ব্যাংক ডেইলি কুইজ [লং কোর্সের অংশ]

Question: The LCM of the two numbers is 12 times their HCF. The sum of HCF and LCM is 403. If one number is 93, find the other =?

Solution:
Let HCF be h and LCM be l
Then l = 12h and
l + h = 403

∴12h + h = 403
⇒ h = 31

So, l = (403 − 31) =372

Hence, the other number = (31 × 372)/93 = 124

Question: One-third of one-fourth of a number is 12. Then the number is:

Solution:
Let the number be p.
We know that the term "of" indicates the sign of multiplication.

Now, it is given that one-third of one-fourth of p is 12, therefore, we have:

1/3 × 1/4 × p = 12
⇒ p/12 = 12
⇒ p = 12 × 12
⇒ p = 144

Hence, the number is 144.

Question: 84 Maths books, 90 Physics books, and 120 Chemistry 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: The HCF and LCM of the two numbers are 21 and 84 respectively. If the ratio of the two numbers is 1 : 4 then the larger of the two numbers is =?

Solution:
We know,
LCM × HCF = 1^st number × 2^nd number

Let 1^st number = P
2^nd number = 4P

P × 4P = 21 × 84
⇒ 4P² = 21 × 84
⇒ P² = 21 × 21
∴ P = 21

Then, the numbers = 21, 84
So, the larger number = 84

Question: In a Mathematics examination, the number scored by 5 candidates is 5 successive odd integers. If their total marks are 185, the highest score is:

Solution:
Let the five successive odd numbers be,
x, x + 2, x + 4, x + 6, x + 8

Then, according to given information,
x + x + 2 + x + 4 + x + 6 + x + 8 = 185 
⇒ 5x + 20 = 185 
⇒ 5x = 165 
⇒ x = 33

∴ Highest number = 33 + 8 = 41

Question: The smallest number, which when increased by 5 is divisible by each of 24, 32, 36 and 64 is =?

Solution:
To find the least common multiple (LCM) of the given numbers (24, 32, 36, 64), we can first find the prime factorization of each number:

24 = 2³ × 3
32 = 2⁵ 
36 = 2² × 3²
64 = 2⁶

Then, we take the highest power of each prime factor that appears in any of the numbers:
The highest power of 2 is 2⁶
The highest power of 3 is 3²

So, the LCM is 2⁶ × 3² = 64 × 9 = 576

∴ The required number is 576 − 5 = 571

Question: The maximum number of students among whom 1001 pens and 910 pencils can be distributed in such a way that each student gets the same number of pens and the same number of pencils is =?

Solution:
Required number of students
= HCF of 1001 and 910
= 91
........................................................
........................................................
HCF of 1001 and 910:

We can find the HCF of 1001 and 910 using prime factorization.
First, let's find the prime factorization of both numbers:

1001 = 7 × 11 × 13
910 = 2 × 5 × 7 × 13

Now, to find the HCF, we take the product of the common prime factors with the lowest exponent:
The common prime factors are 7 and 13.

So, the product of the common prime factors is 7 × 13 = 91

Therefore, the HCF of 1001 and 910 is 91.

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

Solution:
Let the number be x

Then,
x − 12 = 20% of x
⇒ x − 12 = x/5
⇒ x − x/5 = 12
⇒ 4x/5 = 12
⇒ x = (12 × 5)/4 
∴ x = 15

Question: The product of two numbers is 2028 and their HCF is 13. The number of such pairs is =?

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, then,
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.

Question: If the sum of two numbers is 15 and their difference is 5. Find the two numbers.

Solution:
Let the two numbers be x and y. Then,
x + y = 15 .......(1)
x − y = 5 .......(2)

Adding equation (1) and (2), we get,
2x = 20
∴ x = 10

Thus, y = 5
Hence, the required numbers are 10 and 5.

Question: The greatest number which when divides 989 and 1327 leaves the remainder 5 and 7 respectively =?

Solution:
Let us subtract the remainder from the number.
989 − 5 = 984
1327 − 7 = 1320

For the greatest number, let us take the HCF of the numbers.
HCF = (984, 1320) = 24

∴ The greatest number is 24

Question: If the numerator of a fraction is increased by 200% and the denominator is increased by 300%, the resultant fraction is 15/26. What was the original fraction?

Solution:
Let the fraction be x/y

Then,
(x + 200% of x)/(y + 300% of y) = 15/26
⇒ (x + 2x)/(y + 3y) = 15/26
⇒ 3x/4y = 15/26
⇒ x/y = (15/26) × (4/3)
∴ x/y=10/13

Question: If the product of three consecutive integers is 120, then the sum of the integers is:

Solution:
120 = 2 × 2 × 2 × 3 × 5
= (2 × 2) × 5 × (2 × 3)
= 4 × 5 × 6

Clearly, the three consecutive integers whose product is 120 are 4, 5 and 6.

∴ Required sum
= 4 + 5 + 6
= 15