পেট্রোবাংলা প্রিলি ও লিখিত সমন্বিত প্রস্তুতি [Archived]

Question: x, y are positive integers. When x is divided by y, the remainder is 5. If x/y = 5.20, what is the value of x?

Solution: 
দেওয়া আছে 
x/y = 5.20
⇒ x/y = 520/100
∴ x/y = 26/5

এখানে 5 দিয়ে 26  ভাগ করলে ভাগশেষ 1 থাকে 
কিন্তু বলা আছে ভাগশেষ 5 থাকবে 
তাই 
26/5 এর লব ও হরের সাথে 5 গুণ করতে হবে। 
x/y = 26/5 = (26 × 5)/(5 × 5) = 130/25
130 কে 25 দ্বারা ভাগ করলে 5 ভাগশেষ থাকে। 
∴ x এর মান = 130

Question: HCF and LCM of two fractions is 1/35 and 15/1, if one fraction is 3/5, then the second fraction is:

Solution:
Given,
HCF of two fractions = 1/35
LCM of two fractions = 15/1
One fraction = 3/5

HCF of fractions = HCF of numerators/LCM of denominators
LCM of fractions = LCM of numerators/HCF of denominators
LCM × HCF = Product of two fractions

Let second fraction be a/b,
then
a/b × 3/5 = 15/1 × 1/35
⇒ a/b = 15/1 × 1/35 × 5/3
∴ a/b = 5/7

Question: Which of the following numbers will completely divide 7¹⁰ + 7¹¹ + 7¹² + 7¹³?

Solution:
Factors of a number refers to those values that can exactly divide the original number without leaving a remainder. 

7¹⁰ + 7¹¹ + 7¹² + 7¹³ 
= (1 + 7 + 7² + 7³) 7¹⁰
= (1 + 7 + 49 + 343) 7¹⁰
= 400 × 7¹⁰
= 2⁴ × 5² × 7¹⁰
So, the factors are 2, 4, 5, 7, 8, 10 etc.
So, out of given options, required factor = 2 × 7 = 14
∴ 14 will completely divide 7¹⁰ + 7¹¹ + 7¹² + 7¹³.

Question: Four bells ringing together and ring at an interval of 12 sec, 15 sec, 20 sec, and 30 sec respectively. How many times will they ring together in 8 hours?

Solution:
Four bells ringing timing is 12 sec, 15 sec, 20 sec,30 sec
Now we have to take LCM of time interval ⇒ LCM of (12, 15, 20, 30) = 60
Total seconds in 8 hours = 8 × 3600 = 28800

Number of times bell rings = 28800/60
⇒ Number of times bell rings = 480

If four bells ring together in starting ⇒ 480 + 1
∴ The bell ringing 481 times in 8 hours.

Mistake Points: The bells start tolling together, the first toll also needs to be counted, that is the number of times of tolling since the first time.

Question: If the 5-digit number 676xy is divisible by 3, 7 and 11, then what is the value of (3x - 5y)?

Solution:
When 676xy is divisible by 3, 7 &11, it will also be divisible by the LCM of 3, 7 &11.
Dividend = Divisor × Quotient + Remainder 

LCM (3, 7, 11) = 231
By taking the largest 5-digit number 67699 and divide it by 231.
∵ 67699 = 231 × 293 + 16
⇒ 67699 = 67683 + 16
⇒ 67699 - 16 = 67683 (completely divisible by 231)

∴ 67683 = 676xy (where x = 8, y = 3)

(3x - 5y)
= 3 × 8 - 5 × 3
= 24 - 15
= 9
∴ The required result = 9

Question: What is the HCF of 408 and 1032?

Solution:
The prime factorisation of 408 is 2 × 2 × 2 × 3 × 17

The prime factorisation of 1032 is 2 × 2 × 2 × 3 × 43

Thus, the product of the prime factors of 2 × 2 × 2 × 3 is 24

Hence, the HCF of 408 and 1032 is 24.

Question: Find the sum of the factors of 3240.

