Number System, Problems on Number

PrepBank · পাতা ৬ / ১৮ · ৫০১–৬০০ / ১,৭৩৬

৫০১.

If one-third of one-fourth of a number is 15, then five-ninths of that number is-

ক
86
খ
90
গ
100
ঘ
106

ব্যাখ্যা

Question: If one-third of one-fourth of a number is 15, then five-ninths of that number is-

Solution:
Let, the number be = a

Now
(1/3) × (1/4) × a = 15
⇒ a/12 = 15
⇒ a= 12 × 15
∴ a = 180

So, five-ninths of that number will be = (5a/9)
= (5 × 180)/9
= 5 × 20
= 100

৫০২.

A sum of Tk. 312 was divided among 100 boys and girls in such a way that each boys gets Tk. 3.6 and each girl Tk. 2.4. The number of girls is :

ক
ক) 35
খ
খ) 40
গ
গ) 55
ঘ
ঘ) 50

ব্যাখ্যা

ধরি,
বালিকার সংখ্যা =x জন
বালকের সংখ্যা = (100 - x) জন

1 জন বালক পায় 3.60 টাকা
(100-x) জন বালক পায় =3.60 (100 - x) টাকা

1 জন বালিকা পায় = 2.40 টাকা
x জন বালিকা পায় = 2.40x টাকা

প্রশ্নমতে,
3.60(100 - x) + 2.40x=312
360 - 3.60x + 2.40x=312
360 - 1.2x = 312
360 - 312 = 1.2x
1.2x = 48
x = 48/1.2
x = 40

৫০৩.

Find the difference between 5/8 of Tk. 4 and 4/5 of Tk. 2?

ক
ক) Tk. 0.90
খ
খ) Tk. 1.60
গ
গ) Tk. 0.09
ঘ
ঘ) Tk. 1

ব্যাখ্যা

Question: Find the difference between 5/8 of Tk. 4 and 4/5 of Tk. 2?

Solution:
5/8 of Tk. 4 = (4 × 5)/8 = 2.5 Tk.
4/5 of Tk. 2 = (2 × 4)/5 = 1.6 Tk.

∴ Difference = (2.5 - 1.6) = Tk. 0.90

৫০৪.

Which of the following fractions is the largest?

ক
ক) 63/80
খ
খ) 13/16
গ
গ) 31/40
ঘ
ঘ) 7/8

ব্যাখ্যা

63/80 =
13/16 = 65/80
31/40 = 62/80
7/8 = 70/80

সবগুলো হরকে 80 তে রূপান্তর করার পর দেখা যাচ্ছে সবচেয়ে বড় লব হচ্ছে 7/8 এর, তাই এটিই সবচেয়ে বড় সংখ্যা

৫০৫.

If 2^x + 2^y = 2²², what is the value of x + y?

ক
39
খ
40
গ
41
ঘ
42
ঙ
None

ব্যাখ্যা

Question: If 2^x + 2^y = 2²², what is the value of x + y?

Solution:
Here,,
2^x + 2^y=2²²
⇒ 2^x(1 + 2^{y - x})=2²²

2^x must be a power of 2 that divides 2²²
(1 + 2y - x) must be a power of 2
Only way (1 + 2y - x) is a power of 2 is if, y - x = 0

So, y - x = 0
⇒ y = x

∴ 2^x + 2^x = 2²²
⇒ 2.2^x = 2²²
⇒ 2^{1 + x}=2²²
⇒ 1 + x = 22
∴ x = 21
∴ x + y = 21 + 21 = 42

৫০৬.

The average of a non-zero number and its square is 5 times the number. The number is-

ক
9
খ
17
গ
29
ঘ
none of these

ব্যাখ্যা

Question: The average of a non-zero number and its square is 5 times the number. The number is-

Solution:
Let the number be x (x ≠ 0).

According to the question,
The average of the number and its square is 5 times the number.
⇒ (x + x²)/2 = 5x
⇒ x + x² = 10x
⇒ x² + x - 10x = 0
⇒ x² - 9x = 0
⇒ x(x - 9) = 0
So, x = 0 or x = 9
But the number is non-zero, so we discard x = 0.

Therefore, the number is 9.

৫০৭.

Find the three consecutive odd numbers whose sum of the squares is 2531.

ক
19, 21, 23
খ
23, 25, 27
গ
27, 29, 31
ঘ
31, 33, 35

ব্যাখ্যা

Question: Find the three consecutive odd numbers whose sum of the squares is 2531.

Solution:
Let three consecutive odd numbers be x, x + 2, x + 4.
x² + (x + 2)² + (x + 4)² = 2531
Simplifying we get,
⇒ x² + 4x - 837=0
⇒ x² + 31x - 27x - 837=0 [ 27 × 31 = 837 and also the difference between 27 and 31 is 4 ]
⇒ (x + 31) (x - 27)
⇒ x = 27 or x = - 31
Hence, the value of
x = 27
(x + 2) = 27 + 2 = 29
(x + 4) = 27 + 4 = 31

৫০৮.

In a class, 30 students study Mathematics, 20 students study Physics, and 8 students study both. 12 students study neither Mathematics nor Physics. What is the total number of students in the class?

ক
45
খ
50
গ
54
ঘ
60

ব্যাখ্যা

Question: In a class, 30 students study Mathematics, 20 students study Physics, and 8 students study both. 12 students study neither Mathematics nor Physics. What is the total number of students in the class?

Solution:
Number of students who study Mathematics, n(M) = 30
Number of students who study Physics, n(P) = 20
Number of students who study both Mathematics and Physics, n(M ∩ P) = 8
Number of students who study neither = 12

n(M ∪ P) = n(M) + n(P) - n(M ∩ P)
= 30 + 20 - 8 = 42

Total students in the class = students who study Mathematics or Physics + students who study neither
= 42 + 12 = 54

∴ There are 54 students in the class.

৫০৯.

Which of the following numbers is a prime number?

ক
167
খ
213
গ
350
ঘ
437

ব্যাখ্যা

Question: Which of the following numbers is a prime number?

Solution:
Step 1: Find a whole number 'X' for each number such that X² > the number;
13² > 167
15² > 213
19² > 350
21² > 437

Step 2: Get all the prime numbers which are less than 'X'.
Prime numbers less than 13 are 2, 3, 5, 7, and 11
Prime numbers less than 15 are 2, 3, 5, 7, 11, and 13
Prime numbers less than 19 are 2, 3, 5, 7, 11, 13, and 17
Prime numbers less than 21 are 2, 3, 5, 7, 11, 13, 17, and 19

Step 3: Check divisibility of each number with prime numbers which are less than 'X'.
167 is not divisible by any prime number
213 is divisible by 3
352 is divisible by 2 and 11
437 is divisible by 19

∴ 167 is the required prime number as it is not divisible by any prime number.

৫১০.

What must be added to x/y to make 2y/x?

ক
2x²/y
খ
(xy - y²)/2
গ
(2y² - x²)/xy
ঘ
x/y

ব্যাখ্যা

Question: What must be added to x/y to make 2y/x?

Solution:

৫১১.

A number when multiplied by 16 increases by 540. What is the number?

ক
30
খ
36
গ
42
ঘ
46

ব্যাখ্যা

Question: A number when multiplied by 16 increases by 540. What is the number?

Solution:
Let the number is x.

As per question;
16x - x = 540
⇒ 15x = 540
∴ x = 36

৫১২.

Half of 1 percent written as a decimal is ________

ক
ক) 0.2
খ
খ) 0.05
গ
গ) 0.02
ঘ
ঘ) 0.005

ব্যাখ্যা

Question: Half of 1 percent written as a decimal is ________

Solution:
As we know,
1% = 1/100

Hence,
(1/2)% = (1/2) × (1/100) = 1/200 = 0.005

৫১৩.

