Embed Size (px)
DESCRIPTION
maths
Citation preview
STUDY NOTE-1
ARITHMETIC
RATIO AND PROPORTION
1.1 Ratio : The ratio between two quantities a and b of same kind is obtained by dividing a by b
and is denoted by a : b.
Inverse Ratio : For the ratio a : b, inverse ratio is b : a .
A ratio remains unaltered if its terms are multiplied or divided by the same number.
a: b = am : bm (multiplied by m)
)0mbydivided(mb:
mab:a ¹=
Thus 2 : 3 = 2×2 : 3×2 = 4 : 6, again 4 6
4 : 6 : 2 : 32 2
= =
If a = b, the ratio a : b is known as ratio of equality.
If a > b, then ratio a : b is known as ratio of greater inequality i.e. 7 : 4 And for a < b, ratio a : b
will be the ratio of Lesser inequality i.e. 4 : 7.
For solving problems :
1. Reduce the two quantities in same unit.
If a = 2 kg, b = 400 gm, then a : b =2000 : 400 = 20 : 4 = 5 : 1 (here kg is changed to gm)
2. If a quantity increases by a given ratio, multiply the quantity by the greater ratio.
If price of crude oil increased by 4 : 5, which was, Rs. 20 per unit of then present
price = .unitper25.Rs4520 =´
3. If again a quantity decreases by a given ratio, then multiply the quantity by the lesser ratio.
In the above example of the price of oil is decreased by 4 : 3, the present
price = .unitper15.Rs4320 =´
4. If both increase and decrease of a quantity are present is a problem, then multiply the quantity
by greater ratio for inverse and lesser ratio for decrease, to obtain the final result.
Proportion : The equality of two ratios is called the proportion thus 2 : 3 = 8 : 12 is written as
2 : 3 : : 8 : 12 and we say 2, 3, 8, 12 are in proportion.
In proportion the first and fourth terms are known as extremes, while second and third terms
are known as means.
In proportion, product of means = product of two extremes As 2, 3, 8, 12 are in proportion, we
have 2×12 = 3×8 (=24)
Few Terms :
1. Continued proportions : The quantities a, b, c, d, e….. are said to be in continued
proportion of a : b = b : c = …..Thus 1, 3, 9, 27, 81, ….. are in continued proportion as 1 :
3 = 3 : 9 = 9 : 27 = 27 : 81 =……
MATHS 1.01
MATHS1.02
AR
ITH
MET
ICExample : If 2, x and 18 are in continued proportion, find x Now 2 : x = x : 18 or,
6x,or36x,or18x
x2,or
18x
x2 2 ±====
Obs. If a, b, c are in continued proportion , the aca,acb2 ±== .
2. Compound Proportion : If two or more ratios are multiplied together then they are
known as compounded.
Thus a1 a2 a3 : b1 b2 b3 is a compounded ratios of the ratios a1 : b1 ; a2 : b2 and a3 : b3. This method
is also known as compound rule of three.
Example : 10 men working 8 hours a day can finish a work in 12 days. In how many days can 12
men working 5 hours a day finish the same work.?
Men Hours day
Arrangement : 10 8 12
12 5 x
daysx 161210
5812 =´´= Obs : less working hour means more working days, so multiply
by greater ratio 58 . Again more men means less number of days, so multiply by lesser ratio
1210 .
Derived Proportion : Given quantities a, b, c, d are in proportion.
(i) Invertendo : If a : b = c : d then b : a = d : c
(ii) Alternendo : If a : b = c : d, then a : c = b : d
(iii) Componendo and Dividendo
If dcdc
babathen
dc
ba
-+-
-+=
Proof : Let ,kdc
ba == then a = bk, c = dk
L. H. S. = 1k1k
)1k(b
)1k(b
bbkbbk
-+=
-
+=
-+
R. H. S. = 1k1k
)1k(d
)1k(d
ddkddk
-+=
-
+=
-+ . Hence the result.
An Important Theorem
If ,then......fe
dc
ba ==
each ratio =
n/1
nnn
nnn
......rfqdpb
......reqcpa
ïþ
ïýü
ïî
ïíì
++
++where p, q, r,…. are quantities positive or negative.
Let ........fke,dkc,bkathatso,kfe
dc
ba ======
Hence, Pan = p(bk) n = pnkn, qcn = qdn kn, ren = rfnkn, etc.
MATHS 1.03
.k)k(......rfqdpb
......rfqdpb(k
.....rfqdpb
.....reqcpa n
nn
nnn
nnnnn
nnn
nnn==
ïþ
ïýü
ïî
ïíì
+++
++=
ïþ
ïýü
ïî
ïíì
++
++\
1
11
Hence the result.
Cor. 1. Putting n = 1, we get
If ,then......fe
dc
ba == each rati o, =
.....rfqdpb
.....reqcpa
+++
+++
Cor. 2. Puttting p = q= r= ….=1, we find
If If ........fe
dc
ba === each ratio
n1
nnn
nnn
.....fdb
......eca
þýü
îíì
++
++
Cor.3. If ........fe
dc
ba === then each rat io =
a c e .....
b d f ....
± ± ± ±
± ± ± ±
Putting p = .1.Corin......1r,1q,1 ±==±
Note. 1. x y z
a b c= = is sometimes written as x : y : z = a : b : c.
2. If x : y = a : b, it does not mean x = a, y = b. It is however to take x = ka, y = kb.
Solved problems :
1. If ,a2
x3y4
b3
y3z4
c4z3x4 -
=-
=- , show that each ratio is equal to c4b3a2
zyx
++
++.
Each of the given ratio c4b3a2
zyx
a2b3c4
x3y4y3z4z3x4
++
++=
++
-+-+-=
2. If h
g
fe
dc
ba === show that
4 4 4 4
4 4 4 4
aceg a c e g
bdfh b d f g
+ + +=
+ + +
h
g
fe
dc
ba === =k (say), so that a = bk, c = dk, e = fk, g = hk.
L. H. S. = 4kbdfg
hk.fk.dk.bk =
R. H. S. = 4 4 4 4 4 4 4 4
4 4 4 4
b k d k f k h k
b d f h
+ + +
+ + +=
( ) 44444
44444k
hfdb
hfdbk=
+++
+++. Hence the result
3. If a1, a2, ………, an, be continued proportion, show that 1n
2
1
n
1a
a
a
a-
÷ø
öçè
æ=
We have, 31 2 n 1
2 3 4 n
aa a a..... k (say)
a a a a-= = = =
n
1
n
1n
4
3
3
2
2
11n
a
a
a
a....
a
a
a
a
a
ak =´´´´= --
again,
n 1
n 1 1
2
ak
a
-
- æ ö= ç ÷
è ø
sum (or difference)of numerators
sum(or difference)of deno min ators=
4444 hfdb +++( (
4444 hfdb +++
4k 4k=
denomin ators
1
2
11-
÷ø
öçè
æ=\
n
n a
a
a
a
4. abc
xyz
cab
zxy
bca
yzxthatprove
cz
b
y
ax
-
-=
-
-=
-
-==
2
2
2
2
2
2 [I.C.W.A. (F) June 2007]
kcz
b
y
ax === (say); x = ak, y = bk, z = ck
( )( )
22
22
2
2k
bca
bcak
bca
yzx=
-
-=
-
-, Similarly
abc
xyzk
cab
zxy
-
-==
-
-2
22
2
2
(to show in detail) Hence the
result.
5. If ba
rac
q
cb
p
-=
-=
- prove that p + q + r = 0 = pa + qb + rc [I.C.W.A. (F) Dec. 2007]
bar
ac
q
cb
p
-=
-=
-= k (say), p = k (b-c), q = k (c–a), r = k (a–b)
Now p + q + r = k (b – c + c – a + a – b) = k × 0 = 0
And pa + qb + rc = ka (b – c) + kb (c – a) + kc (a – b) = k (ab – ac + bc – ba + ca – cb) = k × 0
= 0. Hence the result.
6. If ba
zac
y
cbx
+=
+=
+ prove that
222222 ba
)yx(z
ac
)xz(y
cb
)zy(x
-
-=
-
-=
-
- [I.C.W.A (F) Dec. 2005]
,kba
zac
y
cbx =
+=
+=
+ x = k (b+c), y = k(c+a), z = k(a+b)
22
k)cb)(cb(
)ac)(cb(k
)cb)(cb(
)baac(k).cb(k
cb
)zy(x-=
-+
-+=
-+
--++=
+
-
Similarly, 22
222 ba
)yx(zk
ac
)xz(y
-
-=-=
-
- (To show in detail) Hence the result.
7. The marks obtained by four examinees are as follows :
A : B = 2 : 3, B : C = 4 : 5, C : D = 7 : 9, find the continued ratio.
A : B = 2 : 3
B : C = 4 : 5 = 4
15:3435:
434 =´´ [for getting same number in B, we are to multiply by
43 ]
C : D = 7 : 9 = 7 ×28
135:4
152815 = [to same term of C, multiply by
2815 ]
\A : B : C : D = 2 : 3 : 4
15 : .135:105:84:5628
135 =
8. Two numbers are in the ratio of 3 : 5 and if 10 be subtracted from each of them, the
remainders are in the ratio of 1 : 5, find the numbers.
Let the numbers be x and y, so that )1...(y3x5,or53
yx ==
Again 51
10y10x =
--
MATHS1.04
or, 5x–y = 40 ….(ii) , Solving (I) & (ii), x = 12, y =20
\ regd. Numbers are 12 and 20.
9. The ratio of annual incomes of A and B is 4 : 3 and their annual expenditure is 3 : 2 . If each
of them saves Rs. 1000 a year, find their annual income.
Let the incomes be 4x and 3x (in Rs.)
Now 23
1000x31000x4 =
-- or, x = 1000 (on reduction)
\ Income of A = Rs 4000, that of B = Rs. 3000.
10. The prime cost of an article was three times the value of material used. The cost of raw
materials was increased in the ratio 3 : 4 and the productive wage was increased in the ratio
4 : 5. Find the present price cost of an article, which could formerly be made for Rs. 180.
[I.C.W.A (F) June 2007]
Prime cost = x + y, where x = productive wage, y = material used.
Now prime cost = 180 =3y or, y = 60, again x + y = 180, x = 180–y = 180–60 = 120
Present material cost = ,3
y4 present wage = ,
4x5
\Present prime cost = .230.Rs1508041205
3604 =+=´+´
SELF EXAMINATION QUESTIONS :
1. The ratio of the present age of a father to that of his son is 5 : 3. Ten years hence the ratio
would be 3 : 2. Find their present ages. [ICWA (F) 84] [Ans. 50,30]
1. The monthly salaries of two persons are in the ratio of 3 : 5. If each receives an increase of
Rs. 20 in salary, the ratio is altered to 13 : 21. Find the respective salaries.
[Ans. Rs. 240, Rs. 400]
3. What must be subtracted from each of the numbers 17, 25, 31, 47 so that the remainders may be
in proportion. [Ans. 3]
4. ba
zac
y
cbxIf
+=
+=
+ show that (b–c)x+(c–a)y + (a–b)z = 0
5. a2
x3y4
b3
y3z4
c4z3x4If
-=
-=- show that each ratio =
c4b3a2
zyx
++
++.
