Fraction and Simplification, Average and Mean

PrepBank · পাতা ৯ / ১০ · ৮০১–৯০০ / ৯৪৮

৮০১.

The average age of A and B was 18 years five years ago. Now, the average age of A, B, and C is 24 years. What will C's age be in five years?

ক
31 years
খ
33 years
গ
37 years
ঘ
41 years

ব্যাখ্যা

Question: The average age of A and B was 18 years five years ago. Now, the average age of A, B, and C is 24 years. What will C's age be in five years?
(৫ বছর আগে A এবং B-এর গড় বয়স ছিল ১৮ বছর। বর্তমানে, A, B, এবং C-এর গড় বয়স ২৪ বছর। ৫ বছর পর C-এর বয়স কী হবে?)

Solution:
5 বছর আগে A ও B এর বয়সের যোগফল = 18 × 2 = 36 years.
A ও B এর বর্তমান বয়সের যোগফল = 5 + 5 + 36 = 46 years.
A, B ও C এর বর্তমান বয়সের যোগফল = 24 × 3 = 72 years.

∴ C এর বর্তমান বয়স = 72 - 46 = 26 years.

∴ 5 বছর পর C এর বয়স = 26 + 5 = 31 years.

৮০২.

If 2 is added to the numerator of a fraction, the fraction becomes 1. If 9 is added to the denominator, the fraction becomes 1/2. Find the fraction.

ক
11/13
খ
12/14
গ
13/15
ঘ
9/11

ব্যাখ্যা

Question: If 2 is added to the numerator of a fraction, the fraction becomes 1. If 9 is added to the denominator, the fraction becomes 1/2. Find the fraction.

Solution:
ধরি,
লব x, হর y
শর্তমতে,
(x +2)/y = 1
⇒ x + 2 = y ................(1)

আবার,
x/(y + 9) = 1/2
⇒ 2x = y + 9
⇒ 2x - 9 = y ..................(2)

(1) ও (2) হতে পাই,
2x - 9 = x + 2
⇒ x = 11

x এর মান (1) নং এ বসিয়ে পাই,
11 + 2 = y
⇒ y = 13

∴ ভগ্নাংশটি 11/13

৮০৩.

The average age of three boys is 15 years. If their ages are in ratio 3 : 5 : 7, the age of the youngest boy is-

ক
ক) 10 years
খ
খ) 15 years
গ
গ) 12 years
ঘ
ঘ) 9 years

ব্যাখ্যা

প্রশ্ন: The average age of three boys is 15 years. If their ages are in ratio 3 : 5 : 7, the age of the youngest boy is-

সমাধান:
The average age of three boys is 15 years
∴ Sum of ages of three boys = 45 years

Their ages are in ratio 3 : 5 : 7
Now,
3x + 5x + 7x = 45
⇒ 15x = 45
∴ x = 3

∴ Age of youngest boy is = 3x
= 3 × 3 years
= 9 years

৮০৪.

A student's marks were wrongly entered as 75 instead of 60. Due to this, the average marks of the class increased by 0.3. Find the number of students in the class.

ক
45
খ
50
গ
55
ঘ
60

ব্যাখ্যা

Question: A student's marks were wrongly entered as 75 instead of 60. Due to this, the average marks of the class increased by 0.3. Find the number of students in the class.

Solution:
Let
the number of students be x
Total increase in marks = x × 0.3
= x × (3/10)
= 3x/10

ATQ,
3x/10 = (75 - 60)
⇒ 3x/10 = 15
⇒ 3x = 150
⇒ x = 150/3
∴ x = 50

So the number of students in the class = 50

৮০৫.

If 3/4 of a number is 7 more than 1/6 of the number then 5/3 of the number is-

ক
18
খ
15
গ
12
ঘ
20

ব্যাখ্যা

Question: If 3/4 of a number is 7 more than 1/6 of the number then 5/3 of the number is-

Solution:
Let the number be x
According to the question,
⇒ (3x/4) - (x/6) = 7
⇒ (9x - 2x)/12 = 7
⇒ 7x = 7 × 12
∴ x = 12

Then 5/3 of the number will be
= x × (5/3)
= (12 × 5)/3
= 20

৮০৬.

The average of 4 terms is 20 and the last term is 1/3 of the remaining terms. What will be the last number?

ক
ক) 15
খ
খ) 20
গ
গ) 25
ঘ
ঘ) 30

ব্যাখ্যা

Question: The average of 4 terms is 20 and the last term is 1/3 of the remaining terms. What will be the last number?

Solution:
The average of 4 terms is 20
total sum = (4 × 20) = 80

let, last term is x
remaining terms is 3x

3x + x = 80
⇒ 4x = 80
∴ x = 20
The last number is 20

৮০৭.

The average age of A, B, C, D and E is 25 years. The average age of A and B is 28 years and the average of C and D is 32 years. Age of E is :

ক
25 years
খ
22 years
গ
15 years
ঘ
5 years

ব্যাখ্যা

Question: The average age of A, B, C, D and E is 25 years. The average age of A and B is 28 years and the average of C and D is 32 years. Age of E is :

Solution :
The average age of A, B, C, D and E is = 25 years.
The total age of A, B, C, D and E is = (25 × 5) = 125 years.

The average age of A and B is = 28 years
The total age of A and B is = (28 × 2) = 56 years

The average of C and D is = 32 years
The total age of C and D is = (32 × 2) = 64 years

∴ Age of E is = (125 - 56 - 64) = 5 years

৮০৮.

The price of 10 chairs is equal to that of 4 tables. The price of 15 chairs and 2 tables together is Tk. 4000. The total price of 12 chairs and 3 tables is?

ক
Tk. 3000
খ
Tk. 3900
গ
Tk. 3360
ঘ
Tk. 3200
ঙ
None of these

ব্যাখ্যা

Question: The price of 10 chairs is equal to that of 4 tables. The price of 15 chairs and 2 tables together is Tk. 4000. The total price of 12 chairs and 3 tables is?

Solution:
Let,
The cost of a chair and that of table be Tk. x and Tk.y respectively.
Then,10x = 4y
⇒ y = (5/2)x

15x + 2y = 4000
15x + 2 × (5/2)x = 4000
15x + 5x = 4000
20x = 4000
x = 4000/20
x = 200

So, y = (5/2) × 200 = 500
Hence, the cost of 12chairs and 3 tables =12x + 3y
= Tk. {(12 × 200) + (3 × 500)}
= Tk. (2400 + 1500)
= Tk. 3900

৮০৯.

The average price of the first five pens out of six is Tk. 12 and the average price of the last five is Tk. 16. If the price of the first pen is Tk. 50, what is the price of the last pen?

ক
Tk. 70
খ
Tk. 68
গ
Tk. 65
ঘ
Tk. 80

ব্যাখ্যা

Question: The average price of the first five pens out of six is Tk. 12 and the average price of the last five is Tk. 16. If the price of the first pen is Tk. 50, what is the price of the last pen?

Solution:
Given that,
P₁ = 50 taka
Average of first 5 = Tk. 12
Average of last 5 = Tk. 16

Let the prices of the 6 pens are- P₁, P₂, P₃, P₄, P₅, P₆

now,
50 taka
Average of first 5 = 12 taka
Average of last 5 = 16 taka

Now,
Sum of first 5 pens is
P₁+ P₂ + P₃ + P₄ + P₅ = 5 × 12 = 60 .......(1)
And,
Sum of last 5 pens is
P₂ + P₃ + P₄ + P₅ + P₆ = 5 × 16 = 80 ........(2)

Now, subtract equation (1) from (2),
(P₂ + P₃ + P₄ + P₅ + P₆) - (P₁+ P₂ + P₃ + P₄ + P₅ ) = 80 - 50
⇒ P₆ - P₁ = 20
⇒ P₆ - 50 = 20
⇒ P₆ = 20 + 50
∴ P₆ = 70

So the price of the last pen is Tk. 70

৮১০.

Stefanie swam four-fifths of a lap in the morning and seven-fifteenths of a lap in the evening. How much farther did Stefanie swim in the morning than in the evening?

ক
1/3
খ
1/5
গ
2/5
ঘ
3/5
ঙ
2/3

ব্যাখ্যা

Question: Stefanie swam four-fifths of a lap in the morning and seven-fifteenths of a lap in the evening. How much farther did Stefanie swim in the morning than in the evening?

Solution:
Stefanie swam in the morning 4/5 of a lap
Stefanie swam in the evening 7/15 of a lap

Here, 4/5 = 12/15
∴ She swam more in morning (12/15 - 7/15)
= (12 - 7)/15
= 5/15
= 1/3

৮১১.

