NPTEL COURSE ON

MATHEMATICS IN INDIA:FROM VEDIC PERIOD TO MODERN TIMES

Lecture 25

Gan. itakaumudı of Narayan. a Pan.d. ita 1

M. S. SriramUniversity of Madras, Chennai.

Outline

◮ Importance of Gan. itakaumudı

◮ Solutions of quadratic equations◮ Double equations of second and higher degree - rational

solutions◮ Determinations pertaining to the mixture of things◮ Interest calculations - payment in installments

Narayan. a Pan. dita’s Gan. itakaumudı

Gan. itakaumudı was composed in 1356 CE by Narayan. a Pan.d. ita

as indicated in the final verses of the work. It is not clear wherehe was born, or where he flourished. It was published byPadmakar Dvivedi in two volumes, based on a singlemanuscript, which belonged to his late father, the legendarySudhakar Dvivedi, in 1930’s. There is another work of Narayan. aPan. d. ita entitled ‘Bıjagan. itavatam. sa’. Only the first portion of thishas been published, based on a single and incompletemanuscript at Benares.

Gan. itakaumudı has been translated with explanatory notes bythe late Paramanand Singh of Bihar. The translation and notesare published in Volumes 20-24 of Gan. ita Bharatı, an Indianjournal devoted to history of mathematics, during 1998-2001.

Post Bhaskara Indian Mathematics

After Bhaskara-II, there were two major developments in Indianmathematics: (i)Narayana Pan.d. ita’s Gan. itakaumudı and (ii)Kerala school of mathematics and astronomy, mainly during14th − 17th centuries, wherein calculus concepts were developed.

Narayan. a Pan.d. ita carries forward the tradition substantially.There are more formulae, generalisations of earlier results, andsystematisation. There is a big leap in the treatment ofcombinatorics and magic squares in chapters 13 and 14, whichwill be dealt with separately. It was meant to a comprehensivetext, covering most of the prevalent mathematical knowledge inIndia at the time of its composition, and making substantialadditions to it.

Contents of Gan. itakaumudı

I will give a brief summary of the 14 chapters below.

◮ Chapter 1 is on measures of weight, length, area, volume,capacity etc., 8 operations namely, addition, subtraction,multiplication, division, square, square root, cube and cuberoot described. Solutions of more complex (compared toearlier works) linear and quadratic equations are discussed.

◮ Chapter 2 is on ‘Vyavahara gan. ita’ or ‘mathematicspertaining to daily life’. Calculation pertaining to mixture ofmaterials, interest on a principal, payment in instalments,mixing gold objects with different purities and otherproblems pertaining to linear indeterminate equations formany unknowns are considered. There is substantialprogress compared to earlier treatments. Example: Interestcalculations where payment by instalments are considered.

Contents of Gan. itakaumudı contd.

◮ Chapter 3 on arithmetic and geometric progressions, wherethe concept of ’sum of sums’ is generalised to the ‘k th sum’of the first n integers for arbitrary k . This is a very importantgeneralisation, which is needed for finding the infinite seriesfor sine and cosine functions, among other things.

◮ Chapter 4 is on plane geometry where properties oftriangles, quadrangles, circles, cyclic quadrilaterals etc., areconsidered. The ‘third diagonal’ associated with a cyclicquadrilateral and the area and circumference of the sameare discussed.

◮ Chapter 5 to 8 on three dimensional geometry, where rulesfor solid figures and capacities of excavated volumes arepresented.

◮ Chapter 9 is on the Kut.t.aka procedure for solving linearindeterminate equations of the first degree.

Contents of Gan. itakaumudı contd.

◮ Chapter 10 is on quadratic indeterminate equations or‘Vargaprakr. ti’. Here a variant of the ‘Cakravala’ procedure ofJayadeva and Bhaskara is also considered.

◮ Chapter 11 and 12 is on the divisors of a number andfractions.

◮ Chapter 13 is on ‘ankapasa or combinatorics, partitions ofnumbers, sequences of binomial and polynomialcoefficients, and various ‘Meru’s.

◮ Chapter 14 is on Magic squares.I take up some important mathematical results ofGan. itakaumudı, pertaining to chapters 1-12 in what follows.

Arithmetical operations and sankraman. a

Chapter 1

After discussing the eight operations, various simplifyingmanipulations involving fractions are discussed. In the earliertexts, ‘sankraman. a’ (or concurrence) problems involved solving

for x and y ; given x + y = a and x − y = b, =⇒ x =a + b

2,

y =a − b

2. In Gan. itakaumudı, problems which can be reduced

to sankraman. a are discussed.

For example. Rule 31: Given x2 − y2 = a, x − y = b, we find

x + y =x2 − y2

x − y=

ab

and x , y can now be solved as we know

x − y and x + y . Similarly, given x2 − y2 = a, x + y = b,

x − y =ab

and solution by sankraman. a.

sankraman. a

Rule 33.va:gRa:sa:ma.a:sa.a:d, ;�a.dõ :gua:Na.a:t,a A:nta.=;va:ga.eRa:a.na:ta.a:t,a :pa:dM ya.ea:gaH Á Á 33 Á Á“The square-root of the difference between ‘twice the sum ofsquares (of two numbers’ and ) the ‘square of (their) difference’ istheir sum.”

If x2 + y2 = a, x − y = b, x + y =√

2a − b2 and solution by sankraman. a.

Rule 34. Given x2 − y2 = a, x2y2 = b, x2 + y2 = [(x2 − y2)2 + 4x2y2]1/2 = c,say. From a and c, x2 and y2 are obtained, and then x and y .

Rule 35. Given x − y = b, xy = c, then x + y = (b2 + 4c)1/2 and solution forx , y by sankraman. a. Similarly, given x + y , xy = c, x − y = (a2 − 4c)1/2 andsolution by sankraman. a.

Rule 37(a). If x + y and x2 + y2 are known, x − y = [2(x2 + y2) −(x + y)2]1/2,

and solution by sankraman. a.

Solution of the quadratic equationsRule 39-40 (a)..�+pa.ea:tTa:&+ta:pa:d.a:g{ea .~ya.a:ta.a:ma:nta.=;va:Da.Ea ta:ta:~ta.a:Bya:m,a Á:pra.a:gva:d, ya.ea:gaH .sa.a:DyaH .~ya.a:ta.Ma .sa:ñÍö�ÅÅ*:" ;a:ma:ta.ea .=:a:Z�a.a Á Á 39 Á Á[a:ya:gea mUa:le :Y:na:�pMa ta:tkx +.t�a.a .=:a:�a.ZaH Á 40 Á

“ ‘Pada’ and ‘agra’ (dr. sya) divided by ‘rupottha’ happens to be the(so called) difference and the (so called) ‘product’ respectively.Calculate the (so called) ‘sum’ from them by the method statedearlier. The numbers are (obtained from them) by the method ofconcurrence. (In case) Pada is negative, the greater number is tobe taken. The square of that is the number.”