6. If 0)zyx(if,21kthatprovek
yxz
xz
y
zyx ¹++==
+=
+=
+
7. If ,21
ba
ba=
+
- prove that ]2001June)F(ICWA[
7391
baba
baba22
22=
+-
++
2 2 2
2 2 2
Hint s :2 a 2 b a b or, a 3 b or,a 9b.
81b 9b bL.H.S &etc.
81b 9b b
é ù- = + = =ë û
+ +=
- +
MATHS 1.05
8. If 2c
cbathatprove,9c
5b
4a =++== [ICWA Dec. 2000]
þýü
îíì ===== .etc&k9c,k4a;k
9c
5b
4a:sintH
9. (i) If c
bab
aca
cb +=+=+ and a + b + c 0¹ then show that each of these ratios is equal to 2.
Also prove that 222 cba ++ = ab + bc + ca. [ICWA (Prel.) Dec. ‘90]
(ii) If a : b = c : d show that xa + yb : b-a+=b-a dc:ydxcba:
10. Given qpprrq -
g=
-
b=
-a , prove that g+b+a==g+b+a rqp0 [ICWA, July, 62]
11. In a certain test, the number of successful candidates was three times than that of unsuccessful
candidates. If there had been 16 fewer candidates and if 6 more would have been unsuccessful, the
numbers would have been as 2 to 1. Find the number of candidates. [Ans. 136]
12. (i) Monthly incomes to two persons are in ratio of 4 : 5 and their monthly expenditures are in
the ratio of 7 : 9. If each saves Rs. 50 a month, find their monthly incomes.
[Ans. Rs. 400, Rs.500]
(ii) Monthly incomes of two persons Ram and Rahim are in the ratio 5 : 7 and their monthly
expenditures are in the ratio 7 : 11. If each of them saves Rs. 60 per month. Find their monthly
income. [ICWA(F) June 2003][Ans. 200, Rs. 280]
úû
ùêë
é=
-- .etc&
117
60x760x5sinth
13. A certain product C is made or two ingredients A and B in the proportion of 2 : 5. The price
of A is three times that of B.
The overall cost of C is Rs. 5.20 per tonne including labour charges of 80 paise per tonne. Find
the cost A and B per tonne. [Ans. Rs. 8.40, Rs. 2.80]
14. The prime cost of an article was three times than the value of materials used. The cost of
raw materials increases in the ratio of 3 : 7 and productive wages as 4 : 9. Find the present prime
cost of an article which could formerly be made for Rs. 18. [Ans. Rs. 41]
15. There has been increment in the wages of labourers in a factory in the ratio of 22 : 25, but
there has also been a reduction in the number of labourers in the ratio of 15 : 11. Find out in what
ratio the total wage bill of the factory would be increased or decreased. [Ans. 6 : 5 decrease]
16. Three spheres of diameters 2, 3 and 4 cms. respectively formed into a single sphere. Find the
diameter of the new sphere assuming that the volume of a sphere is proportional to the cube of its
diameter. [Ans. cm993 ]
MATHS1.06
AR
ITH
MET
IC
MATHS 1.07
OBJECTIVE QUESTIONS :
1. Find the ratio compounded of 3 :7, 21 : 25, 50 : 54 [Ans. 1 : 3]
2. What number is to be added to each term of the ratio 2 : 5 to make to equal 4 :5. [Ans. 10]
3. Find the value of x when x is a mean proportional between : (i) x–2 and x+6
(ii) 2 and 32 [Ans. (I) 3 (ii) ± 8] [ICWA (F)]
4. If the mean proportional between x and 2 is 4, find x [ICWA (F) June 2007] [Ans. 8]
5. If the two numbers 20 and x + 2 are in the ratio of 2 : 3 ; find x
[ICWA(F) Dec.2006] [Ans. 28]
6. If bafind
21
ba
ba=
+
- [I.C.W.A. (F) Dec. 2006] [Ans. 19 ]
7. If 3, x and 27 are in continued proportion, find x [Ans. ± 9]
8. What number is to be added to each term of the ratio 2 : 5 to make it 3 : 4 ? [Ans. 7]
úûù
êëé =
++ .etc&
43
x5x2:sinth
9. If 2baba =
-+ find the value of
22
22
baba
baba
++
+- [Ans. 137 ] [I.C.W.A. (F) Dec.2005]
10. If a2
x3y4
b3
y3z4
c4
y3x4 -=
-=
-show that each ratio is equal to
c4b3a2
zyx
++
++
[I.C.W.A (F) June, 2005]
úû
ùêë
é++
-+-+-= .etc&
a2b3c4
x3y4y3z4z3x4ratioeach:sinth
11. The ratio of the present age of mother to her daughter is 5 : 3. Ten years hence the ratio
would be 3 : 2. Find their present ages. [I.C.W.A. (F) Dec. 2004] [Ans. 50; 30 years)
12. If A : B = 2 : 3, B : C = 4 : 5 and A : C [Ans. 8 :15]
13. If x : y = 3 : 2, find the value of (4x–2y) : (x + y) [Ans. 8 : 5]
14. If 15 men working 10 days earn Rs. 500. How much will 12 men earn working 14 days?
[Ans. Rs. 560]
15. Fill up the gaps : -
==-
=-
=22
11 b–a
ba [Ans. ab, ]orderin,ab,ba,ab 3
16. If x, 12, y and 27 are in continued proportion, find the value of x and y [Ans. 8 ; 18]
17. If ,43
yx = find the value of
yx3
y4x7
+
- [Ans.
135 ]
18. What number should be subtracted from each of the numbers 17, 25, 31, 47 so that the
remainders are in proportion. [Ans 3]
úûù
êëé =
--=
-- 3or
x47x31
x25x17:sinth
19. 10 years before, the ages of father and son was in the ratio 5 : 2; at present their total age is
90 years. Find the present age of the son. [Ans. 30 years]
[Ans. 28]
[Ans. 10]
[Ans. 1 : 3]
[Ans. 8]
MATHS1.08
AR
ITH
MET
IC20. The ratio of work done by (x–1) men in (x+1) days to that of (x+2) men in (x–1) days is
9 : 10 , find the value of x. [Ans. 8]
úû
ùêë
é=
-+
+-.etc&
109
)1x()2x(
)1x()1x(:sinth
1.2 AVERAGE :
The arithmetic average or arithmetic mean or simple mean of a number of quantities is the sum of
the quantities divided by their number.
If x1, x2,….,x3, are the n numbers, then the average X is given by
n
x....xxX n21 +++
=
For example, if there are 5 boys whose height are 50, 54, 52, 56 and 58 inches, then the average
(or mean) .Inches545
5856525450 =++++=
quantitiesofnumber
quantitiesofsumaverage,findweThus =
or, average × number of quantities = sum of quantities.
Note : When quantities x1, x2,….,xn are all different, the average of the numbers is known as
simple average.
WEIGHTED AVERAGE :
If there are n quantities x1, x2,….,xn and w1, w2,….wn are their respective weights, then the
weighted arithmetic average is given by
n21
nn2211w.....ww
xw.............xwxw
+++
+++
For example, a contractor pays wages to his employees at the following rates : Rs. 2.50 per man
per day and Rs. 2.25 per woman per day. If he engages 20 men and 12 women, the total amount
paid per day is 2.50 × 20 + 2.25×12 = 50+27 = Rs. 77.
.).app(41.2.Rs3277
122077averageweightedthe ==+
=\
SOLVED EXAMPLES :
1. The mean of 6 numbers (out of which 2 numbers are equal) is 40. If two equal numbers are
excluded, the mean becomes 56. Find the missing numbers. [(ICWA (Prel.) June 1989]
Total of 6 numbers = 6×40 = 240
Total of 4 numbers when two equal numbers are excluded = 4 × 56 =224; total of equal numbers
= 240 – 224 = 16
So each of them is 8, (as two numbers are equal)
\ reqd. numbers are 8, 8.
2. The average score of girls in the preliminary examination is 75 and that of boys is 70. The
average score of all the candidates in the examination is 72. Find the ratio of number of girls and
boys that appeared in the examination.
Let number of girls be m and that of boys be n so that total numbers of all candidates be (m + n).
Now total score of girls = 75 m
Total score of boys = 70 n
Total score of all candidates = 72 (m + n)
By question, 75m + 70 n = 72 m + 72 n
Or, 3m = 2n or, 32
nm =
.3:2isratio.reqd\
3. In a class there are three divisions. The number of students and the average marks in
mathematics in the three divisions are 30, 40 30 and 40%, 30%, 50% respectively. What is the
average marks in mathematics of the class?
For division I of 30 students, total marks = 30 × 40 = 1200
,, ,, II ,, 40 ,, ,, ,, = 40 × 30 = 1200
,, ,, III ,, 30,, ,, ,, = 30 × 50 = 1500
For the whole class of 100 students total marks = 3900
\ average mark of the class = 1003900 = 39.
4. The average monthly income of 6 employees is Rs. 310. After one of them receives an
increment the average rises to Rs. 312.50. Find the amount of increment.
Total income of 6 employees = 6 × 310 = Rs. 1860
Total income of 6 employees after increment of one employee only = 6 × 312.50 = 1875.00
\ amount of increments = 1875–1860 = Rs. 15.
5. During the year 1932, the bank rates where as follows : 6% for 1 week, 5% for 1 week, 4½%
for 1 week, 4% for 17 weeks, 3½% for 3 weeks, 3 % for 5 weeks and 4% for 24 weeks. What
was the average rate during the year?
Diff. rates are 6% 5% 4½% 4% 3½% 3% 4%
Occurs (time) 1 1 1 17 3 5 24
\ required average rate
%245317111
244533317414151621
21
++++++
´+´+´+´+´+´+´=
%.52493%
50205 ==
Again, the simple average rate 6
33445621
21 +++++
= = %214 (as there are 6 diff. rates)
This average is different from the correct average.
MATHS 1.09
6. The average dividend paid by a company during the last five consecutive years was 10%. The
dividends for the last four years were %218 , %
219 , 11% and 12%. What was the dividend for the
first year?
Total dividends for last four years = %218 + %
219 +11+12 = 41%
Since the average dividend for 5 years was 10%, so the total dividend for 5 years = 5 × 10% =
50%.
\Dividend for the first year = 50–41 = 9%.
7. The weight of a person during July is 85.50 kg. and every following month he reduces his
weight by 1/50 kg. than the previous month. Find out his weight in June and by how much
percent there is a shortfall from his average weight of the year.
From July to June we find 1 year (= 12 months)
His weight in August = 85.50 – 1.50 kg.; that in September = 85.50 –2×1.50 = 82.50 kg.
In the same way his weight in June = 85.50 – 11×1.50 = 85.50 – 16.50 = 69.00 kg.
\ total weight for the whole year will be —
85.50 + 84 + 82.50+81.00+79.50+78.00+76.50+75.00+73.50+72.00+70.50+69.00 = 927.00 kg.
\Average weight = .kg25.7712927 =
Again shortfall = 77. 25 – 69.00 = 8.25 kg.
Av. Weight Shortfall x = .)app(%..
. 68101002577
258 =´
77.25 8.25
100 x \ reqd. percentage = 10.4 (app.)
