Bank Math Master

Question: The sum of four consecutive two-digit odd numbers, when divided by 10, become a perfect square. Which of the following can possibly be one of these four numbers?

Solution: 
Using options,
We find that four consecutive odd numbers are 37, 39, 41 and 43
The sum of these 4 numbers is 160, when divided by 10 we get 16 which is a perfect square.
Thus, 41 is one of the odd numbers

Question: Two alarm clocks ring their alarms at regular intervals of 50 seconds and 48 seconds. If they first beep together at 2 : 30 PM, at what time will they beep again for first time?

Solution: 
They will ring together after,
LCM of 48 and 50 secs.
48 = 2 × 2 × 2 × 2 × 3
50 = 2 × 5 × 5

∴ LCM = 2 × 2 × 2 × 2 × 3 × 5 × 5 = 1200 secs
= 1200/60 = 20 min. [1 min = 60 sec]

∴ They will beep together at 2 : 30 + 20 = 2 : 50 PM

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

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Solution: 

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Solution:

Question: If , the value of x is = ?

Solution: 
Given that, 
52/x = √(169/289)
⇒ 52/x = 13/17
⇒ x = (52 × 17)/13
⇒ x = 4 × 17
∴ x = 68

Question: The number of numbers from 1 to 200 which are divisible by neither 3 nor 7 is-

Solution: 
The required number = Number of numbers, which are (divisible by 3 + divisible by 7 - divisible by 21)

Now,
Number of number divisible by 3,
 = {(198 - 3)/3} + 1
= 65 + 1 = 66 

Number of number divisible by 7
= {(196 - 7)/7} + 1
= 27 + 1 = 28 

And,
Number of number divisible by 21,
= {(189 - 21)/21} + 1
= 8 + 1 =  9

Thus, the divisible value = 66 + 28 - 9 = 85 

Thus, number of numbers which are not divisible by 3 or 7
= 200 - 85 = 115

Question: Find the LCM of 2.5, 0.5 and 0.175.

Solution: 
Given that, 
2.5 = 25/10
0.5 = 5/10
0.175 = 175/1000

Now,
 LCM of Numerators is 25, 5, 175 = 175
and HCF of Denominators is 10, 10, 1000 = 10

We know,
LCM of two or more fractions is given by,
LCM = LCM of Numerators/HCF of Denominators
= 175/10
= 17.5

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Solution:

Question: A sum of Tk. 1360 has been divided among A, B and C such that A gets 2/3 of what B gets and B gets 1/4 of what C gets. B's share is-

Solution:
Let,
C′s share = Tk. x
Then,
B′s share = Tk. x/4,
A′s share = Tk.(2/3) × (x/4) = Tk. x/6

∴ x/6 + x/4 + x = 1360
⇒ 17x/12 = 1360
⇒ x = (1360 × 12)/17
∴ x = Tk. 960

∴ B′s share = Tk.(960/4) = Tk. 240

Question: The least number by which 294 must be multiplied to make it a perfect square is :

Solution:
294 = 7 × 7 × 2 × 3
Here, 2 and 3 have odd exponents.
Multiplying by 2 × 3 = 6 will make 294 a perfect square.

∴ Multiplying by 6 will make 294 a perfect square.

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

Question: The sum of two numbers is 18. The greatest product of these two number can be-

Solution: 
Given that, 
a + b = 18
So, maximum of (a × b) will be only when a = b
Thus, a = b = 9
∴ Maximum of (a × b) = 9 × 9 = 81.

So the greatest product is 81.

Question: What is the difference between the biggest and the smallest fraction among 2/3, 3/4, 4/5 and 5/6 ?

Solution:
Converting each of the given fractions into decimal form, we get,
2/3 = 0.66
3/4 = 0.75
4/5 = 0.8
5/6 = 0.833

Since 0.833 > 0.8 > 0.75 > 0.66
So, 5/6 > 4/5 > 3/4 > 2/3

∴ Required difference = (5/6) - (2/3) = 1/6

Question: 5 - [4 - {3 - (3 - 3 - 6)}] is equal to

Solution:
Given,
5 - [4 - {3 - (3 - 3 - 6)}]
= 5 - [4 - {3 - (-6)}]
= 5 - [4 - {3 +6}]
= 5 - [4 - {9}]
= 5 - [4 - 9]
= 5 - [-5]
= 5 + 5
= 10

Question: The square root of (7 + 3√5) (7 - 3√5) is ?

Solution:
Given that, 
{(7 + 3√5) (7 - 3√5)}
= √{(7)² - (3√5)²}
= √(49 - 45)
= √4
= 2

Question: The value of √ 0.01 + √0.81 + √1.21 + √0.0009 is = ?

Solution:
Given that,
√ 0.01 + √0.81 + √1.21 + √0.0009
= √ (1/100) + √(81/100) + √(121/100) + √(9/10000)
= (1/10) + (9/10) + (11/10)+ (3/100)
= 0.1 + 0.9 + 1.1 + 0.03
= 2.13

Question: The maximum number of students among whom 1001 pens and 910 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is-

Solution:
We use prime factorization,
1001 = 7 × 11 × 13
910 = 2 × 5 × 7 × 13 
∴ Common factors are 7 and 13

∴ HCF = 7 × 13 = 91

∴ Maximum number of students = HCF of 1001 and 910 = 91 

Question: If n is a whole number greater then 1, then n² (n² - 1) is always divisible by -

Solution: 
Given that,
n²(n² - 1)
= 2²(2² - 1)     ; [put n = 2]
= 4(4 - 1)
= 4 × 3
= 12
Check the option it is divisible by 12.
Take n = 3
= 3³(3² - 1)
= 9(9 - 1)
= 9 × 8
= 72
It is divisible by 12

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Solution: