ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

Question: The number of students in 3 classes is in the ratio 3 : 4 : 5. If 10 students are increased in each class this ratio changes to 4 : 5 : 6. The total number of students in the three classes in the beginning was

Solution: 
Let the number of students in the classes be 3x, 4x and 5x respectively;
Total students = 3x + 4x + 5x = 12x
According to the question,
(3x + 10) : (4x + 10) = 4 : 5
or, 5(3x + 10) = 4(4x+10)
or, 15x + 50 = 16x + 40
or, x = 10

Hence,
Original number of students,
12x = 12 × 10
= 120

question: P, Q, R have the total money of Tk. 2800. Q have half of P and R have half of Q. The amount of R is 

Solution: 
Let P have x

then, 
Q have x/2 and
R have x/4


Now
x + (x/2) + (x/4) = 2800
7x/4 = 2800
x = 1600

Hence, 
The amount of R is (1600/4) = 400

Question: The largest 3 digit number exactly divisible by 8 is:

Solution: 
The largest 3 digit number is 999

So, the number is (999 - 7) = 992

Question: The angles of triangle are in the ratio  3 : 4 : 5. The largest angle is 

Solution: 
we know that, the sum of the angles of a triangle is 180°

Hence, 
The largest angle is (180° × 5/12) or, 75°

Question: Two trains, one from Dhaka to Chittagong and one from Chittagong to Dhaka, started simultaneously. After they meet, the trains reach their destinations after 4 hours and 9 hours respectively. What is the ratio of their speed?

Solution: 


ধরি, 
প্রথম ট্রেনের বেগ S₁ এবং এটি ঢাকা থেকে চট্টগ্রাম যাচ্ছে।
দ্বিতীয় ট্রেনের বেগ S₂ এবং এটি  চট্টগ্রাম থেকে ঢাকা যাচ্ছে।
তারা t সময় পর P বিন্দুতে মিলিত হবে।

ঢাকা থেকে P বিন্দুর জন্য
প্রথম ট্রেনের দূরত্ব = S₁t
দ্বিতীয় ট্রেনের দূরত্ব = 9S₂
∴ S₁t = 9S₂ . . . . . .(1)

চট্টগ্রাম থেকে P বিন্দুর জন্য
প্রথম ট্রেনের দূরত্ব = 4S₁
দ্বিতীয় ট্রেনের দূরত্ব = S₂t
∴ S₂t = 4S₁ . . . . . . (2)

(1) ÷ (2)
S₁t/S₂t = 9S₂/4S₁
S₁²/S₂² = 9/4
S₁/S₂ = 3/2
S₁ : S₂ = 3 : 2

Question: Find the greatest number that will divide 45, 93 and 185 so as to leave the same remainder in each case.

Solution: 
Here,
93 - 45 = 48
185 - 93 = 92
185 - 45 = 140

So, the required number = H.C.F of 48, 92 and 140
= 4

Question: A can do a work in 4 hours, B and C together in 3 hours, and A and C together in 2 hours. How long will B take alone to do it? 

Solution: 
A's 1 hour work = 1/4
(B + C)'s 1 hour work = 1/3
(A + C)'s 1 hour work = 1/2

(A + B + C)'s 1 hour work = 1/4 + 1/3 = 7/12

∴ B's 1 hour work is = {(A + B + C)'s 1 hour work} - {(A + C)'s 1 hour work}
= 7/12 - 1/2
= 1/12

so, B alone can do it in 12 hours.

Question: Three times the first of three consecutive even integers is 2 more than twice the third. The third integer is:

Solution: 
Let,
The three consecutive even integers is x, x + 2 and x + 4 
Then,
3x = 2(x + 4) + 2 
⇒ 3x = 2x + 8 + 2
⇒ x = 10

The third integer is 10 + 4 = 14 

Question: 5 people can do a work in 8 days working 7 hours a day. If 2 new people join the team to complete the work in 4 days, they have to work per day for -

Solution: 
প্রশ্নের প্রথম অংশ - ৫ জন লোক একটি কাজ দৈনিক ৭ ঘন্টা কাজ করে ৮ দিনে শেষ করে।
এখানে, 
লোক, M₁ = 5,
কাজ, W₁ = 1,
দিন, D₁ = 8,
সময়, H₁ = 7.