8. The average age of a class of 17 boys at the end of a term was 14 years 3 months. Of these 5 left
of the average age of 15 years 2 months and at the beginning of the next term, 2 months later 6
new boys of average age 13 years 11 months came into the class. What was then to the nearest
month, the average age of the class?
Total age of 17 boys = 14 yrs. 3m. × 17 = 242 yrs. 3m.
Total age of 5 boys left the class = 15 yrs. 2m. × 5 = 75 yrs. 10 m.
Now, total age of (17 – 5=) 12 boys belonging to the class = 242 yrs. 3m. – 75 yrs. 10 m. = 166
yrs. 5 m.
Again, total age of 6 new boys joined the class = 13 yrs. 11 m. × 6 = 83 yrs. 6m.
Now, total age of (12+6=) 18 boys in the class = 166 yrs. 5 m. + 83 yrs. 6 m = 249 yrs. 11 m.
\Average age = .)app.(m11.yrs1318
.m11.yrs249=
9. A school with 78 boys and 72 girls on the books meets 432 times in the year. If each boy loses
one meeting in 9, and each girl one in 8, find the average attendance of each sex for the year.
Out of 9 meetings each boy attends 8 meetings.
MATHS1.10
AR
ITH
MET
IC
\….432…... …… ….. …… .. 43298 ´ = 384 meeting.
78 boys meet = 384 × 78 times and average attendance of boys per meeting = 2169
43278384 =´ i.e.
69 boys.
Similarly, out of 72 girl, each girl attends = 43287 ´ meetings.
\72 girls meet = 7284327 ´´ times
\average attendance of girls per meeting = 634328
724327 =´
´´ girls.
10. A person drove his car for first 20 km and then 30 km at an average speed of 20 km and 30 km
per hour respectively. At what speed must he drive next 50 km if the average speed of the whole
distance of his driving is 40 km per hour? [(ICWA (F)
June 2006]
Time required for first 20 km = 2020 = 1hr. and time for next 30 km. =
3030 = 1hr. lastly time
required for next 50 km = x
50 (where x = reqd. average speed)
Again time for whole journey =40
503020 ++ =40
100 =25
So we get, 1+1+25
x50 = or,
25
x502 =+ or,
25
x50x2 =+
Or, 5x = 4x + 100 or, x = 100.
\ reqd. average speed = 100/hr.
11. The average salary per head of all the works in an institution is Rs. 60. The average per head
of 12 officers is Rs. 400. The average salary per head of the rest is Rs. 56. Find the total number of
workers.
Let total number of workers (including officers) = n and number of others = 12
No. of workers other than officers = n–12.
Now, 60n = 400 × 12 + 12+ 56 (n – 12) or, 60n = 4800 + 56n – 672,
or, 60n – 56n = 4128 or, 4n = 4128 or, 10324
4128n ==
SELF EXAMINATION QUESTIONS :
1. The average height of 50 students of a class is 160 cm. The average height of 30 of them is 155
cm. Find the average height of remaining 20 students. [Ans. 167.5 cm.]
2. The mean of 5 numbers is 27. If one number is excluded, the mean becomes 25. Find the
excluded number. [Ans. 35]
3. The average earning of a man for 9 days from 1st to 9th January is Rs. 5 per day and that for 10
days from 1st to 10th January is Rs. 5.10. What is his earning on the 10th January?
[I.C.W.A (Prel.) June, 1986] [Ans. Rs.6]
MATHS 1.11
4. In a factory out of 60 workers 40 are males and 20 are females. The mean wage of a male
worker is Rs. 70. If the mean wage of a male worker is Rs. 80; find the mean wage of a female
worker.
[ICWA(Prel) Dec.1984] [Ans. Rs. 50]
5. From the following table find the value of x.
Class No of students Average wt. in kgs
VIII 30 35
IX 35 40
X 50 45
XI x 50
Average weight of all students is 42.4. [ICWA(Prel)June1984] [Ans. 30]
6. A candidate obtains following percentage of marks in an interview for a technical job :
(i) Written examination 75
(ii) Practical 80
(iii) Qualification and achievements 70
(iv) Viva 60.
Find the weighted average marks obtained by the candidate of the weights allotted are 3, 2, 2, 2
respectively. [Ans. 70.5]
7. A candidate obtained the following % of marks in a job interview.
Written examination 83, Practicals 75, Viva 71, G. K. 65. Find the weighted mean in the weights
are 3, 2, 3, 2 respectively [ICWA(Prel) Dec. 1984] [Ans. 74.2]
8. A candidate obtained the following percentages of marks in an examination.
English 75, Mathematics 60, G.K. 59, Accountancy 55, Hindi 63.
Find the weighted mean if the weight are 2,1, 3, 3 and 1 respectively.
[ICWA (prel.) June 1984] [Ans. 61.5]
9. The employees of a company consist of its officers and clerks. The average wages of employees
in the company is Rs. 100. The average wages of 150 clerks is Rs. 80 and the average wages of
officers is Rs. 300. Find the number of officers in that company.
[ICWA (Prel) June 1988] [Ans. 15]
10. (I) A truck runs 25 km. at a speed of 50 km.p.h. another 50 km. at a speed of 60 km. p.h. then
due to bad condition of road travels for 12 minutes at a speed of 5 km. p.h. and finally covers the
remaining distance of 24 km. at a speed of 30 km. p.h. show that the average speed of the truck is
7642 km. p.h.
MATHS1.12
AR
ITH
MET
IC
(ii) Form the following data, find the value of x, if the average marks in mathematics for all the
centres together is 52.
Centre No. of student Average mark
Scored in mathematics
Bombay 30 45
Calcutta 30 55
Delhi x 50
Madras 40 60
[ICWA(Prel) June 1990] [Ans. 100]
(iii) The average marks in “Elements of Mathematics” of Preliminary students of 3 centres in
India is 50. The number of candidates in three centres are respectively 100, 120, 150. If the
averages of the first two centres are 70 and 40, find the average marks of the third centre.
ICWA (Prel.) Dec. 1990] [ Ans.44.67 marks]
úû
ùêë
é++
+´+´= .etc&
150120100
x150401207010050:sintH
11. A person drove his car 20 km. at an average speed of 25 km. per hour. At what average speed
for the whole distance is to be 30 km. per hour? [Ans. 37.5 km. per hr.]
12. The average score of boys is 60, that of girls is 70 and that of all the candidates is 64
appearing in mathematics of annual examination. Find the ratio of number of boys and the
number of girls there. If the total number of candidates appearing in mathematics is 150, find the
number of boys there.
[ICWA (F) Dec. 2006] [Ans. 3 : 2 ; 90]
OBJECTIVE QUESTIONS :
1. The average of 3 numbers is 15. With inclusion of a fourth number, the average becomes 17.
Find the included number. [ICWA(F) June 2004] [Ans. 23]
2. Speed of a car to go up a hill us 10 km. per hour and to go down is 20 km per hour. Compule
it average speed. [ICWA(F) June, 2003] [Ans. 3113 km. per. Hr.]
[hints : refer w.o. ex. 12]
3. The mean of 5 numbers is 27. If one number is excluded the mean becomes 25. Find the
excluded number. [Ans. 35]
4. The average height of 50 students in a class is 160 cm. The average height of 30 of them is
155 cm. Find the average height of remaining 20 students. [Ans. 167.5 cm]
5. The mean of 4 numbers is 9. If one number is excluded the mean becomes 8. Find the
excluded number. [Ans. 12]
6. The average weight of 24 students is a class is 35 kg. If the weight of the teacher be included,
the average rises by 400 gms. Find the weight of the teacher. [Ans. 45 kg]
MATHS 1.13
MATHS1.14
AR
ITH
MET
IC1. A batsman scored 87 runs in the 17th innings and thus increased his average by 3. Find his
average after 17th innings. [Ans. 39]
2. The average of 50 numbers is 38. If two numbers, say 45 and 55 are discarded, find average of
remaining numbers. [Ans. 37.5]
3. The average age of three boys is 15 years. If their ages are in the ratio 3 : 5 : 7; Find the age
of youngest boy [Ans. 9 years]
4. The average score of a Cricketer for 10 matches is 43.9 runs. If the average for the first six
matches is 53, find the average for the last four matches. [Ans. 30.25]
5. The average salary of 20 workers in an office is Rs.1900 per month. If the manager’s salary
is added, the average salary becomes Rs. 2000 p/m. Find the manager’s annual salary.]
[Ans. 48,000]
6. The average age of four children in a family is 12 years. If the spacing between their ages is 4
years, find the age of the youngest child. [Ans. 6 years]
1.3 MIXTURE :
Problems of mixture can be classified into two ways—Direct and Inverse.
In case of direct problems, if the individual price of the different ingredients and the proportion in
which they are to mixed are given, we can find the price of mixture.
Formula. Price of mixture (per unit quantity) = quantitytotal
tcostotal.
In case of inverse problems, we are to find the proportions of the ingredients are given.
Example : To find proportion in which two kinds of tea costing Rs. 6 and Rs. 6.30 P per kg are
mixed up to produce a muxture of costing Rs. 6.20 P per kg.
If 10 kg. of first kind are taken, then there is a gain of 10 ×20P = Rs. 2.
Similarly, if 20 kg. of second kind are taken, the loss is 20 × 10 P = Rs. 2.
So, 10 kg. of first kind are to be mixed with 20 kg. of second kind. Hence the mixture should be
10 : 20 = 1 : 2.
Rule. First kind; gain per kg. (i.e. difference) = 20
Second kind per kg. loss = 10
\First : Second = 10 : 20 = 1 : 2
i.e. First : Second = diff. of second : diff. of first.
WORKING RULE :
(i) All the prices must be expressed in the same unit.
(ii) The cost price (C.P.) of the mixture must lie in-between the cost prices of the other
ingredients.
(iii) All prices used should be cost prices. If in some cases, the selling price of the mixture at
profit or loss is given, find the corresponding cost price.
(iv) Now applying the following formula to find the required ratio
kindst1ofthatwithmixtureof.P.Cofdifference
kindnd2ofthatwithmixtureof.P.Cofdifference
kindnd2
kindst1=
Alternatively :
Working Rule
\1st kind : 2nd kind = 2nd difference : 1st difference = (…) : (….)
SOLVED EXAMPLES :
(On direct problems)
1. A dealer blends 75 kg. of salt @ Rs. 1.50 per kg. with 45 kg. of salt @ Rs. 1.30 per kg. what is
the lowest price at which he can sell the mixture so as to gain at least 25%?
Cost price (C.P.) of the mixture 120171.Rs
754550.1530.145 =
+´+´=
C.P. S.P.
100 125 \x = .)app(78.1.Rs100120
171125 =´
´
120171 x
the lowest price per kg = Rs. 1.78.
2. A man mixes 5 gallons of spirit at Rs. 5.50 per gallon,12 gallons at Rs. 6 per gallon and 13
gallons at Rs. 6.50 per gallon. At what price per gallon must he sell mixture so as to gain Rs. 26
on the whole?
C.P. of 5 gallons = Rs. 5.50 × 5 = Rs. 27.50
C.P. of 12 gallons = Rs. 72.00
C.P. of 13 gallons = Rs. 6.50 × 13 = Rs. 84.50
\C.P. of 30 (= 5+12+13) gallons = Rs. (27.50 + 72.00 + 84.50) = Rs. 184.
