Advanced Math Part -02
1
Installment Related Problem
Installment
1. A person takes a loan of Tk. 200 at 5% simple interest. He returns Tk. 100 at the end of
1 year. In order to clear his dues at the end of 2 years, he would pay: (GKRb e¨w³ 5% mij
my‡` 200 UvKv FY †bq| 1 eQi †k‡l †m 100 UvKv wdwi‡q †`q| m¤ú~Y© FY cwi‡kva Kivi Rb¨ Zv‡K 2 eQi ci Av‡iv
KZ UvKv w`‡Z n‡e?) [Aggarwal-83]
(a) Tk. 105
(b) Tk. 110
(c) Tk. 115
(d) Tk. 115.50 Ans: c
Solution: (GB AsKUv‡Z †kLvi welq n‡jv, †h cÖ_g eQi 100 UvKv w`‡q w`‡j c‡ii eQi H UvKvi my` w`‡Z n‡e bv)
Interest after 1 year = 5% of Tk. 200 = 10
Since Tk. 100 is returned in first year, So, interest in 2nd year = 5% of 100 = 5
So, total interest of 2 years = 10+5 = Tk.15 then Total amount to be paid = 200+15=215
Amount to be paid in 2nd year = Tk.(215 – 100 ) = Tk.115
[G¸‡jv me simple interest Gi installment Gi Dci cÖkœ| ZvB Pµe„w×i gZ my‡`i my` †f‡e fzj Ki‡eb bv| ]
Dc‡ii AsKwUI installment Gi As‡Ki gZB GKwU AsK| wKš‘ GLv‡b wKw¯Íi cwigvY mgvb bq|
wKw¯Íi AsK¸‡jvi gyj welqUv Ggb, Avcwb GKUv wd«R wb‡jb| hvi mvaviY gyj¨ 10000 UvKv| GLb Avcwb wKw¯Í‡Z
wb‡j cwi‡kva Ki‡Z n‡e 12000 UvKv| GLb GB AwZwi³ 2000 UvKv wKfv‡e wb‡e? 6 gv‡mi wKw¯Í n‡j cÖwZ gv‡m
120006 = 12000 UvKv Gfv‡e nq bv| (hw`I jv‡fi % Gi nv‡ii GZ RwUj wn‡me A‡b‡K eyS‡Z bv cvivi Kvi‡b
Gfv‡e mnR mij wn‡me K‡i) wKš‘ gyj welqUv wfbœ| wb‡Pi cÖkœwUi mgvav‡bi we¯ÍvwiZ e¨vL¨v co–b|
Simple Interest related:
2. What equal installment of annual payment will discharge a debt which is due as Tk. 848
at the end of 4 years at 4% per annum simple interest
A_©: 848 UvKvi GKwU FY cwi‡kva Kivi Rb¨ evwl©K 4% nv‡i 4 eQ‡ii mgcwigv‡bi 4wU evrmwiK wKw¯Íi cÖwZwUi
cwigvb KZ n‡e?
Solution: Let, the annual installment = Tk. x
ATQ, (x + x 34%) + (x + x 24%) + (x+ x14%) + x = 848 [4wU wfbœ wfbœ my`vm‡ji mgwó =848]
12x  
8x  
4x 
  x 
+ x 
 + x 
 + x = 848
100  
100  
100 

100x  12x  100x  8x  100x  4x  100x
84800

= 848  424x = 84800 x =
= 200
100
424
The annual installment = Tk.200
Ans: Tk.200
3. What annual installment will discharge a debt of Tk. 770 due in 5 years at 5% simple
interest ?(evwl©K 5% nv‡i 5 eQ‡i 770 UvKv my`vmj cwi‡kva Ki‡Z n‡j evrmwiK wKw¯Íi cwigvY KZ?) [Youtube
video]
(a) Tk. 120
(b) Tk. 150
(c) Tk. 140
(d) Tk.145
Ans: c
Solution:
Here interest rate = 5% and time = 5 years total debt = Tk. 770
Let, each installment be Tk.x (GB UvKvi mv‡_B cÖwZevi my‡`i UvKv ¸‡jv 1 eQi K‡i K‡g †hvM n‡e)

 x  4  5  
 x  3  5  
 x  2  5  
 x  1  5 
Then,  x  
   x  770
   x  
   x  
   x  
 100  
 100  
 100  
 100  

