ব্যাংক নিয়োগ বিষয়ভিত্তিক প্রস্তুতি

Question: If two numbers have an HCF of 11 and an LCM of 693, and one of them is 77, find the other number.

Solution:
The relationship between two numbers can be represented by the equation: their product equals the product of their HCF and LCM.

Let the unknown number be denoted by P
Applying the given values, we set up the equation, 77 x P = 11 x 693
⇒ 77P = 11 × 693

By isolating P,
We calculate that,
P = 99

Question: A mixture of 150 liters of wine and water contains 20% water. How much more water should be added so that water becomes 25% of the new mixture?

Solution:
Number of liters of water in 125 liters of the mixture = 20% of 150 = 1/5 of 150 = 30 liters

Let us Assume that another 'P' liters of water are added to the mixture to make water 25% of the new mixture.
So, the total amount of water becomes (30 + P) and the total volume of the mixture becomes (150 + P)

Thus, (30 + P) = 25% of (150 + P)
Solving, we get P = 10 liters

Question: A group of hikers is walking in a line. If Javed is 9^th from the front and 12^th from the back, how many hikers are there in total?

Solution:
Javed is 9^th from the front.
Javed is 12^th from the back.

To find the total number of hikers in the line, we can use the formula:
Total number of hikers = Position of Javed from the front + Position of Javed from the back -1

Substituting the given values:
Total number of hikers = 9 + 12 -1 = 20

Thus, there are 20 hikers in total.

Question: The combined average age of the husband, wife, and their child was 27 years three years ago, and the average age of the wife and child was 20 years five years ago. What is the double of the husband's present age?

Solution:
Sum of the present ages of husband, wife and child =(27 × 3 + 3 × 3) years
= (81 + 9) years
= 90 years

Sum of the present ages of wife and child =(20 × 2 + 5 × 2) years
= (40 + 10) years
= 50 years

∴ Husband's present age =(90 - 50) years = 40 years.

∴ The double of the husband's present age is = 40 × 2 = 80 years

Question: A fraction’s numerator is cut down by 25%, and its denominator grows by 250%. If this produces a fraction of 6/5, what is the original fraction?

Solution:
Let the numerator is x and the denominator is y

x is decreased by 25%.

so, new numerator = x - 25% of x
= x - x/4 = 3x/4 = 0.75x

y is increased by 250%

so, new denominator = y + 250% of y
= y + 2.5y
= 3.5y

ATQ,
0.75x / 3.5y = 6/5
x/y = (6 × 3.5)/(5 × 0.75)
= 28/5

Question: Two persons Choton and Billal started a business in which Choton invested Tk. 50000, Billal invested Tk. 80000, after 4 months Robiul joined them with a certain amount. At the end of the year, a total profit of Tk. 40000 was recorded. Robiul's share in the profit was Tk. 15000, then find Robiul's investment in the business.

Solution:
Let,
Robiul's investment in the business be Tk. 1000x
Profit ratio of Choton, Billal and Robiul = (50000 × 12) : (80000 × 12) : {1000x × (12 - 4)}
= 75 : 120 : x

Total profit = 75 + 120 + x = 195 + x

Robiul's share in profit = x
ATQ,
(195 + x)/x = 40000/15000
⇒ (195 + x)/x = 8/3
⇒ 3(195 + x) = 8x
⇒ 585 + 3x = 8x
⇒ 5x = 585
⇒ x = 117

Investment by Robiul = 1000 × 117 = Tk. 117000

Question: Asif, Sami and Riad started a shop by investing Tk. 2700, Tk. 9000 and Tk. 6300 respectively. At the end of one year, the profit was distributed. If Riad's share was Tk. 2100, what was their total profit?

Solution:
Let the total profit is = x

Here, Asif : Sami : Riad = 2700 : 9000 : 6300
= 3 : 10 : 7

then, Riad's share = (7/20) × x = 7x/20
ATQ,
7x/20 = 2100
⇒ x = (2100 × 20)/7
∴ x = 6000

∴ The total profit = Tk. 6000

Question: If n/23 is 2 more than m/23, then n = ?