Total gain = Rs. 26
\ S. P. of 30 galls. = Rs. (184 + 26) = Rs. 210
\ S. P. per gallon = Rs. 30210 = Rs. 7.
3. A merchant buys sugar at Rs. 4.10, Rs. 3.75 and Rs. 4.50 per kg. and mixes in the proportion
5 : 4 : 1. At what price must he sell the mixture so to gain 25%?
C.P. of the mixture per kg.
4.Rs10
00.40.Rs145
150.4475.3510.4 ==++
´+´+´
Cost Price of ingredients 1st kind 2nd kind
Cost Price of mixture
Difference
(...) (...)
(...) (...)
(...)
MATHS 1.15
C.P. S.P.
100 125 4
X 125 Rs.5.00100
\ = ´ =
4 x
\S. P. per kg. = Rs. 5.00.
4. There are two one-litre decanters containing liquor and water in the ratios 7: 3 and 8 : 2
respectively. The contents of the two decanters are poured into a two litre decanter, and the
mixture is sold at Re. 1 per litre, thereby realising a profit of %2112 on the returns. Find the cost
price of a litre of liquor.
Quantity of liquor the 1st decanter litre127=
Quantity of liquor the 2nd decanter litre108=
Total quantity of liquor in the 2-litre decanter = 1015
108
107 =+ litre.
Now S.P. of 2 litres = Rs. 2, gain = 0.25 P. (= %2112 of Rs. 2)
\ C.P. = 2 – 0.25 = Rs. 1.75
2 litres of mixture contains 1015 litre of liquor
\C.P. of 1015 litre liquor = Rs. 1.75 (as water is free of cost)
C.P. of 1 litre liquor = 1.75 × 1510 = Rs1.16 P. (app.)
5. 30 litres of mixed paint containing 10% oil is to be made into a paste containing 50% oil. Find
the quantity of oil added. If 6 more litres are added, find the percentage of colouring matter in the
mixture.
Quantity of oil in 30 l mixed paint = 33010010 =´ l; so quantity of paint = 30 – 3 = 27l. For 50%
oil in the mixed paint quantity of oil should be made equal to that paint. So quantity of oil = 27 l
i.e. oil to be added = 27 –3 = 24l.
If again 6 l oil being added more, total quantity of oil = 27 + 6 = 33 l
And total quantity of mixed paint = 27 + 27 + 6 = 60 l.
\Reqd. percentage of paint %.451006027 =´=
6. In a liquid mixture 20% is water, and in another mixture water is 25%. These two mixtures are
mixed in the ratio 5 : 3. Find the percentage of water in the final mixture.
In the first pot liquid : water= 80 : 20 = 4 : 1
(as water = 20% so liquid = 80%)
In the second pot liquid : water = 3: 1
MATHS1.16
AR
ITH
MET
IC
Let 5 litres of mixture from first pot be mixed with 3 litres of second pot In
5l mixture, water 15515
411 =´=´+
=
In 3 l mixture, water 433
413
311 =´=´+
=
In 8 l mixture, total water 47
431 =+=
In 100 l mixture total water = 8721
8100
47 =´
\ reqd. percentage = 8721
7. Two types of oil A and B are mixed in the ratio 3 : 2. After selling 25% of this mixture, type A
is mixed to the remaining mixture so as to make a new ratio 5 : 3. If this new mixture is 400 litres
by quantity, find the quantity of original mixture (before selling) and also the quantity of A mixed
later on.
In 400 l mixture, quantity of A = 40035
5 ´+
= 250 l.
And that of B = (400 –250) = 150 l. Let the quantity of mixture (after selling 25%) = x l, quantity
of 5x3A = , quantity
5x2B = .
Now as quantity of B remains fixed, so 5x2 = 150 or, x = 375 l
i.e., quantity of A = 37553 ´ = 150 l
So quantity of A added = 250 – 225 = 25.
Let original mixture = y l.
Now 75 % of y = 375 or, 375y10075 =
or, 50075
100375y =´=
\ reqd. quantity = 500 l.
8. 4 Litres are drawn from a cask full of sepecial liquid, it is then filled with water then 6 litres of
the mixture are drawn and the cask is again filled with water. If the quantity of special liquid now
in the cask be to the quantity of water in it as 1 to 2; how much does the cask hold?
Let x = capacity of cask. After taking 4l of liquid and filling with water, the ratio of liquid and
water is (x–4) : 4. Now in 6 l of drawn mixture, there are ( ) 4
)4x(66
44x4x -
=´+-
- l of liquid.
Remaining liquid in the cask
= (x – 4) – x
24x10xx
24x6x4xx
)4x(6 22 +-=+--=-
MATHS 1.17
remaining water in cask x
24x10xx2 +--= =
x24x10 -
Now 2:1x
24x10:x
24x10x2=-+-
or, 24x10
24x10x2
-+- =
21 or 2(x2 – 10x + 24) = 10x – 24
or, 036x15x2 =+-
or, (x – 3) (x – 12) = 0 or, x = 3, 12
Now x = 3 is impossible, as in the first case 4l of liquid drawn out.
\ reqd. capacity = 12l.
(On inverse problem)
9. How should salts at Rs. 2.50 and Rs. 3.35 per lb. be mixed to produce a mixture worth Rs. 3
per lb.?
In the 1st kind, profit per lb = 3 – 2.50 = 50 P.
In the 2nd loss per lb = 3.35 – 3 = 35 P.
\ 1st : 2nd = 35 : 50 = 7 : 10.
10. A dealer mixes two varieties of grains costing Rs. 6 per quintal and Rs. 15 per quintal in such
a way that can gain 10% by selling the resultant mixture at Rs. 8.25 P. per quintal, What is the
proportion in which the gains are mixed?
On 10% profit, S.P. = 100 + 10 = 110. Now to find the C. P. of mixture.
S.P. C.P.
110 100
8.25 x x = 100× 50.7110
25.8 =
So, for the 1st variety, gain = 7.50 – 6 = 1.50
And for the 2nd loss = 15 – 7.50 = 7.50
\ 1st variety : 2nd variety = 7.50 : 1.50 = 5 : 1
11. A dealer mixes tea costing Rs. 8 per kg, with tea costing Rs. 7 per kg. and thereafter, sells the
mixture at Rs. 8 per kg and earns profit of 7.5% on his sale price. In what proportion does he mix
them? [ICWA (F) Dec. 2003]
To find cost price of mixture. Here gain is on selling price. So
S.P. C.P.
100 92.5
8 ? ? 4.78100
5.92 =´= (C.P. of mixture)
32
60.040.0
40.78740.7
kindnd2
kindst1==
--=
MATHS1.18
AR
ITH
MET
IC
12. A dealer mixed two varieties of teas having costs Rs. 1200 and Rs. 2500 each per kg in such a
way that he can gain 20% by selling the resultant mixture at Rs. 1800 per kg. Find the proportion
in which the two types of tea are mixed. [ICWA (F) Dec. 200]
We are to find cost price, here gains is on cost price So
S.P. C.P.
120 100
1800 ? ? 15001800120100 =´= \Cost price of mixture = Rs. 1500
310
30001000
1200150015002500
kindnd2
kindst1==
--= \reqd proportion is 10 : 3.
13. In mixing tea, 1 kg. in every 100 kgs. is wasted. In what proportion must a dealer mix teas
which cost him Rs.24 and Rs. 18 per kg. respectively, so that the cost is Rs. 20 per kg.
Let us take 100 kg. of two kinds. Then during the time of mixing, 1 kg. is wasted, leaving 99 kg.
only.
C.P. of 99 kg. of the mixture = Rs.99 × 20. Again , C.P. of 100 kg. of two kinds = 100x, where x =
average price, if there would be no loss
\ 99 × 20 =100 x
or, x 80.19100
1980 ==
So, mixtureandkindst1of.)kgper(pricesof.diff
mixtureandkindnd2of)kgper(pricesof.diff
kindnd2
kindst1=
73
20.480.1
80.19241880.19 ==
--= \ reqd. proportion = 3 : 7.
14. A lump of gold and silver weighing 200 gm. is worth Rs. 825, but if the weights of gold and
silver be interchanged it would be worth Rs. 562.50. If the price of 10 gm. of gold be Rs. 67.50;
find the rate per 10 gm. of silver. Also calculate the weight of silver in the lump.
Worth of 200 gm. Gold and Silver = 825
Worth of 200 gm. Gold and Silver = 562.50
(When weight of gold and silver interchanged)
\Worth of 200 gm gold and 200 gm of silver = Rs. (825+562.50) = Rs. 1387.50
50.37.Rssilver.gm200ofworth,gSubtractin
00.1350.Rs)2050.67(gold.gm200ofworthAgain
=
=´=
worth of 10 gm. 88.1.Rs20
50.37 == From 1st lump, 23
52550.787
825135050.37825
silver
gold==
--=
\gold : silver = 3 : 2 Weight of silver = .gm8020052 =´
15. Two vessels contain mixture of milk and water are in the ratio of 5:1 and in the other in the
ratio of 9 : 1. In what proportion the quantities from the two should be mixed together so that the
mixture thus formed may contain milk and water in the ratio 8 : 1?
Let 1 litre mixture from 1st vessel be mixed with k litres mixture of 2nd vessel
MATHS 1.19
MATHS1.20
AR
ITH
MET
ICFrom 1st vessel :
65milk = l,
61water = l, taken
From 2ndvessel : 10
k9milk = l,10kwater = l, taken
Total milk 60
k545010
k965 +=+ l, Total water
60k610
10k
61 +=+=
By condition, 1:860
k610:60
k5450 =++ or, 18
60k5450 =+
or, k = 5 (on reduction)
\ reqd. proportion = 1 : 5.
Mixture of three or more kinds :
Problems regarding the mixture containing three or more kinds are also solved in similar way. At
first, two kinds are to be selected, one of which is at a higher price than the average price while
the other at a lower price. Similar treatment is made for the other two and so on, till all the kinds
are taken at least once.
It may be noted that such problems give infinite number of solutions.
16. The prices of 3 kinds of tea are Rs. 5.00, Rs. 6.00 and Rs. 6.80 respectively. In what
proportion they should be mixed so that the price of the mixture may be Rs. 6.50 per kg?
Let A, B and C are the three kinds.
Taking A and C, at first,
For A, profit per kg. = 6.50 – 5.00 = Rs. 1.50 = 150P.
For C, loss per kg. = 6.80 – 6.50 = Rs. 0.30 = 30 P. \A : C = 30 : 150 = 1 : 5
Taking B and C.
For B, profit per kg. = 6.50 – 6.00 = Re. 0.50 = 50 P
For C, loss per kg. = 6.80 – 6.50 = Re. 0.30 = 30 P. \ A : B : C = 1 : 3 : 10
Note. In the above example, if we take A : C = 3 : 15 and B : C = 6, 10, then A : B : C = 3 : 6 : 25
(a different result from the former)
Alternatively
C.P. of 1st kind = Rs. 5
…….2nd kind = Rs. 6
…….3 rd kind = Rs. 6.80
Let the proportion of mixing the three kinds be x : y : z
\ C.P. per kg. of mixture.
zyx
z5
34y6x5.Rs
++
++=
)zyx(5
z5
34y30x25.Rs
++
++=
Here, C.P. of the mixture = Rs. 6.50 = Rs. 2
13 per kg.
