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Bank Math

মোট প্রশ্ন১৬,১২৪এই পাতা১০০প্রতি পাতা১০০
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Bank Math

PrepBank · পাতা ১৫২ / ১৬১ · ১৫,১০১–১৫,২০০ / ১৬,১২৪

১৫,১০১.
A room 6.2m × 8m is to be carpeted leaving a margin of 10 cm from each wall. If the cost of the carpet is Tk. 15 per sq. meter, the cost of carpeting the room will be :
  1. ক
    ক) Tk. 579.92
  2. খ
    খ) Tk. 682.50
  3. গ
    গ) Tk. 702
  4. ঘ
    ঘ) Tk. 702.60
ব্যাখ্যা

Area of the carpet :
= [(6.20 - 0.20) × (8 - 0.20)] m2 
= (6 × 7.8) m2 
= 46.8 m2 
∴ Cost of carpeting :
= Tk. (46.8 × 15)
= Tk. 702

১৫,১০২.
Liton’s age after 30 years would be equal to 5 times his age 10 years ago. Find his age 6 years hence.
  1. ক
    42 years
  2. খ
    32 years
  3. গ
    30 years
  4. ঘ
    26 years
ব্যাখ্যা
Question: Liton’s age after 30 years would be equal to 5 times his age 10 years ago. Find his age 6 years hence.

Solution: 
Let
Liton’s present age be ‘x’ years.

ATQ,
x + 30 = 5(x - 10)
⇒ x + 30 = 5x – 50
⇒ 4x = 80
⇒ x = 20

∴  Liton’s present age = 20 years
Therefore, Liton’s age 6 years hence = 20 + 6 = 26 years
১৫,১০৩.
The ratio of the cost price and selling price is 5 : 4, the loss percent is =?
  1. ক
    20%
  2. খ
    15%
  3. গ
    17.33%
  4. ঘ
    25%
  5. ঙ
    None
ব্যাখ্যা
Question: The ratio of the cost price and selling price is 5 : 4, the loss percent is =?
 
Solution: 
Let the cost price be Tk. 5x
The selling price is Tk. 4x

Hence, the loss percent = [(5x - 4x)/5x] × 100% 
= (1/5) × 100%
= 20%
১৫,১০৪.
A cylinder has a radius of 5 cm and a height of 7 cm. What is its volume? 
  1. ক
    1040 cm3
  2. খ
    1240 cm3
  3. গ
    540 cm3
  4. ঘ
    550 cm3
ব্যাখ্যা

Question: A cylinder has a radius of 5 cm and a height of 7 cm. What is its volume?

Solution: 
Radius, r = 5 cm 
Height, h = 7 cm

We know, 
Volume = πr2h
= (22/7) × (5)2 × 7
= 550 cm3

১৫,১০৫.
The perimeter of one face of a cube is 36 cm. Its volume must be-
  1. ক
    216 cm3
  2. খ
    592 cm3
  3. গ
    343 cm3
  4. ঘ
    729 cm3
ব্যাখ্যা

Question: The perimeter of one face of a cube is 36 cm. Its volume must be-

Solution:
perimeter of one face is = 36 cm

let, length of one side is = a cm
∴ perimeter = 4a cm

ATQ,
⇒ 4a = 36
⇒ a = 36/4
= 9 cm

∴ volume = a3
= 93
= 729 cm3

১৫,১০৬.
100 cubic centimeters of silver is drawn into a wire 2 mm in diameter. The length of the wire in meters will be:
  1. ক
    31.83 meters
  2. খ
    30 meters
  3. গ
    32.33 meters
  4. ঘ
    84 meters
ব্যাখ্যা
Let the length of the wire be 'h'


Radius of the wire, r = 2/2 mm
                                  = 1 mm
                                  = 1/10 cm

Therefore, πr2h = 100
or, π (1/10)2h =100
or, h = 100 × 100/3.1416
         = 3,183.09 cm
         = 31.83 meters
১৫,১০৭.
A dog takes 3 leaps for every 5 leaps of a hare. If one leap of the dog is equal to 3 leaps of the hare, the ratio of the speed of the dog to that of the hare is:
  1. ক
    ক) 9 : 5
  2. খ
    খ) 2 : 3
  3. গ
    গ) 4 : 7
  4. ঘ
    ঘ) 5 : 6
  5. ঙ
    ঙ) 5 : 9
ব্যাখ্যা

Dog : Hare = (3 × 3) leaps of hare : 5 leaps of hare
= 9 : 5.

১৫,১০৮.
The value of p for which equation x² + (p – 3)x + p = 0 has real and equal roots is-
  1. ক
    ক) 9
  2. খ
    খ) 4
  3. গ
    গ) 3
  4. ঘ
    ঘ) 0
ব্যাখ্যা

Given, x2 + (p − 3)x + p = 0
Here, a = 1,b = (p − 3),c = p
Given that the roots are equal,
So, Discriminant = 0
⇒ b2 − 4ac = 0
Discriminant = (p − 3)2 − 4(1)(p) = 0
⇒ p2 + 9 − 6p − 4p = 0
⇒ p2 − 10p + 9 = 0
⇒ p2 − 9p − p + 9 = 0
⇒ p(p − 9) − 1(p − 9) = 0
⇒ (p − 9)(p − 1) = 0
⇒ p − 9 = 0 or p − 1 = 0
Hence, p = 9 or p = 1

১৫,১০৯.
The average daily wage of 10 workers is Tk. 800. If the lowest wage is Tk. 650, then what is the possible maximum wage?
  1. ক
    1400 Tk.
  2. খ
    1750 Tk.
  3. গ
    2150 Tk.
  4. ঘ
    2600 Tk.
ব্যাখ্যা

Question: The average daily wage of 10 workers is Tk. 800. If the lowest wage is Tk. 650, then what is the possible maximum wage?

Solution: 
10 জন লোকের মোট মজুরি (10 × 800) টাকা = 8000 টাকা
650 টাকা করে 9 জনের মজুরি = (650 × 9) = 5850 টাকা

∴ Possible maximum wage = (8000 - 5850) টাকা = 2150 টাকা।

১৫,১১০.
A tiger is 50 of its own leaps behind a deer. The tiger takes 5 leaps and per minute to the deer's 4. If the tiger and the deer cover 8 m and 5 m per leap respectively, what distance will the tiger have to run before it caches the deer?
  1. ক
    ক) 600 m
  2. খ
    খ) 700 m
  3. গ
    গ) 800 m
  4. ঘ
    ঘ) 1000 m
ব্যাখ্যা

Speed of tiger = 40 m/min
Speed of deer = 20 m/min.
Relative speed = 40 - 20 = 20 m/min.
Initial difference in distance = 50 × 8 = 400 m

Time taken to catch = 400/20 = 20 min.
Distance traveled in 20 min,
= 20 × 40
= 800 m.

১৫,১১১.
A television and a washing machine were sold for Tk. 12500 each. If the television was sold at a gain of 30% and the washing machine at a loss of 30%. Find the overall profit% or loss% on the entire transaction?
  1. ক
    9% profit
  2. খ
    9% loss
  3. গ
    12% profit
  4. ঘ
    12% loss
ব্যাখ্যা
Question: A television and a washing machine were sold for Tk. 12500 each. If the television was sold at a gain of 30% and the washing machine at a loss of 30%. Find the overall profit% or loss% on the entire transaction?

Solution:
Total SP = 2 × 12500 = 25000
CP of TV = [100/(100 + 30)] × 12500 = 12500 × 100/130
CP of Washing machine = [100/ (100 - 30)] × 12500 = 12500 × 100/70
∴ Total CP = 12500 [(100/130) + (100/70)] = 12500 × 200/91 = 2500000/91 = 27472.53
Clearly SP < CP, that is, there is a loss.

Reqiured loss percentage = [{(2500000/91) - 25000}/(2500000/91)] × 100%
= {(2500000 - 2275000)/2500000} × 100%
= (225000/2500000) × 100%
= 9%
Therefore, there is loss of 9%
১৫,১১২.
A pipe can fill a cistern in 16 hours. After half the tank is filled, three more similar taps are opened. What is the total time taken to fill the cistern completely?
  1. ক
    9.5 hours.
  2. খ
    10 hours.
  3. গ
    11 hours.
  4. ঘ
    11.5 hours.
ব্যাখ্যা
Question: A pipe can fill a cistern in 16 hours. After half the tank is filled, three more similar taps are opened. What is the total time taken to fill the cistern completely?

Solution:
Time is taken to fill half of the tank = (1/2) × 16 = 8 hrs

In 1 hour 1 pipe can fill = 1/16 part
∴ In 1 hour 4 pipe can fill = {4 × (1/16)} part
= 1/4 part

4 pipe can fill 1/4 part in 1 hour
∴ 4 pipe can fill 1 part in 4 hour
∴ 4 pipe can fill 1/2 part in (4 × 1/2) hour
= 2 hours

∴ Total time = (8 + 2) = 10 hours.
১৫,১১৩.
A certain sum of money becomes three times of itself in 20 years at simple interest. In how many years does it become double of itself at the same rate of simple interest?
  1. ক
    ক) 5 years
  2. খ
    খ) 10 years
  3. গ
    গ) 12 years
  4. ঘ
    ঘ) 15 years
ব্যাখ্যা
Question: A certain sum of money becomes three times of itself in 20 years at simple interest. In how many years does it become double of itself at the same rate of simple interest?

Solution: 
Let,
the principal be P.
After 20 years, Amount = 3P
Simple interest, I = 3P - P = 2P

We know that,
SI = PRT/100
⇒ 2P = (P × R × 20)/100
⇒ R = 10%

Now,
Amount = 2P
SI = Amount - Principal
= 2P - P
= P

∴ SI = PRT/100
⇒ P = (P × 10 × T)/100
∴ T = 10 years
১৫,১১৪.
A father is five times as old as his son. After eight years, the father will be three times as old as his son will be then. What is the father's present age?
  1. ক
    40
  2. খ
    64
  3. গ
    50
  4. ঘ
    65
ব্যাখ্যা

Question: A father is five times as old as his son. After eight years, the father will be three times as old as his son will be then. What is the father's present age?

