Real Numbers Test

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Real Numbers Test - 15

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Question 1
1 / -0

Find the greatest number that will exactly divide both 360 and 42.

A
2

B
4

C
6

D
12

Solution

Factors of $360 = 2 \times 2 \times 2 \times 3 \times 3 \times 5$
Factor of $42 = 2 \times 3 \times 7$
Common Factor $= 2 \times 3$
Thus, HCF of $360$ and $42 = 6$
Hence, the correct option is (C).
Question 2
1 / -0

Which of the following is the correct prime factorization of 1230?

A
$2 \times 3 \times 5 \times 23$

B
$2 \times 3 \times 5 \times 41$

C
$2^{2} \times 3 \times 5$

D
$2^{2} \times 3^{2} \times 5^{2} \times 41$

Solution

The correct prime factorization of $1230 = 2 \times 3 \times 5 \times 41$
Hence, the correct option is (B).
Question 3
1 / -0

Which of the following expressions on simplification gives a rational number?

A
$\sqrt{7}(2+\sqrt{7})$

B
$\sqrt{3}(2+\sqrt{3})$

C
$\sqrt{2}(\sqrt{8}+\sqrt{2})$

D
None of these

Solution

Using distributive property
$a \cdot(b+c)=a \cdot b+a \cdot c$
For option A:
$\sqrt{7}(2+\sqrt{7})=2 \cdot \sqrt{7}+\sqrt{7} \cdot \sqrt{7}=2 \sqrt{7}+7$
Which is irrational.
For option B:
$\sqrt{3}(2+\sqrt{3})=2 \cdot \sqrt{3}+\sqrt{3} \cdot \sqrt{3}=2 \sqrt{3}+3$
Which is also irrational.
For option C:
$\sqrt{2}(\sqrt{8}+\sqrt{2})=\sqrt{2}(2 \sqrt{2}+\sqrt{2})=2 \sqrt{2} \cdot \sqrt{2}+\sqrt{2} \cdot \sqrt{2}=4+2=6$
Which is rational.
Hence, the correct option is (C).
Question 4
1 / -0

What is the greatest number that will exactly divide 225 and 375?

A
150

B
75

C
5

D
65

Solution

Factors of $225 = 3 \times 3 \times 5 \times 5$
Factor of $375 = 3 \times 5 \times 5 \times 5$
Highest Common Factor $= 3 \times 5 \times 5$
Thus, HCF of $225$ and $375 = 75$
Hence, the correct option is (B).
Question 5
1 / -0

A shopkeeper has 420 Science books and 130 Maths books. He wants to stack them so that each stack has the same number of books and they take up the least area of his shelf. Each stack should have one type of book. How many books can be placed in each stack?

A
10

B
20

C
5

D
30

Solution

HCF of $420$ and $130$ .
$420 = 2^{2} \times 3 \times 5 \times 7$
$130 = 2 \times 5 \times 13$
$∴ H C F = 2 \times 5$
$= 10$
$∴$ Each stock will have $10$ .
Hence, the correct option is (A).
Question 6
1 / -0

Which of following is true for Euclid's division algorithm between two numbers 'a' and 'b'?

A
\(a=b q+r ; 0

B
\(a=b+r g ; 0 \leq r

C
\(a+b q=r ; 0

D
\(a=b q+r ; 0 \leq r

Solution

According to Euclid's Division, if we have two positive integers a and b, then there would be whole numbers q and r that satisfy the equation: a = bq + r, where 0 ≤ r < b. a is the dividend. b is the divisor. q is the quotient and r is the remainder. By using this, we can find the HCF of two numbers.
Hence, the correct option is (D).
Question 7
1 / -0

The LCM of two numbers $7^{2} \times 3^{2} \times 5 \times 2$ and $2^{3} \times 5 \times 7$ is of form $2^{a} \times 3^{b} \times 5^{c} \times 7^{d}$ . What are the respective values of $a, b, c$ , and $d$ ?

A
$1, 0, 1$ and $0$

B
$3, 2, 0$ and $2$

C
$3, 2, 1$ and $2$

D
$3, 2, 2$ and $1$

Solution

The highest power of each number is their respective values.
Highest power of value is:
$2^{a} = 2^{3}$
$3^{b} = 3^{2}$
$5^{c} = 5^{1}$
$7^{d} = 7^{2}$
Values of $a, b, c$ and $d = 3, 2, 1$ and $2$ .
Hence, the correct option is (C).
Question 8
1 / -0

In Euclid's Division Lemma, when a=bq+r where a and b are positive integers then what values r can take?