\ 2
13)zyx(5
z3430x25 =++++
Solving, 15x + 5y = 3z
If we take x = 1 and y = 3, then z = 10 (A)
Thus one possible proportion is 1 : 3 : 10
Such a problem will have infinite number of solutions.
Note. The values of x and y in (A) should be chosen such that after substituting the values of x and
y in (A) . L.H.S. does not become negative.
On withdrawal :
Rule. Original quantity of liquid in vessel = x units (say) from which if y units being withdrawn
and replaced in same quantity of water for n operations (repeated), then amount of liquid left = x
n
x
y1 ÷
ø
öçè
æ- units.
Example . A vessel contains 50 litres of milk. From the vessel 5 litres of milk being withdrawn
and replaced by water by same quantity. The operation being repeated three times. Find the
quantity of milk left in the cask.
Here x = 50, y = 5 , n = 3. Now using the rule, we get quantity of milk left in vessel
33
)10.1(50505150 -=÷
øö
çèæ -´= 45.36729.50)9.0(50 3 =´== l
SELF EXAMINATION QUESTIONS :
1. A milkman mixes 61 litres of water with 349 litres of milk which he buys for Rs. 123. At what
price should he sell a litre so as to save one-third of his outlay? [Ans. 40P]
2. A grocer blends 75kg. salt at Rs. 1.12 P per kg. with 45 kg. of salt at Rs. 1.31 P. per kg. What is
the price at which he should sell the mixture so as to gain 25% [Ans. Rs.1.49P]
3. Three equal jars are filled with mixtures of milk and water, the proportion of milk to water
being at 5 : 2, 4 : 3, and 8 : 3 respectively. The mixture of the jar are poured into a drum. What is
proportion of milk and water in the final mixture? [Ans. 155: 76]
4. There are two one-litre decanters containing liquor and water in the ratios 11 : 4 and 10 : 5
respectively. The contents of the two decanters are poured into a two-litre decanter and the
mixture is sold at Re . 1 per litre thereby realising a profit of %2112 on the returns. Find the cost
price of a litre of liquor. [Ans. Rs. 1.25]
5. A dealer mixes 90 litres of wine containing 10% of water with 60 litres of wine containing 20%
of water. What is the percentage of water in the mixture. [ICWA(F) June, 2005] [Ans. 14%]
6. Two grades of motor oil A and B are mixed in the proportion 3 : 1 to make 96 litres of grade
C. When half of C has been sold, a further quantity of A is added to increase the proportion of A
to B in the resulting mixture to 7 : 2, Find the quantity of last added. [Ans. 6 litres]
7. Two grades C and D of alcohol are mixed in the proportion 3 : 2. After 25% of this has been sold from stock, a sufficient quantity of C is mixed with the remainder to raise the proportion to
MATHS 1.21
5 : 3. If the stock is now 7200 gallons what was the quantity of the original mixture and how much
of grade C was added to make the new mixture? [Ans. 9000 gals. ; 450 gals.]
8. A dealer mixes tea costing Rs. 8. Per kg. with tea costing Rs. 7 per kg. and sells the mixture at
Rs. 8 per kg. with tea costing Rs. 7 per kg. and sells the mixture at Rs. 8 per kg. and earns a profit
of 217 % on his sale price. In what proportion does he mix them? [Ans. 2: 3]
9. A tea dealer mixes two qualities of tea at Rs. 2.56 P. and Rs. 5.53 P. per kg. and sells the
mixture at Rs. 3.41 P. per kg. to make a profit of 10% on his outlay. In what proportion does he
mix the two qualities? [Ans. 9 :2]
10. A dealer buys oranges of two qualities one at Rs. 12 a dozen and the other at Rs. 8 a dozen.
These were mixed up and he sells them at 15 per Rs. 12 and thereby makes 5% on his total
outlay. Find the ratio of the number of oranges of the two kinds. [Ans. 2 : 5]
11. A grocer sells one kind of tea at Rs. 2.70 P. per kg. and loses 10% and another at Rs. 4.50P.
per kg. and gains 2112 %. How the two quantities of tea should be mixed, so that the mixture may
be sold at Rs. 3.95. P. kg. at a profit of 25%. [Ans. 21 : 4]
12. A grocer mixes 4 kinds of rice which cost him Rs. 5, Rs. 4, Rs. 3 and Rs. 2 per kg.
respectively, in the proportion of 2, 3, 4, 7 respectively. Find at what rate he must sell the mixture
so as to gain 25% on the whole. [Ans. Rs. 3.75]
13. A dealer mixes varieties of grains at Rs. 6 per kg. and Rs. 15 per kg. and in such a way that he
can gain 10% by selling the resulting mixture at Rs. 8.25 per kg. What is the proportion in which
the grains are mixed? [Ans. 5 : 1]
14. (i) In mixing tea 1 kg. in every 100 kg. is wasted. In what proportion must a dealer mix teas
which cost him Rs. 42 and Rs. 32 per kg. respectively so as to gain 10% by selling the mixture at
Rs. 40 per kg. [Ans. 2 : 3]
(ii) In mixing tea 1% is wasted. In what proportion must a dealer mix two grades of tea which
cost him Rs. 4.50 and Rs. 7.00 per kg. respectively, so as to gain 8.9% by selling the mixture at
Rs. 5.50 per kg? [Ans. 4: 1]
15. What weight of tea worth Rs. 7.00 per kg. should be mixed with 60 kg. of tea worth Rs. 5.00
per kg. so that when the mixture is sold at Rs. 5.50 per kg. there may be neither gain nor loss?
[Ans. 20 kg]
16. A gold is value at Rs. 10 per gm. An alloy of gold and silver weighs 1 kg. and its value is Rs.
6080. If the weights of gold and silver are interchanged it would worth Rs. 4120. Find the
proportion of gold and silver in and alloy also the price of silver per kg. [Ans. Rs. 200; 3 : 2]
17. With grain worth Rs. 10 per kg. a trader mixes an inferior quality worth Rs. 6 per kg. In what
proportion must he mix them so that by selling the mixture at the higher price he may gain 16%?
[Ans. 19 : 20 ]
MATHS1.22
AR
ITH
MET
IC
18. (i) Two casks A and B are filled with the two kinds of liquids; mixed in cask A in the ratio of
2 : 7 and in cask B in the ratio of 1 : 5. What quantity must be taken from each to form a mixture
which shall consist of 2 litres to one kind and 9 litres of the other? [Ans. 3 : 8]
(ii) Two vessel contain mixtures of milk and water in the proportion 2: 3 and 4 : 3 respectively. In
what proportions should the two mixtures be mixed so as to form a new mixture containing equal
quantities of milk and water? [ICWA (F) Dec, 2005] [Ans. 5 : 7]
19. Three vessels have capacities in the ratio 1 : 2 : 3. All the three glasses are filled with
mixtures of milk and water. The proportion of milk and water in the first glass is 3 : 1, in the
second the proportion is 5 : 3 and in the third it is 9 : 7. If the contents of all the three glasses are
put into a bigger single vessel what will be the proportion of milk and water in the mixture in the
bigger vessel? What is the percentage of milk in the final mixture in the bigger vessel?
[Ans. 59 : 37 : 61.46%]
20. A trader mixes 100 lbs. of tea at one price with 70 lbs. of tea at a dealer price. By selling the
mixture at Rs. 2.12 per lb.he gained 20% on his outlay. Find the value of each kind of tea, the
difference in their prices being Re. 0.36 per lb. [Ans. Rs. 1.62; Rs. 1.98]
21. Nine litres are drawn from a pot of full of milk. It is then filled with water, then nine litres of
the mixture are drawn and the pot is again filled with water. If the ratio of the quantity of milk
now in the pot to the quantity of water in it 16 : 9, how much does the pot hold?
[Ans. 45 litres]
22. Several litres of milk were drawn off a 54 litre vessel full of milk and an equal volume of
water added. Again the same volume of mixture were drawn off. As a result the mixture in the
vessel contains 24 litres of pure milk. How much milk was drawn off initially? [Ans. 18 l ]
úúû
ù
êêë
é=÷
øö
çèæ - .etc&24
54x154Use.Hints
2
23. Four litres are drawn from a cask full of wine. It is then filled with water. Four litres of
mixture are again drawn and the cask is again filled with water. The quantity of wine now left in
the cask is to that of wine in it is 36 : 13. How much does the cask hold? [Ans. 28 l ]
OBJECTIVE QUESTIONS :
1. A mixture of 30 litres contain wine and water in the ratio 7 : 3, how much water must be added
to it., so that the ratio of wine and water may become 3 : 7 ? [Ans. 24 l ]
2. 65 litres of a mixture of wine and water contain 15 litres more wine than water. Find the ratio of
wine to water. [Ans. 8 : 5]
Hints : 65 –15 = 50 litres water = 255021 =´ l.
Total wine = 25 + 15 = 40; ratio = 40 : 25 = 8 : 5]
3. In what ratio salt worth Rs. 6 per kg. be mixed with salt worth Rs. 7.30 per kg. to make mixture worth Rs. 6.50 per kg? [ Ans. 8 : 5 ]
MATHS 1.23
4. In what ratio must wheat at Rs. 16.0 per kg. be mixed with wheat at Rs. 14.50 per kg. so that
the mixture be worth Rs. 15.40 per kg.? [Ans. 3: 2]
5. In what ratio must rice at Rs. 31.0 per kg. be mixed with rice at Rs. 36.0 per kg. So that the
mixture be worth Rs. 32.50 a kg.? [Ans. 7 : 3]
6. A sum of Rs. 4/- was divided among 50 boys and girls. Each boy gets 90 paise and a girl 65
paise. Find the ratio of boys and girls. [Ans. 17 : 8]
Hints : Average money received by each = .P825041 =
[ Each boy rececves 82 65 17
Each girl rececves 90 82 8
-= =
-& etc.]
7. A vessel contains 10 litres of milk. From the vessel 2 litres of milk being withdrawn and
replaced by water by same quantity, the operation being repeated two times. find the quantity of
milk left in the vessel. [Ans. 6.4 litres]
8. How should tea at Rs. 100 and Rs. 120 per kg be mixed to produce a mixture worth Rs. 108
per kg.? [Ans. 3 : 2]
1.4 INTEREST :
The price to be paid for the use of a certain amount of money (called principal) for a certain
period is known as Interest. The interest is payable yearly, half-yearly, quarterly of monthly.
The sum of the principal and interest due at any time, is called the Amount at that time.
The rate of interest is the interest charged for one unit of principal for one year and is denoted by
(i) . If the principal is Rs. 100 then the interest charged for one year is usually called the rate of
interest percent per annum, and is denoted by r.
For example, if the principal is Rs. 100 and the interest Rs. 3, then we say usually that the rate of
interest is Rs. 3 percent annum (or r = 3 %)
Here ).yearoneforrupee1forerestint.e.i(03.0100
3i ==
Simple interest is calculated always on the original principal for the total period for which the
sum (principal) is used.
Let P be the principal (original)
n be the number of years for which the principal is used
r be the rate of interest p.a.