Solution:
Let the son's present age be x years.
Then, the father's present age = 5x years.

After 8 years, 
Son’s age = x + 8
Father’s age = 5x + 8

According to the question,
5x + 8 = 3(x + 8)
⇒ 5x + 8 = 3x + 24
⇒ 5x - 3x = 24 - 8
⇒ 2x = 16
∴ x = 8

So, father’s present age = 5 × 8 = 40 years.

১৫,১১৫.
The mean of the first 15 even natural numbers is-
  1. ক
    12
  2. খ
    14
  3. গ
    16
  4. ঘ
    17
ব্যাখ্যা
Question: The mean of the first 15 even natural numbers is-

Solution:
First 15 even natural numbers = 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30.

Mean = (2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20 + 22 + 24 + 26 + 28 + 30)/15
= 240/15
= 16
১৫,১১৬.
The average of 12 numbers is 16 and the average of the first five is 14. What is the average for the rest?
  1. ক
    17.5
  2. খ
    17.43
  3. গ
    18.26
  4. ঘ
    16.54
ব্যাখ্যা
Question: The average of 12 numbers is 16 and the average of the first five is 14. What is the average for the rest?

Solution: 
Average of twelve numbers = 16
Sum of twelve numbers = 16 × 12 = 192
Average of first five numbers = 14
Sum of first five numbers = 14 × 5 = 70

Total of remaining seven numbers = 192 - 70 = 122
Average of rest = 122/7 = 17.43
১৫,১১৭.
Eight printers working together can print 960 documents in 6 days. If 3 printers stop working, how many documents can the remaining printers print in 9 days?
  1. ক
    900 documents
  2. খ
    1200 documents
  3. গ
    1000 documents
  4. ঘ
    1500 documents
ব্যাখ্যা

Question: Eight printers working together can print 960 documents in 6 days. If 3 printers stop working, how many documents can the remaining printers print in 9 days?

Solution:
8টি প্রিন্টার 6 দিনে প্রিন্ট করে = 960টি ডকুমেন্ট
∴ 1টি প্রিন্টার 6 দিনে প্রিন্ট করে = 960/8 = 120টি ডকুমেন্ট
∴ 1টি প্রিন্টার 1 দিনে প্রিন্ট করে = 120/6 = 20টি ডকুমেন্ট

অবশিষ্ট প্রিন্টার = 8 - 3 = 5টি

∴ 5টি প্রিন্টার 9 দিনে প্রিন্ট করবে = 20 × 5 × 9 = 900টি ডকুমেন্ট

১৫,১১৮.
The percentage profit earned by selling an article for Tk. 1920 is equal to the percentage loss by selling the same article for Tk. 1280. At what price should the article be sold to make 25% profit?
  1. ক
    Tk. 1800
  2. খ
    Tk. 2000
  3. গ
    Tk. 2250
  4. ঘ
    Tk. 2500
ব্যাখ্যা
Question: The percentage profit earned by selling an article for Tk. 1920 is equal to the percentage loss by selling the same article for Tk. 1280. At what price should the article be sold to make 25% profit?

Solution:
Let the cost price of the article be C.
From the given problem, the percentage profit made by selling the article for Tk. 1920 is equal to the percentage loss incurred by selling the article for Tk. 1280. This can be expressed as:
{(1920 - C)/C​} × 100 = {(C - 1280)/C} ​×100
⇒ 1920 - C = C - 1280
⇒ 1920 + 1280 = 2C
⇒ 3200 = 2C
∴ C = 1600

Now, to make a 25% profit, the selling price should be:
Selling Price = C + (25/100) × C = 1600 + (25/100) ×1600 = 1600 + 400 = 2000
১৫,১১৯.
A sum of money at simple interest amounts to Tk. 1665 in 3 years and to Tk. 1782 in 4 years. The sum is :
  1. ক
    1414
  2. খ
    1352
  3. গ
    1374
  4. ঘ
    1314
ব্যাখ্যা
Question: A sum of money at simple interest amounts to Tk. 1665 in 3 years and to Tk. 1782 in 4 years. The sum is :

Solution: 
interest in one year = (1782 - 1665 ) = 117
in three years = ( 117 × 3 ) = 351

∴ the sum of the money = ( 1665 - 351 )
= 1314
১৫,১২০.
Three coins are tossed. Find the probability of exactly 2 heads -
  1. ক
    (3/8)
  2. খ
    (1/2)
  3. গ
    (1/8)
  4. ঘ
    None
ব্যাখ্যা

Question: Three coins are tossed. Find the probability of exactly 2 heads -

Solution: 
n(S) = 23 = 8 
Let E = Event of getting exactly two heads,
= {(H, H, T), (H, T, H), (T, H, H)}
= n(E)
= 3

Required probability = 3/8

১৫,১২১.
A boatman rows 2 km in 6 minutes along the stream, and 7 km in 1 hour against the stream. What is the speed of the stream?
  1. ক
    ক) 5 kmph 
  2. খ
    খ) 6.5 kmph 
  3. গ
    গ) 7 kmph 
  4. ঘ
    ঘ) 8.5 kmph 
ব্যাখ্যা
Question: A boatman rows 2 km in 6 minutes along the stream, and 7 km in 1 hour against the stream. What is the speed of the stream?

Solution: 
Rate of downstream = (2 × 60)/6 kmph = 20 kmph
Rate of upstream = 7 kmph

speed of boat + speed of stream = 20 kmph
speed of boat - speed of stream = 7 kmph
--------------------------------------------------
(-)  2 × speed of stream = 13
⇒ speed of stream = 13/2 = 6.5 kmph
১৫,১২২.
41, 43, 47, 53, 61, 71, 73, 81
  1. ক
    ক) 61
  2. খ
    খ) 71
  3. গ
    গ) 73
  4. ঘ
    ঘ) 81
ব্যাখ্যা
Each of the numbers except 81 is a prime number.
১৫,১২৩.
A cyclist travels the length of a bike path that is 225 miles long, rounded to the nearest mile. If the trip took him 5 hrs, rounded to the nearest hour, then his average speed must be between:
  1. ক
    38 and 50 mph
  2. খ
    40 and 50 mph
  3. গ
    40 and 51 mph
  4. ঘ
    41 and 50 mph
  5. ঙ
    41 and 51 mph
ব্যাখ্যা
Question: A cyclist travels the length of a bike path that is 225 miles long, rounded to the nearest mile. If the trip took him 5 hrs, rounded to the nearest hour, then his average speed must be between:

Solution:
Let,
Range of his average speed R

Length is between 224.5 and 225.4 miles.
Trip is between 4 : 30 and 5 : 29 long

We want to maximize and minimize results.
225.4/4.5 = 50.1
224.5/5.5 = 40.8

40.8 ≤ R ≤ 50.1

So the range is greater than or equal to 40.8 and less than or equal to 50.1.
The only answer choice that allows for all possible values of R is C.
১৫,১২৪.
If m is an even integer and n is an integer (either odd or even), then which of the following will always be even?
i) m2 + n2 + n        ii) (m - n)(n + 1)           iii) m2 - n2 + 1
  1. ক
    Only I
  2. খ
    Only II
  3. গ
    Only III
  4. ঘ
    Both I and II
ব্যাখ্যা

Question: If m is an even integer and n is an integer (either odd or even), then which of the following will always be even?
i) m2 + n2 + n        ii) (m - n)(n + 1)           iii) m2 - n2 + 1

Solution:
এখানে
m একটি জোড় সংখ্যা তাই m = 4 ধরি,
n জোড় সংখ্যা ও হতে পারে আবার বিজোড় সংখ্যাও হতে পারে। 
n জোড় হলে n = 2 এবং n বিজোড় হলে n = 3 ধরি।

i)
m এবং n উভয়ে জোড় হলে
m2 + n2 + n = 42 + 22 + 2 = 16 + 4 + 2 = 22 যা জোড় সংখ্যা

m জোড় এবং n বিজোড় হলে
m2 + n2 + n = 42 + 32 + 3 =  16 + 9 + 3 = 28 যা জোড় সংখ্যা

∴ i) সর্বদা জোড় 

ii)
m এবং n উভয়ে জোড় হলে
(m - n)(n + 1) = (4 - 2)(2 + 1) = 2 × 3 = 6 যা জোড় সংখ্যা

m জোড় এবং n বিজোড় হলে
(m - n)(n + 1) = (4 - 3)(3 + 1) = 1 × 4 = 4 যা জোড় সংখ্যা

∴ ii) সর্বদা জোড় 

iii)
m এবং n উভয়ে জোড় হলে
m2 - n2 + 1 = 42 - 22 + 1 = 16 - 4 + 1 = 13 যা বিজোড় সংখ্যা

m জোড় এবং n বিজোড় হলে
m2 - n2 + 1 = 42 - 32 + 1 = 16 - 9 + 1 = 8 যা জোড় সংখ্যা

∴ iii) সর্বদা জোড় নয়। 

১৫,১২৫.
How many terms of the arithmetic progression 2, 7, 12,... should be taken so that their sum equals 354?
  1. ক
    10
  2. খ
    11
  3. গ
    12
  4. ঘ
    13
ব্যাখ্যা

Question: How many terms of the arithmetic progression 2, 7, 12,... should be taken so that their sum equals 354?

solution:
Given arithmetic progression: 2, 7, 12, …
First term, a = 2
Common difference, d = 5

We know, 
Sum of first n terms, Sn = (n/2) × [2a + (n - 1)d]

ATQ,
(n/2) × [2 × 2 + (n - 1) × 5] = 354
⇒ (n/2) × [4 + 5n - 5] = 354
⇒ (n/2) × (5n - 1) = 354
⇒ n(5n - 1) = 708
⇒ 5n2 - n - 708 = 0
⇒ 5n2 - 60n + 59n - 708 = 0
⇒ 5n(n - 12) + 59(n - 12) = 0
⇒ (n - 12)(5n + 59) = 0
Now, n - 12 = 0
∴ n = 12
Or
5n + 59 = 0
∴ n = - 59/5 ; [not possible, n must be positive]

∴ 12 terms of the arithmetic progression must be taken to result in a sum of 354. 

১৫,১২৬.
Current age of Zamil's father is three times of Zamil. After 16 years from now father's age will be twice of Zamil. What is the current age of Zamil?
  1. ক
    ক) 16
  2. খ
    খ) 32
  3. গ
    গ) 48
  4. ঘ
    ঘ) 64
ব্যাখ্যা
ধরি, জামিলের বয়স x বছর।
তাহলে জামিলের বাবার বয়স 3x বছর৷
প্রশ্নমতে, 2(x+16) = 3x+16
বা, 2x+32 = 3x+16
বা, x = 16
১৫,১২৭.
What is the measure of each interior angle in a regular hexagon?
  1. ক
    80°
  2. খ
    100°
  3. গ
    105°
  4. ঘ
    120°
ব্যাখ্যা
Question: What is the measure of each interior angle in a regular hexagon?

Solution: 
সুষম বহুভুজের বাহুর সংখ্যা n হলে তার কোণগুলোর সমষ্টি (2n - 4) সমকোণ।
সুতরাং সুষম ষড়ভুজের ছয় কোণের সমষ্টি = (2 × 6 - 4) সমকোণ
= (12 - 4) × 90°
= 8 × 90°
= 720°
সুতরাং সুষম ষড়ভুজের একটি শীর্ষ কোণ = 720°/6
= 120°
১৫,১২৮.
How long does a train 1000 m long moving at a speed of 90 km/hr would take to pass through a 500 m long bridge?
  1. ক
    45 sec
  2. খ
    1 minute
  3. গ
    1 minute 15 sec
  4. ঘ
    2 minute
ব্যাখ্যা

Question: How long does a train 1000 m long moving at a speed of 90 km/hr would take to pass through a 500 m long bridge?

Solution:
Here, the time taken by the train to pass the bridge completely would be the time it takes to cover 1000 + 500 = 1500 m at the speed of 90 km/hr
= 90 × (5/18)
= 25 m/sec

Therefore, time required = 1500/25
= 60 sec
= 1 minute

১৫,১২৯.
Tamim has a certain average for 12 innings. In the 13th inning, he scores 120 runs, increasing his average by 5 runs. What is his new average?
  1. ক
    42 runs
  2. খ
    55 runs
  3. গ
    60 runs
  4. ঘ
    50 runs
ব্যাখ্যা
Question: Tamim has a certain average for 12 innings. In the 13th inning, he scores 120 runs, increasing his average by 5 runs. What is his new average?

Solution:
Let Tamim's average be x for 12 innings.
So, Tamim scored 12x run in 12 innings.
In the 13th inning, he scored 120 runs then the average became (x + 5) .
And he scored (x + 5) × 13 runs in 13 innings.
Now,
⇒ 12x + 120 = 13(x + 5)
⇒ 12x + 120 = 13x + 65
⇒ x = 120 - 65
∴ x = 55

∴ New average = (x + 5) = 55 + 5 = 60 runs
১৫,১৩০.
200 ÷ 25 × 4 + 12 - 3 = ?
  1. ক
    ক) 40
  2. খ
    খ) 41
  3. গ
    গ) 42
  4. ঘ
    ঘ) 43
ব্যাখ্যা

200 ÷ 25 × 4 + 12 - 3
= 200/ 25 × 4 + 12 - 3
= 8 × 4 + 12 - 3
= 32 + 12 - 3
= 44 - 3
= 41

১৫,১৩১.
Find the multiple of 11 in the following numbers.
  1. ক
    ক) 112144
  2. খ
    খ) 978626
  3. গ
    গ) 869756
  4. ঘ
    ঘ) 447355
ব্যাখ্যা
Question: Find the multiple of 11 in the following numbers.

Solution:
যদি কোনো সংখ্যার জোড় স্থানের অঙ্ক ও বিজোড় স্থানের অঙ্কের যোগফলের পার্থক্য 0 হয় অথবা 11 দ্বারা বিভাজ্য হয় তবে ঐ সংখ্যাটি 11 দ্বারা বিভাজ্য হবে।
এখানে,
a. (1 + 2 + 4) - (1 + 1 + 4) = 1
b. (9 + 8 + 2) - (7 + 6 + 6) = 0
c. (8 + 9 + 5) - (6 + 7 + 6) = 3
d. (4 + 7 + 5) - (4 + 3 + 5) = 4

∴ 978626 সংখ্যাটি 11 দ্বারা বিভাজ্য।
১৫,১৩২.
What is the value of tan40° tan50° tan60°?
  1. ক
    ক) 1
  2. খ
    খ) 1/√2
  3. গ
    গ) √3
  4. ঘ
    ঘ) - √3
ব্যাখ্যা
tan40° tan50° tan60°
= tan(90°− 50°) tan50° tan60°
=cot50° tan50° tan60°
= (1/ tan50°)tan50° tan60°
= tan60°
= √3
১৫,১৩৩.
Quantity A = Time to travel 95 miles at 50 miles per hour; and Quantity B = Time to travel 125 miles at 60 miles per hour.
  1. ক
    Quantity A is greater
  2. খ
    Quantity A equals Quantity B
  3. গ
    Quantity B is greater
  4. ঘ
    Relationship indeterminate
ব্যাখ্যা
Question: Quantity A = Time to travel 95 miles at 50 miles per hour; and Quantity B = Time to travel 125 miles at 60 miles per hour.

Solution: 
Quantity A = 95/50 = 19/10 = 1.9 hour 

Quantity B = 125/60 = 25/12 = 2.083 hour 

∴ Quantity B is greater.
১৫,১৩৪.
15 workers can build 45 chairs working 6 hours per day. How many extra workers are required to produce 60 chairs working 4 hours per day?
  1. ক
    0 workers
  2. খ
    10 workers
  3. গ
    12 workers
  4. ঘ
    15 workers
ব্যাখ্যা

Question: 15 workers can build 45 chairs working 6 hours per day. How many extra workers are required to produce 60 chairs working 4 hours per day?

Solution:
6 ঘণ্টা কাজ করে 45টি চেয়ার তৈরি করে 15 জন শ্রমিক 
1 ঘণ্টা কাজ করে 1টি চেয়ার তৈরি করে = (15 × 6) /45 জন শ্রমিক 
4 ঘণ্টা কাজ করে 60টি চেয়ার তৈরি করে = (15 × 6 × 60)/(45 × 4) জন শ্রমিক 
= 30 জন

∴ অতিরিক্ত শ্রমিকের সংখ্যা = 30 - 15 = 15 জন

১৫,১৩৫.
The average age of A, B, C, D and E is 40 years. The average age of A and B is 35 years and the average of C and D is 42 years. Age of E is :
  1. ক
    ক) 40 years
  2. খ
    খ) 42 years
  3. গ
    গ) 44years
  4. ঘ
    ঘ) 46 years
ব্যাখ্যা
The average age of A, B, C, D and E is 40 years.
The total age of A, B, C, D and E is 40 × 5 = 200 years.

The average age of A and B is 35 years
The total age of A and B is 35 × 2 = 70 years

The average of C and D is 42 years
The total age of C and D is 42 × 2 = 84 years

Age of E is 200 - 70 - 84 = 46 years
১৫,১৩৬.
A garrison has food for 40 days for 200 soldiers. After 10 days, 50 soldiers leave. How many more days will the remaining food last?
  1. ক
    20 days
  2. খ
    30 days
  3. গ
    35 days
  4. ঘ
    40 days
ব্যাখ্যা

Question: A garrison has food for 40 days for 200 soldiers. After 10 days, 50 soldiers leave. How many more days will the remaining food last?

Solution:
দেওয়া আছে,
40 দিনের খাদ্য আছে = 200 জনের

10 দিন পর 50 জন সৈন্য চলে গেলে,
 বাকি দিন = 40 - 10 = 30 দিন
বাকি সৈন্য = 200 - 50 = 150 জন

বাকি খাদ্যে,
200 জন সৈন্যের চলবে = 30 দিন
∴ 1 জন সৈন্যের চলবে =30 × 200 দিন 
∴ 150 জন সৈন্যের চলবে = (30 × 200)/150 = 40 দিন 

সুতরাং বাকি খাদ্যে 40 দিন চলবে।

১৫,১৩৭.
Rohit is 28th from the left end of a row of 50 students and Shyam is 28th from the right end of the row. How many students are sitting between them?
  1. ক
    ক) 4
  2. খ
    খ) 5
  3. গ
    গ) 2
  4. ঘ
    ঘ) 3
  5. ঙ
    ঙ) None of these
ব্যাখ্যা
রোহিত যদি বা দিক থেকে ২৮ হয় তবে ডান দিক থেকে তার অবস্থান ২৩ এবং শ্যাম যদি ডান দিক থেকে ২৮ হয় তবে বা দিক থেকে তার অবস্থান ২৩। তাদের দুজনের মাঝে ছাত্রের অবস্থান হবে {৫০ - (২৩ + ২৩)} = ৪ জন।
১৫,১৩৮.
A basket contains 6 blue, 2 red, 4 green and 3 yellow balls. If four balls are picked up at random, what is the probability that 2 are red and 2 are green?
  1. ক
    5/197
  2. খ
    3/293
  3. গ
    2/455
  4. ঘ
    4/185
  5. ঙ
    None
ব্যাখ্যা
Question: A basket contains 6 blue, 2 red, 4 green and 3 yellow balls. If four balls are picked up at random, what is the probability that 2 are red and 2 are green?

