Introduction to Real Numbers Important Questions

                                  Important Questions                                  

                        Introduction to Real Numbers

Question 1:

The H.C.F. of the numbers, 55125 and 2250, is of the form a2 × ba.

 What are the respective values of a and b?

Solution :

The numbers, 55125 and 2250, can be prime factorised as 55125 = 3 × 3 × 5 × 5 × 5 × 7 × 7

= 32 × 53 × 72

2250 = 2 × 3 × 3 × 5 × 5 × 5

= 2 × 32 × 53

∴ H.C.F. (55125, 2250) = 32 × 53

Comparing with a2 × ba, we obtain

a = 3 and b = 5

Thus, the value of a and b are 3 and 5 respectively.

Hence, the correct answer is option A.

Option1 :

3 and 5

Option2 :

2 and 5

Option3 :

2 and 3

Option4 :

3 and 7

Option5 :
Question 2:

The steps for finding the H.C.F. of 2940 and 12348 by Euclid’s division lemma are as follows.

12348 = a × 4 + b

a = b × 5 + 0

What are the respective values of a and b?

Solution :

The steps to find the H.C.F. of 12348 and 2940 are as follows.

12348 = 2940 × 4 + 588

2940 = 588 × 5 + 0

Comparing with the given steps, we obtain

a = 2940 and b = 588

Thus, the value of a and b are 2940 and 588 respectively.

Hence, the correct answer is option B.

Option1 :

2352 and 588

Option2 :

2940 and 588

Option3 :

2352 and 468

Option4 :

2940 and 468

Option5 :
Question 3:

If the LCM of the numbers 216 and 576 is of the form a2b × (a + 1)b,

 then what is the values of and b?

Solution :

The numbers 216 and 576 can be prime factorised as

216 = 2 × 2 × 2 × 3 × 3 × 3 = 23 × 33

576 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 = 26 × 32

∴ L.C.M. (216, 576) = 26 × 33 = 22 × 3 × (2 + 1)3

Given, L.C.M. (216, 576) is of the form  a2b × (a + 1)b.

⇒ 22 × 3 × (2 + 1)3 = a2b × (a + 1)b

Comparing the two sides of the equation, we have a = 2 and = 3.

Hence, the correct answer is option C.

Option1 :

6 and 2

Option2 :

3 and 2

Option3 :

2 and 3

Option4 :

6 and 3

Option5 :
Question 4:

Use the following information to answer the next question.

The given steps represent the procedure for finding the H.C.F. of the numbers, 

360 and 252, by using Euclid’s division algorithm.

360 = 252 × 1 + a

252 = a × 2 + b

ab × 3 + 0

What are the respective values of a and b?

Solution :

The H.C.F. of the given positive integers, 360 and 252, can be found by 

using Euclid’s division algorithm. This can be done through the following steps.

Step 1

As 360 > 252, by applying Euclid’s division lemma to the numbers 360 and 252, we obtain

360 = 252 × 1 + 108

Step 2

As the remainder 108 ≠ 0, by applying Euclid’s division lemma to the numbers 252 and 108, 

we obtain

252 = 108 × 2 + 36

Step 3

Again, as the remainder 36 ≠ 0, by applying Euclid’s division lemma to the numbers 108 and 36, 

we obtain

108 = 36 × 3 + 0

Thus, H.C.F. (360, 252) = 36

Hence, the values a and b are 108 and 36 respectively.

The correct answer is B.

Option1 :

30 and 21

Option2 :

108 and 36

Option3 :

72 and 36

Option4 :

30 and 36

Question 5:
On dividing a number X by 126, the remainder is 47. The remainder when the
 same number is divided by 18 is 
Solution :
We have,
Let the number X when divided by 126 gives x as quotient and 47 as the remainder
Thus, = 126x + 47 = 18×7x+18×2+11
X=18×7x+2+11
Thus, when X is divided by 18, it gives 11 as the remainder.
Hence, the correct answer is C.
Option1 :
7
Option2 :
9
Option3 :
11
Option4 :
13
Option5 :
Question 6:

Use the following information to answer the next question.

