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Aptitude Mock Test
This section presents you various set of Mock Tests related to Aptitude. You can download these sample mock tests at your local machine and solve offline at your convenience. Every mock test is supplied with a mock test key to let you verify the final score and grade yourself.
Aptitude Mock Test IV
Q 1 - At what time between 4 and 5 o' clock will the hands of a clock be at right angle?
Answer : B
Explanation
At 4 o' clock, the minute hand is 20 minute spaces behind the hour hand. Now, when the two hands are at right angles, they are fifteen min spaces apart. So, they are at right angles in following two cases. Case I ? When the minute hand is 15 min spaces behind the hour hand. In this case, min hand will have to gain (20-15) = 5 min spaces. = (60/55)*5 = 60/11 ∴ They are at right angle at 60/11 minutes past 4. Case II ? When the minute hand is 15 minute spaces ahead of the hour hand. In this case the minute hand will have to gain (20+15) = 35 min spaces. = (60/55)*35 = 420/11 min ∴ They are at right angles at 420/11 min past 4.
Q 2 - Find the time between 8 and 9 o' clock will the hands of a clock be in the same straight line but not together.
Answer : B
Explanation
At 8 o? clock, the two hands are 20 min spaces apart. To be in the same straight line but not together, they will be 30 min spaces apart. So, the minute hand will have to gain (30-20) = 10 minute spaces over the hour hand. = 60/55 * 10 = 120/11 ∴The hands will be in the same straight line but not together at 120 min past 8.
Q 3 - At what time between 4 and 5 o' clock are the hands of a clock 3 minutes apart?
A - 45.09 minutes past 4 o' clock.
B - 35.09 minutes past 4 o' clock.
Answer : C
Explanation
At 4 o' clock, the minute hand is 20 min spaces behind the hour hand. There are two possible cases for the given scenario. Case I ? Minute hand is 3 min spaces behind the hour hand. In this case the minute hand has to gain (20-3) = 17 minutes. = (60/55) * 17 = (12/11) * 17 = 204/11 ∴ The minute hand will be 3 minute apart at 204/11 min or 18.54 min past 4. Case II ? Minute hand is 3 minutes spaces ahead of the hour hand. In this case the minute hand has to gain (20+3) = 23 minute spaces. = (60/55)*23 = (12/11)*23 = 276/11 = 25.09 minutes ∴ The hands will be 3 minute apart at 25.09 minutes past 4 o' clock.
Q 4 - The angle of elevation of a ladder leaning against a wall is 60° and the foot of the ladder is 4.6 m away from the wall. The length of the ladder is:
Answer : D
Explanation
Let AB be the wall and BC be the ladder. Then, ∠ACB = 45° and AC = 7.5 m AC/BC= Cos (45) =1/√2 BC=7.5√2
Q 5 - A man observes the elevation of a balloon to be 45° at a point A .He then walks towards the balloon and at a certain place B finds the elevation to be 60°. He further walks in the direction of the balloon and finds it to be directly over him at a height of 450 m. Distance travelled from A to B is
Answer : A
Explanation
450/BD= tan (60) =>BD =450/√3 450/AD= tan (30) =>AD= 450√3 AD =BD +AB =>AB=AD-BD= 450√3-450/√3=(450x3-450)/√3=300√3m
Q 6 - A man on the top of a vertical observation tower observes a car moving at a uniform speed coming directly towards it. If it takes 10 minutes for the angle of depression to change from 45° to 60°, how soon after this will the car reach the observation tower?
Answer : B
Explanation
Let AB be the tower and C and D be the two positions of the car. Then,from figure AB/AC=tan 60 =√3 => AB=√3AC AB/AD=tan 45=1 => AB=AD AB=AC+CD CD=AB-AC=√3AC - AC=AC (√3-1) CD = AC (√3-1) =>10 min AC=> ? AC/(AC(√3-1)) x 10=?=10/(√3-1)=13.66=13 min 20 sec(approx)
Q 7 - A vertical pole fixed to the ground is divided in the ratio 1:4 by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a place on the ground, 16 m away from the base of the pole, what is the height of the pole?
