405 math solved by md.mutaher hussain [www.onlinebcs.com]

1. LMixture and allegations:
By:MD MUTAHER HUSSAIN
1. A Container contains 192 liter of Milk. A seller draws out x% of Milk and replaced it
with the same quantity of water. He repeated the same process for 3 times. And thus Milk
content in the mixture is only 81 liter. Then how much percent he withdraw every time?
#Solution:
Initial mixture of milk= 192 L
A seller draws out = x% of milk.
ATQ,
192*(1-x%of 192/192)^3= 81
=> (1-x/100)^3= 81/192
=> (1-x/100)^3= (3*27)/(3*64)
=> (1-x/100)^3= 27/81
=> (1-x/100)^3= (3/4)^3
=> (1-x/100)= 3/4
=> 1-3/4= x/100
=> 1/4= x/100
=> 4x= 100
=> x= 25%
So, initial replace= 25% of milk.[Answer.]
[ ऀ
C=P*(1+r/100)^n
, C= !
P= " # !
n= $( , , % &' %)

2. ( ) * +, # &) ऀ % ) -$
. /01 ऀ , 2 2 3 $ , 4
5/ 3 $ ऀ & C 3&) 4 ! C 6 3 ऀ
C F 7 & ऀ & )8 9 4 ऀ
n 7 )8 $ n 3 - ऀ
3 /:: & 3 -$ ऀ
)% r 7 ) & 3$ ; r 7
& ऀ
& 3 4 % <
F= I*(1-R/I)^n
F= Final amount
I=Initial amount
n=repetition time
R=Replace.
' -' ( 3 $ = & ]
02.
A Jar contains 30 liters mixture of Milk and Water in the ratio of x:y respectively. When
10 liter of the mixture is taken out and replaced it water, then the ratio becomes 2:3. Then
what is the initial quantity of Milk in the Jar?
#Solution:
Here, Milk = x
Water = y
ATQ,
x+y= 30
=> y= 30-x
Now, 10 L mixture is removed and replaced by water,
So, Milk Remain= x-10*x/30= 2x/3

3. W Remain= y-10*y/30+10
=(2y+30)/3
ATQ,
(2x/3)/(2y+30)/3= 2/3
=> 2x/(2y+30)= 2/3
=> 6x= 4y+60
=> 6x= 4*(30-x)+60
=> 6x= 120-4x+60
=> 10x= 180
=> x= 18
So, Initial Milk= 18 L[Answer.]
03.A Container contains ‘X’ Liters of Milk. A thief stole 50 Liters of Milk and replaced it
with the same quantity of water. He repeated the same process further two times. And
thus Milk in the container is only ‘X-122’ liters. Then what is the quantity of water in the
final mixture?
#Solution:
Final mixture= X-122
Initial Mix= X
ATQ
X-122= X(1-50/X)^3
=> X-122/X= (1-50/X)^3
=> (X-122)/X= 1-150/X+7500/X^2-2500*50/X^3
=>X^2(X-122)= X^3-150X^2+7500X-1,25,000
=> X^3-122X^2+150X^2-7500X+125000=0
=> 28X^2-7500X+1,25,000= 0
=> 7X^2-1875X+31,250= 0
=> (X-250)(7X-125)= 0
=> X= 250 L
So, Final M= (250-122)= 128

4. So, Water= (250-128)= 122L(Answer.
04. A Jar contains 100 liters of Milk a thief stole 10 liter of Milk and replaced it with
water. Next, he stole 20 liter of Mixture and replaced it with water. Again he stole 25 liter
of Mixture and replaced with water. Then what is the quantity of water in the final
mixture?
[ 4 )> ( ? $ ( @=]
#Solution:
* & /:: %)! ऀ A!) %)!ऀ
Bऀ # ( ) /: ) ऀ "# + C $ A!) %)! ऀ )
A!) %)! ) ऀ
% /: ) 0: ऀ
= ) 7 ; ऀ
) ऀ & )8 & & 4 D ऀ
%) ! = = 0:
= /:
* & A!) %)! ऀ 6 ऀ & )
%) ! # # 6 ) ऀ > 3 = ' )8 & ऀ ?
) & 3 & ऀ 3 , +, ऀ
% ) %) ! + ) & &=??
= E= # ) ) 3 F + ) & ऀ
= %) ! == /: : 0:
= / : 0
! %)! = 0
=
" G ) 1: ) ऀ
0 + = 1:

5. => = 1
& %) ! = & 1: !'?
%) ! == 0*1= /5
== /*1= 1
& 3 , &> = %)! E =?
3&) /5 ) 3 $ ? ऀ %)! = 0:-/5= H1
= /:-1= 5
) 36$ = B %$ E = = ऀ
E= ) 1: ऀ & 3 &> =
= 5+1:= 15
? > $ @ ऀ = ) # $ ऀ
&> %) ! == H1
= = 15
? E= # ) + ) & ) ऀ
& , %)! : = H1 : 15= / 5 : H
E= # 1I ) ऀ
! , %) ! == /5
= H
H +/5 = 1I
=> = /
& %)! ) 3 /5*/= /5
= H*/= H
( -Jऀ ) # $ 6 K % < ? $ L< ? ऀ
&> %) ! = H1-/5= IM

6. = 15-H+1I= MN
05. In a 250 liter of a mixture of Milk and Water, Water is X%. The milkman sold 50
liters of the mixture and replaced same quantity with water. If the percent of Milk in final
mixture is 64%, then what is the percentage of Milk in the initial mixture?
#Solution:
Final milk= 250*64/100= 160 L
Initial Milk= 250-250*x/100
= 250-5x/2
=(500-5x)/2
ATQ,
(500-5x)/2- 50*(500-5x)/2/250= 160
=> (500-5x)/2-(500-5x)/10= 160
=> (2500-25x-500+5x)/10= 160
=> 2000-20x= 1600
=> 20x= 400
=> x= 20
Initial milk= 500-5*20/2= 200
%= 200*100/250= 80%
06. A jar contains ‘x’ liters of Milk, a seller withdraws 25% of it and sells it at Rs.20 per
liter. He then replaces it water. He repeated the process total three times. Every time while
selling he reduces selling price by Rs.2. After this process Milk left in the mixture is only
108 liters so he decided to sell the entire Mixture at Rs. 15 per liter. Then how much profit
did he earned if bought Milk at Rs.20 per liter?
#Solution:
x*(1-25% of x/x)^3= 108
=> x(1-1/4)^3= 108
=> x= 108*4*4*4/(3*3*3)= 256 L
CP = (256*20)= 5120
SP= 256*15+ 25% of 256 *(20+18+16)

7. = 256*15+64*54= 7296
Profit= (7296-5120)= 2176[Answer.]
07. ‘X’ Liters of the mixture contains Milk and Water in the ratio 4:3. If 13 liters of Water
is added then the ratio becomes 1:1. Then what is the final quantity of water in the
mixture?
#Solution:
Let, Initial milk= 4x
Initial water= 3x
ATQ,
(3x+13)/4x= 1/1
4x= 3x+13
x= 13
So, water at final = 3*13+13= 52 L(Answer.)
08. A jar contains 200 litres of milk a thief stole X litres of milk and replaced it with
water. Next, he stole 40 litres of milk and replaced it with water. Again he stole 50 litres of
milk and replaced with water. If quantity of water in the final mixture is 92 litres. Then
what is the value of X?
#Solution:
Initial mixture= 200L
Final milk= (100-92)= 108L
ATQ,
200*(1-x/200)(1-40/200)(1-50/200)= 108
=>200*(200-x)/200* (4/5)*(3/4)= 108
=> (200-x)*12= 108*20
=> (200-x)*3= 108*5
=> 3x-600= 540
=> 3x= 60

8. => x= 20L
[Answer.]
09. From a 200 L container,a thief has stolen 10 liters of milk and replaced with water.He
repeated the process for three times,then the ratio of milk to water become 343:169.the
initial amount of milk in the container?
#Solution:
Let, intially mix.= x
ATQ,
x*(1-10/x)^3/x= 343/512
(1-10/x)^3= (7/8)^3
1-10/x= 7/8
1/8= 10/x
x= 80L(answer)
Law:x*(1-R/x)^n=Final mixture
X= initial amount
R= Replacement
n= repetition time
10. A Jar contains a mixture of Milk and Water 18 and 12 Liters respectively. When ‘x’
liter of the mixture is taken out and replaced with the same quantity of Water, then the
ratio of Milk and Water becomes 2:3. Then what is the quantity of Water in final
Mixture?
#Solution:
Milk : Water = 18:12
M renain= 18-18x/30= 18-3x/5= (90-3x)/5
W remain = 12-12x/30+x= 12-2x/5+x
= (60+3x)/5
ATQ,
(90-3x)/(60+3x)= 2/3
270-9x= 120+6x

9. 15x= 150
x=10L
Final water = (60+30)/5= 18 L(Answer.)
Mixture and allegation
Part-2
Solved by:MD MUTAHER HUSSAIN
11.Vikram covered 180 km distance in 10 hours. The first part of his journey he covered
by Car, then he hired a Rickshaw. The speed of the car and rickshaw is 25 kmph and 15
kmph respectively. The ratio of the distances covered by the car and the rickshaw is
#Solution:
Total time= 10 hr
Speed of car= 25 km/hr & Rickshaw= 15 km/hr
Let, he travel by car = t hr
Travel by rickshaw= (10-t) hr
ATQ,
t*25+(10-t)*15= 180
25t+150-15t=180
10t=30
t=3
Travel by car:Rickshaw=3*25:7*15=75:105=5:7[Answer.]
Alternative:
Average speed= 180/10= 18km/hr
25_________15
_____18____
3_________7
Distance car:Rickshaw= 3*25:7*15= 5:7
Alternative:
Let cover by car= x km

10. Cover by rickshaw= (180-x) km
ATQ,
x/25+(180-x)/15= 10
(3x+900-5x)= 750
900-2x= 750
150= 2x
x=75
So, cover by car =75 km
And, cover by rickshaw= (180-75)= 105
Raito of distance car:Rickshaw=75:105= 5:7[Answer.]
12.A mixture of wheat is sold at Rs.3 per Kg. This mixture is formed by mixing the Wheat
of Rs.2.10 per kg and Rs.2.52 per kg. What is the ratio of price of cheaper to the costlier
quality in the mixture if the profit of 25% is being earned
#Solution:
Cost price of mixture =3*100/125= 3*4/5= 12/5= 2.40RS/kg
Let, the amount of first part= x kg
Amount of second part= y kg
ATQ,
2.10x+2.52y= 2.4(x+y)
=> 2.4x-2.10x= 2.52y-2.4y
=>. 30x=.12y
=> 30x= 12y
=> x:y= 12:30= 2:5
Alternative:
Let, the amount of first part= x
Amount of second part =y
Sell price of mixture= 3
Since, 25% profit,
SP of first part= 125% of 2.10
= 5*2.10/4= 10.50/4= 2.625
SP of 2nd
part= 125% of 2.52= (5*2.52)/4= 3.15
ATQ,
2.625x+3.15y= 3(x+y)

11. => 3x-2.625x= 3.15y-3y
=>. 375x=. 15y
=> x:y=. 15/.375
=> x:y= 1/2.5
=> x:y= 1:2.5= 10:25= 2:5[Answer.]
13.From the 50 liters of a chemical solution, 5 liters of chemical solution is taken out and
after it, 5 liters of water is added to the rest amount of chemical solution. Again 5 liters of
chemical solution and water is drawn out and it was replaced by 5 liters of water. If this
process is continued similarly for the third time, the amount of chemical solution left after
the third replacement
#Solution:
Initial amount= 50 L
Replacement= 5 L
Final = 50*(1-5/50)^3= 50*45/50*45/50*45/50
= 50*(9/10)*(9/10)*(9/10)= 36.45L(Answer.)
Alternatives:Manual process:
Initial amount = 50 L
Replace by 5 L water.
So, chemical solution:Water= 45:5= 9:1(after first replacement)
2nd
replacement:
C remain= 45-9*5/10= 45-9/2= 45-4.5= 40.5
So, W= 9.5
C:W= 40.5:9.5
3rd
Replacement :
C remain= 40.5-40.5*5/50= 40.5-4.05= 36.45 L(Answer.)
14. From a container of milk, which contains 200 liters of milk, the seller replaces each
time with water when he sells 40 liters of milk(or mixture). Every time he sells out only 40
liters of milk(or mixture). After replacing the milk with water 4th time, the total amount
of water in the mixture is
#Solution:
Initial Amount= 200L

12. Replacement= 40L
Final water = 200-200*(1-40/200)^4
= 200-200*(160/200)^4
= 200-200*(4/5)*(4/5)*(4/5)= 118.08 L[Answer.]
[If you want to do this manually follow the previous math method]
15. A jar was full with Milk. A person used to draw out 20% of the Milk from the jar and
replaced it with water. He has repeated the same process 4 times and thus there was only
512 gm of milk left in the jar, the rest part of the jar was filled with the water. The initial
amount of milk in the jar was:
#Solution:
Let, Initial Milk = x L
Final milk = 512 gm
Repetition time, n= 4
Replacement = 20% of x
ATQ,
I*(1-R/I)^n= 512
x*(1-20%of x/x)^4= 512
x*(1-20/100)^4= 512
x*(1-1/5)^4=512
x*(4/5)^4=512
x= 512*(5/4)*(5/4)*(5/4)*(5/4)
x= 1250 gm or 1.25 kg[Answer]
16. From a container of Milk, a thief has stolen 15 liters of milk and replaced it with same
quantity of water. He again repeated the same process. Thus in three attempts, the ratio of
Milk and water became 343:169. The initial amount of Milk in the container was:
#Solution:
Let, the container contain = x L of milk.

