Number System Questions and Answers – Real Numbers and BODMAS Simplification – Set 2

This set of Aptitude Questions and Answers (MCQs) focuses on “Real Numbers and BODMAS Simplification – Set 2”.

1. When a certain number is multiplied by 21, the product consists entirely of 3’s. What is the minimum number of 3’s in product?
a) 5
b) 6
c) 7
d) 4
View Answer

Answer: b
Explanation: If we keep on dividing 33333…… by 21, till we get 0 as remainder.
We end up with remainder 0 and quotient 15873, when the product 333333 is divided by 21.
Hence number of 3’s in the product is 6.

2. Consider the following statements.
I. There are finite number of rational numbers between 51 and 52.
II. There are infinite number of rational numbers between 99 and 100.
III. There are finite number of irrational numbers between√2 and √3.
Which of the statement(s) given above is/are correct?
a) Only I
b) Only II
c) Only III
d) Both I & III
View Answer

Answer: b
Explanation: We know that between any two rational numbers, there are infinite number of rational and irrational numbers. Hence only Statement II is true.

3. If \(\frac{49}{15}\)=3+\(\frac{1}{x+\frac{1}{y+\frac{1}{z}}}\), where x, y and z are natural numbers, then what is value of z?
a) 1
b) 2
c) 3
d) 0
View Answer

Answer: c
Explanation: \(\frac{49}{15}\)=3+\(\frac{1}{x+\frac{1}{y+\frac{1}{z}}}\)
\(\frac{49}{15}\)-3=\(\frac{1}{x+\frac{1}{\frac{yz+1}{z}}}=\frac{1}{\frac{xyz+x+z}{yz+1}} \)
\(\frac{4}{15}=\frac{yz+1}{xyz+x+z}\)
15(yz+1) = 4(xyz+x+z)
z(4+4xy-15y) = 15-4x
z=\(\frac{15-4x}{4+4x-15y}\)
Since x, y and z are natural numbers. By trial & error, assuming x=3 and y=1, we get,
z = 3
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4. If p and q are distinct natural numbers, then which of the following is/are integer(s)?
I. \(\frac{p}{q}+\frac{q}{p}\)
II. pq(\(\frac{p}{q}+\frac{q}{p}\)) (p2+q2)-1
III. \(\frac{pq}{p^2+q^2}\)
a) I and II
b) Only II
c) II and III
d) Only III
View Answer

Answer: b
Explanation: \(\frac{p}{q}+\frac{q}{p}\) is integer only if p=q. Since p and q are distinct, equation I is not an integer.
\(\frac{pq}{p^2+q^2}\) is a fraction for all values of p and q. Hence, equation III is not an integer.
pq(\(\frac{p}{q}+\frac{q}{p}\)) (p2+q2)-1=pq(\(\frac{p^2+q^2}{pq})(\frac{1}{p^2+q^2})\)=1, which is an integer. Hence only Equation II holds good.

5. What is the value of 12.8- 0.4 of (7.2-3.7)+2.4*3.02?
a) 18.648
b) 18.849
c) 16.848
d) 17.648
View Answer

Answer: a
Explanation: = 12.8- 0.4 of (7.2-3.7)+2.4*3.02
= 12.8- 0.4*3.5+7.248
= 12.8-1.4+7.248 = 18.648

6. What is the value of (3+4-\(\frac{6}{2}\)+2) + \((\frac{\frac{9}{3}+6*5}{11})\)*((4+5-6)+(18+3*4)))?
a) 105
b) 100
c) 110
d) 115
View Answer

Answer: a
Explanation: = (3+4-\(\frac{6}{2}\)+2) + \((\frac{\frac{9}{3}+6*5}{11})\)*((4+5-6)+(18+3*4)))
= (7-3+2) + (\((\frac{3+30}{11})\)*(3+(18+12)))
= 6 + (3*(3+30))
= 6 + (3*33) = 105

7. What is the value of [26 + {42(23 – 20) + 2 – (2(5*80 – 100)) + 16 – 4}]?
a) -434
b) 434
c) -443
d) -453
View Answer

Answer: a
Explanation: = [26 + {42(23 – 20) + 2 – (2(5*80 – 100)) + 16 – 4}]
= [26 + {42*3 + 2 – (2*300) + 16 – 4}]
= [26 + {126 + 2 – 600 + 16 – 4}]
= 26 – 460 = -434
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8. What could be the maximum value of Z in the following equation?
7X9 + 2Y8 + 5Z6 = 1503
a) 8
b) 9
c) 7
d) 5
View Answer

Answer: a
Explanation: On adding digits in units place, 9+8+6=23. 2 is carried forward to tens place.
On adding digits in hundreds place, 7+2+5=14, but in question it is given as 15. Hence, 1 is carried from tens place.
On adding digits in tens place, 2+X+Y+Z=10.
Therefore, the maximum value of Z is 8 when X=Y=0.

9. Find the value of y which satisfy the inequalities |y| ≥ y and 7y – 25 > 3.
a) All positive numbers
b) All positive number greater than 4
c) All numbers greater than -4
d) All rational numbers
View Answer

Answer: b
Explanation: |y| ≥ y is true for all real values of y.
Consider, 7y – 25 > 3
Or 7y > 28
Or y > 4
So, the solution is set of all positive number greater than 4.
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10. A and B are two positive integers such that AB = 81. Which of the following cannot be the value of (A+B)?
a) 30
b) 18
c) 82
d) 81
View Answer

Answer: d
Explanation: The possible values of A and B such that the product is 81 are, (1,81), (9,9), (27,3).
1+81=82
9+9=18
27+3=30
Hence, 81 cannot be the value of (A+B).

To practice all aptitude questions, please visit “1000+ Quantitative Aptitude Questions”, “1000+ Logical Reasoning Questions”, and “Data Interpretation Questions”.

If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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