The initial step for finding the H.C.F. of the numbers 4093 and 741

 by Euclid’s division algorithm is given as 4093 = 741 × 5 + 388

Which alternative correctly shows the remaining steps?

Solution :

The H.C.F. of two positive integers 4093 and 741 can be found by using Euclid’s division algorithm. This can be done through the following steps.

Step 1

As 4093 > 741, by applying Euclid’s division lemma to the numbers 4093 and 741, we obtain

4093 = 741 × 5 + 388

Step 2

As the remainder 388 ≠ 0, by applying Euclid’s division lemma to the numbers 741 and 388, 

we obtain

741 = 388 × 1 + 353

Step 3

Again, as the remainder 353 ≠ 0, by applying Euclid’s division lemma to the numbers 388 and 

353, we obtain

388 = 353 × 1 + 35

Step 4

Similarly, apply Euclid’s division lemma till the remainder is zero. The remaining steps are

353 = 35 × 10 + 3

35 = 3 × 11 + 2

3 = 2 × 1 + 1

2 = 1 × 2 + 0

∴ H.C.F. (4093, 741) = 1

Thus, the remaining steps in the given problem are correctly listed in alternative A.

741 = 388 × 1 + 353

388 = 353 × 1 + 35

353 = 35 × 10 + 3

35 = 3 × 11 + 2

3 = 2 × 1 + 1

2 = 1 × 2 + 0

The correct answer is A.

Option1 :

741 = 388 × 1 + 353

388 = 353 × 1 + 35

353 = 35 × 10 + 3

35 = 3 × 11 + 2

3 = 2 × 1 + 1

2 = 1 × 2 + 0

Option2 :

741 = 388 × 2 – 35

388 = 35 × 12 – 32

35 = 32 × 2 – 29

32 = 29 × 2 – 26

29 = 26 × 2 – 23

26 = 23 × 2 – 20

23 = 20 + 3

Option3 :

741 = 800 – 59

800 = 59 × 14 – 26

59 = 60 – 1

Option4 :

741 = 700 + 41

700 = 41 × 17 + 3

41 = 40 + 1

Question 7:
HCF of 378 and 360 is represented by 2x+53×4y. Find the relation between and y
Solution :
On applying Euclid's division lemma on 378 and 360, we can find the HCF as: 

Since, the HCF is 18, we equate the given expression with 18. 
18=2x+53×4y
18=2x+212y
2x=18-212y
x=9-106y
Hence, the correct answer is option D


Option1 :
x=9y
Option2 :
x=106 y
Option3 :
x=9+106 y
Option4 :
x=9-106y
Option5 :
Question 8:

A rectangular floor of dimensions 280 cm × 320 cm is to be tiled with square shaped tiles. 

What should be the maximum length of the side of each square tile?

Solution :

Side of the square tile must be the common divisor of the dimensions of the floor.

∴ Maximum lengh of the side of a square tile = H.C.F. of 280 cm and 320 cm

To find the H.C.F., let us apply Euclid’s division algorithm.

Here 320 > 280.

∴ 320 = 280 × 1 + 40

280 = 40 × 7 + 0

As the remainder is zero, the H.C.F. of 320 and 280 is 40.

∴ Maximum length of the side of the square tile should be 40 cm.

The correct answer is D.

Option1 :

20 cm

Option2 :

28 cm

Option3 :

32 cm

Option4 :

40 cm

Question 9:

Rahul wants to measure the length of the boundary of a rectangular courtyard exactly

 by using a stick. If the dimensions of the courtyard are 79 m and 42 m, what is the maximum

 length of the stick he should use?

Solution :

Length of the stick must be the common divisor of the dimensions of the courtyard.

∴ Maximum length of the stick = H.C.F. of 79 m and 42 m

To find the H.C.F., let us apply Euclid’s division algorithm.

Here 79 > 42.