13. Replacement= 15 L
Final amount of milk= 343
So, initial amount = 343+169= 512
ATQ,
x*(1-15/x)^3/x= 343/512
(1-15/x)^3= (7/8)^3
(1-15/x)= 7/8
1-7/8=15/x
1/8= 15/x
x=120L[Answer.]
17. The ratio of Solution “A” and Solution “B” in the container is 3:2 when 10 liters of the
mixture is taken out and is replaced by the Solution “B”, the ratio become 2:3. The total
quantity of the mixture in the container is:
#Solution:
Ration of A:B= 3:2
10 L taken out and replaced by B
So, A remain= 3x-3*10/5= 3x-6
And B remain= 2x-2*10/5 +10
= 2x+6
ATQ,
(3x-6)/(2x+6)= 2/3
9x-18= 4x+12
5x=30
x=6
The total quantity of mixture= (3x+2x)= 5x= 5*6= 30 L[Answer.]
18. From a container, 6 liters Solution “A” was drawn out and was replaced by water.
Again 6 liters of the mixture was drawn out and was replaced by the water. Thus the
quantity of Solution “A” and water in the container after these two operations is 9:16. The
quantity of the mixture is:

14. #Solution:
Let, initial mixture= x L
Final amount= x*(1-6/x)^2
Solution A in final:Initial= 9:(9+16)= 9:25
ATQ,
x*(1-6/x)^2/x= 9/25
(1-6/x)^2= (3/5)^2
1-6/x= 3/5
2/5=6/x
2x=30
x=15
19. The diluted Milk contains only 8 liters of Milk and the rest is water. A new mixture
whose concentration is 30%, is to be formed by replacing Milk. How many liters of the
mixture shall be replaced with pure Milk if there was initially 32 liters of water in the
mixture?
#Solution:
Milk = 8 L
Water = 32 L
Let Replacement= x
Milk Remain= 8-8x/40+x= 8-x/5+x= (40-x+5x)/5= (40+4x)/5
Water Remain= 32-32x/40= 32-4x/5
= (160-4x)/5
ATQ,
(40+4x)/(160-4x)= 3/7
280+28x= 480-12x
40x= 200
x= 5 L [Answer.]
20. In a school, the average weight of boys in a class is 30 kg and the average weight of
girls in the same class is 20kg. If the average weight of the whole class is 23.25 kg, what
could be the possible strength of boys and girls respectively in the same class?
#Solution:
Let, the number of boys=x

15. Number of girls=y
ATQ,
30x+20y=23.25(x+y)
30x+20y= 23.25x+23.25y
30x-23.25x= 23.25y-20y
6.75x= 3.25y
x:y= 3.25:6..75= 325:675=13:27
Number of Boys:Girl=13:27[Answer.]
#Mixture_Allegation_Part_3
Solved by::MD MUTAHER HUSSAIN
21. A vessel is filled with liquid, 4 parts of which are water and 5 parts syrup. How much
of the mixture must be drawn o and replaced with water so that the mixture may be half
water and half syrup?
#Solution:
Total part= 4+5= 9
4_______9
___4.5___
4.5______.5
45:5= 9:1
Replace = 1/(1+9)= 1/10[Answer.]
#Solution_2::
Let, Replace = x L
Water remain after replacement = 4/9- x*4/9+x
= (4/9-4x/9+x)
= (4+5x)/9
Syrup remain = 5/9-5x/9= (5-5x)/9

16. ATQ,
(4+5x)= 5-5x
10x= 1
x=1/10[Answer.]
22. Rice worth Rs. 126 per kg and Rs. 134 per kg are mixed with a third variety in the
ratio 1 : 1 : 2. If the mixture is worth Rs. 177 per kg, the price of the third variety per kg
will be:
#Solution:
Let, the price of third vareity = x
ATQ,
126*1+134*1+x*2= 177*4
260+2x= 708
2x=448
x= 224[Answer.]
23. One quantity of rice priced at Rs 9.30 per Kg is mixed with another quality at a
certain rate in the ratio 8:7. If the mixture so formed be worth Rs 10 per Kg, what is
the rate per Kg of the second quality of wheat?, what is the rate per Kg of the second
quality of wheat?
#Solution:
Let, the price of other = x
ATQ,
9.30*8+7*x= 10*(8+7)
7x= 75.6
x= 10.8[Answer.]
#Solution_2::
9.30________X(other price)
______10_____
8__________7

17. ATQ,
7X-70= 5.6
7X= 75.6
X= 10.8[Answer.]
#Solution_3:
Each product less=10-9.3= 0.7rs
So, 8 product less= 0.7*8=5.6 rs
So, Price of another quality
(7*10+5.6)/7=10.8 RS
24. How many kgs of rice of variety-1 costing Rs.42/kg should a shopkeeper mix with 25
kgs of rice of variety-2 costing Rs.24 per kg so that he makes a prot of 25% on selling the
mixture at Rs.40/kg?
#Solution:
Let, 42 kg quantity rice = x
ATQ,
(42*x+25*24)*125/100= 40*(x+25)
(42x+25*24)*5= 160*(x+25)
210x+ 3000= 160x+4000
50x= 100
x= 20 (Answer.)
#Solution_2:
SP given = 40
CP= 40*4/5= 32
42_________24
______32____
8__________10
4:5
5===25

18. 4===20 kg[Answer.]
#Solution_3::
SP 1= 42*5/4= 105/2
SP 2= 24*5/4= 30
Final SP = 40
105/2________30
_____40_______
10_________25/2
20________25
4:5
Now,
5 part ==25
4 part ==20 [Answer.]
25. An alloy contains Brass, Iron and Zinc in the ratio 2:3:1 and another contains Iron,
zinc and lead in the ratio 5:4:3.If equal weights of both alloys are melted together to form
a third alloy, then what will be the weight of lead per kg in new alloy?
#Solution:
B:I:Z= 2:3:1= [6]
I:Z:L= 5:4:3= [12]
LCM of 6& 12= 12
So, first ratio have to multiplied by = 12/6= 2
B:I:Z= 4:6:2
Final; I:Z:L= 5:4:3
So, Total = 24
Led= 3

19. Wt. Of led = 3/24= 1/8[Answer.]
26. A can contains 50 litres of milk. 10 litres of this milk is taken out and replaced with
water. This process is repeated twice. Find the amount of remaining milk in the mixture
#Solution::
We know, F= I*(1-R/I)^n
Where, F= Final mixture
I = Initial
R= Replacement
n= Repetition time
So, 50*(1-10/50)^3= 50*(4/5)*(4/5)*(4/5)= 128/5 L [Answer.[
#Solution_2::
Initially = 10 L Replace
M:W= 40:10
Again 10 L replace;
M= 40-10*40/50= 40-8= 32
W= 18
Again 10 L Replace:
M= 32-32*10/50= 32-32/5= (160-32)/5= 128/5 L (Answe.)
27. A milkman mixes 6 litres of free tap water with 20litres of pure milk. If the cost of
pure milk is Rs.28 per litre the % profit of the milkman when he sells all the mixture at
the cost price is
#Solution:
Total sell = 26 L of Milk
SP = 26*28= 728

20. CP= 28*20= 560
Profit= (768-560)= 168
% profit = 168*100/560= 30%[Answer.]
28. A 144 litres of mixture contains milk and water in the ratio 5 : 7. How much milk need
to be added to this mixture so that the new ratio is 23 : 21 respectively?
#Solution:
Milk= 144*5/12= 60L
Water = (144-60)= 84 L
ATQ,
(60+x)/(144+x)= 23/44
2640+44x= 3312+23x
21x= 672
x= 672/21
x= 32 L
#Solution_2::
ATQ,
(60+x)/84= 23/21
60*21+21x= 23*84
21x= 672
x= 32 L
#Solution_3::
Initial Ratio; Milk:Water= 5:7
Final Ratio; M:W= 23:21
Now, 21 part==84
23 part= 23*4= 92 L
29. In a 100 litre mixture of milk and water, the % of water is only 20%. The milkman
gave 25 litres of this mixture to a customer and then added 25 litres of water to the
remaining mixture. What is the % of milk in the fi nal mixture?

21. #Solution::
Initial Mix= 100 L
Milk = 80 L
Water = 20
M:W= 80:20
Now Replace = 25 L
M= 80-25*80/100= 80-25*4/5= 80-20= 60 L
So, water = 40 L
%= 60*100/100= 60%[Answer.]
30. vessel is lled with 120 litres of Chemical solution, Acid “A”. Some quantity of Acid “A”
was taken out and replaced with 23 litres of Acid “B” in such a way that the resultant
ratio of the quantity of Acid “A” to Acid “B” is 4:1. Again 23 litres of the mixture was
taken out and replaced with 28 litre of Acid “B”. What is the ratio of the Acid “A” to Acid
“B” in the resultant mixture?
#Solution:
A:B= 4:1
1==23
4= 92
A now = 92-92*23/115= 92-92/5= 368/5
B now = 23-23*23/115+28= 51-23/5= 232/5
A:B= 368/5:232/5= 368:232= 46:29[Answer.]
#Mixture_Part_4
#Solved By:Md Mutaher Hussain

22. 31. Two vessels A and B contain a mixture of Milk and Water. In the first vessel (i.e)
Vessel A has the ratio of Milk to water is 8 : 3 and in the second vessel, Vessel B has the
ratio of 5 : 1. A 35 litre capacity vessel is filled from these two vessels so as to contain a
mixture of Milk and water in ratio of 4 : 1. Then how many litres should be taken from
the first vessel.
#Solution_1:
First vessel: M:W= 8:3[11]
2nd vessel: M:W= 5:1[6]
Final: M:W= 4:1[5]
LCM of 11,6,5= 330
M at first = 8*330/11= 240
M at 2nd = 5*330/6= 275
M at final = 4*330/5= 264
275______264______240
(275-264)_______(264-240)
11:24
Taken from first vessel= 35*11/35= 11 L[Answer.]
#Solution_2:
LCM of 11,6,5= 330
W at first = 3*330/11= 90
W at 2nd = 1*330/6= 55
W at final = 1*330/5= 66
55________66________90
11__________________24
Ratio= 11:24
Taken from first vessel = 11*35/11= 11 L
#Solution_3:
Milk at first = 8/11

23. Milk at 2nd = 5/6
Milk at final= 4/5
8/11__________5/6
________4/5_____
1/30__________4/55
1/6___________4/11
11________24
11:24
Taken from vessel A= 35*11/35= 11 L
#Solution_4_Written_Approach_
Let, first vessel contain = x L
2nd vessel contain = y L
Milk at both= (8x/11+ 5y/6)
Water at both= (3x/11+y/6)
ATQ,
(8x/11+ 5y/6)/(3x/11+y/6)= 4/1
=> (48x+55y)/(18x+11y)= 4/1
=> 48x+55y= 72x+44y
=> 24x= 11y
=> x:y= 11:24
Taken from vessel A= 35*11/35= 11 L[Answer.]
#Solution_5_Written_Approach_
Milk at A = 8x/11
Milk at B= 5y/6
ATQ,
(8x/11+5y/6)= 4/5(x+y)

24. => (48x+55y)/66= 4*(x+y)/5
=> 240x+275y= 264x+264y
=> 24x= 11y
=> x:y= 11:24
Taken from vessel A = 11*35/35= 11 L[Answer.]
32. When one litre of water is added to a mixture of milk and water, the new mixture
contains 25% of milk. When one litre of milk is added to the new mixture, then the
resulting mixture contains 40% milk. What is the percentage of milk in the original
mixture?
#Solution:
(W+1)/M= 75/25= 3/1
W+1= 3M
2W+2= 6M
And,
(W+1)/(M+1)= 60/40= 3/2
2W+2= 3M+3
6M= 3M+3
3M= 3
M= 1
And,
(W+1)= 3
W= 2
So, Initially = M+W= 1+2= 3
Milk= 1
%= 1*100/3= 33(1/3)%[Answer.]
33. The price of a box and a pen is Rs.60. The box was sold at a 40% profit and the pen at
a loss of 10%. If the Shop keeper gains Rs.4 in the whole transaction, then how much is
the cost price of Box?
#Solution:

25. B+P= 60______[1]
140% of B-90% of P= 64
7B/5-9P/10= 64
(14B-9P)= 640
14B-9(60-B)= 640
14B-540+9B= 640
5B= 100
B= 100/5= 20
So, price of box = 20 RS [Answer.]
34. A vessel contains a mixture of diesel and petrol in which there is 20% diesel. Five
litres are drawn o and then the vessel is filled with petrol. If the diesel present in the
mixture is now 15% then how much does the vessel hold?
#Solution:
Petrol:Diesel= 8:2= 4:1
5 L is drawn off
Petrol remain = 4x-4*5/5+5= 4x-4+5= 4x+1
Diesel Remain= x-1*5/5= (x-1)
ATQ,
(4x+1)/(x-1)= 85/15= 17/3
12x+3= 17x-17
5x= 20
x= 4
Vessel hold = (4x+x)= 5x= 5*4= 20 L [Answer.]
35. In a lab, two chemical solutions Acid “A” with 90% purity and Acid “B” with 96%
purity are mixed resulting in 24 litres of mixture of 92% purity. How much is the quantity
of the rst solution, Acid “A” in the resulting mixture?
#Solution::
Acid A = 90% purity
Acid B = 96% purity
ATQ,

26. 90% of x+ 96% of y= 92% of (x+y)
90x+96y= 92x+92y
2x= 4y
x:y= 2:1
Acid A in resulting mixture = 24*2/3= 16 L [Answer.]
36. . 60 kg of a certain variety of Sugar at Rs.32 per kg is mixed with 48 kg of another
variety of sugar and the mixture is sold at the average price of Rs.28 per kg. If there be no
profit or no loss due to the new selling price, then what is the price of second variety of
Sugar?
#Solution:
60*32+48*x= 28*(60+48)
1920+48x= 1680+1344
48x= 1104
x= 23 RS [Answer.]
37. Six litre of milk was taken out from a vessel and is then filled with water. This
operation is performed two more times. The ratio of the quantity of milk now left in vessel
to that of the water is 8 : 19.How much is the quantity of the milk contained by the vessel
originally?
#Solution:
Taken out = 6 L
Final milk = x*(1-6/x)^3
Initial milk= x
ATQ,
x*(1-6/x)^3/x= 8/27
x*(1-6/x)^3= (2/3)^3
(1-6/x)= 2/3
1-2/3= 6/x
1/3=6/x
x= 18 L [Answer.]
38. 18 litres of Petrol was added to a vessel containing 80 litres of Kerosene. 49 litres of the
resultant mixture was taken out and some more quantity of kerosene and petrol was
added to the vessel in the ratio 2:1. If the respective ratio of kerosene and petrol in the
vessel was 4:1, what was the quantity of kerosene added in the vessel?

27. #Solution:
Initial solution= (80+18)= 98 L
Petrol:Kerosene = 80:18
Taken out = 49 L
Kerosene Remain= 80-49*80/98= 80-40= 40 L
Petrol remain= 18-49*18/98= 18-9= 9 L
Now, kerosene added= 2x
Petrol added = x
ATQ,
(40+2x)/(9+x)= 4/1
40+2x= 36+4x
2x= 4 L
So, kerosene added = 4 L [Answer.]
39. A vessel which contains a mixture of acid and water in ratio 13:4. 25.5 litres of mixture
is taken out from the vessel and 2.5 litres of pure water and 5 litres of acid is added to the
mixture. If resultant mixture contains 25% water, what was the initial quantity of mixture
in the vessel before the replacement in litres?
#Solution:
Initial A:C= 13:4
Taken out = 25.5 L
Acid remain= 13x-13*25.5/17+5
= 13x – 19.5+5= 13x-14.5
Water Remain= 4x -4*25.5/17+ 2.5
= 4x-3.5
ATQ,
(13x-14.5)/(4x-3.5)= 75/25= 3/1
13x-14.5= 12x-10.5
x= 4

28. Initial Quantity of mixture = 17x= 17*4= 68 L [Answer.]
40. How many kilograms of rice costing Rs. 9 per kg must be mixed with 27kg of rice
costing Rs.7 per kg so that there may be gain of 10% by selling the mixture at Rs.9.24 per
kg?
#Solution:
9*x+27*7= 9.24*10/11(x+27)
9x+ 189= 8.4(x+27)
9x+189= 8.4x+ 226.8
.6x= 37.8
x= 63[Answer.]
41. A trader mixes 6ltr of milk costing 5000 rupees with 7ltr of milk costing 6000 rupees
per litre. The trader also mixes some quantity of water to the mixture so as to bring the
price to 4800 per litre. How many litres of water is added.
#Solution:
Let, Quantity of water = x
ATQ,
(6*5000+6000*7)/(13+x)= 4800
(30,000+42,000)/(13+x)= 4800
72,000= 4800*13+4800x
x= 9600/4800= 2[Answer.]
42. Two cans P and Q contains milk and water in the ratio of 3:2 and 7:3 respectively. The
ratio in which these two cans be mixed so as to get a new mixture containing milk and
water in the ratio 7:4.
#Solution:
Let, P cans contain = x L.
Q can contain= y L.
Milk in total = 3x/5+ 7y/10= (6x+7y)/10

29. Water in total = (2x/5+3y/10)= (4x+3y)/10
ATQ,
4*(6x+7y)= 7*(4x+3y)
24x+28y= 28x+21y
4x= 7y
x:y= 7:4[Answer.]
43. A sample of x litre is replaced from a container containing milk and water in the ratio
2:3 by pure milk. If the container hold 30 litres of the mixture, and after the operation
proportion of milk and water is same. Find the value of X?
#Solution:
Replace= x L
Mixture contains= 30L
Milk= 30*2/5= 12 L
Water= 30*3/5= 18 L
After replacement of x L
Milk= 12-12x/30= 12-2x/5+x= (60+3x)/5
Water= 18-18x/30= 18-3x/5= (90-3x)/5
ATQ,
(60+3x)/(90-3x)= 1/1
60+3x= 90-3x
6x= 30
x= 5 L [Answer.]
44. How many litres of water must be added to 60 litre mixture that contains milk and
water in the 7:3 such that the resulting mixture has 50% water in it?
#Solution’
Milk = 60*3/10= 18 L
Water = 42 L
ATQ,

30. (42+x)/(18)= 1/1
42+x= 18
x= 24 L[Answer.]
45. How many Kgs of rice A costing rupees 20 per kg must be mixed with 20 kg of rice B
costing rupees 32 per kg, so that after selling them at 35 rupees per kg, he gets a prot of
25%.
#Solution:
x*20+20*32= 4/5*35*( x+20)
(20x+640)*5= 140*(x+20)
100x+3200= 140x+2800
40x= 400
x= 400/40= 10kg[Answer.]
46. A milkman mixes water and milk in a container in the ratio 2 : 5,respectively. Initially,
he sells the mixture at Rs. 30 per liter. After selling 70 liters of mixture, he mixes another
20 liters of water and then ratio of water and milk becomes 8 : 15, respectively. The new
mixture is sold by the milk man at the rate of Rs. 25 per liter. If cost of milk is Rs. 27 per
liter then find the profit of milkman in the whole transaction?
#Solution:
W:M= 2x:5x
Let water= 2x
Milk = 5x
So, Total = 7x
He sold = 70 L
Where, Milk= 70*5/7= 50 L
And Water= 20L
Now,
(5x-50)/(2x-20+20)= 15/8
=> 40x-400= 30x

31. => 10x= 400
=> x= 40
So, Initial milk= 5*40= 200L
Milk CP= 200*27
Intially sold= 70*30= 2100 TK
Final W= 2*40= 80 L
Final M= 5*40-50= 150L
Final Total mix= (150+80)= 230 L
Total SP= 230*25+2100= 7850
And CP= 200*27= 5400
Profit= (7850-5400)= 2450[Answer.]
47. Two vessels contain milk and water in the ratio of 7:3 and 2:3 respectively. Find the
ratio in which the contents of both the vessels must be mixed to get a new mixture
containing milk and water in the ratio 3:2.
#Solution:
First vessel; M:W= 7:3[10]
2nd
vessel; M:W= 2:3[5]
Final vessel; M:W= 3:2[5]
LCM of 10,5,5= 10
Milk at first = 10*7/10= 7L
Milk at 2nd
= 10*2/5= 4 L
Final Milk= 10*3/5= 6 L
4===========6==========7
(6-4)=2=============(7-6)= 1
2:1
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48. A 100-litre mixture of milk and water contains 36 litres of milk. 'x' litres of this
mixture is removed and replaced with an equal quantum of water. If the process is
repeated once, then the concentration of the milk stands reduced at 25%. What is the
value of x?
#Solution:
Initial Milk = 36 L

33. Final Milk = 36*(1-x/100)^2
Final/Initial= 36*(1-x/100)^2/36
25/36= (1-x/100)^2
5/6= 1-x/100
x/100=1/6
6x= 100
x= 100/6= 16.667 L[ANSWER.]
49. 20% ammonia solution and 70% ammonia solution were mixed to form 50% ammonia
solution. If 120 litres of 20% ammonia solution were used in the process, what was the
volume of the 50% ammonia solution that was produced?
#Solution:
Let, 20% solution= x
70% solution = y
ATQ,
20% of x+ 70% of y= 50% of (x+y)
20x+70y= 50(x+y)
20x+70y= 50x+50y
30x= 20y
x:y= 2:3
Now, 2 part = 120
5 part = 120*5/2= 300 L [Answer.]
50. Eight litres are drawn off from a vessel full of water and substituted by pure milk.
Again eight litres of the mixture are drawn off and substituted by pure milk. If the vessel
now contains water and milk in the ratio 9:40, find the capacity of the vessel.
#Solution:
Let, Initial Mixture contain = x L of solution(Water.)
Final water = x*(1-8/x)^2
ATQ,
x*(1-8/x)^2/x= 9/49
(1-8/x)^2= (3/7)^2
1- 8/x= 3/7

34. 4/7= 8/x
4x= 56
x= 14 L [Answer.]
51. Each of the cucumbers in 100 pounds of cucumbers is composed of 99% water, by
weight. After some of the water evaporates, the cucumbers are now 98% water by weight.
What is the new weight of the cucumbers, in pounds?
#Solution:
Total = 100 pounds of cucumber
Where 99% water and 1% of non water.
So, original cucumber = 1% of 100= 1 pounds
Now, some of the water evaporate and it becomes 98% of water. So, non water is = 2%,
since only water is evaporated
So, 2% of cucumber = 1 pounds of water
100% = 100/2= 50 pounds of water[Answer.]
#Similar Math practice: (Try yourself)
1. Five kilograms of oranges contained 98% of water. If the next day the concentration
of water decreased by 2%, what was the new weight of the oranges, in
kilograms?[Answer. 2.5 kg]
2. Fresh dates contain 90% water while dry dates contain 28% water . How many kg
of dry dates can be obtained from 36kg of fresh dates ?[Answer. 5 kg]
52. A vessel ( capacity 10 ltrs) is filled entirely with petrol and kerosene in the ratio 3:2.
Two ltrs of this solution is removed and replaced with 2 ltrs of kerosene. This procedure is
repeated thrice. Find the % of petrol in the resulting solution.
#Solution:
Petrol = 10*3/5= 6 L
Initial = 6 L
Final = 6*(1-2/10)^3= 6*(4/5)^3= 3.072
%= 3.072*100/10= 30.72%[Answer.]
53. A solution contains 8 parts of water for every 7 parts of Lemonade syrup. How many
parts of the solution should be removed and replaced with water so that the solution will
now contain 40% lemonade syrup?
#Solution:
Water:Lemonde= 8:7
Let, Replacement = x L.
Water remain = 8/15 -8x/15 +x

35. = (8+7x)/15
Lemonade = 7/15-7x/15
= (7-7x)/15
ATQ,
2*(8+7x)= 3*(7-7x)
16+14x= 21-21x
35x= 5
x= 5/35= 1/7 part [Answer.]
54. If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution,
resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?
#Solution:
Initially = 50% of Alcohol.
Replacement = 25% of Alcohol.
Resulting = 30% of Alcohol.
50____________25
______30______
5___________20
1:4
Now, Replacement = 4 part
Total = 5 part
% = 4*100/5= 80%[Answer.]
55. A dessert recipe calls for 50% melted chocolate and 50% raspberry puree to make a
particular sauce. A chef accidentally makes 15 cups of the sauce with 40% melted
chocolate and 60% raspberry puree instead. How many cups of the sauce does he need to
remove and replace with pure melted chocolate to make the sauce the proper 50% of
each?
#Solution:
Chocolate = 15*40/100= 15*2/5= 6 L.
Raspberry = 9 L
Let, Replacements= x L
Chocolate= 6- 6x/15 +x = (90+9x)/15
Raspberry = 9-9x/15= (135-9x)/15.

36. ATQ,
90+9x= 135-9x
18x= 45
x= 45/18= 5/2= 2.5 Cups [Answer.]
56. Sue's trail mix is 30% nuts and 70% dried fruit. Jane's trail mix is 60% nuts and 40%
chocolate chips. If the combined mixture of Sue and Jane's trails mix contains 50% nuts,
what percent of the combined mixture is dried fruit?
#Solution’
30% of x +60% of y= 50% of (x+y)
20% of x= 10% of y
x:y= 1:2
So, if first mixture contain 1 part; 2nd
mixture = 2 part
First; N:D= 30:70
2nd
N:C=120:80
Total mixture = 30+70+120+80= 300 L
Dry part = 70 L
% = 70*100/300= 23(1/3)%[Answer.]
57. A patient is given exactly 750 milliliters of a mixture of two medications daily.
Medication A contains 40% pain killer and medication B contains 20% pain killer. If the
patient receives exactly 215 milliliters of pain killer daily, how many milliliters of
medication B are in the mixture?
#Solution:
A+B= 750
A= 750-B__________[1]
And,
2A/5+ B/5= 215
2A= 1075-B
A= (1075-B)/2________[2]
From [1] & [2]
750-B= (1075-B)/2
1500-2B= 1075-B

37. B= 1500-1075= 425[Answer.]
58. The ratio of number of Boys to Girls in Class A is 2:3 and the ratio of number of Boys
to Girls in Class B is 3:7. If the number of students in Class A is at least thrice as many as
the number of students in Class B, Find the minimum percentage of Boys when both
classes are considered together.
#Solution:
B:G= 2:3= 12:18[Ist class]
B:G= 3:7
Total = 12+18+3+7= 40
Boys = (12+3)= 15
% of minimum boys= 15*100/40= 37.5%[Answer.]
59. The final exam of a particular class makes up 40% of the final grade, and Moe is
failing the class with an average (arithmetic mean) of 45% just before taking the final
exam. What grade does Moe need on his final exam in order to receive the passing grade
average of 60% for the class?
#Solution::
Let, Total Marks = 100.
He obtained = 45% of 60= 27
Marks needed to be obtained extra= (60-27)= 33
%= 33*100/40= 82.5%[Answer.]
60. A milkman ordered his assistant to first remove 3 liters out of the 8 liters of a solution
of milk and water and then add 3 liters of water to it. Thinking that it would not make any
difference, the lab assistant first added 3 liters of water and then removed 3 liters of the
solution. Find the ratio of the expected concentration of the milk to the actual
concentration of the milk.
#Solution:
First case:
Remove = 3 L of solution.

38. M:W= 5:3
Actually, added first = 3 L of solution(Water)
So, Mixture contain now= (8+3)= 11 L of solution
Where, Milk = 8-24/11[After 3 L replacement]
= 64/11
Expected:Actual = 5:64/11= 55:64[Answer.]
61. A company produces a cleaning solution that has 60% concentration of Chemical K.
The current production batch contains 200L of an ingredient with a 30% concentration of
Chemical K and 150L of an ingredient with a 40% concentration of Chemical K. How
much pure chemical K must be added to the batch to attain the correct concentration of
Chemical K?
#Solution:
Total solution= 150+200= 350
Let, added amount = x L
Solution have already = 200*30/100+150*40/100= 120 L
ATQ,
(120+x)/(350+x)= 60/100
(120+x)/(350+x)= 3/5
600+5x= 1050+3x
2x= 450
x = 225 L[Answer.]
Alternative:
Clearly the solution have two concentration. One is k and other is water.
Final = 40% water[Which will be the same, since no water is added]
Initially mixture contain = 200+150= 350 L of solution
Where water = 350- (200*.3+150*.4)= 230 L
Now, 40%= 230 L
100%= 230*100/40= 575 L of solution
So, pure k needed to be added = 60% of 575- 120

39. = (345-120)L= 225 L[Answer.]
62. Two mixtures of clay and sand contain 40% and 70% sand by weight. When both the
mixtures are mixed, the resultant mixture is 700 gm. If the resultant mixture contains
45% clay by weight, then what is weight of the first mixture?
#Solution:
Two mixture clay and sand contain 40% & 70% of sand.
Let, First mixture contain = x
2nd
mixture contain = y
ATQ,
40% of x+70% of y = 55% of (x+y)
40x+70y = 55x+55y
15x= 15y
x:y= 1:1
So, x+y= 700
Since x= y
So, First mixture weight = 700/2= 350 L [Answer.]
#Alternative:
Since first mixture contain = 40% of sand; so clay= 60%
2nd
mixture sand = 70% so clay = 30%
Final mixture clay contain= 45%
ATQ,
60% of x + 30% of y=45% of (x+y)
60x+30y= 45x+45y
60x-45x=45y-30y
15x=15y
x:y= 1:1
So, first mixture by weight = 700/2= 350 L[Answer.]
63. One cup of nuts that contains exactly half peanuts and half cashews is added to a bowl
of nuts that is exactly one third peanuts, one third cashews, and one third almonds. This
results in a three-cup mixture of nuts. What fraction of the new nut mixture is peanuts?
#Solution:
P:C=1/2:1/2[1 bowl]

40. P:C:A= 2/3:2/3:2/3[2 bowl]
Total Peanut = ½+2/3
=7/6
Fraction = 7/6*3= 7/18[Answer.]
64. A restaurant makes its regular sauce from 5 parts sweet sauce and 3 parts chilli sauce.
Then its spicy sauce is made from 5 parts chilli sauce and 3 parts sweet sauce. How many
cups of chilli sauce must be mixed with a 24-cup regular sauce to turn it to spicy sauce?
#Solution:
Total regular sauce = 24 Cup
Where, Chili = 24*3/8= 9 cup
To make it spicy sauce chili might be = 5 part
Let, chili added = x L
ATQ,
(9+x)/(24+x)= 5/8
72+8x= 120+5x
3x= 48
x= 18 cup
So, chili might be added = 18 cup [Answer.]
#Alternative:
Sweet sauce = 15 L
Which does not added further.
So, 3= 15
8= 40
So, chili might be added= (40-9-15)= 16 Cup[Answer.]
65. If a milkman wishes to earn a profit of 16.66% by selling milk at cost price and mixing
water to the milk. What percent of the mixture has to be water?
#Solution:
Let Milk CP:SP= 100:350/3= 300:350= 6:7
So In total Milk= 6
Water = 7-6= 1