প্রশ্নের দ্বিতীয় অংশ থেকে - যদি অতিরিক্ত ২ জন লোক যুক্ত করা হয় তাহলে একই কাজ ৪ দিনে শেষ করতে সবার প্রতিদিন কতক্ষণ কাজ করতে হবে?
এখানে,
লোক, M₂ = 5 + 2 = 7,
কাজ, W₂ = 1,
দিন, D₂ = 4,
সময়, H₂ = ?

আমরা জানি,
M₁ × W₂ × D₁ × H₁ = M₂ × W₁ × D₂ × H₂
H₂ = (M₁ × W₂ × D₁ × H₁)/ (M₂ × W₁ × D₂)
= (5 × 1 × 8 × 7)/(7 × 1 × 4)
= 280/28
= 10 hours

Question: Bangladesh need 282 runs against Australia. In the first 10 overs the run rate is only 3.2, What should be the run rate in the remaining 40 overs to reach the target?

Solution: 
In the first 10 overs the run rate is 3.2
∴ After 10 overs the total run is = (3.2 × 10) runs
= 32 runs

Remaining = (282 - 32) runs
= 250 runs

The run rate in the remaining 40 overs to reach the target should be =(250 ÷ 40) 
= 6.25 

Question: If a product is sold at 250% profit, then what is the ratio of buying and selling price?

Solution: 
ধরি,
ক্রয়মূল্য = ১০০
২৫০% লাভে বিক্রয়মূল্য = ১০০ + ১০০ এর ২৫০%
= ৩৫০ 

∴ ক্রয়মুল্য : বিক্রয়মুল্য = ১০০ : ৩৫০ 
= ১ : ৩.৫

Question: If 4sin²(2θ) +1 = 4, then θ = ?

Solution: 
4sin²(2θ) +1 = 4
বা, 4sin²(2θ) = 4 - 1
বা, sin²(2θ) = 3/4
বা, sin(2θ) = √3/2
বা, sin(2θ) = sin60°
বা, 2θ = 60° 
∴ θ = 30°

Question: A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?

Solution: 
ধরি,
পাত্রের মিশ্রণের পরিমাণ ৮ লিটার।

মিশ্রণে পানির পরিমাণ ৩ লিটার
মিশ্রণে সিরাপের পরিমাণ ৫ লিটার

পাত্রের পানি ও সিরাপের পরিমাণ অর্ধেক অর্ধেক করতে ক লিটার মিশ্রণ অপসারণ করে পানি দিতে হবে।

ক লিটার মিশ্রণে পানির পরিমাণ ৩ক/৮ লিটার
ক লিটার মিশ্রণে সিরাপের পরিমাণ ৫ক/৮ লিটার

পানি মিশানোর পর,
নতুন মিশ্রণে পানির পরিমাণ হবে  (৩ - ৩ক/৮) + ক লিটার
=(২৪ - ৩ক)/৮ + ক লিটার
= (২৪ + ৫ক)/৮ লিটারনতুন মিশ্রণে সিরাপের পরিমাণ হবে (৫ - ৫ক/৮) লিটার
= (৪০ - ৫ক)/৮ লিটার

শর্তমতে,
(২৪ + ৫ক)/৮ = (৪০ - ৫ক)/৮
বা, ২৪ + ৫ক = ৪০ - ৫ক
বা, ১০ক = ১৬
বা, ক = ৮/৫

৮ লিটার মিশ্রণ ৮ লিটারের সম্পূর্ণ বা ১ অংশ
∴ ৮/৫ লিটার মিশ্রণ ৮ লিটারের (৮/৫)/৮ অংশ
= ১/৫ অংশ

Question: Father is aged four times more than his son. After 8 years, he would be two and a half times of son's age. What is the current age of the father?

