Bangladesh Bank

Mathematics & Aptitude Test

Exam date: 20.05.2016

36. A person has to cover a distance of 6 km in 45 minutes. If he covers one-half of the distance in two-thirds of the total time; to cover the remaining distance in the remaining time, his speed (in km/hr) must be.

(a) 15                 (b) 6                (c) 8                (d) 12

Ans. (d) 12

Solution: One half of the distance of 6 km in 3 km. He has to cover remaining 3 km in one-third of time 45 minutes which is \frac{45}{3} or 15 minutes or \frac{1}{4} hours.

\therefore His speed in km/hr =\frac{3}{\frac{1}{4}}=3\times 4=12

37. Which decimal of an hour is a second?

(a) 0.000126        (b) 0.0025         (c) 0.0256        (d) 0.00027

Ans. (d) 0.00027

Solution: 1 hour=3600 seconds

\therefore 1 second = \frac{1}{3600} hours = 0.00027

38. One pipe can fill a tank three times as fast as another pipe. If together the two pipes can fill the tank in 36 minutes, then the slower pipe alone will be able to fill the tank in: 

(a) 192 min      (b) 81 min        (c) 108 min       (d) 144 min

Ans. (d) 144 min

Solution: Lets assume, slower pipe fills in 3t mins and faster in t mins together in one minutes they fill = \frac{1}{3t}+\frac{1}{t}=\frac{4}{3t} portion

total time taken, \frac{3t}{4}=36

\therefore  t = 48 min

\therefore slower one takes = 3t = 3\times 48=144 mins

39. Anik, Kamal and Jamal invested Tk. 8000, Tk. 4000 and Tk. 8000 respectively in a business. Anik left after six months. If after eight months, there was a gain of Tk. 4005, then what will be the share of Kamal?

(a) Tk.1780      (b) Tk.890        (c) Tk.1335       (d) Tk.1602

Ans. (b) Tk.890

Solution: Anik invested=8000×6=48000 portion

Kamal invested=4000×8=32000 portion

Jamal invested=8000×8=64000 portion

\therefore Total = 48000+32000+64000=144000 portion

\therefore Kamal gets=\frac{4005\times 32000}{144000}=890tk.

40. The ratio of two numbers is 3:4 and their H.C.F is 4. Their L.C.M is:

(a) 48           (b) 12          (c) 16          (d) 24
Ans. (a) 48

Solution: Given, two number’s ratio is 3:4;

Let the numbers be 3x and 4x. Now thier HCF (the common number) is x.

so x=4

We know,

1st number×2nd number=their HCF×their LCM

or, 3x×4x=x × LCM

\therefore LCM=\frac{3\times 4\times 4\times 4}{4}=48

41. If selling price of an article is \frac{4}{3} of its cost price, the profit in the transaction is: 

(a) 33\frac{1}{3} %     (b) 16\frac{2}{3}%     (c) 20\frac{1}{2} % (d) 25\frac{1}{2}%

Ans. (a) 33\frac{1}{3}%

Solution: Let, the cost price be x

\therefore selling price is \frac{4}{3} of x =\frac{4}{3}\times x=\frac{4x}{3}

\therefore Profit%=\frac{selling price-cost price}{cost price}\times 100%

=\frac{\frac{4x}{3}-x}{x}\times 100%

=\frac{x\left ( \frac{4}{3}-1 \right )}{x}\times 100%

=\left ( \frac{4}{3}-1 \right )\times 100%

=\left ( \frac{4-3}{3} \right )\times 100%



42. Tk. 800 becomes Tk. 956 in 3 years at a certain rate of simple interest. If the rate of interest is increased by 4%, what amount will Tk. 800 become in 3 years?

(a) Tk. 1052          (b) Tk. 1020.80        (c) Tk. 1025           (d) None of these

Ans. (a) Tk. 1052

Solution: For Simple interest, A=Principal+interest

\Rightarrow A=P+i

and i=A-P=Pnr

\Rightarrow 956-800=800\times 3\times \frac{r}{100}

\Rightarrow 156=800\times 3\times \frac{r}{100}

\therefore r=\frac{156\times 100}{800\times 3}=6.5

When the interest rate (r) is increased by 4%,

it becomes =6.5+4=10.5

\therefore new i = Pnr=800\times 3\times \frac{10.5}{100}=252

So total amount A=P+i=800+252=1052

43. The smallest 6 digit number exactly divisible by 111 is:

(a) 11111      (b) 110011      (c) 110101      (d) 100011

Ans. (d) 100011

Solution: They asked for the smallest number, so we will start from the smallest number given in the answer options and check if it is divisible by 111.

The smallest is here 100011.

100011 ÷ 111=901

So, 100011 is divisible by 111 and the this number is smallest. That is the correct option.