41. %= 100/7=14.28%[Answer.]
66. . 12.5% of a solution having milk and water in the ratio 24 : 25 is removed and
replaced with water. Find the ratio of milk and water in the solution if this operation is
done a total of three times.
#Solution::
Let, total = 49 L
M left = 24(1-12.5/100)³
= 24 × (7/8) × (7/8) × (7/8)
= 1029/64
W = 49 - 1029/64 = 2107/64
M:W = 1029/64 : 2107/64
= 1029 : 2107 = 21:43
#Alternative:
24/49= Milk/Total
Replace = 12.5% of 24/49
M/T= 24/49*(1- 12.5%)^3
= 24/49*(1-1/8)^3= 24/49 *7/8 *7/8*7/8= 21:64
M:W= 21:(64-21)= 21:43[Answer. ]
67. Papayaya, a popular soft drink, contains only four ingredients. Soda water comprise
4/7 of Papayaya and natural lemon juice makes 1/3 of Papayaya. The amounts of sugar
and papaya puree in Papayaya are equal. Due to a malfunction, the mixing machine mixes
double the regular amount of lemon juice and three times the regular amount of papaya
puree. If no other changes were made to the relative quantities of the ingredients, what is
the fractional portion of soda water in the drink that comes out of the malfunctioning
machine?
#Solution:
S:L:P:Su= 4/7:1/3:1/21:1/21
= 12:7:1:1
Now,
Lemon twice = 7*2= 14
Pure thrice = 3*1= 3

42. New Ration:
S:L:P:Su= 12:14:3:1
Sum = (12+14+3+1)= 30
Soda water = 12/30= 2/5[Answer.]
68. Sixty percent of the rats included in an experiment were female rats. If some of the
rats died during an experiment and 70 percent of the rats that died were male rats, what
was the ratio of the death rate among the male rats to the death rate among the female
rats?
#Solution:
Let, F= 60
M= 40
Let total died = x
F died,30% of x= 3x/10
M died = 70% of x= 7x/10
Rate ratio,
(7x/10)/40:(3x/10)/60= 7:2[Answer.]
69. A dishonest milkman mixes water and milk and claims to sell the milk at cost price,
thereby making a profit of 25%. How much water is there in one litre that he delivers to
the customers?
#Solution:
Milk:Water = 125:25=5:1
Water = 1/5
= 1000/5= 200 mL [Answer.]
70. A dishonest grocer professes to sell pure butter at cost price, but he mixes it with
adulterated fat and thereby gains 25%. Find the percentage of adulterated fat in the
mixture assuming that adulterated fat is freely available?
#Solution:
Butter:Fat= 125:25
= 5:1
% fat = 25*100/125= 20%[Answer.]

43. #Solved and editted by:MD. MUTAHER HUSSAIN
71. There are three vessels of equal capacity .Vessel A contains Milk and water in the ratio
4:3 ;Vessel B contains Milk and water in the ratio 2:1 and Vessel C contains Milk and
water in the ratio 3:2 .If the mixture in all the three vessels is mixed up.What will be the
ratio of milk and water ?
#Solution:
Vessel 1: M:W= 4:3[7]
Vessel 2: M:W= 2:1[3]
Vessel 3: M:W= 3:2[5]
LCM of 7,3,5= 105
Milk in total = (4*105/7+2*105/3+3*105/5)=(60+70+63)= 193
Water in total= (3*105/7+1*105/3+2*105/5)=(45+35+42)=122
Milk:Water= 193:122[Answer.]
#Alternative
M:W=(4/7+2/3+3/5):(3/7+1/3+2/5)
= (60+70+63)/105:(45+35+42)/105
=193/105:122/105
=193:122[Answer]
72. Two quarts containing 2⁄3 water and 1⁄3 formula are mixed with three quarts
containing 3⁄8 water and 5⁄8 formula. Approximately what percent of the combined five-
quart mixture is water?
#Solution:
Let, Initial = 2
Final = 3
ATQ,
(2*2/3+3*3/8)= 5x
59/24= 5x
x= 59/120= 0.5

44. %= 0.5*100= 50%[Answer.]
#Alternative
LCM of 2 & 3= 6
Let, first mix contain= 6 L
2nd
mix contain = 9 L
Total mix = (6+9)= 15 L
Water in total =6*2/3+9*3/8= 4+27/8= 59/8
% water= 59*100/120= 50%[Answer]
73. A tank holds x gallons of a saltwater solution that is 20% salt by volume. One Fourth
of the water is evaporated, leaving all of the salt. When 10 Gallons of water and 20 gallons
of salt are added, the resulting mixture is 33 1/3 % salt by volume. What is the value of x?
#Solution
A tank hold = x gallons
Where salt= x*20/100= x/5
So, water = 4x/5
Now ¼ of the water is evaporated.
Water remain= 4x/5-x/5= 3x/5
Now, Total mixture added = 10+20= 30 L
So, Total mixture = (3x/5+ x/5 + 30)= (4x/5 +30)
ATQ,
(x/5+20)/(4x/5+30)= 1/3
(x+100)/(4x+150)= 1/3
3x+300= 4x+150
x= 150 L[Answer.]
#Alternative
ATQ,
(3x/5+10)/(4x/5+30)=2/3
(3x+50)/(4x+150)= 2/3
9x+150= 8x+300
x= 300-150= 150L [Answer.]
#Alternative:

45. ATQ,
(x/5 +20)/(3x/5+10)= ½
(x+100)/(3x+50)= ½
2x+200= 3x+50
x= 150L[Answer.]
74. A 48 lts container containing liquid A and B contains 25% liquid A. a few lts of the
mixture is released and replaced with equal amount of Liquid B. If this process is repeated
once, the cylinder is found to contain 16% liquid A .How many lts of the mixture was
released each time.?
#Solution:
Liquid A + Liquid B = 48 L
Liquid A initially = 48*1/4= 12 L [25%= 25/100= ¼]
Let, Replacement = x L
Final A= 16% of 48
12*(1-x/48)^2= 16% of 48
(1-x/48)^2= 16% of 4
(1-x/48)^2= 64/100
(1-x/48)^2= (8/10)^2
(1-x/48)^2= (4/5)^2
(1-x/48)= 4/5
x/48= 1-(4/5)
x/48= 1/5
5x= 48
x= 48/5= 9.6 L [Answer.]
75. milk and water mixture are in the ratio 7:3. A 10 L mixture is removed and replaced
with water to bring down the concentration of milk by 10%. The amount, in litres, of
water that needs to be added to the resulting mixture in order to reduce the concentration
of milk to 50% is
#Solution:
Milk:Water = 7:3
Let, Milk = 7x
Water = 3x
Replacement = 10 L
After, Replacement of 10 L

46. Milk = 7x-7*10/7= (7x-7)
Water = (3x-30/10 +10)
= (3x+7)
Milk quantity reduced by 10% which is 60% now,
ATQ,
(7x-7)/(3x+7)= 60/40
(7x-7)/(3x+7)= 3/2
14x-14= 9x+21
14x-9x= 21+14
5x= 35
x = 7
So, Milk = 7*7-7= 42
Water = 3*7+7= 28
Again, y L water is added, Thus milk concentration = 50%
ATQ,
(28+y)/(42)= 1/1
28+y= 42
y= 14 L
So water added finally = 14 L[Answer.]
76. In the first consignment, 12% bulbs were faulty. In the second consignment, 4% bulbs
were faulty.In the two consignments, 7% bulbs were faulty. If the two consignments
combined had 4000 bulbs,how many faulty bulbs did the first consignment have?
#Solution:
Let, First consignment = x bulb
2nd
consignment = y bulb
ATQ,
12% of x + 4% of y= 7% of (x+y)
12x+4y= 7x+7y
12x -7x= 7y-4y
5x= 3y

47. x:y= 3:5
So, 1st
consignment have = 3*4000/8= 1500 bulb
Faulty bulb = 1500*12/100= 180 bulb [Answer.]
77. John has 3 solutions: a 12% saline solution, a 8% vinegar solution, and a 15% alcohol
solution. He mixes 3 liters of the alcohol solution with 3 liters of the vinegar solution. What
is the minimum amount of the saline solution he must add if the resulting mixture must be
at least 2% saline solution?
#Solution::
3L Alcohol + 3 L vinegar = 6 L solution
Let,
He adds x L of saline solution. Where saline = 12% of x
So, Now Total Solution= (6+x)
Saline content = 2% of (6+x)
ATQ,
2% of (6+x)= 12% of x
2*(6+x)= 12x
12+2x= 12x
12x-2x= 12
10x= 12
x = 12/10= 6/5 or, 1.2 L [Answer.]
78. A chemistry student is working with solution M, which has a 20% calcium chloride
concentration. He needs to remove three-fourths of this solution and then add a new
solution, solution N, to create a third solution, solution P, with double the volume of
solution M. What must the calcium chloride concentration of solution N be to create a
calcium chloride concentration of 37.5% in solution P?
#Solution:
Initially solution contains = M L of solution
Where Calcium chloride= M*20/100= M/5 L
He removed 3M/4
So, M remain = (4M-3M)/4= M/4
Where calcium chloride remove = (3M/4)*(1/5)= 3M/20
So Calcium chloride now = M/5 – 3M/20= M/20

48. Now, N solution is added and finally the volume is double which means 2M Solution.
ATQ,
M/4+N= 2M
N= 2M -M/4
N = 7M/4
Finally, Calcium chloride = 2M*3/8= 3M/4 L
So, Calcium chloride must be added = (3M/4)- (M/20)
= (15M-M)/20= 14M/20= 7M/10
Since N= 7M/4
So, Calcium chloride in N= 7M/10
%= (7M*100/10)*(4/7M)
= 40%[Answer.]
79. Grandmother's Punch hangover cure is a mix of two ingredients: a herbal extraction
that contains 20% alcohol and apricot cider that contains 5% alcohol. If a 750-milliliter
Grandmother's Punch bottle has 87 milliliters of alcohol, how much apricot cider does it
contains?
#Solution:
Let, Two ingredients are x & y.
ATQ,
20% of x + 5% of y = 87
20x+5y= 8700
20x= 8700-5y
x= (8700-5y)/20____________[1]
And, x + y= 750
(8700-5y)/20 + y= 750
(8700-5y+20y)= 750*20
(8700+15y)= 15,000
15 y= 15,000-8700
15y = 6300
y = 6300/15= 420
Apricot solution= 420 L [Answer.]

49. 80. An alloy of gold,silverand bronze contain 90% bronze, 7% gold and3% silver.a second
alloy of bronze and silver only is melted with the first and mixture contain 85% of
bronze,5% of gold,10% of silver find the percentage of bronze in second alloy?
#Solution::
Let, 1st
mixture contain = 100 g
Bronze = 90 g
Gold = 7 g
Silver = 3 g
2nd
Mixture: Only Bronze and silver is added.
Gold remain same. Since no gold is added.
Let, Total Mixture = x L[After.added.amount]
Initially gold = 7g
ATQ,
5% of x= 7
x= 7*100/5= 140
So, new amount added = (140-100)= 40 g
So, Bronze in new mixture = 140*85/100= 7*17= 119
So, Amount added = (119-90)= 29
%= 29*100/40= 29*5/2= 145/2= 72.5%[Answer.]
81. . A milk seller has a milk of Rs.100 per litre. In what ratio should water be mixed in
that milk, so that after selling the mixture at Rs.80 per litre, he may get a profit of 50%?
#Solution:
Let, Amount of milk = x
Amount of water = y
ATQ,
3/2*(100*x+0)= 80(x+y)
300x= 160x+160y
140x= 160y
x:y= 160:140= 8:7

50. So, Milk:Water= 8:7[Answer.]
82. 6 litres from a 30 litre mixture of milk and water in the ratio 3 : 2 is replaced with
milk. The operation is repeated once. What will be the ratio of milk and water in the
resultant mixture?
#Solution:
Milk = 30*3/5= 18 L
Water = 12 L
Final water = 12*(1-6/30)^2
= 12*4/5 *4/5
= 192/25= 7.68 L
So, Milk = (30-7.68)= 22.32 L
Milk : Water = 22.32:7.68[Answer]
83. In a vessel there is a certain quantity of mixture of milk and water in the ratio 5 : 1
respectively. 24 litres of mixture is taken out and same quantity of milk is added to the
vessel. The ratio of milk and water now becomes 13 : 2 respectively. Again 15 litres of
mixture is taken out. What is the quantity of milk in the resulting mixture? (in litres)
#Solution:
Let, Initially,
Milk = 5x
Water = x
After Replacing 24 L by milk
Milk remain = 5x -24*5/6 +24= (5x+4)
Water = x- 24/6= (x-4)
ATQ,
(5x+4)/(x-4)= 13/2
10x+8= 13x-52
3x= 60
x = 20L
So, Milk now = 5*20+4= 104
Again, 15 L remove

51. So, Milk = 104-104*15/120= 104-13= 91 L[Answer.]
#Alternative
M:W= 5:1[Initial ]= 80:16
M:W= 13:2[Final]= 104:16
So, Final Milk = 104 L
After 15 L removal
Milk remain = 104-104*15/120= 104-13= 91L[Answer.]
84. A sugar salution of 3 litre contain 60 % sugar. One litre of water is added to this
solution. Then the percentage of sugar in the new solution is:
#Solution:
In 3 L solution sugar is 60%
So, Sugar Amount = 3*3/5= 1.8 L
Now, 1 L water is added.
So, Total solution= (3+1)= 4 L
% sugar = 1.8*100/4= 180/4= 45%[Answer.]
85. A milkman adds 500 ml of water to each litre of milk he has in a container. He sells 30
litre of mixture from container and adds 10 litre milk in the remaining. The ratio of milk
and water in the final mixture is 11:5.Find the initial quantity of milk in the container.
#Solution:
Let, Initially Milk = 10x
So, Water = 5x
So, Total Quantity = 15x
After selling 30 L mixture and adding 10 L of milk,
Mixture = (15x+10-30)= (15x-20)
And Milk = (10x-30*2/3+10)= (10x-10)
ATQ,
(15x-20)/(10x-10)= 16/11
15x*11-220= 160x-160
5x= 60
x = 12

52. Initial quantity of milk = 10*12= 120 L[Answer.]
86. Ramesh has a container filled completely with 80% milk and 20% water. 5 litres of the
solution is removed and replaced with water. Then, 15 litres of this solution is removed
and replaced with water. The milk percentage is now 55%. Which of the following can be
the capacity of the container (in litres)?
#Solution:
Let, Total mixture = x
Milk = 4x/5
ATQ,
4x/5*(1-5/x)*(1-15/x)= 55% of x
4/5{(x-5)(x-15)/x^2)}= 11/20
4*{x^2-15x-5x+75}= 11/4
16*(x^2-20x+75}= 11x^2
16x^2-320x+1200= 11x^2
5x^2-320x+1200=0
5x^2-300x-20x+1200= 0
5x(x-60)-20(x-60)= 0
(5x-20)(x-60)= 0
5x= 20; x= 60
x= 4; x= 60
x = 4 can not be, so capacity of the container = 60 L[Answer.]
87. In a vessel there is a mixture of apple, orange and mango juices in ratio 3:5:4. A
quantity of 12 litre from the mixture is replaced with 8 litre of apple juice. Thereafter the
quantities of apple and orange juices in the resultant mixture became same. Find out the
initial quantity of mixture in the vessel?
#Solution:
A:O:M= 3:5:4
Replace 12 L.
Orange remain = 5x -5*12/12= 5x-5
Apple now= 3x -3*12/12 +8= (3x+5)
ATQ,
5x-5= 3x+5
2x= 10
x = 5

