Mutaher combined math file [www.onlinebcs.com]

Random 122 Math . Solved by:Mutaher Hussain
[Q-01]A bag contains 5 red and 7 white balls. Four balls are drawn out one by one and not replaced.
What is the probability that they are alternatively of different colours?
Solution:
Probability of alternatively different color
=(5/12)×(7/11)×(4/10)×(6/9)+
(7/12)×(5/11)×(6/10)×(4/9)
[Q-02].A and B can finish a work, working on alternate days, in 19 days, when A works on the first day.
However, they can finish the work, working on alternate days, in 19 5/6 days, when B works on the first
day. How many days does A alone take to finish the work?
Solution
10A+9B= 1_____[1]
10B +
59𝐴
6
= 1____[2]
From [1] & (2]
60A+54B= 60B+59A
=> A= 6B
=> A:B= 6:1
ATQ,
69= 6T
=> T= 11.5[Answer]
[Q-03] A and B can complete a work separately in 9 and 12 days respectively. If they work for a day
alternately in how many days the work will be completed.
Solution
A:B= 9:12[T]
A:B= 4:3[E]
Two days work = 4+3= 7 unit
(5×2)= 10 days work = 7×5= 35 unit
Remain= (36-35)= 1 unit
Now A turn,
4 unit in = 1 day
Note:Alternatively different color :
মানে প্রথমটা রেড হনে পনেেটা
রহায়াইট, এভানে রহায়াইনটে পে
রেড
RWRW+WRWR

1 unit in =
1
4
Time needed = 10+
1
4
= 10
1
4
[Q-04] The price of each mango is tk.65 and the price of each guava is tk.45.Aslam buys total 10mabgoes
and guavas at an average price of tk56.how many mangoes must Aslam put back &replace with guavas
so that the average price if the pieces of fruit that he keeps is 52 tk.
Solution:
Apply Role of Allegations
Mango Guava
65 45
56
11 9
And
Mango __________Guava
65_______________45
________52________
7_________________13
So, he put back (11-7) or (13-9)=
4
2
= 2 mangoes and replace with guava[Answer]
[Q-05]A cricket team has won 40 games out of 60 played. It has 32 more games to play. How many of
these must the team win to make it record 70% win for the season??
Alternative
Total Price of 10= 10×56= 560
He saves = (560-52×10)= 40 TK
If he buy two mangoes less then he
could save 40 TK.
Price difference = (65-45)= 20
So, Mangoes less = 40/20= 2[Answer]

Solution
Let he won x match after wining = 40 match
Total match he will play = 60+32= 92
ATQ,
40+𝑥
92
=
70
100
=
7
10
=> 400+10x= 644
=> 10x= 244
=> x= 24.4 or 24 match approximately
[Answer.]
[Q-06] The average weight of 15 boys is decreased by 5 kg when, one of them weighing 112 kg is replaced
by another one. This new one is again replaced by another, whose weight is 15 kg higher than the person
he replaced. What is the overall change due to this dual change?
Solution
Wt decrease by 5 kg that means total decrease = 15×5= 75 kg
So, Replaced man weight = 112-75= 37 kg
Now this man is replaced whose wt is 15 kg lower than the replaced man,
So Replaced man wt. = 37+15= 52 kg
So, Overall wt. Decrease = 112-52= 60 kg
Overall change per person=
60
15
= 4 𝑘𝑔
[Answer]
[Q-07] Lokesh deposits some amount in a bank .He gets simple interest which is 1/8 of the sum. If the
number of years is equal to twice the rate percent per annum, then the rate percent per annum is?
Solution
Let, Sum= P
Simple interest= P/8
Number of years = x
Rate = 2x%
Alternative:
Let, he won x match total
ATQ,
𝑥
92
=
7
10
=> 10x= 644
=> x= 64.4
So, he have to win = (64.4-40)
= 24.4 or 24 match
approximately [Answer ]

ATQ
𝑃
8
=
𝑃2𝑥2
100
=> 100= 16𝑥2
=> x=
10
4
= 2.5% [Answer]
[Q-08] A building worth Tk.133100 is constructed on land worth Tk.72900. After how many years will
the value of both be the same if land appreciates at 10% p.a. and building depreciates at 10% p.a??
Solution:
72900×{1 + (
10
100
)}
𝑛
= 133100×(1 −
10
100
)
𝑛
=> 72900×(
11
10
)
𝑛
= 133100×(
9
10
)
𝑛
=>(
11
10
)
𝑛
× (
10
9
)
𝑛
=
133100
72900
=> n= 3 year
[Answer.]
[Q-09] .Two airticles are sold for Rs.30 each. On one,he losses 30 % & on the others he earns a profit of
30 %.Find his loss /profit (in Rs) in the overall transaction
Solution
Each SP = 30
CP of first = 30 ×
10
7
=
300
7
CP of 2nd
= 30 ×
10
13
=
300
13
Total CP = (
300
7
+
300
13
) =
3900+2100
91
=
6000
91
Total SP= 30+30= 60
Alternative
Percent Loss = 30 -30 –
30×30
100
= 9%
Formula= P-Q-
𝑃𝑄
100
Here, P, Q profit & Loss
Respectively

Loss =
6000
91
− 60 =
(6000−5460)
91
=
540
91
Percent Loss =
540
91
×
91
6000
× 100
= 9%[Answer]
[Q-10] A conical tent is required to accommodate 5 persons and each of them needs 16 𝑪𝒎 𝟐
space on the
ground and 100 cubic metres of air to breathe. Find the vertical height of the tent.
Solution
Area of base = 5×16= 80 𝐶𝑚2
Let, Radius = r
Height = h
So, Area = 80
=> π𝑟2
= 80
And, Volume = 100×5= 500
=> (
𝜋𝑟2ℎ
3
= 500
=> 80 h= 1500
=> h= 18.75 cm[Answer]
[Q-11] Find the cost of carpeting a room 13 m long and 9m broad with a carpet 75cm wide at the rate of
tk 12.40 per sq. metre.
Solution
13×9×(4/3)×12.4= 39×4×12.4= 156×12.4= 1934.4[Answer.]

[Q-12]In a competitive exam, the number of passed students was four times the number of failed
students. If there had been 35 fewer appeared students and 9 more had failed, the ratio of passed and
failed students would have been 2 : 1, then the total number
of students appered for the exam
Solution
P= 4×F
=> P:F= 4:1
Total Students= 5x-35
ATQ,
𝑥+9
5𝑥−35
=
1
3
=> 3x+27= 5x-35
=> 2x= 62
=> x= 31
Appear = 5×31= 155[Answer.]
[Q-13] A & B enter into a partnership business. A supplies whole of the capital amounting to taka 45000
with condition that profits to be equally divided and that B Pays A interest on half of the capital at 10%
per annum, but receives taka 120 per month for carrying the concern. Find the total yearly profit when
B's income is one-half of A's income
Solution
B's income=
𝐴
2
=> A= 2B
=> A:B= 2:1[Income Ratio]
B paid A =
22500
10
= 2250
B receive = 120×12= 1440
Profit will be shared by both in same quantity.
Let, Shared profit = P
Alternative
Pass:Fail= 4:1
Total Students = 5x-35
Fail = x+9
Pass= 5x-35-x-9= 4x-44
ATQ
𝑥+9
4𝑥−44
=
1
2
 2x+18= 4x-44
 2x= 62
 x= 31
So, Total Students = 5×31= 155

A wiil received =
𝑃
2
And,
B will received =
𝑃
2
Since,
B get an interest of 1440 & given A to 2250 then B profit will be=
𝑃
2
+1440-2250
=
P
2
-810
And,
A will be received in Total
=
𝑃
2
+2250
ATQ,
P
2
+2250
P
2
−810
=
2
1
=> P+4500= 2P-3240
=> P= 7740
So, (A+B) profit in total = P+2250-810
= 7740+1440= 9180[Answer]
[Q-14] A boat goes 12 km downstream and return to the strating point in 180 minutes. If the speed of the
stream is 3 km/hr. Then what will be the speed of boat in still water?
Solution
Let, Boat speed, = x
ATQ,
12
𝑥+3
+
12
𝑥−3
= 3
=> (12x-36+12x+36)= 3(𝑥2
− 9)
=> 24x= 3𝑥2
− 27
=> 𝑥2
− 8𝑥 − 9 = 0
=> 𝑥2
− 9x + x − 9 = 0
=> x(x-9)+1(x-9)= 0

=> x= 9
So, Boat speed = 9 km/hr[Answer]
[Q-15] The wages of labourers in a factory increases in the ratio 22:25 and there was a reduction in the
number of labourers in the ratio 15:11. Find the original wage bill if the present bill is Rs. 5000
Solution
Let Previous wage= 22x
So, New wage = 25x
Previous number of workers= 15y
New number of workers = 11y
Total wages of present = 25x×11y= 275xy
Previous wages= 22x×15y= 330xy
ATQ,
275xy= 5000
 xy= 18.18
Previous wages= 330×18.18= 6000[Answer ]
[Q-16]A garments factory employs 3 categories of workers, K, M & N. The efficiency of the 3 categories
of worker to do a specific task is in the ratio of 3:5:6 respectively & the factory employs 15 of category K,
12 of category M & 3 of category N. If the total wages paid to the entire group of workers is tk 492 per
hour,what will be the wages earned by a category N worker for an 8 hour day?
Solution
M:K:N= (15×3:12×5:3×6)= 45:60:18
N in 1 hr= 492×(18/123)= 72
N in 8 hr= 72×8= 576
So, N, 1 worker =
576
3
= 192[Answer]
[Q-17] 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 final mixture
Solution:
Initial Mixture= 100 L
Water = 20 L
Ration of two numbers now=
22×15:25×11
= 2×3:5
= 6:5
Now,
5==5000
6==6000[Answer]

Milk = 80 L
Milk now = 80- 25×
4
5
= 80-20= 60
% of Milk = 60×100/100= 60%[Answer]
[Q-18] Three cats are roaming in a zoo n such a way that when cat A takes 5 steps, B takes 6 steps and C
takes 7 steps. But the 6 steps of A are equal to the 7 steps of B and 8 steps of C. what is the ratio of their
speeds
Solution
A:B:C= 5:6:7
And,
6A= 7B= 8C= x
=> A:B:C=
1
6
:
1
7
:
1
8
= 28:24:21
So, Ration of speed = (5×28):(24×6):(21×7)
= 140:144:147
[Answer]
[Q-19] In a group of 60 probationary officers, 40% of them were prompted , 12 of them were terminated
and rest of them were placed on probation for the second time. What percentage of the probationary
officers placed on probation for second time?
Solution
Promoted = 60×
2
5
=24
Terminated= 12
Remain = 60-36= 24
Percentage= 24×
5
3
= 40%[Answer]
[Q-20] A sum of Rs. 210 was taken as a loan. This is to be paid back in two equal installments.lf the rate
of interest be 10% compounded annually, then the value of each installment is
Solution

Let, Each installment = x
ATQ,
x
(1+
10
100
)
2 +
x
1+
10
100
=210
=> x= 121[Answer.]
[Q-21] In a fort, there are 1200 soldiers. If each soldier consumes 3 kg per day, the provisions available in
the fort will last for 30 days. If some more soldiers join, the provisions available will last for 25 days given
each soldier consumes 2.5 kg per day. Find the number of soldiers joining the fort in that case
Solution1
Let, Extra Men added= M
ATQ,
(1200×3×30)= (M+1200)×25×5/2
=> 2400×90-1200×25×5= 15×5×M
=> M= 528
So, Men added = 528[Answer]
[Q-22] Paint needs to be thinned to a ratio of 2 parts paint to 1.5 parts water. The painter has by mistake
added water so that he has 6 litres of paint which is half water and half paint. What must he add to make
the proportions of the mixture correct
Solution:
Paint:Water = 2:1.5
= 4:3
ATQ,
3+𝑥
3
=
4
3
=> 9+3x= 12
=> 3x= 3
=> x= 1
So, 1 L paint should be added[Answer.]