Answer : C
Explanation
Let CB be the pole and point D divides it such that BD : DC = 1 : 4 = X:4X Given that AB = 16 m Let the the two parts subtend equal angles at point A such that CAD = BAD = Θ =>tan Θ=X/16 =>X=16 tan ( Θ) ------ (1) =>tan( Θ+ Θ)=4X/16 =>16 tan (2 Θ)=4X =>16(2tan ( Θ))/(1-tan ( Θ)2)=4X ------ (2) From eqn 1 & 2 2X/(1-tan ( Θ)2)=4X (X=16tan Θ) 1/(1-(X/16)2)=2 1-(X/16)2=1/2=>162- X2=162/2=>X2=128 =>X=8√2 =>Height of pole BC = X+4X=5X=40√2
Q 8 - An aeroplane when 750 m high passes vertically above another aeroplane at an instant when their angles of elevation at same observing point are 45° and 30&de; respectively. Approximately, how many meters higher is the one than the other?
Answer : C
Explanation
Let C and D be the position of the aeroplanes. Given that CB = 900 m,∠CAB = 60°,∠DAB = 45° From the right △ ABC, Tan45=CB/AB=>CB=AB From the right △ ADB, Tan30=DB/AB=>DB=ABtan30=CBx(1/√3)=750/√3 CB=CD+DB => Required height CD=CB-DB=750-750/√3=250(3- √3)
Q 9 - 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?
Answer : A
Explanation
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 10 - From the top of mast head of height 210 meters of a ship, a boat is observed at an angle of depression of 30° then the distance between them is
Answer : A
Explanation
From the right angled triangle CAB Tan(30) =210/X =>X=210/Tan(30)=210/(1/√3)=210√3
Q 11 - Ramesh , suresh and kumar are partnersin a business. They invested their capital in the ratio of 3:4:7. An amount of Rs. 21000 earned by them at the end of the year, How much share kumar earned in this profit?
Answer : B
Explanation
Ramesh : suresh : kumar = 3:4:7 Share of Kumar = Rs. (21000* 7/14) = Rs. 10500
Q 12 - If a business started by three partners A, B and C with the investment of Rs. 26000 , 34000 and 10000 respectively.An amount of Rs. 3500 earned by them at the end of the year, find out the B share in profit?
Answer : C
Explanation
A: B: C = 26000: 34000: 10000 = 13: 17: 5 Share of B = Rs. (3500* 17/35) = Rs. 1700
Q 13 - With an amount of Rs. 25000 started business by ajay. After 3 month, vijay joined the business by investing amount of Rs. 30000. If profit of the firm is 19000 at the end of the year then what should be the share of Ajay in profit?
Answer : D
Explanation
Ajay : Vijay = (25000*12: 30000*9) = 300000: 270000 = 10: 9 Share of Ajay = Rs. (19000* 10/19) = Rs. 10000
Q 14 - 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?
Answer : A
Explanation
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 15 - Ajay , Bijender and chaman started a business and invested Rs. 20000 each partner .After the period of 4 month, some changes are made in investment just like Ajay withdraw Rs. 5000, Bijender and chaman both add 6000 each in the invested capital. An amount of Rs. 69900 recorded as a profit in the firm. What should be the share of Bijender in profit?
Answer : B
Explanation
Ajay : Bijender: chaman = (20000*5+15000*7) : (20000*5+ 16000*7) :(20000*5+26000*7) = 205000: 212000 : 262000 = 205 : 212: 282 Share of B in profit = Rs. (69900* 212/699) = Rs. 21200
Q 16 - There are three partners in a firm and condition of investment by them is that at the beginning time investment done by A and after the period of 6 month B invested double than A and C invested 8month later but 3 times in the comparision of A. If annual profit Rs. 27000 is announced then what will be the share of profit in the favour of C?
Answer : B
Explanation
If x , 2x, 3x amount invested by A, B and C respectively. A: B: C = (x*12): (2x*6): (3x:4) = 12x:12x: 12x = 1:1:1 Share of C in profit = Rs. (27000* 1/3) = Rs. 9000.
Directions: Study the graph below carefully and answer the following:
Q 17 - On the off chance that the salary in the year 2001 was Rs. 60 crore and the use of that year squares with the use of the year 2003, what was the salary of the organization in crore rupees in 2003?