Solution:
Let,
The current age of son is x years
∴ The current age of the father is 4x years 

ATQ,
4x + 8 = (5/2)(x + 8)
⇒ 8x + 16 = 5x + 40
⇒ 3x = 24
∴ x = 8

∴ The current age of the father is (4 × 8) years
= 32 years 

Question: What is the angle between the hour hand and minute hand at 1 : 20 pm? 

Solution: 
কোণ =  |১১ × মিনিট - ৬০ × ঘণ্টা|°/২
= |১১ × ২০ - ৬০ × ১|°/২
= |২২০ - ৬০|°/২
= ১৬০°/২
= ৮০°

Question: 4 men can do one work in 8 days.  How many men can do the half work in 2 days?

Solution: 
From the first part of the question, we get - 4 men can do one work in 8 days.
here, 
men, M₁ = 4,
work, W₁ = 1,
day, D₁ = 8.

From the second part of the question, we get - How many men can do the half work in 2 days?
here, 
men, M₂ =?
work, W₂ = 1/2,
day, D = 2.

we know,
M₁ × W₂ × D₁ = M₂ × W₁ × D₂
M₂ = (M₁ × W₂ × D₁) / (W₁ × D₂)
= {4 × (1/2) × 8}/ (1 × 2)
= 16/2
= 8 men

Question: In how many different ways can the letters of the word 'CORRUPTION' be arranged so that the vowels always come together?

Solution: 
In the word 'CORRUPTION', we treat the vowels OUIO as one letter.

Thus, we have CRRPTN (OUIO).

This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.

Number of ways arranging these letters =    7!/2!   = 2520.

Now, 4 vowels in which O occurs 2 times and the rest are different, can be arranged in    4!/2!   = 12 ways.

Required number of ways = (2520 x 12) = 30240.

Question: In the cost price of 12 pens is equal to the selling price of 16 pens, find the loss per cent.

Solution:
Let the C.P. of 1 pen =Tk. 1
∴ C.P of 12 pens=Tk. 12×1 =Tk. 12

C.P. of 12 pens = S.P of 16 pens
∴ S.P of 16 pens= Tk. 12
∴ C.P of 16 pens= Tk. 16


∴ Loss = C.P - S.P
=16- 12
=4 Tk.

Loss = {(4/16​) × 100}%
= 25%.

Question: Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 2 and 3?

Solution: 
Number which are multiple of 2 and 3 between 1 to 20  = {6, 12, 18} 

∴ The probability is =  3/20

Question: A circular grassy plot of land, 16m in diameter, has a path 2m wide running round it outside. Find the cost of gravelling the path at Tk. 10 per square metre.

Solution: 
the radius of the plot is = 16/2 = 8m 
the radius with the path is = 8 + 2 = 10m

the area of the path is = π (10² - 8²)
= 3.14 × 36
= 113.04 sq. m

total cost = (113.04 × 10)
= 1130.4 Tk.

Question: It is Tuesday on May 23, 2023. What will be the day of  June 1, 2023? 

Solution: 
মে মাস ৩১ দিনের।
 মে মাসের ২৩ তারিখ থেকে ৩১ তারিখের মাঝে =(৩১ - ২৩) দিন
= ৮ দিন 
১ জুনসহ মোট দিন ৮ + ১ দিন = ৯ দিন 

এখন,
৯ ÷ ৭ = ভাগফল ১, ভাগশেষ ২।

সুতরাং, ৩১ তারিখ হবে মঙ্গলবার + ২ দিন।
= বৃহস্পতিবার

Question: If a big water droplet with diameter of 20m converted to small water droplets of radius of 2m. How many small droplets can be formed from the big droplet?

Solution: 
দেওয়া আছে, 
বড় ফোঁটার ব্যাস = 20m
∴ ব্যাসার্ধ , R = 10m
আয়তন, V = (4/3)πR³

ছোট ফোঁটার ব্যাসার্ধ, r = 2m
আয়তন, v = (4/3)πr³

∴ ছোট ফোঁটার সংখ্যা = বড় ফোঁটার আয়তন / ছোট ফোঁটার আয়তন
= (4/3) πR³ / (4/3) πr³
= R³/r³
= 10³/2³
= 125

Question: Find out the odd-one:
8, 27, 64, 100, 125, 216, 343

Solution: 
Here,
8 = 2³
27 = 3³ 
64 = 4³
100 = 10² 
125 = 5³ 
216 = 6³ 
343 = 7³
 
∴ 100 is the odd-one.

