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IBA ফ্যাকাল্টি ভিত্তিক প্রস্তুতি

পরীক্ষাIBA ফ্যাকাল্টি ভিত্তিক প্রস্তুতিতারিখতারিখ অনির্ধারিতসময়22 minutes
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পরীক্ষা - ২ বিষয়: গণিত - ১ টপিক: Number System; Problems on Number; HCF & LCM
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উত্তর
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IBA ফ্যাকাল্টি ভিত্তিক প্রস্তুতি

IBA ফ্যাকাল্টি ভিত্তিক প্রস্তুতি · তারিখ অনির্ধারিত · ২০ প্রশ্ন

.
If 1050 - 74 is written as an integer in base 10 notation, what is the sum of the digits in that integer?
  1. 433
  2. 467
  3. 424
  4. 440
  5. 449
সঠিক উত্তর:
440
উত্তর
সঠিক উত্তর:
440
ব্যাখ্যা
Question: If 1050 - 74 is written as an integer in base 10 notation, what is the sum of the digits in that integer?

Solution:
1050 has 51 digits (1 followed by 50 zeros).
1050 - 74 has 50 digits: the last 2 digits are 2 and 6 [100 - 74 = 26] and the first 48 digits are 9's
So the sum of the digits is (9 × 48) + 2 + 6 = 440
.
If x is an odd negative integer and y is an even integer, which of the following statements must be true?
  1. (3x - 2y) is odd
  2. xy2 is an even negative integer
  3. (y2 - x) is an odd negative integer
  4. All of the above
  5. None of the above
সঠিক উত্তর:
(3x - 2y) is odd
উত্তর
সঠিক উত্তর:
(3x - 2y) is odd
ব্যাখ্যা
Question: If x is an odd negative integer and y is an even integer, which of the following statements must be true?

Solution:
Let x = - 1, y = 2
Option A: (3x - 2y) = 3(- 1) - 2(2) = -7 is ODD
Option B: xy2 = (-1)(22) = - 4 is EVEN NEGATIVE
Option C. (y2 - x) = 22 - (-1) = 5 is ODD POSITIVE

Since the question involves Even and Odd numbers, let us also consider y = 0.
Option A: (3x - 2y) = 3(-1) - 2(0) = -3 is ODD
Option B: xy2 = (-1)(02) = 0 is EVEN POSITIVE
Option C. This condition was proved false using the above values.

Hence only option A is satisfied.
.
Twin primes are defined as prime numbers that can be expressed as p and (p + 2), and any number p that is a member of such a pair is considered to have a twin. For example, 3 and 5 are twin primes, and 3 has a twin. Each of the following prime numbers has a twin except
  1. 23
  2. 13
  3. 7
  4. 17
  5. 29
সঠিক উত্তর:
23
উত্তর
সঠিক উত্তর:
23
ব্যাখ্যা
Question: Twin primes are defined as prime numbers that can be expressed as p and (p + 2), and any number p that is a member of such a pair is considered to have a twin. For example, 3 and 5 are twin primes, and 3 has a twin. Each of the following prime numbers has a twin except

Solution:
ক) p = 23:
p + 2 = 23 + 2 = 25,যা মৌলিক সংখ্যা নয়।
p - 2 = 23 - 2 = 21, যা মৌলিক সংখ্যা নয়।
তাই, ২৩-এর কোনো যুগল নেই।

খ) p = 13
p + 2 = 13 + 2 = 15, যা মৌলিক সংখ্যা নয়।
p - 2 = 13 - 2 = 11, যা একটি মৌলিক সংখ্যা। 
তাই, ১৩-এর একটি যুগল রয়েছে (১১)।

গ) p = 7:
p + 2 = 7 + 2 = 9 যা মৌলিক সংখ্যা নয়।
p - 2 = 7 - 2 = 5 যা একটি মৌলিক সংখ্যা। 
তাই, ৭-এর একটি যুগল রয়েছে (৫)।

ঘ) p = 17:
p + 2 = 17 + 2 = 19, যা একটি মৌলিক সংখ্যা। 
p - 2 = 17 - 2 = 15, যা মৌলিক সংখ্যা নয়।
তাই, ১৭-এর একটি যুগল রয়েছে (১৯)।

ঙ) p = 29:
p + 2 = 29 + 2 = 31, যা একটি মৌলিক সংখ্যা। 
p - 2 = 29 - 2 = 27, যা মৌলিক সংখ্যা নয়।
তাই, ২৯-এর একটি যুগল রয়েছে (৩১)।
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Find the greatest number that will divide 43, 91, and 183 and leave the same remainder. What is the square root of this number?
  1. 2
  2. 3
  3. 6
  4. 7
  5. None of the above
সঠিক উত্তর:
2
উত্তর
সঠিক উত্তর:
2
ব্যাখ্যা
Question: Find the greatest number that will divide 43, 91, and 183 and leave the same remainder. What is the square root of this number?

