Aptitude Online Quiz

Following quiz provides Multiple Choice Questions (MCQs) related to Aptitude. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using Show Answer button. You can use Next Quiz button to check new set of questions in the quiz.
Q 1 - A is twice as quick as B and B is thrice as quick as C. The excursion secured by C in 42 min. will be secured by A in
A - 7 min.
B - 14 min.
C - 28 min.
D - 63 min.
Let c speed be x meters/min.
Then, B speed=3x meters /min and A speed =6x meters/ min.
Ratio of speed of A and C =ratio of times taken by C and A
6x:x=42:ymin⇒6x/x=42/y⇒y=42/6min=7 min.
Q 2 - 9 men working 7 hours a day can finish a bit of work in 15 days. In how long can 6 men functioning for 9 hours a day, complete the same bit of work?
A - 63/4 days
B - 16 days
C - 67/4 days
D - 35/2 days
(9*7) men working 1 hr a day can finish the work in 15 days.
63 men finish it in 15 days.
1 man can finish it in (15*63) days.
(6*9) men can finish it in (15*63)/ (6*9) days = 35/2 days
Q 3 - Two no. are in the extent 3/2 :8/3, when each of these are extended by 15 , their extent gets the chance to be 5/3 :5/2 . The greater of the no. is :
A - 27
B - 36
C - 48
D - 64
Explanation
Ratio of the given numbers = 3/2:8/3 = 9 :16
Let the numbers be 9x and 16 x, Then,
 (9x+15)/ (16 x+15) = (5/3)/ (5/2) =2/3
=>  3(9x+15)= 2 (16x+15)
=>  5x= 15  =x= 3.
 ∴ Larger no.= (16*3)= 48

Q 4 - 15% of 180 = 50% of?
A - 48
B - 50
C - 52
D - 54
Let 15% of 180=50% of x
Then 15/100*180=50/100*x
=>x=54
Q 5 - The range of a square field is 6050m2. The length of its corner to corner is
A - 110 m
B - 112 m
C - 120 m
D - 135 m
Let the diagonal be d meter. Then,
1/2 d2=6050⇒ d2= 12100 ⇒d=√12100= 110 m.
Q 6 - The average age of five officers in a department is 32 years. If the age of their supervisor is added the average increases by 1. What is the supervisor's age?
A - 32 years
B - 48 years
C - 38 years
D - 42 years
Supervisor?s age = 32 + 6 = 38 years
Q 7 - On solving p/x+q/y = m, q/x+p/y = n, we get:
A - x=(q2-p2)/(mp-nq) , y = (q2-p2)/(np-mq)
B - x=(p2-q2)/(mp-nq), y=(p2-q2)/(np-mq)
C - x=(p2-q2)/(mp-nq) , y= (q2-p2)/(np-mq)
D - x=(q2-p2)/(mp-nq), y = (p2-q2)/(np-mq)
Given equations are p/x+q/y = m...(i),   q/x+ p/y = n ...(ii)
On multiplying (i) by q, (ii) by p and subtracting, we get:
q2/y- p2/y = mq-np
⇒y (mp-np) = (q2- p2)
⇒y = (q2-p2)/(mq- np)
= (p2- q2)/(np-mq)
On multiplying (i) by p, (ii) by q and subtracting, we get:
p2/x - q2/x = mp- nq
⇒ (p2- q2) = x (mp- nq)
⇒x = (p2- q2)/ (mp-nq)
∴ x= (p2-q2)/(mp-nq) , y = (p2-q2)/(np- mq)
Q 8 - When the sun's altitude changes from 45° to 60°, the length of the shadow of a tower decreases by 45m. What is the height of the tower?
A - (45√3)/(√3-1)
B - (45√3)/(√3+1)
C - (45+√3)/(√3-1)
D - (45-√3)/(√3-1)
Let AD be the tower, BD be the initial shadow and CD be the final shadow.
Given that BC = 45 m,  ABD = 45°,  ACD = 60°,
Let CD = x, AD = h
From the right   CDA, tan60=h/x
From the right   BDA, tan45=(45+x)/h=>h=45+x
=>h=45+h/√3
=>h(1-1/√3)=45
=>h=45/(1-1/√3)=(45√3)/(√3-1)
Q 9 - If the the total investment of a business is an amount of rs. 5000. Ajay invested 4000 Rs. more  in the comparision of Bijender and Bijender invested 5000 more in the comparision  of Chman . At the end of the year, Rs. 35000 earned in the form of profit then what should be the share of Ajay?
A - Rs. 14700
B - Rs. 5000
C - Rs. 11000
D - Rs. 12000
If we can assume that the C , B and A  have the investment of Rs. x ,
(x+5000) and Rs. (x+9000)  respectively. Then,
X+(x+5000)+(x+9000)  =50000 ⇒ 3x = 36000 ⇒ x =12000.
C =Rs. 12000 , B = Rs. 17000 and C = Rs. 21000
A : B : C = 21000 : 17000: 12000 =  21: 17: 12
Share of A = Rs. (35000* 21/50) = Rs. 14700
Q 10 - If a monkey can climb 6m on a round pole during 1 min.  but  he  slipped  3 m  in the next attempt or next min. The length of the pole is 21 m then how much time he will take to complete the task?
A - 10 min
B - 11 min
C - 12 min
D - 13 min
 Time taken to climb 6m in the last = 1 min.
remaining distance = (21-6)  = 15 m .
total distance covered in 2 min. = (6-3 ) = 3m .
∴ distance of 15m covered in  ( 2/3* 15) = 10 min.
total time  taken = (10+1) = 11 min.