Evaluate :

ক
ক) 20
খ
খ) 16
গ
গ) 24
ঘ
ঘ) 30

ব্যাখ্যা

Question: Evaluate :

Solution:
√{248 + √(52 + √144)}
= √{248 + √(52 + 12)}
= √(248 + √64)
= √(248 + 8)
= √256
= 16

৫১৪.

If 2/3 of a number is 5 more than 1/4 of the number then 7/2 of the number is-

ক
41
খ
42
গ
43
ঘ
44

ব্যাখ্যা

Question: If 2/3 of a number is 5 more than 1/4 of the number then 7/2 of the number is-

Solution:
Let,
the number be x

According to the question,
⇒ (2x/3) - x/4 = 5
⇒ (8x - 3x)/12 = 5
⇒ 5x = 5 × 12
⇒ 5x = 60
⇒ x = 12

Then 7/2 of the number will be = x × 7/2
= (12 × 7)/2
= 42

৫১৫.

A room is 12.25m long and 7m wide. The maximum length of a square tile to fill the floor of the room with a whole number of tiles should be:

ক
110 cm
খ
175 cm
গ
157 cm
ঘ
150 cm
ঙ
170 cm

ব্যাখ্যা

Length of largest tile:

= H.C.F. of 12.25 m and 7 m

= H.C.F. of 1225 cm and 700 cm

= 175 cm

৫১৬.

Three numbers are in the ratio 3 : 4 : 5, and the sum of their squares is 1250. Find the smallest number.

ক
15
খ
18
গ
21
ঘ
25

ব্যাখ্যা

Question: Three numbers are in the ratio 3 : 4 : 5, and the sum of their squares is 1250. Find the smallest number.

Solution:
Let,
the numbers be 3x, 4x, 5x

ATQ,
(3x)² + (4x)² + (5x)² = 1250
⇒ 9x² + 16x² + 25x² = 1250
⇒ 50x² = 1250
⇒ x² = 25
∴ x = 5

∴ Smallest number = 3x
= 3 × 5
= 15

৫১৭.

What is the value of

ক
6
খ
4
গ
2
ঘ
1

ব্যাখ্যা

Question: What is the value of

Solution:

৫১৮.

The smallest number which is divided by 4, 6, 8, 12 and 16, leaving the remainder 2, is-

ক
ক) 50
খ
খ) 46
গ
গ) 48
ঘ
ঘ) 56

ব্যাখ্যা

প্রশ্ন: The smallest number which is divided by 4, 6, 8, 12 and 16, leaving the remainder 2, is

সমাধান:
4, 6, 8, 12, 16 এর ল.সা.গু = 48

∴ যেহেতু প্রতিক্ষেত্রে 2 ভাগশেষ থাকে তাই সংখ্যাটি = 48 + 2 = 50

৫১৯.

Anita had to do a multiplication. instead of taking 35 as one of the multipliers, she took 53. As a result, the product went up by 540. What is the new product?

ক
ক) 1,050
খ
খ) 1,250
গ
গ) 1,440
ঘ
ঘ) 1,590

ব্যাখ্যা

Question: Anita had to do a multiplication. instead of taking 35 as one of the multipliers, she took 53. As a result, the product went up by 540. What is the new product?

Solution:
ধরি
সংখাটি = x
প্রশ্নমতে
53x - 35x = 540
⇒18x = 540
⇒x = 540/18
⇒ x =30

নতুন গুণফল
53 × 30 = 1590

৫২০.

If m² < 225 and n - m = - 10, what is the difference between the smallest possible integer value of 3m + 2n and the greatest possible integer value of 3m + 2n?

ক
- 190
খ
- 188
গ
- 150
ঘ
- 148
ঙ
- 40

ব্যাখ্যা

Question: If m² < 225 and n - m = - 10, what is the difference between the smallest possible integer value of 3m + 2n and the greatest possible integer value of 3m + 2n?

Solution:
From n - m = - 10 it follows that n = m - 10.
Thus, 3m + 2n = 3m + 2(m - 10) = 5m - 20.
So, we need to find the difference between the smallest possible integer value of 5m - 20 and the greatest possible integer value of 5m - 20.

Now,
lets' work on m² < 225
Take the square root from both sides: |m| < 15
Get rid of the modulus sign: - 15 < m < 15
Multiply all three parts by 5: - 75 < 5m < 75
Subtract 20 from all three parts: - 95 < 5m - 20< 55

From - 95 < 5m - 20 < 55 it follows that the smallest possible integer value of 5m - 20 is - 94 and the greatest possible integer value of 5m - 20 is 54.

Therefore, the difference is: - 94 - 54 = - 148.

৫২১.

How many number of pairs in there, where the pair contains consecutive odd positive integers,both of which are smaller than 10, such that their sum is more than 11.

ক
1
খ
2
গ
3
ঘ
None of the above

ব্যাখ্যা

Question: How many number of pairs in there, where the pair contains consecutive odd positive integers,both of which are smaller than 10, such that their sum is more than 11.

Solution:
Let x the smaller of the two consecutive odd positive integers be = x
Then the other odd integer is x + 2.

It is given that both the integers are smaller than 10 and their sum is more than 11.

∴ x + 2 < 10 and, x + (x + 2) > 11
⇒ x < 10 - 2 and 2x + 2 > 11
⇒ x < 8 and 2x > 9
⇒ x < 8 and x > 9/2
⇒ 9/2 < x < 8
⇒ x = 5, 7

The required pairs of odd integers are (5, 7) and (7, 9).

৫২২.

Let x be a number. GCF of 2/5, 3/5 and 1/4 of that number is 5. find the number.

ক
100
খ
200
গ
120
ঘ
90

ব্যাখ্যা

Question: Let x be a number. GCF of 2/5, 3/5 and 1/4 of that number is 5. find the number.

Solution:
the numbers are 2x/5, 3x/5, x/4.
GCF of 2x, 3x, and x is x
and LCM of 5, 5 and 4 is 20
∴ the GCF of 2x/5, 3x/5, x/4 is x/20

ATQ,
x/20 = 5
x = 100

৫২৩.

The number of students in each section of a school is 24, After admitting new students, three new sections were started. Now, the total number of sections is 16 and there are 21 students in each section. The number of new students admitted is-

ক
ক) 24
খ
খ) 14
গ
গ) 48
ঘ
ঘ) 114
ঙ
ঙ) None of these

ব্যাখ্যা

২৪ ছাত্র নিয়ে section ১৬ - ৩ = ১৩ টি।
তাহলে ১৩ টি সেকশনের ছাত্র ১৩×২৪ = ৩১২ জন।
নতুন ছাত্র ভর্তি হওয়ার পর ২১ জন করে ১৬ সেকশনে মোট ছাত্র ১৬×২১ = ৩৩৬ জন।
∴ নতুন ভর্তি হয়েছে ৩৩৬ - ৩১২ = ২৪ জন।

৫২৪.

For all numbers x and y, x # y = xy + x. What is the value of 5 # 4?

ক
ক) 9
খ
খ) 24
গ
গ) 25
ঘ
ঘ) 36

ব্যাখ্যা

Question: For all numbers x and y, x # y = xy + x. What is the value of 5 # 4?

Solution:
Given,
x # y = xy + x.
∴ 5 # 4 = (5 × 4) + 5= 25

৫২৫.

What value will come in place of question mark in the following equations 0.006 ÷ ? = 0.6

ক
0.01
খ
0.001
গ
0.002
ঘ
0.0001

ব্যাখ্যা

Question: What value will come in place of question mark in the following equations 0.006 ÷ ? = 0.6

Solution:
Let
? = x

Now
0.006 ÷ x = 0.6
0.006/x = 0.6
x = 0.006/0.6
x = 0.01

৫২৬.