44. If m and n are whole numbers such that m^{n}=121, then the value of \left ( m-1 \right )^{n+1} is :

(a) 1000            (b) 1               (c) 10              (d) 121

Ans. (a) 1000

Solution: m^{n}=121=\left ( 11 \right )^{2}

\therefore m=11;n=2

\therefore \left ( m-1 \right )^{n+1}=\left ( 11-1 \right )^{2+1}  =10^{3}=1000

45. The ages of A and B are in the ratio 3:1. Fifteen years hence, the ratio will be 2:1. Their present ages are:

(a) 21 years, 7 years                  (b) 30 years, 10 years

(c) 45 years, 15 years               (d) 60 years, 20 years

Ans. (c) 45 years, 15 years

Solution: Given that, their present age ratio is 3:1;

so assume their present age is 3x and x respectively

After 15 years ratio will be 3x+15: x+15

In 15 years the ratio becomes 2:1

\therefore \frac{3x+15}{x+15}=\frac{2}{1}

\Rightarrow \left (3x+15  \right )\times 1=\left ( x+15 \right )\times 2

\Rightarrow 3x+15= 2x+30

\Rightarrow 3x-2x=30-15

\Rightarrow x=15

\therefore 3x=3×15=45 and x=15

\therefore Their present ages are 45 and 15 years respectively.

46. \left ( 2\sqrt{27}-\sqrt{75}+\sqrt{12} \right ) is equal to:

(a) 4\sqrt3    (b) \sqrt3  (c) 2\sqrt3   (d) 3\sqrt3

Ans. (d) 3\sqrt3

Solution: 2\sqrt27-\sqrt75+\sqrt12






47. 15 men take 21 days of 8 hours each to do a pice of work. How many days of 6 hours each would 21 women take; if 3 women do as much work as 2 men?

(a) 30         (b) 18               (c) 20              (d) 25

Ans. (a) 30


3 women’s work is equivalent to 2 men’s work

\therefore 1 women’s work is equivalent to \frac{2}{3} men’s work

\therefore 21 women’s work is equivalent to \frac{2\times 21}{3}=14 men’s work


15 men working 8 hr takes 21 days

\therefore  1 men working 1 hr takes 15×8×21 days

\therefore 14 men working 6 hr takes \frac{15\times 8\times 21}{14\times 6} days=30 days.

48. If 25% of a number is subtracted from a second number, the second number reduces to its five-sixth. What is the ratio of the first number to the second number?

(a) 3:2            (b) 1:3               (c) 2:3               (d) None of these

Ans. (c) 2:3

Solution: Assume that, the first number be x and the second number be y

here 25%=\frac{25}{100}=\frac{1}{4}



\Rightarrow y-\frac{5}{6}y=\frac{1}{4}x

\Rightarrow \frac{6y-5y}{6}=\frac{x}{4}

\Rightarrow \frac{y}{6}=\frac{x}{4}

\Rightarrow \frac{x}{y}=\frac{4}{6}=\frac{2}{3}

\therefore x:y=2:3

49. A is 30% more efficient than B. How much time will they, working together, take to complete a job which A alone could have done in 23 days? 

(a) 21 days      (b) 11 days         (c) 13 days        (d) None of these

Ans.  (c) 13 days

Solution: Here, 30% more = \frac{130}{100}=1.30

Since A is 30% more efficient than B, B would take 30% more time that is, 23×1.30 days =29.9 \approx30 days to finish the work.

So, \left ( \frac{1}{23}+\frac{1}{30} \right )

or, \frac{\left ( 30+23 \right )}{30\times 23} portion of work is done by both A and B in 1 day

So \frac{53}{30\times 23} portion is done by both in 1 day

\therefore Total work is done by both is \frac{1}{\frac{53}{30\times 23}}=\frac{30\times 23}{53}=13.0188=13days

50. What mathematical operation should come at the place of ‘?’ in the equation: 2 ? 6-12÷4+2=11

(a) ÷              (b) +                       (c) –                    (d) ×

Ans. (d) ×


L.H.S= 2 ? 6-12÷4+2

= 2 ? 6-3+2

so we see that only multiplication can place of ‘?’ for come 11

\therefore 2 × 6-3+2 =11

51. Two numbers are such that the ratio between them is 4:7. If each is increased by 4, the ratio becomes 3: 5. The larger number is: 

(a) 64              (b) 36                   (c) 48                   (d) 56

Ans. (d) 56

Solution: Let the number be 4x and 7x

If increased 4 then ratio will be 4x+4:7x+4

by condition, \frac{4x+4}{7x+4}=\frac{3}{5}

\Rightarrow 20x+20=21x+12

\Rightarrow 21x-20x=20-12

\Rightarrow x=8

\therefore larger number = 7×8=56

52. The percentage increase in the area of a rectangle, if each of its sides is increased by 20% is : 

(a) 46%                 (b) 40%                      (c) 42%                  (d) 44%

Ans. (d) 44%

Solution: Shortcut system:

20+20+\frac{20\times 20}{100}



53. How many seconds will a 500 meter long train take to cross a man walking with a speed of 3 km/hr in the direction of the moving train if the speed of the train is 63 km/hr ?