Solution:
If k = a^x × b^y, then
a, and b must be prime number
Sum of all factors = (a⁰ + a¹ + a² + ….. + a^x) (b⁰ + b¹ + b² + ….. + b^y)

3240 = 2³ × 3⁴ × 5¹
Sum of factors = (2⁰ + 2¹ + 2² + 2³) (3⁰ + 3¹ + 3² + 3³ + 3⁴) (5⁰ + 5¹) 
= (1 + 2 + 4 + 8) (1 + 3 + 9 + 27 + 81) (1 + 5) 
= 15 × 121 × 6 
= 10890

∴ required sum is 10890

Question: (12)³ × 6⁴ ÷ 432 =?

Solution:
(12)³ × 6⁴ ÷ 432
= {(12)³ × 6⁴}/432
= {(12)³ × 6⁴}/(12 × 6²)
= (12)² × 6²
= (72)²
= 5184

Question: Which one of the following is not a prime number?

Solution:
91 is divisible by 7.
So, it is not a prime number.

• ১ এর চেয়ে বড় যে সকল সংখ্যাকে শুধু ১ এবং ঐ সংখ্যা ছাড়া আর কোনো সংখ্যা দ্বারা ভাগ করা যায় না, তাদেরকে মৌলিক সংখ্যা বলে। 
অর্থাৎ মৌলিক সংখ্যার উৎপাদক হবে দুইটি ১ এবং শুধুমাত্র সেই সংখ্যাটি।

Question: Find the LCM of the fractions 5/8, 3/4, 1/2.

Solution:
The fractions are 5/8, 3/4 and 1/2 
LCM of the fraction = LCM of numerator/HCF of denominator

LCM of numerators = LCM of (5, 3 , 1 = 5 × 3 × 1 = 15
HCF of denominators = HCF of ( 8, 4, 2) = 2

∴ LCM of fraction = 15/2

Question: Find which of the following are twin Primes.

Solution:
- A twin prime is a prime number that is either 2 less or 2 more than another prime number.
- The difference between the twin prime number is always two.
- In twin prime number, both the number should be the prime number.
- Twin primes are pairs of successive primes that differ by two.

The primes from 1 to 100 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97

Options:
(37, 41) - Difference between them is 4.
(3, 7) - The difference between them is 4.
(43, 47) - Difference between them is 4.
(71, 73) - Difference between them is 2.

Here, in the given option (71 and 73) are prime numbers and their difference is '2'.

Question: What least number must be added to 1056, so that the sum is completely divisible by 23?

Solution:
23 ) 1056 ( 45
         92
     _______
         136
         115
     _______
           21

∴ Required number = (23 - 21) = 2

Question: Find the least number which is exactly divisible by 12, 15, and 20.

Solution:
Least number = L.C.M. of 12, 15, and 20 = 60
Hence, the required least number = 60

Question: If a number is in the form of 8¹⁰ × 9⁷ × 7⁸, find the total number of prime factors of the given number.

Solution:
If a number of the form x^a × y^b × z^c ...... and so on,
then total prime factors = a + b + c ..... and so on Where x, y, z, ... are prime numbers

The number 8¹⁰ × 9⁷ × 7⁸ can be written as (2³)¹⁰ × (3²)⁷ × 7⁸
= 2³⁰ × 3¹⁴ × 7⁸

Total number of prime factors = 30 + 14 + 8 = 52

∴ The total number of prime factors are 52

Question: The largest 4 digit number exactly divisible by 88 is-

Solution:
88 ) 9999 ( 113
       88
     _______
       119
         88
      _______
          319
          264
       _______
            55

∴ Required number = (9999 - 55) = 9944

Question: H.C.F. of two numbers is 13. If these two numbers are in the ratio of 15: 11, then find the numbers.

Solution:
H.C.F. of two numbers = 13
The numbers are in the ratio of 15 : 11

Let the two numbers be 15y and 11y
H.C.F. is the product of common factors
Therefore, H.C.F. is y.
So y = 13

The two numbers are:
15y = 15 × 13 = 195
11y = 11 × 13 = 143

We can cross-check the answer using the trick. (Product of two numbers = Product of their H.C.F. and L.C.M.)
Product of H.C.F. and L.C.M. = 13 × 2145 = 27885
Product of two numbers = 195 × 143 = 27885
Hence, the calculated answer is correct.

