পেট্রোবাংলা প্রিলি ও লিখিত সমন্বিত প্রস্তুতি [Archived]

Question: Simplify the expression using BODMAS rule:
(3/2) of (4/7){(10 × 3) - (8 × 2)}

Solution:
(3/2) of (4/7){(10 × 3) - (8 × 2)}
= (6/7){30 - 16}
= (6/7)(14)
= 6 × 2
= 12

Question: What is the greatest number which divides 639, 1065 and 1491 exactly?

Solution:
The greatest number will be H.C.F. of 639, 1065 and 1491

H.C.F. of 639 and 1065 is 213.
H.C.F. of 213 and 1491 is 213.

Question: Maria spent 1/3 of the money her grandparents gave her on an adventure book. She also spent 1/9 of the money on a bag of candy. What fraction of the payment has Maria spent?

Solution:
Maria spent on an adventure book 1/3 of the money.
Maria spent on candy 1/9 of the money.

∴ Total spent by Maria = (1/3  + 1/9) of the money
= (3 + 1)/9
= 4/9

Question: The average marks of 13 papers is 40. The average marks of the first 7 papers are 42 and that of the last seven papers is 35. Find the marks obtained in the 7th paper.

Solution:
The average of 13 papers is 40,
so the sum = 13 × 40 = 520

The average of first 7 papers is 42,
so the sum will be = 7 × 42 = 294

The average of last 7 papers is 35,
so the sum will be = 7×35 = 245

So, the marks obtained in the 7th paper will be = 539 - 520 = 19

Question: The ratio of Sara's age 4 years ago and Vaishali's age after 4 years is 1 : 1. Presently, the ratio of their ages is 5 : 3. Find the ratio between Sara's age 4 years hence and Vaishali's age 4 years ago.

Solution:
Currently, the ratio of their ages is 5 : 3.
Suppose, their ages are: 5x and 3x.

Sara’s age 4 years ago = 5x - 4
Vaishali’s age after 4 years = 3x + 4
Ratio of Sara’s age 4 years ago and Vaishali's age after 4 years is 1 : 1
Therefore,
(5x - 4)/(3x + 4) = 1/1
⇒ 5x - 4 = 3x + 4
⇒ 2x = 8
∴ x = 4

We are required to find the ratio between Sara’s age 4 years hence and Vaishali’s age 4 years ago.
Sara's age 4 years hence: (5x + 4)
Vaishali's age 4 years ago: (3x - 4)
Putting the value of x, we get:
(5x + 4)/(3x - 4)
= 24/8
= 3/1
= 3 : 1

Question: The six digit number 54321A is divisible by 9 where A is a single digit whole number. Find A.

Solution:
A number is divisible by 9, when the sum of its digits is divisible by 9.
Here, 5 + 4 + 3 + 2 + 1 + A = 15 + A should be divisible by 9.
Therefore, A = 3 gives 15 + 3 = 18 as the sum of digits, which is divisible by 9.

Question: The H.C.F. of two numbers is 12 and their difference is 12. Which of the following can be the numbers?

Solution:
The difference of requisite numbers must be 12 and each should be divisible by 12. Checking the options given, only the fifth option satisfies.

84, 96
96 - 84 = 12
HCF of 84 and 96 is 12

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

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

Question: Daud is younger to Rohan by 9 years. If their ages are in the respective ratio of 4 : 5, how old is Daud?

Solution:
Let Rohan's age be x years.
Then,
Daud's age = (x - 9) years.

ATQ,
(x - 9)/x = 4/5
⇒ 5x - 45 = 4x
∴ x = 45

Hence, Daud's age = (x - 9) = 45 - 9 = 36 years.

Question: Find the greatest 6-digit number, which is a multiple of 12.

Solution:
Greatest six-digit number is 999999.
Divide this number by 12 and get remainder as 3.
Since the remainder is 3, if you subtract 3 from the number, the remaining number will be a multiple of 12.
So the greatest such number will be 999999 - 3 = 999996

Since the greatest 6-digit number divisible by 12 is 999996, and this option is not listed in the given choices, the correct answer should indeed be: ঙ) None of these.

Question: Find the side of the largest square slab which can be paved on the floor of a room 5 meters 44cm long and 3 meters 74 cm broad.

Solution:
The side of the square slab is the H.C.F. of 544 and 374 cm i.e. 34.

Question: An electrician has three and seven-sixteenths cm of wire. He needs only two and five-eighths cm of wire for a job. How much wire doesn't he need to cut?

Solution:

Question: The average salary of 30 officers in a firm is Tk.120 and the average salary of laborers is Tk. 40. Find the total number of laborers if the average salary of the firm is Tk. 50.

Solution:
The sum of salary of officers will be = 30 × 120 = 3600
Let the number of labourers = X.
ATQ,
3600 + 40X = 50(30 + X)
⇒ 3600 + 40X = 1500 + 50X
⇒ 2100 = 10X
∴ X = 210

Question: A man said to his son, "I was one-third of your present age when you were born". If the present age of the man is 48 years, find the present age of the son.