The rule gives the solution of a quadratic equation of the type ax + b√

x = c. ais the ‘rupottha’, b is the ‘Pada’ and c is the ‘dr. sya’, also ‘agra’. Find b

a and ca .

The root of the equation is√

x . Consider an auxiliary quantity√y , when b is negative. b = −|b|. In this case,

√x −√

y =|b|a,√

x√

y =ca

Quadratic equation

Then, from rule 35,√

x +√

y =(b2 + 4ac)1/2

a√

x =|b|+

√b2 + 4ac2

=−b +

√b2 + 4ac2

In this case,√

x is the greater quantity out of√

x and√

y . When b is positive, itis implied that

√y −

√x =

ba,√

x√

y =ca

and√

x =−b +

√b2 + 4ac2

(In this case√

x is the smaller quantity out of√

x and√

y .)

[In either case,√

x is the correct solution using ‘modern’ standard way. Here c

comes in the RHS, so we get, b2 + 4ac inside the square root, instead of

b2 − 4ac. It appears that Narayan. a is giving only the positive root, so only one

root is presented.]

Example

Example 27.A.a:k+.NyRa ;Dva:a.na:ma:�a.dÒ ;mUa:Á:éÈåîÅéé:aR ;�a.Za: a.Ka:na.eaY:b.d.a:na.Ma .~å.Pu ;=;�a.dõ :dùÅ;au :ta.Mavxa:nd.a:Da.Ra:�a�a:l+va.Ea ta:d:nta.=;. a:tua:Ba.Ra:gEa:�/�a.~:�a:�a.BaH .sMa:yua:ta.Ea ÁA:Dya:DERa:k+.pa:d.a:a.Da:k+Ea na:nxa:ta:tuaH :pr�a.a:tya.aYY:�a:vxa:nd.Ea .sa:Kea.$a.a:tMa ta.�a Za:ta.�a:yMa :pra:va:d mea ta:tpUa:vRa:vxa:ntea k+.a.ta Á Á 27 Á Á“A group of peacocks along with its half, its one-thirdand 3

4 of their difference (i.e, difference between its halfand its one-third) together with 11

2 times the square root(of the group) standing on top of a mountain, hearingthe thunder of clouds surcharged with electricity, aredancing with joy. The peacocks are 300 in all. O friend,tell me (the number of peacocks) in the earlier group.”

Solution

Let x be the number of Peacocks in the ‘earlier’ group. Then

x +12

x +13

x +34

(

12− 1

3

)

x +32

√x = 300

or4724

x +32

√x = 300

or4712

x + 3√

x = 600

Hence

√x =

−3 +√

9 + 4 · 4712 · 600

2 · 4712

=−3 + 97

47· 6 = 12

∴ x = 144

This is the number of peacocks in the earlier group.

Quadratic equations involving successive remainders

Rule 40 b.o+�+.a.na.$a:�a.va:a.Da:va:d:ntya.a:. Ce +Sa:�a.va:Da.Ea .$a.a:ya:tea .=:a:�a.ZaH Á 40 b,a Á“Starting from the end, by the method stated earlier, theprocess (on being performed on) the remainders, givesthe number.”

This involves solving equations of the type:

(x − a√

x)− b(√

x − a√

x)

= c

and more complicated equations. This is solved by puttingx − a

√x = y . Then y − b

√y = c. Then the problem is solved for

y first and substituting this value for y in the equation for x , x issolved.

This is best illustrated by an example.

Example

Example 40.k+a:nta.a:ya.aH .sua.=;ta:pra:sa:ñÍçÅÅ*:+sa:ma:yea ;�a.Ba:�a.a .. a mua:�+a:va:l� ;amua:�+a:na.Ma .. a :pa:d:dõ :yMa ;�a.va:. a.=;NMa Za:yya.a:pa:f:~ya.ea:pa:�a= Áta:. Ce +Sa:~ya :pa:dM ;�a�a:Ba.a:ga:yua:ga:le +na.aYY:Q.�aM ;�a.pra:yea:Na.aYY:&+tMata:. Ce +Sa:~ya :pa:dM ;�a.[a:ta.Ea ;a.na:pa:a.ta:tMa .sUa.�ea dõ :yMa ;�a.kM va:d Á Á“During the amorous tussle, (the beloved’s) garland ofpearls was broken twice the square root less 1

4 (of theroot) of the pearls were on the cover of the bed. Thesquare root of the rest along with 2

3 (of this root) wereseized by the lover. The square root of the rest fell downon the earth and 2 (pearls) were in the string (of thegarland). Tell, how many pearls were in the garland.”

Solution

Solution: Let x be the number of pearls in the garland.

Cover of the bed: =

(

2 −14

)√x =

74

√x = y

So remaining = x − y

Seized by the lover :√

x − y +23√

x − y =53√

x − y = z

So remaining = x − y − z = z ′

On the earth =√

z ′ =√

x − y − z

Remaining (on the string) = 2.

∴

√z ′ + y + z + 2 = x

√z ′ + 2 = x − y − z = z ′

Solution contd.This is solved first:

√z′ =

1 +√

1 + 8

2= 2. ∴ z′ = 4. Now, x − y − z = z′, and

z =5

3

√x − y.

∴ x − y − 5

3

√x − y = 4

Solving for this√

x − y ,

√x − y =

5

3+

√25

9+ 16

2= 3

∴ x − y = 9

Now y =7

4

√x

∴ x − y = x − 7

4

√x = 9.

Solving this,

√x =

7

4+

√49

16+ 36

2=

7

4+

√625

42

=7 + 25

8= 4.

∴ x = 16.

This is the number of pearls originally in the garland.

Remainder type equation again

Rule 41 a..�+pa.ea:tTaÈåî ÁÁ*+.a:dùÅ;a.a:g{Ma ya.ea.$ya.a:ntya.a:g{ea ;�a.va:a.DaH :pra.a:gva:t,a Á Á 41:A Á Á“Multiply the “rupottha” by the first remainder and add the product tothe last remainder. (After that solve the problem) by the methodstated earlier.”