On multiplying a number by 7 all the digit in the product appear as 3's , the smallest such numbers is -

ক
48619
খ
47619
গ
47719
ঘ
47649

ব্যাখ্যা

Question: On multiplying a number by 7 all the digit in the product appear as 3's , the smallest such numbers is -

Solution:
Let's check the options one by one:
47649 × 7 = 333543; This result is not all 3's.
47719 × 7 = 333033; This result is not all 3's.
47619 × 7 = 333333; This result is all 3's!
48619 × 7 = 340333; This result is not all 3's.

৫২৭.

A gardener has to plant trees in rows containing equal number of trees. If he plants in rows of 6, 8, 10 or 12 then five trees are left unplanted. But if he plants in rows of 13 trees each, then no tree is left. What is the number of trees that the gardener plants?

ক
ক) 485
খ
খ) 845
গ
গ) 725
ঘ
ঘ) 625

ব্যাখ্যা

LCM of 6, 8, 10, 12 = 120
∴ Required number is of the from 120k + 5
Least value of k for which (120k + 5) is divisible by 13 is k = 7
∴ Required number
= (120 × 7 + 5)
= 845

৫২৮.

The sum of all prime numbers between 60 and 80 is:

ক
523
খ
272
গ
351
ঘ
414

ব্যাখ্যা

Question: The sum of all prime numbers between 60 and 80 is:

Solution:
According to the list of prime numbers, we know that—
prime numbers between 60 and 80 are:
61, 67, 71, 73, 79

Sum of these number = (61 + 67 + 71 + 73 + 79) = 351

৫২৯.

The number nearest to 320, which is exactly divisible by each of 2, 3, 5 is-

ক
ক) 305
খ
খ) 330
গ
গ) 310
ঘ
ঘ) 270

ব্যাখ্যা

প্রশ্ন: 320 এর নিকটবর্তী কোন সংখ্যাটি 2, 3, 5 দ্বারা নিঃশেষে বিভাজ্য?

সমাধান:
2, 3, 5 এর ল.সা.গু = 30
320 এর কাছাকাছি 30 দ্বারা বিভাজ্য সংখ্যা হলো 300 এবং 330.
এরমধ্যে নিকটবর্তী সংখ্যা 330

৫৩০.

Find the square of a positive number which when decreased by 17 is equal to 60 times the reciprocal of the number.

ক
169
খ
196
গ
225
ঘ
400

ব্যাখ্যা

Question: Find the square of a positive number which when decreased by 17 is equal to 60 times the reciprocal of the number.

Solution:
Let the number be x

Then, x - 17 = 60/x
⇒ x² - 17x − 60 = 0
⇒ x² - 20x + 3x − 60 = 0
⇒ (x - 20)(x + 3) = 0
∴ x = 20, - 3

The positive number = 20
Hence, the square of the positive number = 400

৫৩১.

Which number is odd one in oval A and B respectively?

ক
ক) 42,18
খ
খ) 48, 52
গ
গ) 36, 6
ঘ
ঘ) 42, 52

ব্যাখ্যা

In the first oval every number is divisible by 12 except 42 and in the second oval every number is divisible by 6 except 52.

৫৩২.

Three sets of English, Mathematics, and Science books containing 336, 240, and 96 books, respectively, have to be stacked in such a way that all the books are stored subject-wise and the height of each stack is the same. The total number of stacks will be -

ক
18
খ
14
গ
22
ঘ
24

ব্যাখ্যা

Question: Three sets of English, Mathematics, and Science books containing 336, 240, and 96 books, respectively, have to be stacked in such a way that all the books are stored subject-wise and the height of each stack is the same. The total number of stacks will be -

Solution:
প্রত্যেক তাকে বইয়ের সংখ্যা হবে ৩৩৬, ২৪০ এবং ৯৬ এর গ.সা.গু. এর সমান।
৩৩৬, ২৪০ এবং ৯৬ এর গ.সা.গু = ৪৮

মোট তাক হবে = (৩৩৬ + ২৪০ + ৯৬)/৪৮
= ১৪টি

৫৩৩.

Divide 30 by half and add 37. What do you get?

ক
97
খ
70
গ
83
ঘ
89

ব্যাখ্যা

30/0.5 + 37
= (30 ×10)/5 + 37
= 60 + 37
= 97

৫৩৪.

What least number must be subtracted from 427398 so that remaining number is divisible by 15?

ক
ক) 3
খ
খ) 5
গ
গ) 7
ঘ
ঘ) 9

ব্যাখ্যা

On dividing 427398 by 15 we get the remainder 3, so 3 should be subtracted
Answer : 3

৫৩৫.

Find the greatest number of five digits which is divisible by 15, 21 and 36:

ক
ক) 99540
খ
খ) 99650
গ
গ) 99780
ঘ
ঘ) 99430

ব্যাখ্যা

Greatest number of five digits = 99999.
Required number must be divisible by L.C.M. of 15, 21 and 36, i.e 1260
On dividing 99999 by 1260, we get 459 as a reminder.
∴ Required number = (99999 - 459) = 99540
Answer : 99540

৫৩৬.

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.

ক
2230
খ
2460
গ
2520
ঘ
2680
ঙ
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.

৫৩৭.

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

ক
ক) - 12m - 7n
খ
খ) 12m + 7n
গ
গ) 11m - 7n
ঘ
ঘ) 12m - 7n

ব্যাখ্যা

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

Solution:
- 6m - [3n - {8m - (4n - 10m)}]
= - 6m - [3n - {8m - 4n + 10m}]
= - 6m - [3n - 8m + 4n - 10m]
= -6m - 3n + 8m - 4n + 10m
= 12m - 7n

৫৩৮.

The total of three successive multiples of 3 is 117. Determine the greatest number.

ক
32
খ
36
গ
42
ঘ
45

ব্যাখ্যা

Question: The total of three successive multiples of 3 is 117. Determine the greatest number.

Solution:
Let,
First multiple: 3x
Second multiple: 3(x + 1) = 3x + 3
Third multiple: 3(x + 2) = 3x + 6

ATQ,
3x + (3x + 3) + (3x + 6) = 117
⇒ 9x + 9 = 117
⇒ 9x = 108
⇒ x = 108/9
∴ x = 12

∴ The largest number = 3x + 6
= 3 × 12 + 6
= 36 + 6
= 42

৫৩৯.

Find the number of three-digit numbers which are divisible by 6.

ক
110
খ
120
গ
130
ঘ
150

ব্যাখ্যা

Question: Find the number of three-digit numbers which are divisible by 6.

Solution:
The required three digit numbers would be 102, 108, 114, 120...990 and 996.
The sequence of numbers shows that it is an arithmetic progression, where 'a' = 102, 'd' = 6 and last number = 996
Let the number of terms = n

Applyiong the formula: n = (last term - first term)/d + 1
= (996 - 102)/6 + 1
= 894/6 + 1
= 149 + 1
= 150

৫৪০.

The HCF of two numbers, each having three digits, is 17 and their LCM is 714. The sum of the numbers will be?

ক
ক) 289
খ
খ) 391
গ
গ) 221
ঘ
ঘ) 731
ঙ
ঙ) 121

ব্যাখ্যা

HCF = 17
Let numbers are = 17x, 17y
LCM = 17xy = 714 (given)
xy = 42
Possible pairs are (1, 42), (2, 21), (3, 14), (6, 7)
Possible numbers are (17, 714), (34, 357), (51, 238), (102, 119)
but given that both numbers are of three digits
∴ numbers are = (102, 119)
∴ sum of numbers = 102 + 119 = 221

৫৪১.

