Bangladesh bank written maths 2010 2020

Bangladesh Bank Recruitment Examination 2010-2020
WRITTEN
MATH
Problems & Solutions
|
Sultan Mahmud (SMcxlvii)
Contact: www.facebook.com/SMcxlvii
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-Contents-
OFFICER (GENERAL) UPCOMING BATCH [EXAM HELD ON 24.01.2020] ...........................................................................1
AD (GENERAL) 2019 BATCH [EXAM HELD ON 27/07/2018]..................................................................................................4
OFFICER (GENERAL) 2019 BATCH [EXAM HELD ON 25/05/2018] .......................................................................................6
CASH OFFICER 2018 BATCH [EXAM HELD ON 24/02/2017].................................................................................................8
AD 2018 BATCH [EXAM HELD ON 17/02/2017].....................................................................................................................10
AD (FF) 2017 BATCH [EXAM HELD ON 31/07/2015].............................................................................................................11
OFFICER (G) 2017 BATCH [EXAM HELD ON 08/05/2015]....................................................................................................14
AD 2015 BATCH [EXAM HELD ON 30/05/2014].....................................................................................................................17
AD 2014 BATCH [EXAM HELD IN 2013] ................................................................................................................................18
CASH OFFICER 2013 BATCH [EXAM HELD IN 2011]...........................................................................................................21
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Written Math : Problems & Solutions
Officer (General) Upcoming Batch [Exam held on 24.01.2020]
Problem-1: Find the sum of all the two digit numbers that are divisible by 3.
Solution:
The smallest and the greatest two digit numbers that are divisible by 3 are 12 and 99
respectively. So, we have to find the sum of the following series:
12+15+18+ … + 99.
It is an arithmetic progression where,
the first term, a = 12,
common difference, d = (15-12) = 3,
last term, l = 99,
number of terms n = ?
We know,
l = a+(n-1)d
or, n =
l-a+d
d
=
99-12+3
3
= 30
We know, Sum of first n terms
=
n
2
{2a+ n-1 d}
=
30
2
{2×12+ 30-1 3}
= 15×(24+87)
= 1665.
So, the sum of all the two digit numbers that are divisible by 3 is 1665.
Answer: 1665.
Problem-2: If 2n-1+2n+1= 320, then find the value of n.
Solution:
Given, 2n-1
+2n+1
= 320
or, 2n
(2-1
+21
) = 320
or, 2n
(
1
2
+2)= 320
or, 2n
×
5
2
= 320
or, 2n
= 128
or, 2n
= 27
or, n = 7
So, the value of n is 7.
Answer: 7.
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Problem-3: A man took a loan of Tk. 160000 at 9% compound interest p.a. If he pays
Tk. 60000 at the end of each year, then after paying out the third installment what is
the amount of debt left?
Solution:
Total due amount at the end of 1st year Tk. 160000 + (160000×9%) Tk. 174400
Less: Amount paid by the end of 1st year Tk. 60000
Total due amount at the beginning of the 2nd year Tk. 114400
Total due amount at the end of 2nd year Tk. 114400 + (114400×9%) Tk. 124696
Less: Amount paid by the end of 2nd year Tk. 60000
Total due amount at the beginning of the 3rd year Tk. 64696
Total due amount at the end the 3rd year Tk. 64696 + (64696×9%) Tk. 70518.64
Less: Amount paid by the end of 3rd year Tk. 60000
Total due amount after the 3rd installment paid Tk. 10518.64
So, after paying out the third installment the amount of debt left is Tk. 10518.64.
Answer: Tk. 10518.64.
Problem-4: The ratios of acid to water in two bottles are 2:3 and 1:2 respectively. If a
new mixture is formed by taking liquids from the bottles in the ratio of 1:3, then what
is the ratio of acid to water in the new mixture?
Solution:
Suppose, x liters of liquid from the first bottle and 3x liters of liquid from the second bottle are
taken and mixed to form a new mixture.
So, from the first bottle,
acid taken = x ×
2
(2+3)
=
2x
5
liters
water taken = x ×
3
(2+3)
=
3x
5
liters
And from the second bottle,
acid taken = 3x ×
1
(1+2)
= x liters
water taken = 3x ×
2
(1+2)
= 2x liters
So, the quantity of acid in the new mixture =
2x
5
+ x =
7x
5
liters
And the quantity of water in the new mixture =
3x
5
+ 2x =
13x
5
liters
Required ratio, acid : water =
7x
5
:
13x
5
= 7 : 13.
So, the ratio of acid to water in the new mixture is 7:13.
Answer: 7:13.
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Problem-5: Two straight roads go in such a way that at a common point they make 60°
angle with each other. Two persons started traveling from that common point at the
same time by each of the road at speeds of 7 KMPH and 5 KMPH respectively. After 4
hours of their journey what will be the direct distance between them?
Solution:
Suppose, the common point is A, the person traveling
at 7KMPH reaches B after 4 hours where AB = (7×4)
= 28 KM. The other person travels from A towards D.
In 4 hours he reaches to D and AD = 5×4 = 20 KM.
Let‟s draw a perpendicular BC from Bon AD (or its
extension). So, angle ACB = 90°.