∴ 79 = 42 × 1 + 37

42 = 37 × 1 + 5

37 = 5 × 7 + 2

5 = 2 × 2 + 1

2 = 1 × 1 + 1

1 = 1 × 1 + 0

As the remainder is 0, the H.C.F. of 79 and 42 is 1.

∴ Maximum length of the stick should be 1 m = 100 cm.

The correct answer is D.

Option1 :

79 cm

Option2 :

84 cm

Option3 :

96 cm

Option4 :

100 cm

Question 10:

The rectangular-shaped tiles of dimension 50 cm × 36 cm are to be paved in a

 square-shaped courtyard. Assuming that none of the tiles was broken or overlapped with 

each other, what is the least area of the courtyard?

Solution :

The least length of the square courtyard can be found by finding the L.C.M. of 50 and 36.

Now, 50 = 2 × 5 × 5

36 = 2 × 2 × 3 × 3

∴ L.C.M (50, 36) = 2 × 2 × 3 × 3 × 5 × 5 = 900

Least length of the side of the square courtyard = 900 cm = 9 m

Least area of the square courtyard = 9 m × 9 m = 81 m2

Thus, the least area of the square courtyard is 81 m2.

Hence, the correct answer is option B.

Option1 :

51.84 m2

Option2 :

81 m2

Option3 :

116.64 m2

Option4 :

144 m2

Fundamental Theorem of Arithmetic


Question 1:

The number 75600 can be factorised as a1×b2×c3×d4. What is the value of (a + c)?

Solution :

According to the fundamental theorem of arithmetic, every number can be factorised as

 a product of its primes in a unique way.

Therefore, the number 75600 can be prime-factorised as

75600 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 5 × 5 × 7 = 24 × 33 × 52 × 71

On comparing it to (a1 × b2 × c3 × d4), we obtain

a = 7, b = 5, c = 3, d = 2

Thus, the value of (a + c) is (7 + 3) = 10.

Hence, the correct answer is option D.

Option1 :

5

Option2 :

7

Option3 :

8

Option4 :

10

Option5 :
Question 2:

What is the prime factorisation of 5148?

Solution :

The number 5148 can be prime-factorised as:

5148 = 2 × 2 × 3 × 3 × 11 × 13
         = 22 × 32 × 11 × 13

Hence, the correct answer is option B.

Option1 :

 23 × 32 × 11 × 13

Option2 :

 22 × 32 × 11 × 13

Option3 :

 22 × 32 × 112 × 13

Option4 :

 22 × 32 × 11 × 132

Option5 :
Question 3:

The number 5457375 can be factorised as a1 × b2 × c3 × d4. The value of (a + b + c + d) is

Solution :

According to the fundamental theorem of arithmetic, every number can be factorised as

 a product of its primes in a unique way.

Therefore, the number 5457375 can be prime-factorised as

5457375 = 3 × 3 × 3 × 3 × 5 × 5 × 5 × 7 × 7 × 11 = 34 × 53 × 72 × 111

On comparing it to (a1 × b2 × c3 × d4), we get

a = 11, b = 7, c = 5, d = 3

Thus, the value of (a + b + c + d) is (11 + 7 + 5 + 3) = 26.

Hence, the correct answer is option B.

Option1 :

23

Option2 :

26

Option3 :

28

Option4 :

32

Option5 :
Question 4:

What is the LCM of 36, 96, and 72?

Solution :

The numbers 36, 96, and 72 can be prime-factorised as

36 = 2 × 2 × 3 × 3 = 22 × 32

96 = 2 × 2 × 2 × 2 × 2 × 3 = 25 × 3

72 = 2 × 2 × 2 × 3 × 3 = 23 × 32

LCM is the product of the highest powers of all the prime factors of the numbers.

∴ LCM (36, 72, 96) = 25 × 32 = 288

The correct answer is C.

Option1 :

144

Option2 :

254

Option3 :

288

Option4 :

336

Question 5:

If the number 7875 can be prime factorized as 7875 = 3a × b3 × c1,

 then what are the respective values of ab, and c?

