RS Aggarwal Test: Real Numbers - 2 - Class 10 MCQ

# RS Aggarwal Test: Real Numbers - 2 - Class 10 MCQ

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## 15 Questions MCQ Test Mathematics (Maths) Class 10 - RS Aggarwal Test: Real Numbers - 2

RS Aggarwal Test: Real Numbers - 2 for Class 10 2024 is part of Mathematics (Maths) Class 10 preparation. The RS Aggarwal Test: Real Numbers - 2 questions and answers have been prepared according to the Class 10 exam syllabus.The RS Aggarwal Test: Real Numbers - 2 MCQs are made for Class 10 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for RS Aggarwal Test: Real Numbers - 2 below.
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RS Aggarwal Test: Real Numbers - 2 - Question 1

### If n is a Natural number, then 52n − 22n is divisible by

Detailed Solution for RS Aggarwal Test: Real Numbers - 2 - Question 1

52n −22n is of the form a2n − b2n which is divisible by (a – b).

25− 4n ⇒ a− bn
is always divisible by (a−b)
(25 − 4) = 21
∴ So factors are both 7 and 3.

RS Aggarwal Test: Real Numbers - 2 - Question 2

### Two alarm clocks ring their alarms at regular intervals of 50 seconds and 48 seconds. If they first beep together at 12 noon, at what time will they beep again for the first time?

Detailed Solution for RS Aggarwal Test: Real Numbers - 2 - Question 2

LCM of 50 and 48 = 1200
∴ 1200 sec = 20 min
Hence at 12.20 pm they will beep again for the first time.

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RS Aggarwal Test: Real Numbers - 2 - Question 3

### The product of three consecutive integers is divisible by

Detailed Solution for RS Aggarwal Test: Real Numbers - 2 - Question 3

Let the three consecutive positive integers be n, n+1 and n+2.

Whenever a number is divided by 3, the remainder obtained is either 0,1 or 2.

Therefore, n=3p or 3p+1 or 3p+2, where p is some integer.

If n=3p, then n is divisible by 3.

If n=3p+1, then n+2=3p+1+2=3p+3=3(p+1) is divisible by 3.

If n=3p+2, then n+1=3p+2+1=3p+3=3(p+1) is divisible by 3.

So, we can say that one of the numbers among n,n+1 and n+2 is always divisible by 3 that is:

n(n+1)(n+2) is divisible by 3.

Similarly, whenever a number is divided by 2, the remainder obtained is either 0 or 1.

Therefore, n=2q or 2q+1, where q is some integer.

If n=2q, then n and n+2=2q+2=2(q+1) is divisible by 2.

If n=2q+1, then n+1=2q+1+1=2q+2=2(q+1) is divisible by 2.

So, we can say that one of the numbers among n, n+1 and n+2 is always divisible by 2.

Since, n(n+1)(n+2) is divisible by 2 and 3.

Hence, n(n+1)(n+2) is divisible by 6.

RS Aggarwal Test: Real Numbers - 2 - Question 4

If A = 2n + 13, B = n + 7, where n is a natural number, then HCF of A and B is:

Detailed Solution for RS Aggarwal Test: Real Numbers - 2 - Question 4

Taking different values of n we find that A and B are coprime

e.g. put n=1

we get A= 2(1)+13=15

B=1+7=8
∴ HCF = 1

put n=2

we get A= 2(2)+13=17

B=2+7=9
∴ HCF = 1

RS Aggarwal Test: Real Numbers - 2 - Question 5

The largest number which divides 615 and 963 leaving remainder 6 in each case is

Detailed Solution for RS Aggarwal Test: Real Numbers - 2 - Question 5

we have to  find the largest number which divide 615 and 963 leaving remainder 6 in each case.

so let us subtract 6 from 615 and 963

⇒ 615 − 6 = 609

⇒ 963 − 6 = 957

now lets find the HCF

⇒ prime factorisation of 609=29×3×7

prime factorisation of 957=29×3×11

now lets take out the common factors from both the cases

⇒ 29 and 3

x = 29 × 3 = 87

∴ 87 is the number which will divide 615 and 963 leaving remainder 6 in each case.

RS Aggarwal Test: Real Numbers - 2 - Question 6

There are 576 boys and 448 girls in a school that are to be divided into equal sections of either boys or girls alone. The total number of sections thus formed are:

Detailed Solution for RS Aggarwal Test: Real Numbers - 2 - Question 6

The number 576 can be factorised as,
576 = 2×2×2×2×2×2×3×3
The number 448 can be factorised as,
448=2×2×2×2×2×2×7
Write the common factors of the given numbers.
2×2×2×2×2×22×2×2×2×2×2
Multiply the common factors to determine the highest common factor (HCF) of the given numbers.
2×2×2×2×2×2 = 642×2×2×2×2×2 = 64
Since the highest common factor (HCF) of the given numbers is 64, this implies that each section will have 64 number of students.
Now, we need to find the number of sections formed.
Let us first find the number of sections formed by the total number of boys by dividing 576 by 64.
576/64 = 9
Now, find the number of sections formed by the total number of girls by dividing 448 by 64.
448/64=7
Thus, the total number of sections formed will be 9+7=16
Hence, option B is the correct answer.