53. Initial quantity = 12*5= 60L [Answer]
88. A vessel is filled completely with a solution of spirit and water. The capacity of vessel is
84 litres. 12 litres of this solution is replaced with pure water. The concentration of spirit
in the new solution is 40%. What was the concentration of spirit in the original solution?
#Solution:
Let, Sprit initial = x%
So, Sprit initial = 84x/100
Remove = 12 L
Sprit now = 84x/100- 12x/100= 72x/100
ATQ,
72x/100= 84*40/100
72x= 84*40
x = (84*40)/72= 46.66L [Answer.]
89. By weight, liquid A makes up 8 percent of solution R and 10 percent of solution S. If 3
grams of solution R mixed with 7 grams of solution S, then liquid A accounts for what
percent of the weight of the solution
#Solution:
Liquid A = 8% solution of R & 10% solution of S
In 10 gm, Liquid A = 3*8/100+ 7*10/100= 94/100
%= (94*100/100)*(1/10)= 9.4%[Answer.]
90. Alloy A contains 40% gold and 60% silver. Alloy B contains 35% gold and 40% silver
and 25% copper. Alloys A and B are mixed in the ratio of 1 : 4. What is the ratio of gold
and silver in the newly formed alloy is?
#Solution:
Alloy A: G:S= 40:60[Total 1]
Alloy B: G:S:C= 35:40:25[Total 4]
Gold in total = 40/100+ 4*35/100= 180/100
Silver in total = 60/100+4*40/100= 220/100
Gold:Silver= 180:220= 18:22= 9:11[Answer.]
Previous Question practice:

54. 91. In a mixture of milk & water, their ratio is 4:5 in the first container. And the same
mixture has ratio 5:1 in 2nd
container. In what ratio should the mixture be extracted from
each container and pured into the 3rd
container, so the ratio of milk and water comes to
5:4 in the third container?[Combined 3 bank SO,2018]
#Solution::
Let, First vessel contain = x L
2nd
vessel contain = y L
Milk in total = (4x/9+5y/6)= (8x+15y)/18
Water in total = (5x/9+y/6)= (10x+3y)/18
ATQ,
(8x+15y)/(10x+3y)= 5/4
32x+60y= 50x+15y
18x= 45y
x:y= 45:18= 5:2[Answer.]
92 A vessel contains 28 L of honey and water solution with honey & water in the ratio of
4:3, 21 L of honey water solution is added to this that this honey to water ratio as 2:1,
Again a 51 L honey-water solution that has honey to water ratio as 9:8 is added to this.
After this 10L solution is replaced with pure honey. What is the ratio of water to honey in
the final mixture.[Combined 3 bank cash,2018]
#Solution:
Honey = 28*4/7+21*2/3+51*9/17= 16+14+27= 57
Water = 28*3/7+21*1/3+51*8/17= 12+7+24= 43
Honey:Water = 57:43
10 L.Replace by pure honey.
Honey Remain = 57-57*10/100+10= 57-5.7+10= 67-5.7= 61.3
Water = (100-61.3)= 38.7
Water:Honey= 38.7:61.3= 387:613[Answer.]
93. A bucket contains a mixture of two liquid A and B in the propotion of 7:5. If 9 L.of
mixture is replaced by 9 L of liquid B, then the ration of two liquid becomes 7:9. How
much of the liquid A was there in bucket?[Combined 8 bank SO-2019]
#Solution:
A:B= 7:5

55. Let, A= 7x
B= 5x
A remain = 7x-7*9/12= 7x-21/4= (28x-21)/4
B remain = 5x-5*9/12 +9 = 5x -15/4 +9= (20x-15+36)/4= (20x+21)/4
ATQ,
(28x-21)/(20x+21)= 7/9
28x*9-21*9= 20x*7+7*21
252x-140x= 21*16
112x= 21*16
x= (21*16)/112= 21/7= 3
Initial A= 7x= 7*3= 21 L[Answer.]
94.A can contains milk and water in the ration 3:1. A part of mixture is replaced with
milk, and now the new ratio of milk and water is 15:4. What proportion of original
mixture had been replaced by milk. [Combined 8 bank, 2018]
#Solution:
Milk:Water= 3:1
Let, Replace = x L.of mixture
Milk = 3/4 – 3x/4+x
= (3-3x+4x)/4
= (x+3)/4
Water = 1/4 – x/4
= (1-x)/4
ATQ,
(x+3)/(1-x)= 15/4
4x+12= 15-15x
19x= 3
x = 3/19[Answer.]
………………………… Chapter_2(AUST_Previous Written_2019 _51 Math
AUST All Written Math_2019:

56. SOLVED BY:MD MUTAHER HUSSAIN
01.60 men could complete a piece of work in 250 days. The worked together for
200 days. After that the work had to be stopped for 10 days due to bad weather.
How many more men should be engaged to complete the work in time
#Solution:
60*250=200*60+40*(60+x)
=> 15,000-12000= 2400+40x
=> 40x= 600
=> x= 15
So, More men= 15
02. 12 men can complete a piece of work in 36 days. 18 women can complete the
same piece of work in 60 days. 8 men and 20 women work together for 20 days.
If only women were to complete the remaining piece of work in 4 days, how many
women would be required? [Sonali Bank Officer (FF), 2019]
#Solution:
12M= 1/36
=> 12M*36= 1___________[1]
And 18W= 1/60
=> 18W*60= 1____________[2]
From [1] & [2]
12M*36= 18W*60
=> M:W= 5:2
ATQ,
18*2*60= [8*5+20*2]*20+W*2*4
=> 36*60-80*20= 8W

57. => 8W= 560
=> W= 70[Answer.]
[03]Machine A, working alone at its constant rate, produces x pounds of peanut
butter in 12 minutes. Machine B, working alone at its constant rate, produces x
pounds of peanut butter in 18 minutes. How many minutes will it take machines A
and B, working simultaneously at their respective constant rates, to produce x
pounds of peanut butter? [Sonali Bank Cash Officer (FF), 2019]
#Solution:
Machine A, 12 min= x
1 min= x/12
& Machine B, 18 min= x
=> 1 min= x/18
Together in 1 min= x/12+x/18= 5x/36 part
So,
5x/36 part in= 1 min
x part in= 36/5= 7.2 min[ Answer.]
[Q_4] The ratio of the numbers of boys and girls in a school was 5:3. Some new
boys and girls were admitted to the school, in the ratio 5:7. At this, the total
number of students in the school became 1200, and the ratio of boys to girls
changed to 7:5. What was the number of students in the school before new
admissions? [05 Bank Officer-Cash, 2019]
#Solution:
Ratio of no. Of boy & Girl= 7:5[Final]
Final students = 1200
Where, B= 1200*7/12= 700

58. So, Girl= 500
Let, initial B:G= 5x:3x
Added B:G= 5y:7y
So,
5x+5y= 700
=> (x+y)= 140
=> x= 140-y______[1]
And,
3x+7y= 500
=> 3*(140-y)+7y= 500
=> 4y= 80
=> y= 20
So, x= 140-20= 120
So, Intial students= 5x+3x= 8x= 8*120= 960[Answer.]
[Q_5] . A product is made with three components A, B, C and the ratio of the
prices are 4:3:2. After 1 year price of A increased by 10%, B increased by 8%
and C decreased by 5%. What is the percentage of total increase? [Sonali Bank
Officer (FF), 2019]
#Solution:
A:B:C= 4:3:2
Let, A= 40
B= 30
C= 20
New:

59. A= 110% of 40= 44
B= 108% of 30= 32.4
C= 95% of 20= 19
So, Initial= (40+30+20)= 90
New= (44+32.4+19)= 95.4
Increase= 5.4
%= 5.4*100/90= 6%
[Answer.]
[Q_6] The cost of manufacturing a popular model car is made up of three items-
cost of raw material, labor and overheads. In a year, the cost of three items were
in the ratio of 4:3:2. Next year the cost of the raw material rose by 10%, labor
cost increased by 8% but the overhead reduced by 5%. What is the percentage
increase in the price of the car? [Sonali Bank Officer (FF), 2019]
#Solution:
R:L:O= 4:3:2
Let, R= 40
L= 30
O= 20
New:
R= 110% of 40= 44
L= 108% of 30= 32.4
O= 95% of 20= 19
So, Initial= (40+30+20)= 90
New= (44+32.4+19)= 95.4

60. Increase= 5.4
%= 5.4*100/90= 6%
[Answer.]
[Q_7] . A jar contains 81 liter pure milk. 1/3 of the milk is replaced with water
again 1/3 of the resulting mixture is replaced with water. Find out the ratio of milk
and water? [Sonali Bank Senior Officer (FF), 2019]
#Solution:
81/3=27 L replace with water.
So. milk= 54 L
Water= 27L
Total = 81 L.
Again= 81/3= 27 L water
Milk= 54-27*54/81= 36 L
So, Water= 45 L
M:W= 36:45= 4:5[Answer.]
[Q_8] Amit deposited some money in a bank, which pays 15% interest per annum
compounded yearly. If the bank provides simple interest instead of compound
interest, he receives Tk. 2400 after 2 years. Find the total amount that he received
after 2 years. [Sonali Bank Officer (FF), 2019]
#Solution:
I= pnr
=> 2400= p*2*15/100

61. => p= 8000
Total amount he will get=8000*(1+15/100)^2= 10,580
[Answer.]
[Q_9] The purchase price of a watch and wall clock is Tk. 390. If the watch is sold
at 10% profit and the wall clock at 15% profit, total profit earned is Tk. 51.50.
Find the purchase price difference between the wall clock and the watch. [Sonali
Bank Senior Officer (FF), 2019]
#Solution:
W+C= 390___________[1]
And,
110% of W+115% of C= 441.5
=> 11W/10+23C/20= 441.5
=> 22W+23C= 441.5*20
=> 22*(390-C)+23C= 8830
=> C=8830-8580= 250
So, W= (390-250)= 140
Diff. [250-140]=110[Answer.]
[Q_10] An article manufactured by a company consists of two parts A and B. In
the process of manufacture of part A, 9 out of 100 are likely to be defective.
Similarly, 5 out of 100 are likely to be defective in the process of manufacture of
part B. What will be the probability that the assembled part will not be defective?
[05 Bank Officer-Cash, 2019]

62. #Solution:
91/100*95/100= 0.8645[Answer.]
[Q_11] A machine is made of two components- A and B. Of component A 7 is
defective out of 100 and of B 9 is defective out of 100. What is the probability of
the machine is not defective? [Sonali Bank Senior Officer (FF), 2019]
#Solution:
93/100*91/100= 0.8463[Answer.]
[Q_12] Three runners A, B and C run a race, with runner A finishing 12 m ahead
of runner B and 18 m ahead of runner C, in another race of same type runner B
finished 8 m ahead of runner C. Each runner travels the entire distance at a
constant speed. Find the length of the race. [05 Bank Officer-Cash, 2019]
#Solution:
A= x
B= x-12
C= x-18
And,
B= x
C= x-8
ATQ,
(x-12):x::(x-18):(x-8)
=> x(x-18)= (x-8)(x-12)
=> x^2-18x= x^2-12x-8x-96

63. => 2x= 96
=> x= 48m [Answer.]
[Q_13] . A river is flowing at a speed of 5 kmph in a particular direction. A man,
who can swim at a speed of 20 kmph in still water, starts swimming along the
direction of flow of the river from point A and reaches another point B which is
at a distance of 30 km from the starting point A. On reaching point B, the man
turns back and starts swimming against the direction of flow of the river and
stops after reaching point A. The total time taken by the man to complete his
journey is? [Sonali Bank Officer (FF), 2019]
#Solution:
Stream speed= 5 km/hr
Boat speed= 20 km/hr
So, Downstream speed= 20+5= 25
Upstream = 20-5= 15
Time needed= 30/25+30/15
= 6*60/5+2
= 1 hr 12 min+ 2 hr
= 3 hr 12 min
[Answer]
[Q_14] The distance between two stations ‘X’ and ‘Y’ is 450 km. A train L starts
at 6:00pm from X and moves towards Y at an average speed of 60 kmph. Another
train M starts from Y at 5:20pm and moves towards X at an average speed of 80
kmph. How far from X will the two trains meet and at what time? [Sonali Bank
Officer (FF), 2019]
#Solution:
X_________450___________Y
L>_______________________<M

64. 6:00______________________5:20
Speed of M= 80 km/hr
In 40 min M goes= 80*40/60= 160/3
Now, Distance between them=[450-160/3]= 1190/3
RS= 140
Time= (1190/3)*(1/140)= 17/6 hr
= 2 hr 50 min
So, they meet at= (6:00+2:50)= 8:50
And Distance= 17*60/6= 170 km[Answer.]
[Q_15] Two trains running at the rate of 75 km and 60 km an hour respectively
on parallel rails in opposite directions, are observed to pass each other in 8 seconds
and when they are running in the same direction at the same rates as before, a
person sitting in the faster train observes that he passes the other in 31½ seconds.
Find the lengths of the trains. [Sonali Bank Cash Officer (FF), 2019]
#Solution:
(x+y)= 135*5/18*8= 300
&x= 75*5/18*63/2= 131.25
And L of other= (300-131.25)= 168.75 m
============================
[Q_16]
1. A gardener plants two rectangular gardens in separate regions on his property.
The first garden has an area of 600 feet and a length of 40 feet. If the second
garden has a width twice that of the first garden, but only half of the area, what
is the ratio of the perimeter of the first garden to that of the second garden?
[Sonali Bank Cash Officer (FF), 2019]

65. #Solution:
First Garden Area= 600 feet
L= 40 ft.
So, B= 600/40= 15
Perimeter= 2*(15+40)= 110
2nd garden W= 30
So, L= 300/30= 10
Perimeter= 2*(30+10)= 80
Ratio of perimeter= 110:80= 11:8[Answer.]
[Q_17]A’ began a small business with a certain amount of money. After four
months from the start of the business, ‘B’ joined the business with an amount
which was Tk. 6000 less than ‘A’s initial investment. ‘C’ joined the business after
seven months from the start of the business with an amount which was Tk. 2000
less than A’s initial investment. At the end of the year total investment repented
was Tk. 142000. What will be A’s share in the profit, if B received Tk. 8000 as profit
share? [05 Bank Officer-Cash, 2019]
#Solution:
A:B:C= 12x:8(x-6000):5*(x-2000)
ATQ,
12x+8x-48,000+5x-10,000= 1,42,000
=> x= 8000
A:B:C= 12*8000:8*2000:5*6000
= 96:16:30= 48:8:15
Now,