Solution:
Total number of balls = 6 + 2 + 4 + 3 = 15.
Let E be the event of drawing 4 balls such that 2 are red and 2 are green.
Then, n(E) = (2C2 × 4C2) = 6
And n(S) = 15C4 = 365

∴ P(E) = n(E)/n(S)
= 6/1365
= 2/455
১৫,১৩৯.
If log5(x2 + x) - log5(x + 1) = 2 what is the value of x?
  1. ক
    ক) 9
  2. খ
    খ) 3
  3. গ
    গ) 5
  4. ঘ
    ঘ) 25
ব্যাখ্যা
Question: If log5(x2 + x) - log5(x + 1) = 2 what is the value of x?

Solution
log5(x2 + x) - log5(x + 1) = 2
⇒ log5{(x2 + x)/(x + 1)} = 2
⇒ log5{x(x + 1)/(x + 1)} = 2
⇒ log5x = 2
⇒ x = 52
   x = 25
১৫,১৪০.
The area of a square inscribed in a circle is 140 cm2. What is the area of the semi-circle?
  1. ক
    ক) 220 cm2
  2. খ
    খ) 200 cm2
  3. গ
    গ) 150 cm2
  4. ঘ
    ঘ) 110 cm2
ব্যাখ্যা
Question: The area of a square inscribed in a circle is 140 cm2. What is the area of the semi-circle?

Solution:
The area of a square inscribed in a circle is 140 cm2
side of square = √140 cm = 2√35 cm
diagonal of the square = √2 × 2√35
= 2√70 cm

diameter of circle = 2√70 cm
radius of the circle = √70 cm
∴ area of the circle = π (√70)2 cm2
= (22/7) × 70 cm2
= 220 cm2

area of semi-circle = 220/2 
= 110 cm2
১৫,১৪১.
Two-ninth of half of a number is 20. Find 40% of that number.
  1. ক
    60
  2. খ
    90
  3. গ
    180
  4. ঘ
    72
ব্যাখ্যা

Question: Two-ninth of half of a number is 20. Find 40% of that number.

Solution: 
Let the number be x.

Given that, 
Two-ninth of half of the number is 20.
⇒ (2/9) × (1/2) × x = 20
⇒ (1/9) × x = 20
⇒ x = 20 × 9
∴ x = 180

Now,
Find 40% of that number = 40% of 180
= (40/100) × 180
= 72

So 40% of that number is 72.

১৫,১৪২.
In a box, there are 4 red, 6 white, and 2 black balls. If two balls are drawn one after the other without replacement, what is the probability that the first ball is red and the second ball is white?
  1. ক
    2/11
  2. খ
    3/7
  3. গ
    5/13
  4. ঘ
    2/13
  5. ঙ
    None of these
ব্যাখ্যা

Question: In a box, there are 4 red, 6 white, and 2 black balls. If two balls are drawn one after the other without replacement, what is the probability that the first ball is red and the second ball is white?

Solution:
মোট বলের সংখ্যা = 4 (লাল) + 6 (সাদা) + 2 (কালো) = 12টি।
∴ প্রথম বলটি লাল হওয়ার সম্ভাবনা = 4/12 = 1/3

যেহেতু বলটি প্রতিস্থাপন করা হয়নি (Without replacement),
তাই প্রথম বলটি তোলার পর বক্সে মোট বলের সংখ্যা = 12 - 1 = 11টি।
∴ দ্বিতীয় বলটি সাদা হওয়ার সম্ভাবনা = 6/11

∴ প্রথমটি লাল এবং দ্বিতীয়টি সাদা হওয়ার সম্ভাবনা = (প্রথমটি লাল হওয়ার সম্ভাবনা) × (দ্বিতীয়টি সাদা হওয়ার সম্ভাবনা)
= (1/3) × (6/11)
= 6/33
= 2/11

∴ নির্ণেয় সম্ভাবনা 2/11

১৫,১৪৩.
Out of 9 persons, 8 persons spent Tk. 30 each for their meals. The ninth one spent Tk. 20 more than the average expenditure of all the nine. The total money spent by all of them was-
  1. ক
    Tk. 260
  2. খ
    Tk. 290
  3. গ
    Tk. 292.50
  4. ঘ
    Tk. 400.50
ব্যাখ্যা
Question: Out of 9 persons, 8 persons spent Tk. 30 each for their meals. The ninth one spent Tk. 20 more than the average expenditure of all the nine. The total money spent by all of them was-

Solution:
Let the average expenditure be Tk. x
Then,
9x = 8 × 30 + (x + 20)
⇒ 9x = x + 260
⇒ 8x = 260
⇒ x = 32.50

∴ Total money spent
= Tk. 9x
= Tk. (9 × 32.50)
= Tk. 292.50
১৫,১৪৪.
If , then the value of
  1. ক
    1/3
  2. খ
    1/2
  3. গ
    2
  4. ঘ
    3
ব্যাখ্যা
Question: If , then the value of

Solution:
Given, x + 1/x =1
⇒ (x2 + 1)/x = 1
 ⇒ x2 - x = - 1

Now, 
3/(x2 - x + 7)
= 3/( - 1 + 7) [x2 - x = - 1]
= 3/6
= 1/2
১৫,১৪৫.
What is the H.C.F. of 4/9, 10/21 and 20/63?
  1. ক
    4/189
  2. খ
    6/63
  3. গ
    2/63
  4. ঘ
    20/21
ব্যাখ্যা
Question: What is the H.C.F. of 4/9, 10/21 and 20/63?

Solution:
H.C.F of 4/9, 10/21 and 20/63 = H.C.F of 4,10 and 20 / L.C.M of 9,21 and 63

H.C.F of 4, 10 and 20 = 2
& L.C.M. of 9, 21 and 63 = 63.

∴ Required H.C.F. = 2/63
১৫,১৪৬.
The base of a right-angled triangle is 24 and hypotenuse is 26. What is its area?
  1. ক
    220 sq. meters
  2. খ
    80 sq. meters
  3. গ
    140 sq. meters
  4. ঘ
    120 sq. meters
  5. ঙ
    90 sq. meters
ব্যাখ্যা
Question: The base of a right-angled triangle is 24 and hypotenuse is 26. What is its area?

Solution:
The area of a right angled triangle = (1/2) × base × height

Given that,
Base = 24 and Hypotenuse = 26
Height2 = Hypotenuse2 - Base2
= 262 - 242
= 676 - 576
Height2 = 100
∴ Height = 10

Area = (1/2) × base × height
= (1/2) × 24 × 10
= 120 sq. meters
১৫,১৪৭.
Quadratic equation corresponding to the roots 2 + √5 and 2 - √5 is-
  1. ক
    x2 - 4x + 1 = 0
  2. খ
    x2 - 4x - 1 = 0
  3. গ
    x2 + 4x + 1 = 0
  4. ঘ
    x2 + 4x - 1 = 0
ব্যাখ্যা
Question: Quadratic equation corresponding to the roots 2 + √5 and 2 - √5 is-

Solution:
The quadratic equation is: x2 - (Sum of roots)x + Product of roots = 0

Let the roots of the equation be A and B.
A = 2 + √5 and B = 2 - √5

∴ A + B = 2 + √5 + 2 - √5 = 4
∴ A × B = (2 + √5)(2 - √5) = 4 - 5 = - 1

Then equation is x2 - 4x - 1 = 0
১৫,১৪৮.
If a man were to sell his table for Tk 800, he would lose 20%. To gain 25% he should sell it for -
  1. ক
    ক) 1250 Tk
  2. খ
    খ) 1000 Tk
  3. গ
    গ) 1200 Tk
  4. ঘ
    ঘ) 950 Tk
ব্যাখ্যা
Question: If a man were to sell his table for Tk 800, he would lose 20%. To gain 25% he should sell it for -

Solution:
Let, the cost price is 100 Tk
At 20% loss, Selling price = 100 - 20 = 80 Tk

Original cost price = (100 × 800) / 80 = 1000 Tk

to gain 25%, selling price be = 1000 × 125% = 1250 Tk
১৫,১৪৯.
A number when multiplied by 16 increases by 540. What is the number?
  1. ক
    30
  2. খ
    36
  3. গ
    42
  4. ঘ
    46
ব্যাখ্যা
Question: A number when multiplied by 16 increases by 540. What is the number?

Solution:
Let the number is X.

As per question;
16X - X = 540
⇒15X = 540
∴ X = 36
১৫,১৫০.
An article costs TK 500 and the marked price is mentioned as Tk. 800. What is the profit % for the seller if he sells and offers a discount of 10% on the marked price?
  1. ক
    ক) 30%
  2. খ
    খ) 44%
  3. গ
    গ) 56%
  4. ঘ
    ঘ) 64%
ব্যাখ্যা

10% of 800 = 800 × 10/100 = 80
So, the selling price is = 800 - 80 = 720
Profit = 720 - 500 = 220
So, profit percentage = (220 × 100)/500 = 44%

১৫,১৫১.
5 mat-weavers can wave 5 mats in 5 days. At the same rate, how many mats would be woven by 10 mat-weavers in 10 days? 
  1. ক
    15 mats
  2. খ
    20 mats
  3. গ
    25 mats
  4. ঘ
    30 mats
ব্যাখ্যা
Question: 5 mat-weavers can wave 5 mats in 5 days. At the same rate, how many mats would be woven by 10 mat-weavers in 10 days? 

Solution: 
5 mat weavers in 5 days wave = 5 mats
∴ 1 mat weavers in 1 day wave = 5/(5 × 5) mats
∴ 10 mat weavers in 10 days wave = (5 × 10 × 10)/(5 × 5)  = 20 mats
১৫,১৫২.
The LCM of two numbers is 2852 and their HCF is 4. If one of the number is 124, find the other number.
  1. ক
    92
  2. খ
    96
  3. গ
    104
  4. ঘ
    84
ব্যাখ্যা
Question: The LCM of two numbers is 2852 and their HCF is 4. If one of the number is 124, find the other number.