If x is an integer and y = 7x + 5, which of the following CANNOT be a divisor of y?

ক
ক) 10
খ
খ) 11
গ
গ) 12
ঘ
ঘ) 14

ব্যাখ্যা

Question : If x is an integer and y = 7x + 5, which of the following CANNOT be a divisor of y?

Solution:
ধরি
x = 1, 2, 3, 4,................
x = 1 হলে, y = 7x + 5 = 7 × 1 + 5 = 12 , যা 12 দ্বারা বিভাজ্য
x = 2 হলে, y = 7x + 5 = 7 × 2 + 5 = 19 , যা 19 দ্বারা বিভাজ্য
x = 3 হলে, y = 7x + 5 = 7 × 3 + 5 =26 , যা 13 দ্বারা বিভাজ্য
x = 4 হলে, y = 7x + 5 = 7 × 4 + 5 =33 , যা 11 দ্বারা বিভাজ্য
x = 5 হলে, y = 7x + 5 = 7 × 5 + 5 =40, যা 10 দ্বারা বিভাজ্য
.....................................................................................
...................................................................................

7x + 5 সংখ্যাটি 14 দ্বারা বিভাজ্য নয়

৫৪২.

The number 3 divides 'a' with a result of 'b' and a reminder of 2. The number 3 divides 'b' with a result of 2 and 'a' reminder of 1. What is the value of 'a'?

ক
13
খ
17
গ
23
ঘ
21

ব্যাখ্যা

Question: The number 3 divides 'a' with a result of 'b' and a reminder of 2. The number 3 divides 'b' with a result of 2 and 'a' reminder of 1. What is the value of 'a'?

Solution:
আমরা জানি,
(ভাজ্য - ভাগশেষ) ÷ ভাজক = ভাগফল

১ম শর্তমতে,
(a - 2) ÷ 3 = b
বা, b = (a - 2)/3 -------------- (1)

২য় শর্তমতে,
(b - 1) ÷ 3 = 2
বা, (b - 1)/3 = 2
বা, b - 1 = 6
∴ b = 7

b এর মান (1) নং সমীকরণে বসিয়ে পাই,
(a - 2)/3 = 7
বা, a - 2 = 21
∴ a = 23

৫৪৩.

The average of 5 consecutive numbers is n. If the next two numbers are also included, the average will:

ক
increase by 1.5
খ
increase by 1
গ
remain the same
ঘ
increase by 2
ঙ
None of these

ব্যাখ্যা

Question: The average of 5 consecutive numbers is n. If the next two numbers are also included, the average will:

Solution:
The average of 5 consecutive terms is n, implies that the 3rd term is n.
Now as the next 2 terms are included implies that the new average for 7 terms would be the 4th term.
So, the 4th term would be n + 1.

Example:
(1 + 2 + 3 + 4 + 5)/5
= 15/5
= 3

(1 + 2 + 3 + 4 + 5 + 6 + 7)/7
= 28/7
= 4

∴ The average will increase by 1

৫৪৪.

How many pieces of 85 cm length stick can be cut from a 21.25 meters long stick?

ক
20
খ
25
গ
30
ঘ
35

ব্যাখ্যা

Question: How many pieces of 85 cm length stick can be cut from a 21.25 meters long stick?

Solution:
আমরা জানি,
1 মিটার= 100 সে.মি.
∴ 21.25 মিটার= (100 × 21.25) সে.মি.
= 2125 সে.মি.

টুকরার সংখ্যা হবে = 2125/85 টি
= 25 টি

৫৪৫.

7⁶ⁿ - 6⁶ⁿ, where n is an integer > 0, is not divisible by,

ক
11
খ
559
গ
13
ঘ
127

ব্যাখ্যা

Question: 7⁶ⁿ - 6⁶ⁿ, where n is an integer > 0, is not divisible by,

Solution:
7⁶ⁿ - 6⁶ⁿ
= 7⁶ - 6⁶
= (7³)² - (6³)²
= (7³ + 6³)(7³ - 6³)
= (343 - 216)(343 + 216)
= 127 × 559
= 127 × 13 × 43

Clearly, it is divisible by 127, 13, 559.

৫৪৬.

6 × 3 (3 - 1) is equal to =?

ক
ক) 19
খ
খ) 20
গ
গ) 53
ঘ
ঘ) 36

ব্যাখ্যা

Question: 6 × 3 (3 - 1) is equal to =?

Solution:
6 × 3 (3 - 1)
= 6 × 3 × 2
= 36

৫৪৭.

How many positive integers less than ten thousand are multiples of both eight and eighteen.

ক
ক) 70
খ
খ) 72
গ
গ) 138
ঘ
ঘ) 139

ব্যাখ্যা

৪ ও 18 এর ল.সা.গু = 72
∴ নির্ণেয় পূর্ণ সংখ্যা =10000/72
= 138.8 ≈ 138 টি

৫৪৮.

If a and b are non-negative real numbers such that a + 2b = 6, then the average of the maximum and minimum possible values of (a + b) is

ক
3
খ
3.5
গ
4
ঘ
4.5

ব্যাখ্যা

Question: If a and b are non-negative real numbers such that a + 2b = 6, then the average of the maximum and minimum possible values of (a + b) is

Solution:
a + 2b = 6; the maximum value of b can be 3

⇒ a + b = 6 - b

maximum value of a + b = 6 - 0 = 6
minimum value of a + b = 6 - 3 = 3

average = (6 + 3)/2 = 9/2 = 4.5

৫৪৯.

A forester wants to plant 88 coconut trees, 110 date trees and 132 plum trees in equal sized rows (in terms of number of tress). Also he wants to make distinct rows of trees (i.e only one type of trees in one row). What is the minimum number of rows required?

ক
ক) 12
খ
খ) 13
গ
গ) 15
ঘ
ঘ) 18

ব্যাখ্যা

Question: A forester wants to plant 88 coconut trees, 110 date trees and 132 plum trees in equal sized rows (in terms of number of tress). Also he wants to make distinct rows of trees (i.e only one type of trees in one row). What is the minimum number of rows required?

Solution:
এখানে
88, 110 এবং 132 এর গ.সা.গু = 22

নারিকেল গাছের সারি লাগবে = 88/22 = 4টি
খেজুর গাছের সারি লাগবে = 110/22 = 5টি
পাম গাছের সারি লাগবে = 132/22 = 6টি

মোট সারি লাগবে = 4 + 5 + 6 = 15 টি

৫৫০.

When the positive integer n is divided by 5, the remainder is 2. Which of the following must be true?

ক
n is odd
খ
n is even
গ
n - 1 is divisible by 3
ঘ
None

ব্যাখ্যা

Question: When the positive integer n is divided by 5, the remainder is 2. Which of the following must be true?

Solution:
Given that when the positive integer n is divided by 5, the remainder is 2, we can express n as:
n = 5k + 2 where k is an integer.

Option ক)
n is odd
Since n = 5k + 2 the value of n depends on k. If k is even, n is even; if k is odd, n is odd.
Therefore, n is not necessarily odd.

Option খ)
n is even
Similarly, as discussed above, n = 5k + 2 could be odd or even depending on the value of k. If k is even, n is even, but if k is odd, n is odd.
Therefore, n is not necessarily even.

Option গ)
n - 1 is divisible by 3
Let's calculate n - 1:
n - 1 = 5k + 2 - 1 = 5k + 1
We need to check if 5k + 1 is divisible by 3.
We know that any integer 5k leaves a remainder of 2 when divided by 3
∴ For n - 1 to be divisible by 3, 5k + 1 mod 3 must equal 0. This can occur depending on k, but it's not necessarily true for all k.