In the right angled triangle ABC, the adjacent sides
are AC and BC, the hypotenuse AB = 28 KM, and angle BAC = 60°.
So, cos 60° =
AC
AB
or,
1
2
=
AC
28
or, AC = 14
Again, in triangle ABC,
tan 60° =
BC
AC
or, √3 =
BC
14
or, BC = 14√3
Let‟s join B and D.
In triangle BCD, angle BCD = 90°, BC = 14√3 KM and CD = AD-AC = 20-14 = 6 KM.
According to Pythagorean Theorem, we can write-
BD² = BC² + CD² = (14√3)² + (6)² = 624
So, BD = √624 = 24.98 KM.
So, after 4 hours, the direct distance between them will be 24.98 KM.
Answer: 24.98 KM
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AD (General) 2019 Batch [Exam held on 27/07/2018]
Problem-1: A senior citizen invests Tk. 50 lac in a fixed deposit scheme at 11.5% p.a.
for six months. In every six months he withdraws Tk. 2 lac from his principal plus the
interest earned. What will be his principal amount to invest after two years?
Solution:
Total number of withdrawals in 2 years = (2×12) ÷ 6 = 4.
As the person withdraws Tk. 2 lac of principal plus the interest earned, in two years, with 4
withdrawals he withdraws = Tk. 2 lac × 4 = Tk. 8 lac of principal.
So, after two years, the amount of principal he will have is = Tk. (50-8) lac = Tk. 42 lac.
Answer: Tk. 42 lac.
Problem-2: After traveling 108 km, a cyclist observed that he would have required 3
hours less time if he could have traveled at a speed of 3 kmph more. At what speed
did he travel?
Solution:
Let, the actual speed of the cyclist be x kmph.
So, the actual time he took to travel 108 km is =
108
x
hours.
According to the question,
108
x
-
108
(x+3)
= 3
or,
36x+108-36x
x(x+3)
= 1
or, x²+3x–108 = 0
or, (x+12)(x-9) = 0
or, x = 9 [x = -12 is not acceptable, as the speed value cannot be negative]
So, the cyclist traveled at a speed of 9 kmph.
Answer: 9 kmph.
Problem-3: Three numbers are in A.P. and their sum is 30. Also the sum of their
squares is 308. Find the numbers.
Solution:
Suppose, the numbers are (a-d), a, and (a+d)
(a-d) + a + (a+d) = 30
or, 3a = 30
or, a = 10
Again,
(a-d)² + a² + (a+d)² = 308
or, a² - 2ad + d² + a² + a² + 2ad + d² = 308
or, 3a² + 2d² = 308
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or, 3×10² + 2a² = 308
or, 2a² = 308 - 300 = 8
or, a² = 4
or, a = 2.
So, the numbers are: (a-d) = 10-2 = 8, a = 10, and (a+d) = 10+2 = 12.
Answer: 8, 10 and 12.
Problem-4: A square is inscribed inside a circle. What is the area of the square, if the
radius of the circle is 10 cm?
Solution:
Suppose O is the center of the circle and ABDC is the square
inscribed to it. The diagonals of the square also intersect at O.
According to the question, OA = OB = OC = OD = 10 cm.
We know, the diagonals of a square bisect each other
perpendicularly and divide the square into four right angled
triangles of equal size.
So, Area of the triangle AOB = ¼(The area of the square ABDC)
or, ½×10cm×10cm = ¼(The area of the square ABDC)
or, The area of the square ABDC = 200 sq. cm
So, the area of the square is 200 sq. cm
Answer: 200 sq. cm.
Problem-5: 50 daily workers can complete a dam project in 40 days. If 30 of them work
daily and the rest work in every alternative day, how many more days will be required
to complete the project?
Solution:
According to the question, 30 workers work daily and the rest work in every alternative day.
So, effectively the number of workers working there daily = 30 +
50-30
2
= 40.
So, 50 workers can complete the project in 40 days
1 worker can complete the project in 40×50 days
40 workers can complete the project in
40×50
40
= 50 days.
So, the number of more days required is = 50-40 = 10.
Answer: 10 days.
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Problem-6: 3 coins are tossed at random. Construct the sample space and find the
probability of getting: (i) one head two tails, (ii) one tail, (iii) One tail and two heads.
Solution:
If three coins are tossed at random, the possible outcomes are: HHH, HHT, HTH, THH, HTT,
THT, TTH and TTT, i.e: total 8 types of outcome.
(i) The probability of getting one head and two tails:
In the above sample space, the outcomes with one head and two tails are HTT, THT and
TTH; total three outcomes. So, the probability is =
3
8
(ii) The probability of getting one tail:
In the above sample space, the outcomes with only one tail are HHT, HTH and THH; i.e:
total 3. So, the probability is =
3
8
(iii) The probability of getting one tail and two heads:
In the above sample space, the outcomes with one tail and two heads are HHT, HTH and
THH; i.e: total 3. So, the probability is =
3
8
Answer: Sample space: HHH, HHT, HTH, THH, HTT, THT, TTH and TTT, (i)
3
8
, (ii)
3
8
, (iii)
3
8
.