The average of three consecutive multiples of 7 is 70. What is the largest of these three multiples?

ক
90
খ
87
গ
77
ঘ
67
ঙ
None

ব্যাখ্যা

Question: The average of three consecutive multiples of 7 is 70. What is the largest of these three multiples?

Solution:
Let the three consecutive multiples of 7 be 7x, 7x + 7, 7x + 14.

Average = {(7x + (7x + 7) + (7x + 14))}/3
= (21x + 21)/3

As per the question,
(21x + 21)/3 = 70
⇒ 21x + 21 = 210
⇒ 21x = 189
⇒ x = 189/21
∴ x = 9

So, the largest multiple is 7x + 14 = 63 + 14 = 77.

৮১২.

The simplified value of

ক
2
খ
3
গ
4
ঘ
6

ব্যাখ্যা

Question: The simplified value of

Solution:

৮১৩.

Out of 9 persons, 8 persons spent Tk. 30 each for their meals. The ninth one spent Tk. 20 more than the average expenditure of all the nine. How many total money spent by all of them?

ক
Tk. 262.50
খ
Tk. 282
গ
Tk. 294
ঘ
Tk. 292.50

ব্যাখ্যা

Question: Out of 9 persons, 8 persons spent Tk. 30 each for their meals. The ninth one spent Tk. 20 more than the average expenditure of all the nine. How many total money spent by all of them?

Solution:
Let, the average expenditure be Tk. x.

ATQ,
9x = (8 × 30) + (x + 20)
⇒ 9x = 240 + x + 20
⇒ 9x - x = 260
⇒ 8x = 260
∴ x = 32.5

Total money spent = 9x = 9 × 32.5 = Tk. 292.50

৮১৪.

If 0 < x < 1, which of the following expressions is greatest?

ক
1/√x
খ
√x
গ
(1/π)x
ঘ
x³
ঙ
x⁴

ব্যাখ্যা

Question: If 0 < x < 1, which of the following expressions is greatest?

Solution:
Since x is > 0 and x < 1, it has to be a decimal/fraction.
Take a value that is a perfect square, as some options have a root value. Let's take x = 0.25

A) 1/√x = 1/√0.25 = 1/0.5 = 2
B) √x = √0.25 = 0.5
C) (1/π)x = 0.25/(22/7) = 0.08
D) x³ = (0.25)3 = 0.015
E) x⁴ = (0.25)4 = 0.004

So, If 0 < x < 1, then the expressions 1/√x is greatest.

৮১৫.

Find the average of the numbers 32, 18, 95, 52 and 86.

ক
57.5
খ
65.2
গ
63.5
ঘ
56.6

ব্যাখ্যা

Question: Find the average of the numbers 32, 18, 95, 52 and 86.

Solution:
Average = Sum of all observations/ Number of observations
∴ Average = (32 + 18 + 95 + 52 + 86)/5
= 283/5
= 56.6
∴ The correct answer is 56.6.

৮১৬.

The average temperature for Wednesday, Thursday, and Friday was 38°C. The average for Thursday, Friday, and Saturday was 40° C. If the temperature on Saturday was 41° C, what was the temperature on Wednesday?

ক
35° C
খ
36° C
গ
37° C
ঘ
38° C

ব্যাখ্যা

Question: The average temperature for Wednesday, Thursday, and Friday was 38°C. The average for Thursday, Friday, and Saturday was 40° C. If the temperature on Saturday was 41° C, what was the temperature on Wednesday?

Solution:
Average temperature for Wednesday, Thursday, and Friday = 38° C
∴ Total temperature = 3 × 38 = 114° C

Average temperature for Thursday, Friday, and Saturday = 40° C
∴ Total temperature = 40 × 3 = 120° C

Temperature on Saturday = 41° C

Now,
(Thursday + Friday + Saturday) - (Wednesday + Thursday + Friday) = 120 - 114
= 106° C

Hence, Saturday - Wednesday = 6
∴ Wednesday = 41 - 6 = 35° C

৮১৭.

Ten years ago, the average age of A, B, C and D was 40 yr. with E joining them now, the average of all the five is 49 yr. How old is E?

ক
ক) 44 years
খ
খ) 45 years
গ
গ) 46 years
ঘ
ঘ) 47 years

ব্যাখ্যা

Question: Ten years ago, the average age of A, B, C and D was 40 yr. with E joining them now, the average of all the five is 49 yr. How old is E?

Soultion:
Total present age of A, B, C and D
= (40 + 10) × 4
= 200 years
Total age present age of A, B, C, D and E
= 49 × 5
= 245 years
So, Age of E = 45 years

৮১৮.

The average of the two-digit numbers which remain the same when the digits interchange their positions is-

ক
ক) 33
খ
খ) 44
গ
গ) 55
ঘ
ঘ) 66

ব্যাখ্যা

প্রশ্ন: The average of the two-digit numbers which remain the same when the digits interchange their positions is-

সমাধান:
দুই অঙ্কের একটি সংখ্যার অঙ্কদ্বয় স্থান পরিবর্তন করলে, সংখ্যাটি একই থাকে।
অর্থাৎ, একক ও দশক স্থানীয় অঙ্ক একই হবে।

এমন সংখ্যাগুলো হল ১১, ২২, ৩৩, ৪৪, ৫৫, ৬৬, ৭৭, ৮৮, ৯৯

∴ সংখ্যাগুলোর গড় = (১১ + ২২ + ৩৩ + ৪৪ + ৫৫ + ৬৬ + ৭৭ + ৮৮ + ৯৯)/ ৯
= ৪৯৫/৯
= ৫৫

৮১৯.

For 9 innings, Roman has an average of 65 runs. In the tenth inning, he scores 200 runs, thus increasing his average. His average increased by-

ক
78.5
খ
72
গ
13.5
ঘ
77.5

ব্যাখ্যা

Question: For 9 innings, Roman has an average of 65 runs. In the tenth inning, he scores 200 runs, thus increasing his average. His average increased by-

Solution:

Total score for 9 innings = 65 × 9 = 585

Total score after 10^th innings = 585 + 200 = 785

So, the new average is 785/10 = 78.5

So, the increment is 78.5 - 65 = 13.5

৮২০.

A group of 10 people has their average weight increased by 3 kg after replacing a 50 kg person with someone new. What is the possible weight of the replacement?

ক
75 kg
খ
84 kg
গ
80 kg
ঘ
68 kg

ব্যাখ্যা

Question: A group of 10 people has their average weight increased by 3 kg after replacing a 50 kg person with someone new. What is the possible weight of the replacement?

Solution:
ধরি, 10 জন ব্যক্তির বর্তমান গড় ওজন = x কেজি।
∴ মোট ওজন = 10x কেজি।
50 কেজি ওজনের ব্যক্তি চলে যাওয়ার পর মোট ওজন = (10x - 50) কেজি।

মনে করি, নতুন ব্যক্তির ওজন = y কেজি।
নতুন ব্যক্তি প্রবেশ করার পর নতুন মোট ওজন হবে = (10x - 50 + y) কেজি।
এবং নতুন গড় = (x + 3) কেজি।

প্রশ্নমতে,
(10x - 50 + y)/10 = x + 3
⇒ 10x - 50 + y = 10(x + 3)
⇒ 10x - 50 + y = 10x + 30
⇒ 10x - 10x + y = 30 + 50
⇒ y = 80

সুতরাং, নতুন ব্যক্তির ওজন = 80 কেজি।

৮২১.

ক
1/2
খ
1/3
গ
2
ঘ
10

ব্যাখ্যা

Question:

Solution:

৮২২.

Which of the following is the average of first five prime numbers?

ক
ক) 4.5
খ
খ) 5.6
গ
গ) 7.5
ঘ
ঘ) 8.6

ব্যাখ্যা

First five prime numbers = 2, 3, 5, 7, 11
∴ Average = (2 + 3 + 5 + 7 + 11)/5 = 28/5 = 5.6

৮২৩.

Find the greatest number-

ক
1/9
খ
√0.3
গ
2/5
ঘ
0.3

ব্যাখ্যা

Question: Find the greatest number-

Solution:
(1/9)² = 1/81 = 0.012
(√0.3)² = 0.3
(2/5)² = (0.4)² = 0.16
(0.3)² = 0.09

∴ √0.3 সংখ্যাটি বৃহত্তম।

৮২৪.

The average of runs of a cricket player of 10 innings was 32. How many runs must he make in his next innings so as to increase his average of runs by 4?

ক
81
খ
76
গ
79
ঘ
78

ব্যাখ্যা

Question: The average of runs of a cricket player of 10 innings was 32. How many runs must he make in his next innings so as to increase his average of runs by 4?

