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The document discusses the concepts of alligation and mixtures, focusing on the calculation of weighted averages and the mixing of different substances. It provides examples illustrating how to find average prices and ratios in mixtures, as well as methods for mixing substances with and without replacement. Various scenarios and mathematical approaches are presented to solve problems related to mixtures and their concentrations.
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Save Allegation and Mixtures For Later ALLIGATION AND MIXTURES
Introduction
Welghted Average
Iti observed thatthe average can be calculated only ifthe
weights ofall he factors are same Hence the weighted
zverage lea more generalized form of average Tis canbe
further understood with the following lustration,
Class A Class B
No, ofstudents 10 10
Averageage — I2yrs 16yrs
Now, ifwe combine oth these classes then the average age
ofall the students =*!2 = 22 14 years. This is one standard
example of average.
Letus see another example:
Class A Class B
No.ofstudents 10 4
Averageage = 12yrs ldyrs
Now, if we combine these two classes, then the average can
not be calculated by the above mentioned method, since the
‘weights attached to different averages are different.
Finding Expression for Weighted Average
Lower value
a 4
Weighted average
sum we me
(ataverage A,)
a
‘And we wre this as:
Higher value
(average)
Quanty
(at average As)
e)
mote
Higher prce-nverage price
‘geaniy igh priced) ~ Average price—Lower pre
Itis quite obvious that the ratio ofthe number of persons
items in different groups is proportionate to the deviations
of their average from the average of all the people combined,
‘This average ofall the members combined is known as
weighted average and is denoted by Ay. This process of
‘mixing the two groups is also referred as alligation,
Elements of Weighted Average
‘As we can see from the above derivation, there are five
quantities:
Number of members in Ist group (1)
Number of members in 2nd group (n:)
i. Average of Ist group (A)
Average of 2nd group (Az)
‘Weighted average (An)
Normally, in the case of weighted average, we get questions
in which one of these five elements is missing, and with the
help ofthe remaining four quantities, the value of that,
‘missing quantity is found. Different possibility (situations)
are given below (Y represents — data given, N represents
= data not given):
Sitution Ny
First
Second
‘Third
Fourth
ifith
Sixth
NAL Ae
z24<<<
z Mixture)
23 kg of copper in 25 kg mixture
1kg of copper
=2 tg minture
69 kg of copper
=2x 69 =25%23 =S75ko
Alloy = 575 ke
(d) Let the amount of water be x liter
© seamed ith on Scamner According to the question,
x43 3
z 2
Thus, Quantity of spirit in the mixture
x+3 =643 =Sliters
Alternate:
Spirit : Water
biog
3:2
Lunit more
unit = 3 liters
Bunits=9liters
(a) Water content in 40 liters of mixture
= 40 *10/100 = 4 liters
Thus, the mixtur
Let x litters of water is mixed
4+x 20 1
10-4 = 36 liters
+x 100 5
x =Sliters
Alternate:
10% 100%
(inthis (Pure when
water water is added
| 10%) it 100% water)
0
20% — > (Final water %)
(This mixture is 1 (This is the
£\
°
of aoe asset fe s 1 water added in
ratio)
by 5 tomake it 40
40 5 Liters
Note: You can also solve this problem by ratio
method,
Alternative: Change the percentage into the ratio
Initial Mixture Final mixture
10% water 20% water
{00 “6 oo = 5
(1 >Water, 10 Mixture) (1 Water, 5->
Mixture)
Water
Thus, Se
Water 1
Talk
Now,
Water Milk
Initial Ratio 1 Dag
FinalRatio te9 Aug
Remember water is added not milk, so make
milk equal
Water Milk
Initial Ratio : 36
S units add
Final Ratio 36
> 4436=40
Thus, 40 units = 40 liters
unit = Liter
Sunits =5 liters
a) Milk Water
Initial ratio 743: 3x3
FinalRatio 3.7: Tey
Thus, Remainder water is added, so make milk
equal
ao: 9 149 =30
40
21; 49 = 21449 =70
30 units = 30 liters
Lunits = 1 liters
40 units =40 liters
{b) According to the question
Pure alcohol = 100%
By Allegation,
15% 100%
NZ
32% > Final Alcohol
68 ”
This mixture — 4 1 > Thisis the
quantity
is of 400 mi | | of pure
© seamed ith on Scamner 2
Alcohol in ration
multiply by 100
400 [100]
Alternate:
we
unsaset (
20 units > 400m
1 unit -> 400/20
sunits 3425 = 100
‘hus, Atohol added othe misure = 100ml
(a) Accoring othe question,
4 (Profit, 10 >cP,)
8, =10+1 = 11 units
Thus, 11 units = 68.2
‘Thus, Cost price of 1 kg mixture
Rs. 62
(65-62)=3 : (62-60)=2
Required ra 2
(b) According to the question,
Wine Water Total
Initial Ratio 3 1
FinalRatio 1 9: 1 2x2
i
vo}
(a »
‘unit taken out and added
Hence Required part of quantity =4
Alternative :
Let the quantity of liquid drawn out
1
t-teee 2
12-3x =4— x4 ax
8=6x
xy
Hence required part of quantity = 2 =
Alternative
Let the quantity of lquié drawn out =x
3
ads 4
1
=
1-frtx
12-3x=4-x+4x
&
Hence Required part of quantity,
{(b) By allegation rule
80% 100%
(in the initial
(When pure gold is added
mixture gold
itis 100%)
is 80%) 4 95%
(100-95)=5 = (95-80)
=15
5 : 15
(This is the quantity
a 3
of gold added in ratio
x50
50
50 gm [150
em]
{d) According to the question,
dima = 4x01 = 27
gee = 3
Fal Quant of mik= nal Quant (12
x= Quantity taken out atatime
€ = capaty of ves
n=no, of process
=a (1-2)° = 81 (1 -
saixdxt 236
Thus, Quantity of water= 8136 = 45
Rati of mikand water in final minture
© seamed ith on Scamner 10.
a,
2,
13,
(a) According to the question,
Mille Water
Initial Ratio 7.2 Ba
Final Ratio 2.7 ley
Remainder waters added not milk so make milk equal
1“ 5 = 20unit
Wi?
20units = BOliters
1 unit=4 liters
Water added = 4 liters
() According to the question
wie Water Tota
(Take Lom)
Mixture A 4x2 5x2 =9._ 2 ae
oN
8
Mutues 5x3: 1x3
Final Mire «S82: 4x2
asin
ik + water
MistureA e120
Mintre 8 Bi 3
Final micure 30:8
By alegaton rule (ik)
8 1
5
(Milk in. (Milk in mixture B)
mixture A)
4 ‘\
s ' 2 =
2
(b) According to the question,
va
AN
Sa
# Remember water is added and not milk, so
make milk equal but here milk is already equal
15
Sunits
Total quantity of water in the new mixture.
= 40 liters.
(c) According to the question,
Initial mixture contains 15% water
‘Therefore milk content in the mixture 16.
=(100-15)% = 85%
By Allegation rule:
85% 100%
(Milk in
Mixture Pure milk)
"37.5%
2s 25
5 1
40% 40%
200 40 titers
‘Amount of milk added = 40 liters
{d) Let the price of the water be Rs. 0
‘According to the question
S.P. of the mixture = 20 Rs.
CP of the mixture = 20 x 100/125 =Rs. 16
Now using Allegation method
Chemical
Water
(25-16) =9
Ratio of water to chemical
16
{b) According to the question,
‘Average price of mixed tea
36
= 198012160 _ 120
16
=" 6
Rs, 257.60
Alternate:
Take Difference = 40 divide itin ratio 7:9
40x 22.5
Thus, 280- x = 225
x =280 -225 =Rs. 257.50
{b) According to the question
By allegation:
© seamed ith on Scamner 7,
18.
19.
(Alcohol in
first mixture)
30%
Nd (Alcohol in final mixture)
-
5 15
1 3
{a) According to the question
Milk: Water
7:2
) ai
73
Thus, Remember, water is added not milk, so
‘make milk equal. Here milk is already equal.
9 units 729
1 unit =81 units
The amount of water added =81 liters
(2) According to the question,
(Alcohol in second
mixture)
50%
Mik =: Water
Initial Ratio Fast das
Final Ratio Tes 1a
Thus, Remember, water is added, not milk, so
make milk equal
Initial Ratio 21 °) 24
4ur
Final Ratio 21: 7
24 units =40 liters
1 Unit = # titers
4 units = 2x4
=8=4 = 6
- 2 liters
(2) According to the question
Infirst alloy 1x65: 2x39 65
Second Alloy 2x39: 3 x39
195
Newalloy 5x15: 8x15 =13 15
First alloy 65: 130
Secondalloy 78: ‘117
New alloy 75: 120
Apply allegation
oP?
