ব্যাংক নিয়োগ বিষয়ভিত্তিক প্রস্তুতি

Question: Simplify the expression using BODMAS rule (105 + 206) - 550 ÷ 5² + 10

Solution:
(105 + 206) - 550 ÷ 5² + 10
= 311 - 550 ÷ 25 + 10
= 311 - 22 + 10
= 289 + 10
= 299

Question: If the sum two numbers is 31 and their product is 240, then find the absolute difference between the numbers.

Solution:
Let two numbers be x and y
We are given that, sum of two numbers x + y = 31 and product = xy = 240
Therefore,
x - y = √{(x + y)² - 4xy}
Substituting the values, we get
x - y = √{(31)² - 4 × 240}
= √(961 -  960)
= √1
= 1
The required difference between the numbers is 1.

Question: H.C.F. of 513, 1134 and 1215 is-

Solution:

H.C.F. of 513, 1134 and 1215 = 3 × 3 × 3 = 27

Question: The mean weight of a group of seven boys is 56 kg. The individual weights (in kg) of six of them are 52, 57, 55, 60, 59 and 55. Find the weight of the seventh boy.

Solution:
Mean weight of 7 boys = 56 kg.
Total weight of 7 boys = (56 × 7) kg = 392 kg.

Total weight of 6 boys = (52 + 57 + 55 + 60 + 59 + 55) kg
= 338 kg.

Weight of the 7th boy = (total weight of 7 boys) - (total weight of 6 boys)
= (392 - 338) kg
= 54 kg.

Hence, the weight of the seventh boy is 54 kg.

Question: A’s age after 15 years would be equal to 5 times his age 5 years ago. Find his age 3 years hence.

Solution: 
Let A’s present age be ‘n’ years.
According to the question,
n + 15 = 5(n - 5)
⇒ n + 15 = 5 n – 25
⇒ 4n = 40
⇒ n = 10

∴  A’s present age = 10 years
Therefore, A’s age 3 years hence = 10 + 3 = 13 years

Question: Find the smallest three digit number which is a multiple of 7.

Solution:
Smallest three digit number =100.
Divide this number by 7 and get the remainder as 2.
If you subtract 2 from the number the remaining number will be a multiple of 7, 100 - 2 = 98 which is two digit number.
Now if we add 7 so that we get the smallest three digit number which is a multiple of 7.
Required number = 98 + 7 = 105.

Question: Find the three consecutive odd numbers whose sum of the squares is 2531.

Solution:
Let three consecutive odd numbers be x, x + 2, x + 4.
x² + (x + 2)² + (x + 4)² = 2531
Simplifying we get,
⇒ x² + 4x - 837=0
⇒ x² + 31x - 27x - 837=0         [ 27 × 31 = 837 and also the difference between 27 and 31 is 4 ]
⇒ (x + 31) (x - 27)
⇒ x = 27 or x = - 31
Hence, the value of
x = 27
(x + 2) = 27 + 2 = 29
(x + 4) = 27 + 4 = 31

Question: Find the largest number of 4-digits divisible by 12, 15 and 18.

Solution:
Required largest number must be divisible by the L.C.M. of 12, 15 and 18
L.C.M. of 12, 15 and 18
12 = 2 × 2 × 3
15 =5 × 3
18 = 2 × 3 × 3
L.C.M. = 180
Now divide 9999 by 180, we get remainder as 99
The required largest number = (9999 - 99) = 9900
Number 9900 is exactly divisible by 180.

Question: A cricketer has a mean score of 58 runs in nine innings. Find out how many runs are to be scored by him in the tenth innings to raise the mean score to 61.

Solution:
Mean score of 9 innings = 58 runs.
Total score of 9 innings = (58 × 9) runs = 522 runs.

Required mean score of 10 innings = 61 runs.
Required total score of 10 innings = (61 × 10) runs = 610 runs.

Number of runs to be scored in the 10th innings 
= (total score of 10 innings) - (total score of 9 innings)
= (610 - 522) = 88. 

Hence, the number of runs to be scored in the 10th innings = 88.

Question: The product of the ages of A and B is 240. If twice the age of B is more than A’s age by 4 years, what was B’s age 2 years ago?

Solution: 
Let A’s present age be x years.
Then, B’s present age = 240/x years

So, according to question
2(240/x ) -  x = 4
⇒ 480 - x² = 4x
⇒ x² + 4x - 480 = 0
⇒ (x + 24) (x - 20) = 0
⇒ x = 20

B’s present age = 240/20 = 12 years
Thus, B’s age 2 years ago = 12 - 2 = 10 years

Question: If both 11² and 3³ are factors of the number a × 4³ × 6² × 13¹¹, then what is the smallest possible value of 'a'?