None of the provided options must be true in all cases. Therefore, none of the options is necessarily true based on the information given.

৫৫১.

The difference between two numbers is 1365. When larger number is divided by the smaller one, the quotient is 6 and the remainder is 15. The smaller number is?

ক
200
খ
220
গ
250
ঘ
270

ব্যাখ্যা

Question: The difference between two numbers is 1365. When larger number is divided by the smaller one, the quotient is 6 and the remainder is 15. The smaller number is?

Solution:
let, larger number be x
smaller number be y

x - y = 1365
x = y + 1365

x = 6y + 15
⇒ y + 1365 = 6y + 15
⇒ 6y - y = 1365 - 15 = 1350
⇒ y = 1350/5 = 270

৫৫২.

If 'x' is an odd number and 'y' is an even number, which one of the following must be an even number?

ক
ক) xy + 1
খ
খ) x² + y²
গ
গ) (x + y)² + 1
ঘ
ঘ) x + y

ব্যাখ্যা

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

Solution:
Suppose, x = 1 and y = 2

• xy + 1 = 1 × 2 + 1 = 3
• x² + y²= 1² + 2² = 5
• (x + y)² + 1 = (1 + 2)² + 1 = 10
• x + y = 1 + 2 = 3

৫৫৩.

Find the smallest number of 6 digits which is exactly divisible by 349.

ক
101063
খ
100163
গ
160063
ঘ
None of the above

ব্যাখ্যা

Question: Find the smallest number of 6 digits which is exactly divisible by 349.

Solution:
The smallest 6 digit number is 100000.
When divided by 349, the remainder is 186.
So, Required Number = 100000 + (349 - 186) = 100163

৫৫৪.

(12)³ × 6⁴ ÷ 432 = ?

ক
5184
খ
5060
গ
5148
ঘ
5084

ব্যাখ্যা

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

৫৫৫.

What will be the least number which when doubled will be exactly divisible by 12, 18, 21 and 30?

ক
ক) 630
খ
খ) 1260
গ
গ) 1920
ঘ
ঘ) 320

ব্যাখ্যা

Question: What will be the least number which when doubled will be exactly divisible by 12, 18, 21 and 30?

Solution:
12 = 2 × 2 × 3
18 = 2 × 3 × 3
21 = 3 × 7
30 = 2 × 3 × 5

L.C.M. of 12, 18, 21 30 = 2 × 3 × 2 × 3 × 7 × 5 = 1260
Required number = 1260/2 = 630

৫৫৬.

If the 5^th number in a series of a 5 consecutive integers has the value n + 3, find the 1^st number in the series expressed in terms of n?

ক
0
খ
n - 3
গ
- 3n
ঘ
n - 1
ঙ
1

ব্যাখ্যা

Question: If the 5^th number in a series of a 5 consecutive integers has the value n + 3, find the 1^st number in the series expressed in terms of n?

Solution:
Given that,
A series of 5 consecutive integers.
Let the first number be x
Then the series is,
x, (x + 1), (x + 2), (x + 3), (x + 4)
And
The 5^th number in the series is given as n + 3
So,
⇒ x + 4 = n + 3
⇒ x = n + 3 - 4
∴ x = n - 1

So the first number in terms of nnn is n - 1.

৫৫৭.

If 4 times of a number is subtracted from its 13 times, the result is 171. What is the number?

ক
ক) 17
খ
খ) 19
গ
গ) 29
ঘ
ঘ) 21

ব্যাখ্যা

Question: If 4 times of a number is subtracted from its 13 times, the result is 171. What is the number?

Solution:
Let,
The number is x

ATQ,
13x - 4x = 171
Or, 9x = 171
Or, x = 171/9
∴ x = 19

৫৫৮.

The B.D. and T.D. on a certain sum is Tk. 200 and Tk. 100 respectively. Find out the sum.

ক
ক) 400
খ
খ) 300
গ
গ) 100
ঘ
ঘ) 200

ব্যাখ্যা

F = (BD×TD)/(BD−TD)
= (200×100)/(200−100)
= 200−100/ 100
= Tk. 200

৫৫৯.

In a school, as many students as there are, each gives 10 taka and thus Tk.25000 was collected. What is the number of students?

ক
45
খ
50
গ
55
ঘ
60

ব্যাখ্যা

Question: In a school, as many students as there are, each gives 10 taka and thus Tk.25000 was collected. What is the number of students?

Solution:
ধরি,
শিক্ষার্থী সংখ্যা = ক জন
প্রত্যেকে চাঁদা দেয় = ১০ক টাকা

প্রশ্নমতে,
ক × ১০ক = ২৫০০০
⇒ ১০ক^২ = ২৫০০০
⇒ ক^২ = ২৫০০০/১০
⇒ ক^২ = ২৫০০
⇒ ক = √২৫০০
∴ ক = ৫০

∴ উক্ত শ্রেণিতে শিক্ষার্থী সংখ্যা ৫০ জন।

৫৬০.

The sum of the digits of two-digit number is 10, while when the digits are reversed, the number decrease by 54. Find the changed number.

ক
28
খ
46
গ
19
ঘ
37
ঙ
None

ব্যাখ্যা

Question: The sum of the digits of two-digit number is 10, while when the digits are reversed, the number decrease by 54. Find the changed number.

Solution:
Let number be (10x + y)

According to question,
(10x + y) - (10y + x) = 54
10x - 10y + y - x = 54
⇒ 9x - 9y = 54
⇒ x - y = 6 ................(i)

Sum of digits,
(x + y) = 10 ................. (ii)

(i) - (ii)
x - y - x - y = 6 - 10
⇒ - 2y = - 4
⇒ y = 2 and, x = 8

The required number is
= (10y + x)
= 10 × 2 + 8
= 28

৫৬১.

In a class of 60 students, 20 students like Math, 25 students like English, and 30 students like Science. If 5 students like both Math and English, 7 students like both Math and Science, 8 students like both English and Science, and 3 students like neither of these subjects, how many students like all three subjects?

ক
2
খ
4
গ
6
ঘ
5
ঙ
1

ব্যাখ্যা

Question: In a class of 60 students, 20 students like Math, 25 students like English, and 30 students like Science. If 5 students like both Math and English, 7 students like both Math and Science, 8 students like both English and Science, and 3 students like neither of these subjects, how many students like all three subjects?

Solution:
Total students, n(U) = 60
Number who like Math, n(M) = 20
Number who like English, n(E) = 25
Number who like Science, n(S) = 30
Number who like both Math and English, n(M ∩ E) = 5
Number who like both Math and Science, n(M ∩ S) = 7
Number who like both English and Science, n(E ∩ S) = 8
Number who like neither subject = 3

n(M ∪ E ∪ S) = n(U) - neither
= 60 - 3 = 57

∴ n(M ∪ E ∪ S) = n(M) + n(E) + n(S) - n(M ∩ E) - n(M ∩ S) - n(E ∩ S) + n(M ∩ E ∩ S)
⇒ 57 = 20 + 25 + 30 - 5 - 7 - 8 + n(M ∩ E ∩ S)
⇒ 57 = 75 - 20 + n(M ∩ E ∩ S)
⇒ 57 = 55 + n(M ∩ E ∩ S)
⇒ n(M ∩ E ∩ S) = 57 - 55
⇒ n(M ∩ E ∩ S) = 2

∴ 2 Students like all three subjects.

৫৬২.

How many integers from 1 to 150 are divisible by 5 but not by 6?

ক
20
খ
25
গ
28
ঘ
30

ব্যাখ্যা

Question: How many integers from 1 to 150 are divisible by 5 but not by 6?