Solution:
Present age of the son be P, he was born P years ago.
The age of the man was: (48 - P).
His age when the son was born should be equal to 1/3 of P.
(48 - P) = (1/3)P
⇒ 144 - 3P = P
⇒ 4P = 144
∴ P = 36

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.

Question: 3 different pieces of iron are of varying length are given to a student which are 44cm, 22 cm,55 cm respectively. He has to form rods of maximum length such that no iron waste is left. Find the maximum length of such rod.

Solution:
Maximum possible length of such rod = (H.C.F. of 44, 22, 55) cm = 11 cm.

Question: One half of the students in a school are girls, 3/5 of these girls are studying in lower classes. What fraction of girls are studying in lower classes?

Solution:

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

Question: Sagar is 50 years old and Nazma is 40 years old. How long ago was the ratio of their ages 3 : 2?

Solution:
Here, we have to calculate: How many years ago the ratio of their ages was 3 : 2.
Let us assume x years ago
At present: Sagar is 50 years and Nazma is 40 years
x years ago:
Sagar’s age = (50 - x) and Nazma's age = (40 - x)
Given, the ratio of their ages was 3 : 2
(50 - x)/(40 - x) = 3/2
Solving, we get: x = 20
Therefore, the answer is 20 years.

Question: A boy was asked to multiply a number by 25 but by mistake he multiplied by 45 and the answer was 200 more than the correct answer. What was the number?

Solution:
Let the correct number be x.
The boy was supposed to multiply the number by 25, so the correct answer would be 25x
Instead, he multiplied the number by 45, giving 45x

According to the problem, the wrong answer is 200 more than the correct answer:
45x = 25x + 200
⇒ 45x - 25x = 200
⇒ 20x = 200
∴ x = 10

Question: Find the HCF of 1/3 , 8/7, 9/11.

Solution:
HCF of given Numbers : HCF(1, 8, 9)/LCM(3, 7, 11) = 1/231.

Question: Maddy reads three-fifth of 75 pages of his lesson. How many more pages he need to complete the lesson?

Solution:
Maddy reads = (3/5) of 75
= (3/5) × 75
= 45 pages.

Maddy has to read = 75 - 45.
= 30 pages.

Therefore, Maddy has to read 30 more pages.

Question: Ten years ago, the age of mother was three times the age of her son. After ten years, mother’s age will be twice that of his son. Find the ratio of their present ages.

Solution:
10 years ago, age of mother was three times the age of her son.
Say, the age of son was x and mother's age was 3x.
At present:
Mother's age is (3x + 10)
and son’s age is (x + 10)

After ten years:
Mother's age will be (3x + 10) +10
and son’s age will be (x + 10) + 10.

Given that, mother’s age is twice that of son after ten years.
(3x + 10) +10 = 2 [(x + 10) + 10]
(3x + 20) = 2[x + 20]
Solving the equation, we get x = 20
(3x + 10) : (x + 10) = 70 : 30 = 7 : 3.

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

Solution:
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.

Question: The LCM of the two numbers is 360, and their HCF is 24. If one of the numbers is 120, find the other number.

Solution:
We know that the product of the HCF and LCM of two numbers is equal to the product of the two numbers.
Let the number be a and b.

So, for two numbers a = 120 and b with HCF = 24 and LCM = 360:

HCF × LCM = a × b
⇒ 24 × 360 = 120 × b
⇒  (24 × 360)/120 = b
⇒  24 × 3 = b
∴ b = 72.

Question: A herd of cows gives 4 litres of milk each day. But each cow gives one-third of total milk each day. They give 24 litres milk in six days. How many cows are there in the herd?

Solution:
A herd of cows gives 4 litres of milk each day.
Each cow gives one-third of total milk each day = 1/3 of 4
Therefore, each cow gives 4/3 of milk each day.
Total no. of cows = 4 ÷  (4/3)
= 4 × (3/4)
= 3

Therefore there are 3 cows in the herd.

Question: The average weight of 39 men travelling to Cox's Bazar is 30. If an obese man with weight 130 kg join them. What will be the average weight of the people travelling to Cox's Bazar?

Solution:
If the weight of the man would have been 30,
then the average weight would have been the same. So, the extra 100 kg that the obese man brings with him would be distributed equally amongst all of them, i.e. 100/40 = 2.5
So, the average becomes 30 + 2.5 = 32.5

Question: Nisha is 15 years elder to Romi. If 5 years ago, Nisha was 3 times as old as Romi, then find Nisha’s present age.

Solution:
Let age of Romi be y
Nisha is 15 years elder than Romi = (y + 15).
So Nisha's age 5 years ago = (y + 15 - 5).
Romi's age before 5 years = (y - 5)

5 years ago, Nisha is 3 times as old as Romi
(y + 15 - 5) = 3 (y - 5)
⇒ (y + 10) = (3y - 15)
⇒ 2y = 25
⇒ y = 12.5

Romi's age = 12.5 years
Nisha's age = (y + 15) = (12.5 + 15) = 27.5 years.