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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 105
1) A shopkeeper allows a discount of 10% on the marked price of a camera. Marked price of the camera, which costs him Rs. 600, to make a profit of 20% should be:-
a) Rs. 750
b) Rs. 800
c) Rs. 700
d) Rs. 650
a) Rs. 750 Explanation: Cost price = Rs. 600; Suppose the marked price = Rs. X; X* 0.8 – 500 = 500* 20%; X= 500*1.2/0.8 = Rs. 750;
2) O is the incentre of ∆PQR and ∠QPR = 500, then the measure of ∠QOR is :-
a) 1300
b) 1250
c) 1150
d) 1000
a) 1300
3) If x2 + y2 + z2 = 2(x + z – 1), then the value of x3 + y3 + z3 =?
a) 1
b) 0
c) 2
d) -1
c) 2 Explanation: Take x =1 y =0 z =1
1+0+1= 2
2(1+1-1) =2
So 1+0+1= 2;
4) AC is a transverse common tangent to two circles with centers P and Q and radii 6cm and 3cm at the point A and C respectively. If AC cuts PQ at the point B and AB = 8cm then the length of PQ is:-
a) 12 cm
b) 13 cm
c) 10 cm
d) 15 cm
d) 15 cm
5) A dealer sold a bicycle at a profit of 10%. Had he brought the bicycle at 10% less price and sold it at a price Rs. 60 more, he would have gained 25%. The cost price of the bicycle was-
a) Rs. 2600
b) Rs. 2200
c) Rs. 2400
d) Rs. 2000
c) Rs. 2400Â Explanation: Cost price = Rs. x; SP = 1.1x; 0.9*CP * 125%= 1.1x + 60; 1.25*0.9*x = 1.1x + 60; => x = Rs. 2400;
6) A dealer buys an article listed at 100 and gets successive discounts of 10% and 20%. He spends 10% of the Cost Price on transportation. At what price should he sell the article to earn a profit of 15%?
a) Rs. 91.20
b) Rs. 92.00
c) Rs. 90.80
d) Rs. 91.08
d) Rs. 91.08 Explanation: Resultant successive discount = 20% + 10% – 20% * 10% = 28%; The buying price of the article= 100 * (100-28)% = 100*72% = Rs. 72; Buying price after transportation= 72 + 7.2 = Rs. 79.2; Hence, the selling price = 79.2 * 1.15 = Rs. 91.08;
7) A librarian purchased 50 story-books for his library. But he saw that he could get 14 more books by spending Rs. 76 more and the average price per book would be reduced by Rs. 1. The average price (in Rs.) of each book he bought, was:
a) 25
b) 15
c) 20
d) 10
d) 10Â Explanation: Suppose, Total price of 50 story-books = Rs. x; Average price of 1 book = x/50; Total price of 64 books including 14 books = x + 76; Average price of 1 book = (x + 76)/64; (x+76)/64 = x/50 -1; => Rs. 500; Hence, The average price of 1 book = 500/50 = Rs. 10;
8) The speed of a boat in still water is 6 Km/hr and the speed of stream is 1.5 Km/hr. A man rows to a place at a distance of 22.5 Km and comes back to the starting point. The total time taken by him is:-
a) 6 hours
b) 8 hours
c) 10 hours
d) 4 hours
b) 8 hours Explanation: Total time taken = 22.5/(6 + 1.5) + 22.5/(6 – 1.5) = 3 + 5 = 8 hours;
9) A and B together can do a piece of work in 30 days. B and C together can do it in 20 days. A starts the work and works on it for 5 days, then B takes up and works for 15 days. Finally C finishes the work in 18 days. The number of days in which C alone can do the work when doing it seperately is:-
a) 40 Days
b) 24 Days
c) 120 Days
d) 60 Days
b) 24 Days Explanation: (A + B) ‘s efficieny = 100/30 = 3.33%; (B + C)’s efficiency = 100/20 = 5%; Assume that the efficiency of A, B and C = x%, y%, and z%; (x + y) = 3.33; (y + z) = 5; 5x + 15y + 18z = 100; 5(x + y) + 10(y + z) + 8z =100; => 5*3.33 + 10 * 5 + 8z =100; 16.65 + 50 + 8z =100; => z= 4.168%; Hence, C can alone do that work = 100/4.168 = 24 days;
10) If Tan A + Cot A = 2, then the value of Tan10A + Cot10A is-
a) 1
b) 210
c) 2
d) 4
c) 2Â Explanation: Put A = 45; => 1 + 1 =2; Hence, Tan10A + Cot10A = (1)10 + (1)10 =2;
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