Real Numbers Test

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Real Numbers Test - 6

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following is NOT a non-terminating repeating decimal?

A

B

C

D

Solution

We know that if a rational number (≠ integer) can be expressed in the form where p ∈ Z, n ∈ W and m ∈ W, the rational number will be a terminating decimal. Otherwise, the rational number will be a non terminating, recurring decimal.
= , non terminating

= , terminating

= , non terminating

, non terminating
Hence, only has terminating decimal expansion.
Question 2
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A number (n) when divided by 7 leaves a remainder of 2. What will be the remainder if one more than square of the number is divided by 7?

A
6

B
5

C
3

D
1

Solution

Let the number be n.
n = 7q + 2 (where q is the quotient)
Now,
n² + 1 = (7q + 2)² + 1 = (7q)² + 2² + 28q + 1 = 49q² + 28q + 4 + 1
n² + 1 = 7(7q² + 4q) + 5
When 7(7q² + 4q) + 5 is divided by 7, the remainder will be 5.
Question 3
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The product of three numbers a, b and c is 15360. The values of H.C.F(a, b), H.C.F(a, c) and H.C.F(b, c) are 4, 4 and 8, respectively. What are the L.C.M. and H.C.F. of the three numbers if the product of L.C.M. and H.C.F. is 1920?

A
240, 4

B
240, 8

C
920, 2

D
480, 4

Solution

L.C.M(a, b, c) =

Let L.C.M(a,b,c) be x and H.C.F(a,b,c) be y.
x ÷ y = 120
x = 120y
xy = 1920
(120y)y = 1920
y² = 1920 ÷ 120
y² = 16
y = 4
x = 120 × 4 = 480
L.C.M(a, b, c) = 480
H.C.F(a, b, c) = 4
Question 4
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The fraction is equal to

A

B

C

D

Solution

×

=
= =
Question 5
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Which of the following is a fractional form of ?

A

B

C

D

Solution

Let the number be x.

x =
Multiply the number by 10.

10x = ... (i)

Multiply the equation (i) by 1000.

10000x = ... (ii)

Subtracting (i) from (ii),

10000x - 10x =

9990x = 5180

x = =
Question 6
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+ + = ?

A

B

C

D

Solution

× + × + ×
=
⁼+ +
=
=
=
=
Question 7
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If = a, then what is the value of a²?

A

B

C
5

D
25

Solution

× - × = a

- = a

- = a

- = a

= a

= a

= a

a²= 5
Question 8
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Which of the following prime numbers should be the denominator to get a terminating decimal number?

A
2

B
3

C
5

D
Either 2 or 5

Solution

Fractions with denominator 2 or 5 only have terminating decimals as 2 and 5 completely divide 10 or 100.
Question 9
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+ + + ... + equals

A
9

B
3

C
1

D
0

Solution

x + × + ×+...+ ×
=
= + + + ... +
=
= - ()
= -(4 - 5) = 1
Question 10
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What is the rationalising factor of ?

A

B

C

D

Solution

Power of 'x' required in rationalizing factor = 3 - (1/2) = (5/2)
Power of 'y' required in rationalizing factor = 3 - (-5) = (8)
Power of 'z' required in rationalizing factor = 3 - (-3/4) = (15/4)
∵ × = = xyz

∴ The rationalising factor is .
Question 11
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Which of the following is the greatest surd?

A
$\sqrt[4]{2}$

B
$\sqrt[4]{6}$

C
$\sqrt[3]{3}$

D
$\sqrt[6]{5}$

Solution

= = = = = = = = = = = = = = = = So, is the greatest surd.
Question 12
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What will be the 8^th root of the 32^nd root of ?

A

B

C

D

Solution

32^nd root of = = = =
8^th root of = = = = =
Question 13
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If (a³ + k) is divisible by k, where 'a' is a positive integer and 'k' is a prime number, then which of the following is true?

A
k is a multiple of a.

B
k is a factor of a²but not a.