Solution:
the number is the H.C.F of (91 - 43), (183 - 91) and (183 - 43)
= H.C.F of 48, 92 and 140
= 4

The square root of 4 is √4 = 2
.
Digit 1 is occurring 134 times on writing all of the page numbers of a book. What will be the number of pages in the book?
  1. 192
  2. 193
  3. 194
  4. 195
  5. 196
সঠিক উত্তর:
193
উত্তর
সঠিক উত্তর:
193
ব্যাখ্যা
Question: Digit 1 is occurring 134 times on writing all of the page numbers of a book. What will be the number of pages in the book?

Solution:
From 1 - 99, the digit 1 occurs 20 times,
from 100 - 199, the digit 1 occurs 120 times.
So, from 1 to 199, the digit 1 occurs 20 + 120 = 140 times
According to question 1 is occurring only 134 times, which means we need to remove 194, 195 196, 197, 198, and 199.
So, the required number of pages will be 193.
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A natural number n is such that 120 ≤ n ≤ 240. If HCF of n and 240 is 1, how many values of n are possible?
  1. 48
  2. 40
  3. 32
  4. 28
  5. 24
সঠিক উত্তর:
32
উত্তর
সঠিক উত্তর:
32
ব্যাখ্যা
Question: A natural number n is such that 120 ≤ n ≤ 240. If HCF of n and 240 is 1, how many values of n are possible?

Solution:
240 = 24 × 3 × 5, Therefore n must not be multiple of 2, 3 or 5

Lets calculate the number of multiples of 2,3 & 5 and subtract from total
Number of multiples of, 2 from 120 to 240 (both inclusive) = [{240 - 120}/2] + 1 = 61
3 from 120 to 240 (both inclusive) =[{240 - 120}/]3 + 1 = 41
5 from 120 to 240 (both inclusive) =[{240 - 120}/5] + 1 = 25

Multiples common to 2 & 3, 2 & 5, 3 & 5  twice. The same needs to be subtracted
6 from 120 to 240 (both inclusive) = [{240 - 120}/6] + 1 = 21
10 from 120 to 240 (both inclusive) = [{240 - 120}/10] + 1 = 13
15 from 120 to 240 (both inclusive) = [{240 - 120}/15] + 1 = 9

We need to add the multiple of 2 & 3 & 5 once as the subtraction of cases 2 & 3, 3 & 5, 2 & 5 from 2, 3 & 5 removes all the cases of 2 & 3 & 5.
30 from 120 to 240 (both inclusive) = [{240 - 120}/30] + 1 = 5

Total number of multiples of 2, 3 or 5 from 120 to 240 = 61 + 41 + 25 - 21 - 13 - 9 + 5 = 89
Possible values of n = 121 - 89 = 32
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A 107 digit number is formed by writing first 58 natural numbers next to each other. Find the remainder when number is divided by 8.
  1. 4
  2. 6
  3. 7
  4. 9
  5. 10
সঠিক উত্তর:
6
উত্তর
সঠিক উত্তর:
6
ব্যাখ্যা
Question: A 107 digit number is formed by writing first 58 natural numbers next to each other. Find the remainder when number is divided by 8.

Solution:
Given that the 107-digit number is formed by writing the first 58 natural numbers.
The last few natural numbers are 56, 57, 58.
So, the last three digits of the number are: 56,57,58 which form the number 5758
Thus, the last three digits of the number are 758.
Now, 758 is divided by 8: 758 ÷ 8 = 94 remainder 6
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A merchant has three different types of milk- 324 litres, 351 litres and 459 litres. Find the minimum number of casks of equal size that can store the milk without mixing.
  1. 61
  2. 45
  3. 51
  4. 42
  5. None of the above
সঠিক উত্তর:
42
উত্তর
সঠিক উত্তর:
42
ব্যাখ্যা
Question: A merchant has three different types of milk- 324 litres, 351 litres and 459 litres. Find the minimum number of casks of equal size that can store the milk without mixing.

Solution:
Since a minimum number of casks are required, the size of the cask is greatest.
Also the cask in three cases is of equal size. The size of the cask is the H.C.F. of 324 litres, 351 litres and 459 litres which is 27.

Now, the number of casks required for storing the milk = (324 + 351+ 459)/27 = 42.
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A gardener planted trees in rows and columns such that number of rows is five more than number of columns. If the total number of rows and column is 105, find the number of trees.
  1. 2230
  2. 2460
  3. 2520
  4. 2680
  5. 2750
সঠিক উত্তর:
2750
উত্তর
সঠিক উত্তর:
2750
ব্যাখ্যা
Question: A gardener planted trees in rows and columns such that number of rows is five more than number of columns. If the total number of rows and column is 105, find the number of trees.