Solution:
1st number × 2nd number = L.C. M. × H.C.F

We have,
First number × second number = LCM × HCF
∴ Second number = (2852 × 4)/124
= 92
১৫,১৫৩.
The price of a jewellery, passing through three hands. rises on the whole by 65%. If the first and the second sellers earned 20% and 25% profit respectively, the profit earned by the third seller is-
  1. ক
    20%
  2. খ
    15%
  3. গ
    10%
  4. ঘ
    5%
  5. ঙ
    None of above
ব্যাখ্যা

Let,
Cost price = 100
Selling price of third seller = 100 + 65 = 135
Selling price of 1st seller = 100 × 120% = 120
Selling price of 2nd seller = 120 × 125% = 150
So profit of third seller = 165 - 150 = 15
So profit percent = (15/150) × 100 = 10%

১৫,১৫৪.
If tan45° = 2x, what is the value of x2?
  1. ক
    1/2
  2. খ
    1/4
  3. গ
    1/6
  4. ঘ
    1/8
ব্যাখ্যা
Question: If tan45° = 2x, what is the value of x2? 

Solution: 
tan45 = 2x
⇒ 1 = 2x
⇒ x = 1/2
⇒ x2 = 1/4
১৫,১৫৫.
A man starts climbing a 15m high wall at 8 am. In each minute, he climbs up 3m but slips down 1m. At what time will he climb the wall?
  1. ক
    8 : 05 am
  2. খ
    8 : 07 am
  3. গ
    8 : 09 am
  4. ঘ
    8 : 10 am
ব্যাখ্যা

Question: A man starts climbing a 15m high wall at 8 am. In each minute, he climbs up 3m but slips down 1m. At what time will he climb the wall?

Solution:
দেওয়া আছে, দেয়ালের মোট উচ্চতা = 15 মিটার।

লোকটি প্রতি মিনিটে উপরে ওঠে = 3 মিটার।
লোকটি প্রতি মিনিটে নিচে নেমে যায় = 1 মিটার।

∴ প্রতি মিনিটে তার নিট বা প্রকৃত আরোহণ = 3 মিটার - 1 মিটার
= 2 মিটার।

এখানে, শেষ মিনিটে লোকটি উপরে উঠে যাবে এবং আর পিছলে নামবে না।
তাই, শেষ 3 মিটার বাদ দিয়ে হিসাব করতে হবে।

যে উচ্চতা পর্যন্ত তাকে পিছলে নামতে হবে = 15 - 3 = 12 মিটার।

2 মিটার উঠতে সময় লাগে = 1 মিনিট।
∴12 মিটার উঠতে সময় লাগে = 12/2 মিনিট
= 6 মিনিট।

সর্বশেষ 3 মিটার সে পরের মিনিটে উঠে যাবে এবং দেয়ালের শীর্ষে পৌঁছাবে।
সুতরাং, মোট সময় লাগবে = 6 মিনিট + 1 মিনিট = 7 মিনিট।

যেহেতু লোকটি সকাল 8টায় আরোহণ শুরু করেছিল, তাই সে সকাল 8টা 7 মিনিটে দেয়ালে পৌঁছাবে।

১৫,১৫৬.
A grocer buys some eggs at Tk. 3 each. He finds that 12 of them are broken, but he sells the others at Tk. 4 each and makes a profit of Tk. 96. How many eggs did he buy?
  1. ক
    ক) 140
  2. খ
    খ) 142
  3. গ
    গ) 144
  4. ঘ
    ঘ) 150
ব্যাখ্যা

Let, The grocer buys = x eggs
ATQ,
4(x - 12) - 3x = 96
Or, 4x - 48 - 3x = 96
∴ x = 96 + 48 = 144

১৫,১৫৭.
The H.C.F of two numbers is 24. The number which can be their L.C.M is-
  1. ক
    ক) 84
  2. খ
    খ) 128
  3. গ
    গ) 148
  4. ঘ
    ঘ) 120
ব্যাখ্যা

LCM must be divisible by HCF
Here, only 120 is divisible by 24

১৫,১৫৮.
A box having height 2m, length 10m and width 7m have the top lid open. What is the surface area of the box?
  1. ক
    138
  2. খ
    118
  3. গ
    104
  4. ঘ
    208
ব্যাখ্যা
Question: A box having height 2m, length 10m and width 7m have the top lid open. What is the surface area of the box?

Solution:
বাক্সের দৈর্ঘ্য a = 10 m
বাক্সের প্রস্থ b = 7 m
বাক্সের উচ্চতা c = 2 m

উপরের অংশ বাদে পৃষ্ঠের ক্ষেত্রফল = 2(ab + bc + ca) - ab
= 2(10 × 7 + 7 × 2 + 2 × 10) - 10 × 7
= 2(70 + 14 + 20) - 70
= 2 × 104 - 70
= 208 - 70
= 138
১৫,১৫৯.
A factory produces 500 bottles of soda in 2 hours. How many bottles will it produce in 6 hours, working at the same rate?
  1. ক
    1200 bottles
  2. খ
    1350 bottles
  3. গ
    1500 bottles
  4. ঘ
    1680 bottles
ব্যাখ্যা
Question: A factory produces 500 bottles of soda in 2 hours. How many bottles will it produce in 6 hours, working at the same rate?

Solution:
Production rate = Number of items produced / Time
Production rate = 500 bottles / 2 hours = 250 bottles per hour

Bottles produced in 6 hours = Production rate × Time
Bottles produced in 6 hours = 250 bottles/hour × 6 hours = 1500 bottles
১৫,১৬০.
What is the length of the diagonal of a square whose side is √18 cm?
  1. ক
    √6​ cm
  2. খ
    72 cm
  3. গ
    6​ cm
  4. ঘ
    6​√2 cm
  5. ঙ
    None of these
ব্যাখ্যা
Question: What is the length of the diagonal of a square whose side is √18 cm?

Solution:
We know that,
Diagonal of a square = √2 × each side
= √2 × √18
= √36
= 6

So, the length of the diagonal is 6​ cm.
১৫,১৬১.
C scored 30% marks and failed by 15 marks. D scored 45% marks and obtained 30 marks more than the pass marks. What is the pass percentage?
  1. ক
    30%
  2. খ
    33%
  3. গ
    35%
  4. ঘ
    39%
ব্যাখ্যা

Question: C scored 30% marks and failed by 15 marks. D scored 45% marks and obtained 30 marks more than the pass marks. What is the pass percentage?

Solution:
Let the total marks be x.

Given,
C scored 30% marks and failed by 15 marks:
0.30x + 15 = Pass marks
D scored 45% marks and obtained 30 marks more than pass marks:
0.45x - 30 = Pass marks

Now,
0.30x + 15 = 0.45x - 30
⇒ 0.45x - 0.30x = 15 + 30
⇒ 0.15x = 45
⇒ x = 45/0.15
∴ x = 300

Pass marks = 0.30 × 300 + 15
= 90 + 15 = 105

∴ Pass percentage = (105/300) × 100% = 35%

১৫,১৬২.
A man completes 5/8 of a job in 10 days. At this rate, how many more days will it takes him to finish the job?
  1. ক
    4
  2. খ
    5
  3. গ
    6
  4. ঘ
    7
ব্যাখ্যা
Question: A man completes 5/8 of a job in 10 days. At this rate, how many more days will it takes him to finish the job?

Solution:
Work done = 5/8
Balance work = 1 - (5/8) = 3/8

Let the required number of days be x
Then,
(5/8) : (3/8) :: 10 : x
⇒ (5/8) × x = (3/8) × 10
⇒ x = (3/8) × 10 × (8/5)
∴ x = 6
১৫,১৬৩.
37.5% of 800 - 6.25% of 1600 =?
  1. ক
    200
  2. খ
    250
  3. গ
    275
  4. ঘ
    236
ব্যাখ্যা
Question: 37.5% of 800 - 6.25% of 1600 =?

Solution:
37.5% of 800 - 6.25% of 1600
= (375/1000) × 800 - (625/10000) × 1600
= 300 - 100
= 200
১৫,১৬৪.
2 - 2 + 2 - 2 + …….. 103 terms = ?
  1. ক
    -2
  2. খ
    2
  3. গ
    1
  4. ঘ
    -1
  5. ঙ
    0
ব্যাখ্যা
Clearly, the given series is such that the sum of any odd number of terms is 2 while that of any even number of terms is 0.
Thus, sum of 103 terms is 2.
১৫,১৬৫.
Which of the following fractions is equal to the decimal 0.0625?
  1. ক
    ক) 5/8
  2. খ
    খ) 3/8
  3. গ
    গ) 1/26
  4. ঘ
    ঘ) 1/18
  5. ঙ
    ঙ) 5/80
ব্যাখ্যা

5 / 80

= 0.0625

১৫,১৬৬.
A train was delayed by 30 minutes due to a technical snag. To cover the distance of 100 km on time, the driver had to increase its usual speed by 10 km/h. Find the usual speed of the train in km/h?
  1. ক
    34 km/h
  2. খ
    40 km/h
  3. গ
    54 km/h
  4. ঘ
    60 km/h
ব্যাখ্যা

Question: A train was delayed by 30 minutes due to a technical snag. To cover the distance of 100 km on time, the driver had to increase its usual speed by 10 km/h. Find the usual speed of the train in km/h?