Answer - A
Explanation
Expenditure in 2001 =( 60-25) = 35 crore ∴ income in the year 2003 = (35+40) = 75 crores.
Directions: Study the graph below carefully and answer the following:
Q 18 - What is the inexact % expansion in the benefit of the year 2004. On the off chance that the pay of that year was Rs. 120 crore?
Answer - C
Explanation
Increase% in gain in 2007
= {(75-70)/70*100} %
= (5/70*100) %
= 7%
Directions: Study the graph below carefully and answer the following:
Q 19 - What is the estimated % benefit of the year 2004, if the wage of that year was Rs. 120 crores?
Answer - B
Explanation
Required % gain in 2004 = (50/120*100)% = 47%
Directions: Study the graph below carefully and answer the following:
Q 20 - Deals in the year 2004 for organization A structures what % of aggregate offers of organization A for every one of the years together?
Answer - A
Explanation
Total sale of Company A in 2004 =1100 Total sale of A during These years = (600+900+300+600+1100+1000+1100)= 5600 Required % = (1100/5600*100) % = 19.64%
Q 21 - In 1 minute 3/7 of a basin is filled. Whatever remains of the container can be filled in:
Answer : B
Explanation
Part filled in 1 min. = 3/7. remaining part = (1- 3/7)= 4/7 Let the required time be x min. More part, more time taken. (Direct) 3/7: 4/7:: 1: x ⇒ 3x/7 = (4/7*1) ⇒ x= 4/3 min.
Q 22 - Two funnels can fill a tank in 20 minutes and 30 minutes separately. On the off chance that both the channels are opened at the same time, then the tank will be filled in:
Answer : B
Explanation
Part filled by both pipes in 1 min. =(1/20+ 1/30)= 5/60 = 1/12 Time taken to fill the tank = 12 minutes
Q 23 - A tap can fill storage in 8 hours and another tap can discharge it in 16 hours. In the event that both the taps are open, the time taken to fill the tank will be:
Answer : C
Explanation
Net part filled in 1 hr = (1/8 ? 1/16)= 1/16 Total time taken to fill the tank = 16 hrs.
Q 24 - A pump can fill a tank with water in 2 hours. In light of a break in the tank, it takes 7/3 hours to fill the tank. The hole can discharge the filled tank in:
Answer : D
Explanation
Part filled by the pump in 1 hr = 1/2 Net part filled by the pump and leak in 1 hr = 3/7 Emptying work done by the leak in 1 hr= (1/2 - 3/7)= 1/14 Leak can empty the tank in 14 hours.
Q 25 - A channels can fill a tank in x hours and another funnel can exhaust it in y (y>x) hours. In the event that both the funnels are open, in how long will the tank be filled?
Answer : D
Explanation
Work done by filling pipe in 1 hr = 1/x Work done by emptying pipe in 1 hr = 1/y Net filling work done by both in 1 hr = (1/x- 1/y) = (y-x)/xy ∴The tank will be filled in xy/(y-x) hrs.
Q 26 - A tank can be filled by a tap in 20 minutes and by another tap in an hour. Both the taps are kept open for 10 minutes and after that the first tap is shutoff. After this, the tank will be totally filled in:
Answer : D
Explanation
Part of the tank filled by both in 10 min. = 10*(1/20+1/60)= 40/60 = 2/3 Remaining part = (1-2/3) = 1/3 1/60 part is now filled in 1 min. 1/3 part is now filled in (60*1/3) min. = 20 min.
Q 27 - Pipes A and B can fill a tank in 20 hours and 30 hours separately and channel C can purge the full tank in 40 hours. In the event that every one of the funnels is opened together, what amount of the reality of the situation will become obvious eventually expected to make the tank full?
Answer : C
Explanation
Net part filled in 1 hr = (1/20+ 1/30 ? 1/40)= 7/120 Time needed to make the tank full = 120/7 hrs.
Q 28 - A storage has two funnels. One can fill it with water in 8hours and the other can exhaust it in 5 hours. In how long will the reservoir be purged if both the channels are opened together when 3/4 of the storage is now loaded with water?
