এনএসআই [NSI] নিয়োগ প্রস্তুতি [Archived]

Question: If x is an odd integer, then which of the following is true?

Solution:
Now, suppose x = 1. So,
ক) 5.1 - 2 = 3 ;  odd
খ) 5.1² + 2 = 7 ; odd
গ) 5.1² + 3 = 8 ; even

Shortcut:  x is odd ⇒ x² is odd ⇒ 5x² is odd ⇒ (5x² + 2) is odd.

Question: Find the smallest number of 6 digits which is exactly divisible by 349.

Solution:
The smallest 6 digit number is 100000.
When divided by 349, the remainder is 186.
So, Required Number = 100000 + (349 - 186) = 100163

Question: If a number is decreased by 4 and divided by 6, the result is 8. What would be the result if 2 is subtracted from the number and then it is divided by 5?

Solution:
Let the number be x. 
Then,
(x - 4)/6 = 8
⇒ x - 4 = 48
∴ x = 52

Now, for the second condition
(x - 2)/5
= (52 - 2)5
= 10

Question: Find the number which when multiplied by 15 is increased by 196.

Solution:
Let, 
the number be x. 

Then,
15x - x = 196
⇒ 14x = 196
⇒ x = 14

Question: In a two-digit positive number, the digit in the unit’s place is equal to the square of the digit in ten’s place, and the difference between the number and the number obtained by interchanging the digits is 54. What is 40% of the original number?

Solution:
Let, ten’s digit = x.
Then, unit’s digit = x².
∴ number = 10x + x².

Since x² > x, 
so, the number formed by interchanging the digits is greater than the original number. 

∴ (10x² + x) - (10x + x²) = 54
 ⇒ 9x² - 9x = 54 
⇒ x² - x = 6 
⇒ x² - x - 6 = 0 
⇒ x² - 3x + 2x - 6 = 0
⇒ x(x - 3) + 2(x - 3) = 0
⇒ (x - 3) (x + 2) = 0 
⇒ x = 3. [∵ it is positive number]

 So, ten’s digit = 3, unit’s digit = 3² = 9. 
∴ Original number = 39.
 Required result = 40% of 39
= (40/100) × 39 = 15.6.

Question: The product of two whole numbers is 37. The square root of the difference of the numbers is-

Solution:
Let, the number be ‘a’ and ‘b’.
Then ab = 37.

Since 37 is a prime number.So, it has exactly two factors 1 and 37.
∴ a = 1 , b = 37

So, √(b - a)
= √(37 - 1)
= √36 
= 6

Question: The ages of Nishi and Rimi are in the ratio 6 : 5 respectively. After 9 years, the ratio of their ages will be 9 : 8. What is the difference in their ages now?

Solution:
Let,
Nishi’s age be 6x years.
Then, Rimi’s age = 5x years. 

Now
∴ (6x + 9)/(5x + 9) = 9/8
⇒ 8(6x + 9) = 9(5x + 9) 
⇒ 48x - 45x = 81 - 72 
⇒ 3x = 9 
⇒ x = 3 

∴ Difference in their ages = (6x - 5x) years
= x years
= 3 years.

Question: Sabina 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 3 years ago?

Solution:
Let, Sabina's age 8 years ago be x years.
Then, her present age = (x + 8) years.

according to the question,
x + 8 = (9/7) . x 
⇒ 7x + 56 = 9x 
⇒ 2x = 56 
⇒ x = 28

∴ Sabina’s age now = (x + 8) years
= (28 + 8) = 36 years.

Her daughter’s age now = (1/6) × 36 years
= 6 years.
Her daughter’s age 3 years ago = (6 – 3) = 3 years.

Question: The age of the father ten years ago was thrice the age of his son. Ten years hence, the father's age will be twice the age of his son. What is the ratio of their present ages?

Solution:
Let,
son's age 10 years ago be x years.
Then, father’s age 10 years ago = 3x years.

Son's present age = (x + 10) years
Father's present age = (3x + 10) years.

according to the question,
(3x + 10) + 10 = 2 (x + 10 + 10)
⇒ 3x + 20 = 2 (x + 20) 
⇒ 3x + 20 = 2x + 40 
∴ x = 20

Ratio of present ages of father and the son
= (3x + 10)/(x + 10)
= {(3 × 20) + 10}/(20 + 10)
= 70/30 
= 7 : 3

Question: Four metal rods of lengths 78 cm, 104 cm, 117 cm, and 169 cm are to be cut into parts of equal length. Each part must be as long as possible. What is the maximum number of pieces that can be cut?

Solution:
Maximum length of each part
= H.C.F. of 78 cm, 104 cm, 117 cm, and 169 cm
= 13 cm.