Here author is referring to an equation of the form:

x−a− b

f(x−a)− c

g

{

x − a − b

f(x − a)

}

− e

h

[

x − a − b

f(x − a)− c

g{ }

]

−j√

x = d

We find

x(

1 − b

f

)(

1 − c

g

)

· · · − j√

x = a(

1 − b

f

)(

1 − c

g

)

· · ·(

1 − e

h

)

︸ ︷︷ ︸

Rupottha

+ d

a → first remainder, d → last remainder.

Example

Example 41.ga:Nea:ZMa :pa:�íåÅ+ae :na ;�a�a:na:ya:na:h:�a=;b.ra:Ǒ:�a.d:na:pa.a:n,a;�a.va:l;ea:mEaH Zea:Sa.Ma:ZEaH ;�a.va:Sa:ya:l+va:pUa:vERa:(ãÉa k+.ma:l;a:m,a Á:pa:de :na.aYY:pUa.$yEa:ke +.na .. a gua.�+.pa:d.a:}Ba.ea.$a:yua:ga:lM.sa.=:ea.$ea:na.aYY:. a:[va dÒu ;ta:ma: a.Ka:l+ma:}Ba.ea.$a:a.na:. a:ya:m,a Á Á 41:A Á Á“Gan. esa (was worshipped) with 1 lotus flower. Siva,Hari, Brahma (and) the sun (were worshipped) with15,14,

13

and12

of what remains successively. Kamala

was worshipped with the square root of the lotusflowers (in the beginning) and the two feet of theteacher resembling lotus flowers (were worshipped)with 1 lotus flower. Quickly say the number of lotusflowers in the collection.”

Solution

Here let x be the number of lotus flowers.

Gan. esa → 1

After worshipping Gan. esa → R1 : x − 1.

Siva → 15(x − 1)

After worshipping Gan. esa, Siva → R2 : (x − 1)− 15(x − 1)

Hari → 14

{

(x − 1)− 15(x − 1)

}

Solution contd.

After worshipping Ganesa, Siva, Hari, → R3 : (x−1)−1

5(x−1)−1

4

{

x − 1 − 1

5(x − 1)

}

Brahma → 1

3

[

x − 1 − 1

5(x − 1)− 1

4

{

x − 1 − 1

5(x − 1)

}]

After worshipping Gan. esa, Siva, Hari, and Brahma R4 :

(x − 1)− 1

5(x − 1)− 1

4

{

(x − 1)− 1

5(x − 1)

}

− 1

3[x − 1 − · · · ]

Sun → 1

2R4

After worshipping Gan. esa, Siva, Hari, Brahma and sun → R4 − 12 R4.

Kamala was worshipped with →√

x

Remaining Last remainder = 1

Finding the Solution

∴ (x − 1)− 1

5(x − 1)− 1

4

{

(x − 1)− 1

5(x − 1)

}

− 1

3[ ] − 1

2· · · −

√x = 1

or (x − 1)(

1 − 15

)(

1 − 14

)(

1 − 13

)(

1 − 12

)

−√

x = 1

or x(

1 −1

5

)(

1 −1

4

)(

1 −1

3

)(

1 −1

2

)

︸ ︷︷ ︸

Rupottha=1

5

−√

x = 1 ·

(

1 −1

5

)(

1 −1

4

)(

1 −1

3

)(

1 −1

2

)

︸ ︷︷ ︸

Rupottha=1

5

+1

∴1

5x −

√x =

1

5+ 1 =

6

5

∴

√x =

1 +

√

1 + 4 · 6

5· 1

5

2 · 1

5

=1 +

√25 + 24

5

2 · 1

5

=5 + 7

2= 6

∴ x = 36

So, there were 36 lotus flowers in the original collection.

Another type of quadratic equation

Consider a quadratic equation of the form

x −[

x(m/n)

− h]2

= d .

One can easily check that this reduces to x2 − bx + c = 0,

where b =(

mn

)2+ 2

(

mn

)

h and c =(

mn

)2(h2 + d)

Then, Sum of the roots x + y = b,

Product of the roots xy = c,

Then we obtain the solutions x , y by sankraman. a with

x − y =√

(x + y)2 − 4xy =√

b2 − 4ac,

so that

x =b +

√b2 − 4c2

, y =b −

√b2 − 4c2

Narayan. a callsmn

as the rupahara, h the subtractive, and d , the dr. sya or the

observed quantity and states the solution thus:

Rule

Rule 42 (b) - 43:.~va.Ma:Za:k+.Ba:�+.ZCe +d.ea .�+pa:h.=:a:K.ya.ea ;�a.dõ :a.naéÈåî Åéé:a:h� .a:na:h:taH Á Á 42 b,a Á Á.�+pa:h.=;va:gRa:yua:�+Ea ya.ea:gaH .~ya.a:d, h� .a:na:va:gRa:yua:ta:dx :ZyaH Á.�+pa:h.=;va:gRa:gua:�a.Na:ta.ea ;Ga.a:ta.ea .=:a:�a.Za:�a.dõR :Da.a :pra.a:gva:t,a Á Á 43 Á Á“The divider in one’s own part is called Rupahara. This multiplied bytwice the subtractive is added to the square of rupahara. (Theresult) happens to be the sum. The square of the subtractive isadded to the observed quantity (i.e., added to dr. sya). The (sum)multiplied by the square of rupahara is the product of two numbers.(The numbers should be obtained) by the method stated earlier.”

So, he is stating that,

Sum = x + y = b =(m

n

)2+ 2

(mn

)

h and Product = xy =(m

n

)2(h2 + d).

[It appears that Narayan. a considers both the roots when both are positive.]

ExampleyUa:Ta.a:t,a ;�a.vMa:Za.Ma:Za:k+.~yEa:k+.va:�a$Ra:ta:~ya kx +.a.taH .sa:Kea Á:pra:ya.a:ta.a ma.a:na:sMa hM :sa.aH KMa :pa.úãÁ*.a.ea:na:Za:tMa k+.a.ta Á Á 45 Á Á“The square of

120

of a swarm of swans less 1 flew away in sky

towards Manasarovara and 95 (remained there). O friends, (tell)how many (of them) were there?”

If the number of swans is x ,

x −( x

20− 1

)2= 95

So heremn

= 20,h = 1, d = 95.

b =(m

n

)2+ 2

(mn

)

h = 202 + 2 · 20 · 1 = 440

c =(m

n

)2[h2 + d ] = 202(1 + 95) = 202 · 96

Example

Now x + y = b, xy = c,

∴ x =b +

√b2 − 4c2

, y =b −

√b2 − 4c2

We find b2 − 4c = 4 × 202 × 25 ∴

√b2 − 4c = 200.

∴ x =440 + 200

2= 320, y =

440 − 2002

= 120.