Question: A pipe can fill a tank in 3 hours but an outlet B can empty the tank in 10 hours. If both the pipes are opened simultaneously, then the tank will be filled in -  

Solution: 
in 1 hour, A fills = 1/3
but B reject = 1/10

so, in 1 hour the net fill-up is = 1/3 - 1/10 = 7/30

hence, 
It will take 30/7 hours to fill the tank if both the pipes are opened.

Question: A train running at a speed of 90km/h crosses a platform double its length in 36 seconds. What is the length of the platform in metres?

Solution: 
ধরি, 
ট্রেনের দৈর্ঘ্য = X 
∴ প্লাটফর্মের দৈর্ঘ্য = 2X 
মোট দূরত্ব, D = 3X

ট্রেনের বেগ, S = 90km/h
= (90 × 1000)/3600
= 25 m/s

আমরা জানি,
D = S × T
3X = 25 × 36
3X = 900
X = 300m

∴ প্লাটফর্মের দৈর্ঘ্য = 2 × 300 = 600m

Question: 

In the figure above, what is the value of d?

Question: 
AC² = AB² + BC² 
⇒ (d + 3d)² = 6² + 8²
⇒ (4d)² = 36 + 64
⇒ 16d² = 100
⇒ d² = 100/16
⇒ d = 10/4
∴ d = 2.5 

Question: The compound interest on Tk. 2000 for 2 years at 10% is-

Solution:
Givent that, P = Tk. 2000
n = 2 years
r = 10%
Required amount = P(1 + r) n
= [2000 × {1 + (10/100)²}]
= [2000 × (11/10) × (11/10)]
= 2420

Compound Interest = 2420 - 2000 = 420 Tk.

Question: In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?

Solution: 
We may have,
(1 boy and 3 girls) = (⁶C₁ x ⁴C₃)
= 6 × 4 ways
= 24 ways

(2 boys and 2 girls) = (⁶C₂ x ⁴C₂)
= 15 × 6 ways 
= 90 ways 

(3 boys and 1 girl) = (⁶C₃ x ⁴C₁)
= 20 × 4 ways
= 80 ways

(4 boys) = ⁶C₂ = 15 ways

∴ Required number of ways   
= (24 + 90 + 80 + 15) ways
= 209 ways

Question: A rectangular carpet has an area of 120 sq. metres and a perimeter of 46 metres. What is the length of the diagonal?

Solution: 
Let the length and breadth of the rectangle be x and y metres.
 
given 2(x + y) = 46
x + y = 23
x = 23 - y

and, 
xy = 120
(23 - y) y = 120
23y - y² = 120
y² - 23y + 120 = 0
y² - 15y - 8y + 120 = 0
y(y - 15) - 8(y - 15) = 0
(y - 15) (y - 8) = 0
y = 15 or, 8

∴ length, x = 15 and breadth, y = 8

so, the diagonal is = √{(15)² + (8)²} 
= √289
= 17

Question: Raju drove 8 miles west, 6 miles north, 3 miles east, and 6 more miles north. How far was Raju from his starting place? 

Solution: 

AE = √(AF² + EF²)
= √{(8 - 3)² +(6 + 6)²}
= √( 25 + 144)
= √169
= 13

Question: A square that has a side length of 4m is reduced by 50% of its area. The new side length will be - 

Solution: 
here, the side length is = 4m
the area of the square = 4² =16m²

after reducing the area by 50% the remaining area is = 16 - (50% of 16) 
= 16 - 8 = 8m²

the new side length is = √8 = 2√2 m

Question: If 2√(2x + 1) + 5 = 8, then x = 

Solution: 
Given that,
2√(2x + 1) + 5 = 8
⇒ 2√(2x + 1) = 3
⇒ √(2x + 1) = 3/2
⇒ 2x + 1 = 9/4
⇒ 2x = (9/4) - 1
⇒ 2x = 5/4
∴ x = 5/8

Question: What will be the total worth if 135 Tk. is kept for 2 years at simple interest rate 5%?

Solution: 
Here,
P = 135,
n = 2,
R = 5%
I = ?

I = Pnr
= 135 × 2 × 5%
= 270 × 5/100
= 13.5

total worth = 135 + 13.5 = 148.5 Tk.

Question: A boat went to a place in 7km/h and came back in 8km/h. What is average speed of the boat?

Solution: 
ধরি,
দূরত্ব = x

তাহলে, 
যেতে সময় লাগে x/7 hour
আসতে সময় লাগে x/8 hour

আমরা জানি,
গড় বেগ = (মোট দুরত্ব)/মোট সময়
= 2x/(x/7 + x/8)
= 2x/ (15x/56)
= 112/15
= 7.47km/h

Question: Find the odd-one:
835, 734, 642, 751, 853, 981, 532

Solution: 
In each number except 751, the difference of third and first digit is the middle one.

Question: A bike covers a distance of 750m in just 2 min 30sec. What is the speed in km/h?

Solution: 
here,
distance, D = 750m
time, T = 2min 30sec
= 150sec

We know,
D = S × T
S = D/T
= 750m/150s
= 5m/s
= (5×3600)/1000 km/h
= 5×3.6 km/h
= 18km/h

Question: If x² = 2, what is the value of (x + 1/x)(x - 1/x)?

Solution:
Given that,
x² = 2
∴ x = √2

1/x = 1/√2

Now,
(x + 1/x)(x - 1/x)
= x² - (1/x)² 
= (√2)² - (1/√2)² 
= 2 - (1/2)
= 3/2
= 1.5 

Question: A two-digit number is such that the sum of the digits is 8. When 54 is added to the number, then the digits are reversed. The number is:

Solution: 
Let the ten's and unit digit be x and 8 - x 
∴ The number be 10x + 8 - x = 9x + 8 

ATQ,
9x + 8 + 54 = 10(8 - x) + x
⇒ 9x + 62 = 80 - 10x + x
⇒ 9x + 62 = 80 - 9x
⇒ 18x = 80 - 62
⇒ 18x = 18
∴ x = 1 

∴ The number is 9 × 1 + 8 = 9 + 8 = 17 

Question: What is 25% of 30% of 1/3?

Solution: 
30% of 1/3 = 3/10 of 1/3 = 1/10

25% of 1/10 = 1/4 of 1/10 = 1/40

Question: The angle of elevation of a ladder leaning against a wall is 60° and the foot of the ladder is 4.5 m away from the wall. The length of the ladder is:

Solution:
Let,
AB be the wall and BC be the ladder.
Then, ∠ACB = 60° and AC = 4.5 m.

Here,
AC/BC = cos60°
⇒ AC/BC = 1/2
⇒ BC = 2 × AC 
⇒ BC = 2 × 4.5
∴ BC = 9 

∴ The length of the ladder is 9 m.

Question: If a : b = 7 : 3 then the value of (ab + b²)/(a² - b²)

Solution:
a : b = 7 : 3
a = 7x
b = 3x

 (ab + b²)/(a² -  b²) = 7x × 3x + (3x)²/{(7x)² - (3x)²}
= (21x² + 9x²)/(49x² - 9x²)
= 30x²/40x²
=3/4

Question: A grocer has a sale of Tk. 6435, Tk. 6927, Tk. 6850, Tk. 7226 and Tk. 6562 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Tk. 6500?

Solution: 
Total sale after 5 months (6435 + 6927 + 6850 + 7226 + 6562)  Taka 
= 34000 Taka

The average sale after 6 months is 6500 Taka 
∴ Total sale after 6 months (6500 × 6) Taka
= 39000 Taka

Sale must he have in the sixth month is (39000 - 34000) Taka 
= 5000 Taka