78
SN
(c) According to the question
10 liters of mixture taken out
Ratio will alsobe 4 : 1
Thus, Liquid Bis poured, there is no change in A
make A equal.
Ao: B
ao: 3
2a Baa
a7 eS unk
{10 liters ‘of tiquid)
Thus, Sunt» 10 liters
unit =2 ters
S units = 1Oliters
But j10 liters were initially taken out
Thus, Initia moctre = 10 +10 = 20 ers
© seamed ith on Scamner a.
2.
2a.
24,
auantty =£%20 = 16 liters
[erate Le the intl quantity of quid A and B
x and x
According to question
4x-8 2
¥-2+10
1x -24
> Initial quantity = 4x
(c} According to the question,
on
east
ja
3 units (added) =3 «25
(6) According to the question
AlloyA> 5.9: 3.2 =B.a ee
5 liters
AlloyB > 5 : 11 =165 Mixed
AlloyA > 10 : 6 =16
AlloyB > 5 : 11 =16
15:17
(2) According to the question
By Alligation
Gold silver
=e ye
(23) : (8)
(b) According to the question
Gold Copper = Tin Total
De2 : 3x2) 1x2 = 6x2 (Make
uentity
4 6 2 12 Equal)
Copper: Tin + Lead Tota
s 4 2 2
Weightoflead = 2 2 =Eke
25.
27.
{a} According to the question,
5. of mixed tea
s. 194.40
Profit = 20% == (1 > Profit, 5>CP.)
SP. =Gunits
6 unts= 29 = 324
S units - 5324 =Rs, 162
By Alligation
~ SL
\
(162-150)
2
(192-162)
{d)_ According to the question
When there is no profit no loss, here
cP =SP
Thus, Cost price of mixed sugar
Apply Alligation
s. 16/kg,
(16-15)
=1 =
> Ratio
{c)_ By Alligation
Initial Mixture
Water
© seamed ith on Scamner 29.
30.
40 [titers]
(b) According to the question
Wine Water
Former 3x18 2x18
18
latter 4x10; 5x10
90
Final 1x45 1x45
Wine Water
34 36
40 50
45 45
\ 4 40
48
a.
J °
Pail of forner andar = 8
Sunits 33
Lunt > 4
ants pt wl on
imesh
eta laa edsurepera unter
40% 400%
5 (Water in final Mixture)
Je 2
50% 10%
3 7
16 x16
20 16
(Liters) (Liters)
(d) According to the question,
Mixture ~ 60 ites
tatio of 2m EW aunt
Lunits > 20 liters
rh i= sles
x40 +
1 2
x40
80 Liters
Thus, 20 +x =80
x =60 liters
Water added = 60 liters
4). According to the question
SP of amixture of Tea
CP of amixture of Tea = 324x100
‘Thus, Now using Aligation,
Tea-1 Tea-2
320 250
{Ratio of
Quantity)
(270-250)=20: (320-270)
2 : 5
{c} According to the question,
Container:~
5:32:
Container (
me's
Werer = 3
Container (ii):
Milk 2
Water
Container (ii): ~
Mi 3 ag
water = 3 =
Container (iv): -
Milk 7
Water 4 475
Thus, The quantity of mik relative to water
minimum in container I
© seamed ith on Scamner 33.
35.
36.
(b) According to the question,
Mature =
ED aunts
Ratio of
Biycerine
4units > 240
1 unit > 60
Thus, ze
‘Giycerine ~ 3x60 ~ 300
Water Glycerine
60 180,
2 3
4x60
120
Thus, 60+x
Quantity of water added = 60
(a) According to the question,
Acid Water
Initial te =2. Baa
solution 1 unit
Final 145= 3) Qe
Solution
Final Solution
=3+6 =9 units
‘As we know that only acid is added so water is
same initially and finally
We know 5 liters acid is added so,
1 unit 5 liters
9 liters > 5x9 =4S5 liters
Thus, Final mixtures = 45 liters
Final mixtures = 45 liters
(2) According to the question,
Misture= 25 liters
Ratioof AM = 2S. 5 units
S units > 25 liters
4 unit 5 liters
Acid 8 _ 20
Weer ~~
Acid Water
Initial «20:5
——
Final Ratio 20 3
bo: 4a
(a) According to the question,
Water Milk Total
Vessel-1 3.6=24 day Tae
Vessel-2 5, Be
Final Rati 33
Note: If two different solution are mixed then.
37.
38,
39,
censure that the quantity of both solution are same.
