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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.
ExamsCart is providing free quizzes to help all the aspirants with the SSC & Railway Quantitative Aptitude section by providing links to attempt quizzes. You can attempt quizzes as per the exam interface before the actual exam by attempting the quizzes on the web browser. It’s time to pull up your socks and burn the midnight oil by practicing the SSC Quantitative Aptitude quiz daily.
Quantitative Aptitude Quiz: Set 104
1) The internal bisector of the ∠B and ∠C of the ΔABC, intersect at O. If ∠A = 1000, then the measure of ∠BOC is:-
a) 1100
b) 1300
c) 1400
d) 1200
c) 1400
2) A conical iron piece having diameter 28 cm and height 30cm is totally immersed in the rise of water level by 6.4cm. The diameter, in cm, of the vessel is:-
a) 3.5
b) 32
c) 35
d) 35/2
c) 35 Explanation: Let the radius of the vessel = r cms; Volume of displaced water = volume of conical iron piece; Pi* r2 * 6.4 = 1/3 * (14)2 * 30; => r = 17.5 cms; Hence, the diameter of the vessel = 35 cms.
3) The value of the following is 3(Sin4Θ + Cos4Θ) + 2(Sin6Θ + Cos6Θ) + 12Sin2ΘCos2Θ
a) 3
b) 0
c) 5
d) 2
c) 5 Explanation: = 3(Sin4Θ + Cos4Θ) + 2(Sin6Θ + Cos6Θ) + 12Sin2ΘCos2Θ; = 3(Sin4Θ + Cos4Θ) + 2(Sin2Θ + Cos2Θ) (Sin4Θ + Cos4Θ – Sin2Θ Cos2Θ) + 12Sin2ΘCos2Θ; = 5(Sin4Θ + Cos4Θ) + 10Sin2ΘCos2Θ; = 5(Sin4Θ + Cos4Θ + 2Sin2ΘCos2Θ); = 5(Sin2Θ + Cos2Θ)2; =5;
4) If the area of the base, height and volume of a right prism be (3√3/ 2) P2 cm2, 100√3 cm and 7200 cm3 respectively, then the value of P will be?
a) 2/√3
b) √3
c) 3/2
d) 4
d) 4 Explanation: Volume of prism = area of base * height; 7200 = (3√3/ 2) P2 * 100√3; => P = 4;
5) If the discount of 10% is given on the marked price of a radio, the gain is 20%. If the discount is increased to 20%, the gain is:-
a) 5%
b) 6.67%
c) 7.62%
d) 6.25%
b) 6.67% Explanation: Suppose marked price = Rs. 100; Price after discount = Rs. 90; Since, the realized gain = 20%; Hence, the cost price = Rs. 75; If the discount price = 20%, then the price after discount = Rs. 80; Hence, the required answer = 5*100/75 = (20/3) %=6.67%;
6) If 4a – 4/a + 3 = 0 then the value of a3 – 1/a3 + 3 =?
a) 3/16
b) 21/64
c) 7/16
d) 21/16
d) 21/16 Explanation: a – 1/a = -3/4; Cubing both sides- a3 – 1/a3 -3*( a – 1/a) = -27/64; a3 – 1/a3 = -27/64 + 3 *(-3/4); a3 – 1/a3 = -171/64; a3 – 1/a3 + 3 = -171/64 + 3 = 21/16;
7) O is the circumcentre of ∆ABC. If ∠BAC = 850 ∠BCA = 750, the ∠OAC is equal to:-
a) 600
b) 700
c) 500
d) 400
b) 700
8) If A, B and C can complete a work in 6 days. If A can work twice faster than B thrice faster than C, then the number of days C alone can complete the work is:
a) 22 Days
b) 44 Days
c) 33 Days
d) 11 Days
c) 33 Days Explanation: (A + B + C)’s efficiency = (100/6)%; A’s efficiency = 2 * B’s efficiency = 3* C’s efficiency; 3C + 3C/2 + C = 100/6; => 11C/2 = 100/6; C’s efficiency = 100/(3*11); Hence, C alone can finish the work in = 33 days;
9) A circular swimming pool is surrounded by a concrete wall 4m wide. If the area of the concrete wall surrounding the pool is 11/25 that of the pool, then the radius (in m) of the pool is:
a) 8
b) 20
c) 16
d) 30
b) 20 Explanation: Let the radius of the swimming pool = R meter; Radius of pool with wall = (R + 4) meters; Pi* [(R + 4)2 – R2] = 11/25 * pi* R2; 8(R + 2) = 11/25 * R2; After solving this equation- we get, R = 20 meter.
10) Two pipes A and B can fill a tank with water in 30 minutes and 45 minutes respectively. The water pipe C can empty the tank in 36 minutes. First A and B are opened. After 12 minutes C is opened. Total time (in minutes) in which the tank will be filled up is:-
a) 12
b) 30
c) 36
d) 24
d) 24 Explanation: A’s efficiency to fill the tank in every minute= (100/30)% =3.33% B’s efficiency to fill the tank in every minute = (100/45)% = 2.22% and C’s efficiency to empty the tank in every minute = (100/36) = 2.78%; In 12 minutes, A and B will fill the tank in a minute = 5.55*12 = 66.6%; In 1 minute, A and B fill the tank = 5.55%; Therefore, A, B and C will fill the tank in 1 minute = 5.55 – 2.78 = 2.77%; Remaining tank to be filled = 100 -66.6 = 33.4%; Time taken to fill the empty tank = 33.4/2.77 = 12.05 minutes; Total time = 12 + 12 = 24 minutes;
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