RS Aggarwal Test: Real Numbers - 2 - Question 7

The largest number which divides 70 and 125 leaving remainders 5 and 8 respectively is

Detailed Solution for RS Aggarwal Test: Real Numbers - 2 - Question 7

The largest number by which x , y

divisible and gives the remainder a ,

and b is

the HCF of ( x - a ) and ( y - b)
According to the given problem ,

The largest number which divides

70 and 125 leaving remainders 5 and

8 respectively are

HCF of ( 70 - 5 ) = 65 and

( 125 - 8 ) = 117

65 = 5 × 13

117 = 3 × 3 × 13

HCF ( 65 , 117 ) = 13

Required number is 13.

RS Aggarwal Test: Real Numbers - 2 - Question 8

The HCF of 2472, 1284 and a third number N is 12. If their LCM is 23 x 32 x 5 x 103 x 107, then the number N is :

Detailed Solution for RS Aggarwal Test: Real Numbers - 2 - Question 8

Given that, LCM (2472, 1284, and n) is 2× 3× 5 × 103 × 107

Let us express the numbers 2472, and 1284 as a product of prime numbers.

2472 = 2 × 2 × 2 × 3 × 103  (23 × 3 × 103)

1284 = 2 × 2 × 3 × 107 (2× 3 × 107)

HCF (2472, 1284)=  2 × 2 × 3 (2× 3)

'n' should also have one of the factors as HCF, and another factor as the missing element of other numbers from the LCM (i.e., 3 × 5)

Therefore,

n =  22 × 3 × (3 × 5)

n = 22 × 32 × 5

n = 4 × 9 × 5

n = 180

Therefore, if the HCF of 2472 and 1284 and a third number 'n' is 12 and if their LCM is 23 × 3× 5 × 103 × 107, then the number 'n' is 180 (22 × 32 × 5)

RS Aggarwal Test: Real Numbers - 2 - Question 9

The product of a non zero rational and an irrational number is

Detailed Solution for RS Aggarwal Test: Real Numbers - 2 - Question 9

The product of a non zero rational and an irrational number is always irrational

RS Aggarwal Test: Real Numbers - 2 - Question 10

Two natural numbers whose difference is 66 and the least common multiple is 360, are:

Detailed Solution for RS Aggarwal Test: Real Numbers - 2 - Question 10

Let the two numbers be x and x+66.

Since, the LCM is 360, there must be 5,2 and 3 as the prime factors of the two numbers.

So, one pair is 24 & 90, because 90−24=66.

Also, 24=2×2×2×3

90=2×3×3×5

∴ LCM=2×2×2×3×3×5=360.

So, the numbers are 24 and 90.

RS Aggarwal Test: Real Numbers - 2 - Question 11

The decimal expansion of the rational number 14587/1250 will terminate after

Detailed Solution for RS Aggarwal Test: Real Numbers - 2 - Question 11

RS Aggarwal Test: Real Numbers - 2 - Question 12

Which of the following rational numbers have a terminating decimal expansion?

Detailed Solution for RS Aggarwal Test: Real Numbers - 2 - Question 12

The denominator 26 x 52 is of the form 2m x 5n, where m and n are non-negative integers. Hence, it is a terminating decimal expansion.

RS Aggarwal Test: Real Numbers - 2 - Question 13

If two positive integers a and b are written as a = x3y2 and b = xy3 ; x, y are prime numbers, then HCF (a, b) is

Detailed Solution for RS Aggarwal Test: Real Numbers - 2 - Question 13

It is given that,

a = x3y2 = x × x × x × y × y

b = xy3 = x × y × y × y

The HCF (Highest Common Factor) of two or more numbers is the highest number among all the common factors of the given numbers.

As HCF is the product of the smallest power of each common prime factor involved in the numbers.

HCF of a and b = HCF (x3y2, xy3)

= x × y × y

= xy2

Therefore,HCF (a, b) is xy2

RS Aggarwal Test: Real Numbers - 2 - Question 14

The LCM of two numbers is 1200. Which of the following cannot be their HCF?

Detailed Solution for RS Aggarwal Test: Real Numbers - 2 - Question 14

*Answer can only contain numeric values
RS Aggarwal Test: Real Numbers - 2 - Question 15

Evaluate:

Detailed Solution for RS Aggarwal Test: Real Numbers - 2 - Question 15

## Mathematics (Maths) Class 10

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## Mathematics (Maths) Class 10

126 videos|477 docs|105 tests