I be the simple interest
i be the rate of interest per unit (i.e. interest on Re. 1 for one year)]
Now I = P.i.n, where 100
ri =
Amount A = P + I = P + P. i. n = P (1+ i.n) i.e. A = P (1 + n .i)
Observation. So here we find four unknown A, P, i., n, out of which if any there are known, the
fourth one can be calculated.
MATHS1.24
AR
ITH
MET
IC
SOLVED EXAMPLES :
1. Amit deposited Rs. 1200 to a bank at 9% interest p.a. find the total interest that he will
get at the end of 3 years.
Here P = 1200, ,09.0100
9i == n = 3, I = ?
I = P. i.n = 1200 ×0.09 ×3 = 324.
Amit will get Rs. 324 as interest.
2. Sumit borrowed Rs. 7500 at 14.5% p.a. for 212 years. Find the amount he had to pay
after that period.
P = 7500, ,145.0100
5.14i == ?A,5.2212n ===
A = P (1+ in) = 7500 (1+0.145×2.5) = 7500 (1+0.3625)
= 7500 × 1.3625 = 10218.75
\reqd. amount = Rs. 10218.75.
3. Find the simple interest on 5600 at 12% p.a. from July 15 to September 26, 1993.
Time = number of days from July 15 to Sept. 26
= 16 (July) + 31 (Aug.) + 26 (Sept.)= 73 days.
P = 5600, 10012i = =0.12, .yr
36573n = = .yr
51
S.I. = P. i.n. = 5600 × 0.12 × 51 =134.4
\ reqd. S.I. = Rs. 134.40.
(In counting days one of two extreme days is to be excluded, Usually the first day is
excluded).
To find Principle :
4. What sum of money will amount to Rs. 1380 in 3 years at 5% simple interest?
Here A = 1380, n = 3, ?P,05.0100
5i ===
From A = P (1 + 0.05.3) or, 1380 = P (1+0.15)
Or, 1380 = P (1.15) or, P 1380
12001.15
= =
\reqd. sum = Rs. 1200
5. What sum of money will yield Rs. 1407 as interest in 211 year at 14% p.a. simple interest.
Here S.I. = 1407, n = 1.5, I = 0.14, P = ?
S.I. = P. i.n. or, 1407 = P × 0.14 × 1.5
Or, 670021.0
14075.114.1
1407P ==´
=
\ reqd. amount = Rs. 6700.
MATHS 1.25
6. What principal will produce Rs. 50.50 interest in 2 years at 5% p.a. simple interest.
S.I. = 50.50, n = 2, i = 0.05, P = ?
S.I. = P. i.n. or, 50.50 = P × 0.05 × 2 = P × 0.10
or, 50510.050.50P ==
\ reqd. principal = Rs. 505.
Problems to find rate % :
7. A sum of Rs. 1200 was lent out for 2 years at S. I. The lender got Rs. 1536 in all. Find the
rate of interest p.a.
P = 1200, A = 1536, n = 2, i = ?
A = P (I + ni) or 1536 = 1200 (1 +2I) = 1200 + 2400 I
or, 2400i = 1536 – 1200 = 336 or, 14.02400336i ==
\ reqd. rate = 0.14×100 = 14%.
8. At what rate percent will a sum, become double of itself in 215 years at simple interest?
A = 2P, P = Principal, 215n = , i = ?
A = P (1 + ni) or, 2P = P (1+ 2
11 i)
or, 2 = 1+ 2
11 i or, 112i =
or, 18.18100112n =´= (app.); \reqd. rate = 18.18%.
Problems to find time :
9. In how many years will a sum be double of itself at 10% p.a. simple interest.
A = 2P, P = Principal, 10010i = = 0.10, n= ?
A = P (1 + ni) or, 2P = P [1 + n(.10)] or, 2 = 1 + n (.10)
or, n (.10) = 1 or, 1010.1n ==
\Reqd. time = 10 years.
10. In what time Rs. 5000 will yield Rs. 1100 @ 215 %?
P = 5000, S.I. = 1100, ,055.0100
5i 2
1
== n = ?
S.I.= P. ni or, 1100 × 5000 × n × (0.055) = 275 n
or, .4275
1100n == \reqd. time = 4 years.
MATHS1.26
AR
ITH
MET
IC
Miscellaneous worked out Examples :
11. In a certain time Rs. 1200 becomes Rs. 1560 at 10% p.a. simple interest. Find the
principal that will become Rs. 2232 at 8% p.a. in the same time.
In 1st case : P = 1200, A = 1560, i = 0.10, n = ?
1560 = 1200 [1+n (.10)] = 1200 + 120 n
or, 120 n = 360 or, n = 3
In 2nd case : A = 2232, n = 3, i = 0.08, P = ?
2232 = P (1+3×0.08) = P (1 + 0.24) = 1.24 P
or, .180024.1
2232P ==
12. A sum of money amount to Rs. 2600 in 3 years and to Rs.2600 in 3 years and to Rs.
2900 in 214 years at simple interest. Find the sum and rate of interest.
Amount in 214 yrs. = 2900
Amount in 3 years = 2600
\ S. I. for 211 yrs. = 300
S.I. for 1 yr. 20032300
1
300
21
=´==
and S.I. for 3 years.= 3×200=600
\ Principal = 2600 – 600 = Rs. 2000
P = 2000, A = 2600, n = 3, i = ?
2600 = 2000 (1+ 3 i) = 2000 + 6000 i
or, 6000 i = 600 or 101
6000600i == or, 10
10100r ==
\ reqd. rate = 10%.
Alternatively. 2600 = P(1+3 i)….(i), 2900 = P (1+ 4.5 i)………….(ii)
Dividing (ii) by (I),, i31
i5.41
)i31(P
)i5.41(P
26002900
+
+=
+
+=
or, i31
i5.41
2629
+
+= or, i = 0.10 (no reduction)
or, r = 0.10 × 100 = 10%
From (i) 2600 = P (1+3×0.10) = P (1+0.30) = P (1.30)
.200030.1
2600P ==\
MATHS 1.27
13. A person lent some amount at 12% p.a. for 212 years and some amount at 12.5% p.a. for
2 yrs. If he had amount of Rs. 10,000 in hand and on such investment earned Rs. 2700 in all,
find the amount he invested in each case.
In 1st case let the investment be Rs. x, then in 2nd case, it would be Rs. (10000 – x)
In 1st case, interest earned 10
x3212
10012x =´´=
In 2nd case, interest earned = (10000 – x) 12.5 x(1000 x) 2 2500
4100= - ´ ´ = -
By question, 27004x2500
10x3 =-+
or, 4000xor20020xor200
4x
10x3 ===-
\ reqd. investment in 1st case = Rs. 4000.
And that in 2nd case = Rs. (10000 – 4000) = Rs. 6000.
14. Divide Rs. 2760 in two parts such that simple interest on one part at 12.5% p.a. for 2
years is equal to the simple interest on the other part at 12.5% p.a. for 3 years.
Investment in 1st case = Rs. x (say)
Investment in 2nd case = Rs. (2700 – x)
Interest in 1st case = x × 5x2
10010 =´
Interest in 2nd case 8x310353
1005.12)x2760( -=´´-=
By question, 5x =1035 –
8x3 or, 1035
8x3
5x =+
or, 103540
x15x8 =+ or, 103540
x23 = or, x = 1800
\ Investment in 2nd case = Rs. (2760 – 1800) = Rs. 960.
15. A person borrowed Rs. 8,000 at a certain rate of interest for 2 years and then Rs. 10,000
at 1% lower than the first. In all he paid Rs. 2500 as interest in 3 years. Find the two rates
at which he borrowed the amount.
Let the rate of interest = r, so that in the 2nd case, rate of interest will be (r–1). Now
25001100
)1r(000,102
100r8000 =´
-´+´´
or, 160r + 100 r – 100 = 2500 or, r = 10
\In 1st case rate of interest = 10% and in 2nd case rate of interest = (10 – 1) = 9%
Calculations of interest on deposits in a bank : Banks allow interest at a fixed rate on deposits
from a fixed day of each month up to last day of the month. Again interest may also be calculated
by days.
MATHS1.28
AR
ITH
MET
IC
16. A man deposited Rs. 5000 on 20th April in a Co., paying interest at 2% p.a. He
withdraws Rs. 3000 on 15th may and deposited Rs.400on 6th June. How much interest was
due to him on 30th June following?
From 21st April to 14th May = (10 +14) = 24 days. For investment of Rs. 5000 for 24 days, the
corresponding interest (l2) due
,73
480.Rs= From I = n.r.
P.100
From 15 th may to 5th June = (17+5) = 22 days.
For the investment of Rs. (5000 – 3000) for 22 days,
The interest (I2)= Rs. 176
73
From 6th June to 30th June = 25 days For the investment of Rs. (2000 + 4000) for 25 days
The interest (I3) = Rs. 73
600
Total interest I1+I2+I3 = 73
60073
17673
480 ++ = Rs. 17.21 (approx).
Interest on instalment basis :
For purchasing costly goods, instead of each down price, instalment payment is introduced. In that
case, the seller charges some interest, calculation of which will be clear from the following
example:
17. If I buy a watch with cash down, I pay Rs. 150. If I pay it on instalment basis, I pay
Rs.10 cash down and 10 monthly instalment of Rs. 15. What rate of interest is calculated in
the second case ?
The price of the watch = Rs. 150. Now Rs. 10 is paid in cash and the balance Rs. 140 is to
cleared up with interest in 10 monthly instalment of Rs. 15 each. So the amount of Rs. 140 for 10
months = the sum of amount of the successive instalments of Rs. 15 each, with the respective
earning of interest. Taking i = interest of Re. 1 for one month, in this case
140(1+10 i) = 15 (1 + 9 i) + 15 (1+8 i) + …….+ 15 (1 + i) + 15
= 15 × 10 + 15 i (9 + 8+ …..+1)
= 150 + 15 i × 45 = 150 + 675 i
or, 140 + 1400 i = 150 + 675i or, 725 i = 10 or, 72510i =
Now interest for Rs. 100 for 1 year 29161612100
72510 =´´=
\ reqd. rate of interest .a.p%291616=
MATHS 1.29
18. A Pressure cooker is available for Rs. 350 cash or Rs. 80 down-payment and Rs. 50 per
month for 6 months.
Find (I) total amount paid (ii) rate of interest charged.
Principal left for 1st month = 350 – 80 = Rs. 270
…… …… 2nd …. = 270 – 50 = Rs. 220
…… …... 3rd…… = 220 – 50 = Rs. 170
…. ….. 4th…… = 170 – 50 = Rs. 120
….. ….. 5th……. = 120 – 50 = Rs. 70
…. ….. 6th…… = 70 – 50 = Rs. 20
Total = Rs. 870
Instalment charge = amount paid in all – cash prige
= (50 × 6 + 80) – 350 = Rs. (380 – 350) = Rs. 30
Thus Rs. 30 is the interest on Rs. 870 for 1 month
Now reqd. interest .%38.411210087030 =´´=
SELF EXAMINATION QUESTIONS :
1. What sum will amount to Rs. 5,200 in 6 years at the same rate of simple interest at which
Rs. 1,706 amount to Rs. 3,412 in 20 years? [Ans. Rs. 4000]
2. The simple interest on a sum of money at the end of 8 years is 52 th of the sum itself. Find the
rate percent p.a. [Ans. 5%]
3. A sum of money becomes double in 20 years at simple interest. In how many years will it be
treble? [ICWA (F) Dec. 2004] [Ans. 40 yrs.]