Solution:
150 পর্যন্ত সংখ্যাগুলোর মধ্যে-
5 দ্বারা বিভাজ্য সংখ্যা = ( 150 ÷ 5) = 30 টি
5এবং 6 এর লসাগু = 30
এখন, 150 ÷ 30 = 5
∴ 5 এবং 6 উভয় দ্বারা বিভাজ্য সংখ্যা = 5টি
সুতরাং, 5 দ্বারা বিভাজ্য কিন্তু 6 দ্বারা বিভাজ্য নয় এমন সংখ্যা = (30 - 5) = 25 টি সংখ্যা

৫৬৩.

If p and q are even numbers, which of the following is always even?

ক
p + q + 3
খ
pq + 5
গ
3p + q
ঘ
p² + q + 3

ব্যাখ্যা

Question: If p and q are even numbers, which of the following is always even?

Solution:
Take p = 2 and q = 6 (both even)
a) p + q + 3 = 2 + 6 + 3 = 11 → Odd
b) pq + 5 = (2 × 6) + 5 = 12 + 5 = 17 → Odd
c) 3p + q = (3 × 2) + 6 = 6 + 6 = 12 → Even
d) p² + q + 3 = (2)² + 6 + 3 = 4 + 6 + 3 = 13 → Odd

Answer: c) 3p + q is always even.

৫৬৪.

If the sum of the 3 consecutive integers is 240, then the sum of the two larger integers is:

ক
ক) 79
খ
খ) 159
গ
গ) 169
ঘ
ঘ) 161
ঙ
ঙ) None of these

ব্যাখ্যা

Question: If the sum of the 3 consecutive integers is 240, then the sum of the two larger integers is:

Solution:
Let,
Three numbers are , x - 1, x , x + 1

ATQ,
x - 1 + x + x + 1 = 240
⇒ 3x = 240
⇒ x = 240/3
∴ x = 80

∴ the sum of the two larger integers is = x + x + 1
= 80 + 80 + 1
= 161

৫৬৫.

What is the sum of the first 110 natural numbers?

ক
ক) 5010
খ
খ) 5515
গ
গ) 6050
ঘ
ঘ) 6105

ব্যাখ্যা

আমরা জানি,
n সংখ্যক স্বাভাবিক সংখ্যার যোগফল = {n(n + 1)/2}
∴ 110 টি স্বাভাবিক সংখ্যার যোগফল = {110(110 + 1)/2} = 6105

৫৬৬.

The average of 7 consecutive numbers is 20. The largest of these number is -

ক
ক) 23
খ
খ) 22
গ
গ) 20
ঘ
ঘ) 24

ব্যাখ্যা

ধরি, সংখ্যাগুলো, x - 3, x - 2, x - 1, x + 1, x + 2, x + 3
(x - 3 + x - 2 + x - 1 + x + x + 1 + x + 2 + x + 3)/7 = 20
Or, 7x/7 = 20
Or, x = 20
∴ বৃহত্তম সংখ্যাটি = x + 3 = 20 + 3 = 23

৫৬৭.

(1/2) × (3/4) × (8/4) ÷ (8/2) × (1/2) =?

ক
ক) 3/32
খ
খ) 3/8
গ
গ) 2/3
ঘ
ঘ) 1
ঙ
ঙ) 3/2

ব্যাখ্যা

Question: (1/2) × (3/4) × (8/4) ÷ (8/2) × (1/2) =?

Solution:
(1/2) × (3/4) × (8/4) ÷ (8/2) × (1/2)
= (1/2) × (3/4) × (8/4) × (2/8) × (1/2)
= 3/32

৫৬৮.

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.

ক
4
খ
6
গ
7
ঘ
9
ঙ
10

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

৫৬৯.

A collection of books went on sale and 2/3 of them was sold for Taka 2.30 each. If none of the 36 remaining books were sold, what was the total amount received for the books that were sold?

ক
ক) 165.6
খ
খ) 180
গ
গ) 135.6
ঘ
ঘ) 90

ব্যাখ্যা

Question: A collection of books went on sale and 2/3 of them was sold for Taka 2.30 each. If none of the 36 remaining books were sold, what was the total amount received for the books that were sold?

Solution:
মনেকরি
মোট বই = x টি
বিক্রিত বই = 2x/3 অংশ
বই বিক্রয় হয়নি = x - (2x/3) অংশ
= (3x - 2x)/3
= x/3

এখন
x/3 অংশ বই = 36
2x/3 অংশ বই = 36 × 2
= 72 টি

বিক্রয়কৃত বইয়ের মূল্য= (72 × 2.30) টাকা
= 165.6 টাকা

৫৭০.

In a quiz competition, Ms. Fatima is the 20th highest and the 15th lowest in the rankings. How many contestants were in the competition?

ক
36
খ
35
গ
32
ঘ
34

ব্যাখ্যা

Question: In a quiz competition, Ms. Fatima is the 20th highest and the 15th lowest in the rankings. How many contestants were in the competition?

Solution:
Given that,
Ms. Fatima is the 20th highest,
and the 15th lowest ranked contestant.

∴ Total = people above + Ms. Fatima + people below = 19 + 1 + 14 = 34

৫৭১.

The sum of first 40 natural numbers is -

ক
800
খ
810
গ
820
ঘ
840

ব্যাখ্যা

Question: The sum of first 40 natural numbers is -

Solution:
We know that,
The sum of the first n natural numbers is n(n+1)/2

Sum of first 40 natural numbers is = 40(40 + 1)/2
= 820

৫৭২.

We reverse a number and form a new one. The old number is 45 less than the new number. The sum of the digits of the old number is 9. What is the new number?

ক
a. 36
খ
b. 54
গ
c. 72
ঘ
d. 81

ব্যাখ্যা

Let the two digits be X and Y.
Let the older number be A and the newer one be B.
A = 10X + Y
∴ B = 10Y + X

From given, B = 45 + A = 45 + 10X + Y
10Y + X = 45 + 10X + Y
⇒ 9Y - 9X = 45
⇒ 9(Y - X) = 45
⇒ Y - X = 45/9
⇒ Y - X = 5

Y - X = 5 ----------------- (1)
X + Y = 9 ---------------- (2)

Solving (1) and (2),
Y - X + X + Y = 5 + 9
⇒ 2Y = 14
⇒ Y = 7
∴ X = 9 - Y
= 9 - 7 = 2

So, A = 27; B = 72

৫৭৩.

666 ÷ 6 ÷ 3 = ?

ক
37
খ
333
গ
111
ঘ
84

ব্যাখ্যা

Question: 666 ÷ 6 ÷ 3 = ?

Solution:
666 ÷ 6 ÷ 3
=(666/6) ÷ 3
= 111 ÷ 3
= 37

৫৭৪.

If √3ⁿ= 729, then the value of n is ?

ক
15
খ
6
গ
9
ঘ
12

ব্যাখ্যা

Question: If √3ⁿ= 729, then the value of n is ?

Solution:
Given that,
√3ⁿ = 729
⇒ √3ⁿ = 3⁶
⇒ (√3ⁿ)² = (3⁶)²
⇒ 3ⁿ = 3¹²
∴ n = 12

৫৭৫.

The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:

ক
ক) 74
খ
খ) 94
গ
গ) 184
ঘ
ঘ) 364

ব্যাখ্যা

L.C.M. of 6, 9, 15 and 18 is 90.
Let required number be 90k + 4, which is multiple of 7.
Least value of k for which (90k + 4) is divisible by 7 is k = 4.
Required number = (90 x 4) + 4 = 364.

৫৭৬.

The difference of two numbers is 11 and one-fifth of their sum is 9. Find the numbers.