[Q-23] Mr. Kamal and his wife get a rise of 8% percent annually in their income.Kamal get a raise of
800tk while his wife get 840 tk.What is difference
between their income after the raise?
Solution
Let Kamal Initial Income =k
Rise his income = 8% of k=
8k
100
=
2k
25
ATQ,
2k
25
=800
 2k= 25×800
 k= 10,000
Let, Kamal Wife Income = W
Rise of Income = 8% of W=
8W
100
=
2W
25
ATQ,
2W
25
= 840
 2W= 840×25
 W= 10,500
Difference = (10,500-10,000)= 500[Answer.]
Q-24] 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
A::- G:S= 40:60
B::-G:S:C= 35:40:25
New, G:S= (1×40+4×35):(40×4+1×60)
= 180:220
= 18:22=9:11[Answer. ]
[Q-25] A man saves 200 Tk at the end of each year and lends the money at 5% compound interest. How
much will it becomeat the end of 3 years?
Solution
He lends money each year means,
First 200 Tk. for 3 years +2nd 200 tk for 2 years + third 200 tk for one year .
8%= 800
=> 100%= 10000
And,
8%= 840
=> 100%= 10,500
Diff. = (10,500-10000)= 500[Answer] [

It will become=
200×(
105
100
)
3
+200×(
105
100
)
2
+200×
105
100
= 662.03[Answer. ]
[Q-26] The CI on 20480 Tk at 6(1/4)% per annum for 2 years 73 days is?
Solution:
1 year=365 days
73 days =
73
365
=
1
5
Interest Rate =
25
4
So, CI = 20,480×(
425
400
)
2
×
405
400
= 2929 TK [Answer. ]
[Q-27]The compound interest on 2800TK for 18 month at10% interest per annum.
Sollution:
Time, n= 18 month=
18
12
=
3
2
CI = 2800×(
110
100
×
105
100
-2800
= 434 [Answer. ]
[Q-28] What would be the compound interest accrued on an amount of Rs. 8000 at the rate of 15% p.a. in
three years?
[Grab your reader’s attention with a great
quote from the document or use this space
to emphasize a key point. To place this text
box anywhere on the page, just drag it.]
Note, 1 year Interest rate = 25/4%
1/5 year interest rate = (25/4)×(1/5)=
(5/4)%
[Basic Formula:
If the interest rates are R1, R2, R3 for ist,
2nd &3rd year respectively
Then Amount A=
P×(1+R1%)(1+R2%)(1+R3%)
Here 18 month interest means interest
for 1 year and 6 month.

#Solution
Let, the principle= P
Interest rate=r%
ATQ,
P×(1 +
𝑟
100
)
𝑛
-P= 8000×(1 +
15
100
)
3
-8000
=12167-8000=4167[Answer. ]
[Q-29] Two pipes can fill a tank in 20 and 24 minutes respectively and a waste pipe can empty 3 gallons
per minute. All the three pipes working together can fill the tank in 15 minutes. The capacity of the tank
is
#Solution
Ist pipe:2nd pipe:Combined 3= 20:24:15[T]
Ist pipe:2nd pipe:Combined 3= 6:5:8[Eff.]
So, Waste pipe [E]= (6+5-8)= 3
Let, Capacity = 20×6= 120
ATQ,
3T= 120
=> T= 40 min
So, The capacity of Tank = 40×3= 120 gallons[Answer.]
[Q-30] A leak in the bottom of a tank can empty the full tank in 8 hours. An inlet pipe fills water at the
rate of 6 liters a minute. When the tank is full, the inlet is opened and due to the leak, the tank is empty in
12 hours. How many liters does the tank hold
[Basic formula:
If the principle is P, interest r%, Amount
after n years ,
A= P×((1 +
𝑟
100
)
𝑛
Interest = P×(1 +
𝑟
100
)
𝑛
−P [when,
Compounded annuallay ]
Alternative
In 15 min both pipe fill the cistern =
15
20
+
15
24
=
90+75
120
=
165
120
=
33
24
So waste pipe enpty =
33−24
24
=
9
24
So,
9
24
= 45
 9= 45×24
 1=5×24= 120 L
So, Capacity of Tank = 120 L

#Solution
(Inlet+Outlet):Outlet= 12:8[T]
(Inlet+Outlet):Outlet= 2:3[Eff.]
Inlet eff. = (3-2)= 1
Let, Capacity = 24 Unit
ATQ,
1×T= 24
=> T= 24 hr
So, Tank hold= 24×60×6= 24×360= 8640L
[Answer]
[Q-31] A person sold a tube light at tk 85.25 in such a way that his percentage profit is the same as the
cost price of the Tube light. If he sells it at twice the percentage profit of his previous percentage of the
profit then the new selling price will be.
#Solution:
%Profit= CP
=>
(SP−CP)100
𝐶𝑃
= CP
=> (85.25-CP)×100= 𝐶𝑃2
=> CP2
+ 100CP − 8525 = 0
=> CP2
+ 155CP − 55CP − 8525 = 0
=> CP= 55
So, CP= 55
New % Profit= 110%
New SP= (
21×55
10
=115.5[Answer.]
[Q-32] An elevator starts with 4 passenger and stops at 7 floors of an apartment.Find the probability that
all 4 passengers travel to different floor?
#Solution
7th floor =7/7
6 th floor = 6/7

5 th floor = 5/7
4th floor = 4/7
So, P= (7×6×5×4)/(7×7×7×7)
[Answer]
[Q-33] The ratio of number of girls to the number of boys is the 5: 2 in a class of 21 students. A group of
three students is to be selected at random amongst them. What is the probability that the selected group
of students contain one boy and two girls?
#Solution
Number of boys= 21×
2
7
=6
Number of girls = 21-6= 15
Probability of one boy and two girls =
6C1×15C2
21C3
= 45/266[Answer]
[Q-34]Two joggers left Delhi for Noida simultaneously. The first jogger stopped 42 min later when he was
1 km short of Noida and the other one stopped 52 min later when he was 2 km short of Noida. If the first
jogger jogged as many kilometers as the second and the second as many kilometers as the first, the first
one would need 17 min less than the second. Find the distance between Delhi and Noida.
#Solution:
Let,
Total distance = x km
1st jogger speed = v=
d
t
=
𝑥−1
42
60
2nd jogger speed =
x−2
52
60
Atq,
2nd jogger cross =(x-1) km
1st jogger cross = (x-2) km
Now,
(x-1)/[(x-2)/ (52/60)] -(x-2)/[(x-1)/(42/60)] =17/60
=> (x-1)*52/(x-2)- 42*(x-2)/(x-1)= 17

=> (x-1)^2*52-42*(x-2)^2= 17*(x-1)(x-2)
=> 52x^2-104x+52-42x^2+168x-168= 17(x^2-2x-x+2)
=> 10x^2+64x-116= 17x^2-51x+34
=> 7x^2-51x-64x+34+116= 0
=> 7x^2-115x+150= 0
=> 7x^2-115x +150 =0
=> 7x^2-105x-10x+150=0
=> (x-15)(7x-10)=0
=> x=15 km ( Ans)
[Q-35] A man spent 1/2 of his money and lost 1/4 of the remainder. He was left with taka 3,600. How
much did he start with?
#Solution:
Let, Total Money = 24
Spent = 24/2= 12
Lost = 12/4= 3
So, Remain = 24-15= 9
Now,
9===3600
24==24×400= 9600[Answer]
[Q-36]. Tap A fills a tank in 4 hours whereas tap B empties the full tank in 24 hours. A and B are opened
alternately for 1 hour each. Every 2 hour the level of water is found to increase by . 5m. The depth of the
tank is
#Solution
Let, Capacity = 24 L
Eff. A:B= 6:1
In 2 hr = (6-1)= 5 unit
Now,
5 ===0.5
24==24×0.5/5= 12/5= 2.4 m
[Answer.]
Alternative
In 2 hr A+B can fill
=
2
4
−
2
24
=
12−2
24
=
5
12
So,
5
12
=
1
2
 1=4.8

[Q-37].A train passes two bridges of lengths 500 m and 250 m in 100 seconds and 60 seconds respectively.
The length of the train is
#Solution:
Let, Length of Train = L
ATQ,
L+500
100
=
L+250
60
=> 3L+1500= 5L+1250
=> 2L= 250
=> L= 125 m
So, Length of Train = 125 m[Answer]
[Q-38] A cistern has an inlet pipe and an outlet pipe.The inlet pipe fills the cistern completely in 1 hour 20
minutes when the outlet pipe is plugged. The outlet pipe empties the tank completely in 6 hours when the
inlet pipe is plugged.If there is a leakage also which is capable of draining out the water from the tank at
half of tth rate of the outlet pipe, then what is the time taken to fill the empty tank when both pipe are
opened.
#Solution:
Inlet takes = 80 min
Outlet takes= 360 min
Leakage = 720. Min
Inlet:Outlet:L= 80:360:720[T]
Inlet:Outlet= 9:2:1
Let, Capacity= 360×2= 720 L
ATQ,
(9-2-1)×T= 720
=> T= 720/6= 120 min or 2 hours [Answer.]
[Q-39] Two pipes A and B can fill a cistern in 12 minutes and 15 minutes respectively while a third pipe
can empty the full cistern in 6 minutes. A and B are kept open for 5 minutes in the beginning and then C
is also opened. In what time is the cistern emptied
Alternative
Inlet Pipe take = 80 min
Outlet = 360 min
Leakage= 720
So, In 1 hr fill=
1
80
−
1
360
−
1
720
=
6
720
So, 1 part in = 120 min