4. At what simple interest rate percent per annum a sum of money will be double of it self in 25
years? [ICWA(F) June, 2005] [Ans. 4%]
5. A certain sum of money at simple interest amounts to Rs. 560 in 3 years and to Rs. 600 in 5
years. Find the principal and the rate of interest. [Ans. Rs. 500; 4%]
6. A tradesman marks his goods with two prices, one for ready money and the other for 6 month’s
credit. What ratio should two prices bear to each other, allowing 5% simple interest.
[Ans. 40 : 41]
7. A man lends Rs. 1800 to two persons at the rate of 4% and 214 % simple interest p.a.
respectively. At the end of 6 years, he receives Rs. 462 from them. How much did he lend to each
other? [Ans. Rs. 800 ; Rs. 1000]
8. A man takes a loan of Rs. 10,000 at the rate of 6% S.I. with the understanding that it will be
repaid with interest in 20 equal annual instalments, at the end of every year. How much he is to
pay in each instalment? [Assam H.S. 1986) [Ans. 700.64]
[Hints. Refer W.O. ex. 4. 10,000 (1+20 ×.06) = P (1+19×.06) + P(1+18×.06) + …P]
MATHS1.30
AR
ITH
MET
IC
MATHS 2.31
9. The price of a watch is Rs. 500 cash or it may be paid for by 5 equal monthly instalments of
Rs. 110 each, the first instalment to be paid one month after purchase. Find the rate of interest
charged. [Ans. 7642 %]
10. Divide Rs. 12,000 in two parts so that the interest on one part at 12.5% for 4 years is equal to
the interest on the second part at 10% for 3 years. [Ans. Rs. 4500; Rs. 7500]
11. Alok borrowed Rs. 7500 at a certain rate for 2 years and Rs. 6000 at 1% higher rate than the
first for 1 year. For the period he paid Rs. 2580 as interest in all. Find the two of interest.
[Ans. 12%; 13%]
12. If the simple interest on Rs. 1800 exceeds the interest on Rs. 1650 in 3 years by Rs. 45, find
the rate of interest p.a. [Ans. 10%]
1. A certain sum put out at S.I. amounts to Rs. 708 in 3 years. If the rate of interest increased by
one third, it will amount to Rs. 744 in the same time. Find the sum and rate on interest.
[Ans. Rs. 600; 6%]
úúú
û
ù
êêê
ë
é
÷÷÷
ø
ö
ççç
è
æ
+=÷øö
çèæ += )ii......(
1003r4.3
1p744);i.....(100
r31P708.sintH
úúú
û
ù
êêê
ë
é
=
÷øö
çèæ +
÷øö
çèæ +
= .etc&Pfind),i(from%;6r,or
100r31p
100r41p
708744,or
14. A certain sum at a certain rate of interest p.a. S.I. becomes Rs. 1150 in 3 years and Rs. 1250 in
5 years. Find the rate percent p.a. [Ans. 5%]
15. Mr. X deposited a total of Rs. 95000 in two different banks which give 5% and 217 % interest.
If the amounts repayable by the two banks at the end of 7 years are to be equal. Find the individual
amount of deposit. [Ans. Rs. 5700; Rs. 3800]
16. A man left Rs. 130000 for his two sons aged 10 years and 16 years with the direction that the
sum should be divided in such a way that the two sons got the same amount when they attain the
age of 18 years. Assuming the rate of simple interest is 2112 % p.a. calculate how much the elder
son got in the beginning. [Ans. Rs. 80,000]
[Hints. n (for elder son) = 18 – 16 = 2, n (for younger son)
= 18 – 10 = 8 ; x = elder son’s sum. Now
÷øö
çèæ ´+-=÷
øö
çèæ ´+ 8
1005.121)x130000(2
1005.121x or, x = 80000]
13.
MATHS1.32
AR
ITH
MET
IC17. What annual installment will discharge a debt of Rs. 3094 due in 4 years at 7% S.I.?
[Ans. Rs. 700]
[Hints. (amount of x for 3 yrs) + (x for 2 years) + (x for 1 yr.)
18. A dealer of radio offers radio for Rs. 2700 cash down or for Rs. 720 cash down and 24
monthly instalments of Rs. 100 each. Find the rate of simple interest charged per annum.
[ICWA (F) June, 2007] [Ans. 10.06%(opp)]
[Hints : rafer solved ex. 17]
19. A dealer offers an item for Rs. 270 cash down or Rs. 30 cash down and 18 equal monthly
instalments of Rs. 15. Find the rate of simple interest charged. [ICWA (F) June, 2004]
[Hints : refer solved problem no. 17] [Ans. %9717 ]
OBJECTIVE QUESTION :
1. At what rate of S.I. will Rs. 1000 amount to Rs. 1200 in 2 years? [Ans. 10%]
2. In what time will Rs. 2000 amount to Rs. 2600 at 5% S.I.? [Ans. 6 yrs.]
3. At what rate per percent will S.I. on Rs. 956 amount to Rs. 119.50 in 212 years? [Ans. 5%]
4. The S.I. on a sum of money at the end of 8 years is 52 th of the sum itself. Find the rate percent
p.a. [Ans. 5%] 5. To repay a sum of money borrowed 5 months earlier a man agreed to pay Rs. 529.75. Find the
amount borrowed it the rate of interest charged was 214 % p.a. [Ans. Rs. 520]
6. What sum of money will amount to Rs. 5200 in 6 years at the same rate of interest (simple) at
which Rs. 1706 amount to Rs. 3412 in 20 years? [Ans. Rs. 4000]
7. A sum money becomes double in 20 years at S.I., in how money years will it be triple?
[Ans. 40 years]
8. A certain sum of money at S.I. amount to Rs. Rs. 560 in 3 years and to Rs. 600 in 5 years. Find
the principal and the rate of interest. [ Ans. Rs. 500; 4%]
9. In what time will be the S.I. on Rs. 900 at 6% be equal to S.I. on Rs. 540 for 8 years at 5%.
[Ans. 4 years]
10. Due to fall in rate of interest from 12% to 2110 % p.a.; a money lender’s yearly income
diminishes by Rs. 90. Find the capital. [Ans. Rs. 6000]
11. A sum was put at S.I. at a certain rate for 2 years. Had it been put at 2% higher rate, it would
have fetched Rs. 100 more. Find the sum. [Ans. Rs. 10,000]
9. Complete the S. I. on Rs. 5700 for 2 years at 2.5% p.a.
[ICWA(F) June, Rs. 2007] [Ans. Rs. 285]
13. What principal will be increased to Rs. 4600 after 3 years at the rate of 5% p.a. simple
interest? [ICWA (F) Dec. 2006] [Ans. Rs. 4000]
14. At what rate per annum will a sum of money double itself in 10 years with simple interest ?
[ICWA (F) Dec. 2005] [Ans. 10%]
diminishes by Rs. 90. Find the capital.
15. The simple interest on Rs. 300 at the rate of 4% p.a. with that on Rs. 500 at the rate of 3% p.a.
both for the same period, is Rs. 162. Find the time period. [ICWA (F) June, 2005] [Ans. 6 years]
16. Calculate the interest on Rs. 10,000 for 10 years at 10% p.a.
[ICWA(F) Dec. 2003] [Ans. Rs. 10,000]
17. A person deposited Rs. 78,000 in Post office monthly interest scheme (MIS) after retirement at
8% p.a. Calculate his monthly income. [ICWA (F) June, 2003] [Ans. Rs. 520]
[Hints : Income (monthly) = .]etc&121
100878000 ´´
1.5 DISCOUNTING OF BILLS :
Few Definitions :
Present Value (P.V.) : Present value of a given sum due at the end of a given period is that sum
which together with its interest of the given period equals to the given sum i.e.
P.V. + Int. on P.V. = sum due Sum due is also known as Bill Value (B.V.)
Symbols : If A = Sum due at the end of n years, P = Present value, i = int. of Re. 1 for 1 yr.
n= unexpired period in years, then A = P+P n i == P(1+n i)…….(i)
or, ni1
AP+
=
True Discount (T.D) :
True discount of a given sum due at the end of a given period, is the interest on the present value
of the given sum i.e. T.D. = P n i…………..(ii)
T.D.= Int. of P.V. = amount due – Present value i.e. T.D. = A – P………(iii)
Again T.D. = )iv..(..........ni1
Anini1
AA+
=+
-
Example : Find P.V. and T.D. of Rs. 327 due in 18 months hence at 6% S.I..
3 6327
Ani 2 100T.D 27,3 61 ni 12 100
´ ´= = =
+ + ´
here A = 327, n = 18 m = 3/2 yrs. i= 6/100.
We know P.V. +Int. on PV (i.e.T.D)= sum due (i.e.B.V)
Or, P.V. = B.V.– T.D. = 327 – 27 = Rs. 300.
SOLVED EXAMPLES :
(T.D; n; i are given, to find A)
1. The true discount on a bill due 6 months hence at 8% p.a. is Rs. 40, find the amount of the bill.
In the formula, ni1
AniD.T+
= , T.D. = 40, 21
126n == , i = 8/100 = 0.08
)08.0(211
)08.0.(21.A
40+
=\ , or, A = 1040 (in Rs.)
2 : (T.D.; A, i are given, to find n)
Find the time when the amount will be due if the discount on Rs. 1,060 be Rs. 60 at 6% p.a.
MATHS 1.33
)06.0(n1
)06.0.(n.106060,or,
ni1Ani.D.T
+=
+= or,
32n = yrs.
3. : (T.D.; A; n are given, to find i)
If the discount on Rs. 11,000 due 15 months hence is Rs. 1,000, find the rate of interest,
Here, A =11,000, 45
1215n == , T.D. =1000, i = ?
So we have, ,ori.
451
i.45,11000
1000+
= 12500 i = 1000 or i = 0.08. %8r =\
4. (If A, n; i; are given to find T.D.)
Find the T.D. on a sum of Rs. 1750 due in 18 months and 6% p.a.
06.0100
6i;yrs.23.yrs
1718n;1750AHere =====
So we get )app(50.14409.01
09.01750
06.0231
06.0231750
.D.T =+
´=´+
´´=
.50.144.Rs.D.T.reqd =\
5. Find the present value of Rs. 1800 due in 73 days hence at 7.5% p.a. (take 1 year = 365 days)
Here : A = 1800, 51
36573n == years; 075.0
1005.7i ==
60.26015.127
015.01015.01800
)075.0(511
)075.0.(51.1800
.D.T ==+
´=+
=
We know P.V. = A – T.D. \ P.V. = 1800 – 26.60 = 1773.40
6. The difference between interest and true discount on a sum due in 5 years at 5% per annum is
Rs. 50. Find the sum. [ICWA (F) Dec. 2005]
Let sum = Rs. 100, Interest = 100×5×0.05 = Rs. 25, 05.0100
5i ==
Again ;20Re05.1
2505.01
05.05100ni1
Ani.D.T ==+
´´=+
= So difference = 25 – 20 = 5
Diff sum ? = ,1000505
100 =´ .1000.Rssum.reqd =\
5 100
50 ?