ক
ক) 28 & 16
খ
খ) 28 & 17
গ
গ) 28 & 18
ঘ
ঘ) 28 & 19
ঙ
ঙ) 28 & 12

ব্যাখ্যা

Let, The numbers are x & y,
therefore,
x - y = 11 ------ (1) and
1/5(x + y) = 9 or, x + y = 45 ------ (2)
adding two equation we got,
2x = 56 or, x = 28,
putting the value of x in equation 1,
we get, y = 17

৫৭৭.

Two times a number added to another number is 25. Three times the first number minus the other number is 20. Find the numbers.

ক
ক) 8,11
খ
খ) 9,12
গ
গ) 9, 7
ঘ
ঘ) 6, 8

ব্যাখ্যা

Question: Two times a number added to another number is 25. Three times the first number minus the other number is 20. find the numbers.
Solution:
প্রশ্নমতে,
2x + y = 25 ----- (1)
3x - y = 20 -----(2)

সমীকরণ দুটি যোগ করে পাই,
5x = 45
x = 9

x এর মান (1)নং এ বসিয়ে পাই,
2x + y = 25
বা, 2(9) + y = 25
বা, 18+ y = 25
বা, y = 7

৫৭৮.

Which of the following fraction is the largest?

ক
7/8
খ
13/16
গ
31/40
ঘ
63/80

ব্যাখ্যা

Question: Which of the following fraction is the largest?

Solution:
L.C.M. of 8, 16, 40 and 80 = 80.

7/8 = 70/80; 13/16 = 65/80; 31/40 = 62/80

Since, 70/80 > 65/80 > 63/80> 62/80,
so 7/8> 13/16 > 63/80 > 31/40

So, 7/8 is the largest.

৫৭৯.

The product of two different irrational numbers is always -

ক
ক) rational
খ
খ) irrational
গ
গ) both of above
ঘ
ঘ) none

ব্যাখ্যা

Product of two different irrational numbers is sometimes irrational and sometimes rational.
For example, product of √2 and √3 is √6, which is irrational
but product of √3 and √12 is √36, which is rational number 6.

৫৮০.

If (2^p + 1) is a prime number, which one of the following digits could be the value of p?

ক
6
খ
5
গ
4
ঘ
3

ব্যাখ্যা

Question: If (2^p + 1) is a prime number, which one of the following digits could be the value of p?

Solution:
If P= 6, we will get; 2⁶ + 1 = 64 + 1= 65

If P = 5, we will get; 2⁵ + 1 = 32 + 1 = 33

If P = 4, we will get; 2⁴ + 1 = 16 + 1 = 17

If P = 3, we will get; 2³ + 1 = 8 + 1= 9

Out of the four results, only 17 is the prime number. So, the required value of the P is 4.

৫৮১.

A student scored 30% marks and failed by 12 marks. Another student scored 55% marks and secured 38 marks more than the pass marks. What is the pass percentage?

ক
33%
খ
37.5%
গ
42%
ঘ
36%

ব্যাখ্যা

Question: A student scored 30% marks and failed by 12 marks. Another student scored 55% marks and secured 38 marks more than the pass marks. What is the pass percentage?

Solution:
Let total marks = x

According to the question,
30% of x + 12 = 55% of x - 38
⇒ 0.3x + 12 = 0.55x - 38
⇒ 0.55x - 0.3x = 12 + 38
⇒ 0.25x = 50
⇒ x = 50/0.25
∴ x = 200

∴ Pass marks = 30% of x + 12
= 0.3 × 200 + 12
= 60 + 12
= 72

∴ Pass percentage = (72/200) × 100
= 36%

৫৮২.

Of the following list of numbers, which has the greatest standard deviation?

ক
ক) 1, 2, 3
খ
খ) 6, 8, 10
গ
গ) 2, 4, 6
ঘ
ঘ) 4, 7, 10

ব্যাখ্যা

অপশন ক এর ক্ষেত্রে
x x²
1 1
2 4
3    9
∑x = 6 ∑x² =14

SD =√{(∑x²/n) - (∑x/n)²}
= √{(14/3) - (6/3)²}
=√(4.67 - 4)
= 1.63

অপশন খ এর ক্ষেত্রে
x x²
6 36
8 64
10 100
∑x = 24 ∑x² =200

SD =√{(∑x²/n) - (∑x/n)²}
= √{(200/3) - (24/3)²}
=√(66.666 - 64)
= 1.63

অপশন গ এর ক্ষেত্রে
x x²
2 4
4 16
6 36
∑x = 12 ∑x² =56

SD =√{(∑x²/n) - (∑x/n)²}
= √{(56/3) - (12/3)²}
=√(18.67 - 16)
= 1.63

অপশন ঘ এর ক্ষেত্রে
x x²
4 16
7 49
10 100
∑x = 21 ∑x² =165

SD =√{(∑x²/n) - (∑x/n)²}
= √{(165/3) - (21/3)²}
=√(55 - 49)
= 2.45

৫৮৩.

√0.01 + √0.81 +√1.21+√0.0009 = ?

ক
2.13
খ
2.03
গ
3.03
ঘ
3.01
ঙ
2.06

ব্যাখ্যা

The sum of
√0.01 + √0.81 +√1.21+√0.0009
= 0.1 + 0.9 + 1.1 + 0.03
= 2.13

৫৮৪.

A rectangular block 6 cm by 12 cm by 15 cm is cut up into an exact number of equal cubes. Find the least possible number of cubes.

ক
ক) 30
খ
খ) 40
গ
গ) 50
ঘ
ঘ) 60

ব্যাখ্যা

Volume of the block = (6 × 12 × 15) cm³
= 1080 cm³

Side of the largest cube
= H.C.F of 6 cm, 12 cm, 15 cm
= 3 cm.

Volume of this cube = (3 × 3 × 3) cm³
= 27 cm³

Number of cubes = 1080/27
= 40.

৫৮৫.

A pendulum strikes 5 times in 3 seconds and another pendulum strikes 7 times in 4 seconds. If both pendulums start striking at the same time, how many clear strikes can be listened to in a minute?

ক
195
খ
199
গ
200
ঘ
205

ব্যাখ্যা

First pendulum strikes once in 3/5 seconds.
Second pendulum strikes once in 4/7 seconds
L.C.M of 3/5 and 4/7
= (L.C.M of 3 and 4)/(H.C.F of 5 and 7)
= 12.
So, they strike together after every 12 seconds.
Thus,
they strike together {(60/12) + 1}
= 6 times in 1 minute.
∴ Total number of clear strikes heard
= [{60/(3/5)} + {60/(4/7)}] - 6
= {60 × (5/3) + 60 × (7/4)} - 6
= (100 + 105) - 6
= 199.

৫৮৬.

Three-fifth of the square of a certain number is 126.15. What is the numbers?

ক
ক) 13.5
খ
খ) 14.5
গ
গ) 75.5
ঘ
ঘ) 145

ব্যাখ্যা

Let the number be x
Then,⇔3/5 x²= 126.15
⇔x²=(126.15× 5/3)
⇔x²=210.25
⇔x= √ 210.25
⇔x=14.5

৫৮৭.

What is the sum of first five prime numbers?

ক
21
খ
23
গ
28
ঘ
32

ব্যাখ্যা

Question: What is the sum of first five prime numbers?

Solution:
First five prime numbers are 2, 3, 5, 7, 11

sum of first five prime numbers = 2 + 3 + 5 + 7 + 11 = 28

৫৮৮.

A number whose fifth part increased by 4 is equal to its fourth part reduced by 10. Find the number.

ক
ক) 260
খ
খ) 270
গ
গ) 280
ঘ
ঘ) 290

ব্যাখ্যা

Let
the number be x
According to problems condition
(x/5) + 4 = (x/4) - 10
(x/5) - (x/4) = - 10 - 4
(4x - 5x)/20 = - 14
-x/20 = - 14
x = 280

৫৮৯.