#Solution
A:B:C= 12:15:6[T]
A:B:C= 5:4:10[E]
Let, Capacity = 12×5= 60 L
In 5 min (A+B) fill = 5×(5+4)= 45 unit
ATQ,
T×(5+4-10)= 45
=> T= 45 min
So, Total Time needed= 45+5 min= 50 min
[Answer]
[Q-40] A, B and C enter into partnership by making investments in the ratio 3: 5:7. After a year, C
invests another Tk. 337600 while A withdraws Tk.45600. The ratio of investments then changes to
24:59:67. How much does A invest initially?
#Solution
Let, A:B:C= 3x:5x:7x
After 1 year C investment = 7x+337600
After 1 year A investment = 3x-45,600
ATQ,
3x−45600
5x
=
24
59
=> 59×(3x-45,600)= 120x
=> 177x-120x= 45600×59
=> 57x= 45600×59
=> x= 800×59= 47,200
A initial investment = 47,200×3= 141600[Answer.]
[Q-41]:A starts from a place at 11.00 A.M. and travels at a speed of 4 kmph, B starts at 1.00 P.M. and
travels with speeds of 1 kmph for 1 hour, 2 kmph for the next 1 hour, 3 kmph for the next 1 hour and so
on. At what time will B catch up with A?]
#Solution
A start 2 hr earlier. In that time A goes =4×2= 8 km
B start at 1.00 P.M.
In 1.00-2.00 distance between them = 8+4-1= 11

In 2.00-3.00 distance between them =
11+4-2= 13 km
In 3.00-4.00 distance between them = 13+4-3= 14 km
Now Remain distance = 4 km
Relative speed= (9-4)= 5 km/hr
T= 4/5= 48 min
So, they meet at 9:48 P.M[Answer.]
[Q-42] An express train travelled at an average speed of 100 kmph, stopping for 3 minutes after 75 km.
A local train travelled at a speed of 50 kmph, stopping for 1 minute after every 25 km. If the trains began
travelling at the same time, how many kilometres did the local train travel in the time it took the express
train to travel 600 km?
#Solution
Express train time needed = (600/100)+7×3 min
= 6 hr 21 min= 381 min
Local train speed = 50 km/hr
In every 31 min it goes= 25 km
So, in (31×12)= 372 min it goes = 25×12= 300 km
Remain Time = 9 min
In that time it goes= 9×50/60= 45/6= 7.5 km
So, Total distance = 300+7.5= 307.5 km
[Answ
[Q-43] A motorboat in still Water travels at a speed of 36 km/ hr. It goes 56 km upstream in 1 hour 45 minutes.
The time taken by it to cover the same distance down the stream will be :
#Solution_
Let, boat speed = x
Stream speed = y

56 km upstream in = 1(3/4)= 7/4 hr
Speed = 56×4/7= 32 km/hr
So, x-y= 32
=> 36-y= 32
=> y= 4 km/hr
So, Downstream speed= 36+4= 40 km/hr
So, Time needed= 56/40= 14/10=7/5= 1(2/5)hr or 1 hr 24 min[Answer.]
[Q-44] A, B and C can do a work together in a certain number of days. If A leaves after half the work is done,
then the work takes 4 more days for completion, but if B leaves after half the work is done, the work takes 5
more days for completion. If A takes 10 more days than B to do the work alone, then in how many days can C
alone do the work?
[ A,B,C একটট কাজ নেনদিষ্ট নদনে রেষ কনে। যনদ A অনধ িক কাজ রেষ কোে পে চনে যায় োনক কাজ সম্পন্ন হনে চােনদে রেনে সময়
োনে নকে্েু যনদ B অনধ িক কাজ রেষ কোে পে চনে যায় োহনে কাজটট সম্পন্ন কেনে আনো পাাঁচনদে রেনে সময় োনে। যনদ সম্পূর্ ি
কাজটট কেনে A, B এে রথনক ১০ নদে রেনে সময় রেয়, োহনে C সম্পূর্ িকাজ একা কেনদনে রেষ কেনে পােনে?]
#solution_
Let,
A+B+C= x days
(B+C), 1/2 work in = (x/2+4)days
Full work= (x+8) days
Similarly
(A+C), 1/2 work= x/2+5 days
Full work = (x+10) days
Since,
1/A+1/B+1/C= 1/x
=> 1/A= 1/x-1/x+8
= (x+8-x)/x*(x+8)= 8/x*(x+8)
So A, 8/x*(x+8) part in= 1 days
1 part in = x*(x+8)/8 days
Similarly,
1/B= 1/x-1/(x+10)= 10/x*(x+10)
: B= x*(x+10)/10

ATQ,
x*(x+8)/8- x*(x+10)/10= 10
=> [5x(x+8)-4x(x+10)]= 400
=> 5x^2+40x-4x^2-40x= 400
=> x^2= 400
=> x= 20
So, A= 20*(20+8)/8= 70
B= 20*(20+10)/10= 60
Now,
1/A+1/C= 1/30
=> 1/30-1/70= 1/C
=> 7-3/210= 1/C
=> 1/C= 4/210
C, 4/210 part in= 1 days
1 part in= 210/4= 52.5 days[Answer.]
.[Q-45] Mr X signed a contract for building a road for 1920 meters long within 120 days.He employed 160
workers for this task.But after 24 days he found that only 1/8 of the task has been finished .If Mr.X wants to
finish the road in time how many additional workers he has to employ ?
#Solution:
Total work = (24×160×8)
Let, Extra Men = M
ATQ,
24×160+96×(M+160)= 24×160×8
=> 86M = 24×160×8-24×160-96×160
=> M= 120
Additional Men= 120[Answer]
[Q-46]If A completes 2/3 part of a work in 6 days, C completes 1/4 part of the work in 3 days and B completes
1/3 part of the work in 2 days. Then find in how many days A and C together will complete 2/3 part of the
work?
#Solution
A= 6×3/2= 9 days

C= 4×3= 12 days
A:C= 4:3[Eff.]
ATQ,
(4+3)×T= 36×2/3
=> 7T= 24
=> T= 24/7 days[Answer]
[Q-47] A man covers a distance on scooter by 3 kmph faster and takes 40 min less. But if he decrease his speed
by 2 kmph then he is 40 min late. Find the distance.
#Solution:
Let, Distance= D km
Speed = V km/hr
ATQ,
D/(V) - D/(V+3)= 2/3_____[1]
And,
D/(V-2)-D/V= 2/3_________[2]
From [1] &[2]
D/V-D/(V+3)= D/(V-2)-D/V
=> 2/V= (V+3+V-2)/(V+3)(V-2)
=> 2(V^2-2V+3V-6)= V×(2V-1)
=> 2V^2+2V-12-2V^2-V= 0
=> V= 12 km/hr
Put this value into [1]
D/12 -D/15= 2/3
=> D/60= 2/3
=> 3D= 120
=> D= 40 km[Answer]
[Q-48] ramesh has a container filled completely with 80 % milk and 20 % water. 5 litres of thesolution is
removed and replaced with water. then, 1 5 litres of this solution is removed andreplaced with water. the milk
percentage is now 55%. which of the following can be thecapacity of the container (in litres)
#Solution:
Let, initial = x

Milk Initial = 4x/5
ATQ,
4x/5*(1- 5/x)(1-15/x)= 55% of x
=> 4/5*(x-5)(15-x)= 11x^2/20
=> 16(x^2-15x-5x+75)= 11x^2
=> 16(x^2-20x+75)= 11x^2
=> 16x^2-320x+ 1200-11x^2= 0
=> 5x^2-320x+1200= 0
=> 5x^2-300x-20x+1200= 0
=> x= 4,60
4 can not be, since already added 20 L, so solution must be greater than 20.
So, initial = 60 L[answer.]
[Q-49]P, Q and R are three typists who working simultaneously can type 216 pages in 4 hours. In one hour, R
can type as many pages more than Q as Q can type more than P. During a period of five hours, R can type as
many pages as P can during seven hours. How many pages does each of them type per hour ?
#Solution:
P+Q+R= 216/4= 54______[1]
R-Q= Q-P
=> R+P= 2Q
2Q+Q= 54
=> Q= 18 page/hr
And,
5R= 7P
=> R= (7P/5) ; put this value into [1]
P+18+(7P/5)= 54
=> 12P/5= 36
=> P= 15 page/hr
So, R= 7×15/5= 21 page/hr

[Q-50] To do a piece of work, B takes 3 times as long as A and C together and C twice as long as A and B
together. If A, B and C together can complete the works in 10 days, how long would A take alone to complete
it?
#Solution
B:(A+C)= 1:3[eff.]= 3:9
C:(A+B)= 1:2[eff.]= 4:8
A eff. = (8-3)= 5
ATQ,
A×5= 10×(3+9)
=> A= 120/5= 24 days[Answer]
#Alternative
Let, B takes= 3x days
(B+C) takes = x days
ATQ,
1/3x +1/x= 1/10
=> 4/3x= 1/10
=> x= 40/3 days
So, B takes = (40/3)×3= 40 days
Again,
Let, C takes = 2y days
(A+B) takes = y days
ATQ,
1/2y+1/y= 1/10
=> 3/2y= 1/10
=> 2y= 30
So, C takes = 30 days
Lcm of 10, 30,40= 120
Eff.
(A+B+C):B:C= 12:3:4
So, A= (12-3-4)= 5

Time needed= 120/5= 24 days[Answer]
.[Q-51] A milk vendor has 2 cans of milk. The first contains 25% water and the rest milk. The second contains
50% water. How much milk should he mix from each of the containers so as to get 12 litres of milk such that
the ratio of water to milk is 3 : 5?
#Solution;
Milk : Water = 75:25[1st]
Milk :Water = 50:50[2nd]
Milk:Water= 3:5= 37.5:62.5[Final]
(75-37.5):(62.5-25)= 37.5:37.5= 1:1
So, 12×1/2= 6 L each[Answer ]
[Q-52] Devide tk 1050 among A, B and C such that A receives 2/5 as much as B and C together and B receives
3/7 as much as A and C together
#Solution
A:(B+C)= 2:5
A got = 1050×2/7= 300
B:(A+C)= 3:7
B got = 1050×(3/10)
= 315
C got = 1050-615=435
A+C=(300+435)= 735[Answer.]
.[Q-53]A tap can fill a tank in 6 hours. After half the tank is filled, three more similar taps are opened. What is
the total time taken to fill the tank completely
#Solution:
Let The capacity of tank = 6 L
So, Eff. = 1
Half fill in = 6/2= 3 hr
Now, pipe = 4
Per hr can fill = 4 L
Total time = 3+(3/4)= 3 hr 45 min[Answer.]
[Q-54] A shopkeeper marks up his goods by 20% and then gives a discount of 20%. Besides he cheats both his
supplier and customer by 100 g, i.e., he takes 1100 g from his supplier and sells only 900 g to his customer.
What is his net proﬁt percentage? (Rounded off to two decimal points)