Solution:
Let, the number of columns = x.
Number of rows = x + 5.
ATQ,
x + x + 5 = 105
⇒ 2x + 5 = 105
⇒ 2x = 100
∴ x = 50.

So, the number of columns = 50.
Hence, the number of rows = 50 + 5 = 55.
Hence, the number of trees = 55 × 50 = 2750.
১০.
The traffic lights at three different road crossings change after every 24 sec., 36 sec. and 72 sec. respectively. If they all change simultaneously at 8 : 20 : 00 hrs; then they will again change simultaneously at
  1. 8 : 21 : 12 hrs
  2. 8 : 27 : 48 hrs
  3. 8 : 28 : 22 hrs
  4. 8 : 27 : 36 hrs
  5. None of these
সঠিক উত্তর:
8 : 21 : 12 hrs
উত্তর
সঠিক উত্তর:
8 : 21 : 12 hrs
ব্যাখ্যা
Question: The traffic lights at three different road crossings change after every 24 sec., 36 sec. and 72 sec. respectively. If they all change simultaneously at 8 : 20 : 00 hrs; then they will again change simultaneously at

Solution:
Interval of change = (L.C.M. of 24, 36, 72) sec. = 72 sec. So, the lights will change after every 72 seconds, i.e. 1 min. 12 sec.
So, the next simultaneous change will take place at 8 : 21 : 12 hrs.
১১.
Three times the first of three consecutive odd integers is 3 more than twice the third. The third integer is:
  1. 15
  2. 29
  3. 36
  4. 12
  5. None of the above
সঠিক উত্তর:
15
উত্তর
সঠিক উত্তর:
15
ব্যাখ্যা
Question: Three times the first of three consecutive odd integers is 3 more than twice the third. The third integer is:

Solution:
Let the three odd integers be x, x + 2 and x + 4.

Then,
3x = 2(x + 4) + 3
⇒ 3x = 2x + 8 + 3
∴ x = 11

Third integer = x + 4 = 11 + 4 = 15
১২.
Which of the following is the least number which will leave the remainder 5, when divided by 8, 12, 16, and 20?
  1. 235
  2. 245
  3. 255
  4. 265
  5. None of the above
সঠিক উত্তর:
245
উত্তর
সঠিক উত্তর:
245
ব্যাখ্যা
Question: Which of the following is the least number which will leave the remainder 5, when divided by 8, 12, 16, and 20?

Solution:
First we need to find the least number, so we have to find out the LCM of 8, 12, 16, and 20.
8 = 2 x 2 x 2
12 = 2 x 2 x 3
16 = 2 x 2 x 2 x 2
20 = 2 x 2 x 5

LCM = 2 x 2 x 2 x 2 x 3 x 5 = 240

240 is the least number that is exactly divisible by 8, 12, 16, and 20.

So, the required number that will leave remainder 5 is -
240 + 5 = 245
১৩.
Given n = 1 + x and x is the product of four consecutive integers. Then which of the following us true?
I. n is an odd integer.
II. n is prime.
III. n is a perfect square
  1. Only I is correct
  2. Only II is correct
  3. Only III is correct
  4. Both I and II are correct
  5. Both I and III are correct
সঠিক উত্তর:
Both I and III are correct
উত্তর
সঠিক উত্তর:
Both I and III are correct
ব্যাখ্যা
Question: Given n = 1 + x and x is the product of four consecutive integers. Then which of the following us true?
I. n is an odd integer.
II. n is prime.
III. n is a perfect square

Solution:
Out of four consecutive integers, two are even and therefore, their product is even, and on adding 1 to it, we get an odd integer. So, n is odd. Some possible values of n are as follows:
n = 1 + (1 × 2 × 3 × 4) = (1 + 24) = 25 = 52
n = 1 + (2 × 3 × 4 × 5) = (1 + 120) = 121 = 112
n = 1 + (3 × 4 × 5 × 6) = (1 + 360) = 361 = 192
n = 1 + (4 × 5 × 6 × 7) = (1 + 840) = 841 = 292
And so on.....
Hence, n is odd and a perfect square.
১৪.
The product of two numbers is 4107. If the L.C.M of these numbers is 111, then the H.C.F is:
  1. 28
  2. 37
  3. 49
  4. 43
  5. None of the above
সঠিক উত্তর:
37
উত্তর
সঠিক উত্তর:
37
ব্যাখ্যা
Question: The product of two numbers is 4107. If the L.C.M of these numbers is 111, then the H.C.F is:

Solution:
product of two numbers = H.C.F × L.C.M
⇒ 4107 = 111 × H.C.F
⇒ H.C.F = 4107/111
= 37

Therefore, the H.C.F is = 37
১৫.
The difference between a two-digit number and the number obtained by interchanging the digits is 36. What is the difference between the sum and the difference of the digits of the number if the ratio between the digits of the number is 2 : 1?
  1. 8
  2. 6
  3. 9
  4. 4
  5. 5
সঠিক উত্তর:
8
উত্তর
সঠিক উত্তর:
8
ব্যাখ্যা
Question: The difference between a two-digit number and the number obtained by interchanging the digits is 36. What is the difference between the sum and the difference of the digits of the number if the ratio between the digits of the number is 2 : 1?