7. If the interest on Rs. 800 is equal to the true discount on Rs. 848 at 4% When the later amount
be due?
T.D. = A – P.V. = 848 – 800 = Rs. 48, here A = 848, P.V. 800
Again T.D. = P. n.i or, 48 = 800 × n × 0.04, or, 211n = yrs.
MATHS1.34
AR
ITH
MET
IC
MATHS 1.35
SELF EXAMINATION QUESTIONS :
1. The true discount on a bill due 211 years hence %
214 p.a. is Rs. 54. Find the amt. of the bill?
[Ans. Rs. 854]
2. The true discount on a bill due 146 days hence at %214 p.a. is Rs. 17. Find the amount of the
bill (take 1 year = 365 days) [Ans. Rs. 1017]
3. When the sum will be due if the present worth on Rs. 1662.25 at 6%p.a. amount to Rs. 1,525.
[Ans. 211 yrs.]
4. Find the time that sum will be due if the true discount on Rs. 185.40 at 5% p.a. be Rs. 5.40
(taking 1 year = 365 days) [Ans. 219 days]
5. If the true discount on Rs. 1770 due212 years hence, be Rs. 170, find the rate percent.[Ans.
214 % p.a]
6. If the present value of a bill of Rs. 1495.62 due 411 years hence, be Rs. 1424.40; find the rate
percent. [Ans. 4% p.a.]
7. Find the percent value of Rs. 1265 due 212 years at 4% p.a. [Ans. Rs.1150]
8. The difference between interest and due discount on a sum due 73 days at 5% p.a. is Re.1. Find
the sum. [Ans. Rs. 10,000]
9. The difference between interest and true discount on a sum due 212 years at 4% p.a. is Rs.
18.20. Find the sum. [Ans. 20,000]
10. If the interest on Rs. 1200 in equal to the true discount on Rs. 1254 at 6%, when will the later
amount be due ? [ Ans. 9 months]
BILL OF EXCHANGE :
This is a written undertaking (or document) by the debtor to a creditor for paying certain sum of
money on a specified future date.
A bill thus contains (i) the drawer (ii) the drawee (iii) the payee. A specimen of bill is as follows
Stamp Address of drawer
percent.
C.K. Roy
1/1 K.K. Bose Rd.
Kolkata 700 012A.
P. Dhar
(drawer)
Bill of Exchange is two kinds
(i) Bill of exchange after date, in which the date of maturity is counted from the date of
drawing the bill.
(ii) Bill of exchange after sight, in which the date of maturity is counted from the date of
accepting the bill.
The date on which a bill becomes due is called nominal due date. If now three days, added with
this nominal due date, the bill becomes legally due. Thus three days are known as days of grace.
Banker’s Discount (B.D.) & Banker’s Gain (B.G.):
Banker’s discount (B.D.) is the interest on B.V. and difference between B.D. and T.D. is B.G.
i.e. B.D. = int. on B.V. = Ani ………(v)
B.G. = B.D. – T.D. and B.G. = interest on T.D.
)vi....(..........ni1
)ni(A.G.B
2
+=
B.V. – B.D. = Discounted value of the bill……(vii)
SOLVED EXAMPLES :
Problems regarding T.D. and B.D.
8. A bill for Rs. 1224 is due in 6 months. Find the difference between true discount and banker’s
discount, the rate of interest being 4% p.a.
.2402.148.24
)04.0.(211
)04.0.(21.1224
ni1AniD.T ==
+=
+= B.D. = Ani = 1224.
21 .(0.04) = 24.48;
B.D. – T.D. = 24.48 – 24= 0.08; \reqd. difference = Rs. 0.48 [ This difference is B.G. (0.48)
Again Int. on 24 (i.e., T.D.) ,48.0)04.0.(21.24 == i.e, B.G. = Int. on T.D.]
9. If the difference between T.D. and B.D. on a sum due in 4 months at 3% p.a. is Rs. 10, find the
amount of the bills.
B.G. = B.D. – T.D. = 10; 3112/4n == yrs.; 03.0
1003i == ; A = ?
B.D. = A ni, T.D. = Pni, B.G. = B.D. – T.D.= A ni – Pni = (A –P) ni
Now, (A – P) ni = 10 or (A – P) 31 . (0.03) = 10,
or, (A–P) 10001.0
10 ==
or, T.D. = 1000 (as T.D. = A – P.) Again T.D. = P ni,
or, 21.P (0.03) = 1000 or,
01.01000P =
\P = 1,00, 000. Now, A = P.V. + T.D. = 1,00,000 + 1000 = 1,01,000
\ reqd. amount of the bill = Rs. 1,01,000.
MATHS1.36
AR
ITH
MET
IC
Aliter : ni1
)ni(A.G.B
2
+= ,
03.311
03.31.A
10
2
´+
÷øö
çèæ ´
= , A = 1,01,000.
10 : The T.D. and B.G. on a certain bill of exchange due after a certain time is respectively Rs. 50
and Re. 0.50. Find the face value of the bill.
We know B.G. = Int. on T.D. or, 0,50 = 50 × n × i or, 01.05050.0ni ==
Now, B.D. = T.D + B.G. or, B.D. = 50 + 0.50 = 50.50
Again B.D. int. on B.V. (i.e., A)
or, 50.50 = A. ni or, 50.50 = A.. (0.01) or, 505001.050.50A ==
\reqd. face value of the bill is Rs. 5050.
Problems regarding discounting of bills.
11 : A bill of Exchange drawn on 4.1.81 at 5 months was discounted on 26-3-1981. If the banker’s
discount at 3% be Rs. 603.60; find the value of the bill and also B.G.
Unexpired days from 26 March to 7 June, 1981.
(M) (A) (M) (J) drawn on 4. 1. 81
5+30+31+7 = 73 period 5
(excluding 26 – 3 = 81) normally of grace 4. 6. 81
days of grace 3
leqully due on 7. 6. 81
Let the face value of the bill = Rs. 100, here 100
3i = = 0.03, 51
36573n == yr.
B.D. = int. on B.V. = 100. 1/5. (0.03) = 0.60
B.D. B.V
0.60 100 600,00,160.60360.0
100x =´=
603.60 x Hence, reqd. face value = Rs. 1,00,600
Now, 03.0
511
3.051100600
nii
)ni(A.G.B
2
2
´+
÷øö
çèæ ´
=+
= 6.30060.1
000036.0100600 =´=
Hence, B.G. = Rs. 3.60
[Observation : In this case, the discounted value of the bill = 1,00,600 – 603.60 = Rs.99,996.40
(i.e., this amount the holder of the bill will receive.]
12 : A bill of exchange drawn on 5.1.1983 for Rs. 2,000 payable at 3 months was accepted on the
same date and discounted on 14.1.83, at 4% p.a. Find out amount of discount.
[ICWA (F) June, 2004]
Unexpired number of days from 14 Jan to 8 April = 17 (J) + 28 (F) + 31 (M) + 8 (A) = 84
(excluding 14.1.83)
MATHS 1.37
1983 is not a leap year, Feb. is of 28 days. 41.18100
4365842000.D.B =´´= (after reduction)
Hence, reqd. discount = Rs. 18.41.
Drawn on period 5.1.83
Period 3
Nominally due on 5.4.83
Days of grace 3
Legally due 8.4.83
SELF EXAMINATION QUESTIONS :
Problems regarding T.D. B.D. B.G.
1. At the rate of 4% p.a. find the B.D., T.D. and B.G. on a bill of exchange for Rs. 650 due 4
months hence. [Ans. Rs. 8.67; Rs. 8.55; Re. 0.12]
2. Find the difference between T.D. and B.D. on Rs. 2020 for 3 months at 4% p.a. Show that the
difference is equal to the interest on the T.D. for three months at 4%. [Ans. Re. 0.20]
1. Find the T.D. and B.D. on a bill of Rs. 6100 due 6 months hence, at 4% p.a.
[Ans. Rs. 119.61;Rs.122]
4. Find out the T.D. on a bill for 2550 due in 4 months at 6% p.a. Show also the banker’s gain in
this case. [Ans. Rs. 50, Re.1]
5. Find the T.D. and B.G. on a bill for Rs. 1550 due 3 months hence at 6% p.a.
[Ans. Rs. 22.9]; Re.0.34]
6. Calculate the B.G. on Rs. 2500 due in 6 months at 5% p.a.
[ICWA(F) June, 2003] [Ans. Rs. 1.54]
7. If the difference between T.D. and B.D. on a bill to mature 2 months after date be Re. 0.25 at
3% p.a.; find. (i) T.D. (ii) B.D. (iii) amount of the bill. [Ans. Rs. 50; Rs. 50.25; Rs.10,050]
8. If the difference between T.D. and B.D.(i.e. B.G.) on a sum of due in 6 months at 4% is Rs. 100
find the amount of the bill. [ICWA(F) June, 2007] [Ans. Rs. 2,55,000]
[hints : refer solved problem no 9]
9. If the difference between T.D. and B.D. of a bill due legally after 73 days at 5% p.a. is Rs.10,
find the amount of the bill. [Ans.1,01,000]
10. If the banker’s gain on a bill due in four months at the rate of 6% p.a. be Rs. 200, find the bill
value, B.D. and T.D. of the bill. [Ans. (F) Dec. 2006][Ans.5,10,000; Rs. 10,200; Rs.10,000]
11. A bill for Rs. 750 was drawn on 6th March payable at 6 months after date, the rate of discount
being 4.5% p.a. It was discounted on 28th June. What did the banker pay to the holder of the bill?
[Ans. Rs. 743.62]
12. A bill of exchange for Rs. 846.50 at 4 months after sight was drawn on 12.1.1956 and
accepted on 16th January and discounted at 3.5% on 8 th Feb.1956. Find the B.D. and the
discounted value of the bill. [Ans. Rs. 8.18; Rs. 838.32]
MATHS1.38
AR
ITH
MET
IC
MATHS 1.39
13. A bill of exchange for Rs. 12,500 was payable 120 days after sight. The bill was accepted on
2nd Feb.1962 and was discounted on 20 th Feb. 1962 at 4%. Find the discounted value of the bill.
[Ans. 12,356.17]
14. A bill for Rs. 3,225 was drawn on 3rd Feb. at 6 months and discounted on 13th March at 8%
p.a. For what sum was the bill discounted and how much did the banker gain in this?
[Ans. Rs. 3121.80; Rs. 3.20]
Problems regarding rate of interest :
15. What is the actual rate of interest which a banker gets for the money when he discounts a bill
legally due in 6 months at 4% p.a. [Ans. 4.08% app]
16. What is the actual rate of interest which a banker gets for the money when he discounts a bill
legally due in 6 months at 5%. [Ans. %935 ]
17. If the true discounted of a bill of Rs. 2613.75 due in 5 months be Rs. 63.75; find rate of
interest. [Ans. 6%]
BOOK FOR REFERENCE :
1. Basic Mathematics and Statistics – N. K. Nag
2. A Text Book of ICSE Mathematics – O. P. Sinhal
NOTES