Advanced Math Part -02
2
Installment Related Problem
x 
3x  
x 
x 
6 x 23x 11x 21x

x    x 



 x  770
  x    x 
  x  770 
5 
20  
10  
20 
5
20
10
20

24 x  23x  22 x  21x  20 x
110 x
20
= Tk. 140

 770 
 770 x = 770 
20
20
110
[g~j m~Î: As‡Ki cÖ_g jvBbUv B g~j m~Î| hZ eQ‡ii K_v ejv n‡e, Zvi †_‡K 1 K‡i Kg wb‡q my` Avm‡ji mv‡_ hy³
K‡i me¸‡jv eQ‡ii my`vm‡ji †hvMdjwUB n‡e †gvU cwi‡kvwaZ UvKv †hUv due n‡e 5 eQ‡ii g‡a¨]
we¯ÍvwiZ †evSvi Rb¨ GB †UwejwU †`Lyb| me wKQz wK¬qvi n‡q hv‡e|
2010
2011
2012
2013
2014
1g wKw¯Í
2q wKw¯Í
3q wKw¯Í
4_© wKw¯Í
5g wKw¯Í
cÖwZ wKw¯Í 140+(5%4) 140+(5%3) 140+(5%2) 140+(5%1)
700
140
140
my` n‡e c‡ii
my` n‡e
my` n‡e
my` n‡e
‡k‡li wKw¯Íi w`bB
4 eQi
c‡ii 3 eQi
c‡ii 2 eQi
c‡ii 1 eQi
me FY cwi‡kva |
†Kvb my` n‡e bv|
‡kva n‡”Q 168
161
154
147
140
770
MCQ cixÿvq KL‡bvB Dc‡ii we¯ÍvwiZ wbq‡g mgvavb Kiv hv‡e bv| eis wb‡Pi wbq‡g K‡qK †m‡K‡Û mgvavb Ki‡Z n‡e|
Shortcut: r% {(n-1)(n-2)(n-3) (n-4)….} + (n100%) = Total debt (g~L¯’ bv K‡i eyS‡Z mgq w`b)
5%(4+3+2+1) = 5%10 = 50% A_©vr 5 eQ‡ii me¸‡jv wKw¯Í 100% K‡i †gvU 5100% = 500% Gi mv‡_
770  100
50% †gv‡Ui Dci my` = 500+50 = 550% = 770 n‡j 100% =
= 140 UvKv|
550
 Lye mn‡RB fzj n‡Z cv‡i †hfv‡e fve‡j: 770 UvKv FY wb‡q 5% nv‡i 5 eQ‡i †h my`vmj n‡e Zv †ei K‡i 5 w`‡q fvM
K‡i cÖwZ wKw¯Íi gvb †ei Ki‡Z †M‡j Kbwd‡W›Uwj fzj DËi †`qv n‡e| KviY GLv‡b 770 UvKv FY †bqv †evSv‡”Q bv| cÖkœwU
Avevi fv‡jvfv‡e co–b| GLv‡b 770 UvKv FY cwi‡kva n‡q hv‡e, ( A_©vr GKUv Fb †bqv n‡qwQj hv wKw¯Í‡Z ax‡i ax‡i †kva
Ki‡j †gv‡Ui Dci 770 UvKv †kva Ki‡Z n‡e| GLb Avcwb †h wKw¯Í ¸‡jv w`‡Z _vK‡eb, Zvi mv‡_I my‡`i wn‡me evi evi
hy³ n‡e| hv Dc‡ii †Uwe‡j †`Lv‡bv n‡q‡Q|
4. What annual payment will discharge a debt of Tk.6450 due in 4 years at 5% simple
interest? [careerbless.com]+ [Competoid.com]
A_©: 6450 UvKvi GKwU FY cwi‡kva Kivi Rb¨ evwl©K 4% nv‡i 4 eQ‡ii mgcwigv‡bi 4wU evrmwiK wKw¯Íi cÖwZwUi
cwigvb KZ n‡e?
Solution: Let, the annual installment = Tk. x
ATQ, (x + x  3  5%) + (x + x  2  5%) + (x + x 15%) + x = 6450
15x  
10x  
5x 

