IBA ফ্যাকাল্টি ভিত্তিক প্রস্তুতি

Question: B is twice as old as A. C is twice as old as B. If the difference between ages of A and C is 12 years, find the age of B.

Solution:
Let A’s age = x years
Then,
B’s age = 2x
And C’s age = 2 × B = 2 × 2x = 4x

Given condition, Difference between ages of A and C = 12 years
C - A = 12
⇒ 4x - x = 12
⇒ 3x = 12
∴ x = 4

∴ B’s age = 2x = 2 × 4 = 8 years

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

Question: What will come at the place of question mark ?
7, 26, 63, 124, 215, 342, ?

Solution:
The terms are given in a series
(2³ - 1) = 7
(3³ - 1) = 26
(4³ - 1) = 63
(5³ - 1) = 124
(6³ - 1) = 215
(7³ - 1) = 342
So, the missing term is,
(8³ - 1) = 511.

Question: X’s age is 20% more than Y’s age. If the sum of their ages if 20 more than the difference of their ages, find their respective ages.

Solution: 
Let X's age be x and Y's age be y

First condition,
X’s age is 20% more than Y’s age.
 ⇒ xx = y + 0.20y
 ⇒ x = 1.20y
 ∴ x = (6/5)y.......(1)

Second condition,
The sum of their ages is 20 more than the difference of their ages.
x + y = (x - y) + 20
⇒ x + y = x - y + 20
⇒ 2y = 20
⇒ y = 20/2
∴ y = 10 years

Now that we have Y's age, substitute  y = 10 into Equation 1
⇒ x = (6/5)y = (6/5) × 10 = 12
∴ x = 12 years

X's age is 12 years and Y's age is 10 years.

Question: A man earns N dollars a month and spends S dollars a month on rent. If he then spends 3/8 of the remainder on food, how much, in dollars, is left over for other expenses, in terms of N and S?

Solution:
Given that,
Monthly income = N dollars
Rent = S dollars

∴ Remaining after rent = N - S

And, He spends 3/8​ of the remainder on food.
∴ Food expense = (3/8)(N - S)

∴ Left for other expenses = Remaining after rent - Food expense
= (N - S) - {(3/8)(N - S)}
= (8/8)(N - S) - {(3/8)(N - S)}
= {(8 - 3)/8}(N - S)
= (5/8)( N - S)

Question: The average temperature for Wednesday, Thursday and Friday was 40°C. The average for Thursday, Friday and Saturday was 41° C. If temperature on Saturday was 44° C, what was the temperature on Wednesday?

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

Average temperature for Thursday, Friday and Saturday = 41° C
∴ Total temperature = 41 × 3 = 123° C

And,
Temperature on Saturday = 44° C

Now,
(Thursday + Friday + Saturday) - (Wednesday + Thursday + Friday) = 123 - 120
⇒ Saturday - Wednesday = 3
∴ Wednesday = 44 - 3 = 41° C

Question: What will come at the place of question mark ?
8, 28, 116, 584, ?

Solution:
1^st term = 8
2^nd term = (8 × 3) + 4 = 28 
3^rd term = (28 × 4) + 4 = 116 
4^th term = (116 × 5) + 4 = 584 
5^th term = (584 × 6) + 4 = 3508

Question: Age of Kamal will be 4 times the age of Rina in 6 years from today. If ages of Kamal and Mahi are 7 times and 6 times the age of Rina respectively, what is present age of Kamal?

Solution:
Let,
Rina's present age = R years
Then,
Kamal's present age = 7 × R = 7R
Mahi's present age = 6 × R = 6R

Age of Kamal will be 4 times the age of Rina in 6 years.
Kamal’s age after 6 years = 7R +6
Rina’s age after 6 years = R + 6

ATQ,
7R + 6 = 4(R + 6)
⇒ 7R + 6 = 4R + 24
⇒ 7R - 4R = 18
⇒ R = 18/3
∴ R = 6

∴ Kamal's present age = 7 × R = 7R = 7 × 6 = 42 years

Question: In a certain office, 1/3 of the workers are women, 1/2 of the women are married and 1/3 of the married women have children. If 3/4 of the men are married and 2/3 of the married men have children, what part of workers are without children?

Solution:
Given that,
Total women = 1/3
Married women = 1/2 of 1/3 = 1/6
Women who have a child = 1/3 of married women

∴ Women who has child = 1/3 of 1/6 = 1/18

And,
Total men = 2/3       (∵ 1/3 are women)
Married man = 3/4 of the total man
Married man = 3/4 of 2/3 = 1/2
Man who has a child = 2/3 of a married man

∴ Men who has child = 2/3 of 1/2 = 1/3

Now,
Men + Women (​​​​having child) = (1/18) + (1/3) = 7/18

∴ Part that don't have child = 1 - (7/18) = 11/18

∴ 11/18 workers don't have children.

Question: Find the average of first 97 natural numbers.

Solution:
We know,
Average = Sum of first n natural numbers​/n

And the sum of first n natural numbers = n(n + 1)/2

∴ Average = {n(n + 1)/2}/n = (n + 1)/2
= (97 + 1)/2  ; [Substitute n = 97]
= 98/2
= 49

Question: A boy agrees to work at the rate of one Taka on the first day, two Taka on the second day, and four Taka on third day and so on. How much will the boy get if he started working on the 1^st of February and finishes on the 20^th of February?