Officer (General) 2019 Batch [Exam held on 25/05/2018]
Problem-1: A piece of stone fell from a balloon when it was flying in the upward
direction with a velocity of 20m/sec. What will be the height of the balloon when the
stone hit the ground in 10 seconds?
Solution:
As the balloon was flying in the upward direction, the stone was going upward with it at the
speed of the balloon i.e: 20m/sec. When it fell from the balloon, it took 10 seconds (given) to
hit the ground. So, from the ground level, the distance of the position where the stone was
detached from the balloon is
= {-(20×10) + (½×9.8×10²)} meters = (-200 + 490) meters = 290 meters.
By the time the stone hits the ground, the balloon goes further upward by
= (20×10) = 200 meters.
So, the height of the balloon when the stone hit the ground is = (290+200) meters = 490
meters.
Answer: 490 meters.
Problem-2: Find the maximum value of z = 6x+2y, subject to the conditions x≥0, y≥0,
x+y=5, x≤2, y≤4.
Solution:
Given, 0 ≤ x ≤2, 0 ≤ y ≤ 4, and x+y = 5.
We have to find the maximum value of z = 6x+2y.
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Here, the coefficient of x is greater than that of y. In order to maximize the value of z, we
should take the maximum value of x.
Maximum value of x satisfying the above conditions is = 2.
On the other hand, the value of y = 5-2 = 3.
Therefore, the maximum possible value of z = 6x + 2y = 6×2 + 2×3 = 18.
Answer: 18.
Problem-3: The sum of the digits of a two-digit number is subtracted from the number.
How many such two-digit numbers can be formed so that the digit in the unit place of
the resulting number is 6?
Solution:
Suppose, the unit digit of the number is y and the tens digit is x. So, the number is (10x+y).
If the sum of the digits is subtracted from the number, we get = (10x+y) - (x+y) = 9x.
So, the resulting number is a multiple of 9.
According to the question, the resulting number has 6 in its unit place. The only two digit
number that is a multiple of 9 and has 6 in its unit digit position is 36.
So, 9x = 36, or x = 4.
So, the possible such numbers can be: 10×4 + y = 40+y.
Here, the value of y has no effect as it gets cancelled out when the sum of the digits is
subtracted from the number. So, y can take any possible value.
So, the possible numbers can be: 40, 41, 42, 43, 44, 45, 46, 47, 48 and 49.
So, there can be 10 such numbers.
Answer: 10.
Problem-4: How many ways are there to divide 50 people into 3 groups so that each
group contains members equal to a prime number?
Solution:
The sum of three numbers will be even if :
(i) all the three numbers are even
(ii) one of them is even and the other two are odd numbers.
According to the question, the number of members must be prime numbers and we have
only one even prime number. So, the above mentioned combination (i) is not possible.
Hence, there must be an even prime number (i.e: 2) and the other two must be odd prime
numbers.
So, the possible combinations are:
1. 2 + 5 + 43
2. 2 + 7 + 41
3. 2 + 11 + 37
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4. 2 + 17 + 31
5. 2 + 19 + 29
Therefore, there are five ways to divide 50 people into 3 groups so that each group contains
members equal to a prime number.
Answer: 5.
Problem-5: A semicircular sheet of metal of diameter 28 cm is bent into an open
conical cup. Find the depth and capacity of the cup.
Solution:
If the semicircular sheet is bent into an open conical shape, the radius of the sheet will be
the slant height of the cone and the length of the curved side of the sheet will be the
circumference of the base of the cone.
Given, the diameter of the semicircular sheet = 28 cm.
So, the radius is = (28÷2) cm = 14 cm.
So, the length of the curved side of the sheet = 14π cm.
So, the circumference of the base of the cone is = 14π cm, and the slant height of the cone
is = 14 cm.
So, the radius of the base of the cone is = 14π † 2π = 7 cm
Therefore, the height of the cone is = 14²-7² cm = 147 cm =7 3 cm.
And, the capacity i.e.: the volume of the cone is =
π×7²×7√3
3
cubic cm =
343π
√3
cubic cm.
Answer: Depth 7√3 cm, capacity
343π
√3
cubic cm.
Cash Officer 2018 Batch [Exam held on 24/02/2017]
Problem-1: Shakil started a business investing Tk. 25,000 in 2009. In 2010, he invested
an additional amount of Tk. 10,000 and Raihan joined him with an amount of Tk.
35,000. In 2011 Shakil invested another additional amount of Tk. 10,000 and Jafor
joined them with an amount of Tk. 35,000. What will be Raihan’s share in the profit of
Tk. 1,50,000 earned at the end of 3 years from the start of the business in 2009?
Solution:
Time-weighted average investments of the partners:
Shakil = Tk. 25000 ×
3
3
+ Tk. 10000 ×
2
3
+ Tk. 10000 ×
1
3
= Tk.
105000
3
.
Raihan = Tk. 35000 ×
2
3
= Tk.
70000
3
.
Jafor = Tk. 35000 ×
1
3
= Tk.
35000
3
.
So, ratio of their time-weighted investments,
Shakil :Raihan : Jafor =
105000
3
:
70000
3
:
35000
3
= 3 : 2 : 1
So, the profit share of Raihan = Tk. 150000 ×
2
3+2+1
= Tk. 50,000.