66. 8===8000
48===48,000[Answer]
[Q_18] A and B started a business with initial investments in the respective ratio
of 18:7. After four months from the start of the business, A invested Tk. 2000
more and B invested Tk. 7000 more. At the end of one year, if the profit was
distributed among them in the ratio of 2:1 respectively, then what was the total
initial investment with which A and B started the business? [Sonali Bank Officer
(FF), 2019]
#Solution:
A= 18x*4+8*(18x+2000)
= 216x+16,000
B= 4*7x+8(7x+7000)
= 84x+56,000
(216x+16,000)/(84x+56000)= 1/2
=> 216x-168x= 96,000
=>x= 2000
Initial investment= 25*2000= 50,000
[Answer]
[Q_19] A, B and C are partners. ‘A’ whose money has been in the business for 4
months claims 1/8 of the profit; ‘B’ whose money has been in the business for 6
months claims 1/3 of the profit. If ‘C’ had Tk. 1560 in the business for 8 months,
how much money did A and B contribute to the business? [Sonali Bank Cash
Officer (FF), 2019]
#Solution:

67. Since total boys and girls = 16
Boys must multiple of 5 and girls must multiple of 3 so
Total boys = 10
Shorter boys = 10*1/5 = 2
Total girls = 6
Taller girls = 6*1/3 = 2
ATQ,
Shortest girl = 1 and Tallest boy = 1
Shorter boys = 2 and Taller girls = 2
So
Taller than shortest girls and shorter than tallest boy =(16-2-2-1-1)=10
10*100/16 = 62.5%
Ans: 62.5%
21. 6 men can complete a piece of work in 12 days. 8 women can complete the same piece
of work in 18 days and 18 children can do it in 10 days. 4 men, 12 woman and 20 children
do the work for 2 days. If the remaining work be completed by men only in 1 day, how
many men will be required?
#Solution:
6M= 1/12
8W= 1/18
So,
6M*12= 18*8W
=>12M= 24W
=> M:W= 2:1= 10:5
And,
18C*10= 8*18W

68. => 10*C= 8*W
=> C:W= 4:5
So, M:W:C= 10:5:4
ATQ,
36*10*2= [4*10+12*5+20*4]*2+10*M
=> 720-360= 10*M
=> M= 36[Answer.]
22.A bag contains 4 red and 3 black balls. A second bag contains 2 red and 4 black balls.
One bag is selected at random. If from the selected bag one ball is drawn, then what is the
probability that the ball is drawn is red?
#Solution:
1/2*4/7+1/2*2/6
= 2/7+1/6
= (12+7)/42= 19/42[Answer]
23. A cow was standing on a bridge, 5m away from the middle of the bridge. A train was
coming towards the bridge from the ends nearest to the cow. Seeing this cow ran towards
the the train and managed to escape when the train was 2m away from bridge. If it had
run in the opposite direction, it would hit by the train 2m before the end of the bridge.
What is the lenght of the bridge in meters assuming the speed of the train is 4 times that of
cow ?
#Solution:
(x-2)/4y= (B/2-5)/y

69. 2*(x-2)= 4*(B-10)
2x= 4B-36________[1]
And,
(B/2+5-2)/y= (x+B-2)/4y
2B+28= 2x____________[2]
From (1) & (2)
2B+28= 4B-36
2B= 64
B= 32
So, Length of Bridge= 32 m
[Answer.]
24. A man went downstream for 28 km in a motor boat and immediately returned. It took
the man twice as long to make the return trip. If the speed of the river flow were twice as
high, the trip downstream and back would take 672 minutes. Find the speed of the boat in
still water and the speed of the river flow.
Solution:
(x+y)= 2(x-y)
=> x:y= 3:2
x+y= 5k
x-y=k
28/5k+28/k= 11 1/5= 56/5

70. K=3
Still water= 3*3= 9
Speed= 3*1=3
25.A, B and C started a business by investing Rs. 24,000, Rs. 32000 a6nd Rs.18000
respectively. A and B are active partners and get 15% and 12% of total profit and
remaining profit is to be distributed among them in the ratio of their investment. If C got
total Rs.65700 as a profit, what was the total amount of profit ?
#Solution:
A:B:C= 24:32:18= 12:16:9
Atq,
9x= 65700
=> x= 7300
Remaining profit= 37*7300= 270100
Atq,
(100-27)%= 2,70,100
=> 100%= 2,70,100*100/73= 3,70,000TK
[#SO_[8 bank] _Re written]
26. A rectangular plot has a concrete path running in the middle of the plot parallel to the
breadth of the plot. The rest of the plot is used as a lawn, which has an area of 240 sq. m.
If the width of the path is 3m and the length of the path is greater than its breadth by 2 m,
what is the area of the rectangular plot? (in sq m)
#Solution:

71. Let, length of rectangle = x
Width = (x+2)
Given, width of path= 3m
Area of path= (x+2)*x-3x
ATQ,
x*(x+2)-3x= 240
=> x^2-x-240= 0
=> x^2-16x+15x-240= 0
=> x(x-16)+15(x-16)= 0
=> x= 16
So, L= (16+2)= 18
W= 16
Area of rectangle= 16*18= 288 m^2[Answer.]
27. A pipe can fill a tank in 0.9 hours and another pipe can empty in 0.7 hours. If tank is
completely filled and both pipes are opened simultaneously then 450 liters of water is
removed from the tank is 2.5 hours. What is the capacity of the tank?
#Solution:
Pipe one can fill 0.9hr or (9/10)hr = 1 part
So, in 1 hr = 10/9 part
And, other can empty (7/10)hr = 1 part
So, in 1 hr = 10/7 part
So, in 1 hr total empty= 10/7-10/9= 20/63 part
So, in 2.5 hr = 50/63 part
ATQ,

72. 50/63 part = 450 L
=> 1 part = 450*63/50= 567 L[Answer.]
28. A sum of Rs. 3,240 was fixed to complete a work. 54 workers completed the work in 8
days and the sum was divided equally among the workers. If the work was to be
completed in 3 days then how much less money each worker would receive compared to
when the work was completed in 8 days (sum is divided equally among the workers)?
#Solution:
54 worker 8 days salary= 3240 RS
1 worker 8 days salary= 3240/54= 60 RS
1 work 3 days salary= 60*3/8=22.5 RS
Diff. = (60-22.5)= 375[Answer.]
29. Two trains running at the rate of 75 km and 60 km an hour respectively
on parallel rails in opposite directions, are observed to pass each other in 8
seconds and when they are running in the same direction at the same rates as
before, a person sitting in the faster train observes that he passes the other in
31½ seconds. Find the lengths of the trains.
#Solution:
(x+y)= 135*5/18*8= 300
&x= 75*5/18*63/2= 131.25

73. And L of other= (300-131.25)= 168.75 m[Answer.]
30. : A,B and C can do a piece of work in 12,18, and 24 days respectively. they work at it
together . B stops the work before 2 days and C is called off 4 days before the work is done
.In what approximate time the work is finished?
#Solution:
LCM of 12,18,24= 72
A= 6
B= 4
C= 3
ATQ,
6*T+(T-2)*4+(T-4)*3= 72
=> 13T=92
=> T= 92/13= 7(1/13) days[Answer]
31.An engineer undertakes a project to build a road 15 km long in 300 days and employs
45 men for the purpose. After 100 days, he finds only 2.5 km of the road has been
completed. Find the (approximate) number of extra men he must employ to finish the
work in time.
[ & & _SO(2018)]
#Soultion:
100 days 1/6 part by = 45 men
100 days 1 part by= 45*6 men
1 days 1 part by= 45*6*100
Total unit = 45*6*100
ATQ,

74. 100*45+200*(45+x)= 45*6*100
=> 200*45+200x= 22,500
=> 200x= 13,500
=> x= 67.5 or 68 (approximately) [Answer.]
32. A bucket contains a mixture of two liquids A and B in the proportion 7:5. If 9 litres of
the mixture is replaced by 9 litres of liquid B, then the ratio of the two liquids becomes
7:9. How much of the liquid A was there in the bucket?
==================================
#Solution:
A= 7x
B= 5x
9 L replaced.
A remain = 7x-9*7/12= 7x-21/4= (28x-21)/4
B remain = 5x-5*9/12+ 9
= 5x-15/4 +9
= (20x-15+36)/4= (20x+21)/4
ATQ,
(28x-21)/(20x+21)= 7/9
=> 252x-189= 140x+147
=> 112x = 336
=> x= 3
A was in the bucker = 7*3= 21 L
==================================
33. Three partners shared the profit in a business in the ratio 5 : 7 : 8. They had partnered
for 14 months, 8 months and 7 months respectively. What was the ratio of their
investments?

75. ==================================
#Solution:
14x/8y= 5/7
98x= 40 y
x:y= 20:49
And,
8y/7z= 7/8
64y= 49z
y:z= 49:64
x:y:z= 20:49:64
#AUST_Written_Previous
34. A finance company declares that, at a certain compound interest rate, a sum of money
deposited by anyone will become 8 times in 3 years. If the same amount is deposited at the
same compound rate of interest, then in how many years will it become 16 time?
[ & & _SO 8bank]
#Solution:
P*(1+r)^3= 8P
(1+r)^3= 8
(1+r)^3= 2^3
(1+r)= 2__________[1]
Again,
P*(1+r)^n= 16*P
(1+r)^n= 16
(1+r)^n= 2^4
(1+r)^n= (1+r)^4[since, (1+r)= 2]
n= 4 year[Answer.]
#AUST_Written_previous

76. 35
Two pipes can separately fill a tank in 20 hours and 30 hours respectively. Both the pipes
are opened to fill the tank, but when the tank is 1/3 full a leak develops in the tank
through which 1/3 of the water supplied by both the pipes leak out. What is the total time
taken to fill the tank?
[ & & _SO(18)]
#Solution:
Pipe one can fill a tank in = 20 hr
Pipe other can fill a tank in= 30 hr
So, (1/20+1/30) part in = 1 hr
=> 1/12 part in = 1 hr
=> 1 part in = 12 hr
Now, 1/3 water goes out which means extra time need = 12/3= 4 hr
So, total time = (12+4)= 16 hr[Answer.]
Alternative:
LCM of 20 & 30= 60
A= 3
B= 2
Total time need = 60/(3+2)= 12 hr
1/3 goes out means = 60/3= 20 L goes out.
Now, extra time= 20/(3+2)= 4 hr
So, total time = 12+4 = 16 hr[Answer.]
By:MD.MUTAHER HUSSAIN

77. Probashi Kalllyan Bank Written Question
Post Name: Assistant Programmer
Exam Date: 09 November, 2019
Exam Taker: AUST
Solved by:MD. MUTAHER HUSSAIN
36. Two pipes A and B can fill a tank in 20 and 30 hours respectively. Both the pipes are
opened to fill the tank but when the tank is one - third full, a leak develops in the tank
through which one - third water supplied by both the pipes gose out. The total time taken
to fill the tank is?
#Solution:
LCM of 20 & 30= 60
A= 3
B= 2
1/3 rd full means = 60/3= 20
Time = 20/5= 4 hrs
1/3 rd water goes out.
So extra time needed= 60/5= 12 hr
Total Time= (12+4)= 16 hr
Alternative:
1/20+1/30= 1/x
5x= 60
x= 12
Extra time = 12/3= 4 hr
So, Total time= (12+4)= 16 hr[Answer.]
37. In an examination, 34% failed in Mathematics and 42% failed in English. If 20%
failed in both the subjects, the percentage of students who passed in both subjects was?
.#Solution:
n(AUB)= n(A)+n(B)-n(AnB)

78. n(AUB)= 34+42-20
n(AUB)= 56%
Passed in both subject= (100-56)= 44%[Answer]
PKB[EO_Cash]
Exam taker::AUST
#Solved by_Md Mutaher Hussain
38. Q.A and B can do a piece of work in 30 days while B and C can do the same work in 24
days and C and A in 20 days .they all work together for 10 days when B and C leave .how
many days more will A take to finish the work?(PKB EO Cash)
#Solution:
2*(A+B+C)= 1/30+1/24+1/20
= (4+5+6)/120
= 1/8
(A+B+C)= 1/16
10 days work = 5/8
Remain= 3/8
A = 1/16-1/24= 1/48
1/48===_1
3/8===48*3/8= 18 days
#Alternative:
Eff. (A+B):(B+C):(C+A)= 4:5:6
Total Work = 4*5*6= 120
ATQ

79. Eff. Of (A+B+C)= 15/2= 7.5
A= 7.5-5= 2.5
ATQ,
7.5*10+2.5*A= 120
=> 25A= 450
=>A= 450/25= 18 [Answer.]
39. Q.If x^3+3/x=4*(a^3+b^3);
3x+1/x^3=4(a^3-b^3)
Value of a^2-b^2=?(Ans:1)
(PKB_EO_Cash)
#Solution:
By adding and subtracting both
a= (x^2+1)/2x
b = (x^2-1)/2x
a^2-b^2= (a+b)*(a-b)= x*(1/x)= 1
40. A container contains 40 litres of milk. From this container 4 litres of milk was taken
out and replaced by water. This process was repeated further two times. How much milk
is now contained by the container?
#Solution:
Initial Milk = 40 L
We know that,
Final mixture = I*(1-R/I)^n
Here, I= Initial Mixture

80. R= Replacement
n= Repetition time
So,
40*(1-4/40)^3= 29.16 L
41. finance company declares that, at a certain compound interest rate, a sum of
money deposited by anyone will become 8 times in 3 years. If the samamount is
deposited at the same compound rate of interest, then in how many years will it
become 16 time?
#Solution:
P*(1+r)^3= 8P
=> (1+r)^3= 8
=> (1+r)^3= 2^3
=> (1+r)= 2__________[1]
Again,
P*(1+r)^n= 16*P
=> (1+r)^n= 16
=> (1+r)^n= 2^4
=> (1+r)^n= (1+r)^4
=> n= 4 years [Answer.]
42. If age of P and R are added to twice the age of Q, the total becomes 59. It the ages of Q
and R are added to thrice the age of P, the total become 68. And if the age of P is added to
thrice the age of Q and thrice age of R, the total becomes 108, what is the total of ages of
P?
#Solution:
P+R+2Q= 59
Q+R+3P= 68

81. => Q+R= 68-3P_______[1]
And
P+3Q+3R=108
=> 3*(Q+R)= 108-P
=> Q+R= (108-P)/3_____[2]
From [1] & [2]
(108-P)/3= (68-3P)
=> 108-P= 204-9P
=> 8P= 96
=> P= 12[Answer.]
43. boat running upstream takes 8 hours 48 minutes to cover a certain distance, while it
takes 4 hours to cover the same distance running downstream. What is the ratio between
the speed of the boat and speed of the water current respectively?
#Solution:
44/5 *(x-y)= 4*(x+y)
=> 44x -44y = 20x+20y
=> 24x= 64 y
=>x:y= 8:3
44. In a particular school, out of total 640 boys that include all the age-groups, 60% of the
boys below 18 were enrolled. Out of them, 50% attended the school irregularly. If 120
boys among them were regular, how many boys were there in the school who are above
18?[Pkb cash]
Question wrong

