ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

Question: Digit 1 is occurring 136 times on writing all of the page numbers of a book. What will be the number of pages in the book?

Solution:
From 1 - 99, the digit 1 occurs 20 times, 
from 100 - 199, the digit 1 occurs 120 times.
So, from 1 to 199, the digit 1 occurs  20 + 120 = 140 times
According to question 1 is occurring only 136 times, which means we need to remove 196, 197, 198, and 199.
So, the required number of pages will be 195.

Question: If x, y, and z are said to be the real numbers, then what is the value of (x - y)³ + (y - z)³ + (z - x)³?

Solution:
Suppose a = (x - y), b = (y - z), and c = (z - x)
On adding a, b, and c we will get
a + b + c = x - y + y - z + z - x
∴ a + b + c =0

So, a³ + b³ + c³ = 3abc [because if a + b + c = 0, then a³ + b³ + c³ = 3abc]

We can say that (x - y)³ + (y - z)³ + (z - x)³ = 3(x - y)(y - z)(z - x)

Question: Which of the following is the least number which will leave the remainder 5, when divided by 8, 12, 16, and 20?

Solution:
First we need to find the least number, so we have to find out the LCM of 8, 12, 16, and 20.
8 = 2 × 2 × 2
12 = 2 × 2 × 3
16 = 2 × 2 × 2 × 2
20 = 2 × 2 × 5

LCM = 2 × 2 × 2 × 2 × 3 × 5 = 240

240 is the least number that is exactly divisible by 8, 12, 16, and 20.
So, the required number that will leave remainder 5 is -
240 + 5 = 245

Question: Between any two numbers, there are-

Solution:
There are infinite rational numbers in between any two numbers.

Question: Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.

Solution:
Required number = H.C.F. of (91 - 43), (183 - 91) and (183 - 43)
= H.C.F. of 48, 92 and 140 = 4.

Question: Which of the following is the unit digit in the product of 853 × 452 × 226 × 1346?

Solution:
Pick up the unit digit of each number and multiply them;
3 in 853
2 in 452
6 in 226
6 in 1346
∴ 3 × 6 × 2 × 6 = 216 (consider the unit digit in the product)

So, the unit digit in the product of 853 × 452 × 226 × 1346 is 6.

Question: The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is-

Solution:
Clearly, the numbers are (23 × 13) and (23 × 14).

∴ Larger number = (23 × 14) = 322.

Question: The sum of odd numbers up to 240 is -

Solution:
Number of odd numbers up to 240 = 240/2 = 120
Sum of first n odd numbers = n²
n = 120
∴ Required sum = 120² = 14400

Question: A man has some hens and cows. If the number of heads be 48 and the number of feet equals 140, then the number of hens will be-

Solution:
Let the number of hens be x and the number of cows be y.
Then, x + y = 48 .... (i)
and
2x + 4y = 140
x + 2y = 70 .... (ii)

Solving (i) and (ii) we get:
x = 26, y = 22.

The required answer = 26.

Question: What is the H.C.F. of 4/9, 10/21 and 20/63?

Solution:
H.C.F of 4/9, 10/21 and 20/63 = (H.C.F of 4,10 and 20)/(L.C.M of 9,21 and 63)

H.C.F of 4, 10 and 20 = 2
L.C.M. of 9, 21 and 63 = 63.

∴ Required H.C.F. = 2/63

Question: Which of the following number is divisible by 9?

Solution:
A number is divisible by 9 if the sum of its digits is divisible by 9.

The Sum of the digits of the given numbers are;
5+6+7+8+5= 31

4+5+6+7+8= 30

6+5+8+8+9= 36

6+7+5+7+8= 33

The sum of the digits of the number 65889 is 36, which is divisible by 9, so the correct answer is 65889.

Question: What is the square root of (8 + 2√15)?

Solution:

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 fourth option satisfies.

66, 77
77 - 66 ≠ 12

70, 84
84 - 70 ≠ 12

94, 108
108 - 94 ≠ 12

84, 96
96 - 84 = 12 and both 84 and 96 divisible by 12.

Question: What smallest number should be subtracted from 9805 so that it is divisible by 8?

Solution:
On dividing 9805 by 8, the remainder is 5. So, 5 is the smallest number which should be subtracted from 9805 to make it divisible by 8.

Question: If (64)² - (36)² = 20x, then what is the value of x?

Solution:
Here, we can apply the formula of a² - b² = (a + b) (a - b)

So, after applying the formula, we will get
(64)² - (36)² = 20x
⇒ (64 + 36) (64 - 36) = 20x
⇒ 100 × 28 = 20x
⇒ x = (100 × 28)/20
∴ x = 140

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

Solution:
long = 5 meters 44 cm = 500 + 44 cm = 544 cm
broad = 3 meters 74 cm = 300 + 74 = 374 cm

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

Question: If the two-third of three - fourth of a number is 34, what will be the 20% of that number?

Solution:
Let the number be X.

According to the question,
(2/3) × (3/4) × X = 34
⇒ (1/2) × X = 34
⇒ X = 68

Now, 20% of 68 is = 68 × (20/100) = 13.6

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

Question: Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together?

Solution:
L.C.M. of 2, 4, 6, 8, 10, 12 is 120.
So, the bells will toll together after every 120 seconds = 2 minutes

∴ In 30 minutes, they will toll together 30/2 + 1 = 15 + 1 = 16 times.

Question: 7X2 is a three-digit number in which X is a missing digit. If the number is divisible by 6, the missing digit is-

Solution:
The given number is divisible by 6, so it would be divisible by 2 and 3. As the last digit is 2, whatever be the value of X, it would be divisible by 2.
Now, 7 + X + 2 = 9 + X, must be divisible by 3.
So, X = 3 makes the number divisible by 3, so 3 is the required digit.

Question: The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then sum of the number is-

Solution:
Let the numbers be 2x and 3x.
Then, their L.C.M. = 6x.
So,
6x = 48 
∴ x = 8.

The numbers are 16 and 24.

Hence, required sum = (16 + 24) = 40.

Question: If the sum of two numbers is considered as 'a' and their product is considered as 'b', then what will be the sum of their reciprocals?

Solution:
Suppose the numbers are P and Q

So, according to the question -
P + Q = a
PQ = b
Sum of reciprocals of P and Q is = 1/P + 1/Q
= (Q + P)/PQ
= a/b