Both are solutions.

“Double equations of second and higher order degreeequations: Rational solutions”

Rules 45-57 give rational solutions of some second and higher degree equationsinvolving two unknown quantities, x, y . In the following, m, n, p, u, v , a, b, etc., areintegers.

Rule 45. A solution of x2 + y2 + 1 = u2, x2 − y2 + 1 = v2 is x =m2

2, y = m.

Rule 46. A solution of x2 ± y2 − 1 = u2 or v2 is x =m4

2+ 1, y = m3.

Rule 47 A solution of x + y = u2, x − y = v2, xy + 1 = w2 isx = 2(m4 + m2), y = 2(m4 − m2).

Rule 48. A solution of x + y = u2, x − y = v2 is x = m2 + n2, y = 2mn.

Rule 49. A solution of x + y = u2, x − y = v2, xy = w3 is

x =(m2 + n2)pb

[2mn(m2 + n2)]2, y =

2mnpb

[2mn(m2 + n2)]2

Rule 50. A solution of x2 + y2 = u3, x3 + y3 = v2 is x =mb

25, y =

2mb

25.

(

& u =m4

5and v =

3m3

25

)

Double equations, contd.

Rule 51. A solution of ax + 1 = u2, bx + 1 = v2 is x =8(a + b)(a − b)2

Rule 52. x =

[

12

(

a + bm

− m)]2

+ b is a solution of x + a = u2, x − b = v2

[

with u = 12

(

a+bm

)

+ 12 m, v = 1

2

(

a+bm − 1

2 m)]

Rule 53. A solution of x + a = u2, x + b = v2 is x =

(

a + b + 12

)2

− a.

Rule 54. A solution of x − a = u2, x − b = v2 is x =

(

a − b − 12

)2

− a.

Rule 55. A solution of x2 + xy + y2 = u2 is x =12

(

3m2

4n− n − m

)

, y = m

[with u =3m2

8n+

12

n].

Rule 56. A solution of x + y = xy is x =m + n

m, y =

m + nn

.

Rule 57. A solution of x + y = x2 + y2 is x =m(m + n)(m2 + n2)

, y =n(m + n)(m2 + n2)

.

Rule

Rule 58.;Ga:na:yua:a.ta:Ba:�e kx +.a.ta:yua:a.ta yua:a.ta:kx +.a.ta:Ga.a:ta.a:h:tea ;��a.tva:�e Á;Ga:na:yua:a.ta:yua:a.ta:Ga:na:tua:�ya.a kx +.a.ta:yua:a.ta:yua:a.ta:kx +.a.ta:va:Da.Ma .=:a:Zya.eaH Á Á 58 Á Á“Divide the sum of squares, square of the sum and theproduct of any two assumed numbers by the sum of (their)cubes and the cube of (their) sum and (then) multiply by thetwo numbers. The (results will) be the two numbers, the sumof (whole) cubes and the cube (of whose) sum will be equalto the sum of their squares, the square of (their) sum (andtheir) product.”

Rational solution for Rule: 58

Let m and n be any two assumed numbers. Then according to the rule,solution of

1) x3 + y3 = x2 + y2 is x = (m2+n2)mm3+n3 , y = (m2+n2)n

m3+n3

2) x3 + y3 = (x + y)2 is x = (m+n)2mm3+n3 , y = (m+n)2n

m3+n3

3) x3 + y3 = xy is x = m2+nm3+n3 , y = m+n2

m3+n3

4) x3 + y3 = x2 + y2 is x = (m2+n2)m(m+n)3 , y = (m2+n2)n

(m+n)3

5) (x + y)3 = (x + y)2 is x = (m+n)2m(m+n)3 , y = (m+n)2n

(m+n)3

6) (x + y)3 = xy is x = m2n(m+n)3 , y = mn2

(m+n)3

Narayan.a considers the ‘Rule of inversions’ (Remember Lılavatı) inRule 59, and Rules of 3, 5, 7, 9, 11 and inverse rules in Rule 60onwards. Barter of commodities and the sale of living beings etc., arealso considered.

Chapter 2 on determination pertaining to mixture ofthings

Rule 1 (a).:pra:[ea:pa.a:~ta:dùÅ;au :a.ta:&+ta:�a.ma:(rea:Na h:ta.aH :pxa:Ta:k, :P+l;a:a.na .~yuaH Á Á 1a Á Á“The contributions severally divided by their sum (and)multiplied by the mixed amount happen to be (the respective)fruits.”

This is understood with the solution for the following problem:

Example

Example 3..sa:Kea .. a:tua:Na. a va:�a.Na.$a.Ma kÒ +.mea:Na :pa.úãÁ*.a.a:�a.d:k+a.(õ;Ea:�a.dR :va:sEaH :Sa:d.(õ;EaH Á.sMa:. a.a:�a=;tEaH [ea.�a:Da:nMa :pa:Na.a:na.Ma .sa:h:~åò:a:mea:kM va:d ;�a.kM :pxa:Ta:k, .~ya.a:t,a Á Á“Four traders obtain 1000 pan. as by tilling a land for 6etc., days with 5 etc., horses in order. O friend, tell whatare (their shares) respectively.”

Solution

Solution:

ContributionTrader 1: 6 days 5 horses → 6 × 5 = 30

2: 7 days 6 horses → 7 × 6 = 423: 8 days 7 horses → 8 × 7 = 564: 9 days 8 horses → 9 × 8 = 72

Sum = 200

So shares:

Trader 1:30

200× 1000 = 150; Trader 2:

42200

× 1000 = 210.

Trader 3:56200

× 1000 = 280;Trader 4:72200

× 1000 = 360.

Use of rule of proportion

In rules 1(b)-3(a), rule of proportion is used cleverly to solveproblems. Suppose there is one item and x quantity ofsomething is purchased at r quantity / coin and sold at r

′

quantity / coin.

∴ Principal =xr=

xrr ′

· r ′

. Sold: Principal + Profit = Amount A =xr ′

=xrr ′

· r .

Profit I = A − P =xr ′

− xr=

x(r − r ′)rr ′

∴ Principal, P : Amount, A: Profit, I :: r ′ : r : (r − r ′)

That is why, the rule which is applicable when several items arethere, also states:

Rule

Rule 1(b)-2:;�a.ma:(rMa Ba:vea:t,a kÒ +.ya.a:GRaH kÒ +.ya:(ãÉa ;�a.ma:(rMa tua ;�a.va:kÒ +.ya.ea mUa:l+m,a Á Á 1 Á ÁmUa:lM .. a ;�a.va:kÒ +.ya.a:Ga.eRa ;�a.va:kÒ +.ya:h� .a:naH kÒ +.ya.ea l;a:BaH Á¼ea:ya:ma:nua:pa.a:ta:�a.va:a.Da:na.a ya:dùÅ;a:d:�a.va:�a.d:tMa :P+lM ta:t,a ta:t,a Á Á 2 Á Á“The rate of purchase is the mixed (amount) and thepurchase-price is also the mixed amount. Thesale-price is the principal and the rate of sale is also theprincipal. The sale-price less the purchase-price is theprofit. The (unknown) fruit should be obtained from theknowns by the process of proportion (and from themixed amount) and the method of partnership statedearlier.”

Mixed quantities and rule of proportionIt is written somewhat confusingly, but refers to the proportionality among P,A, and Iwhich has been stated by us. Now, if different types of objects are involved. Let us say:

x1 amount of 1 is purchased at the rate of r1 and sold at rate r ′1.

x2 amount of 2 is purchased at the rate of r2 and sold at rate r ′2, etc.,

Then Principals of 1, 2, 3, · · · are: Pi are:x1

r1r ′1· r ′1,

x2

r2r ′2· r ′2, · · ·

Amounts (Principal + Profit) odf 1, 2, 3, · · ·Ai are:x1

r1r ′1· r1,

x

r2r ′2· r2, · · ·

Profits = Ai − Pi :x1

r1r ′1(r1 − r ′1),

x2

r2r ′2(r2 − r ′2), · · ·

For different items,

Principal

Amount=

Principal

Principal + Profitare in ratio :

r ′1r1

:r ′2r2, · · ·

Principal

Profitare in ratio :

r ′1r1 − r ′1

:r ′2

r2 − r ′2:

r ′3r3 − r ′3

· · ·

Principal

Principal -Profitare in ratio :

r ′12r ′1 − r1

:r ′2

2r ′2 − r2: · · ·

Example

Example 5.Za.a:a.l+ga.ea:DUa:ma:ku +.�ma.a:Sa:Ka.a:yRaH .sa:Kea .=:a:ma:ba.a:Na.aY:�a.dÒ ;sMa:K.ya.aH kÒ +.ya.aH .sa:�a:a.taH Á.�+pa:h� .a:na.a ;Da:nMa ;�a.va:kÒ +.ya.a BUa:k+=:a:çÉîå+;a�ua:�/////�a.n}å.a:ta.a:~tua:�ya:l;a:BMa ;Da:nMa ;�a.kM :pxa:Ta:k, Á Ál;a:Ba:yua:�+a:a.na tua:�ya.a:a.na ;�a.va.�a.a:a.na va.a l;a:Ba:h� .a:na.a:a.na va.a .~yuaH k+.TMa b.rUa:�a.h mea Á“Sali rice, wheat and a half-ripe pulse were purchased in proportionof 3, 5, and 7 (per coin) for an amount of 69 (and then) sold in theproportion of 1, 2, and 3 (per coin). If the profits be equal or theprofit added. or subtracted from the principals be equal, tell theprincipal separately.”

SolutionNow, Principal

Profit are in the ratio :r ′1

r1−r ′1:

r ′2r2−r ′2

:r ′3

r3−r ′3(a) If the profits are equal,

Principals are themselves are in this ratio. Here r1 = 3, r2 = 5, r3 = 7,r ′1 = 1, r ′2 = 2, r ′3 = 3.

So, here: Principal of 1 : 2 : 3 ::r ′1

r1−r ′1:

r ′2r2−r ′2

:r ′3

r3−r ′3

=1

3 − 1:

2

5 − 2:

3

7 − 3=

1

2:

2

3:

3

4

Sum =6 + 8 + 9

12=

23

12

Total Principal = 69.

∴ Principal 1 =6923

12

× 1

2= 18

Principal 2 =6923

12

× 2

3= 24

Principal 3 =6923

12

× 3

4= 27.

Solution contd.

(b) Now if Principal + Profit = Amount are equal,

Principal 1 : 2 : 3 ::r ′1r1

:r ′2r2

:r ′3r3

=13:

25:

37

Sum =35 + 42 + 45

105=

122105

Total Principal = 69

∴Total Principal

Sum=

69122

× 105

Principal 1 =69122

× 105 ×13=

69122

× 35

Principal 2 =69122

× 105 ×25=

69122

× 42

Principal 3 =69122

× 105 × 37=

69122

× 45

Solution contd.

(c) If Principal − Profit are equal. Then

Principal 1 : 2 : 3 ::r ′1

r1 − 2r ′1:

r ′2r2 − 2r ′2

:r ′3

r3 − 2r ′3=

11:

21:

31

Total = 6.

Principal 1 =696×1 =

232, Principal 2 =

696×2 = 23, Principal 3 =

696×3 =

692

Several interesting problems involving interest for a principal are discussed.

Mixed problem involving interest

Let the principals (capitals) P1 and P2 be lent out for t1 and t2months at the same rate of interest (r percent per month) and letthe corresponding interests be I1 and I2. Then out of the sevenquantities (P1,P2, r , t1, t2, I1, I2), two can be determined from thesum of them and other four (r consider fixed).In particular, Rule 5(b)-7(a) says:;�a.va:�a.h:tea ;�a.va:pa.=� :a:ta:P+le :pa:[a:�a.dõ :ta:yea ;�a.ma:Ta:~tua :P+l+ya.ea:va.Ra Á Á 5 Á Á;Da:na:ya.ea:(ãÉa k+a:l+ya.ea:va.Ra ;Da:na:P+l+ya.eaH k+a:l+P+l+ya.ea:va.Ra ÁmUa:l+Da:na:k+a:l+ya.ea:va.Ra ;�a.ma:(rMa ya:�a.d dx :Zya:tea ta:d.a ta.�a Á Á 6 Á Áta:tpa:[a:ya.ea:(ãÉa ;Ga.a:ta.Ea ta.a:Bya.Ma :pra:[ea:pa:ta.ea .$a.a:t�a.a Á Á 7a Á Á

Mixed problem involving interest

“In two prescribed different sides, if the sum of eitherthe interest or the capitals or the times or the capitaland the time (of different sides) or the capital and theinterest or time and the interest (of the same side) isobserved as mixed, their shares in the sum can bedetermined, separately by the method of partnershipconsidering the products of two (known ingredients) inthe two sides as contribution.”

So this rule says: when any one sum out ofI1+I2, t1+t2,P1+P2,P1+t2,P2+t1, I1+t1+I2+t2, I1+P1 or I2+P2

is known and the rest 4 ingredients are known, then two(comprising the sum) can be found separately.

ExampleGiven I1 + I2,P1, t1,P2, t2 to find I1 and I2 separately. Now,

I1 =P1t1r100

, I2 =P2t2r100

∴I1I2

=P1t1P2t2

∴I1 + I2

I2=

P1t1 + P2t2P2t2

∴ I2 =(I1 + I2)P2t2(P1t1 + P2t2)

.

Similarly I1 =(I1 + I2)p1t1P1t1 + P2t2

Example.: Suppose P2, t1, I1, I2 and P1 + t2 are given:

Now,I1I2

=P1t1P2t2

P1

t2=

I1P2

I2t1

P1 + t2t2

=I1P2 + I2t1

I2t1

∴ t2 =(P1 + t2)I2t1I1P2 + I2t1

, P1 = (P1 + t2)− t2

Example

Example 17.ma.a:sea:na Za:ta:~ya ;�a.kÒ +.ya:t,a :Sa:�e :vRa:SRa:~ya ya:t,a :P+lM :P+l+ya.eaHya.ea:gea .. a:tva.a:�a=M ;Za:dÒU ;pa:yua:tMa mea :P+lM k+.Ta:ya Á Áta.a:Bya.Ma :pa:[a:�a.dõ :ta:yea ;�a.ma:Ta.ea ;�a.va:�a.ma:(rea :pxa:Ta:k, kx +.tea b.rUa:�a.h Á Á 17 Á Á“The sum of interest of 100 in a month, added to that of 60 in a year(at the same rate) is 41. Tell me the interest separately.”

Here P1 = 100,P2 = 60, t1 = 1, t2 = 12. Given I1 + I2 = 41.

We have I2 =(I1 + I2)P2t2P1t1 + P2t2

=41 × 60 × 12

100 + 60 × 12=

41 × 60 × 12820

= 36

.I1 = (I1 + I2)− I2 = 41 − 36 = 5.

Now let any one out of P1 + I2,P2 + I1, t1 + I2, t2 + I1,P1 + t1P2 + t2 be known

and the other four quantities are known separately. Then the product of the two

quantities can be obtained and the quantities themselves can be determined

according to Rule 7(b)-8.

Example

Example. a) P2, t1, t2, I1 and P1 + I2. To find P1, I2.

Now as the rates of interest are equal,

I1I2

=P1t1P2t2

∴ P1I2 =I1P2t2

t1

P1 − I2 =√

(P1 + I2)2 − 4P1I2

From, P1 + I2 and P1 − I2, P1 and I2 are found from ‘sankraman. a’.

(b) Similarly, Let P1,P2, t2, I1 and I2 + t1 are given. Then

I2t1 =I1P2t2

P1∴ I2 − t1 =

√

(I2 + t1)2 − 4I2t1

From, I2 + t1 and I2 − t1, I2 and t1 can be determined.

Example

Example 18.ma.a:�a.sa Za:ta:~ya :P+lM ya:d, va:SeRa:Na .. a :Sa.æö�ÅÅ*:x +a.taH :P+lM ya:~ya Áta:dùÅ;a.ea:gea :pa.úãÁ*.a:gua:Na.aH �a:ya.ea:d:Za .sa:Kea :pxa:Ta:k, k+.Ta:ya Á Ák+a:le ;a.na.$a:Da:na:yua:�e .sa:h:P+l+yua:�e ;Y:Ta:va.a ga:�a.Na:ta:m,a Á Á 18 Á Á“The sum of the interest of 100 for a month and thecapital (principal) whose interest in a year is 36, is 65.O friend, tell (them) separately (knowing) the sum of thecapital and its time or that of the interest of 100 and thetime (tell them separately).”

Solution: Here P1 = 100, I2 = 36, t1 = 1, t2 = 12, andP2 + I1 = 65.

Now, P2I1 =I2P1t1

t2=

36 × 100 × 112

= 300.

Solution

∴ P2 − I1 =√

(P2 + I1)2 − 4P2I1 =√

652 − 1200 = 5√

132 − 48 = 55

∴ P2 + I1 = 65,P2 − I1 = 55 ∴ P2 = 60, I1 = 5.

Now P2, t2, I1, I2 and P1 + t1 be given. To find P1, t1.

Now, P1t1 =I1P2t2

I2∴ P1 − t1 =

√

(P1 + t1)2 − 4P1t1

In this example, P1t1 = 101,P1t1 =5 × 60 × 12

36= 100.

∴ P1 − t1 =√

1012 − 400 =√

1002 − 200 + 1 =√

(100 − 1)2 = 99

∴ P1 + t1 = 101,P1 − t1 = 99.

P1 = 100, t1 = 1.

Payment in Instalments

The following rule pertains to the payment in instalments of adebt, based on simple interest.

Rule 10..~k+.nDa:k+.k+a:l+k+.l;a:nta.= h� .a:na:~k+.nDea:na Ba.a:�a$a:tea ;�a.va.�a:m,a Á.~k+.nDa:k+.k+a:l+�a.va:gua:�a.Na:tea ;a.na:ya:tMa ;a.na:mRua:�+.k+a:l.H .~ya.a:t,a Á Á 10 Á Á“The capital (principal) is multiplied by the fixed periodof instalment (the product is) divided by the amount ofinstalment less the interest (on the total amount) for theperiod of instalment. (The quotient) determines thetime of being free from debt.”

Amount lent = P, Amount of instalment = a. Period of instalment(No. of months / instalments) = t Total time for being free fromdebt = T = nt let us say.

Payment in Instalments

Interest / period of Instalment (for time t) = I′

∴ Interest for total duration of debt (time T = nt) = nI′.

Now, the amount due after time T = nt is P + nI′ and amountpaid is na. Equating them,

P + nI′ = na

∴ n =P

a − I′

∴ T = nt =Pt

a − I′

This is what is stated in the rule.

Example

Example 20.ma.a:sea:na :pa.úãÁ*.a:k+.Za:tea:na Za:tMa d:Za.ea:nMad.�Ma ku +.sMa:k+.f I+h.a:pya:Da:ma:NRa:k+a:ya Áma.a:sa:dõ :yMa :pra:a.ta .sa:Kea d:Za:pa.úãÁ*.a:yua:�+a:t,a.~k+.nDMa dÒu ;tMa k+.Ta:ya mea :pa:�a=;mua:�+.k+a:l+m,a Á Á 20 Á Á“In his bad days a debtor was given an amount (principal P) of 90 atthe rate of 5 percent per month to be paid back in bimonthly (t = 2)instalments of 15 (each). O friend, tell me quickly, the time of beingfree from the debt.”

Here P = 90, t = 2 months (period of instalments)

I′ (payment interest per period of instalment) = 5100 × 90 × 2 = 9, Instalment

amount, a = 15.

∴ T =Pt

a − I′=

90 × 215 − 9

=90 × 2

6= 30 months

This is the time for being free from debt.

Payment in instalments: Time for being free from debt

The next rule gives the expression for the Principal (capital) P, in terms of the instalmentamount, a, time being free from debt, T , the instalment period, t and the rate of interest(in fact interest on unity) for the time of being free from debt.

Rule 11.;a.na:mRua:�+.k+a:l+vxa.;dÄùÅ;a.a .�+pa:~ya ;�a.h .sEa:k+.ya.a h:tea:na Ba.$ea:t,a Á.~k+.nDa:k+.k+a:le +na .. a ga:ta:k+a:l+~k+.nDa.a:h:a.ta:mRUa:l+m,a Á Á 11 Á Á“The time of being free from debt is multiplied by the amount of instalment.(The product) is divided by (the product of the sum of) the interest on unityfor the time of being free from debt added to 1, multiplied by the period ofinstalment. The (quotient) is the capital.”

We had P + nI′ = na

∴ P(

nI′

P+ 1

)

= na =aT

t

∴ P =aT

(n

pI′ + 1

)

t

This is what being stated, when it is realised thatnI′

Pis the interest for the period T = nt.

ExampleExample 21.:pa.úãÁ*.a:k+.Za:tea:na ;�a.va.�Ma ma.a:sa:�a.dõ :ta:yea:na .sa:d:le +na Á.~k+.nDaH :pa.úãÁ*.a:d:Za.aY:Ta ;�a�Ma:Za.a:n}å.a.a:sa.a ;�a.va:a.na:mRua:�:H Á Ák+a:l+�/////////�a.~tva:h va:d mUa:lM ;�a.kM vxa:a;dÄâ H k+a .. a ya:�a.d :vea:�////�a.tsa Á Á 21 Á Á

“An amount is lent at the the rate of 5 percent per month. Theperiod of instalment is 2 1

2 months, the amount of interest is 15, andthe time of being free from debt is 30 months. If you know, tell thecapital and the interest.”

Solution: Here t = 2.5, a = 15,T = 30. Rate of interest = 5 percent per month.

Interest on unity for T =5

100× 30 = 1.5

(

=nI′

P

)

∴ P =15 × 30

(1.5 + 1)× 2.5= 72.

Total interest paid = 1.5 × 72 = 108

Alternatively, total interest = na − P =302.5

× 15 − 72 = 108.

Payment in instalments: More realistic computation

The above is based on simple interest. The debtor owes the amountP + nI′ at the end of the time nt = T based on simple interest, but he isactually paying an instalment amount a after each instalment period t.He is not getting any benefit for paying back the debt in instalments atearlier times. The next few rules take this into account.

The rule in 12-14(a) pertains to the instalment amount a beingadjusted only towards payment of capital. The rule gives theexpression for the amount of interest to be paid in addition to the‘monthly instalments’ being paid to clear the capital.

Let amount lent (Principal or capital) = P, Instalment amount = a,Instalment time period = t months. Rate of interest = r percent per

month. If the amount is cleared in n instalments P = n′a or n′ =Pa

.

∴ Total time for being free from debt = n′t =Pta

Payment in instalments contd.Here n′ indeed not be an integer. Let

n′ = n +fa

∴Pa

= n +fa

or P = na + f , where f < a.

Consider the amount, P as made up of na and f . Consider the integralpart na, first. At time t instalment amount a is being paid. The interestshould be the entire amount, na.

∴ Interest at time t = na · rt100

Here an amount a is paid towards clearing the principal. So the amountdue at the beginning of the second instalment is na − a = (n − 1)a. So,at time 2t, we have to calculate interest for the amount (n − 1)a.

∴ Interest at time 2t = (n − 1)a · rt100

Payment in instalments: Interest to be paidProceeding in this manner,

Interest at time rt = [n − (r − 1)] · a · rt

100

Interest at time nt = a · rt

100

Hence, total interest for the integral part

= [n + (n − 1) + · · ·+ 2 + 1]a · rt

100=

n(n + 1)

2a · rt

100

At the end of time duration nt, Principal amount due is f . (Integral part na having been

cleared). This will be cleared in timef

at . But the debtor has to pay the interest for f for a

period nt +f

at =

P

at , which is f · Pt

a· r

100. Hence,

Total interest to be paid =

[n(n + 1)

2a +

Pf

a

]rt

100

n is called the ‘pada’. f/a is the ‘agra.[

n(n + 1)

2a +

Pf

a

]

is called ‘mula pin. d. a’.

The total interest to be paid and the time for being free from debt are stated in the

following rule:

Rule

Rule 12-14 a:.~k+.nDa:k+.Ba:�M ;�a.va.�Ma l+b.DMa :pa:d:sMa:¼a:kM .. a Zea:Sa.Ma:ZaH ÁA:g{a.a:K.yaH :pa:d:va:gRaH :pa:d:yua:k, .~k+.nDa.a:DRa:sMa:gua:Na.ea yua:�:H Á Á 12 Á ÁA:g{a.Ma:ZaÈåî ÁÁ*+:Ga:nea:na :pra.$a.a:ya:tea mUa:l+�a.pa:Nq;a:K.yaH Áta:~ya .~k+.nDa:k+.k+a:l;a:t,a .sa:ma.a:na:yea:d, vxa:a;dÄâ :ma.a:na:ma:Ta Á Á 13 Á Á.~k+.nDa:k+.k+a:l:ÈåîÁÁ*+:Da:nea .~k+.nDa:&+tMa mua:K.ya:k+a:l.H .~ya.a:t,a Á Á 14 A Á Á

Total interest and time for being free from debt

“The amount (capital or principal lent) is divided by the amount ofinstalment. (The integral part of) the quotient is called ‘Pada’ andthe remaining part, ‘Agra’. ‘Pada’ is added to its square. (The sum)is multiplied by half the amount of instalment. (The product) isadded to (the product) of ‘agra’ (i.e. the remaining part) multipliedby amount (lent. The sum) is called ‘Mula Pin. d. a’. The interestshould be obtained from that (i.e. from the mula pin. d. a) by takingthe time equal to the period of instalment. After that, the period ofinstalment multiplied by the amount (lent, the product) divided bythe amount of instalment happens to be the time of being free fromthe debt.”