সমাধান:
ধরি, ট্রেনটির স্বাভাবিক গতিবেগ ছিল x কিমি/ঘন্টা।
তাহলে, বর্ধিত গতিবেগ হবে (x + 10) কিমি/ঘন্টা।

স্বাভাবিক গতিতে 100 কিমি যেতে সময় লাগে = 100/x ঘন্টা।
বর্ধিত গতিতে 100 কিমি যেতে সময় লাগে = 100/(x + 10) ঘন্টা।

সময়ের পার্থক্য 30 মিনিট বা 30/60 = 1/2 ঘন্টা

প্রশ্নমতে, 
(100/x) - {100/(x + 10)} = 1/2
⇒ {100(x + 10) - 100x}/x(x + 10) = 1/2
⇒ 1000/x(x + 10) = 1/2
⇒ (x2 + 10x) = 2000
⇒ x2 + 10x - 2000 = 0
⇒ x2 + 50x - 40x - 2000 = 0
⇒ x(x + 50) - 40(x + 50) = 0
⇒ (x - 40)(x + 50) = 0

যেহেতু গতিবেগ ঋণাত্মক হতে পারে না, তাই x = 40
∴ ট্রেনটির স্বাভাবিক গতিবেগ হলো 40 কিমি/ঘন্টা।

১৫,১৬৭.
The price of a product is reduced by 25%, but the daily sale of the article is increased by 30%. Find the net effect on the daily sale.
  1. ক
    2.5% increase
  2. খ
    1.5% decrease
  3. গ
    2% increase
  4. ঘ
    2.5% decrease
ব্যাখ্যা
Question: The price of a product is reduced by 25%, but the daily sale of the article is increased by 30%. Find the net effect on the daily sale.

Solution:
We can say that,
Total Sale = the price of one unit × no. of units sold

Let
the initial price of product = 100, sale =100 items.
Then, Total Sale = 100 × 100 = 10000

ATQ,
price reduces to 25%.
So, the new price = 100 - 25 = 75
Sale increase to 30%,
So, the new sale = 100 + 30 = 130
Now, the new total sale = 75 × 130 = 9750

Difference = 10000 - 9750 = 250
% net effect = (diff/original)× 100
= (250/10000) × 100 = 2.5
So, the net effect on daily sale = 2.5% decrease
১৫,১৬৮.
All possible three-digit numbers are formed by the digits 1, 2, 3, 5, 6 (without repetition). If one number is chosen randomly, what is the probability that it is divisible by 5?
  1. ক
    1/3
  2. খ
    2/5
  3. গ
    5/12
  4. ঘ
    1/5
ব্যাখ্যা

Question: All possible three-digit numbers are formed by the digits 1, 2, 3, 5, 6 (without repetition). If one number is chosen randomly, what is the probability that it is divisible by 5?

Solution:
প্রদত্ত অঙ্কগুলো: 1, 2, 3, 5, 6

তাহলে, তিন অঙ্কের মোট সংখ্যা = 5P3 = 5 × 4 × 3 = 60 টি।

একটি সংখ্যা 5 দ্বারা বিভাজ্য হবে যদি তার শেষ অংকটি 5 হয়।
• শেষ অংকটি 5 (১টি উপায়)।
• বাকি চারটি অংক (1, 2, 3, 6) থেকে প্রথম দুটো স্থান পূরণ করতে হবে।
• প্রথম দুটো স্থান পূরণ করার উপায় = 4P2 = 4 × 3 = 12 টি।

সুতরাং, 5 দ্বারা বিভাজ্য মোট সংখ্যা = 12 টি।

অতএব, সংখ্যাটি 5 দ্বারা বিভাজ্য হবার সম্ভাবনা = (অনুকূল ফলাফল)/(মোট ফলাফল)
= 12/60
= 1/5

১৫,১৬৯.
In a mixture 60 litres, the ratio of milk and water 2 : 1. If this ratio is to be 1 : 2, then the quantity of water to be further added is-
  1. ক
    20 litres
  2. খ
    30 litres
  3. গ
    40 litres
  4. ঘ
    60 litres
ব্যাখ্যা
Question: In a mixture 60 litres, the ratio of milk and water 2 : 1. If this ratio is to be 1 : 2, then the quantity of water to be further added is-

Solution:
Quantity of milk = 60 × (2/3) litres = 40 litres.
Quantity of water in it = (60 -  40) litres = 20 litres.
New ratio = 1 : 2
Let quantity of water to be added further be x litres.
Then,
milk : water = 40/(20 + x)
Now,
40/(20 + x) = 1/2
⇒ 20 + x = 80
∴ x = 60.
Quantity of water to be added = 60 litres.
১৫,১৭০.
A man sells two articles at TK 99 each. He gains 10% on one and loses 10% on the other. Then on overall basis he -
  1. ক
    Loss Tk. 2
  2. খ
    Loss Tk. 4
  3. গ
    Gains Tk. 1
  4. ঘ
    Gains Tk. 5
ব্যাখ্যা
Question: A man sells two articles at TK 99 each. He gains 10% on one and loses 10% on the other. Then on overall basis he -

Solution:
Total Selling Price =Tk. (2 × 99)  = Tk.198

Now,
C.P. of first article = Tk. {(100/110) × 99}
= Tk. 90

C.P. of second article =Tk. {(100/90) × 99}
= Tk. 110

∴ Total C.P. = Tk. (90 + 110) = Tk. 200

So, Loss = Tk. (200 - 198)
= Tk. 2
১৫,১৭১.
If the diagonal and the area of a rectangle are 25 m2 and 168 m2, what is the length of the rectangle?
  1. ক
    ক) 12 m
  2. খ
    খ) 17 m
  3. গ
    গ) 24 m
  4. ঘ
    ঘ) 31 m
ব্যাখ্যা

Let the length of the rectangle be x metre.
Then, a breath of the rectangle = (168/x)

∴ √{(x)2 + (168/x)2} = 25
⇒ √{(x2 + (28224/x2)} = 25
⇒ {(x2 + (28224/x2)} = 625
⇒ x4 - 625x2 + 28224 = 0
⇒ x2(x2 - 576) - 49(x2 - 576) = 0
⇒ (x2 - 576)(x2 - 49) = 0
⇒ x2 = 576 or x2 = 49
⇒ x = 24 or x = 7

Hence, length = 24 cm. and breadth = 7 m.

১৫,১৭২.
The ratio of two numbers is 4 : 5 and their HCF is 4. Their LCM is - 
  1. ক
    120
  2. খ
    40
  3. গ
    80
  4. ঘ
    240
ব্যাখ্যা
Question: The ratio of two numbers is 4 : 5 and their HCF is 4. Their LCM is - 

Solution: 
let, 
the numbers are 4x and 5x
their HCF is = x
∴ x = 4

so, the numbers ar 16, 20.

the LCM is = 80
১৫,১৭৩.
When x3 - 3x2 + x + 2 is divided by polynomial g(x), then quotient and remainder are x - 2 and - 2x + 4, respectively, g(x) is:
  1. ক
    x2 - x + 1
  2. খ
    x2 + x + 1
  3. গ
    x2 + x - 1
  4. ঘ
    x2 - x - 1
ব্যাখ্যা
Question: When x3 - 3x2 + x + 2 is divided by polynomial g(x), then quotient and remainder are x - 2 and - 2x + 4, respectively, g(x) is:

Solution:
According to the division algorithm,
Dividend = Divisor × Quotient + Remainder

We have,
Dividend = x3 - 3x2 + x + 2,
Divisor = g(x),
Quotient = x - 2
and Remainder = -2x + 4

Put the given values in the below equation and simplify it, to get the value of g (x).

Dividend = Divisor × Quotient + Remainder
(x3 - 3x2 + x + 2) = g(x) × (x - 2) + (- 2x + 4)
⇒ (x3 - 3x2 + x + 2) - (- 2x + 4) = g(x) × (x - 2)
⇒ (x3 - 3x2 + x + 2x + 2 - 4) = g(x) × (x - 2)
⇒ (x3 - 3x2 + 3x - 2) = g(x) × (x - 2)
⇒ g(x) = (x3 - 3x2 + 3x - 2)/(x - 2)

Therefore, g(x) = x2 - x + 1
১৫,১৭৪.
A sum of money is divided among 6 males and some females in the ratio of the total money received by males to total money received by females as 3 : 1. If each male gets Tk. 600 and each female gets Tk. 1200, how many females are there? 
  1. ক
    1
  2. খ
    2
  3. গ
    3
  4. ঘ
    4
ব্যাখ্যা

Question: A sum of money is divided among 6 males and some females in the ratio of the total money received by males to total money received by females as 3 : 1. If each male gets Tk. 600 and each female gets Tk. 1200, how many females are there?

Solution:
Let the number of females be x.

Then,
(600 × 6)/1200x = 3/1
Or, 6/2x = 3/1
Or, 3/x = 3/1

So, 3x = 3
∴ x = 1

১৫,১৭৫.
Four boys are sitting on a bench. Rasel is to the left of Tofail. Rakib is to the left of Hasib. Rakib is between Hasib and Tofail. Who would be second from the left on the bench?
  1. ক
    ক) Rasel
  2. খ
    খ) Tofail
  3. গ
    গ) Rakib
  4. ঘ
    ঘ) Hasib
ব্যাখ্যা
Question: Four boys are sitting on a bench. Rasel is to the left of Tofail. Rakib is to the left of Hasib. Rakib is between Hasib and Tofail. Who would be second from the left on the bench?

Solution:
Rasel is to the left of Tofail.
Rasel ⇔ Tofail

Rakib is to the left of Hasib.
Rakib ⇔ Hasib

Rakib is between Hasib and Tofail.
Tofail ⇔ Rakib ⇔ Hasib

∴ Rasel ⇔ Tofail ⇔ Rakib ⇔ Hasib.

Tofail would be second from the left on the bench.
১৫,১৭৬.
If x2 is an odd number, determine the nature of x2 - x.
  1. ক
    Even
  2. খ
    Odd
  3. গ
    Prime
  4. ঘ
    A perfect square
ব্যাখ্যা
Question: If x2 is an odd number, determine the nature of x2 - x.