Solution:
Given that,
1st term, a = 1
Common ratio, r = 2   ; r > 1

We know, 
Sum Sn = a × {(rⁿ - 1)/(r - 1)}
= 1 × {(2²⁰ - 1)/(2 - 1)}. ; [Putting, a = 1, r = 2 and n = 20]
= 2²⁰ - 1

So the boy will get 2²⁰ - 1 Takas if he works from February 1^st to February 20^th.

Question: Father is aged three times more than his son Rakib. After 8 years, he would be two and a half times of Rakib's age. After further 8 years, how many times would he be of Rakib's age? 

Solution:
Let,
Rakib's present age be x years.
Then, father's present age =(x + 3x) years
= 4x years.

ATQ,
(4x + 8) = {5(x + 8)}/2
⇒ 8x + 16 = 5x + 40
⇒ 3x = 24
∴ x = 8

Now,
(4x + 16)/(x + 16)  = 48/24 = 2

∴ Father would be 2 times of Rakib's age.

Question: 


Solution:

Question: The average wages of a worker during a fortnight comprising 15 consecutive working days was Tk. 90 per day. During the first 7 days, his average wages was Tk. 87 per day and the average wages during the last 7 days was Tk. 92 per day. What was his wage on the 8^th day?

Solution:
The total wages earned during the 15 days that the worker worked ,
= 15 × 90
= Tk. 1350

The total wages earned during the first 7 days
= 7 × 87
= Tk. 609

The total wages earned during the last 7 days
= 7 × 92
= Tk. 644

Total wages earned during the 15 days,
= wages during first 7 days + wage on 8th day + wages during the last 7 days.

ATQ,
1350 = 609 + wage on 8^th day + 644
⇒ wage on 8^th day = 1350 - 609 - 644 = Tk. 97
∴ wage on 8^th day = Tk. 97

Question: The sum of the first 16 terms of an Arithmetic Progression(AP) whose first term and third term are 5 and 15 respectively is-

Solution:
1^st term = 5
3^rd term =15
∴ 5 + d + d = 15
⇒ 2d = 10
∴ d = 5

16^th term = a + 15d
= 5 + 15 × 5
= 80

∴ The sum of the first 16 terms = (n/2)[2a + (n - 1)d]
= (16/2)[2 × 5 + (16 - 1)5]
= 8 × (10 + 75)
= 8 × 85
= 680

Question: Fuad, Rubel and Pavel are cousins. Fuad’s age is one-third of Rubel and Pavel is five years elder than Rubel. If the sum of the age of the cousins is 40, find the ages of Fuad.

Solution:
Let,
Rubel’s age = x
Fuad’s age = x/3
Pavel’s age = x + 5

ATQ,
Fuad + Rubel + Pavel = 40
(x/3) + x + (x + 5) = 40
⇒ (x/3) + 2x + 5 = 40
⇒ (x + 6x)/3 = 40 - 5
⇒ 7x/3 = 35
⇒ x = (35 × 3)/7 = 15
∴ x = 15 years

∴ Fuad age is = 15 × (1/3) = 5 years.

Question: If the fractions 7/13, 2/3, 4/11, 5/9 are arranged in ascending order, then the correct sequence is ?

Solution:
Given that,
(7/13) = 0.538
(2/3) = 0.666
(4/11) = 0.3636
(5/9) = 0.5555

Out of 2/3, 7/13, 4/11, 5/9

2/3 is the largest number followed by 5/9 then 7/13 and the smallest is 4/11.

∴ The ascending order will be 4/11, 7/13, 5/9, 2/3. 

Question: Which one of the following numbers can be removed from the set S = {1, 2, 3, 4, 5, 6, 7} without changing the average of set S?

Solution:
Given the set is  S = {1, 2, 3, 4, 5, 6, 7}
Sum of elements  = 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28

There are 7 numbers in the set.
 Average = Sum of elements/Number of elements
= 28/7 = 4

And,
Removing a number that is equal to the current average will not change the average of the remaining numbers. The average of the set is 4, which is an element in the set S.

If 4 is removed, the new set is  {1, 2, 3, 5, 6, 7}
∴ New sum  = 1 + 2 + 3 + 5 + 6 + 7 = 24
And New number of elements  = 6

∴ New average = 24/6 = 4

The number that can be removed from the set S = {1, 2, 3, 4, 5, 6, 7} without changing the average of the set is 4.

Question: Find the wrong term in the following series.
1200, 1188, 1164, 1116, 1020, 828, 484

Solution:
Given series:
1200, 1188, 1164, 1116, 1020, 828, 484
The series decreases with multiples of 12, doubling each time.
1^st term = 1200
2^nd term = 1200 - 12 = 1188
3^rd term = 1188 - 24 = 1164
4^th term = 1164 - 48 = 1116
5^th term = 1116 - 96 = 1020
6^th term = 1020 - 192 = 828
7^th term = 828 - 384 = 444

The series has 484 as the last term, but it should be 444.

 Hence the wrong term in the series is 484.