[Note: Letrt

100= r ′. Probably, the debtor is paying a + nar ′ at time

t , a + (n − 1)ar ′ at time 2t , · · · , a + ar ′ at time nt , and at time nt +fa

t , he is

paying f +Afa

r ′. In all these, the first part is payment towards the principal and

the second part is the interest.]

ExampleExample 22.:pa.úãÁ*.a:k+.Za:tea:na d.�va.a :pa.úãÁ*.a:yua:ta.aH .sa:�a:a.taH :ke +.na Á.~k+.nDa:na .. a :pra:ya:. C+a.ta ma.a:sa.a:Bya.Ma g{a.a:h:k+.vxa:a;dÄâ :Ba:ya.a:t,a Á Ávxa:a;dÄâ M ;�a.va:mua:a.�+.k+a:lM k+.Ta:ya .sa:Kea tvMa :pua.=:a :vea:�////�a.tsa Á Á 22 Á Á

“Somebody lent 75 at the rate of 5 percent per month. The debtor, for fear of(high) interest, returns (the amount) in instalments of 9 (along with theinterest due), bimonthly. O friend, if you know, tell the interest and the timeof being free from debt.”

Solution: Here P = 75, r = 5, instalment a = 9, t =1

2.

A

a=

75

9= 8 +

3

9(f = 3, a = 9)

So, n = 8.

∴ Interest paid =

[8 × 9

2× 9 + 75 × 1

3

]

× 1

2× 5

100

= (36 × 9 + 25)× 2 · 5

100= 8

29

40

Time =75

9× 1

2= 4

1

6months

Equated monthly instalments

Now we consider the case when the instalment a is the interest on theoutstanding amount plus the payment of a part of the principal. Thenthe following rule gives the time of being free from the debt.

Rule 16 b - 17.:pra:a.ta:ma.a:�a.sa:k+.P+l+Zua.;dÄâ .Ea mUa:lM mUa:l;a:t,a :pxa:Ta:k, :pxa:Ta:g,a .$a:hùÅ:a.a:t,a Á Á 16 b,a Á ÁZea:Sa:~ya ma.a:�a.sa:k+.P+lM ;�a.va:Za.ea:Da:yea:d, ma.a:�a.sa:k+ea:pa:na:ya.a:t,a ÁZea:Sea:Na.aY:nea:na mUa:l+�a.va:Zea:Sa:ma.a:�Ma tua ma.a:sa:yua:k, k+a:l.H Á Á 17 Á Á“Subtract the principal part of each monthly instalment free from thecapital, successively. (This gives the number of complete months anda residue of the capital, if any). Subtract the interest of residue for amonth from the amount of monthly instalment. Divide the residue ofthe capital by the remainder (The quotient) added to the number of(complete) months is the time (of being free from the debt).”

Equated monthly instalments

Explanation: Lent amount,principal = P. Rate of interest = r percent

per month. Let r ′ =rt

100, where t is the period of instalment. At the

time of payment of first instalment, interest due is r ′P. Monthlyinstalment = a. ∴ Principal part of monthly instalment = a − r ′P.Hence, Amount due after first instalment paid P1 = P − (a − r ′P) orP1 = P(1 + r ′)− a.

The interest for this at the time of second instalment = r ′P1. Amountpaid = a. ∴ Principal part of monthly instalment = a − r ′P1 Amountdue after payment of second instalment= P1 − (a − r ′P1) = (1 + r ′)P1 − a = (1 + r ′)2P − (1 + r ′)a − a.

Amount due after payment of n instalmentsPn = (1 + r ′)nP − [(1 + r ′)n−1 + · · ·+ 1]a. According to the rule, we goon till R = Pn < a. R is the remainder or the ‘Residue’ he talks about.It takes less than an instalment period = t to clear R with its interest.Let this be Rt. (or R is the fraction of the instalment period).

Equated monthly instalments

The interest on R for the fraction, f of the instalment period = R · R · r .The amount the debtor pays for the fraction of the instalment periodRa. Equating these,

R + RRr ′ = Ra.

∴ R =R

a − Rr ′=

Ra − I′

where I′ = Rr ′ is the interest on R for one instalment. This is what hasbeen stated.

[Though the rule appears to state the result for the case when theinstalment period is a month, it is clearly applicable for any value of theperiod, t ].

Narayan.a’s concept of instalments is very similar to the modernconcept of equated monthly instalments. The only difference is that inmodern times the time for clearing the debt is fixed, and the equatedinstalment worked out, whereas here, the monthly instalment is fixedand the time for clearing the debt is worked out.

ExampleExample 24.d.�Ma d:Za:k+.Za:tea:na .. a Za:tMa .. a k+.~ya.a:�a.pa :ke +.na:a.. a.;dÄâ :a.na:na.a Á:pra:a.ta:ma.a:�a.sa:k+.P+l+sa:�a.h:ta.a :pa.úãÁ*.a.a:Za:t,a .~k+.nDa:kM :pra:ya:. C+a.ta .. a ÁA:nxa:N�a.a k+a:le +na .sa:Kea :ke +.na Ba:vea:d, g{a.a:h:k+.~ya va:d Á Á 24 Á Á

“A rich man lent somebody 100 at the rate of 10 percent per month.(The debtor) gives a monthly instalment of 50 including the interest.O friend, tell me the debtor’s time of being free from debt.”

Solution: P = 100, r ′ = 10100 = 1

10 , a = 50.

At the time of first monthly instalment: Interest = 100 × 110 = 10. Payment

a = 50.

Payment towards principal part = 50 − 10 = 40

Amount due after first instalment = 100 − 40 = 60

Interest on this for month 6010100

= 6

Payment = 50Payment towards principal part 50 − 6 = 44

Solution

Amount due after payment of second instalment = 60 − 44 = 16 = R

as it is less than a = 50.

Interest on R = 16, for a month 16 × 10100

= 1.6

Fraction of a month required to clear this debt with interest, R.

R =16

50 − 1.6=

1648.4

=4

12.1

This is the fraction of a month required to clear the debt. At this time

the debtor pays R× a =4

12.1× 50 to clear the debt completed.

References

1. Gan. itakaumudı of Narayan. a Pan.d. ita, Ed. by Padmakara

Dvivedi, 2 Vols, Varanasi, 1936, 1942.

2. Gan. itakaumudı of Narayan. a Pan.d. ita, Eng. Tr. with Notes,Paramanand Singh, Gan. ita Bharatı 20, 1998, pp. 25-82; ibid.21, 1999, pp. 10-73; ibid. 22, 2000, pp. 19-85; ibid. 23, 2001,pp. 18-82; ibid. 24, 2002, pp. 34-98.

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