C
k is a factor of a.

D
k is a factor of a³and a but not a².

Solution

If (a³ + k) is divisible by 'k', then a³ is also divisible by 'k'.
Hence, 'k' is a factor of a³.
As 'k' is a prime number, 'a' needs to be the multiple of 'k'.
So, 'k' is a factor of a^b, where 'b' is any positive integer.
Question 14
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A real number , where 'k' is a three-digit number with digits 2, 7 and 3 in any order, is expressed as a decimal number. Which of the following is true for the number?

A
It is a non-terminating repeating decimal number.

B
It is a terminating decimal number.

C
It is a non-terminating, non-repeating decimal number.

D
Any of the above statement could be true.

Solution

Sum of the digits of the number (k) = 2 + 7 + 3 = 12
12 is divisible by 3, hence the number (k) is a multiple of 3.
Hence, 150 ÷ 3 = 50
As the denominator can be reduced to 50, the number is a terminating decimal number because 50 = 2 × 5²and if the prime factorisation of the denominator has only powers of 2 or 5 or both, then the number is a terminating decimal.
Question 15
1 / -0

Evaluate:

A

B

C

D

Solution

=

=

= (a² - b² = (a + b)(a – b))

=

=

= (a² - b² = (a + b)(a – b))

=

Hence, option 2 is the answer.
Question 16
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There are three watches, out of which one loses 4 seconds every minute and one gains 4 seconds every minute. If all the watches are set at 8 a.m. and an alarm is activated, each watch beeps for a second when the time turns 1 minute in that watch. At what time will they beep together again?

A
9:52 a.m.

B
9:48 a.m.

C
9:58 a.m.

D
10:02 a.m.

Solution

In 1 minute, the first clock will cover 56 seconds.
The second one will cover 64 seconds and the third one will cover 60 seconds.
The time when they beep together will be the LCM of 56, 60 and 64, i.e. 6720.
The time they will beep together is 6720 seconds or 112 minutes (at 9:52 a.m.).
Question 17
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There are three wheels in three different machines A, B, and C. These three wheels are rotating at different speeds. The diameters of the wheels of these three machines are the same. A's wheels make a revolution inminutes, B's wheels inminutes, and C's wheels inminutes. Three markings, one on each wheel, are at the points of contact of the wheels with the plane when they start off. What is the shortest interval after which the markings on the wheels are again at the points of contact of all the 3 wheels simultaneously?

A
hours

B
hours

C
hours

D
hours

Solution

According to the question, we have to calculate the shortest interval of time in which all the wheels complete their revolutions simultaneously.
As the distance covered by all the wheels in 1 revolution is the same and it only depends on time, the shortest interval after which the markings on the wheels are again at the points of contact of all the 3 wheels simultaneously (T) = LCM of time taken by the wheels to complete the revolutions individually.
T = LCM (,,)
= LCM (,,)
LCM of fractions =

LCM of 20, 25, and 16 = 4 × 4 × 5 × 5 = 400
HCF of 3, 7, and 7 = 1
LCM of () = 400 ÷ 1 = 400
400 minutes = 400 ÷ 60 =hours
Question 18
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A and B think about two numbers 'a' and 'b'. They say that the sum of their numbers is 270. If the HCF of the two numbers is 18, then what is the maximum number of such pairs which will satisfy these conditions?

A
10

B
9

C
6

D
4

Solution

Since 18 is the HCF of the two numbers, let the numbers be a = 18x and b = 18y, such that x and y are relatively prime.
a + b = 18x + 18y = 18(x + y) = 270
So, x + y = 15
We need to find all pairs of numbers such that they are relatively prime and their sum is 15.
(2, 13), (4, 11), (7, 8), and (14, 1) are the pairs satisfying the given conditions.
Question 19
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Sham had a certain number of chocolates which he wanted to share among students of a school. At first he tried to distribute equally between two students, then among three, four, five and six students, but every time he was left with one chocolate. When he tried to distribute them among 7 students, he succeeded. What is the smallest number of chocolates he could have?