Solution:
Average = 32
So, total = 32 × 10 = 320.

Now increase in average = 4
So, new average = 32 + 4 = 36

Total runs after next innings = 36 × 11 = 396
∴ Runs made in the 11th inning = 396 - 320 = 76

৮২৫.

The average age of any and Any's twin two sisters is 8 years. The age of each of twin two sisters is 6 years. What is the any's age?

ক
ক) 9 years
খ
খ) 12 years
গ
গ) 13 years
ঘ
ঘ) 16 years

ব্যাখ্যা

The total age of twin two sisters 6 × 2 = 12 years
The total age of Any and Any's twin two sisters is 8 × 3 = 24 years.
Any's age = (24 - 12) years = 12 years
--------------------------------------------------------
এনি ও তার যমজ দুই বোনের গড় বয়স ৮ বছর। যমজ দুই বোনের প্রত্যেকের বয়স ৬ বছর হলে, এনির বয়স কত?

যমজ দুই বোনের প্রত্যেকের বয়স ৬ বছর হলে, তাদের মোট বয়স ৬ × ২ বছর = ১২ বছর।
এনি ও তার যমজ দুই বোনের গড় বয়স ৮ বছর হলে, তাদের মোট বয়স ৮ × ৩ বছর = ২৪ বছর
এনির বয়স (২৪ - ১২) বছর = ১২ বছর

৮২৬.

The average age of A, B, C, D and E is 48 years. The average age of A and B is 40 years and the average of C and D is 50 years. Age of E is :

ক
55 years
খ
60 years
গ
65 years
ঘ
70 years

ব্যাখ্যা

Question: The average age of A, B, C, D and E is 48 years. The average age of A and B is 40 years and the average of C and D is 50 years. Age of E is :

Solution:
A + B + C + D + E = 48 × 5 = 240
A + B = 40 × 2 = 80
C + D = 50 × 2 = 100
Therefore,
E = (A + B + C + D + E) - (A + B + C + D)
E = 240 - 80 - 100
E = 60 years

৮২৭.

The average of nine numbers is 60, that of the first five numbers is 55 and the next three is 65. The ninth number is 10 less than the tenth number. Then, tenth number is-

ক
80
খ
76
গ
82
ঘ
78

ব্যাখ্যা

Question: The average of nine numbers is 60, that of the first five numbers is 55 and the next three is 65. The ninth number is 10 less than the tenth number. Then, tenth number is-

Solution:
Average of nine numbers = 60
Average of first five numbers = 55 and
average of next three numbers = 65
Tenth number = Ninth number + 10

The sum of nine numbers = 60 × 9 = 540
The sum of the first five numbers = 55 × 5 = 275
The sum of the next three numbers = 65 × 3 = 195
Ninth number = (540 - 275 - 195) = (540 - 470) = 70

∴ Tenth number = 70 + 10 = 80

৮২৮.

1/4, 1/2, 3/4 এর গড় কোনটি?

ক
5/4
খ
2/3
গ
1/2
ঘ
3/4

ব্যাখ্যা

প্রশ্ন: 1/4, 1/2, 3/4 এর গড় কোনটি?

সমাধান:
1/4, 1/2, 3/4 এর যোগফল = (2 + 4 + 6)/8
= 12/8
= 3/2

1/4, 1/2, 3/4 এর গড় = (3/2)/3
= (3/2) × (1/3)
= 1/2

৮২৯.

ক
2
খ
1
গ
3
ঘ
4

ব্যাখ্যা

Question:

Solution:

৮৩০.

The average age of husband, wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 20 years. The present age of the husband is :

ক
ক) 35
খ
খ) 40
গ
গ) 50
ঘ
ঘ) None

ব্যাখ্যা

3 year ago the age was = 27×3 = 81
Currently sum of their age is = 81 + 9 = 90
5 year ago = 20×2 = 40
Now, the sum is = 40 + 10 = 50

So, Present age of the husband = 90 - 50 = 40

৮৩১.

For the past n days, the average (arithmetic mean) daily production at a company was 50 units. If today's production of 90 units raises the average to 55 units per day, what is the value of n?

ক
11
খ
12
গ
9
ঘ
7

ব্যাখ্যা

Question: For the past n days, the average (arithmetic mean) daily production at a company was 50 units. If today's production of 90 units raises the average to 55 units per day, what is the value of n?

Solution:
total production in n days is = 50n

ATQ,
50n + 90 = (n + 1)55
or, 50n + 90 = 55n + 55
or, 55n - 50n = 90 - 55
or, 5n = 35
∴ n = 7

৮৩২.

The mean of 8 numbers is 22. Five of them are 20, 18, 24, 16, and 26. Find the mean of the remaining three numbers.

ক
20
খ
22
গ
23
ঘ
24

ব্যাখ্যা

Question: The mean of 8 numbers is 22. Five of them are 20, 18, 24, 16, and 26. Find the mean of the remaining three numbers.

Solution:
Mean of 8 numbers = 22
Total sum of 8 numbers,
= 8 × 22
= 176

Sum of the given 5 numbers,
= 20 + 18 + 24 + 16 + 26
= 104

Sum of the remaining 3 numbers:
= 176 - 104
= 72

∴ Mean of the remaining 3 numbers,
= 72/3
= 24

৮৩৩.

Three-fifth of one-third of four-ninth of a number is 16. What is one-fourth of that number?

ক
72
খ
45
গ
86
ঘ
55

ব্যাখ্যা

Question: Three-fifth of one-third of four-ninth of a number is 16. What is one-fourth of that number?

Solution:
Let
the number is A

ATQ,
A × (4/9) × (1/3) × (3/5) = 16
⇒ 12A/135 = 16
⇒ 12A = 16 × 135
⇒ A = (16 × 135)/12
∴ A = 180

∴ One-fourth of that number is = 180/4 = 45

৮৩৪.

The average of a,b,c is 6 and a - b = 4, ab = 21, What is the value of c?

ক
6
খ
7
গ
8
ঘ
5
ঙ
4

ব্যাখ্যা

(a + b + c)/ 3 = 6
a + b + c = 18 ---------- (1)

a - b = 4
(a - b)²= 16
(a + b)²- 4ab = 16
(a + b)² - 4.21 = 16
(a + b)² = 100
a + b = 10
From (1) we get,
C = 18 - 10 = 8

৮৩৫.

The average age of 12 children is 15 years. If another child comes, the average age comes to 13. What is the age of the new child?

ক
5 years
খ
7 years
গ
9 years
ঘ
11 years

ব্যাখ্যা

Question: The average age of 12 children is 15 years. If another child comes, the average age comes to 13. What is the age of the new child?

Solution:
Sum of 12 children ages = 12 × 15 = 180
Sum of the 13 children ages = 13 × 13 = 169
So age of new children = 180 - 169
= 11

[বাস্তবিকভাবে নতুন একজনের বয়স যুক্ত হওয়ার পর বয়সের গড় কম-বেশি হতে পারে তবে ১২ জনের মোট বয়স (১৮০ বছর) অপেক্ষা ১৩ জনের মোট বয়স (১৬৯ বছর) কম হতে পারে না। এই প্রশ্নটি যেহেতু জব সল্যুশনের প্রশ্ন তাই গাণিতিক নিয়ম অনুযায়ী ১১ বছর উত্তর রাখা হয়েছে]

৮৩৬.

In a class average age of 15 boys is 11. If 5 boys each of age 9 years are added, what would be the new average?

ক
15 years
খ
12.5 years
গ
10.5 years
ঘ
16 years

ব্যাখ্যা

Question: In a class average age of 15 boys is 11. If 5 boys each of age 9 years are added, what would be the new average?

Solution:
Sum of ages of 15 boys = 15 × 11= 165
Sum of ages of 5 boys = 5 × 9 = 45
Total age of 20 boys = 165 + 45 = 210
∴ Average of ages of 20 boys = 210/20 = 10.5 years

৮৩৭.

A cake is divided into 24 pieces. If Kamal takes 1/4 of the cake and Jamal takes 1/3 of the rest that are left, how many pieces are still left?

ক
18
খ
16
গ
12
ঘ
8

ব্যাখ্যা

Question: A cake is divided into 24 pieces. If Kamal takes 1/4 of the cake and Jamal takes 1/3 of the rest that are left, how many pieces are still left?

Solution:
The cake is divided into 24 pieces, and Kamal takes 1/4 of the cake is = (1/4) × 24 = 6 pieces
After Kamal takes 6 pieces, there are,
24 − 6 = 18 pieces.

Jamal takes 1/3 of the remaining pieces,
(1/3) × 18 = 6 pieces.
So Jamal takes 6 pieces.