#Solution:
Applying MF;
100×(6/5)×(4/5)×(11/10)×(10/9)-100= 117.33-100= 17.33%[Answer.]
===
[Q-55] A 72 liters of mixture of water and milk in the ratio of 4: 5. Then adding some amount of water and the
volume of the mixture is increased by 88 liters. The quantity of the milk is how much percentage more or less
than of water in the final mixture?
#Solution:
===
Milk = 72×5/9= 40
Water = 72×4/8= 32
New volume = (72+88)= 160 L
Water there = (32+88)= 120 L
Milk less than water = (120-40)= 80 L
% less = 80×100/120= 2×100/3= 200/3% [Answer.]
==
[Q-56] Three pipes A,B and C attached to a cistern. A can fill it in 10 min, B in 15 min, C is a waste pipe for
emptying it. After opening both the pipes A and B, a man leaves the cistern and returns when the cistern should
have been just full. Finding, however, that the waste pipe had left open, he closes it and the cistern now gets
filled in 2 min. In how much time the pipe C, if opened alone, empty the full cistern?
#Solution:
===
A & B together can fill the tank = 30/(3+2)= 6 min
So, 6 min = 1 part
2 min = 1/3 part
So, 1/3 part by Pipe C in = 6 min
=> 1 part by pipe C in = 6×3= 18 min[Answer]
===
[Q-57] ravi can walk a certain distance in 40days when he rests 9hours a day.how long will he take to walk
twice the distance, twice as fast and rest twice as long each day?
#Solution:

===
Let, Ravi speed per hr = x km/hr
In 40 days he took 9 hr rest every day so total time of travelling = 40×(24-9)×x
= 600x
Now, Distance is twice that means = 1200 x
Travel per day = (24-18)= 6hr
Speed twice that means = 2x
So time needed = 1200x/12x= 100[Answer.]
===
.[Q-58]If two pipes function simultaneously, the reservoir will be filled in 12 hours. The second pipe fills the
reservoir 10 hrs faster than the first. How many hrs does it take the second pipe to fill the reservoir?
#Solution:
===
Let, Capacity of reservoir = V unit
Both pipe can fill per hour = V/12 unit
First pipe can fill in 1 hr = V/t
2nd pipe can fill in 1 hr = V/(t-10)
ATQ,
V/t+ V/(t-10)= V/12
=> (t-10+t)×12= t×(t-10)
=> 24t-120= t^2-10t
=> t^2-34t+120= 0
=> t= 30,4
If t= 30, (t-10)= 20
If t= 4; (t-4)= 0[Can not be possibe]
So, second pipe takes = 20 hr[Answer.]
===
.[Q-59] Tank A is filled to 1/4th of its capacity while Tank B is filled to 1/8th of its capacity. The content from
both tanks are transferred to Tank C which is filled to 1/2th of its capacity. If twice the content of Tank B is
equal to the content in tank A, what is the ratio of the capacity of Tank A to Tank C?

#Solution:
===
A contents capacity = 2×B contents capacity
=> A:B= 2:1
So,C= (2+1)= 3
Capacity of C= 3×2= 6
A Capacity = 2×4=8
A:C= 8:6= 4:3
[Answer.]
===
[Q-60]while covering a distance of 24kn,a man noticed that after walking for 1 hour and 40 minutes, the
distance covered by him was 5/7 of the remaining distance. what was his speed in meters per second?
#Solution:
===
Remain distance = (24-D)
Initially cover = D
In 5/3 hour cover,D= 5/7(24-D)
= (120-5D)/7
So,
7D= 120-5D
=> 12D= 120
=> D= 10 km
So, (5/3) hr = 10 km
In 1 hr= 10×3/5= 6 km/hr= 6×(5/18)= 5/3 m/s[Answer]
===
.[Q-61] A farmer wishes to start a 100 sq.m rectangular vegetable garden. Since he has only 30 m barbed wire,
he fences three sides of the garden letting his house compound wall act as the fourth side fencing. The
dimension of the garden is
#Solution:
===

Let, Length = L
Breadth = B
ATQ,
L×B= 100_____[1]
And,
2L+B= 30
Or,
2B+L= 30
If 2L+B= 30
Then, B=30-2L
(30-2L)×L=100
=> 30L-2L^2-100= 0
=> 2L^2-30L+100=0
=> L^2-15L+50=0
=> L=10,5
If L=10, B= 10
If L= 5 ;B= 20[Breadth can not be greater than Length ]
Again,
If, 2B+L= 30
=> L= 30-2B
B×(30-2B)= 100
=> 30B-2B^2-100= 0
=> 2B^2-30B+100= 0
=> B^2-15B+50=0
=> B= 10,5
If B= 10,L= 10
If B= 5, L= 20
The Dimension of garden be, (10,10) or (20,5)[Answer.]

===
[Q-62] A bag contains some white and Black Balls.The probability of picking two white balls one after other
without replacement from that bag is 14/33.Then what will be the probability of picking two black balls from
that bag if bag can hold maximum 15 balls only?
#Solution:
===
Total ball = (W+B)
ATQ,
WC2/(W+B)C2= 14/33= 7*2/3*11
=> W*(W-1)/(W+B)(W+B-1)= (8/12)*(7/11)
So, W/W+B= 8/12
W+B= 12
W=8
B= 4
Probability of Black = (4/12)*(3/11)= 1/11[Answer.]
===
[Q-63]A person leaves his home everyday at 11:00 am and reaches his office at 12:00 pm. One day he left his
house at normal time but traveled the first half of the distance at speed of 2/3 of the normal speed. What should
be the speed of second half so that he reaches at the same time?
#Solution:
===
Let, Distance = 6 km
Speed= 6 km/hr.
First half distance = 6/2 = 3 km
(2/3) normal speed =6×(2/3)= 4 km/hr
So, 3 km in = 3/4 hr
Time remain = 1/4 hr
Distance Remain= 3 km
Speed =3×4 = 12 km/hr[Answer.]

So speed is twice than the first
===
[Q-64]A man rides a bicycle at the speed of 20 kmph and reaches the office at 15 minutes late. If he increases
his speed by 5 kmph and reaches the office in time, then what is the distance between house and office?
#Solution:
===
Let, The distance between home & office= D
ATQ,
D/20-15/60= D/25
=> D/20-D/25= 1/4
=> D/100= 1/4
=> 4D= 100
=> D= 25 km[Answer ]
===
#Alternatives:
===
Let, Actual time= t
ATQ,
20×(t+1/4)= 25t
=> 20t+5= 25t
=> 5t= 5
=> t= 1 hr
So, Distance = 25×1=25 km[Answer.]
===
.[Q-65]The length of train X is 600 m . And it crosses train Y running in opposite direction in 36 seconds and
train X also crosses a man standing in a platform in 36 seconds. What is the speed of train Y if the length of
train Y is half of train X?
#Solution:
===
Length of Train X= 600 m
Length of Y= 600/2= 300 m

Let, Speed of Train X= X
Speed of Train Y= Y
ATQ,
(600+300)= (X+Y)×(5/18)×36
=> (X+Y)×10= 900
=> (X+Y)= 90_____[1]
And,
600= X×(5/18)×36
=> 10X= 600
=> X= 60
So, Speed of Y= 90-60= 30 km/hr[Answer.]
===
[Q-66] What is the rate of interest on the amount Rs. 20000 invests in the compound interest scheme for 2 years
and the amount obtained after 2 years is Rs. 26450?
#Solution:
===
Let, Rate of interest = R
ATQ,
20,000×(1+R)²= 26,450
=> (1+R)²= 26,450/20,000
=> (1+R)²= 2645/2,000
=> (1+R)²= 529/400
=> 1+R= 23/20
=> R= 3/20= 3×(100/20)%= 15%
[Answer.]
===
[Q-67]A road was constructed along the fence of a rectangular plot,measuring 20 ft.long and 16ft.wide.The
width of the road was 2ft.Trees were planted 2ft.apart on both sides of the road. How many trees were planted?
[Dutch Bangla BanK PO - 2003 ]

#Solution:
===
Number of trees = 2×22+2×18
= 44+36= 80
===
[Q-68]After working for 15 days Adam realises that only 60% of the work is over and he employs Pat who is
half as efficient as Adam to help him. When will the task be completed
#Solution:
===
Adam can do the work in = 15×(5/3)= 25 days
So Pat = 50 days
Adam:Pat = 2:1[Eff.]
ATQ,
(2+1)×T=(2/5)×50
=> 3T = 20
=> T= 20/3 days
Total time taken = 15 +20/3
= 65/3 days
===
[Q-69]If P and Q work together, they will complete a job in 7.5 days.However, if P works alone and completes
half the job and then Q takes over and completes the remaining half alone, they will be able to complete the job
in 20 days. How long will Q alone take to do the job if P is more efficient than Q ?
#Solution:
===
1/P+1/Q= 2/15
=> (P+Q)×15= 2PQ----[1]
And,
P/2+Q/2= 20

=> P+Q= 40
=> P= 40-Q
(40-Q+Q)×15= 2Q×(40-Q)
=> 600= 80Q-2Q^2
=> Q^2-40Q+300= 0
=> Q^2-30Q-10Q+300= 0
=> Q= 30,10
Since Q is less efficient Q take = 30 days
[Answer.]
===
.[Q-70]If 12 candies are sold for tk. 10 then there is a loss of X%. If 12 candies are sold for Tk. 12 then there is
a profit of x%. What is the value of x?
#Solution:
===
Let, CP = CP
Profit and loss percentage in both case is same.
ATQ,
(CP-10)/CP= (12-CP)/CP
=> CP^2-10CP= 12CP-CP^2
=> 2CP^2= 22CP
=> CP= 11
Value of x= (11-10)×100/11= 100/11%
===
Mental Approach:
Since Profit and loss be same. Then CP value must be lies between 10 to 12.There is only value between 10-12
is 11.
So, x= (11-10)×100/11= 100/11%
[Answer ]

===
.[Q-71] A boat takes 19 hours for travelling downstream from point A to point B and coming back to a point C
midway between A and B. If the velocity of the stream is 4 kmph and the speed of the boat in still water is 14
kmph, what is the distance between A and B??
#Solution:
===
A========C=======B
Let, AB= 2D
AC= D
ATQ,
2D/(4+14) + D/(14-4)= 19
=> D/9+D/10= 19
=> 19D= 19×90
=> D= 90 km
Distance between A&B= 90×2= 180 km [Answer.]
===
[Q-73] 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 every third day. Find the number of days required to complete the work?(NM#19)
#Solution:
===
A:B:C= 12:15:25[T]
A:B:C= 25:20:12[Eff.]
Let, Total work = 12×25= 300
Working scheduled for 3 days = 2(A+B)+A+B+C
= 2×(25+20)+25+12
= 147 unit in _3 days
So, 294 unit in = 6 days
Remain 6 unit in = 6/45= 2/15
Time total = 6(2/15)
===

[Q-73] Profits of a business are divided among three partenrs A,B,C in such a way that 4 times the amount
received by A is equal to 6 times of B and 11 times of C.The ratio in which the three received the amount is--
#Solution:
===
4A= 6B=11C
=> A:B:C= 1/4:1/6:1/11= 132/4:132/6:132/11
= 33:22:12[Answer]
===
[Q-74]The probability of rolling any number on a weighted 6 sided die, with faces numbered 1 through 6, is
directly proportional to the number rolled. What is the probability of getting 5, if the die is rolled only once?
#Solution:
===
probability = individual probability / total probability = 5/(1+2+3+4+5+6)= 5/21
[N.B: Weighted probability means :: For example, if you were rolling a single six-sided die, you would have the
same probability of rolling a one as rolling any other number because each number will come up one out of six
times
Now,
Probability of rolling a 1 = 1 * m ( let m be any multiplier)
similarly rolling 2 = 2* m
..
rolling 6 = 6 * m
These are the possible outcomes and since the probability must be 1

so add these possible outcomes : 1m + 2m + . . . + 6m = 1
or m = 1/21]
[Q-75]Three acids A, B and C are mixed in the ratio 4:5:6 respectively, to from a chemical for washing white
clothes. The quantity of chemical is altered so that the ratio of acids A&B is 1/3 and the ratio of the acids B&C
is made 2/3, the new ratio of all three can now be used to form a chemical for washing colored clothes. What is
the total volume of new chemical solution formed if it contains 30ml of acid B? (Math#18)
#Solution:
===
A:B:C= 4:5:6[15]
= 68:85:102
New,
A:B= 1:3= 2:6
B:C= 2:3= 6:9
A:B:C= 2:6:9[17]
= 30:90:135
90==30
255= 255/3= 85 mL
===
[Q-76]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? Math#১৯(Ans : 82.5%)
#Solution:
===
Let, Total Marks = 100
Final marks = 40
Remain = 60

Moe obtains= 60×(9/20)= 27
Have to obtain = (60-27)= 33 out of 40
%= 33×(5/2)= 165/2= 82.5%
[Answer.]
===
[Q-77]Suppose you deposit tk 10000 on January 1.2013 at 12.50% interest rate for 1years. On july 1 2013 tk
15000 at 12 interest rate for 6 months and on October 1, 2013'tk 20000 at 11.5% interest ratevfor 3 months.
Supposed you withdrawal all deposits including due interest on December 31.2013. Calculated over all annual
rate of interest you have received??
#Solution:
===
Investment of TK 10,000 for 12 months
Investment of TK 15000/2 for 12 month
Investment of TK 2000/4 for 12 month
So, Total Amount= 10,000+7500+500= 22,500 gives
Interest of = 10,000×1×(1/8)+ 7,500×12/100
+ 5000×(11.5/100)
= 1250+ 900+ 575= 2725 interest
Percent interest = 2725×100/22500= 12.11%[Answer]
===
[Q-78]There are three boxes each containing 3 Pink and 5 Yellow balls and also there are 2 boxes each
containing 4 Pink and 2 Yellow balls. A Yellow ball is selected at random. Find the probability that Yellow ball
is from a box of the first group?
#Solution:
===
No.of pink in first group = 3*3=9
No. of yellow in first group=3*5=15
No. Of pink in 2nd group = 2*4= 8

No. Of yellow in 2nd group = 2*2= 4
Total box= 5
Favourable outcome = 3/5*15/24= 3/8
Total outcome
3/5*15/24 + 2/5 * 2/6= 61/120
Probability = (3/8)/(61/120)
= 3/8 * 120/61= 45/61
===
[Q-79]A dealer offers a cash discount of 20% and still makes a profit of 20% when he further allows 16 articles
to be sold at the cost price of 12 articles to a particular sticky bargainer. How much percent above the cost price
were his articles
#Solution:
===
12 article CP= 1200[Assume]
Per piece CP= 100
12 articles at DP= 1200×(4/5)
= 960
Since he Sold 16 articles instead of 12
Price per each = 960/16= 60 and there gain by 20%
CP= 60×(5/6)= 50
Actual CP = 50
Above cost price = 50×100/50= 100%
===
[Q-80]The ratio of women to children at a certain party is 2 to 5 and the ratio of children to men is 3 to 4. If
there are more than 13 and less then 20 women at the party, what is the number of men at the party?
(NM#18)

#Solution:
===
W:C= 2:5= 6:15
C:M= 3:4= 15:20
More than 13 less than 20
Means 14 to 19
18 is divisible by 6
So women = 18
6==18
20==60
===
[Q-81]Cost price of 12 oranges is equal to the selling price of 9 oranges and the discount on 10 oranges is equal
to the profit on 5 oranges. What is the percentage point difference between the profit percentage and discount
percentage?
#Solution:
===
12CP= 9SP
=> CP:SP= 9:12
And, CP= (3SP/4)
Profit percentage = 9×100/12= 33(1/3)%
And,
10D= 5P
=> P= 2D
=> P = 2(MP-SP)
=> SP-CP= 2(MP-SP)
=> SP-(3SP/4)= 2(MP-SP)
=> (4SP-3SP)/4= 2MP-2SP
=> SP= 8MP-8SP

=> 9SP= 8MP
=> MP:SP= 9:8
So, DP= (9-8)×100/9= 11.11%
Diff. = (33.33-11.11)= 22.22%
===
[Q-82]Three men A, B, C working together can do a job in 6 hours less time than A alone, in one hour less time
than B alone and in one half the time needed by C when working alone. Then A and B together can do the job
in:
#Solution:
===
Let, (A+B+C)= t
A= (t+6)
B= (t+1)
C= 2t
ATQ,
1/(t+6) +1/(t+1) + 1/2t= 1/t
=> (t+1+t+6)/(t+1)(t+6)= (2-1)/2t
=> (2t+7)×2t= t^2+6t+t+6
=> 4t^2+14t= t^2+7t+6
=> 3t^2+7t-6= 0
=> 3t^2+9t-2t-6= 0
=> t= 2/3
A= (2/3)+6=20/3
B= (2/3)+1= 5/3
LCM of 20/3. 5/3= 20/3
Eff. A:B= 1:4
ATQ,
5 T= 20/3
=> T= 4/3 hr

===
[Q-83]A bottle contains (3/4) of milk and the rest water. How much of the mixture must be taken away and
replaced by an equal quantity of water so that the nixtude has half milk and half water?
#Solution:
Milk : Water = 3/4:1/4
= 3:1
Let Replace = x
Milk now = 3/4- 3x/4
= (3-3x)/4
Water now = 1/4-x/4+ x
= (1-x+4x)/4= (3x+1)/4
ATQ,
3-3x= 3x+1
=> 6x= 2
=> x= 1/3
===
[Q-83]One test tube contains some acid and another test tube contains an equal quantity of water. To prepare a
solution, 20 g of the acid is poured into the second test tube. Then, two-thirds of the so formed solution is
poured from the second test tube into the first. If the fluid in the first test tube is four times that in second, what
quantity of water was taken initially.
#Solution:
===
Let, First & 2nd test tube each contain= x gm
Step 1:
20 gm is poured into second.
First now= x-20
2nd now= x+20
Step 2:
First now= x-20+(x+20)×(2/3)

2nd now = (x+20)/3
ATQ,
(x-20) +(x+20)×(2/3)= (4/3)×(x+20)
=> (3x-60+2x+40)/3= (4x+80)/3
=> (5x-20)= (4x+80)
=> x= 100 gm
So, Initial water= 100 gm
[Answer.]
===
Profit & Loss; Source:Examveda;Section-01
[Q-84] David sells his Laptop to Goliath at a loss of 20% who subsequently sells it to Hercules at a Profit of
25%. Hercules, after finding some defect in the laptop, returns it to Goliath but could recover only Rs. 4.50 for
every Rs. 5 he had paid. Find the amount at Hercules' loss if David had paid Rs. 1.75 lakh for the laptop.
#Solution
===
David CP = 1.75 lakh
David SP= 1.75×(4/5)= 1.4 lakh
Golith SP= (5/4)×1.4= 1.75 lakh
Hercules CP = 1.75 lakh
Harcules SP = 4.5×1.75/5= 1.575 lakh
Loss = (1.75-1.575)=.175 lakh=. 175×100000= 17,500[Answer]
===
[Q-85] A vendor sells his articles at a certain profit percentage. If he sells his article at (1/3) of his actual selling
price, then he incurs a loss of 40%. What is his actual profit percentage?
#Solution
===
Let, Cost Price = CP
Selling Price = SP

ATQ,
CP- (1/3)SP= 40% of CP
=> CP -(1/3) SP= 2CP/5
=> 15CP-5SP= 6CP
=> 9CP= 5SP
=> CP:SP= 5:9
% profit = (9-5)×100/5= 80%
[Answer]
===
.[Q-86] Arun bought toffees at 6 for a rupee. How many for a rupee he should sell to gain 20%?
#Solution
===
6×(5/6)= 5[Answer]
==
[Q-87] The cost of servicing of a Maruti car at Maruti care Pvt. Ltd. is Rs. 400. Manager of service centre told
me that for the second service within a year a customer can avail a 10% discount and further for third and fourth
servicing he can avail 10% discount of the previous amount paid, within a year. Further if a customer gets more
than 4 services within a year, he has to pay just 60% of the servicing charges on these services. A customer
availed 5 services from the same servicing station, what is the total percentage discount fetched by the
customer?
#Solution:
===
The cost of servicing a car with discount=
{400+400×(9/10)+400×(9/10)×(9/10)+400×(9/10)×(9/10)×(9/10)+400×(3/5)}
=400+360+324+ 291.6+240= 1615.6
Without Discount = 400×5= 2000
So, Discount = (2000-1615.6)= 384.6
% Discount = 384.6×100/2000= 19.22%[Answer.]
===