Solution :

By the fundamental theorem of arithmetic, it is known that every composite number can be

 factorized as the product of primes in a unique way apart from the order in which the prime

 factors occur.Therefore, the number 7875 can be prime factorized as

7875 = 3 × 3 × 5 × 5 × 5 × 7

⇒ 7875 = 32 × 53 × 71

On comparing it with 7875 = 3a × b3 × c1, we obtain a = 2, b = 5, and c = 7

Thus, the respective values of ab, and c are 2, 5, and 7.

Hence, the correct answer is option D.

Option1 :

2, 3, and 1

Option2 :

2, 2, and 5

Option3 :

2, 3, and 5

Option4 :

2, 5, and 7

Option5 :
Question 6:

Which of the following alternatives represents a correct statement?

Solution :

Consider the statement given in alternative C.

The number 135s can be expressed as 33× 5 × s.

Let s be an even number.

i.e., s is a multiple of 2, so we can write s = 2n, where n is a natural number

Now, 135s can be expressed as 33× 5 × 2n, where n is a natural number.

⇒ 135s = 33× (5 × 2) × n

⇒ 135s = 33× 10 × n

∴ 135s will always end with zero if s is an even number.

The correct answer is C.

Option1 :

The number 2p × 3q will always end with zero if p and q are natural numbers.

Option2 :

The number 2n × 5m will always end with zero if n and m are integers.

Option3 :

The number 135s will always end with zero if s is an even number.

Option4 :

The number 1113p will always end with zero if is an odd number.

Question 7:

Use the following information to answer the next question.

The floor of Mayank’s drawing room is 306 inches long and 136 inches wide.

 He wishes to tile the floor with square tiles such that the side of each tile is x inches,

 where x is a natural number.

If Mayank uses identical tiles, then what is the minimum number of tiles used in tiling the floor?

Solution :

It is given that all tiles are identical and minimum number of tiles are used to tile the floor.

Therefore, side of each tile, x is the H.C.F. of the dimensions of the room.

The number 306 can be prime factorised as: 

306 = 2 × 3 × 3 × 17

The number 136 can be prime factorised as: 

136 = 2 × 2 × 2 × 17

∴ H.C.F. of 306 and 136 = 2 × 17 = 34

⇒ Side of each tile = 34 inches

∴ Number of tiles used in tiling the floor  


Thus, Mayank can tile the floor with a minimum of 36 tiles.

The correct answer is C.

Option1 :

31

Option2 :

34

Option3 :

36

Option4 :

39

Question 8:

Use the following information to answer the next question.

The least common multiple and the greatest common divisor of two numbers

are 336 and 8 respectively.

If one of the numbers is 48, then the second number is

Solution :

Least common multiple × greatest common divisor = product of the two

numbers

Let the second number be n.

⇒ 8 × 336 = n × 48

⇒ n = 


The correct answer is B.

Option1 :

54

Option2 :

56

Option3 :

62

Option4 :

64

Question 9:

If n is any natural number, then which of the following expressions ends in 0?

Solution :

If a number ends in 0, then it must be divisible by 10 i.e., it must have 2 and 5 as its factors.

Thus, (2 × 5)n can end in 0 for any natural number n.

The correct answer is C.

Option1 :

(3 × 2)n

Option2 :

(4 × 3)n

Option3 :

(2 × 5)n

Option4 :

(6 × 2)n

Question 10:

If the H.C.F. of two numbers a and b is 7, then what is the L.C.M of a and b?

Solution :

It is known that: Product of two numbers = Product of the H.C.F. and L.C.M. of the two numbers

∴ ab = 7 × L.C.M.

⇒ L.C.M. =
Thus, L.C.M of a and b is  
                   

The correct answer is C.

Option1 :

ab

Option2 :

7ab

Option3 :
             
Option4 :
            

Applications of Euclid's Division Lemma


Question 1:

Which of the following sets of numbers can be expressed in the form 4q + 1 or 4q + 3,

 where q is a whole number?