Solution:
n/23 = (m/23) + 2
⇒ n/23 = (m + 46)/23
∴ n = m + 46

Question: A sum of money at compound interest doubles itself in 15 years. It will become four times of itself in-

Solution:
let the sum P

2P = P (1 + r)¹⁵
⇒ (1 + r)¹⁵ = 2

let, sum will 4 times in n years
4P = P(1 + r)ⁿ
⇒ 4 = (1 + r)ⁿ
⇒ 2² = (1 + r)ⁿ
⇒ ((1 + r)¹⁵)² = (1 + r)ⁿ
⇒ (1 + r)³⁰ = (1 + r)ⁿ
∴ n = 30 years

Question: The ratio of three numbers is 5 : 6 : 7, and their L.C.M. is 210. Find their H.C.F. -

Solution:
Let the numbers be 5x, 6x and 7x
Then, their L.C.M. = 210x

So,
⇒ 210x = 210
∴ x = 1

∴ The HCF of the numbers is x = 1

Question: The average weight of 50 students in a class is 55 kg. If three students weighing 60 kg, 65 kg, and 70 kg leave the class, what is the new average weight of the remaining students?

Solution:
Total weight = 50 × 55 = 2750 kg

Students leaving = 60 kg + 65 kg + 70 kg = 195 kg
Remaining total weight = 2750 - 195 = 2555 kg

Remaining students = 50 - 3 = 47 students
New average = 2555/47
= 54.4 kg

Therefore, the new average weight is approximately 54.4 kg

Question: The product of two co-prime numbers is 540. Then their LCM is =?

Solution:
If HCF of two or more numbers is 1, then two or more numbers are co-prime numbers.
HCF of co-prime number is always 1

∴ Product of number = LCM × HCF
⇒ LCM × 1 = 540
⇒ LCM = 540

Question: A cloth merchant sold half of his cloth at 20% profit, half of the remaining at 20% loss and the rest was sold at the cost price. Calculate the overall profit% or loss%?

Solution:
Let the CP of the whole cloth be x
CP of 1/2 cloth = x/2

CP of half of the remaining = x/4

SP of first half of cloth = (120 × x/2) ÷ 100

SP of the half of the remaining cloth = (80 × x/4) ÷ 100

SP of the remaining half = x/4

Total SP = (120 × x/2)/100 + (80 × x/4)/100 + x/4
= 3x/5 + x/5 + x/4
= (12x + 4x + 5x)/20
= 21x/20

Gain = 21x/20 - x = x/20

Gain % = (x/20)/x × 100% = 5%

Question: B is twice the age of C, and A is 2 years older than B. With their ages adding up to 42, what is A's age?

Solution:
Let,
C is = x years old
B is 2x years old
A is (2x + 2) years old

Now,
2x + 2x + 2 + x = 42
⇒ 5x + 2 = 42
⇒ 5x = 40
∴ x = 8
A is (2 × 8) + 2 years old
= 16 + 2 years old
= 18 years old

Question: The average weight of A, B, and C is 60 kg, and that of B and C is 62 kg. A’s present weight is-

Solution:
The average of A, B and C is 60

So, the sum of their weights = 180
The sum of the weights of B and C is = 62 × 2 = 124

So, A’s present weight is = 180 - 124 = 56 kg

Question: A acquired an article at Tk. 200 and sold it to B with a 20% markup. B then sold it to C at a 10% markup. Find the amount C paid.

Solution:
Price paid by B = 200 + (200/100 × 20) = 200 + 40 = 240

∴ Price paid by C = 240 + (240/100 × 10) = 240 + 24 = 264

Hence, Thus, the price at which C bought the article from B is: Tk. 264

Question: The number is such that when its square is multiplied by three and then reduced by four times the number, the result is fifty greater than the number itself. What is the number?

Solution:
Let the number be x

Then,
3x² - 4x = x + 50
⇒ 3x² - 4x - x - 50 = 0
⇒ 3x² - 5x - 50 = 0
⇒ 3x² - 15x + 10x - 50 = 0
⇒ 3x(x - 5) + 10(x - 5) = 0
⇒ (x - 5)(3x + 10) = 0

∴ x = 5

Hence, the number is 5

Questoin: A sum was put at simple interest at a certain rate for 3 years. Had it been put at 1% higher rate, it would have fetched 510 taka more. The sum is -

Solution:
Let the sum be Tk x and the original rate be R%
Then, ((x × (R +1) × 3)/100) - {(x × R × 3)/100} = 510
⇒ 3Rx + 3x - 3Rx = 51000
⇒ 3x = 51000
⇒ x = 17000

Hence, sum = Tk 17,000

Question: When two values are 25% and 75% of a third number, how much of the second value is the first value as a percentage?

Solution:
Let the third value p
∴ First value = (25p)/100 = p/4
∴ Second value = (75p)/100 = 3p/4

Now,
(First Value/Second value) × 100
= (p/4) × (4/3p) × 100
= 33.33%

Question: Five siblings, born three years apart, have a combined age of 100 years. What is the age of the youngest one?

