/100 + y*11*2/100, = 28x/100+22y/100=3508, 28x+22y=350800, Substitution of y frm 1 28x+22(13900-x) = 350800 now solve.

 Sree said: (May 10, 2018) Good explanation. Thank you all.

 Priya said: (Jul 8, 2018) Thank you so much @Jitendar.

 Mangesh said: (Aug 4, 2018) P=13900 A=14% B=11% So,total SI= 3508. Hence we need to fine out B scheme. Use option given 1st option is 6400, 6400*2 year*11%÷100=1408, And remaining intrest is 3508-1408=2100, Now A scheme, 13900-6400=7500, 7500*2year*14%÷100=2100.

 Swati said: (Sep 5, 2018) Why it has not taken X for B's principal? Please explai me.

 Umesh Gandla said: (Sep 12, 2018) x x 14 x 2 + (13900 - x) x 11 x 2 = 3508. 100 100 28x - 22x = 350800 - (13900 x 22) 6x = 45000 x = 7500. So, sum invested in Scheme B = Rs. (13900 - 7500) = Rs. 6400.

 Vikas said: (Apr 3, 2019) step 1= p*r*t/100. step 2=A=X & B=Y, step 3= (X*14/100*2/2)+(Y*11/100*2/2)=3508/2. step 4=X+y=3508, After calculation we get x and y value.

 Hari Priyan said: (Apr 18, 2019) Assume that the principal invested in scheme A be rupees x. Then according to question the principal invested in scheme B is (13900-x) rupees. Number of years: 2 years. Rate of interest in scheme A:14%. Rate of interest in scheme B: 11%. Total interest earned : rs 3508. SI: PNR/100 formula. (x*2*14/100)+((13900-x)*2*11/100)=3508. (28x/100)+((305800-22x)/100)=3508. (28x-22x)+305800=3508*100. 6x+305800=350800, 6x=350800-305800, 6x=45000, x=7500. Now we got x ie.. principal invested in scheme A. Now we can find the principal invested in scheme B. 13900-7500= 6400. Principal invested in scheme B is rupees 6400.

 Monica Williams said: (Apr 26, 2019) 3508 *100 (÷ will become*) so that's how we got 350800 in the 2nd step?

 Anuj said: (May 18, 2019) With alligation method. B = 11%. A = 14%. Calculate Rate for 1 year on 100*1754/13900*1 = 12.61%. 1754 is one-year int. mean is 12.61%. then 14%-12.61% = 1.38%. 12.61-11 = 1.61. 1.38/3*13900 = 6400.

 B Krishna said: (May 29, 2019) How did you Assume that 7500 here, in what basis did you get that, can we get a Clear Explanation on this?

 Sravs said: (Sep 13, 2019) Let calculate SI for 14% per anum si = 13900*1*14/100 = 1946, Caluculate si for 11% per anum si = 13900 * 1 * 11/100 = 1529, Given intrerest for 2 year so 3508/2 = 1754 p.a. By using alligation: 1956-1754 = 192 1754 - 1529 = 225. Tatio;-B:A = 192 : 225. Sum of B = 13900 * 192/417 = 6400.

 Priya said: (Nov 14, 2019) How this is calculated? I didn't get please explain it.

 Ajay said: (Sep 12, 2020) By using alligation method, we get; 13900/30*14 = 6500.

 Dorji Bhai said: (Oct 3, 2020) How denominator is getting cancelled while adding? Please explain about it.

 Siva said: (Oct 5, 2020) I can't understand. If we use the formula ptr/100, how to identify 'r' in that? Please explain me.

 K. Uma said: (Oct 18, 2020) Let us assume that the scheme A be as X & scheme B be as Y. The total amount invested is 13900. So, X+Y=13900 X=13900-Y--------(1). Now, we use simple interest formula SI=PRT/100 Time T=2years Per annum R=14% & 11% Now, substitute the values in SI formula. Amount invested scheme B= [Xx14x2/100]+[Yx11x2]=3508 [28X/100] +[22Y/100]=3508 28X+22Y=3508x100 28X+22Y=350800--------------> (2) Now substitute eq(1) in eq(2). 28(13900-Y) + 22Y = 350800 389200-28Y + 22Y = 350800 - 28Y + 22Y = 350800 - 389200 - 6Y = - 38400. 6Y = 38400. Y = 38400/6. Y = 6400. Amount invested scheme B= 6400.

 Yogeswara Rao said: (Jan 12, 2021) Invested money = 13900. Divided into two schemes A and B. Let the scheme A = x. Scheme B = 13900-x. SI = PTR/100. SI = x.2.14/100, SI = 28x/100. 2nd case SI = (13900-x).2.11/100. SI = 3508, 3508 = 28x/100+ (13900-x).2.11/100, 350800= 28x +305800-22x, -45000 = -6x. X = 45000/6. X = 7500. B = 13900 - 7500. B = 6400.