Question: The difference between the local value and the face value of 7 in the numeral 32675149 is-

Solution:
32675149

(Local value of 7) - (Face value of 7)
= 70000 - 7
= 69993

Question: The difference of two numbers is 1365. On dividing the larger number by the smaller, we get 6 as quotient and the 15 as remainder. What is the smaller number?

Solution:
Let the smaller number be x.
Then larger number = (x + 1365).

ATQ,
x + 1365 = 6x + 15
⇒ 5x = 1350
∴ x = 270

∴ Smaller number = 270.

Question: The difference between a positive proper fraction and its reciprocal is 9/20. The fraction is-

Solution:
Let the required fraction be x.
Then,
1/x - x =9/20
⇒ (1 - x²)/x = 9/20
⇒ 20 - 20x² = 9x
⇒ 20x² + 9x - 20 = 0
⇒ 20x² + 25x - 16x - 20 = 0
⇒ 5x(4x + 5) - 4(4x + 5) = 0
⇒ (4x + 5)(5x - 4) = 0
4x + 5 = 0   or  5x - 4 = 0
⇒ x = - 5/4  or x = 4/5  [x = - 5/4 not acceptable]
∴ x = 4/5

Question: Find the greatest number, which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively.

Solution:
The number on dividing 1657 and 2037 leaves remainders 6 and 5 respectively.
Hence, make the dividend completely divisible by the divisor. This is possible, if we subtract remainder from the dividend.
Therefore,
1657 - 6 = 1651
2037 - 5 = 2032

H.C.F. of 1651 and 2032 is 127. 127 is the common factor.
127 × 13 = 1651
Thus by adding 6, we get 1651 + 6 = 1657
127 is the correct answer.

Question: How many multiples of both 3 or 4 are there from 1 to 100 in total?

Solution:
On dividing 100 by 3 we get a quotient of 33 The number of multiple of 3, n(A) = 33
On dividing 100 by 4 we get a quotient of 25 The number of multiple of 4, n(B) = 25
LCM of 3 and 4 is 12
On dividing 100 by 12 we get a quotient of 8 The number of multiple of 12, n(A∩B) = 8

The number which is multiple of 3 or 4 = n(A∪B)
Now,
n(A∪B) = n(A) + n(B) - n(A∩B) 
= 33 + 25 - 8 
= 50

∴ The total number multiple of 3 or 4 from 1 to 100 is 50

Question: If n is a natural number, then (6n² + 6n) is always divisible by-

Solution:
(6n² + 6n)
= 6n(n + 1), which is always divisible by 6 and 12 both, since n(n + 1) is always even.

Question: The traffic lights at three different road crossings change after every 40 sec, 72 sec and 108 sec respectively. If they all change simultaneously at 5 : 20 hours, then find the time at which they will change simultaneously.

Solution:
Traffic lights at three different road crossings change after every 40 sec, 72 sec and 108 sec respectively.
Therefore, find the L.C.M. of 40, 72 and 108.
L.C.M. of 40, 72 and 108 = 1080
The traffic lights will change again after 1080 seconds = 18 min
The next simultaneous change takes place at 5 : 38 hrs.

Question: The sum of first five prime numbers is -

Solution:
Required sum = (2 + 3 + 5 + 7 + 11)
= 28.

Question: Which of the following numbers is a divisor of (49¹⁵ - 1)?

Solution:
aⁿ - bⁿ is divisible by (a + b) when n is an even positive integer.
Here, a & b should be prime number.

(49¹⁵ - 1)
⇒ {(7²)¹⁵ - 1)
⇒ (7³⁰ - 1)
Here, 30 is a positive integer.
According to the concept, (7³⁰ - 1) is divisible by (7 + 1) i.e., 8.
∴ 8 is a divisor of (49¹⁵ - 1). 

Question: John, Smith and Kate start at same time, same point and in same direction to run around a circular ground. John completes a round in 250 seconds, Smith in 300 seconds and Kate in 150 seconds. Find after what time will they meet again at the starting point?

Solution:
L.C.M. of 250, 300 and 150 = 1500 sec
Dividing 1500 by 60 we get 25, which mean 25 minutes.
John, Smith and Kate meet after 25 minutes.