Quantitative Aptitude Quiz
Quantitative Aptitude Quiz for Railway and SSC covers an important part of the exam and hence it must be prepared thoroughly. SSC, Railways & many other government job exams are scheduled to be held in the upcoming months. It needs complete understanding of the basic concepts along with thorough understanding. This page will provide you all the quizzes of the Arithmetic section of various exams such as RRB NTPC, SSC CHSL, SSC MTS etc. Practice the quantitative aptitude quizzes given below and surpass the high cut off marks in the exam.
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Quantitative Aptitude Quiz: Set 106
1. A train runs at an average speed of 75 km/hr. If the distance to be covered is 1050 km. How long will the train take to cover it?
a) 13 hrs
b) 14 hrs
c) 12 hrs
d) 15 hrs
b) 14 hrs Explanation: The time taken by train = Covered distance/ Average Speed; = 1050/75 = 14 hrs.
2. G is the centroid of ∆ABC. The medians AD and BE intersect at right angles. If the lengths of AD and BE are 9 cm and 12 cm respectively; then the length of AB (in cm) is?
a) 10
b) 10.5
c) 9.5
d) 11
a) 10
3. The minimum value of 2sin2θ + 3cos2θ is
a) 1
b) 3
c) 2
d) 4
c) 2 Explanation: 2sin2θ + 3cos2θ = 2*(sin2θ + cos2θ) + cos2θ = 2 + 0 =2;
4. If number of Vertices: edges and faces of a rectangular parallelopied are denoted by v, e and f respectively, the value of (v – e + f) is
a) 4
b) 2
c) 1
d) 0
b) 2 Explanation: The rectangular parallelepiped has 8 vertices, 12 edges, and 6 faces. Hence, v – e + f = 8 – 12 + 6 = 2;
5. 5 persons will live in a tent. If each person requires 16m2 floor area and 100m3 space for air then the height of the cone of smallest size to accommodate these persons would be?
a) 18.75 m
b) 16m
c) 10.25 m
d) 20 m
a) 18.75 m Explanation: The required area = 16 sq. m.; Suppose the radius of tent = r meter; => pi*r2 = 5*16; => r = 5.04 meter. (given) The volume of air = 5*100 m3; 1/3 * pi * r2 * h = 500; => h= 18.75 cms
6. If the altitude of an equilateral triangle is 12√3 cm, then its area would be:
a) 12 sq. cms.
b) 72 sq. cms.
c) 36√3 sq. cms.
d) 144√3 sq. cms.
d) 144√3 sq. cms. Explanation: The altitude of the equilateral triangle = (√3* side)/2; => Side = 24 cms. The required area = (√3* side * side)/4 = (√3* 24 * 24)/4 =144√3 sq. cms.
7. The difference between successive discounts of 40% followed by 30% and 45% followed by 20% on the marked price of an article is Rs. 12. The marked price of the article is:
a) Rs. 400
b) Rs. 200
c) Rs. 800
d) Rs. 600
d) Rs. 600 Explanation: Suppose, the marked price of the article= Rs. x; The price after 40% and 30% successive discounts = x*0.60*0.70 = 0.42x; The price after 45% and 20% successive discounts = x*0.55*0.80 = 0.44x; (Given), 0.02x = 12; => x = Rs. 600;
8. The area of the triangle formed by the graphs of the equation x=0, 2x + 3y=6 and x + y = 3 is:
a) 1 sq. unit
b) 1.5 sq. unit
c) 1 sq. unit
d) 4.5 sq. unit
b) 1.5 sq. unit
9. Among the equations x + 2y + 9 = 0; 5x – 4 = 0; 2y – 13 = 0; 2x – 3y = 0, the equation of the straight line passing through origin is-
a) 2x – 3y = 0
b) 5x – 4 = 0
c) x + 2y + 9 = 0
d) 2y – 13 = 0
a) 2x – 3y = 0 Explanation: x + 2y + 9 =0; (this line will intersect both the axes) 5x – 4 =0; (This line will be parallel to Y-axis) 2y – 13 =0; (This line will be parallel to X-axis) 2x – 3y =0; (This line will pass through the origin)
10. The HCF of x8 – 1 and x4 + 2×3 – 2x – 1 is:
a) x2 + 1
b) x + 1
c) x2 – 1
d) x – 1
c) x2 – 1 Explanation: x2 -1 =(x +1)*(x-1); => x = -1, 1; Both the values of x will satisfy the other equation; Hence, (x2 -1) will be the appropriate answer.
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