Bangladesh Bank Assistant Director Recruitment Examination, 2016

Solutions to the Mathematics & Aptitude Test Section

Prepared by:

Sultan Mahmud Cxlvii

Admin, Banking Career in Bangladesh

Date: 22 April, 2016

01] A person has to cover a distance of 6km 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 kmph must be: ?

Solution:

He covered 6/2 = 3km in 45×2/3 = 30 minutes.

So, he must cover in 15 minutes = 3 km.

Required speed = 3km/15min = 3km/15/60 hr = 6kmph. ans.

02] What decimal of an hour is a second?

Solution:

1 hour = 3600 second

1 second as a fraction of ah hour = 1/3600 = 1/36 × 1/100 = 0.027' × 1/100 = 0.00027'. ans.

03] 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:

Solution:

Suppose, the slower pipe can fill in 1 minute = p fraction of the tank

So, the faster pipe can fill in 1 minute = 3p fraction of the tank.

So, in 1 minute, together they can fill = 4p fraction of the tank.

According to the question:

36×4p = 1

p = 1/144

So, the slower pipe can fill in 1 minute = 1/144 of the tank

Therefore, the slower pipe can fill the entire tank in 1/(1/144) = 144 minutes. ans.

04] Aunik, Kamal and Jamal invested Tk 8000, Tk 4000, and Tk 8000 respectively in a

business. Aunik left after six months. If after eight months, there was a gain of Tk 4005, then

what will be the share of Kamal?

Solution:

WAV Ratio of investment in business = A : K : J = (8×6) : (4×8) : (8×8) = 3 : 2 : 4

Kamal's share of gain = 4005 × 2/(3+2+4) = 890. ans.

05] The ratio of two numbers is 3 : 4, and their HCF is 4. Their LCM is:

Solution:

Suppose, the numbers are 3p and 4p.

Their HCF = p, and their LCM = 12p

So, 12p = 12×4 = 48. ans.

06] If the selling price of an article is 4/3 of its cost price, the profit in the transaction is:

Solution :

Cost = P

Selling Price = 4P/3

Profit = 4P/3 - P = P/3

Profit percentage = ((P/3)×100/P)% = 33 1/3 % or, [33.33'%] ans.

07] 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?

Solution:

At 4% rate of interest, total interest earned in 3 years = 800 × 4/100 × 3 = 96.

So the amount will become = 956+96 = 1052. ans.

08] The smallest 6-digit number exactly divisible by 111 is:

Solution:

smallest six digit number = 100000.

100000 divided by 111 = quotient is 900, and remainder is 100.

So, the smallest number is = (900+1) × 111 = 100011. ans.

09] If m and n are whole numbers such that mⁿ = 121, then the value of (m-1)n+1

is:

Solution:

Only two possible values of m and n in mⁿ = 121 is possible. they are:

1211 = 121, or 11

2 = 121

i.e: (m,n) = (121,1) or (11,2)

So, (m-1)n+1

=

Either, (121-1)1+1

= 121²

Or, (11-1)2+1

= 1000

Among the given options we get a value = 1000. ans.

10] The age of A and B are in the ratio 3 : 1. Fifteen years hence, the ratio will be 2 : 1. Their

present ages are:

Solution:

Suppose, the present ages of A and B are 3p and p years respectively.

According to the question:

(3p+15)/(p+15) = 2

3p + 15 = 2p + 30

p = 15.

So, their present ages are = 45years and 15 years. ans.

11] (2√27 - √75 + √12) is equal to:

Solution;

= 2×3√3 - 5√3 + 2√3 = 6√3 - 5√3 + 2√3 = 3√3. ans.

12] 15 men take 21 days of 8 hours each to do a piece of work. How many days of 6 hours

each would 21 women take if 3 women do as much as 2 men?

Solution:

3 women's work = 2 men's work. So, 21 women's work = 2×21/3 = 14 men's work.

15 men can do the work in 21×8 hours.

14 men can do the work in 21×8×15/14 hours = 180 hours.

Required number of days = 180/6 = 30 days. ans.

13] 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?

Solution:

The second number reduces to its 5/6th. i.e the reduction is = 1 - 5/6 = 1/6.

25% of the first number = 25/100 of the first number = 1/4 of the first number.

According to the question:

1/4 of the first number = 1/6 of the second number

First number : second number = (1/6)/(1/4) = 2 : 3. ans.