Solution:
যেহেতু x2 বিজোড় তাই x ও বিজোড় হবে। 

এখন,
x2 - x
= x(x - 1)
= (x - 1)x
∴ (x - 1) এবং x দুইটি ক্রমিক সংখ্যা।

x বিজোড় সংখ্যা হলে (x - 1) অবশ্যই জোড় সংখ্যা হবে।
কারণ দুইটি ক্রমিক সংখ্যার মধ্যে একটি বিজোড় হলে অন্যটি জোড় হবে।

সুতরাং, x ও (x - 1) এর গুনফল = x(x - 1) = x2 - x একটি জোড় সংখ্যা। [জোড় × বিজোড় = জোড়]
১৫,১৭৭.
If a + b + c = 6 and ab + bc + ca = 10 then the value of a3 + b3 + c3 - 3abc is?
  1. ক
    ক) 36
  2. খ
    খ) 48
  3. গ
    গ) 42
  4. ঘ
    ঘ) 40
ব্যাখ্যা

a + b + c = 6
ab + bc + ca = 10
∴ (a + b + c)2= 36
⇒ a2+ b2+ c2+ 2ab + 2bc + 2ca = 36
⇒ a2+ b2+ c2+ 2(ab + bc + ca) = 36
⇒ a2+ b2+ c2+ 2 × 10 = 36
⇒ a2+ b2+ c2= 16
As we know
a3 + b3 + c3 - 3abc/(a2 + b2 + c2 - ab - bc - ca) = a + b + c
⇒a3 + b3 + c3 - 3abc/16 - (ab + bc + ca) = 6
⇒a3 + b3 + c3 - 3abc/(16 - 10) = 6
⇒a3 + b3 + c3 - 3abc = 6× 6
⇒a3 + b3 + c3 - 3abc = 36.

১৫,১৭৮.
If the volume of a sphere is divided by its surface area, the result is 25 cm. the radius of the sphere is - 
  1. ক
    50 cm
  2. খ
    70 cm
  3. গ
    81 cm
  4. ঘ
    75 cm
ব্যাখ্যা

Question: If the volume of a sphere is divided by its surface area, the result is 25 cm. the radius of the sphere is -

Solution:
Let,
the radius of the sphere is r cm
∴ the volume of a sphere = (4/3)πr3
∴ the surface area of a sphere = 4πr2

ATQ,
{(4/3)πr3}/(4πr2) = 25
⇒ r/3 = 25
∴ r = 75

∴ the radius of the sphere is 75 cm

১৫,১৭৯.
A rectangular carpet has an area of 120 sq. meters and a perimeter of 46 meters. The length of its diagonal is-
  1. ক
    17 m
  2. খ
    20 m
  3. গ
    15 m
  4. ঘ
    16 m
ব্যাখ্যা
Question: A rectangular carpet has an area of 120 sq. meters and a perimeter of 46 meters. The length of its diagonal is-

Solution:
Let, the length of carpet be x m and breadth the y m.

ATQ,
2(x + y) = 46
x + y = 23
and xy = 120

Diagonal = √(x2 + y2)
= √{(x + y)2 - 2xy}
= √(232 - 2 . 120)
= √(529 - 240)
= √289
= 17 m
১৫,১৮০.
If the difference between the circumference and diameter of a circle is 60 cm, then the radius of the circle is- 
  1. ক
    ক) 6 cm
  2. খ
    খ) 7 cm
  3. গ
    গ) 14 cm
  4. ঘ
    ঘ) 20 cm
ব্যাখ্যা
Question: If the difference between the circumference and diameter of a circle is 60 cm, then the radius of the circle is- 

Solution:
ধরি,
বৃত্তের ব্যাসার্ধ r 
বৃত্তটির ব্যাস = 2r
বৃত্তটির পরিধি = 2πr

প্রশ্নমতে,
2πr -  2r = 60
2r(π - 1) = 60
2r{(22/7) - 1} = 60
2r{(22 - 7)/7} = 60
2r(15/7) = 60
r = (60 × 7)/30
r = 14
১৫,১৮১.
The difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 4% per annum is Tk. 1. The sum is:
  1. ক
    ক) 650
  2. খ
    খ) 640
  3. গ
    গ) 635
  4. ঘ
    ঘ) 625
ব্যাখ্যা
Question: The difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 4% per annum is Tk. 1. The sum is:

Solution:
Let the sum be Tk. x.

Compound interest = x{1 + (4/100)}2 - x
= x{1 + (1/25)}2 - x
= x(26/25)2 - x
= (676x/625) - x
= (676x - 625x)/625
= 51x/625

Simple interest = (x × 4 × 2)/100
= 2x/25

ATQ,
(51x/625) - (2x/25) = 1
⇒ (51x - 50x)/625 = 1
∴ x = 625
১৫,১৮২.
If 5 men can colour 50-meter long cloth in 5 days, in many days 4 men can color a 40-meter long cloth?
  1. ক
    5 days
  2. খ
    6 days
  3. গ
    4 days
  4. ঘ
    3 days
ব্যাখ্যা
Question: If 5 men can colour 50-meter long cloth in 5 days, in many days 4 men can color a 40-meter long cloth?

Solution:
M1D1W2 = M2D2W1
⇒ 5 × 5 × 40 = 4 × D2 × 50
⇒ D2 = (5 × 5 × 40)/(4 × 50)
∴ D2 = 5
১৫,১৮৩.
112 + 122 + 132 + ........ + 202 = ?
  1. ক
    2655
  2. খ
    2485
  3. গ
    2225
  4. ঘ
    2535
  5. ঙ
    None
ব্যাখ্যা
প্রশ্ন: 112 + 122 + 132 + ........ + 202 = ?

সমাধান:
112 + 122 + 132 + ........ + 202
= (12 + 22 + 32 + .... + 202) - (12 + 22 + 32 + .... + 102)
= [{20(20 + 1)(2 × 20 + 1}/6] - [{10(10 + 1)(2 × 10 + 1)}/6]
= {(20 × 21 × 41)/6} - {(10 × 11 × 21)/6}
= 2870 - 385
= 2485
১৫,১৮৪.
Sajid types 450 words in 30 minutes. How many words would he type in 7 minutes?
  1. ক
    ক) 100 words
  2. খ
    খ) 105 words
  3. গ
    গ) 115 words
  4. ঘ
    ঘ) 125 words.
ব্যাখ্যা
Question: Sajid types 450 words in 30 minutes. How many words would he type in 7 minutes?

Solution: 
Words per minute= (Number of words) / (Time in minutes)
Words per minute = 450 words / 30 minutes 
= 15 words/minute

 number of words  in 7 minutes:
Number of words = Words per minute × Time in minutes 
= 15 words/minute × 7 minutes
= 105 words

Sajid would type 105 words in 7 minutes.
১৫,১৮৫.
What is the ratio of simple interest earned on certain amount at the rate of 12% per annum for 9 years and that for 12 years?
  1. ক
    1 : 2
  2. খ
    2 : 3
  3. গ
    3 : 4
  4. ঘ
    4 : 5
ব্যাখ্যা
Question: What is the ratio of simple interest earned on certain amount at the rate of 12% per annum for 9 years and that for 12 years?

Solution:
ধরি,
আসল = P টাকা
১২% হার বার্ষিক মুনাফায় P টাকার ৯ বছরের মুনাফা = P × ৯ × (১২/১০০) টাকা
= (১০৮P)/১০০ টাকা 

১২% হার বার্ষিক মুনাফায় P টাকার ১২ বছরের মুনাফা = P × ১২ × (১২/১০০) টাকা
= (১৪৪P)/১০০ টাকা

অনুপাত = (১০৮P)/১০০ : (১৪৪P)/১০০
= ১০৮ : ১৪৪
= ৯ : ১২
= ৩ : ৪
১৫,১৮৬.

In the figure above, how many of the points on line segment PQ have coordinates that are both integers?
  1. ক
    5
  2. খ
    8
  3. গ
    10
  4. ঘ
    11
ব্যাখ্যা
Question:

In the figure above, how many of the points on line segment PQ have coordinates that are both integers?

Solution:
The equation of a straight line passing through points P(x1, y1) and Q(x2, y2) is: (y − y1)/(x - x1) = (y1 - y2)/(x1 - x2)
or P(0, 30) and Q(50, 0):
(y - 30)/(x - 0) = (30 - 0)/(0- 50)
⇒ 50(y - 30) = 30x
⇒ 50y - 1500 = 30x
⇒ 3x + 5y = 150

If x is a multiple of 5, then y will be an integer. x ranges from 0 to 50, inclusive. There are total of 11 multiples of 5 in this range: 0, 5, 10, 15, 20, 25, 30, 35, 40, 45, and 50.
১৫,১৮৭.
In an examination, 35% of total students failed in Bangla, 45% failed in English and 20% failed in both. Find the percentage of those students who passed in both the subjects?
  1. ক
    60%
  2. খ
    50%
  3. গ
    40%
  4. ঘ
    35%
ব্যাখ্যা
Question: In an examination, 35% of total students failed in Bangla, 45% failed in English and 20% failed in both. Find the percentage of those students who passed in both the subjects?

Solution:
Failed students in Bangla = 35%
Failed students in English = 45%
Student failed in both subject Bangla and English = 20%

Student only fail in Bangla = (35 - 20)% = 15%
Student only fail in English = (45 - 20)% = 25%

Total fail students = (15 + 25 + 20)%
= 60%

Percentage of passed students in both subjects = (100 - 60)% 
= 40%
১৫,১৮৮.
P is shorter than N. R is shorter than P. S is taller than R. F is taller than N. N is taller than S. Who is the tallest man?
  1. ক
    N
  2. খ
    S
  3. গ
    F
  4. ঘ
    R
ব্যাখ্যা
Question: P is shorter than N. R is shorter than P. S is taller than R. F is taller than N. N is taller than S. Who is the tallest man?