{b) According to the question,
‘Acid Water
VesselA 4: 3
Vessel 5: 3
Now using alligation,
~
Fie»
35
Fnatnavo : [7
fa) Acid
Vessel A 3 1
Vessel B'S. 238
Use Alligation
Ratio of 1 2
{a) According to the question,
Acid Water
VesserA 5: 2
Vessel-B |B 5
Now using Alligation,
A 8
© seamed ith on Scamner a1.
42.
=
*
&
a
Quantity [7 : 2)
(a) According to the question,
saa ere
vem “a 2 “3
vows 2 23
nov eiecaneatin
ri °
£ t
a
12)
G-3)
feet
Quantity (7: 3] 45,
(c) According to the question,
feeureet "coon
wcure-2) = 36ers
Sranus-i aawat
EE 1S, tountts
InMinture2 ratioot
Bee 2S 12 ats
Teunts 50 iter
Loni > 2 ters
aunts > 36
Lunt > 3iters
In Mixture -1 222
irae f
us, InMiture -2 gee 22 28
Thus ee 2 Water ba
Ratio of spirit and water
Teens 27:28 46.
(2) ‘Acid: Water
Vessel A 20: 3
Vessel B 4 3
Now, using alligation,
a 8
NZ)
4 doa
G @7G-
Ratioof 10: 14
quantity 215 7)
{a) According to the question,
‘Acid : Water
Vessel A 5: 3
Vessel B 5: 4
Now, using alligation,
G
Ratioo "24 36
quantity 31: 3)
(6) According to the question
Water Syrup_—_Total
vi rr)
Nia) seme
+ dase =10
Water added in final = 2 = 2
According to the question,
ilk Water
VesselA 4 : 3
Vessel 8 2 3
Now using Alligation,
8
Final
Ratio (7 : 5]
{€)_ According to the question,
Initial quantity = 60 kg
Asweknow that,
Final quantity = Initial Quantity (1—
Yotume taken out
Talla quatlty
Where 'r’ number of times volume taken out
Final quantity
© seamed ith on Scamner a7.
49.
=60(1-£)° = 60x2
Final quantity of milk
= 43.74 kg
(d) According tothe question,
Milk Water
VesselA 8 5
VeselB «5s 2
New mixture containing 69:9
Milkie, = 20> = milk
Thus, Now using Alligation,
51
7] > Final Ratio oe.
(d) According to the question,
Gold Copper
Tal 2a a4
Total
9.2218,
)
AD
Same
B> 2 ua 18
7 7 5]
(c) According to the question,
AB Total
sane untaasca
tunis 99 ter
unit 92 titers
12 units 92x 12 = 27 liters
=gx 36 = 21 liters
53,
(b) According to the question,
Milk: Water
Vessel A 3 : 2
VesselB 7 a 3
Now using alligation,
Vessel-1
3
5 6
Vessel -2
Quantity [2 7 2
{d) According to the question,
CP ofthe mixture
215 x29+25%20 =Rs. 935
SP of the mixture
22740 =Rs. 1080
Profit= SP —CP
1080-935 =Rs. 145
{d) According to the question,
(b) According to the question,
Ga eG
Quantity [4 : n
(d) According to the question,
<<
© seamed ith on Scamner 55.
56.
Finally > 4 5
2units > 5 liters
unit ers se
units 32x 4 = 10 ters
anti of alcoho
sites
(b) According to the question,
Mixtures = 15 liters
Ratio of Het => Sumies
Alcohol Water =
iniaty 3 2
a
Final 53 re
Percentage of alcohol in new mixture
2x 100 = 162%
(d) Capacities of Vessels
Milk Water Total
Mixture
va 6 2 es
vas a 1 Sha
v3 4 a Sher
Equate the Mixture
ilk Water
vas (5 10 as
V2> (28 7 35, eo
v3> (8 7 - 3
Capacities Mo: W fotal Mixture
vad (75 30 105
v2> (66 “4 70
va> (8 7 35
Water taken out
> of waterin -1)+4 of materia V-2) +2
water in {V - 3)
3 xaod deity
3 1b+761 =18
Sirry mixture wl be
£105 +370 +2% 35
37 108 FRE GX
2X 100 24%
% ofwater =22
(2) According to the question By using Aligation
method
15%
30%
\
40% a
(40-30) = 10
Ratio 2:3
(a) According to the question
CP of the mixture
30x9.5 +40%85
=285 +340 = Rs. 625
SP. of the mixture = 8.90% 70
fs. 623
Loss =CP-SP,
Loss =625 623 =Rs.2
(a) According to the question,
SP. of the mixture = 40P per ke
CP. of the mixture =40 x22 =32P perks
Now using Alligation.
(30-15)=15
(32-24)=8 — (42~32) = 10
Ratio of Quantity 4 :
5 units > 25 kg.
1 unit > 5 kg
Aunits 5x4 =20kg
Salt at 42 P per kg was
(b) Let the Price of Water
=Rs. 0
According to the solution,
CP of Pure Milk =Rs. 1.80
CP of the Mixture = Rs. 0.90
Now using Alligation method.
Milk Water
1.80 °
\Z
090
iN
(090-0) (2.8-09)
=0.90 =0.90
1
0 ke
x16 1x16 16 liters (Given)
Ratio of
quantity [16 liters)
{a) According to the questions
By using Alligation method.
Gold-1 Gold-2
12 16
aN
© seamed ith on Scamner 62,
63.
6s.
(16-15)-1 (15 -12)=3
1 3 Ratio of Quantity
(c) According to the questions
Sugar Solution = 300 kg.
300 = 120kg
sugar
120
wf
T 1
x 180
___180
Let the sugar added =e,
120+ x= 180 kg,
x = 60kg
(@ np &
Mik Water
7 1-9 units = 729
567 162 Lunit=81
o.
Mo: ow
nial > 7 0:2
after unit
adding
Water 7 0:3
‘Always milk will be same
ie. Lunit of water will be added = 1
unit 84 mil iter
68,
@ oo woe
res 2 7)xSx3
we 4 1 5)x7x2
aa 1 = 5)x7x1
oF, (75:30) 2 (56 :14)x3 (28 £7 pt
25 10
3B: 7
aoa
37 18 =75
% of water in mixture
=! 100 = 24%
(a) Alloy a Alloy 8
60 kg 100kg.
3 2 1 4
36 kg 24kg 20k 80kg.
(Lead) (Tin) (Tin) (Copper)
Total
Tin 24420 =
e
(c) Let first blend is 2 kg and second blend is 3 kg.
Total cost price = (35 2) + (40 x3) =70+120 >
as. 190
Total selling price = (1 x46)+ (455) =266
:
or sk
Profit percent = “eterel
= “po = 40%
(b) Acid: Water
Glsst of 23 =
Total Pott « 199
Mixture
5) x63
Glas? (3 4
Glass 1 126 : 189
Glass 2 135; 180 = 315
Glass 3.140 : 175
[401] [544]
(d) According to the question,
SP ofthe mixture =Rs. 320
Gain= 20%
Thus, CP. of the mixture = 320 x 292
= as,
Now, Using alligation method.
Tea-1 Tea -2
180 280
© seamed ith on Scamner 70.
n.
200-8 ME 49 = 288
Ratioof-> 40: 260
Quantity 2: 13
(c) According to the question,
Mixture of copper and aluminium = 2000 gm
30% is copper means = 2 x 2000
‘Aluminium
70
80
‘Aluminium
> Lunit = 200m
1:4 94
We have to equal Copper amount because only
Aluminium is added.
Copper: Aluminium
3 : 7
>) + Sunits
(b) cy
Quantity 3:2 >5 units
According to question
SP. of Lunit= Rs. 11
‘Then CP of S units = Rs, 50
Let Y's Cost =2 Rs.
X's cost=2+2 Rs.
CP, = Quantity x Cost
50= 3 (2+2)+22
50 =32+6+2z
BO Rs.
Cost of X= 8.80+2 > 10.80 Rs.
(2) Total milk = 60 liters
Drawn off = 12 liters
rR.
2.
74,
Font Quanity, _ (4
Tritial Quantity ~
x = Replaced Quantity
apacity
lumber of process
Final Quant 2)
Taitiat Quantity ~ 460,
Ratio of milk and water in the resultant mixture =
16:9
fa) 1%. Vessel 2" Vessel
S¢
« S,
Petrol :Kerosene ==
ibs 8 =7:4
(4) According tothe question.
A 8
initil = 7+
fal 7
Solution
units 9
run-> 2
16 unite 9216 =
The capacity ofthe can = 36 ters
(c) Sprit Milk =Total Capacity Ratio
14 42 2
nom +4 3
Wo7 + 3 =10 4
Sprit Milk Total
| 42848 Lzel2 Soa
W166 40224 15.203
7.1284 3.12236 10,304
Total Ratio =198 : 72
m4
JK
© seamed ith on Scamner