x 
Ans: Tk. 1500
 + x 
 + x 
 + x = 6450 Then, x = Tk. 1500
100  
100  
100 

5. What annual installment will discharge a debt of Tk.1092 due in 3 years at 12% simple
interest?
x  12  2  
x  12  1 

Let each installment be Tk. x. ATQ,  x 
  x 
  x  1092 x = 325
100  
100 

Advanced Math Part -02
3
Installment Related Problem
Compound Interest related:
Installment A_© n‡jv wKw¯Í, A_©vr †Kvb UvKv A‡bK †ewk mg‡qi Rb¨ FY wb‡q GKmv‡_ cwi‡kva bv K‡i av‡c
av‡c evrmwiK wKw¯Í‡Z cwi‡kva KivB n‡”Q Installment |
g‡b ivL‡eb, GKmv‡_ 2-3 eQi ci FY cwi‡kva Ki‡Z †M‡j †gvU hZUvKv Pµe„w× my`vmj w`‡Z n‡e| cÖwZ eQ‡i
wKw¯Í‡Z w`‡Z _vK‡j Zvi †_‡K Kg UvKv w`‡Z n‡e| KviY cÖ_g wKw¯Íi mv‡_ my‡`i mv‡_ Avmj I wKQzUv cwi‡kva nq|
Finding the installment amount:
Installment Gi cwigvY †ei Kivi myÎ: If the installment = x then
n 1
n 2
n
r 
r 
r 



x+x 1 
= Principal  1 
 +x 1 

 [Gfv‡e cÖwZevi 1 K‡i K‡g †gvU eQi ch©šÍ]
 100 
 100 
 100 
 GB m~ÎUv cÖvKwUKvwj †evSvi Rb¨ e¨vL¨v †evSv Lye Riæix:
Wvb cv‡k ‡gvU FY 3 eQi ci GKvmv‡_ cwi‡kva Ki‡Z †M‡j 3 eQ‡ii Pµe„w× my`vmj GKmv‡_ hv w`‡Z n‡e, evg
cv‡k 1 eQi ci †h wKw¯Í w`‡Z n‡e Zvi cwigvY x | GB x wKš‘ my`vmj| GLb Gfv‡e 3Uv x UvKvi wKw¯Í hZ UvKv Kg
my` w`‡Z n‡”Q †mB UvKvUv †hvM K‡i w`‡j Dfq cv‡ki my`vmj mgvb n‡e| ‡h‡nZz 1g eQ‡ii my` mn x UvKv cwi‡kva
Kiv n‡q‡Q ZvB c‡ii eQi¸‡jv‡Z 1 eQi K‡i Kg K‡i †gvU my` †hvM K‡i mgvb Ki‡Z n‡e| c‡ii mgvavbwU †`Lyb|
6. A sum of Tk. 1260 is borrowed from a money lender at 10% p.a compounded annually.
If the amount is to be paid in two equal annual installments, find the annual
installments? (BASIC Bank (AO) Cash-2013 (Written)
A_©: evwl©K 10% Pµe„w× - gybvdvq 1,260 UvKv avi †bqv nq Ges hw` `ywU mgvb wKw¯Í‡Z 2 eQ‡i m¤ú~Y© UvKv cwi‡kva
Ki‡Z nq Zvn‡j cÖwZ eQi †k‡l wK cwigvb UvKv cwi‡kva Ki‡Z n‡e?
Solution: Let, the amount of installment = x
1
2
10 
10 


ATQ, x+ x  1 
  1260  1 
 (cÖ_g wKw¯Í+cÖ_g wKw¯Íi Dci c‡ii eQ‡ii my` = Avm‡ji
 100 
 100 
Dci `yeQ‡ii my`vmj ]
11x
11 11
21x
11 11
11 11 10
 1260  
 1260  

 x+
x = 1260    = Tk.726
10
10 10
10
10 10
10 10 21
 GB ai‡Yi As‡Ki mgxKiY ˆZixi ev¯Íe m¤§Z e¨vL¨v:
hw` wKw¯Í‡Z bv w`‡q GKmv‡_ 2 eQ‡ii my`vmj †`qv n‡Zv Zvn‡j 1260 UvKvi 10% nv‡i 2 eQ‡ii my`vmj n‡Zv,
1260 Gi 110% Gi 110% = 1524.6 UvKv|
GLb 1 eQi ci wKw¯Í‡Z UvKv cwi‡kva Ki‡j 726 UvKvi 2wU mgvb wKw¯Í‡Z †gvU cwi‡kva Ki‡Z n‡e, 726+726 =
1452 UvKv| 1260 UvKv 1 eQi ci my‡` Avm‡j †`qvi K_v 1260 wQj 1260 Gi 110% = 1326 UvKv|
Av‡iv GKeQi A‡cÿv Ki‡j GB cy‡iv 1326UvKvi Dci my` w`‡Z n‡e| wKš‘ cÖ_g eQ‡ii wKw¯Í 726UvKv w`‡q †`qvq
UvKv Kg w`‡Z n‡e, 726 Gi 10% =72.6 UvKv| Avi GB msL¨v¸‡jvi mgš^‡qB m~Î ˆZix nq Gfv‡e,
(726+726+726 Gi 10%) = 1260 Gi 110% Gi 110% A_©v; 726+726+76.2 = 1524.6 `ycv‡k mgvb|
cÖ_‡g 726 Gi gvb Rvbv bv _vKvq 726 Gi RvqMvq x ewm‡qB Dc‡ii mgxKiY ˆZix nq|
Alternative Solution:
Let, amount of equal installment = x (G‡Z Avmj Av‡Q + my`I hy³ Av‡Q wKš‘ †Kvb eQ‡ii m¤ú~Y© my` bq)
4
Installment Related Problem
Advanced Math Part -02
ATQ, x + 110% of x = 110% of 110% of 1260 [GLv‡b, cÖ_g mgvavbUv B ïay m~Î ev` w`‡q ‡jLv]
11 11 10
11x
11 11
21x
11 11
 x
= 1260 
= 1260 

x = 1260    = Tk.726
10
10 10
10
10 10
10 10 21
 Confusion Clear: cÖPwjZ wKQz eB I I‡qemvB‡U GB cÖkœwUi DËi 762.3 †`qv _vK‡j Zv fzj| KviY
ïay Pµe„w× my‡`i mvaviY myÎ cÖ‡qvM K‡i 2 eQ‡ii my`vmj 1524.6 UvKv †ei Kivi ci Zv‡K 2 w`‡q fvM K‡i 762.3
UvKv evrmwiK wKw¯Í ej‡j fzj n‡e| †Kb? KviY Gi d‡j 1260 UvKvi cÖ_g eQi ‡k‡l my`vmj ‡_‡K cÖ_g wKw¯Í
cwi‡kva Kivi ci FY K‡g hvq ZvB c‡ii eQi my` Kg w`‡Z n‡e| wKš‘ GKmv‡_ 1260 UvKvi Dci `y eQi wn‡me
Ki‡j cÖ_g eQi †k‡l †kva n‡q hvIqv UvKvi DciI Avevi my` w`‡Z nq †hUv mwVK bq|
7. John borrowed Tk. 2,10,000 from a bank at an interest rate of 10% per annum,
compounded annually. The loan was repaid in two equal instalments, the first after one
year and the second after another year. The first instalment was interest of one year
plus part of the principal amount, while the second was the rest of the principal amount
plus due interest thereon. Then each instalment, in Tk. is?
A_©: Rb 10% Pµe„w× nv‡i 210000 UvKv e¨vsK †_‡K FY †bq hv `ywU mgvb wKw¯Í‡Z cwi‡kva‡hvM¨ | cÖ_g eQi c‡i
1g wKw¯Í Ges 2q eQi c‡i 2q wKw¯Í cÖ`vb Ki‡Z n‡e| cÖ_g wKw¯Í wQj 1 eQ‡ii my` Ges Avm‡ji wKQz Ask | wØZxq
wKw¯Í wQj Avm‡ji evwK Ask Ges Avm‡ji evwK As‡ki Ii Av‡ivwcZ my`| GKwU wKw¯Íi cvwigvY KZ?
Solution: Let, the amount of installment = x
1
2
10 
10 


ATQ, x+ x  1 
  210000  1 

 100 
 100 
11x
11 11
21x
10
 210000  
 x+

= 2100121 x = 2100121  = Tk.1,21,000
10
10 10
10
21
8. A sum of tk 8400 was taken as loan .this is to be paid in two equal annual installment.if
the rate of interest be 10% compound annually,then the value of each installment is
[selfstudy365.com]
Ans:Tk.4840
9. A sum of Tk.11000 was taken as loan. This is to be paid in two equal annual
instalments.If the rate of interest be 20% compounded annually, then the value of each
instalment is [competoid.com]
Hints: Let, the amount of installment = x [GB GKUv m~Î w`‡qB me AsK Kiv hvq, c‡ii wbqg †`Lyb]
x
x
ATQ,
+
= 11000 then, x = 7200
Ans:Tk.7200
1
2
20 
20 


1 

1 

 100 
 100 
Finding the debt: GB m~ÎwUi Wvb cv‡k gvb ewm‡q w`‡q Dc‡ii AsK¸‡jvI mgvavb Kiv hv‡e|
G‡ÿ‡Î m~Î Formula:
x
r 

1 

 100 
n 2
+
x
r 

1 

 100 
n 1
+
x
r 

1 

 100 
n
= Total loan
[ Here x= installment, r = rate of interest and n = time, Gfv‡e hZ eQi _vK‡e 1 K‡i Kg‡e ]
m~ÎwUi e¨vL¨v cÖ_g cÖkœwUi mgvavb †`‡L †evSvi †Póv Kiæb|
5
Installment Related Problem
Advanced Math Part -02
10. A sum of Tk. 1640 is borrowed from a money lender at 5% p.a compounded annually.
If the amount is to be paid in two equal annual installments, find the annual
installments.
A_©: hw` 5% Pµe„w× my‡` 1640 UvKv FY †bqv nq Ges hw` 2 eQ‡i m¤ú~Y© UvKv `ywU mgvb wKw¯Í‡Z cwi‡kva Ki‡Z nq
Zvn‡j cÖ‡Z¨K eQi KZ UvKv cwi‡kva Ki‡Z n‡e ?
Solution: Let, amount of equal installment = x
1
5 
5 


ATQ,x+x 1 
 = 1640 1 

 100 
 100 
2
2
21
21x
21 21
41x
21 21
 21 


= 1640    x 
= 1640

= 1640
20
20
20 20
20
20 20
 20 
21 21 20


 x = 1640
= Tk. 882 So, the amount of installment = Tk.882 Ans: Tk.882
20 20 41
 x+x
11. A sum of money is borrowed and paid back in two annual installments of TK.882 each
allowing 5% compound interest . The sum borrowed was : [Janata Bank (EO)-2017
(Morning)]
A_©: 5% Pµe„w× nv‡i wKQz UvKv FY †bqvi cÖwZwU 882 UvKvi K‡i `ywU evrmwiK wKw¯Í‡Z cwi‡kva Kiv n‡j|
KZUvKv FY †bqv n‡qwQj?
x
x
x
Solution: We know that
r 

1 

 100 
882
n 2
+
n 1
+
n
= Total loan
r 
r 


1 

1 

 100 
 100 
882
882
882
20
20 20
+
=
+
= 882 
+ 882  
= 840+800 = 1640
Here,
1
2
1
2
21
21 21
5 
5 


 21 
 21 
1 

1 

 
 
