Practice Model question set 1 - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) What is {[(A ∪ B)' ∩ A]} - (A - B) equal to?
(a)
(b)
(c)
(d)
{(A ∪ B)' ∩ A} - (A - B)
= {(U – (A ∪ B)) ∩ A} – (A – B)
= {(U ∩ A) – {(A ∪ B) ∩ A}} – (A – B)
= {A – A} – (A – B)
= Φ – (A – B) = Φ
Q-2) Consider the following for the next 04 (four) items that follow :In an examination of Class XII, 55% students passed in Biology, 62% passed in Physics, 60% passed in Chemistry, 25% passed in Physics and Biology, 30% passed in Physics and Chemistry, 28% passed in Biology and Chemistry. Only 2% failed in all the subjects.What percentage of students passed in all the three subjects?
(a)
(b)
(c)
(d)
Total passed student = 98%
7 + x + 2 + x + 2+ x + 30 – x + 25 – x + 28 – x + x = 98
94 + x = 98 ; x = 4%
Q-3) If A is a non-empty subset of a set E, then what is E ∪ (A ∩ Φ) – (A – Φ) equal to?
(a)
(b)
(c)
(d)
E ∪ (A ∩ Φ) – (A – Φ)
= E ∪ Φ – A = E – A = A'
Q-4) In the quadratic equation $x^2$ + ax + b = 0, a and b can take any value from the set {1, 2, 3, 4}. How many pairs of values of a and b are possible in order that the quadratic equation has real roots?
(a)
(b)
(c)
(d)
| $x^2$ + ax + b = 0 | value 0f (a, b) |
| $a^2$ - ab ≥ 0 | [1, 2, 3, 4] |
| for real roots | |
| $1^2$ - 4 × 1 ≥ 0 | (1, 1) |
not possible
(2, 1) → $2^2$ – 4 × 1 ≥ 0 possible
(3, 1) → possible
(3, 2), (4, 1), (4, 2) (4, 3) (4, 4) → possible
So → possible values can be possible
Q-5) In a school there are 30 teachers who teach Mathematics or Physics. Of these teachers, 20 teach Mathematics and 15 teach Physics, 5 teach both Mathematics and Physics. The number of teachers teaching only Mathematics is
(a)
(b)
(c)
(d)
Total number of teachers = 30.
Number of teachers who teaches only Math
= 20 – 5 = 15.
Q-6) Out of 105 students taking an examination English and Mathematics, 80 students pass in English, 75 students pass in Mathematics 10 students fail in both the subjects. How many students fail in only one subject?
(a)
(b)
(c)
(d)
Number of students failing in Mathematics
= 105 – 75 = 30
Number of students failing in English
= 105 – 80 = 25
∴ Number of students failing in 1 subject
= (25 + 30) – 10 = 45
Q-7) Consider the following for the next 04 (four) items that follow : In an examination of Class XII, 55% students passed in Biology, 62% passed in Physics, 60% passed in Chemistry, 25% passed in Physics and Biology, 30% passed in Physics and Chemistry, 28% passed in Biology and Chemistry. Only 2% failed in all the subjects.What percentage of students passed in exactly one subject?
(a)
(b)
(c)
(d)
Percentage of students passed in exactly one subject
= 7 + x + 2 + x + 2 + x = 11 + 3x = 11 + 12 = 23%
Q-8) In an examination, 52% candidates failed in English and 42% failed in Mathematics. If 17% failed in both the subjects, then what percent passed in both the subjects ?
(a)
(b)
(c)
(d)
venn diagram of no. of failed students
No. of students failed in English only = 52 – 17 = 35
No. of students failed in maths only = 42 – 17 = 25
Total no. of failed students in either of the subjects
= 35 + 17 + 25 = 77
No. of passed student in both subjects = 100 – 77
= 23
Q-9) Consider the following for the next 04 (four) items that follow : In an examination of Class XII, 55% students passed in Biology, 62% passed in Physics, 60% passed in Chemistry, 25% passed in Physics and Biology, 30% passed in Physics and Chemistry, 28% passed in Biology and Chemistry. Only 2% failed in all the subjects.If the number of students is 360, then how many passed in at least two subjects?
(a)
(b)
(c)
(d)
Passed in at least two subject
= ${25 - x + 30 - x + 28 - x + x}/{100}$ × 360
= ${83 - 2x}/{100}$ × 360 = 270
Q-10) In a competitive examination, 250 students have registered. Out of these, 50 students have registered for Physics, 75 students for Mathematics and 35 students for both Mathematics and Physics. What is the number of students who have registered neither for Physics nor for Mathematics?
(a)
(b)
(c)
(d)
Number of students registered for
Physics n(P) = 50
Mathematics n(M) = 75
Number of students registered for both subjects n(P ∩ M) = 35.
Number of students, registered for either physics or mathematics
n(P ∪ M) = n(P) + n(M) – n(P ∩ M).
= 50 + 75 – 35 = 90
∴ Number of students registered neither for physics nor for mathematics
n($\ov{P ∪ M}$) = 250 – 90 = 160.