Solution:
Let the two digit number be 10x + y
And the number obtained after interchanging be 10y + x
Difference = 9(x - y) = 36
⇒ x - y = 4
Possible combinations are (5, 1) (6, 2) (7, 3) (8, 4) (9, 5)
Also, given that the ratio of the digits is 2 : 1
Only combination possible is (8, 4)
Sum of the digits = 8 + 4 = 12
Difference of the digits = 8 - 4 = 4
Difference between these two is 12 - 4 = 8
১৬.
The H.C.F of 24 × 32 × 53 × 7, 23 × 33 × 52 × 72 and 3 × 5 × 7 × 11 is = ?
  1. 95
  2. 115
  3. 120
  4. 105
  5. None of the above
সঠিক উত্তর:
105
উত্তর
সঠিক উত্তর:
105
ব্যাখ্যা
Question: The H.C.F of 24 × 32 × 53 × 7, 23 × 33 × 52 × 72 and 3 × 5 × 7 × 11 is = ?

Solution:
H.C.F. = Product of lowest powers of common factors
= 3 × 5 × 7
= 105
১৭.
A student multiplied 765 by a certain number and got 448835 as their answer. If in the answer both 8 is wrong, but the other digits are correct, then what will be the correct?
  1. 446435
  2. 445935
  3. 444635
  4. 442935
  5. None of the above
সঠিক উত্তর:
442935
উত্তর
সঠিক উত্তর:
442935
ব্যাখ্যা
Question: A student multiplied 765 by a certain number and got 448835 as their answer. If in the answer both 8 is wrong, but the other digits are correct, then what will be the correct?

Solution:
The answer is divisible by 765.
So we can use the hit and trial method to find out the number divisible by 765 from the given choices.
446435/765 gives a remainder not equal to 0
445935/765 gives a remainder not equal to 0
444635/765 gives a remainder not equal to 0
But 442935/765 gives 0 as a remainder (equals 579). Hence this is the answer.
১৮.
What is the least number which when divided by the number 3, 5, 6, 8, 10 and 12 leaves in each case a remainder 2 but when divided by 22 leaves no remainder?
  1. 188
  2. 224
  3. 234
  4. 246
  5. None of the above
সঠিক উত্তর:
None of the above
উত্তর
সঠিক উত্তর:
None of the above
ব্যাখ্যা
Question: What is the least number which when divided by the number 3, 5, 6, 8, 10 and 12 leaves in each case a remainder 2 but when divided by 22 leaves no remainder?

Solution:
LCM of 3, 5, 6, 8, 10, 12 = 3 × 5 × 2 × 4
= 120

Required number is
(120K + 2)/22 = (10K + 2)/22
at K = 2; (10K +2)/22 ⇒ Remainder = 0

The given condition satisfied = 120K + 2
= (120 × 2) + 2
= 242
১৯.
What least digit should come in place of # in the 9 digit number 15549#325, for which the number is divisible by 3?
  1. 1
  2. 5
  3. 2
  4. 4
  5. Both B & C
সঠিক উত্তর:
2
উত্তর
সঠিক উত্তর:
2
ব্যাখ্যা
Question: What least digit should come in place of # in the 9 digit number 15549#325, for which the number is divisible by 3?

Solution:
A number if divisible by 3 if sum of digits is multiple of 3. In 15549#325,
Sum of the digits= 1 + 5 + 5 + 4 + 9 + # + 3 + 2 + 5 = 34 + # = 34

Now,
34 + 1 = 35 which is not divisible by 3.
34 + 2 = 36 which is divisible by 3.
২০.
A wholesale coffee dealer has 24 kilograms, 120 kilograms and 72 kilograms of three different qualities of coffee. He wants it all to be packed into various boxes of equal size without mixing. Find the capacity of the largest possible box.
  1. 50
  2. 36
  3. 24
  4. 45
  5. None of the above
সঠিক উত্তর:
24
উত্তর
সঠিক উত্তর:
24
ব্যাখ্যা
Question: A wholesale coffee dealer has 24 kilograms, 120 kilograms and 72 kilograms of three different qualities of coffee. He wants it all to be packed into various boxes of equal size without mixing. Find the capacity of the largest possible box.

Solution:
The capacity of the box is H.C.F. of 24, 120 and 72
H.C.F. of 24, 120 and 72 = 24.