ব্যাংক ডেইলি কুইজ [লং কোর্সের অংশ]

Question: If one-fifth of one-half of a number is 12, then two-thirds of that number is: 

Solution:
Let the number be X.
(1/5) × (1/2) × X = 12
⇒ X = 12 × 5 × 2
⇒ X = 120
Two-thirds of the number:
(2/3) × 120 = 80

Question: Which of the following numbers is not divisible by 3?

Solution:
A number is divisible by 3 if the sum of its digits is divisible by 3.
Checking each number:
729: 7 + 2 + 9 = 18, and 18/3 = 6 → divisible 
567: 5 + 6 + 7 = 18, and 18/3 = 6 → divisible 
1458: 1 + 4 + 5 + 8 = 18, and 18/3 = 6→ divisible 
1376: 1 + 3 + 7 + 6 = 17, and 17/3 = 5.66 → not fully divisible

∴ 1376 can not be divided by 3

Question: (x - 35) is divisible by 36, 48, 60. Find the value of x.

Solution:
Find the LCM of 36, 48, and 60.
Prime factorization
36 = 2² × 3²
48 = 2⁴ × 3¹
60 = 2² × 3¹ × 5¹

LCM formula
LCM = 2⁴ × 3² × 5 = 16 × 9 × 5 = 720
Let
x - 35 = 720
x = 720 + 35
x = 755

Question: The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is: 

Solution:
Let, the two numbers be 23 x and 23 y.
Since the other two factors of LCM are 13 and 14, the values of x and y are 13 and 14, respectively.
So, the largest number is 23y,
23y = 23 × 14 = 322

Question: What is the greatest 4-digit number which is divisible by 15, 25, 40, and 75? 

Solution:
Greatest number of 4-digits is 9999.
L.C.M. of 15, 25, 40 and 75 is 600.
On dividing 9999 by 600, the remainder is 399.
∴ Required number (9999 - 399) = 9600.

Question: he difference between a two-digit number and the number obtained by reversing its digits is 54. If the ratio of the tens digit to the units digit of the number is 3 : 1, what is the difference between the sum and the difference of its digits?

Solution:
Let the digits of the number be:

Tens digit = x
Units digit = y
So, the number = 10x + y
The reversed number = 10y + x

Given condition: The difference between the number and its reverse number is 54:
(10x + y) - (10y + x) = 54
9x - 9y = 54
⇒ x - y = 6 ............... (1)

Given ratio of digits: Tens : Units = 3 : 1
x/y = 3/1  
⇒  x = 3y.................(2)

Solve the two equations:

Substitute x = 3y into x - y = 6:
3y - y = 6  
⇒ 2y = 6 
⇒  y = 3
Then x = 3y = 9

Digits found: Tens = 9, Units = 3
Sum and difference of digits:

Sum = x + y = 9 + 3 = 12
Difference = x - y = 9 - 3 = 6

Difference between sum and difference of digits = 12 - 6 = 6

Question: Find a number that is divisible by 7, but leaves remainder 5 when divided by 8, 12, and 20.

Solution:
Let the number be N.
It satisfies:
N ≡ 0 (mod 7)

If N leaves remainder 5 when divided by 8, 12, and 20, then N - 5 is divisible by all three numbers:
N - 5 ≡ 0 (mod 8, 12, 20)

Find LCM of 8, 12, 20:
8 = 2³
12 = 2² × 3
20 = 2² × 5
LCM = 2³ × 3 × 5 = 120

N must also be divisible by 7:
120k + 5 ≡ 0 (mod 7)  
⇒ 120k ≡ -5 ≡ 2 (mod 7)
(since -5 ≡ 2 mod 7)

Smallest positive k = 2 →
N = 120 × 2 + 5 = 240 + 5 = 245

Question: Find the number of factors of 360.

Solution:
Factorize 360 into prime factors:
360 = 2³ × 3² × 5¹
The formula for the number of factors of a number n =p^aq^br^c…n is:
Number of factors = (a + 1) (b + 1)(c + 1)
Apply the formula:
(3 + 1) (2 + 1) (1 + 1) = 4 × 3 × 2 = 24

Question: How many positive integers less than 100 are multiples of both 2 and 3?

Solution:
A number that is a multiple of both 2 and 3 is a multiple of their LCM:
LCM(2, 3) = 6

Find how many multiples of 6 are less than 100: 
Multiples of 6: 6, 12, 18, …

Largest multiple less than 100: 96
Number of terms in this sequence (arithmetic progression with first term a₁=6, common difference d = 6, last term a_n = 96):
n = (a_n - a₁)/d + 1
= (96 - 6)/6+1
= 90/6 + 1
= 15 + 1
= 16

Question: Two numbers have LCM = 180 and HCF = 6. If one number is 30, find the other.

Solution:
Let the numbers be a=30 and b=?, with HCF = 6 and LCM = 180.
Use the formula connecting LCM and HCF:
a × b = HCF × LCM
30 × b = 6 × 180
b = 1080/30 
b = 36

Question: Find HCF of 84, 126, and 210.

Solution:
Factorize each number into prime factors:
84 = 2² × 3 × 7
126 = 2 × 3² × 7
210 = 2 × 3 × 5 × 7

HCF is the product of common prime factors with smallest powers:
Common primes: 2, 3, 7
Smallest powers: 2¹, 3¹, 7¹
HCF = 2 × 3 × 7 = 42

Question: The sum of digits of a 3-digit number is divisible by 7. Which of these numbers satisfies it?

Solution:
A number is divisible by 7 in terms of digit sum if the sum of its digits is divisible by 7.
Check each number:
234 → 2 + 3 + 4 = 9 → 9/7 = 1 remainder 2 
351 → 3 + 5 + 1 = 9 → 9/7 = 1 remainder 2 
142 → 1 + 4 + 2 = 7 → 7/7 = 1 
429 → 4 + 2 + 9 = 15 → 15/7 = 2 remainder 1

142 satisfies it.

Question: The ratio of two numbers is 15 : 28 and their HCF is 17. Find the numbers.

Solution:
Let the numbers be a and b.
Since the HCF is given as 17, we can write the numbers as:
a = 17 × 15 = 255,
b = 17 × 28 = 476

Check the ratio:
a : b = 255 : 476 =15 : 28 (justified)

Question: Find the smallest number divisible by 7, 8, and 9.

Solution:
The smallest number divisible by several numbers is their LCM.
Factorize the numbers:
7 = 7
8 = 2³
9 = 3²
LCM = product of highest powers of all prime factors:
LCM = 2³ × 3² × 7
= 8 × 9 × 7
= 504

Question: Three lights blink at intervals of 18 sec, 24 sec, and 32 sec. If they blink together now, how many times will they blink together in 6 hours?

Solution:
The lights will blink together at LCM of their intervals.
Intervals: 18, 24, 32 seconds

Factorize:
18 = 2 × 3²
24 = 2³ × 3
32 = 2⁵

LCM = product of highest powers of all primes:
LCM = 2⁵ × 3² =3² × 9 = 288 seconds
Convert 6 hours to seconds:
6 hours = 6 × 60 × 60 = 21600 seconds

Number of times they blink together:
Times = 21600/288 + 1 (+1 because they blink together now, at time 0)
= 75 + 1
= 76

Note: If you only count after the first blink at time 0, it is 75 times; including the initial moment, it is 76 times.