Solution :

Let a be any positive odd integer and let b = 4.

Then, by Euclid’s division lemma, whole numbers q and r exist such that

a = 4q + r              0 ≤ r < 4

Then, r = 0 or 1 or 2 or 3.

Thus, a can be 4q or 4q + 1 or 4q + 2 or 4q + 3.

Since is odd, a cannot be 4q or 4q + 2, as they are divisible by 2.

Thus, a can be expressed in the form 4q + 1 or 4q + 3.

Thus, every positive odd integer can be expressed in the form 4q + 1 or 4q + 3, 


where q is a whole number.

The correct answer is C.

Option1 :

Cubes of integers

Option2 :

Squares of even integers

Option3 :

Positive odd integers

Option4 :

Positive even integers

Question 2:

If k is any whole number, then any perfect square can be written in which of the following forms?

Solution :

Let a be any positive integer. On applying division algorithm on and b = 2,

 we obtain a = 2q + r; 0 ≤ < 2

Hence, the possible remainders are 0 and 1.

This means a is of the form 2q or 2q + 1.

Thus, all positive integers can be written in the form 2q or 2q + 1.

On squaring 2q and 2q + 1, we obtain 4q2 and (4q2 + 4q + 1).

We can write them as 4(q2) and [4(q2 + q) +1] or, 4k and 4k +1, where k is a constant.

Thus, any perfect square can be written in the form 4or 4k + 1.

The correct answer is A.

Option1 :

4k, 4k + 1

Option2 :

4k + 1, 4k + 3

Option3 :

4k + 2, 4k + 3

Option4 :

4k, 4k + 2

Question 3:

Use the following information to answer the next question.

Three statements are given as
I. All positive integers 
can be expressed in the form 

3m or 3m + 1 or 3m +2, where m is a whole number.

II. The square of any positive integer can be expressed

 in the form 3m or 3m + 2, where m is a whole number.

III. The cube of any positive integercan be expressed in the

 form 3m or 3m + 1,where m is a whole number.

Among the given statements,

Solution :

Statement I

Let a be any positive integer and let b = 3

Then, by Euclid’s division lemma, whole numbers q and r exist such that

a = 3q + r                      0 ≤ r < 3

Then, r = 0 or 1 or 2.

Thus, a can be 3q or 3q + 1 or 3q + 2.

Thus, every positive integer can be expressed in the form 3m or 3m + 1 or 3m + 2,

 where m is a whole number.

Thus, statement I is correct.

Statement II

As proved in statement I, any positive integer can be expressed in

 the form 3m or 3m + 1 or 3m + 2, where m is a whole number.

Consider their squares i.e., (3q)2, (3q + 1)2, and (3q + 2)2.

Case (i)

(3q)2 = 9q2 = 3(3q2) = 3m, where m = (3q2) is a whole number.

Case (ii)

(3q + 1)2 = 9q2 + 6q + 1 = 3(3q2 + 2q) + 1 = 3m + 1,

 where m = (3q2 + 2q) is a whole number.

Case (iii)

(3q + 2)2 = 9q2 + 12q + 4 = 9q2 + 12q + 3 + 1 = 3(3q2 + 4q + 1) + 1

= 3m + 1, where m = (3q2 + 4q + 1) is a whole number.

Thus, the square of every positive integer can be expressed in the form 3m or 3m + 1,

 where m is a whole number.

Thus, statement II is incorrect.

Statement III

As proved in statement I, any positive integer can be expressed in the

 form 3m or 3m + 1 or 3m + 2, where m is a whole number. Consider their cubes i.e., (3q)3,

 (3q + 1)3, and (3q + 2)3.

Case (i)

(3q)3 = 27q3 = 3(9q3) = 3m, where m = (9q3) is a whole number.