A
r = 1

B
r ≥ 0

C
r = b

D
r = a

Solution

Euclid’s division lemma:
It tells us about the divisibility of integers. It states that any positive integer ‘a’ can be divided by any other positive integer ‘ b’ in such a way that it leaves a remainder ‘r’.
Euclid's Division Lemma states that for any two positive integers ‘a’ and ‘b’ there exist two unique whole numbers ‘q’ and ‘r’ such that, a = bq + r, where 0≤ r < b.
Here, a= Dividend, b= Divisor, q= quotient and r = Remainder.
Thus, the values 'r’ can take 0≤ r < b.
Hence, the correct option is (B).
Question 9
1 / -0

Which of the following is an irrational number?

A

$\frac{\sqrt{45}}{\sqrt{5}}$

B
$\frac{\sqrt{5}}{\sqrt{45}}$

C
$\sqrt{5}+\sqrt{45}$

D
$\sqrt{5} \times \sqrt{45}$

Solution

In mathematics, the irrational numbers are all the real numbers which are not rational numbers, the latter being the numbers constructed from ratios of integers. Hence option C is correct.
Question 10
1 / -0

Which of the following statements is/are true?
(i) The sum of two irrational numbers is always an irrational number.
(ii) The sum of a non-zero rational number and an irrational number is an irrational number.

A
Both (i) and (ii) are true.

B
Only (i) is true.

C
Only (ii) is true.

D
Both (i) and (ii) are false.

Solution

(i) The sum of two irrational numbers is not always an irrational number.

(ii) The sum of a non-zero rational number and an irrational number is an irrational number.

Let r be rational and let i be irrational.

Now assume that s : = r + i is rational.

Subtracting r on both sides we find s − r = i

But s and r are both rational and it is well known that in that case s−r is a rational number.

But also s − r = i hence is irrational.

This cannot go together so actually, we deduced a contradiction.
Thus, the sum of two irrational numbers is not always an irrational number.
Then we are allowed to conclude that our assumption is false which means exactly that s is irrational.
Sum of two irrational may not be irrational, as we have an example of $\sqrt{2}$ and $- \sqrt{2}$ which add up to $0$ . So,
(i) Is not correct.

If a is rational and b is irrational then a+b must be irrational because if say $a + b = c$ is rational then $b = c - a$ now a rational number from a field so the addition of two rational only gives rational, hence b is irrational which is against our assumption. So (ii) is always true.
Hence, the correct option is (C).
Question 11
1 / -0

Which of the following is/are true?
(i) Every composite number can be expressed as a product of prime numbers.
(ii) Prime factorisation of a natural number is unique, but its order may vary.

A
Both (i) and (ii)

B
Only (i)

C
Only (ii)

D
Neither (i) nor (ii)

Solution

Any composite number can be expressed in terms of its prime factor. The prime factors for $2$ numbers can be the same but their combination is unique.
Example,
$12 = 2^{2} \times 3$ and $18 = 2 \times 3^{2}$
Both have prime factors as $2$ and $3$ but the combination that $12$ has is different from that of $18$ .
Thus, both (i) and (ii) are true.
Hence, the correct option is (A).
Question 12
1 / -0

A wholesale dealer has stocks of three different varieties of tea weighing 132 kg, 156 kg and 84 kg. He wants to pack all of these into boxes of equal size, without mixing. What is the capacity of the largest possible box such that all the boxes are completely filled?

A
12 kg

B
24 kg

C
36 kg

D
50 kg

Solution

LCM of $84$ and $132$ and $156$ .
$84 = 2 \times 2 \times 3 \times 7$
$132 = 2 \times 2 \times 3 \times 11$
$156 = 2 \times 2 \times 3 \times 13$
$2, 2, 3$ are the common numbers.
LCM of $132, 156$ and $84 = 12$
Hence, the correct option is (A).
Question 13
1 / -0

Which of the following fractions will have a non-terminating decimal expansion?