So the cake are left = 24 - 6 - 6 = 12 pieces.

∴ After Kamal and Jamal take their portions, 12 pieces of the cake are still left.

৮৩৮.

Which of the following is the largest?

ক
9/12
খ
10/15
গ
5/8
ঘ
7/10

ব্যাখ্যা

Question: Which of the following is the largest?

Solution:
ক) 9/12 = 0.75

খ) 10/15 = 0.66

গ) 5/8 = 0.62

ঘ) 7/10 = 0.70

So 9/12 = 0.75 is clearly the biggest.

৮৩৯.

If 0.13 ÷ p² = 13, then p equals:

ক
0.1
খ
0.01
গ
0.001
ঘ
0.3

ব্যাখ্যা

Question: If 0.13 ÷ p² = 13, then p equals:

Solution:
0.13 ÷ p² = 13
⇒ p² = 0.13/13
⇒ p² = 1/100
⇒ p = √1/100
⇒ p = 1/10
⇒ p = 0.1

৮৪০.

The average weight of 3 friends is 33 kg. None of the friends weights less than 31 kg. What can be the maximum weight of any three friends?

ক
32
খ
33
গ
35
ঘ
37

ব্যাখ্যা

Question: The average weight of 3 friends is 33 kg. None of the friends weights less than 31 kg. What can be the maximum weight of any three friends?

Solution:
তিনজনের গড় ওজন ৩৩ কেজি
মোট ওজন ৩৩ × ৩ কেজি
= ৯৯ কেজি

প্রতিজনের ওজন সর্বনিম্ন ৩১ কেজি হতে পারে
∴ দুজনের সর্বনিম্ন ওজন হবে ৩১ × ২ কেজি
= ৬২ কেজি

∴ একজনের সর্বোচ্চ ওজন হতে পারে = ৯৯ - ৬২ কেজি
= ৩৭ কেজি

৮৪১.

Two bags together contain 900 marbles. If one-sixth of the marbles in the second bag are moved to the first bag, both bags have the same number of marbles. How many marbles are in the second bag initially?

ক
360
খ
480
গ
540
ঘ
380

ব্যাখ্যা

Question: Two bags together contain 900 marbles. If one-sixth of the marbles in the second bag are moved to the first bag, both bags have the same number of marbles. How many marbles are in the second bag initially?

Solution:
Let the first bag have x marbles and the second bag have y marbles.
Than,
x + y = 900 ..........(1)
And
After transferring one-sixth of the marbles from the second bag to the first. than,
⇒ x + (y/6) = y - (y/6)
⇒ x + (y/6) = 5y/6
⇒ x = (5y/6) - (y/6)
⇒ x = 4y/6
∴ x = 2y/3

From (1),
⇒ (2y/3) + y = 900
⇒ 5y/3 = 900
⇒ y = (900 × 3)/5
∴ y = 540

So the second bag initially contains 540 marbles.

৮৪২.

The average of 2, 7, 6, and m is 5 and the average of 18, 1, 6, m and n is 10. What is the value of n?

ক
10
খ
15
গ
20
ঘ
25

ব্যাখ্যা

Question: The average of 2, 7, 6, and m is 5 and the average of 18, 1, 6, m and n is 10. What is the value of n?

Solution:
Given that
average of 2, 7, 6, m is 5

Therefore,
5 = (2 + 7 + 6 + m)/4
⇒ 20 = 15 + m
⇒ m = 20 - 15
∴ m = 5

Therefore,
10 = (18 + 1 + 6 + m + n)/5
⇒ 50 = 25 + 5 + n
⇒ n = 50 - 30
∴ n = 20

৮৪৩.

In a class of 40 students, the average marks in an exam was 60. The average marks of the students who passed is 65 and the average marks of the students who failed is 40. How many students failed in the exam?

ক
10
খ
9
গ
4
ঘ
8
ঙ
None

ব্যাখ্যা

Question: In a class of 40 students, the average marks in an exam was 60. The average marks of the students who passed is 65 and the average marks of the students who failed is 40. How many students failed in the exam?

Solution:
Let the number of students who failed = x
Then, the number of students who passed = 40 - x

According to the question,
65(40 - x) + 40x = 40 × 60
⇒ 2600 - 65x + 40x = 2400
⇒ -25x = 2400 - 2600
⇒ -25x = -200
∴ x = 200/25
∴ x = 8

∴ Number of students who failed = 8.

৮৪৪.

Consider that w + x = - 4, x + y = 25 and y + w = 15. Then the average of w, x, y is -

ক
ক) 3
খ
খ) 4
গ
গ) 5
ঘ
ঘ) 6

ব্যাখ্যা

w + x + x + y + y + w = - 4 + 25 + 15
⇒ 2 (w + x + y) = 36
⇒ w + x + y = 18
So, average of w, x, y = 18/3 = 6

৮৪৫.

The average height of 30 boys was calculated to be 150 cm. It was detected later that one value of 165 cm was wrongly copied as 135 cm for the computation of the mean. Find the correct mean.

ক
155 cm
খ
153 cm
গ
151 cm
ঘ
149 cm

ব্যাখ্যা

Question: The average height of 30 boys was calculated to be 150 cm. It was detected later that one value of 165 cm was wrongly copied as 135 cm for the computation of the mean. Find the correct mean.

Solution:
Calculated average height of 30 boys = 150 cm.
Incorrect sum of the heights of 30 boys
= (150 × 30)cm
= 4500 cm.

Correct sum of the heights of 30 boys
= (incorrect sum) - (wrongly copied item) + (actual item)
= (4500 - 135 + 165) cm
= 4530 cm.

Correct mean = correct sum/number of boys
= (4530/30) cm
= 151 cm.

∴ Hence, the correct mean height is 151 cm.

৮৪৬.

Shakib's bowling average is 12.4 runs per wicket takes 5 wickets for 26 runs and thereby decreases his average by 0.4. The number of wickets taken by him till the last match was -

ক
65
খ
75
গ
85
ঘ
90

ব্যাখ্যা

Question: Shakib's bowling average is 12.4 runs per wicket takes 5 wickets for 26 runs and thereby decreases his average by 0.4. The number of wickets taken by him till the last match was -

Solution:
Let, Shakib take x wickets before the last match.

Total run = 12.4x + 26
New average = (12.4x + 26)/(x + 5)

ATQ,
(12.4x + 26)/(x + 5) = 12.4 - 0.4
⇒ (12.4x + 26) = 12(x + 5)
⇒ 12.4x - 12x = 60 - 26
⇒ 0.4x = 34
⇒ x = 34/0.4
∴ x = 85

The number of wickets taken by him till the last match was = 85 wickets

৮৪৭.

The average of runs of a cricket player of 10 innings was 32. How many runs must he make in his next innings so as to increase his average of runs by 4?

ক
76
খ
79
গ
85
ঘ
87

ব্যাখ্যা

Question: The average of runs of a cricket player of 10 innings was 32. How many runs must he make in his next innings so as to increase his average of runs by 4?

Solution:
Average = 32
So, total = 32 × 10 = 320.

Now increase in average = 4
So, new average = 32 + 4 = 36

Total runs after next innings = 36 × 11 = 396
∴ Runs made in the 11th inning = 396 - 320 = 76

৮৪৮.

The total marks obtained by a student in English, Biology, and History together is 150 more than the marks obtained by him in Biology. What is the average marks obtained by him in English and History together?

ক
60
খ
75
গ
90
ঘ
120

ব্যাখ্যা

Question: The total marks obtained by a student in English, Biology, and History together is 150 more than the marks obtained by him in Biology. What is the average marks obtained by him in English and History together?

Solution:
Let English = E, Biology = B, History = H

Given,
E + B + H = B + 150
⇒ E + H = 150

∴ Average of English and History
= (E + H)/2
= 150/2
= 75

৮৪৯.

A batsman makes a score of 84 runs in the 21^st inning and thus increases his average by 2 runs. His average after 21^st inning is-

ক
ক) 44
খ
খ) 38
গ
গ) 42
ঘ
ঘ) 48

ব্যাখ্যা

The average for 20 innings be x

Now
(20x + 84)/21 = x + 2
20x + 84 = 21x + 42
21x - 20x = 84 - 42
x = 42

∴ Average after 21^st innings
= 42 + 2
= 44

৮৫০.

The average salary of 30 officers in an office is 120 tk and the average salary of laborers is 40 tk. Find the total number of laborers if the average salary of the office is 50 tk.

ক
180
খ
220
গ
210
ঘ
160

ব্যাখ্যা

Question: The average salary of 30 officers in an office is 120 tk and the average salary of laborers is 40 tk. Find the total number of laborers if the average salary of the office is 50 tk.

Solution:
The sum of the salary of officers will be = 30 × 120 = 3600
Let the number of laborers = L

ATQ,
3600 + 40L = 50(30 + L)
⇒ 2100 = 10L
∴ L = 210

৮৫১.

The average marks of a student in 4 subjects is 75. If the student obtained 85 marks in the fifth subject, then the new average is

ক
ক) 76
খ
খ) 77
গ
গ) 78
ঘ
ঘ) 79

ব্যাখ্যা

Sum of marks in 4 Subjects = 75 × 4 = 300
Sum of marks in 5 Subjects = 300 + 85 = 385
New Average = 385/5 = 77

৮৫২.

What is the median of the modes in the dataset {-5, 4, 3, 7, 2, 1, 3, 4, 5,-1, 7, 8, -4, 2, 6}?

ক
ক) 2
খ
খ) 2.5
গ
গ) 3
ঘ
ঘ) 3.5

ব্যাখ্যা

Question: What is the median of the modes in the dataset {- 5, 4, 3, 7, 2, 1, 3, 4, 5,- 1, 7, 8, - 4, 2, 6}?

Solution:
প্রদত্ত ডাটা সেটকে আমরা ক্রমানুসারে সাজিয়ে পাই,
- 5, - 4, -1, 1, 2, 2, 3, 3, 4, 4, 5, 6, 7, 7, 8

2, 3, 4, 7 প্রত্যেকে দুইবার করে বিদ্যমান।
প্রচুরক গুলো হলো: 2, 3, 4, 7

এখানে
n = 4
মধ্যক = {4/2 তম ও ((4/2) + 1) তম পদের মানের যোগফল} / 2
= {2 ও 3 তম পদের মানের যোগফল}/2
= (3 + 4)/2
= 3.5

৮৫৩.

The average of runs of a cricket player of 10 innings was 32. How many runs must be made in his next innings so as to increase his average of runs by 4?

ক
20
খ
76
গ
86
ঘ
29

ব্যাখ্যা

Question: The average of runs of a cricket player of 10 innings was 32. How many runs must be made in his next innings so as to increase his average of runs by 4?

Solution:

Average after 11 innings = 36

Required number of runs,

= ( 36 × 11) - (32 × 10)

= 396 - 320

= 76

৮৫৪.

The average monthly income of Abir and Sabbir is Tk. 8000. The average monthly income of Sabbir and Rana is Tk. 9500 and the average monthly income of Abir and Rana is Tk. 9000. What is the monthly income of Abir?

ক
Tk. 7000
খ
Tk. 7500
গ
Tk. 6500
ঘ
Tk. 6000
ঙ
None of these

ব্যাখ্যা

Question: The average monthly income of Abir and Sabbir is Tk. 8000. The average monthly income of Sabbir and Rana is Tk. 9500 and the average monthly income of Abir and Rana is Tk. 9000. What is the monthly income of Abir?

Solution:
Abir + Sabbir (total income) = (8000 × 2) = 16000 .......... (1)
Sabbir + Rana (total income) = (9500× 2) = 19000 ........ (2)
Abir + Rana (total income) = (9000 × 2) = 18000 ......... (3)

(1) + (2) + (3) ⇒
2(Abir + Sabbir+ Rana) = 16000 + 19000 + 18000
⇒ 2(Abir + Sabbir + Rana) = 53000
⇒ Abir + Sabbir + Rana = 26500 ....... (4)

(4) - (2) ⇒
Abir = 26500 - 19000 = 7500
∴ Abir's monthly income Tk. 7500.

৮৫৫.

What is the average (arithmetic mean) of the values 36, 42, 40, 38, 44, 40, and 40?

ক
25
খ
40
গ
37
ঘ
17

ব্যাখ্যা

Question: What is the average (arithmetic mean) of the values 36, 42, 40, 38, 44, 40, and 40?

Solution:
Find the sum of all values:
36 + 42 + 40 + 38 + 44 + 40 + 40 = 280

Divide the sum by the total number of values:
Average = 280 ÷ 7 = 40

৮৫৬.

The average of 2, 7, 6, and x is 5 and the average of 18, 1, 6, x and y is 10. What is the value of y?

ক
ক) 5
খ
খ) 10
গ
গ) 15
ঘ
ঘ) 20

ব্যাখ্যা

Given that
average of 2, 7, 6, x is 5

Therefore,
5 = (2 + 7 + 6 + x)/4
⇒ 20 = 15 + x
⇒ x = 20 - 15
x = 5

Therefore,
10 = (18 + 1+ 6 + 5 + y)/5
50 = 30 + y
y = 50 - 30
y = 20

৮৫৭.

The average marks of 50 students was found to be 60. Later, it was found that a student’s marks were wrongly entered as 80 instead of 50. What is the correct average?

ক
59
খ
59.4
গ
58.5
ঘ
58

ব্যাখ্যা

Question: The average marks of 50 students was found to be 60. Later, it was found that a student’s marks were wrongly entered as 80 instead of 50. What is the correct average?

Solution :
Firstly, wrong total = 50 × 60 = 3000
An error happened in one student's mark = 80 - 50 = 30.
Correct total = 3000 - 30 = 2970
Correct average = 2970 ÷ 50 = 59.4

৮৫৮.

The average age of 16 members of a club is 22. If the minimum age requirement for being a member is 19 years. What is the possible maximum range of the ages?

ক
ক) 48
খ
খ) 53
গ
গ) 67
ঘ
ঘ) 71
ঙ
ঙ) None

ব্যাখ্যা

Total age of the members = 16 × 22 = 352 years
Total age of 15 members at least = 15 × 19 = 285 years
∴ Possible maximum age = 352 - 285 = 67 years

৮৫৯.

In a class, the average marks of 40 students was calculated to be 52.15. It was later discovered that the marks of a student were taken to be 49, instead of 85. Find the real average of the class.

ক
53.05
খ
53.15
গ
52.85
ঘ
52.95
ঙ
None of these

ব্যাখ্যা

Question: In a class, the average marks of 40 students was calculated to be 52.15. It was later discovered that the marks of a student were taken to be 49, instead of 85. Find the real average of the class.

Solution:
Average marks for 40 students is equal to 52.15 , the marks were taken as 49 instead of 85 so there will be an increase of 36 which is now to be distributed equally amongst 40 students , so 36/40 = 0.9 which is to be distributed amongst all.
So, new average stands out to be 52.15 + 0.9 = 53.05

৮৬০.

Sajid travels for 2 hours at 30 miles an hour and he covers 60 miles in the next 3 hours. What is the average speed per hour for the entire trip?

ক
20 mile/hr
খ
22 mile/hr
গ
24 mile/hr
ঘ
25 mile/hr

ব্যাখ্যা

Question: Sajid travels for 2 hours at 30 miles an hour and he covers 60 miles in the next 3 hours. What is the average speed per hour for the entire trip?

Solution:
সাজিদ প্রথম 2 ঘণ্টায় অতিক্রম করে 30 × 2 মাইল
= 60 মাইল

সে পরবর্তী 3 ঘণ্টায় অতিক্রম করে = 60 মাইল

∴ গড় গতিবেগ = (60 + 60)/(2 + 3) মাইল /ঘণ্টা
= 120/5
= 24 মাইল/ঘণ্টা

৮৬১.

The average of six numbers is p and the average of three of these is q. If the average of the remaining three is r, then which one is correct?

ক
p = q + r
খ
p = 2q + 2r
গ
2p = q + r
ঘ
None of these

ব্যাখ্যা

Question: The average of six numbers is p and the average of three of these is q. If the average of the remaining three is r, then which one is correct?

Solution:
ATQ,
6p = 3q + 3r
⇒ p = 3(q + r)/6
⇒ p = (q + r)/2
∴ 2p = q + r

৮৬২.

What is the missing number in the following expression?
(5 + 2) ÷ 7 - 3 × 2 = 1 + ?

ক
- 6
খ
- 5
গ
- 4
ঘ
- 7

ব্যাখ্যা

Question: What is the missing number in the following expression?
(5 + 2) ÷ 7 - 3 × 2 = 1 + ?

Solution:
(5 + 2) ÷ 7 - 3 × 2 = 1 + ?
⇒ 7 ÷ 7 - 3 × 2 = 1 + ?
⇒ 1 - 3 × 2 = 1 + ?
⇒ 1 - 6 = 1 + ?
⇒ - 5 = 1 + ?
⇒ ? = - 5 - 1
∴ ? = - 6

৮৬৩.

The average marks of four subjects is 120. If 33 was misread as 13 during the calculation, what will be the correct average?