[Q-88] The cost price of 19 articles is same as the selling price of 29 articles. What is loss percentage?
#Solution
===
CP of 19 = SP of 29
=>CP:SP= 29:19
Loss percentage = (29-19)×100/29= 34.48%[Answer.]
===
[Q-89]The price of an article reduces to 576 after two successive discounts. The markup is 80% above the cost
price of Rs. 500. What is the new profit percentage if instead of two successive discounts the markup price was
further increased successively two times by the same percentage?
#Solution
===
CP= 500
MP= 500*9/5= 900
ATQ,
(100-D)% of (100-D)% of 900= 576
=> (100-D)^2×900/10,000= 576
=> (100-D)^2×9/100= 576
=> (100-D)^2= 57600/9
=> (100-D)^2= 6400
=> (100-D)=√6400= 80
=> D= 100-80= 20%
New MP= (6/5)×(6/5)×900= 1296
Profit = 796
%= 796×100/500= 159.2%[Answer]
===
[Q-90] A dishonest trader marks up his goods by 80% and gives discount of 25%. Besides he gets 20% more
amount per kg from wholesaler and sells 10% less per kg to customer. What is the overall profit percentage?
#Solution:

===
Applying MF,
100×(9/5)×(3/4)×(6/5)×(10/9)-100= 180-100= 80%[Answer.]
===
[Q-91] The profit percentage on three articles A, B and C is 10%, 20%, and 25% and the ratio of the cost price
is 1 : 2 : 4. Also the ratio of number of articles sold of A, B and C is 2 : 5 : 2, then overall profit percentage is:
#Solution
===
Let, CP are 10,20,40
Total CP = 2×10:5×20:2×40[Since numbers of articles 2:5:2]
= 20:100:80
Total CP = 200
SP = {20×(11/10)+100×(6/5)+80×(5/4)}
= (22+120+100)
= 242
Profit = (242-200)= 42
% profit = 42×100/200= 21%[Answer.]
===
[Q-92] The accountants of a company show sales of Rs. 12,600. The primary cost is 35% of sales and trading
cost accounts for 25% of the gross profit. Gross profit is arrived at by excluding the primary cost plus the cost
of advertising expenses of Rs. 1400, director's salary of Rs. 650 per annum plus 2% annual sales as
miscellaneous costs. Find the percentage profit (approx) on a capital investment of Rs. 14,000?
#Solution
===
Total sales = 12,600
Primary cost = 12,600×(7/20)= 4410
Gross profit = 12,600-4410-1400-650-(1/50)×12,600
= 12,600-4410-1400-650-252= 5888
Trading Cost = 5880×(1/4)= 1472
So, Profit = 5880-1472= 4416

% Profit = 4416×100/14,000= 31.54%[Answer]
===
[Q-93] A dishonest shopkeeper, at the time of selling and purchasing, weighs 10% less and 20% more per
kilogram respectively. Find the percentage profit earned by treachery. (Assuming he sells at Cost Price)
#Solution:
===
Applying MF,
100×(10/9)×(6/5)-100= 133.33-100= 33.33%[Answer.]
===
[Q-94] A pharmaceutical company made 3000 strips of tablets at a cost of Rs. 4800. The company gave away
1000 strips of tablets to doctors as free samples. A discount of 25% was allowed on the printed price. Find the
ratio profit if the price is raised from Rs. 3.25 to Rs. 4.25 per strip and if at the latter price, samples to doctors
were done away with. (New profit / Old profit).
#Solution
===
Old profit = (3000-1000)×(3/4)×3.25-4800= 4875-4800= 75
New Profit = 3000×(3/4)×4.25-4800=9562.5-4800= 4762.5
Ratio= 4762.5/75= 63.5:1= 635/10:1
= 635:10= 127:2[Answer]
===
[Q-95] An article costing Rs. 20 was marked 25% above the cost price. After two successive discounts of the
same percentage, the customer now pays Rs. 20.25. What would be the percentage change in profit had the
price been increased by the same percentage twice successively instead reducing it?
#Solution:
==
Here CP= 20
MP= 20×(5/4)= 25
ATQ,
25×(100-x)/100× (100-x)/100= 20.25
=> (100-x)^2= 20.25×400= 8100
=> 100-x= 90

=> x= 10%
Profit here= (20.25-20)= 0.25
New MP = 25×(11/10)×(11/10)= 30.25
Profit= (30.25-20)= 10.25
Difference of Profit= 10.25-0.25= 10
Percentage change in Profit = 10×100/0.25
= 10×100×4= 4000%[Answer.]
===
[Q-96] A pen company produces very fine quality of writing pens. Company knows that on average 10% of the
produced pens are always defective so are rejected before packing. Company promises to deliver 7200 pens to
its wholesaler at Rs. 10 each. It estimates the overall profit on all the manufactured pens to be 25%. What is the
manufactured cost of each pen?
#Solution
===
7200×10×(4/5)×(1/8000)
= 7.2[Answer]
===
[Q-97] A company charges a fixed rental of Rs. 350 per month. It allows 200 calls free per month. Each call is
charge at Rs. 1.4 when the number of calls exceed 200 per month and it charges Rs. 1.6 when the number of
calls exceeds 400 per month and so on. A customer made 150 calls in February and 250 calls in march. By how
much percent each call is cheaper in March than each call in February.
#Solution:
===
Price of each call in February = 350/150= 7/3= 2.33
Price of each call in March = (350+50×1.4)/250= 1.68
Cheaper = (2.33-1.68)×100/2.33= 27.9%[Answer]
===
[Q-98] In the Bargaining Bazar everyone purchase with a fair bargaining, so the traders markup the prices too
much. A trader marked up an article at Rs. M expected huge profit if it is sold on marked price. But a customer
purchased it at M/2 with his fine bargaining skills, so the expected profit of the trader diminished by 66.66%.
What is the percentage discount fetched by the customer through bargaining?
#Solution
===

Let, M= 100
ATQ,
(100-x)% of 100= 100/2= 50
=> 100-x=50
=> x= 50%[Answer]
===
[Q-99] An egg seller sells his eggs only in the packs of 3 eggs, 6 eggs, 9 eggs, 12 eggs etc., but the rate is not
necessarily uniform. One day Raju (which is not the same egg seller) purchased at the rate of 3 eggs for a rupee
and the next hour he purchased equal number of eggs at the rate of 6 eggs for a rupee. Next day he sold all the
eggs at the rate of 9 eggs for Rs. 2. What is his percentage profit or loss?
#Solution:
===
3 eggs CP = 1
=> 1 egg CP= 1/3
In the next hour he purchase 6 egges per rupee.
1 egg CP= 1/6
So, 2 egg CP = (1/3+1/6)= 3/6= 1/2
=> 1 egg CP = 1/4
Again,
9 eggs SP = 2
=> 1 eggs SP = 2/9
Loss= (1/4-2/9)= (9-8)/36
%Loss = (1/36)×4×100= 100/9= 11.11%[Answer]
===
[Q-100] Find the selling price of goods if two salesmen claim to make 25% profit each, one calculating it on
cost price while another on the selling price, the difference in the profits earned being Rs. 100 and selling price
being the same in both the cases?
#Solution
===
CP1:SP1= 100:125= 400:500
CP2:SP2= 75:100= 375:500

ATQ,
(400-375)x= 100
=> 25x= 100
=> 500x= 2000
So, Selling Price= 2000 [Answer]
===
[Q-101] A watch costing Rs. 120 was sold at a loss of 15%. At what price was it sold?
#Solution
===
SP= 120×(17/20)= 102[Answer]
===
[Q-102] The cost of setting up a magazine is Rs. 2800. The cost of paper and ink etc is Rs. 80 per 100 copies
and printing cost is Rs. 160 per 100 copies. In last month 2000 copies were printed but only 1500 copies could
be sold at Rs. 5 each. Total 25% profit on the sale price was realized. There is one more resource of income
from magazine which is advertising. What sum of money obtained from the advertising in magazine?
#Solution
===
Total Cost = 2800+ 20×80+20×160= 7600
Let, Advertising Cost = C
SP of 1500= 1500×5= 7500
ATQ,
SP - CP = Profit
=> (7500+C)-7600= (1/4)×7500
=> C-100= 1875
=> C= 1975[Answer]
===
[Q-104] A person bought a certain quantity of rice at the rate of Rs. 150/quintal. 10% of the rice was spoiled. At
what rate(per quintal) should he sell the remaining rice to earn 20% profit
#Solution:
===

1 Quintal CP= 150
But 10 % spoliled, thus remain Quantity = 0.9 Qunital
At 20% profit SP = 150×(6/5)= 180
So, Now 0.9= 180
1= 180×1×10/9= 200 Rs [Answer.]
===
Q-105] A trader sells two brands of petrol; one is Extra Premium and other one is speed. He mixes 12 litres
Extra Premium with 3 litres of speed and by selling this mixture at the price of Extra Premium he gets the profit
of 9.09%. If the price of Extra Premium Rs. 48 per litre, then the price of Speed is:
#Solution
===
Let, Price of speed = S
ATQ,
(12+3)×48= 109×(12×48+3×S)/100
=> 15×48×100= 109×12×48+109×3S
=> S= 28.18[Answer.]
===
[Q-106]Kamal bought a house, whose sale price was Rs. 8 lakh. He availed 20% discount as an early bird offer
and then 10% discount due to cash payment. After that he spent 10% of the cost price in interior decoration and
lawn of the house. At what price should he sell the house to earn a profit of 25%?
#Solution
===
Kamal Bought house at = 8 lakh
Two successive discount thus CP
= 8×(8/10)×(9/10)= 5.76 Lakh
And then spent 10% thus his cost will be= 5.76×(11/10)= 6.34 Lakh
At 25% Profit he sold it = 6.34 ×(5/4) Lakh= 7.92 Lakh[Answer.]
===

[Q-107] A dishonest dealer marks up the price of his goods by 20% and gives a discount of 10% to the
customer. Besides, he also cheats both his supplier and his buyer by 100 grams while buying or selling 1
kilogram. Find the percentage profit earned by the shopkeeper?
#Solution:
===
Applying MF
= 100×(6/5)×(9/10)×(10/9)×(11/10)-100
= 132-100= 32%[Answer ]
===
[Q-108] When an article was sold for Tk. 696, percent profit earned was P%. When the same article was sold
for Tk.841, percent profit earned was (P + 25%). What is the value of p?
#Solution
===
CP+ P% of CP= 696
=> CP(1+P/100)= 696
=> CP= 696×100/(100+P)_____[1]
And,
CP+ (P+25)% of CP= 841
=> CP(100+P+25)= 841×100
=> CP= 841×100/(125+P)_____[2]
From [1]&[2]
Now,
696/(100+P)= 841/(125+P)
=> 696×125+696P= 841×100+841P
=> 2900= 145P
=> P= 20[Answer.]
#Alternative