Answer: Tk. 50,000.
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Problem-2: If 9 engines consume 24 metric tonnes of coal, when each is working 8
hours a day, how much coal will be required for 8 engines, each running 13 hours a
day, it being given that 3 engines of former type consume as much as 4 engines of
latter type?
Solution:
On the consumption scale,
4 engines of latter type = 3 engines of former type
8 engines of latter type = 3 ×
8
4
= 6 engines of former type.
9 engines of former type consume in 8 hours = 24 metric tonnes of coal
6 engines of former type consume in 13 hours =
24×6×13
9×8
= 26 metric tonnes of coal.
So, 8 engines of latter type, each running 13 hours a day will consume 26 metric tonnes of
coal.
Answer: 26 metric tonnes.
Problem-3: Dawood invested certain amount in three different schemes A, B and C
with the rate of interest 10% p.a., 12% p.a and 15% p.a. respectively. If total interest
accrued in one year was Tk. 3200 and the amount invested in Scheme C was 150% of
the amount invested in Scheme A and 240% of the amount invested in Scheme B,
what was the amount invested in Scheme B?
Solution:
According to the question, investment ratio:
A : C = 100 : 150 = 2 : 3
B : C = 100 : 240 = 5 : 12
So, A : B : C = 8 : 5 : 12
Suppose, Dawood invested Tk. 8x, Tk. 5x and Tk. 12x in the schemes A, B and C
respectively.
So, (8x×10%) + (5x×12%) + (12x×15%) = 3200
or, 3.2x = 3200
or, x = 1000
So, the amount invested in Scheme B was = Tk. 5x = Tk. 5×1000 = Tk. 5000.
Answer: Tk. 5000.
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AD 2018 Batch [Exam held on 17/02/2017]
Problem-1: A man sells an article at a profit of 25%. If he had bought it at 20% less and
sold it for Tk. 10.50 less, he would have gained 30%. Find the cost price of the article.
Solution:
Suppose, the cost price of the article is Tk. x.
So, at 25% profit, the selling price of it is Tk. 1.25x
(1.25x–10.50) - 0.8x = 0.8x × 30%
or, 0.45x - 0.24x = 10.50
or, 0.21x = 10.50
or, x = 10.50÷0.21 = 50
So, the cost price of the article is Tk. 50.
Answer: Tk. 50.
Problem-2: A and B can do a piece of work in 18 days, B and C can do it in 24 days, A
and C can do it in 36 days. In how many days will A, B and C finish it, working
together and separately?
Solution:
A and B working together can do in 18 days = 1 work
A and B working together can do in 1 day =
1
18
work --- (i)
B and C working together can do in 24 days = 1 work
B and C working together can do in 1 day =
1
24
work --- (ii)
C and A working together can do in 36 days = 1 work
C and A working together can do in 1 day =
1
36
--- (iii)
(i)+(ii)+(iii)=>
2(A, B and C) working together can do in 1 day = (
1
18
+
1
24
+
1
36
) =
4+3+2
72
=
4+3+2
72
=
1
8
work
So, A, B and C working together can do in 1 day =
1
8×2
=
1
16
work --- (iv)
Now, A, B and C working together can do
1
16
work in = 1 day
A, B and C working together can do 1 or full work in = 16 days.
(iv)-(ii)=> A can do in 1 day =
1
16
-
1
24
=
1
48
work
A can do
1
48
work in = 1 day
A can do 1 or full work in = 48 days.
(iv)-(iii)=> B can do in 1 day =
1
16
-
1
36
=
5
144
work
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B can do
5
144
work in = 1 day
B can do 1 or full work in =
144
5
days = 28.8 days.
(iv)-(i)=> C can do in 1 day =
1
16
-
1
18
=
1
144
work
C can do
1
144
work in = 1 day
C can do 1 or full work in = 144 days.
Answer: Together 16 days, separately A in 48 days, B in 28.8 days and C in 144 days.
AD (FF) 2017 Batch [Exam held on 31/07/2015]
Problem-1: A bus was hired at the cost of Tk. 2400 and it was decided that every
student would share the cost equally. But 10 more students joined and as a result the
fare decreased by Tk. 8 per person. How many students were travelling in the bus?
Solution:
Suppose, the number of students expected to travel was = x
So, the actual number of students travelling in the bus is = (x+10)
2400
x
-
2400
x+10
=8
or,
2400x+24000-2400x
x x+10
=8
or, x²+10x = 3000
or, x² + 10x - 3000 = 0
or, x² + 60x - 50x - 3000 = 0
or, x(x+60) - 50(x-60) = 0
or, (x+60)(x-50) = 0
or, x = 50 [x = -60 is not acceptable, as the number of students cannot be a negative
number]
So, the number of students actually travelling in the bus was = x+10 = 50+10 = 60.
Answer: 60.
Problem-2: The average age of students of a class is 15.8 years. The average age of
boys in the class is 16.4 years and that of the girls is 15.4 years. Find the ratio of the
number of boys to the number of girls in the class.
Solution:
Suppose, the number of boys in the class is x and the number of girls is y.
So, the sum of the ages of all boys in the class is = 16.4x years
the sum of the ages of all girls in the class is = 15.4y years
And the sum of the ages of all the students in the class is = 15.8(x+y) = 15.8x+15.8y
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16.4x+15.4y = 15.8x+15.8y
0.6x = 0.4y
x : y = 0.4 : 0.6 = 2 : 3
So, the ratio of the number of boys to the number of girls in the class is 2 : 3.
Answer: 2 : 3.
Problem-3: If x+
1
x
=3, then the value of x6
+
1
x6 =?
Solution:
Given, x+
1
x
=3
or, x+
1
x
2
=32
or, x2
+2x
1
x
+
1
x2 =9
or, x2
+
1
x2 =9-2
or, x2
+
1
x2 =7
or, (x2
+
1
x2 )
3
=73
or, x6
+
1
x6 +3x2 1
x2 x2
+
1
x2 =343
or, x6
+
1
x6 +3×7=343
or, x6
+
1
x6 =343-21=322.
Answer: 322.
Problem-4: The percentage of profit earned by selling an article for Tk. 1920 is equal
to the percentage of loss incurred by selling the same article for Tk. 1280. At what
price should the article be sold to make 25% profit?
Solution:
Suppose, the cost of the article is Tk. x.
So, the profit earned by selling it at Tk. 1920 is = Tk. (1920-x)
And, the loss incurred by selling it at Tk. 1280 is = Tk. (x-1280)
1920-x
x
×100 %=
x-1280
x
×100 %
or, 1920-x = x-1280
or, 2x = 3200
or, x = 1600
So, the cost of the article is Tk. 1600.
At 25% profit, the selling price is = Tk. 1600 + (Tk. 1600×25%) = Tk. (1600+400) = Tk. 2000.
Answer: Tk. 2000.
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Problem-5: A can do a piece of work in 10 days, while B alone can do it in 15 days.
They work together for 5 days and the rest of the work is done by C in 2 days. If they
get Tk. 4500 for the whole work, how should they divide the money?
Solution:
A can do in 10 days = 1 or full work
A did in 5 days =
5
10
=
1
2
of the work
B can do in 15 days = 1 or full work
B did in 5 days =
5
15
=
1
3
of the work
C did in 2 days = 1-
1
2
-
1
3
=
1
6
of the work
So, ratio of their work, A : B : C =
1
2
:
1
3
:
1
6
= 3 : 2 : 1
Therefore, they should divide the money in the ratio 3 : 2 : 1.
So, A will get = Tk. 4500 ×
3
3+2+1
= Tk. 2250.
B will get = Tk. 4500 ×
2
3+2+1
= Tk. 1500.
and, C will get = Tk. 4500 ×
1
3+2+1
= Tk. 750.
Answer: A : B : C = 3 : 2 : 1, A = Tk. 2250, B = Tk. 1500, C = Tk. 750.
Problem-6: ABCD is a square and one of its sides AB is also a chord of the circle as
shown in the figure. What is the area of the square?
Solution:
In the figure AOB is a right angle triangle with adjacent sides AO = BO = 3 units.
So, the length of the hypotenuse AB = AO²+BO² = 3²+3² = 3 2 units.
Also, AB is a side of the square ABCD.
Therefore, the area of the square = AB² = (3 2)² = 18 square units.
Answer: 18 square units.
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Officer (G) 2017 Batch [Exam held on 08/05/2015]
Problem-1: A shop stocks four types of caps, there are
1
3
as many red caps as blue
caps and
1
2
as many green caps as red caps. There are equal numbers of green caps
and yellow caps. If there are 42 blue caps, then what percent of the total caps in the
shop are blue?
Solution:
Given, the number of blue caps in the shop is = 42
So, the number of red caps = 42 ×
1
3
= 14
So, the number of green caps = 14 ×
1
2
= 7
So, the number of yellow caps = 7
So, total number of caps in the shop is = 42+14+7+7 = 70
So, the percentage of blue caps in the shop =
Number of blue caps×100
Total number of caps
%=
42×100
70
%=60%
Answer: 60%
Problem-2: The annual incomes and expenditures of a man and his wife are in the
ratios of 5:3 and 3:1 respectively. If they decide to save equally and find a balance of
Tk. 4000 at the end of the year, what was their income?
Solution:
Suppose, the income and expenditure of the man are 5x and 3x and those of his wife are 3y
and y respectively.
According to the question, their savings are equal.
So, 5x–3x = 2x = 3y–y = 2y
So, x = y
So, their total income is = 5x+3y = 5x+3x = 8x.
At the end of the year, they found a savings balance of Tk. 4000.
So, 2x+2y = Tk. 4000
or, 2x+2x = Tk. 4000 [as x = y]
or, 4x = Tk. 4000
or, x = Tk. 1000
So, 8x = Tk. 1000×8 = Tk. 8000.
Answer: Tk. 8000.
[The query is “what was their income?” not “were or incomes”. It indicates, the problem
requires us determine their combined income amount.]
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Problem-3: A person sells two articles, each for the same price Tk. 1040. He incurs
20% loss on the first and 10% loss on the second. Find his overall percentage of loss.