ক
ক) 125
খ
খ) 134
গ
গ) 167
ঘ
ঘ) 178

ব্যাখ্যা

The average marks of four subjects is 120.
Difference of 33 and 13 is 20.
That means 20 must be added to total.
Then average = 20/4 = 5
Correct average = 120 + 5 = 125

৮৬৪.

The average of x and y is 7, and z = 3x + 2. What is the average of y and z?

ক
ক) 2x + 8
খ
খ) 2x - 8
গ
গ) 2x + 16
ঘ
ঘ) x + 8

ব্যাখ্যা

Question: The average of x and y is 7, and z = 3x + 2. What is the average of y and z?
Solution:
x এবং y এর গড় = 7
x + y = 7 × 2
x + y = 14
y = 14 - x

(y + z)/2 = (14 - x + 3x + 2)/2
= (16 + 2x)/2
= 2(8 + x)/2
= 8 + x

৮৬৫.

ক
98
খ
35
গ
350
ঘ
None

ব্যাখ্যা

Question:

Solution:

৮৬৬.

The average weight of 18 boys was recorded as 60 kg. If the weight of the teacher was added, the average increased by 2 kg. What was the teacher's weight?

ক
75
খ
86
গ
98
ঘ
105

ব্যাখ্যা

Question: The average weight of 18 boys was recorded as 60 kg. If the weight of the teacher was added, the average increased by 2 kg. What was the teacher's weight?

Solution:
Average weights of 18 boys = 60 kg
Total weights of 18 boys = 18 × 60 kg = 1080 kg

The weight of the teacher was added then average increase by 2 kg
So, new average weight is 62 kg
Total people including teacher = 19

∴ total weight including teacher
= 62 × 19 kg
= 1178 kg

∴ Weight of teacher = 1178 – 1080 = 98 kg

৮৬৭.

The sum of the ages of the father and daughter is 90 years. If five years ago the age of father was four times that of his daughter, find the present age of the daughter.

ক
ক) 19 years
খ
খ) 31 years
গ
গ) 28 years
ঘ
ঘ) 21 years

ব্যাখ্যা

Question: The sum of the ages of the father and daughter is 90 years. If five years ago the age of father was four times that of his daughter, find the present age of the daughter.

Solution:
Let the age of daughter 5 years ago = x
The age of father 5 years ago = 4x

ATQ,
x + 5 + 4x + 5 = 90
⇒ 5x + 10 = 90
⇒ 5x = 80
∴ x = 16

Daughter present age = 16 + 5 = 21 years.

৮৬৮.

The average of x, y and z is 45. x is as much more than the average as y is less than the average. Find the value of z.

ক
45
খ
35
গ
60
ঘ
15
ঙ
None of these

ব্যাখ্যা

Question: The average of x, y and z is 45. x is as much more than the average as y is less than the average. Find the value of z.

Solution:
As the average is 45
Sum of x, y and z will be 135.
and Also given that ,
x - 45 = 45 - y
∴ x + y = 90,

therefore, 90 + z = 135
∴ z = 45

৮৬৯.

The average weight of 3 friends is 33 kg. None of the friends weights less than 31 kg. What can be the maximum weight of any three friends?

ক
38 kg
খ
37 kg
গ
35 kg
ঘ
34 kg

ব্যাখ্যা

Question: The average weight of 3 friends is 33 kg. None of the friends weights less than 31 kg. What can be the maximum weight of any three friends?

Solution:
তিনজনের গড় ওজন ৩৩ কেজি
মোট ওজন = ৩৩ × ৩ কেজি
= ৯৯ কেজি

প্রতিজনের ওজন সর্বনিম্ন ৩১ কেজি
দুজনের সর্বনিম্ন ওজন = (৩১ × ২) কেজি
= ৬২ কেজি

একজনের সর্বোচ্চ ওজন হতে পারে = (৯৯ - ৬২) কেজি
= ৩৭ কেজি

৮৭০.

The average of the three numbers x, y and z is 45, x is greater than the average of y and z by 9. The average of y and z is greater than y by 2. Then the difference of x and z is.

ক
ক) 3
খ
খ) 5
গ
গ) 7
ঘ
ঘ) 11

ব্যাখ্যা

Question: The average of the three numbers x, y and z is 45, x is greater than the average of y and z by 9. The average of y and z is greater than y by 2. Then the difference of x and z is.

Solution:
প্রশ্নমতে,
⇒ (x + y + z)/3 = 45
⇒ x + y + z = 135 ...................(1)

⇒ x = (y + z)/2 + 9
⇒ 2x - y - z = 18 .........................(2)

(1) + (2) নং যোগ করে পাই,
x + y + z = 135
2x - y - z = 18
3x = 153
∴ x = 51

(1) নং হতে পাই,
y + z = 135 - 51
y + z = 84 ........................(3)

আবার,
(y + z)/2 = y + 2
⇒ y + z = 2y + 4
⇒ z - y = 4 .................... (4)

(3) + (4) হতে পাই,
y + z = 84
z - y = 4
2z = 88
∴ z = 44

x ও z এর মধ্যে পার্থক্য = 51 - 44 = 7

৮৭১.

In a football team, the average of 11 players is 28 years. Out of these, the average ages of three players each are 25 years, 28 years and 30 years respectively. If in these groups, the captain and the youngest player are not included, and the captain is 11 years older than the yougest player, what is the age of the youngest player?

ক
23
খ
24
গ
25
ঘ
26

ব্যাখ্যা

প্রশ্ন: In a football team, the average of 11 players is 28 years. Out of these, the average ages of three players each are 25 years, 28 years and 30 years respectively. If in these groups, the captain and the youngest player are not included, and the captain is 11 years older than the yougest player, what is the age of the youngest player?

সমাধান:
11 জনের মোট বয়স = 11 × 28 = 308 বছর
3 জনের 3টি গ্রুপের 9 জনের মোট বয়স = (3 × 25) + (3 × 28) + (3 × 30) = 249 বছর
∴ ক্যাপ্টেন এবং সবচেয়ে জুনিয়র খেলোয়াড়ের বয়স = 308 - 249 = 59 বছর

ধরি,
ক্যাপ্টেনের বয়স = a
সবচেয়ে জুনিয়র খেলোয়াড়ের বয়স = a - 11

প্রশ্নমতে,
a + a - 11 = 59
⇒ 2a = 70
∴ a = 35
অতএব, সবচেয়ে জুনিয়র খেলোয়াড়ের বয়স = 35 - 11 = 24 বছর

৮৭২.

The average of the largest and smallest 3 digits numbers formed by 0, 3 and 8 would be-

ক
538
খ
598
গ
569
ঘ
583

ব্যাখ্যা

Question: The average of the largest and smallest 3 digits numbers formed by 0, 3 and 8 would be-

Solution:
largest = 830
smallest = 308

Average = (830 + 308)/2
= 1138/2
= 569

৮৭৩.

Five numbers are in an arithmetic progression with common difference 3. Find the average of these five numbers if the largest number is 40% more than the smallest number.

ক
30
খ
33
গ
36
ঘ
39

ব্যাখ্যা

Question: Five numbers are in an arithmetic progression with common difference 3. Find the average of these five numbers if the largest number is 40% more than the smallest number.

Solution:
Let,
The five numbers will be x, x + 3, x + 6, x + 9, x + 12

ATQ,
x × (140/100) = x + 12
⇒ 7x = 5x + 60
⇒ 2x = 60
∴ x = 30

∴ The numbers are = 30, 33, 36, 39, 42

∴ The average of the five numbers = (30 + 33 + 36 + 39 + 42)/5 = 180/5 = 36

৮৭৪.

ক
0.25
খ
0.04
গ
0.03
ঘ
0.3
ঙ
None

ব্যাখ্যা

Question:

Solution:

৮৭৫.

The total marks obtained by a student in Physics, Chemistry and Mathematics together is 120 more than the marks obtained in Chemistry. What are the average marks obtained by him in Physics and Mathematics together?

ক
ক) 40
খ
খ) 60
গ
গ) 120
ঘ
ঘ) Cannot be determined

ব্যাখ্যা

P + C + M = C + 120
⇒ P + M = 120
∴ Required average = (P + M)/2 = 120/2
= 60

৮৭৬.

The average age of husband, wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 20 years. What is the present age of the husband?

ক
35 years
খ
40 years
গ
42 years
ঘ
50 years

ব্যাখ্যা

Question: The average age of husband, wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 20 years. What is the present age of the husband?

