Bank Math Master

Question: The sum of the digits of a two-digit number is 12 and the difference between the two digits of the two-digit number is 6. What is the two-digit number?

Solution:
Let, the two-digit number be 10a + b where a > b.

ATQ,
a + b = 12 ------- (1)
a - b = 6 --------- (2)

On adding equation (1) & (2) 
21 =18
∴ a = 9

Putting this value in (1) we get,
9 + b = 12
∴ b = 3

So the number is 10a + b = 9 . 10 + 3 = 93

When, a < b, Then the required number is 39

Question: Find the HCF of 210, 385, and 735.

Solution:
HCF of 210, 385, and 735.

Factor of 210 = 2 × 3 × 5 × 7
Factor of 385 = 5 × 7 × 11
Factor of 735 = 3 × 5 × 7 × 7 
∴ HCF of (210, 385 and 735) = 35 

Question: If 'a' is an odd number and 'b' is an even number, which one of the following must be an even number?

Solution:
Suppose, a = 1 and b = 2

a + b = 1 + 2 = 3
ab + 1 = 1 × 2 + 1 = 3
a² + b² = 1² + 2² = 5
(a + b)² + 1 = (1 + 2)² + 1 = 10

Question: The difference between the two numbers is 20% of the larger number. If the smaller number is 12, find the larger number.

Solution:
Let the number be x
ATQ,
x - 12 = 20% of x
⇒ x - 12 = x/5
⇒ x - x/5 = 12
⇒ 4x/5 = 12
⇒ x = (12 × 5)/4 
∴ x = 15

Question: 84 English books, 90 Mathematics books, and 120 Bangla books have to be stacked topicwise. How many books will be there in each stack so that each stack will have the same height too?

Solution:
As the height of each stack is the same, the required number of books in each stack
= HCF of 84, 90 and 120

84 = 2 × 2 × 3 × 7
90 = 2 × 3 × 3 × 5
120 = 2 × 2 × 2 × 3 × 5

∴ HCF = 2 × 3 = 6

Hence, The required number of books in each stack is 6.

Question: An officer was appointed on maximum daily wayes on contract money of Tk. 4956. But on being absent for some days, he was paid Tk. 3894. For how many days was he absent?

Solution:
Maximum daily wages of the officers = H.C.F of Tk. 4956 and Tk. 3894
H.C.F of 4956 & 3894 = 354
So, he was appointed for 4956/354 = 14 days
But he was present = 3894/354 = 11 days.

So, he was absent for (14 -11) days = 3 days

Question: 180 mangoes are distributed among 70 men and women such that each men gets 2 and each woman gets 3 mangoes. The number of men is - 

Solution:
Let, the number of men be x.
The number of women = 70 - x

ATQ,
2x + 3(70 - x) = 180
⇒ 2x + 210 - 3x = 180
⇒ x = 210 - 180
∴ x = 30

∴ The number of men is 30.

Question: What is the value of the following expression?
12 ÷ (1/2) + [(35 ÷ 7) × 40] + 20 - (15 × 10)

Solution:
12 ÷ (1/2) + [(35 ÷ 7) × 40] + 20 - (15 × 10)
= 12 ÷ (1/2) + 5 × 40 + 20 - 150
= 12 × 2 + 200 + 20 - 150
= 244 - 150
= 94

Question: The sum of the L.C.M. and H.C.F. of two numbers is 1260, and the L.C.M. is 900 more than the H.C.F. What is the product of these two numbers?

Solution:
Let the HCF be x

LCM = HCF + 900
LCM = x + 900 ...............(1)

And, LCM + HCF = 1260
LCM + x = 1260 .................(2)

From (1) and (2) equation,
(x + 900) + x = 1260
⇒ 2x + 900 = 1260
⇒ 2x = 1260 - 900
⇒ 2x = 360
⇒ x = 360/2
∴ x = 180

∴ HCF = 180
And, from equation (1),
LCM = HCF + 900
LCM = 180 + 900
∴ LCM = 1080

By formula, the product of the numbers is equal to the product of their HCF and LCM.

Product of numbers = HCF × LCM = 180 × 1080
∴ Product = 194400

Question: A box contains a total of 300 coins, some worth 25 paise and others worth 50 paise. If the total value of these coins is Tk 120, how many 50 paise coins are there?

Solution:
Let the number of 50 paise coins be = x
So, the number of 25 paise coins is = 300 - x

According to the question,
50x + {25 × (300 - x)} = 120 × 100
⇒ 50x + 7500 - 25x  = 12000
⇒ 25x = 4500
∴ x = 180

Question:

Solution: 

Question: 7 is added to a certain number; the sum is multiplied by 5, the product is divided by 9 and 3 is subtracted from the quotient. The remainder left is 12. Find the number.

Solution:
Let the original number be x

ATQ,
{5(x + 7)/9} - 3 = 12
⇒ {5(x + 7) - 27}/9 = 12
⇒ 5(x + 7) - 27 = 108
⇒ 5x + 35 - 27 = 108
⇒ 5x + 8 = 108
⇒ 5x = 100
∴ x = 20

Question: - 6m - 2n - [3n - {8m - (4n - 10m)}] - 6m simplifies to

Solution: 
- 6m - [3n - {8m - (4n - 10m)}] - 6m
= - 6m - 2n - [3n - {8m - 4n + 10m}] - 6m
= - 6m - 2n - [3n - 8m + 4n - 10m] - 6m
= - 6m - 2n - 3n + 8m - 4n + 10m  - 6m
= 6m - 9n

Question: The sum of three consecutive odd natural numbers each divisible by 3 is 63. Find the largest number among them.

Solution: 
let, the numbers 3x, 3x + 6, 3x + 12

3x + 3x + 6 + 3x + 12 = 63
⇒ 9x + 18 = 63 
⇒ 9x = 45
∴ x = 5

The largest among them is = (3 × 5) + 12
= 15 + 12
= 27

Question: Find the greatest common divisor (GCD) of the numbers 0.9, 0.36, and 1.08

Solution:
প্রদত্ত সংখ্যাগুলো হলো 0.9, 0.36, 1.08
0.90 × 100 = 90 
0.36 × 100 = 36
1.08 × 100 = 108

90, 36, 108, 36,  এর গ.সা.গু = 18
∴ 0.9, 0.36, 1.08 এর গ.সা.গু = 0.18

Question: If the sum of a number and its reciprocal be 2. What is the number?

Solution:
Let the number be = a
The reciprocal of the number is = 1/a

According to the question,
a + (1/a) = 2
⇒ a² + 1 = 2a
⇒ a² - 2a + 1 = 0
⇒ (a - 1)² = 0
∴ a = 1

Hence, the number = 1

Question: Find the greatest number-

Solution: 
(1/9)² = 1/81 = 0.012
(√0.3)² = 0.3
(2/5)² = (0.4)² = 0.16
(0.3)² = 0.09

∴ √0.3 সংখ্যাটি বৃহত্তম।

Question: Two numbers have a product of 2028 and a highest common factor (HCF) of 13. How many such pairs of numbers exist?

Solution:
Let the two numbers be x and y respectively.
It is given that the product of the two numbers is 2028, therefore, xy = 2028

Also, 13 is their HCF, thus both numbers must be divisible by 13.

So, let x = 13a and y = 13b, 

ATQ,
13a × 13b = 2028
⇒ 169ab = 2028
⇒ ab = 2028
∴ ab = 12

Therefore, the required possible pair of values of x and y which are prime to each other are (1, 12) and (3, 4).
Thus, the required numbers are (12, 156) and (39, 52).
Hence, the number of possible pairs is 2.