Solution:
a × 4³ × 6² × 13¹¹ can be expressed in terms of its prime factors as a × 2⁸ × 3² × 13¹¹

11² is a factor of the given number.
If we do not include 'a', 11 is not a prime factor of the given number.
If 11² is a factor of the number, 11² should be a part of 'a'

3³ is a factor of the given number.
If we do not include 'a', the number has only 3² in it.
Therefore, if 3³ has to be a factor of the given number 'a' has to contain 31 in it.

Therefore, 'a' should be at least 11² × 3 = 363 if the given number has 11² and 3³ as its factors.

The question is "what is the smallest possible value of 'a'?"
The smallest value that 'a' can take is 363

Question: The sum of squares of three numbers is 138 and the sum of their products taken two at a time is 131. Find their sum.

Solution:
Let the three numbers be x, y and z.
Sum of squares of three numbers is 138 and sum of their products taken two at a time is 131
Therefore,
x² + y² + z² = 138
xy + yz + zx = 131


(x + y + z)² = x² + y² + z² + 2(xy + yz + zx)
(x + y + z )² = 138 + 2(131)
(x + y + z )² = 400
Hence, (x + y + z) = 20

Question: Calculate H.C.F. of 2/3 , 16/81 , 8/9.

Solution:
H.C.F. of 2/3 , 16/81 , 8/9 = H.C.F of (2, 16, 8)/L.C.M of (3, 81, 9)
= 2/81

Question: The mean of five numbers is 28. If one of the numbers is excluded, the mean gets reduced by 2. Find the excluded number.

Solution:
Mean of 5 numbers = 28.
Sum of these 5 numbers = (28 × 5) = 140.

Mean of the remaining 4 numbers = (28 - 2) =26.
Sum of these remaining 4 numbers = (26 × 4) = 104.

Excluded number
= (sum of the given 5 numbers) - (sum of the remaining 4 numbers)
= (140 - 104)
= 36. 

Hence, the excluded number is 36.

Question: The present age of a mother is 3 years more than three times the age of her daughter. Three years hence, the mother’s age will be 10 years more than twice the age of the daughter. Find the present age of the mother.

Solution: 
Let the daughter’s present age be ‘n’ years.
∴ Mother’s present age = (3n + 3) years

So, according to the question
(3n + 3 + 3) = 2 (n + 3) + 10
⇒ 3n + 6 = 2n + 16
⇒ n = 10

Hence, mother’s present age = (3n + 3) = ((3 × 10) + 3) years = 33 years

Question: How many different positive integers exist between 10⁶ and 10⁷, the sum of whose digits is equal to 2?

Solution:
Between 10 and 100, that is 10¹ and 10², we have 2 numbers, 11 and 20.
Between 100 and 1000, that is 10² and 10³, we have 3 numbers, 101, 110 and 200.

Therefore, between 10⁶ and 10⁷, one will have 7 integers whose sum will be equal to 2.

Alternative:
All numbers between 10⁶ and 10⁷ will be 7 digit numbers.
There are two possibilities if the sum of the digits has to be '2'.

Possibility 1:
Two of the 7 digits are 1s and the remaining 5 are 0s.
The left most digit has to be one of the 1s. That leaves us with 6 places where the second 1 can appear.
So, a total of six 7-digit numbers comprising two 1s exist, sum of whose digits is '2'.

Possibility 2:
One digit is 2 and the remaining are 0s.
The only possibility is 2000000.

Total count is the sum of the counts from these two possibilities = 6 + 1 = 7

Question: Find a positive number which when increased by 11 is equal to 60 times the reciprocal of the number.

Solution:
Let the number be x.
A positive number (x) increased by 11 is equal to 60 times the reciprocal of the number (1/x)

x + 11 = 60/x
⇒ x² + 11x - 60 = 0
⇒ x² + 15x - 4x - 60 = 0
⇒ x(x + 15) - 4(x + 15) = 0
⇒ (x + 15)(x - 4) = 0
∴x = 4

The positive number (x) = 4

Question: Find L.C.M. of 2/3, 8/9, 64/81, 10/27.

Solution:
L.C.M. = L.C.M. of Numerator/H.C.F. of Denominator

L.C.M. of numerators (2, 8, 64, 10)
2 = 2¹
8 = 2³
64 = 2⁶
10 = 2 × 5

L.C.M of 2, 8, 64, 10 = 2⁶ × 5 = 320

H.C.F. of denominators = 3, 9, 81, 27
3 = 3¹
9 = 3²
81 = 3⁴
27 = 3³

H.C.F. of 3, 9, 81, 27 = 3

∴ L.C.M. of 2/3, 8/9, 64/81, 10/27 = 320/3

Question: The average height of 30 boys was calculated to be 150 cm. It was detected later that one value of 165 cm was wrongly copied as 135 cm for the computation of the mean. Find the correct mean.