(a) 45                (b) 25                 (c) 30                (d) 40

Ans. (c) 30

Solution: As the direction of train and man is same,

the relative velocity of the train would be =(63-3)=60 km/hr. =\frac{60\times 1000}{3600}m/s

\therefore time need to pass

=\frac{500}{\frac{60\times 1000}{3600}}

=\frac{500\times 3600}{60\times 1000}=30 seconds

54. A person’s present age is two-fifth of the age of his mother. After 8 years, he will be one-half of the age of his mother. How old is the mother at present?

(a) 48 years (b) 32 years (c) 36 years (d) 40 years

Ans. (d) 40 years

Solution: Lets the mother’s present age be x years

then the person’s present age is \frac{2}{5}x  years

After 8 years mother’s age will be (x+8)

and person’s age will be \left (\frac{2}{5}x+8  \right )

Now, \left ( \frac{2}{5}x+8 \right )=\frac{\left ( x+8 \right )}{2}

\Rightarrow \frac{4}{5}x+16=x+8

\Rightarrow 4x+5\times 16=5x+5\times 8

\Rightarrow 5x-4x=5\left ( 16-8 \right )

\therefore x=5\times 8=40

\therefore the mother’s present age is 40 years

55. The average of five consecutive odd numbers is 61. What is the difference between the highest and lowest numbers? 

(a) 8               (b) 2                    (c) 5                  (d) None of these

Ans. (a) 8

Solution: For average of consecutive number, the average value is always the middle number. So, the consecutive odd numbers are 57,59,61,63,65

And the difference between highest and lowest number = (65-57)=8

56. If A + B = 2C and C + D =2A, then

(a) A + C = 2B       (b) A + C = B + D   (c) A + C = 2D    (d )A + D = B + C

Ans. (b) A + C = B + D

Solution: Given,

A + B = 2C

C + D = 2A

A + B + C + D = 2C + 2A (adding)

B + D = 2C + 2A – A – C

B + D = C + A

\therefore A + C = B + D

57. A, B and C are sisters. D is the brother of E and E is the daughter of B. How is A related to D ? 

(a) Aunt              (b) Sister              (c) Cousin             (d) Niece

Ans. (a) Aunt.

58. In a class of 60, where girls are twice that of boys, Kamal ranked seventeenth from the top. If there 9 girls ahead of Kamal, how many boys are after him in rank? 

(a) 23       (b) 3           (c) 7              (d) 12

Ans. (d) 12

Solution: Let, number of girls be 2x and boys x

\therefor 2x+x=3x=60

\Rightarrow x=20

\therefore boys 20 and girls 40

Now, Kamal is 17th, Ahead of him are 16 people. Among them, 9 are girls. So no. of boys ahead = (16-9)=7 boys including him, 8 boys upto his position. So no. of boys after him is (20-8)=12 boys.

59. Choose the number pair/group which is different from others.

(a) 63-77            (b) 50-66             (c) 32-48             (d) 64-80

Ans. (a) 63-77

Solution: Here only option A is odd number pair and the other options has even number pair.

60. If A =26, SUN = 27, then CAT = ?

(a) 58          (b) 24            (c) 27           (d) 57

Ans. (d) 57

Solution: Here, Reverse number assigned to letters.

A = 26, B=25, C=24 , …………..

Y=2, Z=1.


So, CAT=C+A+T=24+26+7=57

61. Choose out the odd one.

(a) WHO            (b) IMF                (c) SAARC       (d) UNICEF

Ans. (c) SAARC

Solution: All other’s are Global. SAARC is not .

62. Arrange the following in a logical order:

1.Euphoria, 2. Happiness,3.Ambivalence, 4.Ecstasy, 5.Pleasure

(a) 4,1,3,2,5    (b) 1,4,2,5,3         (c) 2,1,3,4,5      (d) 3,2,5,1,4

Ans. (d) 3,2,5,1,4

Solution: All the given words stand for joy, but these increase in the order Ambivalence, happiness, pleasure, Euphoria, Ecstasy

63. Cyclone is related to Anticyclone in the same way as Flood is related to…………..?

(a) Drought         (b) Devastation      (c) Havoc      (d) River

Ans. (a) Drought

Solution: Flood and Drought is opposite of each other, exact like Cyclone and Anticyclone.

64. How many sons does X have? 

I. Q and U are brother of T.                II. R is sister of P and U.

III. R and T are daughters of X

(a) All I, II and III     (b) Only I and II

(c) Only II and III          (d) I,II and III together are not sufficient

Ans. (d) I,II and III together are not sufficient

Solution: Here, Q,U are sons, T, R are daughters. But we can not determine P’s gender. So not sufficient information.

65. Root: Stem: Branch: ?

(a) Fertiliser            (b) Wood         (c) Leaf         (d) Tree

Ans. (c) Leaf

Solution: As root caries stem of a tree, branch caries the leaf of a tree.