Solution:
In case of the article sold at 20% loss,
If the selling price is (100-20) or 80, the cost is = 100
If the selling price is 1040, the cost is =
100×1040
80
= 1300
In case of the article sold at 10% loss,
If the selling price is (100-10) or 90, the cost is = 100
If the selling price is 1040, the cost is =
100×1040
90
=
10400
9
So, total cost = Tk. (1300 +
10400
9
) = Tk.
22100
9
Total Selling price = Tk. (1040+1040) = Tk. 2080
Total loss = Tk.
22100
9
- 2080 = Tk.
3380
9
So, loss percentage =
Profit×100
Cost
%=
3380
9
×100
22100
9
%= 15.29% (approximately)
Answer: 15.29% (approximately)
Problem-4: If the sum of five consecutive integers is S, what is the largest of those
integers in terms of S?
Solution:
Suppose, the integers (from the largest to the smallest) are: x, x–1, x–2, x–3 and x–4.
x + x–1 + x–2 + x–3 + x–4 = S
or, 5x–10 = S
or, 5x = S+10
or, x =
S+10
5
Answer:
S+10
5
.
Problem-5: The difference between two numbers is five and the difference between
their squares is 65. What is the larger number?
Solution:
Suppose, the larger number is x. So, the smaller number is (x–5).
x² – (x–5)² = 65
or, x²–x²+10x–25 = 65
or, 10x = 90
or, x = 9
So, the larger number is 9.
Answer: 9.
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Problem-6: Robi drove 100 miles to visit a friend. If he had driven 8 miles per hour
faster than he did, he would have arrived in
𝟓
𝟔
of the time he actually took. How many
minutes did the trip take?
Solution:
Suppose, the trip took x hours.
Had he driven 8 miles per hour faster, the trip would have taken =
5x
6
hours.
100
5x
6
-
100
x
= 8
or,
600
5x
-
100
x
= 8
or,
600-500
5x
= 8
or, 40x = 100
or, x =
100
40
hours =
100×60
40
minutes = 150 minutes.
So, the trip took 150 minutes.
Answer: 150 minutes.
Problem-7:
2+x+ 2-x
2+x- 2-x
=2. Find the value of x.
Solution:
2+x+ 2-x
2+x- 2-x
= 2
or,
2+x+ 2-x+ 2+x- 2-x
2+x+ 2-x- 2+x+ 2-x
=
2+1
2-1
or,
2+x
2-x
= 3
or, 2+x = 18-9x
or, 10x = 16
or, x =
16
10
= 1.6
Answer: 1.6
Problem-8: Of the three numbers, second is twice the first and is also thrice the third.
If the average of the three numbers is 44, then what will be the largest number?
Solution:
Suppose, the second number is = x
So, the first number is =
x
2
, and the third number is =
x
3
.
So, x >
x
2
>
x
3
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x +
x
2
+
x
3
= 44×3
or,
6x+3x+2x
6
= 132
or, x = 132 ×
6
11
= 72.
So, the largest number, x = 72.
Answer: 72.
AD 2015 Batch [Exam held on 30/05/2014]
Problem-1: a, b, c, d and e are five consecutive numbers in increasing order of size.
Deleting one of the five numbers from the set decreased the sum of the remaining
numbers in the set by 20%. Which one of the numbers was deleted from a, b, c, d and
e?
Solution:
Here, b = a+1, c = a+2, d = a+3 and e = a+4
So, the sum of the five numbers = a + b + c + d + e = a + a+1 + a+2 + a+3 + a+4 = 5a+10.
20% of the sum = (5a+10)×20% = 5(a+2) ×
20
100
= a+2 which is the value of c.
Therefore, if the number „c‟ is deleted from the set, the sum of the remaining numbers in the
set will decrease by 20%.
Answer: c.
Problem-2: Rahim bought two varieties of rice, costing Taka 5 per kg and Taka 6 per
kg each, and mixed them in some ratio. Then he sold the mixture at Taka 7 per kg,
making a profit of 20 percent. What was the ratio of the mixture?
Solution:
Suppose, the ratio of the mixture is x : y.
1.2(5x+6y) = 7(x+y)
or, 6x+7.2y = 7x+7y
or, x = 0.2y
or, x : y = 0.2 : 1 = 1 : 5.
So, the ratio of the mixture was 1 : 5.
Answer: 1 : 5.
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Problem-3: 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:
Suppose, working alone 1 man can do x part and 1 woman can do y part of the work.
3(2x+5y) =
1
4
or, 2x+5y =
1
12
--- (i)
Again,
2(3x+5y) =
1
4
or, 3x+5y =
1
8
--- (ii)
(ii)–(i)=>
x =
1
8
-
1
12
=
1
24
So,
1
24
of the work can be done in 1 day by = 1 man
1 or full of the work can be done in 4 days by =
24
4
= 6 men.
Answer: 6 men.
AD 2014 Batch [Exam held in 2013]
Problem-1: A, B and 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?