A
$\frac{25}{5^{2} \times 3^{2} \times 7}$

B
$\frac{19}{5^{3} \times 2^{4}}$

C
$\frac{41}{2^{2} \times 5^{3}}$

D
$\frac{49}{2^{2} \times 5^{2} \times 7^{2}}$

Solution

Non-terminating decimals: Non-terminating decimals are those which keep on continuing after decimal point (i.e. they go on forever). They don't come to end or if they do it is after a long interval.
A rational number can be expressed as a terminating decimal if the denominator has the factors of $2$ or $5$ only.
A rational number can be expressed as a non-terminating repeating decimal if the denominator has factors other than $2$ or $5$ .
In option (A)
$\frac{25}{5^{2} \times 3^{2} \times 7} = \frac{1}{3^{2} \times 7}$
Thus, making it a non-terminating decimal expansion
Hence, the correct option is (A).
Question 14
1 / -0

Traffic lights at three consecutive road crossings change after every 48 seconds, 72 seconds and 108 seconds. If the lights change simultaneously at 7:00:00 a.m, at what time will they again change simultaneously?

A
7:07:12 a.m.

B
7:07:36 a.m.

C
7:07:48 a.m.

D
7:07:24 a.m.

Solution

Find the Prime Factor:
$48 = 2^{4} \times 3$
$72 = 2^{9} \times 3^{2}$
$108 = 2^{2} \times 3^{3}$
$L C M = 2^{4} \times 3^{3}$
$L C M = 16 \times 27 = 432$
Convert $432 s e c$ to $m i n$ :
$432 s e c = 432 \div 60 m i n$
$432 s e c = 7.2 m i n$
$7.2 m i n = 7 m i n 12 s e c$
Find the time it will change together again:
$7 a m + 7 m i n 12 s e c = 7.07.12 a m$
Hence, the correct option is (A).
Question 15
1 / -0

Find the greatest number that will divide $90$ and $107$ and leave remainders $2$ and $3$ , respectively.

A
$8$

B
$10$

C
$40$

D
$16$

Solution

Since the number that will divide $90$ leaving a reminder $2$ and $107$ leaving a remainder $3$ , the number should divide numbers $(90 - 2) = 88$ and $(107 - 3) = 104$ i.e., it is the HCF of $88$ and $104$ .

$88 = 8 \times 11$
$104 = 8 \times 13$

HCF $= 8$
The greatest number which divides $90$ and $107$ leaving remainders $2$ and $3$ respectively is $8$ .
Hence, the correct option is (A).

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Real Numbers Test - 15

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Real Numbers Test - 15

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Question 1

2

4

6

12

Solution

Question 2

2×3×5×23

2×3×5×41

22×3×5

22×32×52×41

Solution

Question 3

\(\sqrt{7}(2+\sqrt{7})\)

\(\sqrt{3}(2+\sqrt{3})\)

\(\sqrt{2}(\sqrt{8}+\sqrt{2})\)

None of these

Solution

Question 4

150

75

5

65

Solution

Question 5

10

20

5

30

Solution

Question 6

\(a=b q+r ; 0

\(a=b+r g ; 0 \leq r

\(a+b q=r ; 0

\(a=b q+r ; 0 \leq r

Solution

Question 7

1, 0, 1 and 0

3, 2, 0 and 2

3, 2, 1 and 2

3, 2, 2 and 1

Solution

Question 8

r = 1

r ≥ 0

r = b

r = a

Solution

Question 9

\(\frac{\sqrt{45}}{\sqrt{5}}\)

\(\frac{\sqrt{5}}{\sqrt{45}}\)

\(\sqrt{5}+\sqrt{45}\)

\(\sqrt{5} \times \sqrt{45}\)

Solution

Question 10

Both (i) and (ii) are true.

Only (i) is true.

Only (ii) is true.

Both (i) and (ii) are false.

Solution

Question 11

Both (i) and (ii)

Only (i)

Only (ii)

Neither (i) nor (ii)

Solution

Question 12

12 kg

24 kg

36 kg

50 kg

Solution

$2 \times 3 \times 5 \times 23$

$2 \times 3 \times 5 \times 41$

$2^{2} \times 3 \times 5$

$2^{2} \times 3^{2} \times 5^{2} \times 41$

$1, 0, 1$ and $0$

$3, 2, 0$ and $2$

$3, 2, 1$ and $2$

$3, 2, 2$ and $1$

$\frac{25}{5^{2} \times 3^{2} \times 7}$

$\frac{19}{5^{3} \times 2^{4}}$

$\frac{41}{2^{2} \times 5^{3}}$

$\frac{49}{2^{2} \times 5^{2} \times 7^{2}}$

$8$

$10$

$40$

$16$