Bank Math Master

Question: Present ages of Abir and Rubel are in the ratio of 5 : 4 respectively. Three years hence, the ratio of their ages will become 11 : 9 respectively. What is Rubel's present age in years?

Solution:
Let,
the present ages of Abir and Rubel be 5a years and 4a years respectively

Then,
(5a + 3)/(4a +3) = 11/9
⇒ 9(5a + 3) = 11(4a + 3)
⇒ 45a + 27 = 44a + 33
⇒ 45a - 44a = 33 - 27
⇒ a = 6

∴ Rubal's present age = 4a = 4 × 6 = 24 years

Question: If log₅(a² + a) − log₅(a + 1) = 2, then the value of a is -

Solution:
log₅(a² + a) − log₅(a + 1) = 2
⇒ log₅{(a² + a)/(a + 1)} = 2
⇒ log₅{a(a + 1)/(a + 1)} = 2
⇒ log₅a = 2
⇒ a = 5²
∴ a = 25 

Question: In doing a question of division with zero remainder, a candidate took 12 divisor instead of 15. The quotient obtained by him was 35. The correct quotient is-

Solution:
Divisor taken = 12
Quotient obtained = 35,
Remainder = 0
∴ Dividend = (12 × 35) = 420

Now, Dividend = 420,
Divisor = 15
Remainder = 0
∴ Quotient = 420/15
= 28

Question: The sum of the three consecutive even numbers is 48 more then the average of these numbers. Which of the following is the third largest of these numbers?

Solution: 
Let, the numbers be = x, (x + 2) and (x + 4)

Then,
(x + x + 2 + x + 4) - (x + x + 2 + x + 4)/3 = 48
⇒ (3x + 6) - (3x + 6)/3 = 48
⇒ 9x + 18 - 3x - 6 = 144
⇒ 6x + 12 = 144
⇒ 6x = 144 - 12
⇒ 6x = 132
∴ x = 22

The third largest of these numbers is = x + 4 = 22 + 4 = 26

Question: Ten years ago, a man was six times as old as his son. Two years hence, twice his age will be equal to four times the age of his son. What is the presesnt age of the son?

Solution:
Let,
Son's age 10 years ago be a years.
then, man's age 10 years ago = 6a years

∴ Son's present age = (a + 10) years
And man's present age = (6a + 10) years

ATQ, 2{(6a + 10) + 2} = 4{(a + 10) + 2}
⇒ 2(6a + 12) = 4(a + 12)
⇒ 12a + 24 = 4a + 48
⇒ 12a - 4a = 48 - 24
⇒ 8a = 24
⇒ a = 3

So, son's present age = (a + 10) = 3 + 10 = 13 years.

Question: log₃x + log₉x² + log₂₇x³ = 9, then x equals to-

Solution:

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

Question: 5^{3x - 2} = 625, find the value of x.

Solution:
5^{3x - 2} = 625
⇒ 5^{3x - 2} = 5⁴
⇒ 3x - 2 = 4
⇒ 3x = 6
∴ x = 2

Question: Tania got married 8 years ago. Today her age is 9/7 times her age at the time of her marriage. At present her daughter's age, is one-sixth of her age. What was her daughter's age 2 years ago?

Solution: 
Let,
Tania's age 8 years ago be = a years 
Now, her present age = (a + 8) years

then,
a + 8 = (9/7)a
⇒ 9a = 7a + 56
⇒ 9a - 7a = 56
⇒ 2a = 56
∴ x = 28
Tania's age now = a + 8 = 28 + 8 = 36 years
and her daughter's age now = (1/6) × 36 = 6 years

∴ Tania's daughter's age 3 years ago = (6 - 2) = 4 years

Solution: One year ago, Muna was four times as old as her daughter. Six years hence, Muna's age will exceed her daughter's age by 9 years. The ratio of the present ages of Muna and her gaughter is-

Solution:
Let,
daughter's age one years ago be = x years
Then, Muna's age one years ago be = 4x years

ATQ,
(4x + 1) + 6 = (x + 1 + 6) + 9
⇒ (4x + 7) = (x + 16)
⇒ 4x - x = 16 - 7
⇒ 3x = 9
∴ x = 3

Ratio of Muna and her daughter age now = (4x + 1)/(x + 1)
= (4 · 3 + 1)/(3 + 1) = 13/4
= 13 : 4

Question: The value of (6log₁₀1000)/(3log₁₀100) is equal to-

Solution:
(6log₁₀1000)/(3log₁₀100)
= (6log₁₀10³)/(3log₁₀10²)
= {(6×3)log₁₀10}/{(3×2)log₁₀10}
= 18/6
= 3

Question: Find the number of factors of 600.

Solution:
600= 2³ × 3¹ × 5²
Number of factors = (3 + 1)(1 + 1)(2 + 1)
= 4 × 2 × 3
= 24

So, the number of factors of 600 is 24.

প্রশ্ন: The average expenditure of a man for the first five months is TK.1200 and for the next seven months is TK.1300. If he saves TK.2900 in that year, his monthly average income is 

Solution:
Total annual expenditure = TK.(5 × 1200) + (7 × 1300)
= TK.(6000 + 9100)
= TK. 15100
His total annual income = Total expenditure + Total savings
= (15100 + 2900)
= TK. 18000

∴ Average monthly income = 18000/12 = TK. 1500

Question: If 5^{(a + 3)} = 25^{(3a - 4)} then the value of a is = ?

Solution:
5^{(a + 3)} = 25^{(3a - 4)}
⇒ 5^{(a + 3)} = (5²)^{(3a - 4)}
⇒ 5^{(a + 3)} = 5^{(6a - 8)}
⇒ a + 3 = 6a - 8
⇒ 5a = 11
∴ a = 11/5

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

Question: log₂√6 + log₂(√2/3) = ?

Solution:
log₂√6 + log₂(√2/3)
= log₂[√{6 · (2/3)}]
= log₂√(2 · 2)
= log₂√(2²)
= log₂2
= 1

Question: Mr. P has three daughters, S, M and T. Three years ago, when Mr. P was twice as old as T, he was 30 years older than M. Now he is 47 years older than S. In 4 years, S will be half as old as T. What is the age of S?

Solution:
ATQ,
K - 3 = 2(T - 3) .......... (1)
K = M + 30 ......... (2)
K = S + 47 ............ (3)
∴ 2(S + 4) = T + 4
⇒ 2s + 8 = T + 4
⇒ 2s + 8 - 4 = T
⇒ 2s + 4 = T
From (1) we get K - 3 = 2(2S + 4 - 3)
⇒ K - 3 = 4S + 2
⇒ K = 4S + 5

From (3) we get, 4S + 5 = S + 47
⇒ 4S - S = 47 - 5
⇒ 3S = 42
∴ S = 14

Question: The difference of the number consisting of two digits and the number formed by interchanging the digit is always divisible by-

Solution:
Let the ten's digit be x and the unit's digit be y.
So, the number = 10x + y
After interchanging the positions of the number's digits, the number will be = 10y + x

∴ Difference = (10x + y) - (10y + x) 
= 10x + y - 10y - x
= 9x - 9y
= 9(x - y); which is divisible by 9.

Question: If 2^{n - 1} + 2^{n + 1} = 160, then the value of n is = ?

Solution:
2^{n - 1} + 2^{n + 1} = 160
⇒ 2^{n - 1} (1 + 2²) = 160
⇒ 2^{n - 1} · 5 = 160
⇒ 2^{n - 1} = 160/5
⇒ 2^{n - 1} = 32
⇒ 2^{n - 1} = 2⁵
⇒ n - 1 = 5
∴ n = 6