Solution:
Calculated average height of 30 boys = 150 cm.
Incorrect sum of the heights of 30 boys
= (150 × 30)cm
= 4500 cm.

Correct sum of the heights of 30 boys
= (incorrect sum) - (wrongly copied item) + (actual item)
= (4500 - 135 + 165) cm
= 4530 cm.

Correct mean = correct sum/number of boys
= (4530/30) cm
= 151 cm.

∴ Hence, the correct mean height is 151 cm.

Question: The ratio of present ages of A and B is 6 : 7. Five years hence, this ratio would become 7 : 8. Find the present age of A.

Solution: 
Let
A’s present age = 6n years
B’s present age = 7n years

So, according to the question
(6n + 5)/(7n + 5) = 7/8
⇒ 48 n + 40 = 49 n + 35
⇒ n = 5

Thus, A’s present age = 6n = 30 years

Question: A number when divided by a divisor leaves a remainder of 24. When twice the original number is divided by the same divisor, the remainder is 11. What is the value of the divisor?

Solution:
Let the original number be 'a'.
Let the divisor be 'd'.
Let the quotient of dividing 'a' by 'd' be 'x'.
and the remainder is 24.
∴ a = dx + 24
⇒ 2a = 2(dx + 24) 
⇒ 2a = 2dx + 48

Twice the original number is divided by 'd' means 2a is divided by d.
When (2dx + 48) is divided by 'd' the remainder is 11.
2dx is divisible by 'd' and will therefore, not leave a remainder.
The remainder of 11 would be the remainder of dividing 48 by d.

The question essentially becomes "What number will leave a remainder of 11 when it divides 48?"
When 37 divides 48, the remainder is 11.

Hence, the divisor is 37.

Question: 5 times a positive number is less than its square by 24. What is the integer?

Solution:
Let the unknown number be x.
5 times a positive number = 5x
5 times a positive number is less than its square by 24
x² - 5x = 24
⇒ x² + 3x - 8x - 24 = 0
⇒ x(x + 3) - 8(x + 3) = 0
⇒ (x - 8)(x + 3) = 0
∴ x = 8

8 is the required integer.

Question: Find the sum of two numbers, which are greater than 29 and have H.C.F. and L.C.M. of 29 and 4147 respectively.

Solution:
Product of two numbers = Product of their H.C.F. and L.C.M.
Product of two numbers = 29 × 4147 = 120263

Two numbers are greater than 29.
Therefore, let the two numbers be 29x and 29y.
So, 29x × 29y = 120263
xy = 143

Factors of 143 are: 1, 11, 13, and 143
Case: 1) If we consider factors of 143 as 1 and 143 (co-primes), then we get the value of two numbers x and y = (29 and 4147) ------ (Which is wrong: As it is given that, the numbers are greater than 29)

Case: 2) If we consider factors of 143 as 11 and 13 (co-primes), then we get the value of two numbers x and y = (319, 377) ------ (These two values are greater than 29. So, it is the correct answer)
Therefore, the two numbers are 319 and 377.

Sum of two numbers = 319 + 377 = 696

Question: The mean of 16 items was found to be 30. On rechecking, it was found that two items were wrongly taken as 22 and 18 instead of 32 and 28 respectively. Find the correct mean.

Solution:
Calculated mean of 16 items = 30.
Incorrect sum of these 16 items = (30 × 16) = 480.

Correct sum of these 16 items
= (incorrect sum) - (sum of incorrect items) + (sum of actual items)
= [480 - (22 + 18) + (32 + 28)]
= 500.

Therefore, correct mean = 500/16 = 31.25.
Hence, the correct mean is 31.25.

Question: Age of mother 10 years ago was 3 times the age of her son. After 10 years, mother’s age will be twice that of his son. Find the ratio of their present ages.

Solution:
We are given that, age of mother 10 years ago was 3 times the age of her son
So, let age of son be x and as mother’s age is 3 times the age of her son, let it be 3x, three years ago.
At present: Mother’s age will be (3x + 10) and son’s age will be (x + 10)
After 10 years: Mother’s age will be (3x + 10) +10 and son’s age will be (x + 10) + 10

Mother’s age is twice that of son
(3x + 10) +10 = 2 [(x + 10) + 10]
(3x + 20) = 2[x + 20]
Solving the equation, we get x = 20

We are asked to find the present ratio.
(3x + 10) : (x + 10) = 70 : 30 = 7 : 3