82. [ /5, R /5, B # & /5 $ ( ऀ
& 1M: " 3 6 $ऀ ' # F ऀ ' S) $(
T$ # 1M: 3 & ]
PKB EO (General)
Exam Taker:AUST
Solved by:MD MUTAHER HUSSAIN
45. bucket contains a mixture of two liquids A and B in the proportion 7:5. If 9 litres of
the mixture is replaced by 9 litres of liquid B, then the ratio of the two liquids becomes
7:9. How much of the liquid A was there in the bucket?
#Solution:
A= 7x
B= 5x
9 L replaced.
A remain = 7x-9*7/12= 7x-21/4= (28x-21)/4
B remain = 5x-5*9/12+ 9
= 5x-15/4 +9
= (20x-15+36)/4= (20x+21)/4
ATQ,
(28x-21)/(20x+21)= 7/9
252x-189= 140x+147
112x = 336
x= 3
A was in the bucker = 7*3= 21 L
46.A tank is filled by three pipes with uniform flow. The first two pipes
operating simultaneously fill the tank in the same time during which the tank is filled by
the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4
hours slower than the third pipe. The time required by the first pipe is:
#Solution:
First take = T

83. 2nd
take= (T-5)hr
3rd
take = T-9
ATQ,
1/T+1/(T-5)= 1/(T-9)
(T-5+T)(T-9)= T*(T-5)
(2T-5)*(T-9)=T^2-5T
2T^2-18T-5T+5T+45-T^2=0
T^2-18T+45=0
T^2-15T-3T+45=0
T= 15, 3
3 can not be possible. So first pipe take=15 hr[Answer.]
47. A bag contains 5 red and 3 green balls. Another bag contains 4 red and 6 green balls.
If one ball is drawn from each bag what is the probability that one ball is red and one is
green?
#Solution::
5/8 *6/10 +3/8 * 4/10=21/40
48. Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation
of the top of the lighthouse is observed from the ships are 30 degree and 45 degree
respectively. If the lighthouse is 100 m high, the distance between the two ships is:
#Solution::
Tan 30°= 100/L1
L1/√3= 100
L1= 100*√3
And
Tan 45°= 100/L2
L2= 100
Total Distance, L= L1+L2= 100+100√3= 100*(1+√3)=Answer.
05.
49. A train overtakes two persons walking along a railway track. The first one walks at
4.5 km/hr. The other one walks at 5.4 km/hr. The train needs 8.4 and 8.5 seconds

84. respectively to overtake them. What is the speed of the train if both the persons are
walking in the same direction as the train?
#Solution::
Let, Train speed = x
ATQ,
(x-4.5)*8.4= (x-5.4)*8.5
8.4x -4.5*8.4= 8.5x -5.4*8.5
.1x=. 81
x= 81 km/hr [Answer.]
50. X= (√5+1)/(√5-1) and Y= (√5-1)/(√5+1) then find the value of (x^2+xy+ y^2)/(x^2-
xy+y^2)
#Solution:
X+Y= (√5+1)/(√5-1) +(√5-1)/(√5+1)
=3
XY=1
X²+XY+Y²
=X²+2XY+Y²-XY
=(X+Y)²-XY
=3²-1
=8
X²-XY+Y²
=X²+2XY+Y²-3XY
=(X+Y)²-3XY
=3²-3×1
=6
(X²+XY+Y²)/(X²-XY+Y²)
=8/6
=4/3[Answer.]
51. ab+bc+ca=0, then What's the value of 1/(a^2-bc)+1/(b^2-ca)+1/(c^2-ab)=?
#Solution:

85. ab+bc+ca=0
-bc=ab+ca
-ca=.ab+bc
-ab=bc+ca
1/a^2+ab+ca+1/b^2+bc+ab+1/(c^2+bc+ca)
1/a(a+b+c)+1/b(a+b+c)+1/c*(a+b+c)
=(ab+bc+ca)/abc(a+b+c)
=0/abc(a+b+c)
=0 .
#BCB_Random_32_Math
01.Selling a watch at BDT 90 incurs some losses but selling it at BDT 118 will earn 3/4 of
the loss amount. What is the buying price of the watch?
#Solution:
Let, CP = x
Loss = y
ATQ,
x-90= y
And,
118-x= 3y/4
118-x= 3*(x-90)/4
472-4x = 3x-270
7x= 742
x = 742/7= 106
So, Buying price = 106 TK [Answer.]
02. The compound interest on a sum of money for 2 years is 832tk and the simple interest
on the same sum for the period is 800 tk. The difference between the compound interest
and the simple interest for 3 years will be..?
#Solution:

86. Let, principal = P
Interest rate = r
ATQ,
SI for first year and CI for first year will be same which is 400.
CI for 2nd
year= (832-400)= 432
So, Interest Rate= 32*100/400= 8%
Diff. For third year = 832*8/100= 66.56
Total difference = 66.56+32= 98.56[Answer.]
03. A bag contains 4 red and 3 black balls. A second bag contains 2 red and 4 black balls.
One bag is selected at random. If from the selected bag one ball is drawn, then what is the
probability that the ball is drawn is black?
#Solution:
Selected ball is drawn is black = ½*(3/7+ 4/6)
= 1/2 (3/7+2/3)
= 1/2 * (9+14)/21
= 23/42[Answer]
04. Today is X 12th birthday and his father 40th birthday. How many years from today
Will X father be twice as old as X at that Time
#Solution:
X=12
F= 40
Let, after x year X father will be twice than X
ATQ,
2*(12+x)= 40+x
24+2x= 40+x
x= 16 year[Answer.]
05. To do a pice of work, B takes 3 times as long as A and B together and C twice as long
as A and B together. If A,B and C together can complete the work in 10 days,how long
would A take alone to complete the work it?
#Solution:
Let, B took = 3x days

87. A+ B took = x days
C take = 2y days
A+B took y days
ATQ,
1/y + 1/2y = 1/10
3/2y= 1/10
2y= 30
y= 15
So, (A+B) took 15 day
So, B took = 45 days
ATQ,
1/45+1/30+1/A= 1/10
1/A= 1/10-(2+3)/90
1/A= (9-5)/90
4A= 90
A= 45/2= 22.5 days [Answer.]
06.A, b, and c can complete a work in 12, 15 and 25 days respectively. A and b started
working together whereas c worked with them in every third day. Find the number of
days required to complete the work?
#Solution:
Let, Total work = 30
A= 25
B= 20
C= 12
(25+20)*3+ 12= 147 unit in 3 days
294 unit in = 6 days
Remain = 6 unit in = 6/45= 2/15 days
Time required = 6(2/15) days[Answer]
07.10 women and 8 men do a work in 1 days. 1 men and 5 women can do that work in 4
days. Numbers of women required to complete the job in 3 days with 2 men??
#Solution:

88. (10W+8M)=(M+5W)*4
10 W+ 8M = 4M +20W
4M= 10 W
M:W= 5:2
ATQ,
3*(2*5+2W)= 10*2+8*5
30+6W= 60
6W= 30
W= 5[Answer]
08. A and b working separately can do a piece of work in 9 and 12 days respectively, if
they work for a day alternatively, a beginning, in how many days, the work will be
completed??
#Solution:
Let, Total work = 36
A:B = 4:3
(4+3)*5= 35 unit in = 5*2= 10 days
Remain = 1 in ¼ days
Total time = 10(1/4) [Answer]
09. When an article was sold for Rs. 696, percent profit earned was P%. When the same
article was sold for Rs. 841, percent profit earned was (p + 25%). What is the value of P?
#Solution:
Profit diff. = (841-696)= 145
ATQ,
25% = 145
100%= 145*4= 580
Value of P = 116*100/580= 20%[Answer.]
10. A can do a piece of work in 30 days, b can do it in 15 days and c can do it in 10 days. If
in every second day b and in every third day c help a in doing the work, how many days
will be required to complete the whole work?
#Solution:
Let, Total work = 30 unit
A:B:C= 1:2:3[eff.]

89. Now 3 days work, 1*3+2*1+3*1= 8 unit
9 days work = 24 unit
In 10 days = 25 unit
In 11 days = 25+3= 28
Remain 2 unit in = 2/4 days
Total time = 11(1/2) days[Answer.]
11. In how many ways can 5 books be selected from a bundle of 11 books if 3 particular
books are always excluded?
#Solution:
(11-3)C5= 8C5
12. Rahim undertakes to do a work in 40 days. He engages 100 men at the beginning and
100 men after 35 days and completes the work in stipulated time. If he had not engaged
the additional men, how many days behind schedule would it be finished?
#Solution:
ATQ,
100*35+200*5=100*D
100D= 4500
D= 45
Extra days = (45-40)= 5 days[Answer.]
13. A,b,c do a job alone in 20, 30 and 60 days respectively. In.how many days can a do the
job if he is assisted by b and c in every third day?
#Solution:
Let, Total work = 60 unit
A:B:C= 3:2:1
3 days work = 3*3+2+1= 12 unit in = 3 days
60 unit in = 3*5= 15 days[Answer]
14. If x+1/x=5 then the value of x/x^2+x+1?
#Solution:
x+1/x= 5
x^2+1= 5x

90. x/(x^2+1+x)= x/6x= 1/6[Answer]
15. At the end of a bouquet 50 people hand shake with each other. How many hand shake
will there be in total?
#Solution:
Hand shake in total = 50C2[Answer.]
16. A team of 2 men and 5 women completed one -fourth of a job in 3 days. after 3 days
another man joined the team and they took 2 days to complete another one- fourth of the
job. how many men can complete the whole job in 4 days??
#Solution:
(2M+5W), 1/4th
in = 3 days
(2M+5W), 1th work in = 12 days
Again, (3M+5W), 1 th work in = 8 days
So,
2M+5W= 1/12
3M+5W= 1/8
M= 1/8-1/12= 1/24
Men required = 24/4= 6 men[Answer]
17. . If 5% is gained by selling an article for BDT 350 than selling it for BDT 430,the cost
of the article is?
#Solution:
5%= 80
100%= 1600[Answer]
18. A bag contains 10 balls numbered from 0 to 9. the balls are such that the person
picking a ball out of the bag is equally likely to pick anyone of them. A person picked a
ball and replaced it in the bag after noting its number. He repeated this process 2 more
times. What is the probability that the ball picked first is numbered higher than the ball
picked second and the ball picked second is numbered higher than the ball picked third?
#Solution:
Threee distinct ball can be picked in (10×9×8) ways.
The order of a,b and c can be as follows:
(i). a > b > c ;

91. (ii). a > c > b ;
(iii). b >c >a ;
(iv). b > a > c ;
(v). c > a > b ;
(vi). c > b > a
They will occur equal number of times.
Thus the number of ways in which (a > b > c):
=(16)×(10×9×8)=120
Required probability =12010×10×10
= 3/25[Answer]
19. A and B invest in a business in the ratio 3:2. If 5% of the total profit goes to charity
and A's share is tk.855, the total profit is:
#Solution:
95%= 855*5/3= 285*5= 1425
100%= 1425*20/19= 1500[Answer]
20. In a box, there are 8 red,7 blue and 6 green balls.one ball is picked up randomly.
What is the probability that it is neither red nor green
#Solution:
Neither red nor green = 7
Total = 21
Probability = 7/21= 1/3[Answer]
21. A bath can be filled by the cold water pipe in 10 minutes and by the hot water pipe in
15 minutes.A person leaves the bathroom after turning on both the pipes .He returns just
when the bath should have been full.Finding however the waste pipe was open he closes
it.In 4 minutes more the bath is full.In what time will the waste water pipe empty it?
#Solution:
Let, Total Capacity = 60 L
C:H= 6:4
Time required = 60/10= 6 min
In 4 min =4*(6+4)= 40 unit

92. So, 40 unit = 6 min
60 unit = 6*60/40= 6*3/2= 9 min[Answer]
22. UV $ 1:% W $ I W 4 B + ( X:% W $ W
$ 1: W - $, & 3 Y UV $ & -& & W " $ ( ?
#Solution:
Let, Total mark = x
Pass mark = y
So, 20% of x +5 = y
And, 30% of x -20= y
ATQ,
20% of x +5= 30% of x -20
10% of x = 25
x = 25*100/10= 250
Pass mark = 20% of 250+5= 50+5= 55[Answer.]
23. pipe p can fill an empty tank in 4 hours but pipe q can completely empty the same
tank in 8 hours, Both the pipes were opened alternately after every two hours starting
with P then in how many hours, the tank was completely filled?
#Solution:
Let, Total capacity = 8 L
F:E= 2:1[Eff.]
First 2 hr = 2*2= 4 unit
2nd
2hr = 2*-1=-2 unit
3rd
2 hr= 4+2= 6 unit
4th
2 hr = 6-2= 4 unit
5th
2 hr = 4+4= 8 unit[Filled]
So, Time taken = 10 hr. [Answer.]
24. A thief steals a car at 2:30 PM and drives it at 60km/h. The thief is discovered at 3 PM.
And the owners sets off in another car at 75 kilometers per hour. When will he overtake
the thief
#Solution:

93. Thief speed = 60 km/hr
In 30 min thief goes = 30 km
ATQ,
(x+30)/75= x/60
(x+30)/5= x/4
4x+120= 5x
x = 120
Time needed for car owner= 150/75= 2 hr
So, overtake thief = 3+2= 5:00 PM(Answer)
25. Of the three numbers, the second is twice the first and is also thrice the third. If the
average of three numbers is 44, what is the largest number
#Solution:
Let, 2nd
number = 6x
Ist number = 3x
3rd
number = 2x
ATQ,
6x+3x+2x= 44*3
11x= 44*3
x = 4*3= 12
So, Largest number= 12*6= 72[Answer.]
26. The average of five consecutive numbers is 48 What is the product of the first and five
numbers?
#Solution:
Let, the number, x, x+1, x+2,x+3,x+4
ATQ
5x+(1+2+3+4)= 5*48
5x + 10= 240
5x= 230
x = 46
So, first number = 46 & last number =50
So, product of first and last = 46*50= 2300[Answer]

94. 27. A is thrice as good a workman as B and so takes 60 days less than B for doing a
job,.The time in which they can do the job together is?
#Solution::
A:B= 3:1[Eff.]
A:B= 1:3[Time]
3x-x= 60
2x= 60
X= 30
So, A= 30
B= 90
ATQ,
90= (3+1)*D
90/4= D
D= 22.5 days [Answer]
28. There are 81 litres of pure milk in a container. One third of the milk is replaced by
water in the container. Again one-third of the mixture is replaced with an equal amount of
water. What is the ratio of milk to water in the new mixture?
#Solution:
Initially M= 81 L
81/3= 27 L replace by water
So, Milk:Water = 54:27
Again, 27 L replace
M= 54- 54*27/81= 54-18= 36
W= (81-36)= 45
M:W= 36:45= 4:5[Answer.]

