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
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
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
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
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
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
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
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
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
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
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
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.
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
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).
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