Solution:
Working alone, A can do in 20 days = 1 work
Working alone, A can do in 1 day =
1
20
of the work
Working alone, B can do in 30 days = 1 work
Working alone, B can do in 1 day =
1
30
of the work
Working alone, C can do in 60 days = 1 work
Working alone, C can do in 1 day =
1
60
of the work
So, if A is assisted by B and C, work done in 1 day =
1
20
+
1
30
+
1
60
=
6
60
=
1
10
of the work
So, they can complete
1
10
of the work in = 1 day
They can complete 1 or full work in = 10 days.
Answer: 10 days.
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Problem-2: A bus is travelling with 52 passengers. When it arrives at a stop, Y
passengers get off and 4 get in. At the next stop, one-third of the passengers get off
and 3 get on. There are now 25 passengers. Find out how many passengers got off at
the first stop.
Solution:
(52-Y+4) 1-
1
3
+ 3 = 25
or,
2(56-Y)
3
= 22
or, 56–Y = 33
or, Y = 56–33 = 23
So, in the first stop 23 passengers got off.
Answer: 23.
Problem-3: An Eskimo leaves its igloo and travels 3 kilometers north, then 8
kilometers east and finally 3 kilometers north to reach the North Pole. How many
kilometers does he have to travel to return to his igloo in a straight line?
Solution:
Suppose, the igloo is at A, the Eskimo travels from A to
B then B to C and finally from C to D to reach the North
Pole. So, AB = CD = 3 km, and BC = 8 km. The direct
distance from the North Pole to his igloo is AD.
Let‟s draw a perpendicular line from A on the extension
of DC. Suppose, the line intersects the extension of DC
at E. So, CE = 3 km and AE = 8 km.
Therefore, DE = (3+3) km = 6 km
In right angled triangle ADE, AE = 8 km, DE = 6 km.
So, AD = 8²+6² km = 64+36 km = 100 km = 10 km.
So, the Eskimo will have to travel 10 km to return to his igloo in a straight line.
Answer: 10 km.
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Problem-1: A series has three numbers a, ar and ar². In the series, the first term is
twice the second term. What is the ratio of the sum of the first two terms to the sum of
last two terms in the series?
Solution:
a = 2ar
or, r =
1
2
The sum of the first two terms in the series = a+ar = a(1+r) = a(1 +
1
2
) =
3a
2
The sum of the last two terms in the series = ar+ar² = ar(1+r) =
1
2
× a(1+r) =
1
2
×
3a
2
=
3a
4
So, the required ratio =
3a
2
:
3a
4
=
3a
2
×
4
3a
= 2 : 1
Answer: 2 : 1
Problem-2: See Problem-1 of AD Batch 2015.
Problem-3: Two alloys A and B are composed of two basic elements. The ratios of the
composition of two basic elements in the two alloys are 5:3 and 1:2 respectively. A
new alloy X is formed by mixing the two alloys A and B in the ratio 4:3. What is the
ratio of the composition of the two basic elements in the alloy X?
Solution:
Suppose,
the basic elements are p and q.
4x units of Alloy-A and 3x units of Alloy-B is mixed together to form the new Alloy-X.
In 4x units of Alloy-A, the quantity of p =
4𝑥×5
5+3
=
20𝑥
8
units
In 4x units of Alloy-A, the quantity of q =
4𝑥×3
5+3
=
12𝑥
8
units
In 3x units of Alloy-B, the quantity of p =
3𝑥×1
1+2
=
3𝑥
3
units
In 3x units of Alloy-B, the quantity of q =
3𝑥×2
1+2
=
6𝑥
3
units
So, in the new alloy, X, total quantity of p =
20𝑥
8
+
3𝑥
3
=
84𝑥
24
units
So, in the new alloy, X, total quantity of q =
12𝑥
8
+
6𝑥
3
=
84𝑥
24
units
Therefore, the ratio of p and q in Alloy-X =
84𝑥
24
:
84𝑥
24
= 1 : 1
Answer: 1 : 1.
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21
SMcxlvii
Cash Officer 2013 Batch [Exam held in 2011]
Problem-1: The area of a rectangular plot is 323 square meters. Its perimeter is 72
meters. Find the length and breadth of the plot.
Solution:
Suppose, the length of the plot is x meters.
So, the breadth of the plot is =
323
x
meters.
2 x+
323
x
= 72
or,
x2+323
x
= 36
or, x²-36x+323 = 0
or, x²-19x-17x+323 = 0
or, (x-19)(x-17) = 0
either x = 19 or x = 17.
For x = 19, the breadth is = 323÷19 = 17.
For x = 17, the breadth is = 323÷17 = 19.
As, usually length of a rectangle is greater than the breadth, from the above discussion, we
can say, the length is 19 meters and the breadth is 17 meters.
Answer: 19 meters and 17 meters.
Problem-2: Mr. Hasan has few notes of Tk. 10 and Tk. 50. A total of his 150 notes
amount to Tk. 5100. What is the number of each kind of note?
Solution:
Suppose, the number of Tk. 10 notes is x. So, the number of Tk. 50 notes is (150-x).
10x + 50(150-x) = 5100
or, 10x + 7500 - 50x = 5100
or, 40x = 2400
or, x = 60
So, the number of Tk. 10 notes is = 60, and the number of Tk. 50 notes is = 150-60 = 90
Answer: The number of Tk. 10 notes is 60 and that of Tk. 50 notes is 90.