Solution:
ধরি,
স্বামীর বয়স = ক বছর
স্ত্রীর বয়স = খ বছর
সন্তানের বয়স = গ বছর

5 বছর পূর্বে ,
স্ত্রীর বয়স = (খ - 5) বছর এবং সন্তানের বয়স = (গ - 5) বছর

প্রশ্নমতে,
(খ - 5) + (গ - 5)/2 = 20
⇒ খ + গ - 10 = 40
⇒ খ + গ = 40 + 10
⇒ খ + গ = 50

আবার , 3 বছর পূর্বে তাদের 3 জনের বয়স ছিলো যথাক্রমে (ক - 3) বছর, (খ - 3) বছর এবং (গ - 3) বছর

প্রশ্নমতে,
{(ক - 3) + (খ - 3) + (গ - 3)}/3 = 27
⇒ ক + খ + গ - 9 = 27 × 3
⇒ ক + 50 - 9 = 81
⇒ ক + 41 = 81
⇒ ক = 81 - 41
⇒ ক = 40

∴ স্বামীর বয়স = 40 বছর।

৮৭৭.

The average age of a group of 10 students was 20. The average age increased by 2 years when two new students joined the group. What is the average age of the two new students who joined the group?

ক
ক) 22 years
খ
খ) 30 years
গ
গ) 40 years
ঘ
ঘ) 32 years

ব্যাখ্যা

Question: The average age of a group of 10 students was 20. The average age increased by 2 years when two new students joined the group. What is the average age of the two new students who joined the group?

Solution:
10 জন ছাত্রের গড় বয়স = 20 বছর
10 জন ছাত্রের মোট বয়স = (20 × 10) বছর
= 200 বছর

12 জন ছাত্রের গড় বয়স = 22 বছর
12 জন ছাত্রের মোট বয়স = (22 × 12) বছর
= 264 বছর
2 জন ছাত্রের মোট বয়স = (264 - 200) বছর
= 64 বছর
2 জন ছাত্রের গড় বয়স = 64/2 = 32 বছর

৮৭৮.

If the average (arithmetic mean) of x and y is 60 and the average (arithmetic mean) of y and z is 80. What is the value of (z - x)?

ক
70
খ
40
গ
20
ঘ
None

ব্যাখ্যা

Question: If the average (arithmetic mean) of x and y is 60 and the average (arithmetic mean) of y and z is 80. What is the value of (z - x)?

Solution:
(x + y)/2 = 60
∴ x + y = 120 ...............(1)

(y + z)/2 = 80
∴ y + z = 160 .................(2)

From (2) - (1) we get,
y + z - x - y = 160 - 120
∴ z - x = 40

৮৭৯.

The average weight of 16 students in a class is 50.5 kg, and that of the remaining 8 students is 45.5 kg. Find the average weight of all the students in the class.

ক
38.56
খ
56.33
গ
48.83
ঘ
44.43

ব্যাখ্যা

Question: The average weight of 16 students in a class is 50.5 kg, and that of the remaining 8 students is 45.5 kg. Find the average weight of all the students in the class.

Solution:
Required average = (50.5 × 16 + 45.5 × 8)/(16 + 8)
= (808 + 364)/24
= 1172/24
= 48.83

৮৮০.

Tanzid has a certain average for 9 innings. In the tenth innings, he scores 100 runs thereby increasing his average by 8 runs. His new average is-

ক
18
খ
21
গ
33
ঘ
20
ঙ
28

ব্যাখ্যা

Question: Tanzid has a certain average for 9 innings. In the tenth innings, he scores 100 runs thereby increasing his average by 8 runs. His new average is-

Solution:
Let Tanzid’s average for 9 innings = x
Then total runs in 9 innings = 9x

And,
In the 10th inning, he scores 100 runs
So, total runs after 10 innings = 9x + 100

∴ New average = x + 8 ; (because his average increased by 8)

ATQ,
(9x + 100)/10 = x + 8
⇒ 9x + 100 = 10x + 80
⇒ x = 100 - 80
∴ x = 20

∴ New average = x + 8 = 20 + 8 = 28

৮৮১.

The numerator of a fraction is decreased by 25% and the denominator is increased by 250%. If the resultant fraction is 6/5, what is the original fraction?

ক
24/5
খ
28/9
গ
28/5
ঘ
51/10

ব্যাখ্যা

Question: The numerator of a fraction is decreased by 25% and the denominator is increased by 250%. If the resultant fraction is 6/5, what is the original fraction?

Solution:
let the numerator is a and the denominator is b
a is decreased by 25%.
so, new numerator = a - 25% of a
= a - a/4 = 3a/4 = 0.75a
b is increased by 250%
so, new denominator = b + 250% of b
= b + 2.5b
= 3.5b

ATQ,
0.75a / 3.5b = 6/5
a/b = (6 × 3.5)/(5 × 0.75)
= 28/5

৮৮২.

The average of the ages of a man and his daughter is 34 years. If the respective ratio of their ages four years from now is 14 : 5,what is daughter's present age ?

ক
ক) 12 years
খ
খ) 16 years
গ
গ) 19 years
ঘ
ঘ) None of these

ব্যাখ্যা

After 4 years, the total age of man & daughter is
= {(34 × 2) + 4 + 4}
= 76 years

After 4 years their age ratio is 14 : 5
So, 4 years after the daughter age will be
= 76 × 5/19
= 20 years

∴ Daughter present age
= 20 - 4
= 16 years

৮৮৩.

If each of 4 subsidiaries of Corporation R has been granted a line of credit of Tk. 700,000 and each of the other 3 subsidiaries of Corporation R has been granted a line of credit of Tk. 112,000, what is the average (arithmetic mean) line of credit granted to a subsidiary of Corporation R?

ক
Tk. 1,568,000
খ
Tk. 448,000
গ
Tk. 406,000
ঘ
Tk. 313,600

ব্যাখ্যা

Question: If each of 4 subsidiaries of Corporation R has been granted a line of credit of Tk. 700,000 and each of the other 3 subsidiaries of Corporation R has been granted a line of credit of Tk. 112,000, what is the average (arithmetic mean) line of credit granted to a subsidiary of Corporation R?

Solution:
4 subsidiaries of Corporation R has been granted a line of credit of Tk. 700,000
3 subsidiaries of Corporation R has been granted a line of credit of Tk. 112,000

∴ The average (arithmetic mean) line of credit granted to a subsidiary of Corporation R = (700000 × 4 + 112000 × 3)/7
= (2800000 + 336000)/7
= 3136000/7
= 448000

৮৮৪.

The average of nine numbers is 60, that of the first five numbers is 55 and the next three is 65. The ninth number is 10 less than the tenth number. Then, tenth number is-

ক
80
খ
70
গ
75
ঘ
85

ব্যাখ্যা

Question: The average of nine numbers is 60, that of the first five numbers is 55 and the next three is 65. The ninth number is 10 less than the tenth number. Then, tenth number is-

Solution:
Average of nine numbers = 60
Average of first five numbers = 55 and
average of next three numbers = 65
Tenth number = Ninth number + 10

The sum of nine numbers = 60 × 9 = 540
The sum of the first five numbers = 55 × 5 = 275
The sum of the next three numbers = 65 × 3 = 195
Ninth number = (540 - 275 - 195) = (540 - 470) = 70

∴ Tenth number = 70 + 10 = 80

৮৮৫.

ক
36
খ
37
গ
38
ঘ
39

ব্যাখ্যা

Question:

Solution:

৮৮৬.

What decimal of an hour is a second ?

ক
0.00027
খ
0.0025
গ
0.0256
ঘ
0.000126

ব্যাখ্যা

Question: What decimal of an hour is a second ?

Solution:
There are 60 seconds in a minute and 60 minutes in an hour.
So, 1 hour = 60 × 60 = 3600 seconds
Therefore, 1 second = 1/3600 = 0.00027 of an hour.

৮৮৭.

The average of 5 consecutive integers starting with m as the first integer is n. What is the value of n?

ক
ক) m
খ
খ) m + 1
গ
গ) m + 2
ঘ
ঘ) m + 4

ব্যাখ্যা

If the first integer is m,
then (m + m + 1 + m + 2 + m + 3 + m + 4)/5 = n
⇒ 5m + 10 = 5n
⇒ m + 2 = n
⇒ n = m + 2
----------------------------------------------
m দিয়ে শুরু হয় এমন ৫ টি ক্রমিক পূর্ণ সংখ্যার গড় n হলে n এর মান কত?

m দিয়ে শুরু হয় এমন ৫ টি ক্রমিক পূর্ণ সংখ্যার গড় n হলে, প্রথম পূর্ণ সংখ্য m
অতএব, (m + m + 1 + m + 2 + m + 3 + m + 4)/5 = n
⇒ 5m + 10 = 5n
⇒ m + 2 = n
⇒ n = m + 2

৮৮৮.

The average of x and y is 7, and z = 3x + 2. What is the average of y and z?

ক
ক) x + 8
খ
খ) 2x - 5
গ
গ) 2x + 11
ঘ
ঘ) 3x - 7

ব্যাখ্যা

(x + y)/2 = 7
⇒ x + y = 14
∴ y = 14 - x

Therefore,
(y + z)/2
= (14 - x + 3x + 2)/2
= (2x + 16)/2
= x + 8
------------------------------
x ও y গড় 7 এবং z = 3x + 2 হলে, y ও z এর গড় কত?

x ও y গড় 7 হলে,
(x + y)/2 = 7
⇒ x + y = 14
∴ y = 14 - x

অতএব,
y ও z এর গড়
= (y + z)/2
= (14 - x + 3x + 2)/2
= (2x + 16)/2
= x + 8
সুতরাং y ও z এর গড় x + 8

৮৮৯.

1 - [2 - {3 - (4 - 5) + 6} + 7] = ?

ক
1
খ
2
গ
- 2
ঘ
0

ব্যাখ্যা

Question: 1 - [2 - {3 - (4 - 5) + 6} + 7] = ?

Solution:
1 - [2 - {3 - (4 - 5) + 6} + 7]
= 1 - [2 - {3 - ( - 1) + 6} + 7]
= 1 - [2 - {3 + 1 + 6} + 7]
= 1 - [2 - 10 + 7]
= 1 - [- 1]
= 1 + 1
= 2

৮৯০.

Fresh apples contain 85% water, while dry apples contain 15% water. If the weight of dry apples is 300 kg, what was the total weight of the apples when fresh?

ক
1700 kg
খ
1570 kg
গ
1200 kg
ঘ
2000 kg

ব্যাখ্যা

Question: Fresh apples contain 85% water, while dry apples contain 15% water. If the weight of dry apples is 300 kg, what was the total weight of the apples when fresh?

Solution:
Given that,
Dry apples contain 15% water
∴ they contain 85% solid matter.
Fresh apples contain 85% water
∴ they contain 15% solid matter.

The amount of solid matter (pulp) never changes during drying.

Let the weight of fresh apples = x kg

Solid matter in fresh apples = 15% of x = 0.15x kg
Solid matter in dry apples = 85% of 300 kg = (85/100)× 300 = 255 kg

Since the solid matter remains the same,
0.15x = 255
⇒ x = 255 /0.15
⇒ x = (255 × 100)/15
∴ x = 1700 kg

The total weight of the apples when fresh was 1700 kg.

৮৯১.

The average of ten number is 7. If each number is multiplied by 10, then the average of the new set of number is-

ক
40
খ
50
গ
60
ঘ
70

ব্যাখ্যা

Question: The average of ten number is 7. If each number is multiplied by 10, then the average of the new set of number is-

Solution:
let, 7 numbers are a1, a2, a3,.......,a7
so, (a1 + a2 + a3+.......+a7)/10 = 7
a1 + a2 + a3+.......+a7 = 70

If each number is multiplied by 10, Then sum = (a1 × 10) + (a2 × 10) + (a3 × 10) + .... + (a7 × 10)
= 10 (a1 + a2 + a3+.......+a7)
= 10 × 70
= 700

then average will be = 700/10
= 70

৮৯২.

The average of 7 consecutive numbers is 20. What is the largest of these numbers?

ক
23
খ
22
গ
24
ঘ
21
ঙ
25

ব্যাখ্যা

Question: The average of 7 consecutive numbers is 20. What is the largest of these numbers?

Solution:
Let the 7 consecutive numbers be x, x + 1, x + 2, x + 3, x + 4, x + 5 and x + 6,

As per the given condition;

[x + (x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 5) + (x + 6)]/7 = 20

⇒ 7x + 21 = 140

⇒ 7x = 119

⇒ x =17

∴ The largest number = x + 6 = 17 + 6 = 23

৮৯৩.

The average daily wage of 10 workers is Tk. 800. If the lowest wage is Tk. 650, then what is the possible maximum wage?

ক
1400 Tk.
খ
1750 Tk.
গ
2150 Tk.
ঘ
2600 Tk.

ব্যাখ্যা

Question: The average daily wage of 10 workers is Tk. 800. If the lowest wage is Tk. 650, then what is the possible maximum wage?

Solution:
10 জন লোকের মোট মজুরি (10 × 800) টাকা = 8000 টাকা
650 টাকা করে 9 জনের মজুরি = (650 × 9) = 5850 টাকা

∴ Possible maximum wage = (8000 - 5850) টাকা = 2150 টাকা।

৮৯৪.

The mean of the first 15 even natural numbers is-

ক
12
খ
14
গ
16
ঘ
17

ব্যাখ্যা

Question: The mean of the first 15 even natural numbers is-

Solution:
First 15 even natural numbers = 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30.

Mean = (2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20 + 22 + 24 + 26 + 28 + 30)/15
= 240/15
= 16

৮৯৫.

The average of 12 numbers is 16 and the average of the first five is 14. What is the average for the rest?

ক
17.5
খ
17.43
গ
18.26
ঘ
16.54

ব্যাখ্যা

Question: The average of 12 numbers is 16 and the average of the first five is 14. What is the average for the rest?

Solution:
Average of twelve numbers = 16
Sum of twelve numbers = 16 × 12 = 192
Average of first five numbers = 14
Sum of first five numbers = 14 × 5 = 70

Total of remaining seven numbers = 192 - 70 = 122
Average of rest = 122/7 = 17.43

৮৯৬.

Tamim has a certain average for 12 innings. In the 13th inning, he scores 120 runs, increasing his average by 5 runs. What is his new average?

ক
42 runs
খ
55 runs
গ
60 runs
ঘ
50 runs

ব্যাখ্যা

Question: Tamim has a certain average for 12 innings. In the 13th inning, he scores 120 runs, increasing his average by 5 runs. What is his new average?

Solution:
Let Tamim's average be x for 12 innings.
So, Tamim scored 12x run in 12 innings.
In the 13th inning, he scored 120 runs then the average became (x + 5) .
And he scored (x + 5) × 13 runs in 13 innings.
Now,
⇒ 12x + 120 = 13(x + 5)
⇒ 12x + 120 = 13x + 65
⇒ x = 120 - 65
∴ x = 55

∴ New average = (x + 5) = 55 + 5 = 60 runs

৮৯৭.

The average age of A, B, C, D and E is 40 years. The average age of A and B is 35 years and the average of C and D is 42 years. Age of E is :

ক
ক) 40 years
খ
খ) 42 years
গ
গ) 44years
ঘ
ঘ) 46 years

ব্যাখ্যা

The average age of A, B, C, D and E is 40 years.
The total age of A, B, C, D and E is 40 × 5 = 200 years.

The average age of A and B is 35 years
The total age of A and B is 35 × 2 = 70 years

The average of C and D is 42 years
The total age of C and D is 42 × 2 = 84 years

Age of E is 200 - 70 - 84 = 46 years

৮৯৮.

Out of 9 persons, 8 persons spent Tk. 30 each for their meals. The ninth one spent Tk. 20 more than the average expenditure of all the nine. The total money spent by all of them was-

ক
Tk. 260
খ
Tk. 290
গ
Tk. 292.50
ঘ
Tk. 400.50

ব্যাখ্যা

Question: Out of 9 persons, 8 persons spent Tk. 30 each for their meals. The ninth one spent Tk. 20 more than the average expenditure of all the nine. The total money spent by all of them was-

Solution:
Let the average expenditure be Tk. x
Then,
9x = 8 × 30 + (x + 20)
⇒ 9x = x + 260
⇒ 8x = 260
⇒ x = 32.50

∴ Total money spent
= Tk. 9x
= Tk. (9 × 32.50)
= Tk. 292.50

৮৯৯.

37.5% of 800 - 6.25% of 1600 =?

ক
200
খ
250
গ
275
ঘ
236

ব্যাখ্যা

Question: 37.5% of 800 - 6.25% of 1600 =?

Solution:
37.5% of 800 - 6.25% of 1600
= (375/1000) × 800 - (625/10000) × 1600
= 300 - 100
= 200

৯০০.

Find out the geometric mean of '3, 25, and 45'.

ক
ক) 15
খ
খ) 85
গ
গ) 1125
ঘ
ঘ) 3375

ব্যাখ্যা

Question: Find out the geometric mean of '3, 25, and 45'.

Solution:
The geometric mean of '3, 25, and 45'

= (3 × 25 × 45)^1/3
= (3375)^1/3
= (15³)^1/3
= 15