Solution:
Let the ages of siblings be x, (x + 3), (x + 6), (x + 9) and (x + 12) years.

Then,
x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 100
⇒ 5x = 100 - 30
⇒ 5x = 70
⇒ x = 14

∴ Age of the youngest sibling = x = 14 years.

Question: What percentage of numbers from 1 to 50 has 3 or 7 in the unit's digit?

Solution:
We can list the numbers from 1 to 50 and identify those that have 3 or 7 in the unit's place:

Numbers with 3 in the unit's digit:3, 13, 23, 33, 43
Numbers with 7 in the unit's digit: 7, 17, 27, 37, 47

So, the numbers with 3 or 7 in the unit's digit are:
3, 7, 13, 17, 23, 27, 33, 37, 43, 47

There are 10 such numbers.

There are 50 numbers in total from 1 to 50. The percentage of numbers that have 3 or 7 in the unit's digit is: (10/50) × 100%
= 20%

Question: The students in three batches in school is in the ratio of 2 : 3 : 5. If 20 students in each batch are increased than the ratio changes to 4 : 5 : 7. The total number of students in the three before the increase was-

Solution:
Let,  the students in the three before the increase were 2x, 3x, 5x

After increase, 2x + 20, 3x + 20, 5x + 20

(2x + 20)/ (3x + 20) = 4/5
⇒ 10x + 100 = 12x + 80
⇒ 2x = 20
⇒ x = 10

The total number of students in the three before the increase was = (2x + 5x + 3x)
= 10x
= 10 × 10
= 100

Question: At every stop after the first, half of the passengers leave the bus, and no new passengers board after the first stop. If only 4 people get off at the fourth stop, how many passengers boarded the bus at the first stop?

Solution:
After stop 4: The number of people on the bus = 4
After stop 3: The number of people on the bus = 8
After stop 2: The number of people on the bus = 16
After stop 1: The number of people on the bus = 32

Hence, the number of passengers boarded the bus at the first stop = 32

Question: A man buys Tk. 20 shares paying 9% dividend. The man wants to have an interest of 12% on his money. The market value of each share is-

Solution:
Dividend on Tk. 20 = Tk. (9/100) × 20)
= Tk 9/5

Tk.12 is an income on Tk.100.
∴ Tk. 9/5 is an income on = Tk. (100/12 × 95)
= Tk.15

Question: The average of 60 numbers is 25. If two numbers 40 and 50 are discarded, find the average of the remaining numbers.

Solution:
Given,
Average of 60 numbers = 25
Sum of 50 numbers = 25 × 60 = 1500
Sum of discarded numbers = 40 + 50 = 90
Sum of remaining numbers = 1500 - 90 = 1410
Now, total remaining numbers = 60 - 2 = 58

Average of remaining numbers = 1410/58 = 24.31

Question: A lemonade stand sold only small and large cups of lemonade on Tuesday. 3/5 of the cups sold were small and the rest were large. If the large cups were sold for 7/6 as much as the small cups, what fraction of Tuesday's total revenue was from the sale of large cups?

Solution:
Let, the total cups sold 15
small cups = (3/5) × 15 = 9
large cups = 15 - 9 = 6

let, small cups were sold 6 taka each, then large cups were sold 7 taka each.

large cup's revenue = 7 × 6 = 42 taka
small cup's revenue = 6 × 9 = 54 taka

fraction of Tuesday's total revenue was from the sale of large cups = 42/(42 + 54)
= 42/96
= 7/16

Question: What is the profit percentage if the price for 120 bananas equals the selling price of 100 bananas?

Solution:
Let the C.P. of 120 bananas be Tk. 120
As per question,
S.P. of 100 bananas = Tk. 120

C.P. of 100 bananas would be = Tk. 100
∴ Profit = (S.P. - C.P.) = 120 - 100 = 20

Profit percent = (20/100) × 100%
= 20%

Question: A father is 25 years older than his son. In 5 years, his age will be three times his son's age. What is the present age of the son?

Solution:
Let the son’s present age be x years.
Then, the father’s present age is = x + 25 years.

ATQ
In 5 years, the father’s age will be three times his son's age.
So, (x + 25) + 5 = 3 (x + 5)
⇒ 3x + 15 = x + 30
⇒ 3x - x = 30 - 15
⇒ 2x = 15
∴ x = 7.5

Therefore, the present age of the son is 7.5 years.