That means, if we take the H.C.F the length we get can be used to cut all 4 rods equally, so that each piece will have the same length.

∴ Number of pieces = (78 + 104 + 117 + 169)/13
= 468/13 
= 36

Question: What is the least number which, when doubled, will be exactly divisible by 12, 18, 21, and 30?

Solution:
Let, the least number be x
Then, 2x must be divisible by 12, 18, 21, and 30.

∴ 2x = LCM(12,18,21,30)

L.C.M. of 12, 18, 21, 30  =  2 × 2 × 3 × 3 × 7 × 5
= 1260

∴ Required number = 1260/2 = 630

Question: If the sum of two numbers is 36, and their HCF and LCM are 3 and 105 respectively, what is the sum of the reciprocals of the two numbers?

Solution:
Let, the numbers be a and b.

Then, a + b = 36 and ab =  3 × 105 = 315 [∵ Product of the numbers = HCF×LCM]

∴ sum of their reciprocals
= (1/a) + (1/b)
= (a + b)/ab
= 36/315
= 4/35

Question: The sum of two numbers is 528, and their HCF is 33. How many pairs of numbers satisfy these conditions?

Solution:
Let,
the required numbers be 33a and 33b.
where, a and b are coprime integers (i.e., HCF of a and b is 1).

Then,
33a + 33b = 528
⇒ a + b = 16.

Now, co-primes with sum 16 are (1, 15), (3, 13), (5, 11) and (7, 9).

∴ Required numbers are:
(33 × 1, 33 × 15), (33 × 3, 33 × 13), (33 × 5, 33 × 11), (33 × 7, 33 × 9).

 So, the number of such pairs is 4.

Question: Which of the following numbers does not lie between 4/5 and 7/13?

Solution:
4/5 = 0.8, and 7/13 = 0.53

Now checking the options,

1/2 = 0.5
2/3 = 0.66
3/4 = 0.75
5/7 = 0.714

Clearly, 0.5 does not lie between 0.53 and 0.8

∴ 1/2 does not lie between 4/5 and 7/13.

Question: If 39/x = √(169/289), then what is the value of x?

Sol:
39/x = √(169/289)
⇒ 39/x = 13/17
⇒ x = (39 × 17)/13
∴ x = 51

Question: 555.05 + 55.5 + 5.55 + 5 + 0.55 =?

Solution:
           

Question: When is converted into fraction, the result will be?

Solution:

Question: The total marks obtained by a student in English, Biology, and History together is 150 more than the marks obtained by him in Biology. What is the average marks obtained by him in English and History together?

Solution:
Let English = E, Biology = B, History = H

Given, 
E + B + H = B + 150 
⇒ E + H = 150 

∴ Average of English and History
= (E + H)/2
= 150/2
= 75

Question: The sum of three consecutive even numbers is 60 more than the average of these numbers. Which of the following is the 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 = 60
⇒ (3x + 6) - (3x + 6)/3 = 60
⇒ {3(3x + 6) - (3x + 6)}/3 = 60
⇒ (9x + 18 - 3x - 6) = 180
⇒ (6x + 12) = 180
⇒ x = 168/6
∴ x = 28

∴ Largest number = x + 4
= 28 + 4 = 32.

Question: The average age of all the students in a class is 22 years. The average age of the boys in the class is 25 years, and that of the girls is 18 years. If the number of girls in the class is 24, find the number of boys in the class.

Solution:
Let,
the number of boys in the class be x. 

Then,
22 (x + 24) = 25x + (18 × 24) 
⇒ 22x + 528 = 25x + 432
⇒ 3x = 96
⇒ x = 32

∴ The number of boys in the class is 32

Question: The average weight of 18 boys was recorded as 60 kg. If the weight of the teacher was added, the average increased by 2 kg. What was the teacher's weight?

Solution:
Average weights of 18 boys = 60 kg
Total weights of 18 boys = 18 × 60 kg = 1080 kg

The weight of the teacher was added then average increase by 2 kg
So, new average weight is 62 kg
Total people including teacher = 19

∴ total weight including teacher
= 62 × 19 kg
= 1178 kg

∴ Weight of teacher = 1178 – 1080 = 98 kg

Question: What is the next number in the following sequence?
12, 41, 169, 850, ?

Solution: 
(12 × 3) + 5 = 41
(41 × 4) + 5 =169
(169 × 5) + 5 = 850
(850 × 6) + 5 = 5105

Question: What will come at the place of question mark?
6, 24, 60, 120, 210, ?

Solition:
2³ - 2 = 6
3³ - 3 = 24
4³ - 4 = 60
5³ - 5 = 120
6³ - 6 = 210
7³ - 7 = 336