Solution:
P is shorter than N ⇒ N > P
R is shorter than P ⇒ P > R
S is taller than R ⇒ S > R
F is taller than N ⇒ F > N
N is taller than S ⇒ N > S

Here,
P > R and S > R
Both P & S are taller than R.

N > P and N > S
N is taller than both P & S.

F > N
F is taller than N.
So, F is the tallest man.
১৫,১৮৯.
  1. ক
    ক) 1/4
  2. খ
    খ) 1/2
  3. গ
    গ) 1/8
  4. ঘ
    ঘ) 1/16
ব্যাখ্যা
Question:


Solution:

১৫,১৯০.
A tradesman gives 4% discount on the marked price and gives 1 article free for buying every 15 articles and thus gains 35%. The marked price is above the cost price by-
  1. ক
    30%
  2. খ
    40%
  3. গ
    45%
  4. ঘ
    50%
  5. ঙ
    None of these
ব্যাখ্যা
Question: A tradesman gives 4% discount on the marked price and gives 1 article free for buying every 15 articles and thus gains 35%. The marked price is above the cost price by-

Solution:
Let, the C.P. of each article be = Tk. 100

Then,
Cost Price of 16 articles = Tk. (100 × 16)
= Tk. 1600

Selling Price of 15 articles,
= Tk.{1600 × (135/100)}
= Tk. 2160

S.P. of each article = Tk.2160/15
= Tk. 144

If S.P. is Tk. 96, marked price = Tk. 100
If S.P. is Tk. 144, marked price = Tk.{(100/96) × 144}
= Tk. 150

∴ Marked price = 50% above C.P
১৫,১৯১.
If x is an odd negative integer and y is an even integer, which of the following statements must be true?
  1. ক
    (3x - 2y) is odd
  2. খ
    xy2 is an even negative integer
  3. গ
    (y2 - x) is an odd negative integer
  4. ঘ
    All of the above
  5. ঙ
    None of the above
ব্যাখ্যা
Question: If x is an odd negative integer and y is an even integer, which of the following statements must be true?

Solution:
Let x = - 1, y = 2
Option A: (3x - 2y) = 3(- 1) - 2(2) = -7 is ODD
Option B: xy2 = (-1)(22) = - 4 is EVEN NEGATIVE
Option C. (y2 - x) = 22 - (-1) = 5 is ODD POSITIVE

Since the question involves Even and Odd numbers, let us also consider y = 0.
Option A: (3x - 2y) = 3(-1) - 2(0) = -3 is ODD
Option B: xy2 = (-1)(02) = 0 is EVEN POSITIVE
Option C. This condition was proved false using the above values.

Hence only option A is satisfied.
১৫,১৯২.
To produce an annual income of TK 800 from a 8% stock at 80, the amount of stock needed is-
  1. ক
    10500 TK
  2. খ
    10,000 TK
  3. গ
    15000 TK
  4. ঘ
    12000 TK
ব্যাখ্যা

Question: To produce an annual income of TK 800 from a 8% stock at 80, the amount of stock needed is-

Solution:
since face value is not given, take it as TK 100
As it is a 8% stock, income(dividend) per stock = TK 8
ie, For an income of TK 8, the amount of stock needed = TK 100
For an income of TK 800, the amount of stock needed = (100 × 800)/8 = 10,000

১৫,১৯৩.
In how many ways can a group of 4 men and 3 women be made out of a total of 6 men and 5 women?
  1. ক
    120 ways
  2. খ
    150 ways
  3. গ
    210 ways
  4. ঘ
    90 ways
ব্যাখ্যা

Question: In how many ways can a group of 4 men and 3 women be made out of a total of 6 men and 5 women?

Solution:
There are 6 men and 5 women. We have to select 4 men out of 6 and 3 women out of 5.

Number of ways to select 4 men from 6 = 6C4 = 6!/(4! × 2!)
= (6 × 5)/(2 × 1) = 15

Number of ways to select 3 women from 5 = 5C3
= 5!/(3! × 2!)
= (5 × 4)/(2 × 1)
= 10

∴ The number of ways of making the selection = 15 × 10 = 150 ways

১৫,১৯৪.
A student first reduced a number by 20 percent and then increased it again by 20 percent. If the difference between the two new numbers was 8, then what is the original number?
  1. ক
    ক) 40
  2. খ
    খ) 50
  3. গ
    গ) 75
  4. ঘ
    ঘ) 100
ব্যাখ্যা
Question: A student first reduced a number by 20 percent and then increased it again by 20 percent. If the difference between the two new numbers was 8, then what is the original number?

Solution: 
ধরি, প্রকৃত সংখ্যাটি x

২০%  কমালে, সংখ্যাটি = x - x এর ২০% 
= x - ০.২x
 = o.৮x 

২০% বাড়ালে = ০.৮x + ০.৮x এর ২০%
= ০.৮x + ০.৮x × ০.২ 
= ০.৮x +০.১৬x
= ০.৯৬x 

প্রশ্নমতে, 
 ০.৯৬x - ০.৮x = ৮
⇒ ০.১৬x = ৮ 
 ⇒ x = ৮/০.১৬
= ৫০ 

অতএব, প্রকৃত সংখ্যাটি ৫০।
১৫,১৯৫.
Six identical machines can produce 540 articles in 12 hours. How many articles would 8 such machines produce in 15 hours? 
  1. ক
    700 articles
  2. খ
    900 articles
  3. গ
    800 articles
  4. ঘ
    500 articles
ব্যাখ্যা

Question: Six identical machines can produce 540 articles in 12 hours. How many articles would 8 such machines produce in 15 hours?

Solution: 
Total articles produced by 6 machines in 12 hours = 540.
Articles produced by 1 machine in 12 hours = 540/6
Articles produced by 1 machine in 1 hour = 540/(6×12) = 7.5 articles

So, Articles produced by 8 machines in 15 hours = 7.5 × 8 × 15 
= 900 articles

১৫,১৯৬.
A trader sells his goods at a discount of 20%. He still makes a profit of 25%. If he sells the goods at the marked price only, his profit will be:
  1. ক
    ক) 50.50%
  2. খ
    খ) 54.25%
  3. গ
    গ) 56.25%
  4. ঘ
    ঘ) 57.50%
ব্যাখ্যা

ধরি,
দ্রব্যটির বাজার মূল্য = 100 টাকা
∴ 20% ছাড়ে বিক্রয়মূল্য = (100 - 20) = 80 টাকা।
আবার, ধরি,
দ্রব্যটির উৎপাদন খরচ = x টাকা।
∴ x + x এর 25/100 = 80
⇒ x + x/4 = 80
⇒ 5x/4 = 80
⇒ x = 64

∴ Marked price (বাজার মূল্য) - এ পন্যটি বিক্রি করলে,
লাভ হয় - (100 - 64) = 36 টাকা।
∴ 64 টাকায় লাভ হয় = 36 টাকা
∴ 100 টাকায় লাভ হয় = (36 × 100)/64
                            = 56.25%

১৫,১৯৭.
What will come at the place of question mark ?
8, 28, 116, 584, ?
  1. ক
    1752
  2. খ
    3504
  3. গ
    3508
  4. ঘ
    3502
  5. ঙ
    2428
ব্যাখ্যা

Question: What will come at the place of question mark ?
8, 28, 116, 584, ?

Solution:
1st term = 8
2nd term = (8 × 3) + 4 = 28 
3rd term = (28 × 4) + 4 = 116 
4th term = (116 × 5) + 4 = 584 
5th term = (584 × 6) + 4 = 3508

১৫,১৯৮.
When the sun's altitude changes from 30° to 60°, the length of the shadow of a tower decreases by 70m. What is the height of the tower?
  1. ক
    ক) 60.55m
  2. খ
    খ) 140 m
  3. গ
    গ) 35 m
  4. ঘ
    ঘ) 20.2 m
ব্যাখ্যা


Let AD be the tower, BD be the initial shadow and CD be the final shadow.
Given that BC = 70 m
∠ABD = 30°
∠ACD = 60°
Let CD = x, AD = h
From the right ⊿CDA
tan 60° = AD/CD
√3 = h/x ......eq(1)
From the right ⊿BDA
tan30° = AD/BD
1/√3 = h/(70 + x) ........ eq(2)
eq(1)/eq(2)
√3/(1/√3) = (h/x)/{h/(70+x)}
3 = 70 + x/x
2x = 70
x = 35
Substituting this value of x in eq:1, we have
√3 = h/35
h = 35√3
= 60.55 m

১৫,১৯৯.
A bag contains 30 balls numbered 1 to 30. Two balls are drawn at random. What is the probability that the balls drawn contain a number which is multiple of 4 or 6 but not a multiple of both?
  1. ক
    ক) 1/8
  2. খ
    খ) 1/4
  3. গ
    গ) 1/3
  4. ঘ
    ঘ) None
ব্যাখ্যা
Question: A bag contains 30 balls numbered 1 to 30. Two balls are drawn at random. What is the probability that the balls drawn contain a number which is multiple of 4 or 6 but not a multiple of both?

Solution:
Total outcomes are 30
LCM of 4 and 6 is 12.
Common numbers are 30/12 ≈ 2
Now, favourable outcomes are (28/4) - 2 + (30/6) - 2 = 8

So, Probability is (8/30) × (7/29) = 28/435
১৫,২০০.
One ball is picked up randomly from a bag containing 8 yellow, 7 blue and 6 black balls. What is the probability that it is neither yellow nor black?
  1. ক
    1/2
  2. খ
    1/3
  3. গ
    1/4
  4. ঘ
    3/4
  5. ঙ
    3/5
ব্যাখ্যা

Total number of balls, n(S) = 8 + 7 + 6 = 21
n(E) = Number of ways in which a ball can be selected which is neither yellow nor black = 7 (∵ there are only 7 balls which are neither yellow nor black)
P(E) = n(E)/n(S) = 7/21 = 1/3