Case (ii)

(3q + 1)3 = 27q3 + 27q2 + 9q + 1 = 3(9q3 + 9q2 + 3q) + 1

= 3m + 1, where m = (9q3 + 9q2 + 3q) is a whole number.

Case (iii)

(3q + 2)= 27q3 + 54q2 + 36q + 8 = 3(9q3 + 18q2 + 12q + 2) + 2

= 3m + 2, where m = (9q3 + 18q2 + 12q + 2) is a whole number.

Thus, the cube of any positive integer can be expressed in the form 3m or 3m +1 or 3m + 2.

Thus, statement III is incorrect.

Hence, among the given statements, only statement I is correct.

The correct answer is B.

Option1 :

statements I and II are correct

Option2 :

only statement I is correct

Option3 :

statements I and III are correct

Option4 :

onlystatement II is correct

Question 4:

1051 bottles of soft drink have to be kept in crates. Each crate can hold a maximum of 28 bottles.

 How many empty spaces will be left in the last crate?

Solution :

We first need to find the number of bottles that have to be kept in the last crate.


Using Euclid’s division lemma, we can write 1051 as

1051 = 28 × 37 + 15

Hence, 15 bottles have to be kept in the last crate.


It is given that one crate can hold a maximum of 28 bottles.

Therefore, number of empty spaces in the last crate = (28 − 15) = 13

Hence, the correct answer is option A.

Option1 :

13

Option2 :

14

Option3 :

15

Option4 :

16

Option5 :
Question 5:

Use the following information to answer the next question.

The H.C.F. of the numbers 8008 and 700 is calculated by Euclid’s division lemma.

 The first two steps of the method are given as:

8008 = A × 11 + B

A = B × 2 + C

What are the respective values of AB, and C?

Solution :

The H.C.F of 8008 and 700 can be found by Euclid’s division lemma as:

Since 8008 > 700, applying Euclid’s division lemma to 8008 and 700:

8008 = 700 × 11 + 308           ...(1)

Since remainder 308 ≠ 0, on applying Euclid’s lemma again:

700 = 308 × 2 + 84                 ...(2)

Comparing equations (1) and (2) with the given steps, the values of AB, and C are obtained as:

A = 700, B = 308, and C = 84

Thus, the values of AB, and C are 700, 308, and 84 respectively.

The correct answer is C.

Option1 :

308, 84, and 11

Option2 :

308, 56, and 18

Option3 :

700, 308, and 84

Option4 :

700, 448, and 28

Question 6:
Let x and y be two consecutive positive integers such that 
x<y
Then the difference between their squares is always equal to
Solution :
Since x and y are two consecutive positive integers,
Therefore, one of them is even and other is odd.
By Euclid's algorithm, we know that every positive even integer is of form 2k,
 and that every positive odd integer is of form 2k + 1, where k is some integer.
 
x=2k, y=2k+1 kI

Then 
y2-x2=2k+12-2k2=4k+1 kI
.
Now 
4k+1=2k+2k+1=x+y

Hence, the correct answer is option A.
Option1 :
the sum of the two consecutive positive integers
Option2 :
the difference of the ​two consecutive positive integers
Option3 :
the product of the ​two consecutive positive integers
Option4 :
the quotient of the ​two consecutive positive integers
Option5 :
Question 7:

The square of every positive odd integer can be expressed in terms of m (where m is a whole number) as

Solution :

Let a be any positive odd integer and let b = 6.

Then, by Euclid’s algorithm, whole numbers q and r exist such that

a = 6q + r             0 ≤ r < 6

Thus, the possible values of remainder r are 0, 1, 2, 3, 4, and 5.

Thus, a can be 6q, 6q + 1, 6q + 2, 6q + 3, 6q + 4, or 6q + 5.

Since the squares of positive odd integers are being considered, will always be odd.

Thus, a cannot be 6q, 6q + 2, or 6q + 4.

∴ a = 6q + 1 or 6q + 3 or 6q + 5

Consider the squares of 6q + 1, 6q + 3, and 6q + 5.