If the sum of two numbers is 26 and their H. C. F and L. C. M are 1 and 120 respectively, the sum of the reciprocals of the two numbers is-

ক
13/60
খ
13/62
গ
11/60
ঘ
17/60

ব্যাখ্যা

Question: If the sum of two numbers is 26 and their H. C. F and L. C. M are 1 and 120 respectively, the sum of the reciprocals of the two numbers is-

Solution:
Let the two numbers are x and y then
x + y = 26
and
xy = H. C. F × L. C. M = 1 × 120 = 120

Sum of their reciprocals = (1/x) + (1/y)
= (x + y)/xy
= 26/120
= 13/60

৫৯০.

Think of a number and then double the number. Add 6 and then multiply the number by 10. Now divide the number by 20, then subtract the number you first thought of. What is the result?

ক
ক) 5
খ
খ) 4
গ
গ) 3
ঘ
ঘ) 2

ব্যাখ্যা

ধরি,
সংখ্যাটি 10x

প্রশ্নমতে,
10x × 2 = 20x
আবার,
পরবর্তী সংখ্যা = 20x + 6

এখন
{10(20x + 6)/20} - 10x
= {(20x + 6)/2} - 10x
= (20x + 6 - 20x)/2
= 6/2
= 3

৫৯১.

What is the average of the following numbers: 35.5, 32.5, 34.0, 35.5 and 34.5?

ক
ক) 33.0
খ
খ) 33.3
গ
গ) 34.0
ঘ
ঘ) 34.4

ব্যাখ্যা

Question: What is the average of the following numbers: 35.5, 32.5, 34.0, 35.5 and 34.5?

Solution:
Average = (35.5 + 32.5 + 34.0 +35.5 + 34.5) / 5
= 172/5
= 34.4

৫৯২.

How many prime number are in between 45 to 72?

ক
ক) 5
খ
খ) 6
গ
গ) 7
ঘ
ঘ) 8

ব্যাখ্যা

45 থেকে 72 পর্যন্ত মৌলিক সংখ্যা 47, 53, 59, 61, 67, 71

45 থেকে 72 পর্যন্ত মৌলিক সংখ্যা = ৬টি

৫৯৩.

If n = (33)⁴³ + (43)³³, what is the units digit of n?

ক
ক) 0
খ
খ) 2
গ
গ) 4
ঘ
ঘ) 6

ব্যাখ্যা

First of all, the units digit of (33)⁴³ is the same as that of 3⁴³ and the units digit of (43)³³ is the same as that of 3³³. So, we need to find the units digit of 3⁴³ + 3³³.

Next, the units digit of 3 in positive integer power repeats in blocks of four {3, 9, 7, 1}:
3¹= 3 (the units digit is 3)
3²= 9 (the units digit is 9)
3³= 27 (the units digit is 7)
3⁴= 81 (the units digit is 1)
3⁵= 243 (the units digit is 3 again!)
...

Thus:
The units digit of 3⁴³ is the same as the units digit of 3³, so 7 (43 divided by the cyclicity of 4 gives the remainder of 3).
The units digit of 3³³ is the same as the units digit of 3¹, so 3 (33 divided by the cyclicity of 4 gives the remainder of 1).

Therefore, the units digit of (33)⁴³ + (43)³³ is 7 + 3 = 0.

৫৯৪.

At the beginning of a class period, half of the students in a class go to the library. Later in the period, half of the remaining students go to the computer lab. If there are 8 students remaining in the class, how many students were originally in the class?

ক
ক) 12
খ
খ) 16
গ
গ) 24
ঘ
ঘ) 32

ব্যাখ্যা

Let the initial number of students = x
Goes to library = x/2
So, remaining students are = x - x/2 = x/2
Then, goes to computer lab = (x/2)/2 = x/4
ATQ, x/2 - x/4 = x/4 = 8
∴ x = 32
The initial number of students were 32

৫৯৫.

The sum of squares of 3 consecutive integers is less than 97. What is the greatest possible value of the smallest one?

ক
3
খ
4
গ
5
ঘ
7
ঙ
6

ব্যাখ্যা

Let's assume two sets of integers are {4, 5, 6} and {5, 6, 7}
4²+ 5² + 6²= 16 + 25 + 36 = 77
5² + 6² + 7² = 25 + 36 + 49 = 110
As there sum should be less than 97, so the least number of these 3 consecutive integers is 4.

৫৯৬.

The square root of (3 + 2√5)(3 - 2√5) is:

ক
i√3
খ
√10
গ
i√7
ঘ
√11
ঙ
i√11

ব্যাখ্যা

Question: The square root of (3 + 2√5)(3 - 2√5) is:

Solution:
√{(3 + 2√5)(3 - 2√5)}
= √{3² - (2√5)²}
= √{9 - (4 × 5)}
= √{9 - 20}
= √(- 11)
= √{11(- 1)}
= √11 × √(- 1)
= i√11 [যেখানে i² = - 1]

৫৯৭.

The next number in the sequence 3, 8, 18, 33, 53, ... ... ... is

ক
78
খ
83
গ
88
ঘ
83

ব্যাখ্যা

3, 8, 18, 33, 53, ... ... ...
8 - 3 = 5
18 - 8 = 10
33 - 18 = 15
53 - 33 = 20
The next number is 53 + 25 = 78

৫৯৮.

Kamal gets 3 marks for each correctly done question but loses 2 marks for each wrongly done question. He attempts 30 questions and gets 40 marks. How many questions he has attempted correctly?

ক
20
খ
25
গ
26
ঘ
30
ঙ
32

ব্যাখ্যা

Question: Kamal gets 3 marks for each correctly done question but loses 2 marks for each wrongly done question. He attempts 30 questions and gets 40 marks. How many questions he has attempted correctly?

Solution:
Marks for a correct answer = 3
Marks lost for a wrong answer = 2
Total questions attempted = 30
Total marks obtained = 40

Let the number of correct answers be x,
and the number of wrong answers be (30 - x).
Total marks = 3x - 2(30 - x)
∴ 3x - 2(30 - x) = 40
⇒ 3x - 60 + 2x = 40
⇒ 5x - 60 = 40
⇒ 5x = 100
⇒ x = 100/5
⇒ x = 20
∴ Kamal attempted 20 questions correctly.

৫৯৯.

What is the greatest number that can be subtracted from 10,000 so that the remainder may be divisible by 32, 36, 48, and 54?

ক
ক) 8200
খ
খ) 9136
গ
গ) 9200
ঘ
ঘ) 9228

ব্যাখ্যা

Question: What is the greatest number that can be subtracted from 10,000 so that the remainder may be divisible by 32, 36, 48, and 54?

Solution:
L.C.M of 32, 36, 48, and 54 is = 864

So, required number is = 10000 - 864 = 9136

৬০০.

After distributing the chocolates equally among 25 kids, 8 chocolates remain. Had the number of children been 28, 22 chocolates would have been left after equally distributing. Find the total number of chocolates?

ক
ক) 315
খ
খ) 358
গ
গ) 368
ঘ
ঘ) 322

ব্যাখ্যা

Question: After distributing the chocolates equally among 25 kids, 8 chocolates remain. Had the number of children been 28, 22 chocolates would have been left after equally distributing. Find the total number of chocolates?

Solution:
Let M number of the chocolates distributing each student. Therefore,
total number of chocolates is 25M + 8 when it distributing among 25 children and 28M - 22 when it distributing among 28 children.

Therefore,
28M - 22 = 25M + 8
=> 28M - 25M = 8 + 22
=> 3M = 30
=> M = 10

Therefore total number of sweets is 25 × 10 + 8 = 258