14] A is 30% more efficient that B. How much time will they, working together, take to

complete a job which A alone could have done in 23 days?

Solution:

A can do in 1 day = 1/23 of the work.

B can do in 1 day = 1/(23×1.3) of the work.

Together they can complete the work in = 1/(1/23 + 1/(23×1.3)) = 23×1.3/2.3 = 13 days. ans.

15] What mathematical operation should come at the place of '?' in the equation : 2 ? 6 - 12 ÷

4 + 2 = 11.

Solution:

=> 2 [?] 6 - 3 + 2 = 11 ; => 2 [?] 6 - 1 = 11

=> 2 [?] 6 = 12

We know, 2 × 6 = 12

So, we can conclude, at the place of "?" we should place the multiplication symbol i.e: (×).

ans.

16] 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:

Solution:

Suppose, the numbers are 4p and 7p respectively

According to the question:

(4p+4)/(7p+4) = 3/5

21p + 12 = 20p + 20

p = 8

The larger number = 7×8 = 56. ans.

17] The percentage increase in the area of a rectangle, if each of its sides is increased by 20%

is:

Solution:

Increase in area = [20 + 20 + 20×20/100]% = 44%. ans.

18] How many seconds will a 500 meter long train take to cross a man walking with a speed

of 3 kmph in the direction of the moving train if the speed of the train is 63 kmph?

Solution:

The man is walking in the direction of the moving the train, so, the relative speed of the train

= 63-3 = 60 kmph.

Required time to cross = (500/1000)/(60) = 1/120 hrs = 3600/120 seconds = 30 seconds. ans.

19] 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?

Solution:

Suppose, the mother's present age is = 5p years.

So, the person's present age is = 2p years.

According to the question:

(2p+8)/(5p+8) = 1/2

5p + 8 = 4p + 16

p = 8

5p = 40. ans.

20] The average of five consecutive odd numbers is 61. What is the difference between the

highest and the lowest numbers?

Solution:

FLAT RULE: The difference between the first and the fifth number in a series of consecutive

odd integers is = (5-1)×2 = 8. ans.

21] If A+B = 2C, and C+D = 2A, then,

Solution:

C+D = 2A. => A = C+D-A.

A+B = 2C

C+D-A + B = 2C

D+B = 2C-C+A

B+D = A+C. ans.

22] 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?

Solution:

Generations:

A↔B↔C

↓

D↔E

So, A is the aunt of D. ans.

23] In a class of 60, where girls are twice that of boys. Kamal ranked seventeenth from the

top. If there are 9 girls ahead of Kamal, how many boys are after him in the rank?

Solution:

Number of girls in the class = 60×2/(2+1) = 40.

There are 9 girls in the range of students from 1st to 17th.

So, there are 31 girls in the range from 18th to 60th.

Number of boys after Kamal = (60-18+1) - 31 = 43 - 31 = 12. ans.

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

A. 63-77 ; B. 50-66 ; C. 32-48 ; D. 64-80

Solution:

A. result is = -14

B, C, D. each, result is = -16.

Different one is A. 63-77. ans.

25] If A = 26, SUN = 27, then CAT = ?

Solution:

Label the letters as A = 26, B = 25, .... Z = 1.

Now, SUN = S + U + N = 8 + 6 + 13 = 27. [matched √√]

So, CAT = 24 + 26 + 7 = 57. ans.

26] Choose out the odd one: WHO, IMF, SAARC, UNICEF

Solution:

Range of activities of SAARC = South-east Asia.

Range of activities of WHO, IMF, UNICEF = All over the world.

Answer is SAARC.

27] Arrange the following in a logical order: Euphoria, Happiness, Ambivalence, Ecstasy,

Pleasure

Solution:

Intensity of Joy in an increasing order : 3 2 5 1 4 . ans.

28] Cyclone is related to Anticyclone in the same way as Flood is related to: ?

Solution:

Cyclone and Anticyclone are reversely related.

ans is Drought.

29] How many sons does X have?

i. Q and U are brothers of T

ii. R is sister of P and U

iii. R and T are daughters of X

Solution:

Surely R and T are the daughters of X

And also, Q and U are the sons of X

But what about P, is he a boy or a girl? We don't know.

So, we are unable to answer the question as I, II and III together are not sufficient. ans.

30] Root : Stem : Branch : ?

Solution:

If you observe a tree, you will notice the mentioned parts in an order.

Root > Stem > Branch > Leaf. ans.