95. 29. A, B & C are three partners of a business. A receives 2/3rd of the profit and the
remaining profit is equally shared by B and C. If A's income increases by tk 400 when the
profit of the business rises from 5% to 7%. Find the profit received by B?
#Solution:
Suppose,
Total Profit = x
→ So profit of A = 2x/3
→ Remaining Profit = x/3
→→ Profit of B = 1/2 × x/3 = x/6
→→ Profit of C = 1/2 × x/3 = x/6
So, ratio of profit, A : B : C = 2x/3 : x/6 : x/6 = 4 : 1 : 1
As, if nothing is clearly mentioned, profit is always distributed in the ratio of investment,
we can say, in the business the ratio of the investments of A, B, and C is = 4 : 1 : 1
Suppose,
A's investment is 4a. So, B's investment is a, and C's investment is a.
So, total capital of the business = 4a+a+a = 6a
If profit of the business rises from 5% to 7%, increase in the amount of profit = 6a* 2% =
0.12a
So, profit of A increases = 0.12a * 2/3 = 0.08a
According to the question,
0.08a = Tk. 400
a = 5000.
So, total capital in the business = 6a = 6×5000 = Tk. 30000
Total profit of the business, x = Tk. 30000 × 7% = Tk. 2100
Profit attributable to B = x/6 = 2100/6 = Tk. 350.
Answer: Tk. 350.[Answer.]
30. A takes 5 days more than B to do a certain job 9 days more than C;A & B together can
do the job in the same time as C.How many days A would take to do this?

96. #Solution:
C= T
A= (T+9)
B= (T+4)
ATQ,
1/(T+4)+1/(T+9)= 1/T
T(2T+13)= (T+4)*(T+9)
2T^2+13T= T^2+13T+36
T^2= 36
T= 6
A take = (9+6)= 15 days[Answer]
31. Two trains Ajanta express and Taj express simultaneously started on two parallel
tracks from Meerut to Nagpur, which are 390 km a part. The ratio of the speed of Ajanta
express and Taj express is 6 : 7.After how long (in kms) travelling. Taj express exchanges
the speed with Ajanta express so that both the trains reach at their destination
simultaneously ?
#Solution:
Meerut_________390_______Nagpur.
Speed ratio= 6:7[13]
13= 390
=> 1= 30
=> 7= 210 km
So, After 210 km the exchange their speed [Answer.]
32. A does half as much work as B in three -fourth of the time. If together they take 18
days to complete the work, how much time shall B take to do it?
#Solution:
Let, B takes = x
A takes =2 x*3/4= 3x/2

97. A:B= x:3x/2= 2:3
ATQ,
(2+3)*18= 3*D
=> D= 6*5= 30 days [Answer.]
#AUST RECENT MCQ SOLUTION(MATH)
(AEO)
Math solution
Exam taker:AUST
Solved by: Md Mutaher Hussain
35. What would be the measure of the diagonal of a square whose area is equal to 578 sq
cm?
#Solution:
a^2= 578
=> a= 24.04
Diagonal = 1.41*24.04= 34 cm[Answer]
#Alternatives:
Area = 1/2* digaonal^2
Diagonal ^2= 2*578
=> Diagonal^2= 1156
=> Diagonal= √1156= 34[Answer.]
36. the cost price of table and chair together is 690. If the table cost 30% more than chair,
then find the cost of table and chair?
#Solution:

98. T:C= 13:10
T= 13*690/23= 13*30= 390
C= 300
37. 6(3/4) * 13 (1/3) + 2^2*?= 30% of 520
#Solution:
(27/4)* (40/3)+ 4x= 3*52
=> 90+4x= 156
=> 4x= 66
=> x= 33/2= 16.5
38. I) 6x^2-11x+4= 0
ii) 50y^2-25y+3= 0
#Solution:
i) 6x^2-11x+4=0
=> 6x^2-8x-3x+4= 0
=> 2x(3x-4)-1(3x-4)= 0
=> x= 1/2 or 4/3
ii) 50y^2-25y+3= 0
=> 50y^2-15y-10y+3= 0
=> 5y(10y-3)-1(10y-3)= 0
=> y= 1/5 or 3/10
So, x>y

99. 39. There are 3 green, 4 orange and 5 white color bulbs in a bag. If a bulb is picked at
random, what is the probability of having either a green or a white bulb
#Solution:
Total = 12
Either a green or white = 8
Probability = 8/12= 2/3[Answer]
40. The ratio of pens and pencil in a stationary shop is 3 : 2, respectively, The average
number of pens and pencils is 180. What is the of number of pencils in the shop ?
#Solution:
Ratio of pen :Pencil= 3:2
ATQ,
5x/2= 180
=> 5x= 360
=> x= 72
Pencil = 144
41. P q and r start a business. p invests 3 times as much as q invests and q invests 2/3 rd as
much as r invests. find the ratio of capitals of p q and r
#Solution:
P:Q= 3:1= 6:2
Q:R= 2/3 :1= 2:3
P:Q:R= 6:2:3
42. A dishonest milkman professes to sell his milk at cost price but he mixes it with water
and thereby gains 25%. The percentage of water in the mixture is:

100. #Solution:
100:25= 4:1
%= 100/5= 20%[Answer.]
43. How many iron rods, each of length 7 meters and diameter 2cm can be made out of
0.88 cubic metre of iron?
#Solution:
(22/7)*7*(1/100)*(1/100) *n= 0.88
=> 22n= 8800
=> n= 400[Answer.]
44.A is three times as old as B. C was twice-as old as A four years ago. In four years' time,
A will be 31. What are the present ages of B and C ?
#Solution:
A= 3B
C-4= 2*(A+4)
=> C-4= 2*(A-4)
A+4= 31
=> A= 27
B= 9
So, C-4= 2*23
=> C= 4+46= 50[Answer.]
45. If 3√5+√125
= 17.88, then what will be the value of √80+6√5

101. #Solution:
3√5+5√5= 17.88
=> 8√5= 17.88
=> √5= 1788/800= 2.24
√80+6√5= 4√5+ 6√5= 10*2.24=22.4
46.What will be simple interest for 1 yr and 4 months on a sum of RS. 25800 at the rate of
14% per annum?
#Solution:
I= 25,800*14*16/100*12= 4816[Answer]
47. An error 2% in excess is made while measuring the side of a square. The percentage of
error in the calculated area of the square is:
#Solution:
2+2+2*2/100= 4.04[Answer.]
48.A man took loan from a bank at the rate of 12% p.a. simple interest. After 3 years he
had to pay Rs. 5400 interest only for the period. The principal amount borrowed by him
was:
#Solution::
36%= 5400
=> 100%= 15000[Answer.]
49. A boat covers a certain distance downstream in 1 hour, while it comes back in 1(1/2)
hours. If the speed of the stream be 3 kmph, what is the speed of the boat in still water?
#Solution:

102. U:D= 1:3/2= 2:3
B:S= 5/2:1/2= 5:1
1==3
5==15 km/hr[Answer]
50. Tk 385 has been divided among A, B, C in such a way that A receives 2/9th of what B
and C together receive. Then A's share is ?
#Solution:
A:(B+C)= 2:9
A= 385*2/11= 70[Answer.]
51. Mother’s age today is thrice as her daughter’s. After 10 years it would be just double.
How old is the daughter today ?
#Solution:
M= 3x
D= x
(3x+10)= 2*(x+10)
=> 3x-2x= 10
=> x= 10[Answer]
52. A boat can travel with a speed of 13 km/hr in still water. If the speed of the stream is 4
km/hr, find the time taken by the boat to go 68 km downstream.
#Solution:
68/(13+4)= 4 hr[Answer.]
53. In what time will the simple interest on Rs 400 at 10% per annum be the same as the
simple interest on Rs 1000 for 4 year at 4 % per annum?

103. #Solution:
400*t*10/100= 160
=> 40t= 160
=> t= 4 year [Answer]
54. A student multiplied a number by 3/5 instead of 5/3.What is the percentage error in
the calculation?
#Solution: Let the number= 15
Error = 15*5/3 - 15*3/5= 16
%= 16*100/25= 64%[Answer]
55. A car covers four successive 7 km distances at speeds of 10 km / hour, 20 km / hour,
30 km / hour and 60 km / hour respectively. Its average speed over this distance is
#Solution:
T= 7/10+7/20+7/30+7/60
= (42+21+14+7)/60
= 84/60= 7/5
D= 28
S= D/T= 28*5/7= 20 km/hr[Answer]
56. The total age of A and B is 12 years more than the total age of B and C. C is how
many year younger than A?
#Solution:
A+B-B-C= 12

104. => A-C= 12[Answer]
RBL: (SO)
Math Solution:
Solved by: Md Mutaher Hussain
33. A man can row at 5 kmph in still water. If the velocity of current is 1 kmph and it
takes him 1 hour to row to a place and come back, how far is the place?
#Solution::
Let, The Distance = D
ATQ,
D/6+D/4= 1
=> 5D= 12
=> D= 2.4 km[Answer]
34. Two men P and Q start from a place walking at 5 km/h and 6.5 km/h, respectively.
What is the time they will take to be 92 km apart, if they walk in opposite the direction?
#Solution::
RS= (5+6.5)= 11.5
Time = 92*10/115= (4*10/5)= 8 hr [Answer]
35. Two men, A and B, run a 4 km race on a circular course of 1/4 km. If their speeds are
in the ratio of 5:4, how often does the winner pass the other?
#Solution::
Total Race = 4 km
Course = 1/4
Speed ratio= 5:4

105. Winner meet once in every five rounds which is 5/4 km
Time = 4*4/5= 16/5[3 times]
[Answer]
36. A sum of money amounts to Rs.6690 after 3 years and to Rs.10,035 after 6 years on
compound interest.find the sum.
#Solution::
P*(1+r)^3= 6690______[1]
P*(1+r)^6= 10,035_____[2]
[2]/[1]
(1+r)^3= 10,035/6690= 3/2
Put this value into [1]
P*(3/2)= 6690
=> P= 6690*2/3= 4460[Answer]
37.One card is drawn at random from a pack of 52 cards. What is the probability that the
card drawn is a face card (Jack, Queen and King only)?
#Solution:
Favorable = 3*4= 12
Probability = 12/52= 3/13[Answer]
38. If two times of the daughter’s age in years is included to the mother’s age, the total is
70 and if two times of the mother’s age is included to the daughter’s age, the total is 95. So
the Mother’s age is,
#Solution::
Let Daughter age = x

106. Mother age = y
ATQ
2x+y= 70
=> y= 70-2x_______[1]
And,
2y+x= 95
=> y= (95-x)/2_______[2]
From [1] & [2]
(95-x)= 2*(70-2x)
=> 95-x= 140-4x
=> 3x= 45
=> x= 15
So, Mother age= (95-15)/2= 40 years [Answer]
39. A student multiplied a number by 3/5 instead of 5/3, What is the percentage error in
the calculation ?
#Solution:
LCM of 5 & 3= 15
Let, the number be= 15
Wrong result = 15*3/5= 9
Right result = 15*5/3= 25
% error = 16*100/25= 4*16= 64%[Answer]
40. Ten years ago A was half of B in age. If the ratio of their present ages is 3 : 4, what will
be the total of their present ages

107. #Solution:
Ten years ago A age = x
B age = 2x
ATQ,
(x+10)/(2x+10)= 3/4
=> 4x+40= 6x+30
=> 2x= 10
=> x= 5
Total.age = 3*5+20= 35[Answer]
41. The size of the wooden block is 5 x 10 x 20 cm .How many such blocks will be required
to construct a solid wooden cube of minimum size?
#Solution:
HCF of 5,10,20= 5
Minimum size= (5*10*20)/(5*5*5)
= 2*4= 8[Answer]
42. A tank is 25m long 12m wide and 6m deep. The cost of plastering its walls and bottom
at 75 paise per sq m is
#Solution::
Area of 4 wall and bottom = 2*(L+B)*H+L*B
= 2*(25+12)*6+25*12
= 2*37*6+ 300= 744 m^2
Cost of plastering = 744*3/4= 558[ Answer]

108. 43. The diference of two numbers is 20% of the larger number, if the smaller number is
20, then the larger number is :
#Solution::
Let, the number be = x and y
ATQ,
x-20= x/5
=> 4x/5= 20
=> 4x= 100
=> x= 25[Answer]
44. 12 buckets of water fill a tank when the capacity of each tank is 13.5 liters. How many
buckets will be needed to fill the same tank,if the capacity of each bucket is 9 liters?
#Solution::
ATQ,
9*B= 12*13.5
=> B= 12*135/90= 18[Answer]
45. The sum of three numbers is 98. If the ratio of the first to second is 2 :3 and that of the
second to the third is 5 : 8, then the second number is:
#Solution:
First:2nd = 2:3= 10:15
2nd:3rd= 5:8= 15:24
First:2nd:3rd= 10:15:24
2nd Number = 98*15/49= 30[Answer]

109. 46. If k:l= 4:3 and l:m= 5:3, then find k:l:m=?
#Solution:
K:l= 4:3= 20:15
l:m= 5:3= 15:9
K:l:m = 20:15:9[Answer]
48. The value of √{10+√25+√108+√154+√225}
#Solution:
√{10+√25+√108+√(154+15)
=√{10+√25+√(108+13)
= √{10+√36
= √16= 4[Answer]
49. A 270 metres long train running at the speed of 120 kmph crosses another train
running in opposite direction at the speed of 80 kmph in 9 seconds. What is the length of
the other train?
#Solution::
270+x= 200*5/2
=> 540+2x= 1000
=> x= 230 [Answer]
50. Three partners shared the profit in a business in the ratio 5 : 7 : 8. They had partnered
for 14 months, 8 months and 7 months respectively. What was the ratio of their
investments?
#Solution:
14x/8y= 5/7

110. => 98x= 40y
=> x:y= 20:49
And,
8y/7z= 7/8
=> 64y=49z
=> y:z= 49:64
x:y:z= 20:49:64[Answer]
51. An aeroplane flies along the four sides of a square at a speed of 200, 400, 600 and 800
km/h, respectively. What is the average speed of the plane in its flight around the square?
#Solution:
Let Total distance = 4D
Time = D/200+D/400+D/600+D/800
= (12D+6D+4D+3D)/2400
= 25D/2400
= D/96
Avg Speed = 4D*96/D
= 4*96= 384 km/hr
52. X can do a certain work in the same time in which Y and Z together can do it. If X and
Y together could do it in 10 days and Z alone in 15 days, then Y alone could do it in :
#Solution:
(X+Y):Z= 3:2
Time = 30/5= 6 days
(Y+Z):X= 1:1
So, X= 12 days
So, Y= 60/(10-9)= 60 days

111. 53. 98.98/ 11.03+7.014*15.99
= 99/11 +7*16
= 9+112= 121
54. A and B start a business with initial investment in the ratio 12 : 11 and their annual
profits were in the ratio 4 : 1. If A invested the money for 11 months . B invested the
money for ?
#Solution::
12*11/11*x= 4/1
=> 12/x= 4
=> x= 3
[Answer]
55. If m= 7-4√3 then (√m+1/√m)=?
#Solution:
1/m= 1/(7-4√3)
= (7+4√3)/1
= 7+4√3
So, m+1/m= 7-4√3+7+4√3
= 14
Now,
(√m+1/√m)^2= m+1/m +2= 14+2
=> √m+1/√m= 4[Answer]