Diff. Of two profit = P+25-P= 25%
ATQ,
25% of CP= (841-696)=145
=> CP= 145×4= 580
When SP 696 Profit is P%
Profit = (696-580)= 116
% Profit, P= 116×100/580= 20%[Answer.]
===
[Q-109] Weights of two friends Ram and Shyam are in the ratio 4 : 5. If Ram's weight is increased by 10% and
total weight of Ram and Shyam become 82.8 kg, with an increases of 15%. By what percent did the weight of
Shyam has to be increased?
#Solution
===
Wt before = 82.8×(20/23)= 72
Ram = 72×(4/9)= 32
Shyam = (72-32)= 40
New wt. Of Ram = 32×(11/10)= 35.2
Shyam wt Now = 82.8-35.2= 47.6
Age Increase = 47.6-40= 7.6
% Increase = 7.6×100/40= 19%[Answer.]
===
[Q-110] A trader sells goods to a customer at a profit of k% over the cost price, besides it he cheats his
customer by giving 880 g only instead of 1 kg. Thus his overall profit percentage is 25. Find the value of k?
#Solution:
===
Applying MF
100×(100+k)/100×100/88-100= 25
=> (100+k)×(25/22)= 125
=> (100+k)×25= 125×22

=> 2500+25k= 125×22
=> 25k= 125×22-2500= 250
=> k= 10
So the value of k= 10[Answer.]
===
[Q-111] A tradesman fixed his selling price of goods at 30% above the cost price. He sells half the stock at this
price, one-quarter of his stock at a discount of 15% on the original selling price and rest at a discount of 30% on
the original selling price. Find the gain percentage altogether?
#Solution
===
Let, MP = 130
Half stock SP = 130/2= 65
one-quarter SP= (1/4)×(17/20)×130= 27.625
Rest SP = (1/4)×(7/10)×130= 22.75
Total SP = 22.75+27.625+65= 115.375
Percentage profit = 115.375-100= 15.375%[Answer.]
===
[Q-112]Ajay bought a motor cycle for Rs. 50,000. 2 years later he sold it to Vijay at 10% less of the cost price.
Vijay spend 5% of the purchasing price on its maintenance. Later Vijay displayed the sale price of his
motorcycle Rs. 50,000. Chetan wanted to purchase it at 15% discount but Vijay gave him two successive
discounts of 10% and 5% instead of 15% in one time. What is the actual discount availed by Chetan?
#Solution:
===
Vijay Give Discount of = 10% and 5% Successively.
Actual Discount given = 100- 100×(9/10)×(19/20)
= 100- 85.5= 14.5%[Answer.]
===
[Q-113] A retailer increase the selling price by 25% due to which his profit percentage increase from 20% to
25%. What is the percentage increase in cost price ?

#Solution
===
Let, Initial SP = 120
Intial CP = 120×(5/6)= 100
New SP = 120×(5/4)= 150
So, New CP = 150×(4/5)= 120
CP increase = 120-100= 100
Percentage Increase = 20×100/100= 20%[Answer.]
===
[Q-114] A driver of auto rickshaw makes a profit of 20% on every trip when he carries 3 passengers and the
price of petrol is Rs. 30 a litre. Find the % profit for the same journey if he goes for 4 passengers per trip and
the price of petrol reduces to Rs. 24 litres? (revenue per passenger is same)
#Solution
===
Makes 20% profit = 30×6/5= 36
1 passenger pay= 36/3= 12
Now, 4 passengers
Total Paid = 4×12= 48
Profit = 48-24= 24
Percentage Profit= 24×100/24= 100%[Answer.]
===
[Q-115] A dishonest dealer marks up the price of his goods by 20% and gives a discount of 10% to the
customer. He also uses a 900 gram weight instead of a 1 kilogram weight. Find his percentage profit due to
these maneuvers?
#Solution
Applying MF,
100×(6/5)×(9/10)×(10/9)-100= 120-100= 20%[Answer]

===
[Q-116] The cost of setting up the type of a magazine is Rs. 1000. The cost of running the printing machine is
Rs. 120 per 100 copies. The cost of paper, ink and so on is 60 paise per copy. The magazines are sold at Rs.
2.75 each. 900 copies are printed, but only 784 copies are sold. What is the sum to be obtained from
advertisements to give profit of 10% on the cost?
#Solution
===
Total CP = 1000+(900/100)×120+ (900/100)×60= 1000+1080+ 540= 2620
Total SP = 2620×(11/10)= 2882
SP of 784 pieces = 784×2.75= 2156
Sum of advertisements = 2882-2156= 726[Answer.
[Q-117]A man buys a chair and table for Rs. 6000. He sells the chair at a loss of 10% and the table at gain of
10%. He still gains Rs. 100 on the whole. Cost price of chair is:
#Solution
Let Price of Chair = C
Price of Table = (6000-C)
ATQ,
(9C/10)+ (6000-C)×(11/10)= 6100
=> (9C/10)+ (66,000-11C)/10= 6100
=> 9C+66,000-11C= 61,000
=> 2C= 5000
=> C= 2500
So, Price of Chair = 2500[Answer.]
[Q-118] By selling a bicycle for Rs. 2,850, a shopkeeper gains 14%. If the profit is reduced to 8%, then the
selling price will be:
#Solution
CP of bicycle = 2850×(100/114)= 2500
New profit = 8%
New SP = 2500×(27/25)= 2700[Answer.]
[Q-119] By selling an article, a man makes a profit of 25% of its selling price. His profit percent is:

#Solution
SP:CP= 100:75
Profit = 25×100/75= 33(1/3)%[Answer.]
01. The average price of 10 books is Rs.12 while the average price of 8 of these books is Rs.
11.75. Of the remaining two books, if the price of one book is 60% more than the price of
the other, what is the price of each of these two books?
#Solution: [Examveda;Section-1]
The Total Price of 10 books =10×12= 120
Eight Price Total = 8×11.75=8×11(3/4)
= 8×47/4= 94
Rest Price = (120-94)= 26
Now, Let, One price = P.
So, Other Price = P×8/5
ATQ,
P+8P/5= 26
 13P= 26×5
 P= 10
So, Price of two books 10 & 16[Answer.]
02. Average of ten positive numbers is x. If each number is increased by 10%, then x?
Solution [Examveda; Section-01]
Let x= 10
Summation of 10 number = 10×10= 100
Each number = (11/10)×10= 11
Increased By= (11-10)×100/10= 10%[Answer]
03. Average cost of 5 apples and 4 mangoes is Rs. 36. The average cost of 7 apples and 8
mangoes is Rs. 48. Find the total cost of 24 apples and 24 mangoes.
Solution [Examveda,Section-1]
5A+4M= 36×9= 324_______[1]
7A+8M= 48×15= 720______[2]
Remain Price of two books
= (120-94)= 26
Price Ration= 160;100= 8:5
Price of two = 26×(8/13)= 16
& 26×(5/13)= 10

[2]+[1]
12A+12M = 324+720= 1044
 24A+24M= 1044×2= 2088[Answer.]
4. The average wages of a worker during a fortnight comprising 15 consecutive working
days was Rs.90 per day. During the first 7 days, his average wages was Rs.87/day and the
average wages during the last 7 days was Rs.92 /day. What was his wage on the 8th day?
Solution
[Examveda;Section 1]
Total Weight of students= 15×90= 1350
Total Sum of first 7 days= 7×87= 609
Total Sum of last 7 days= 7×92= 644
Wage on 8th
day= 1350-(644+609)
= 1350-1253= 97[Answer.]
05. When a student weighing 45 kgs left a class, the average weight of the remaining 59
students increased by 200g. What is the average weight of the remaining 59 students?
Solution [Examveda,Section1]
Let, Average weight of the class= T year
60 students average weight = 60 T
Due to one students left avg. Wt =(T+2)
59 students avg. wt. = 59×(T+.2)
ATQ
60T- 59×(T+.2)= 45
 60 T-59T-11.8= 45
 T= 56.8
So, New Avg. Weight = (56.8+.2)= 57 kg
[Answer.]
Average= Summation of
Total/Total Number
 Summation of Total
= Total Number×Average.
Alternative
Average weight of new
= 45+60×.2
= 45+12= 57 kg
[Answer]

06. The difference between two angles of a triangle is 24°. The average of the same two
angles is 54°. Which one of the following is the value of the greatest angle of the triangle?
Solution [Examveda Section-1]
Let, Two angles are A & B
A-B=24°
A+B= 54×2= 108°
Greater Angle= (108+24)/2= 132/2= 66°
Since, Summation of three= 180°
So, Largest angle Can be= 180-108= 72°[Answer.]
7. The average age of a family of 5 members is 20 years. If the age of the youngest member
be 10 years then what was the average age of the family at the time of the birth of the
youngest member?
Solution [Examveda;Section1]
Sum of Total Age = 5×20= 100
When Younger Member born the sum is= 100-10-40= 50
So, Average = 50/4= 12.5 [Answer.]
8. Distance between two stations A and B is 778km. A train covers the journey from A to
B at 84km per hour and returns back to A with a uniform speed of 56km per hour. Find
the average speed of train during the whole journey.
Solution [Examveda; Section-01]
Distance Between A & B = 778 km
Ongoing Average Time = 778/84= 9.26
hr
Returned average Time = 778/56= 13.89
hr
Total Distance = 778×2= 1556 km
Total Time = 9.26+13.89= 23.15 hr
Average Speed = 1556/23.15= 67.2 km/hr
[Answer]
Alternative
Avg. Speed Formula = 2AB/(A+B)
Here, A, B= Avg. Speed of two
Avg. Speed = 2×84×56/(56+84)= 67.2 km/hr
[Answer.]