Case (i)

(6q + 1)2 = 36q2 + 12q + 1

= 6(6q2 + 2q) + 1

= 6m + 1, where m is a whole number and m = (6q2 + 2q)

Case (ii)

(6q + 3)2 = 36q2 + 36+ 9

= 6(6q2 + 6q + 1) + 3

= 6m + 3, where m is a whole number and m = (6q2 + 6q + 1)

Case (iii)

(6q + 5)2 = 36q2 + 60+ 25

= 6(6q2 + 10q + 4) + 1

= 6m + 1, where m is a whole number and m = (6q2 + 10q + 4)

Thus, the square of every positive odd integer can be written in the form

6+ 1 or 6m + 3, where m is a a whole number.

The correct answer is C.

Option1 :

5m + 1 or 5m + 2

Option2 :

3m + 1 or 3m + 2

Option3 :

6m + 1 or 6m + 3

Option4 :

7m + 1 or 7m + 3

Question 8:

Use the following information to answer the next question.

Vijay is finding the H.C.F. of (33976, 18495).

 He starts the process correctly but commits a mistake in the third step. 

However, he manages to get the correct answer.

Ist step: 33976 = 18495 × 1 + 15481

IInd step : 18495 = 15481 × 1 + 3014

IIIrd step: 15481 = 3014 × 7 + 1233

IVth step: 3014 = 1233 × 2 + 137

Vth step: 1233 = 137 × 9 + 0

After the IInd step, the correct steps are

Solution :

It is given that the first two steps are correct.

Ist step: 33976 = 18495 × 1 + 15481

IInd step: 18495 = 15481 × 1 + 3014

On applying Euclid’s division algorithm, we get

15481= 3014 × 5 + 411

3014 = 411 × 7 + 137

411 = 137 × 3 + 0

Hence, the correct steps are shown in table B.

The correct answer is B.

Option1 :

15481= 3014 × 5 + 548

3014 = 548 × 5 + 137

548 = 137 × 4 + 0

Option2 :

15481= 3014 × 5 + 411

3014 = 411 × 7 + 137

411 = 137 × 3 + 0

Option3 :

15481= 3014 × 6 + 274

3014 = 274 × 11 + 137

274 = 137 × 2 + 0

Option4 :

15481= 3014 × 4 + 685

3014 = 685 × 4 + 137

685 = 137 × 5 + 0

Option5 :
Question 9:

Use the following information to answer the next question.

The given steps show the procedure for finding

 the H.C.F. of the numbers 133763 and 10185

 by using Euclid’s division algorithm.

133763 = 10185 × a + b

10185 = b × 7 + c

b =c× 2 + 0

The values of ab, and c are respectively

Solution :

The H.C.F. of two positive integers 133763 and 10185 can be found by using Euclid’s division

 algorithm. This can be done through the following steps.

Step 1

As 133763 > 10185, by applying Euclid’s division lemma to the numbers 133763 and 10185, 

we obtain 133763 = 10185 × 13 + 1358

Step 2

As the remainder 1358 ≠ 0, by applying Euclid’s division lemma to the numbers 10185 and 1358,

 we obtain 10185 = 1358 × 7 + 679

Step 3

Again as the remainder 679 ≠ 0, by applying Euclid’s division lemma to the numbers 1358 and

 679, we obtain 1358 = 679 × 2 + 0

Thus, H.C.F. (133763, 10185) = 679

Hence, the values ab, and c are 13, 1358, and 679 respectively.

The correct answer is C.

Option1 :

14, 1738, and 368

Option2 :

14, 1247, and 368

Option3 :

13, 1358, and 679

Option4 :

13, 1358, and 543

Question 10:
If LCM(70, 125) = 17
k+1-1
. The value of k is
Solution :
LCM(70, 125) = 1750 = 
17×103-1

Comparing it with 17
k+1-1
, we get 
k+1=103k=102

Hence, the correct answer is option C.
Option1 :
100
Option2 :
101
Option3 :
102
Option4 :
1