 100 
 100 
 20 
 20 
evsjvq e¨vL¨v: awi, cÖ_g eQi †k‡l cÖ`Ë wKw¯Í 882 †Z Avm‡ji cwigvb x |
100
cÖkœg‡Z, x Gi 105% = 882 ev x = 882
= 840 UvKv|
105
Avevi, wØZxq eQ‡ii cÖ`Ë wKw¯Í‡Z Avmj x ai‡j,
100 100
20 20
cÖkœg‡Z, x Gi 105% Gi 105% = 882 ev, x = 882
= 882  = 800

105 105
21 21
Ans:1640
myZivs me©‡gvU gyjab: = 840+800 = 1640UvKv|
GLv‡b GB GKevi 105% Ges Av‡iKevi 2evi 105% †jLvi welqUv B myÎ †jLvi mgq 1 eQi, 2eQi wn‡m‡e ‡jLv nq|
882  100
882
882 882  100
‡hgb: 105% = 882 n‡j 100% =
m~‡ÎI GKB
=
=
1
105
105
105
5 

1 

100
 100 
6
Installment Related Problem
Advanced Math Part -02
12. What annual payment will discharge a debt of Rs 1025 due in 2 years at the rate of 5%
compound interest? [careerbless.com]
A_©: 1025 UvKvq evwl©K 5% Pµe„w× gybvdvq 2 eQ‡ii Rb¨ evrmwiK KZ UvKv K‡i cwi‡kva Ki‡Z n‡e?

5 

100 
2 1
2
5 
= 1025  1 

 100 

21 21
21 21
21 21 20
21x
41x
x+
= 1025


= 1025 

x= 1025


= 551.25
20
20 20
20 20
20 20 41
20
Solution: Annual payment be = Tk. x ATQ, x+x 1 
13. A TV set is available for Tk. 19650 cash payment or for Tk. 3100 cash down payment
and three equal annual installments. If the shopkeeper charges interest at the rate of
10% per annum compounded annually. Calculate the amount of each installment.
[doubtnut.com]
A_©: GKwU wUwf †m‡Ui bM` g~j¨ 19650 UvKv, A_ev ïiæ‡Z 3100 UvKv cwi‡kva K‡i evKx UvKv evrmwiK wZbwU
mgvb wKw¯Í‡Z cwi‡kva Kiv hvq| hw` ‡`vKvb`vi evwl©K 10% nv‡i my` PvR© K‡i| Zvn‡j cÖ‡Z¨K wKw¯Íi cwigvY †ei
Kiæb|
Solution: Due after giving down payment Tk. 3100 = Tk. (19650-3100) = Tk. 16550
Let, the value of each installment = Tk. x
x
x
x
x
x
x
ATQ,
+
+
=16550
+
+
= 16550
3
1
2
11
121
1331
10 
10  
10 


1 
 1 

1 

10 100 1000
 100   100 
 100 
1210 x  1100 x  1000 x
= 16550
1331
16550  1331
 3310x = 16550  1331  x =
= 6655Each installment =Tk. 6655 (Ans)
3310
14. A man buys a scooter on making a cash down payment of Tk. 16224 and promises to
pay two more yearly installment of equivalent amount in next two years. If the rate of
interest is 4% per annum compounded yearly, the cash value of the scooter is:

10 x 100 x 1000 x
+
+
= 16550
11
121
1331

[competoid.com]
A_©: GK e¨vw³ GKwU ¯‹zUv‡ii g~j¨ eve` bM` 16224 UvKv cÖ`vb Ges evwK UvKv cieZx© 2 eQ‡i 2wU mgvb evwl©K
wKw¯Í‡Z cwi‡kva k‡Z© µq K‡i| evwl©K 4% Pµe„w× my‡` ¯‹zUviwUi cÖKZ
… g~j¨ KZ?
Solution:
16224
16224
16224 16224
Here,
+
=
+
=
1
2
2
26
4 
4 
 26 


 
1 

1 

25
 25 
 100 
 100 
25
25 25
+ 16224

= 15600+15000 = 30600
26
26 26
So, cash value of the scooter is = Tk. ( 16224+30600) = Tk. 46824
=16224
Ans: Tk. 46824