09. The average of 50 numbers is 35. If two numbers, 35 and 40 are discarded, then the
average of the remaining numbers is nearly:
Solution
Sum of Total Numbers = 50×35= 1750
Two Discarded Number Sum = 35+40= 75
(50-2)=48 Numbers Sum = (1750-75)
= 1675
Average = 1675/48= 34.89 [Answer.]
10. The average score of a cricketer for ten matches is 38.9 runs. If the average for the
first six matches is 42, then find the average for the last four matches.
Solution
Total Score in 10 Matches = 10×38.9= 389
First 6 Match total Score= 6×42= 252
So last Four Matches Total Run = 389-252= 137
Average of last four Match= 137/4= 34.25[Answer.]
11. 3 years ago the average of a family of 5 members was 17 years. A baby having been
born, the average age of the family is the same today. The present age of the baby is:
Solution
3 years Ago Five members Total Age = 5×17= 85
Average Age Today = 17
So, Total Age of Today = 6×17= 102
Without baby Total Age = 5×(17+3)= 100
Baby Age = (102-100)= 2 year [Answer.]
12. A man spends his three months income in four month time. If his monthly income is
Rs. 1000 then his annual savings is ?
Solution
Four Month Income Total = 4×1000= 4000
Spend Total = 3×1000= 3000
Savings = (4000-3000)= 1000

So, 4 month Savings = 1000
12 Month Savings = 3×1000= 3000[Answer]
13. Ajit has a certain average for 9 innings. In the tenth innings, he scores 100 runs
thereby increasing his average by 8 runs. His new average is:
Solution
Let Initial Average = T
Final Average = (T+8)
ATQ,
10(T+8)-9T= 100
 10T+80-9T= 100
 T= 100-80
 T= 20
New Average = (20+8)= 28
[Answer.]
14. The average weight of 39 Students in a class is 23. Among them Sita is the heaviest
while Tina is the lightest. If both of them are excluded from the class still the average
remains same. The ratio of weight of Sita to Tina is 15:8.Then what is the weight of the
Tina? [Affairscloud]
Solution
When Sita and Tina excluded the average age will be the same.
So, Average age of two = 2×23= 46
Let, Sita = 15x
Tina = 8x
ATQ,
15x+8x= 46
 23x= 46
 x= 2
So, Sita Age= 15×2= 30
Tina Age = 8×2= 16[Answer.]
Alternative
Average score Increase By= 10×8= 80
So, Average Original = (100+80)/9= 20
New Average = 20+8= 28 year [Answer.]
Alternative
Total Age Excluded = 23×2= 46
Sita Age= 46×(15/23)= 30
Tina Age = 46×(8/23)= 16
[Answer.]

15. The ages of Four members of a family are in the year 2010 are ‘X’,’X+12’,’X+24’ and
‘X+36’. After some years Oldest among them was dead then average reduced by 3. After
how many years from his death, the average age will same as in 2010? [Affairscloud]
Solution
Total Age of Famly Members = X+X+12+X+24+X+36
= 4X+72
Average of 4 = 4(X+18)/4= (X+18)
New Average = (X+18)-3= X+15
Let After T years the average age will be the same.
ATQ,
3(X+15)+3T= 3×(X+18)
 3T= 3X+54-3X-45
 3T= 9
 T= 3
So, 3 years from his death the average age will be the same[Answer]
16. The average of Four numbers is 24.5. of the four numbers, the first is 1.5 times the
second, the second is 1/3 rd of the third, and the third is 2 times the fourth number. Then
what is smallest of all those numbers? [Affairs-Cloud]
Solution
Total Sum of Four Number = 4×24.5= 98
Let, Fourth Number = x
Third Number = 2x
2nd
Number =
2𝑥
3
First Number= (
2𝑥
3
) × (
3
2
)
=x
Now, 𝑥 + 2𝑥 +
2𝑥
3
+ 𝑥=98
 (3x+6x+2x+3x)= 98×3
 x=21
Alternative
By taking LCM
Let, Fourth= 12
3rd
= 24
2nd
= 8
First = 12
Smallest = 98×8/56
= 14

Smallest Number = 2×
21
3
= 14 [Answer.]
17. There are 459 students in a hostel. If the number of students increased by 36, the
expenses of the mess increased by Rs .81 Per day while the average expenditure per head
reduced by 1. Find the original expenditure of the mess?
Solution [AffairsCloud]
Here Total Students = 459
Newly Added = 36
Total Students = (459+36)= 495
Let Average expense per head = H
ATQ,
495(H-1)-459×H= 81
 36H = 576
 H= 16
Original Expenditure = 459×16= 7344[Answer.]
18. The average cost 32 different Mobiles is Rs. 9000. Among them, Oppo which is the
costliest is 70% higher price than the cheapest Mobile Lava. Excluding those both
mobiles, the average of the Mobiles is Rs.8880. Then what is the cost of Oppo Mobile?
Solution
Cost of Oppo and Lava = (32×9000-30×8880)
= 21,600
Let Lava's Price = x
Oppo Price = 170% of x= 17x/10
ATQ,
x+17x/10= 21,600
 27x= 21,600×10
 x= 800
So, Oppo price = (21,600-8000)=
13,600[Answer.]
Alternative
Cost of Oppo+Lava= (32×90,00-
30×8880)= 21,600
Oppo:Lava= 17:10
Oppo Price = 21600×17/27
= 13600
[Answer.]

19. The average age of a family of 9 members is 22 years. Surya is the youngest and his
age is 6 years, then what was the average age of the family just before Surya was born
Solution
The Total Age = 22×9= 198
Suraya Age = 6 year
When he was born the total Age of his family
= 198-6-8×6
= 198-54= 144
So, Average Age = 144/8= 18 year [Answer.]
20. Dhoni scored 8000 runs in a certain number of innings. In the next five innings, he was
out of form and hence, could make only 85 runs, as a result his average reduced by 1 run.
How many innings did he play in total?
Solution
Let, Number of innings = N
Average = 8000/N[Orginal]
New Average = 8085/(N+5)
ATQ,
8000/N -8085/(N+5)= 1
 8000(N+5)-8085N=N(N+5)
 8000N+40000-8085N= (𝑁)2
+ 5𝑁
 N2
+ 90N − 40000=0
 N2
+ 250N − 160N − 40,000=0
 N= 160
Total Innigs he play = 160+5= 165[Answer.]
21. The weights of 19 people are in Arithmetic progression. The average weight of them is
19. If the heaviest is 37 Kgs. Then what is the weight of the Lightest? [Affairscloud]
Solution
Question Asked:
Number of Innings =?
In order To solve this question you
have to draw an equation.
Total Run/No. of innings =
Average
So,
Difference of two average = 1

Let, First term = a
2nd
term = a+d
37th
Term = a+18d
a+18d= 37________[1]
Series Looks like= a+d+a+2d+….+a+36d=
19×19=361
Sum =19/2×(2a+18d)= 19×(a+9d)
So19×(a+9d)= 361
 a+9d= 19_________[2]
[2]×2-[1]
a= 38-37= 1[Answer]
22. The average weight of 40 Students is 32. If the Heaviest and Lightest are excluded the
average weight reduces by 1. If only the Heaviest is excluded then the average is 31. Then
what is the weight of the Lightest? [Affairscloud]
Solution
Average Weight Total = 40×32= 1280
When heaviest excluded ;Total Age = 39×31= 1209
So, Heaviest weight = (1280-1209)= 71 year
Heaviest and Lightest Weight = (1280-38×31)= 102
So Lightest weight = (102-71)= 31 year [Answer]
23. The average price of 80 mobile phones is Rs.30,000. If the highest and lowest price
mobile phones are sold out then the average price of remaining 78 mobile phones is Rs.
29,500. The cost of the highest mobile is Rs.80,000. The cost of lowest price mobile is?
Solution
Price of two Mobile phone = (30,000×80-78×29,500)= 99000
Cost of highest Mobile Phone = 80,000
Cost of lowest Mobile phone = (99000-80,000)= 19,000[Answer.]
24. Pranav went to the bank at the speed of 60 kmph while returning for his home he
covered the half of the distance at the speed of 10 kmph, but suddenly he realized that he
Alternative
Here Total Term = 19
Since Largest Term = 37
So,
Series Can be
1+2+3+4+…..+37
So lightest = 1[Answer.]

was getting late so he increased the speed and reached the home by covering rest half of
the distance at the speed of 30 kmph.The average speed of the Pranav in the whole length
of journey is?
Solution [AffairsCloud]
Let Total Distance = D km.
Time while going = D/60
While Returning Half distance
Time = D/20
Remain Time = D/60
So Total Time = D/60
+D/20+D/60
= 5D/60= D/12
Average Speed = 2D×12/D= 24 [Answer.]
25. The average expenditure of Sharma for the January to June is Rs. 4200 and he spent
Rs. 1200 in January and Rs.1500 in July. The average expenditure for the months of
February to July is:
Solution
January To June expenditure = 4200×6= 25,200
Expenditure from February to july = 25,200-1200+1500= 25,500
Average Expenditure = 25,500/6= 4250[Answer.]
26. the average of 6 men 8 women and 1 boy is 35. if the average of men is 40 years and the
women average age is 34 years what is the average age of boy [IFIC Bank, TSO,18]
Solution
Total Age = (6+8+1)×35= 15×35= 525
Women Total Age = 8×34= 272
Men Total Age= 6×40= 240
So, Boy Age = 525-(272+240)= 13 years [Answer.]
Alternative
Let Distance = 60 km
Ongoing Time = 60/60= 1 hr
Returning Time = 30/10+30/30= 4 hr
Total Distance = 120 Km
Average Speed = 120/5= 24 km/hr[Answer.]

27. Average Mark obtained by 300 students in a class is 80. When 10% of the students
marks are excluded from the group the average mark of the remaining students will be
85.Find the average mark of 10% excluded students [Bank Asia,16_Written]
Solution
Total Mark of 300 students = 300×80= 240,00
Total Mark of 90% of 300= 270 students = 270×85= 22,950
10% Excluded or 30 students total Marks = (24000-22950)= 1050
Average Mark =
1150
30
=35[Answer.]
28. The Average weight of three men A, B, C is 84 kg. A new comer D joins the group
thus average age will become 80 kg. If another man E. whose weight is 3 kg more than D
replace A. Then The average age of B, C,D,E will becomes 79.The Age of A
Solution JanataBank AEO,17
Total Age of A,B,C= 84×3= 252
Total Age of A,B,C,D = 80×4= 320
So, D age = (320-252)= 68
So, E age = 68+3= 71
So, B, C,D,E age Total = 4×79= 316
So, B, C age = 316-(71+68)= 316-139= 177
So, A age = 252-177= 75 years [Answer.]
29. A worker is paid x dollars per hour for the first 5 hours he works each day. He is paid
y dollars per hour for each hour he works in excess of, 5 hours. During one week he works
8 hours on Saturday , 11 hours on Sunday, 12 hours on Monday, 10 hours on Tuesday,
and 9 hours on Wednesday. What is his average daily wage in dollars for the 5-day week?
Solution
In Saturday he earned = 5x +3y
In Sunday he earned = 5x+6y

Mutaher combined math file [www.onlinebcs.com]

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