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SMcxlvii
Problem-1: Two cars race around a circular track in opposite directions at constant
rates. They start at the same point and meet every 30 seconds. If they move in the
same direction, they meet every 120 seconds. If the track is 1800 meters long, what is
the speed of each car?
Solution:
Suppose, the speeds of the cars are x mps and y mps respectively where x > y.
30x + 30y = 1800
or, x + y = 60 --- (i)
Again,
120x - 120y = 1800
or, x - y = 15 --- (ii)
(i)+(ii)=>
2x = 75
or, x = 75÷2 = 37.5
Again, (i)-(ii)=>
2y = 45
or, y = 45÷2 = 22.5
So, the speeds of the cars are 37.5 mps and 22.5 mps.
Answer: 37.5 mps and 22.5 mps.
Problem-2: A printer quotes a price of Taka 7500 for printing 1000 copies of a book
and Taka 15000 for printing 2500. Assuming a linear relationship and 2000 books are
printed, find (a) the variable cost per book, (b) the average cost per book, and (c) the
fixed cost.
Solution:
Suppose, the total fixed cost of printing the books is Tk. x and the variable cost of printing
per book is Tk. y.
x + 2500y = 15000 --- (i)
x + 1000y = 7500 --- (ii)
(i)-(ii)=>
1500y = 7500
or, y = 7500÷1500 = 5
Substituting the value of y in (i), we get,
x + 2500×5 = 15000
or, x = 15000 - 12500 = 2500
So, (a) The variable cost per book is, y = Tk. 5.
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(b) Average cost per book for printing 2000 books is = Tk. (x + 2000y) ÷ 2000
= Tk. (2500+2000×5) ÷ 2000
= Tk. 6.25
(c) The fixed cost, x = Tk. 2500
Answer: (a) Tk. 5, (b) Tk. 6.25, (c) Tk. 2500.
Problem-3: The length of a rectangle is twice its width. If the length is increased by 4
inches and the width is decreased by 3 inches, a new rectangle is formed whose
perimeter is 62 inches. What is the length of the original rectangle?
Solution:
Suppose, the width of the original rectangle is x inches.
So, the length of the original rectangle is 2x inches.
2{(2x+4)+(x-3)} = 62
or, 2x+4+x-3 = 31
or, 3x = 30
or, x = 10
or, 2x = 10×2 = 20.
So, the length of the original rectangle is 20 inches.
Answer: 20 inches.
AD 2011 Batch [Exam Held in 2010]
Problem-1: Tanim bought some oranges. He gave
1
2
of them to his sister,
1
4
of the
remainder to his neighbor,
3
5
of those left to his children and had 6 left in the end. How
many oranges did Tanim buy?
Solution:
Suppose, Tanim bought x oranges.
So, he gave to his sister = x ×
1
2
=
x
2
oranges.
The remaining number oranges is = x -
x
2
=
x
2
.
He gave to his neighbor =
x
2
×
1
4
=
x
8
oranges.
The remaining number of oranges is =
x
2
-
x
8
=
3x
8
He gave to his children =
3x
8
×
3
5
=
9x
40
oranges.
The remaining number of oranges is =
3x
8
-
9x
40
=
6x
40
=
3x
20
3x
20
= 6
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SMcxlvii
or, x = 6×
20
3
= 40
So, Tanim bought 40 oranges.
Answer: 40.
Problem-2: The length of rectangular plot is greater than its breadth by 20 meters. If
the perimeter of the plot is 160 meters, what is the area of the plot in square meters?
Solution:
Suppose, the breadth of the rectangular plot is x meters.
So, the length of the plot is (x+20) meters.
2(x+20+x) = 160
or, 2x = 60
or, x = 30
So, the area of the plot is = x(x+20) = 30×(30+20) = 1500 square meters.
Answer: 1500 square meters.
Problem-3: Three partners A, B, and C start a business. Twice the capital of A is equal
to thrice the capital of B, and the capital of B is 4 times the capital of C. They share
the profit in the ratio of their capital. In a particular year, the gross profit is Tk. 250000
and the administrative expenses are 20% of the gross profit. Find the share of profit of
each partner.
Solution:
2(Capital of A) = 3(Capital of B)
or, (Capital of A) : (Capital of B) = 3 : 2
Again,
(Capital of B) = 4(Capital of C)
or, (Capital of B) : (Capital of C) = 4 : 1
So, (Capital of A) : (Capital of B) : (Capital of C) = (3×4) : (4×2) : (2×1) = 12 : 8 : 2 = 6 : 4 : 1.
Net profit of the business = (Gross Profit) - (Administrative Expenses)
= Tk. 250000 - (20% of Tk. 250000)
= Tk. 250000 - Tk. 50000
= Tk. 200000
So, the profit share of A = Tk. 200000 ×
6
6+4+1
= Tk. 109090.91
The profit share of B = Tk. 200000 ×
4
6+4+1
= Tk. 72727.27
The profit share of C = Tk. 200000 ×
1
6+4+1
= Tk. 18181.82
Answer: Profit share of A is Tk. 109090.91, B is Tk. 72727.27 and C is Tk. 18181.82.
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