A
301

B
308

C
350

D
413

Solution

L.C.M of 2, 3, 4, 5, and 6 = 60
The number of chocolates would be of the form 60K + 1.
We have to try various values of K, so as to make it divisible by 7.
Starting from K = 1, we need to check unless we get a multiple of 7.
K = 5 makes it 301, which is the answer.
Question 20
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There are three wheels A, B and C. Wheel A can complete 300 revolutions in 5 minutes. Wheel B can complete 72 revolutions in 2 minutes. Wheel C can complete 96 revolutions in 4 minutes. On completing one revolution, the starting point touches the ground. Two coins are tossed when the starting marks of all the wheels touch the ground together. How much time would have passed for the two coins to have been tossed 50 times each?

A
4 minutes 10 seconds

B
5 minutes 15 seconds

C
7 minutes 5 seconds

D
8 minutes 1 second

Solution

Time taken by wheel A to complete one revolution = = 1 second
Time taken by wheel B to complete one revolution = = seconds
Time taken by wheel C to complete one revolution = = seconds
To find out the time in which the three wheels complete the revolution together, we need to find the L.C.M of 1, , and .
L.C.M of the fractions = L.C.M of 1, 5, and 5 is 5.

H.C.F of 1, 3, and 2 is 1.

L.C.M of 1, , and = = 5
Hence, three wheels will complete revolution together after every 5 seconds.
For 50 tosses they need to complete the revolution together 50 times.
Total time required = 5 × 50 = 250 seconds = 4 minutes 10 seconds
Question 21
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Which of the following statements is correct?

A. If number X = pⁿ × q^m × r and number Y = p^n-1 × q^m+1 × r, then L.C.M(X, Y) = pⁿ × q^m+1 × r
B. Square root of any non-square number would be an irrational number.
C. H.C.F(1970, 218) = H.C.F(218, 8)
D. The product of primes is equal to a composite number.

A
A, C and D

B
B and D

C
Only C and D

D
Only A and D

Solution

A. Number X = pⁿ × q^m x r and number Y = p^n-1 × q^m+1 × r
Greatest power of p in both numbers is n.
Greatest power of q in both numbers is m + 1.
Greatest power of r in both numbers is 1.
Hence, L.C.M(X, Y) = pⁿ × q^m+1 × r
So, statement A is correct.

B. Square root of any non-square number can be a rational number or an irrational number, but the square root of a non-square integer would be an irrational number. Hence, statement B is incorrect.
For example: Square root of 0.04 is 0.2, which is a rational number.

C. If a = bq + r, then H.C.F(a, b) = H.C.F(b, r), where r is less than b but greater than or equal to zero.
1970 = 218 × 9 + 8
Therefore, H.C.F(1970, 218) = H.C.F(218, 8)
Hence, statement C is correct.
D. The product of primes is equal to a composite number.
A composite number has more than 2 factors. When two prime numbers are multiplied, the outcome would have 4 factors, which include the outcome number itself, 1 and the two prime numbers.
Hence, the product of the primes is equal to a composite number.
So, statement D is correct.
Question 22
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Find the correct terms which will replace the letters M, N, O, and P in the given sentences.

A. Zero is not included in the set of …M… numbers.
B. The product of the …N… power of each prime factor involved in the numbers is called …O….
C. Square of an irrational number is …P… a rational number.

A
M = natural, N = greatest, O = L.C.M, P = not necessarily

B
M = real, N = smallest, O = H.C.F, P = definitely

C
M = natural, N = smallest, O = L.C.M, P = definitely not

D
M = real, N = greatest, O = H.C.F, P = definitely

Solution

A. Natural numbers start from 1 and go till infinity. Zero is not included in the set of natural number, but it is included in real numbers.

B. The product of the greatest power of each prime factor involved in the numbers is called L.C.M

C. Square of an irrational number is not necessarily a rational number. For example: is an irrational number but its square, which is 2, is a rational number. But, square of π stays irrational.

Question 23

1 / -0

Match the following:

Column I	Column II
1.	a.
2.	b.
3.	c.
4.	d.

1 - c, 2 - b, 3 - d, 4 - a

1 - c, 2 - b, 3 - a, 4 - d

1 - b, 2 - c, 3 - a, 4 - d

1 - b, 2 - c, 3 - d, 4 - a

Solution

1. Let x =
100x =
100x – x = 18
99x = 18
x =
=
Let x =
100x =
1000x =
1000x - 100x = -
900x = 75
x =
x =
Now, ^_- = ^_-
=

=

2. We know that 0.5 =
Now, let x =
10x =
10x – x = 8
9x = 8
x =
So, +

=

=

3. We know that 0.45 =
=
Now, let x =
10x =
10x – x = 4
9x = 4
x =
So, ×
=
=
4. We know that 0.4 =
=
Now, let x =
10x =
10x – x = 3
9x = 3
x =
=
So, ÷
=×
=

Question 24
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A shop owner has four different types of chemicals A, B, C, and D. He has 806 litres of chemical A, 868 litres of chemical B, 930 litres of chemical C, and 992 litres of chemical D. He wants to fill these chemicals in containers to sell them. He prepares least number(a) of containers with equal amount of chemical in them, without mixing these chemicals. There is b litres of chemical in each container.

(1) What is the value of a?
(2) What is the value of b?

A
a = 58, b = 62

B
a = 60, b = 62

C
a = 62, b = 64

D
a = 58, b = 31

Solution

For least possible number of bottles of equal size, the size of the bottle must be of the greatest volume.

806 = 2 × 13 × 31
868 = 2 × 2 × 7 × 31
930 = 2 × 3 × 5 × 31
992 = 2 × 2 × 2 × 2 × 2 × 31

H.C.F(806, 868, 930, 992) = 2 × 31 = 62

So, each bottle must contain 62 litres of chemical, which is the value of b.

Required number of bottles,

Number of bottles of chemical A == 13
Number of bottles of chemical B == 14
Number of bottles of chemical C == 15
Number of bottles of chemical D == 16
Total number of bottles = 13 + 14 + 15 + 16 = 58
Hence, the value of a = 58
Question 25
1 / -0

Read the statements carefully and state 'T' for true and 'F' for false.

(I)is an irrational number.
(II) The LCM of two numbers, 17²× 13²× 7 × 23 and 23³× 7 × 17, is in the form of 17^a× 13^b× 7^c× 23^d. So, the values of a, b, c, and d are 2, 2, 1 and 3, respectively.
(III) The fractional representation of is.
(IV) The product of two numbers is (a⁴ – b⁴). If the LCM of these two numbers is (a + b), then the HCF will be (a³ + ab² + a²b + b³).

A

(I) (II) (III) (IV)

F T F F

B

(I) (II) (III) (IV)

T T F T

C

(I) (II) (III) (IV)

F F T T

D

(I) (II) (III) (IV)

T F F F

Solution

(I)
According to the identity,
(a + b)² = a² + b² + 2ab
Here, a = , b =
= ++ 2 ×
= 2 + 50 + 2 × 10
= 52 + 20
= 72 --- which is a rational number
So, it is false.

(II) Let x = 17² × 13² × 7 × 23
y = 23³ × 7 × 17
LCM (x, y) = 17² × 13² × 7¹ × 23³
Now comparing it with 17^a × 13^b × 7^c× 23^d,
a = 2, b = 2, c = 1, and d = 3
So, it is true.

(III) Let y =--- (1)
100y =--- (2)
On subtracting eq(1) from (2)
99y = 35
y =
This is false.

(IV) HCF × LCM = Product of two numbers
HCF × (a + b) = (a⁴– b⁴)

HCF =

=

=

=

= (a³ + ab² - a²b - b³)

This is false.