Answer : C
Explanation
Net part emptied by both in 1 hr = (1/5-1/8)= 3/40 3/40 part is emptied in 1 hr. 3/4 part will be emptied in (40/3*3/4) hrs = 10 hrs.
Directions: Study the Pie-charts carefully and answer the questions that follow:
Q 29 - Numbers of ladies working in the account and IT office together frame what% of aggregate number of Employees working in that office?
Answer - A
Explanation
Finance Department: Aggregate no. of workers = (14/100*8450) = 1183 Number of ladies = (25/100*3500) = 875 Number of men = (1183-875) = 308 I T Department: Aggregate no. of workers = (34/100*8450) = 2873 Number of ladies = (21/100*3500) = 735 Number of men = (2873-735) = 2138 Number of ladies working in the account and IT division = (875+735) =1610 Aggregate number of workers = 8450 Required rate = (1610/8450*100) % = 19.05%
Directions: Study the Pie-charts carefully and answer the questions that follow:
Answer - C
Explanation
Marketing office: Aggregate no. of workers = (18/100*8450) = 1521 Number of ladies = (17/100*3500) = 595 Number of men = (1521-595) = 926 Number of men working in Marketing division =926
Directions: The diagram given below depicts the sourced and uses of funds in a public Sector Enterprise. The total outlay is Rs. 4000 crore:
Q 31 - On the off chance that the working capital must be overseen out of the advance assets, than what % of credit assets ought to be separate for this reason?
Answer - C
Explanation
Loan reserves = 36.2% of 400 crores = Rs. (36.2/100*4000) crore = 1448 crore. Working capital = (20.3% of 4000) crore = Rs. (20.3/100*4000) crores = 812 crores. Required % = (812/1448*100) % = 56%
Directions: The diagram given below depicts the sourced and uses of funds in a public Sector Enterprise. The total outlay is Rs. 4000 crore:
Q 32 - The aggregate sum which has been utilized for purchasing area, apparatus, setting plants and capital works is roughly.
Answer - B
Explanation
Required Amount = 75.6% of 4000 crores. = (75.6/100*4000) = 3024 crore= Rs. 3000 crore.
Directions: The diagram given below depicts the sourced and uses of funds in a public Sector Enterprise. The total outlay is Rs. 4000 crore:
Q 33 - The aggregate money credits procured by the organization are around.
Answer - D
Explanation
Total money credits =7.1% of Rs 4000 crores= (7.1/100*4000) crores= 284 crores =285 crores.
Directions: The diagram given below depicts the sourced and uses of funds in a public Sector Enterprise. The total outlay is Rs. 4000 crore:
Q 34 - The organization needs all the more working capital. The amount of capital it can secure by reclaiming its venture?
Answer - C
Explanation
Capital obtained by recovering investment =4.1% of 4000 crore =Rs (4.1/100*4000) crores= Rs 164 crores.
Directions: The diagram given below depicts the sourced and uses of funds in a public Sector Enterprise. The total outlay is Rs. 4000 crore:
Q 35 - In the event that the organization were to deal with its working capital from interior assets alone, then the amount of asset from this assets will in any case be left for other use?
Answer - A
Explanation
Funds lefts =(27.5-20.3)% of 4000 crores = 7.2% of Rs 4000 crores =Rs (7.2/100*4000) crores=Rs 288 crores.
Q 36 - The aggregate surface territory of a strong barrel is 231 cm2. On the off chance that its bended surface zone is two ?third of its aggregate surface zone, then its Volume is:
Answer : A
Explanation
(2πrh+2πr2) =231 and 2πrh= 2/3*231= 154 ∴154+ 2πr2 =231 ⇒2*22/7*r2= (231-154) =77 ⇒r2 = (77*7/44)= 49/4 ⇒ r =7/2 cm ∴2*22/7*7/2*h= 154 ⇒ h= 7cm ∴ Volume = πr2h = (22/7 *7/2 * 7/2 * 7) cm3 = 539/2 =269.5 cm3
Q 37 - The total of the range of base and the tallness of a barrel is 37 m. On the off chance that the aggregate surface region of the chamber is 1628m2, then its volume is:
Answer : B
Explanation
Given: (r+h) = 37m Total surface area = 2πr (h+r) ∴2πr* 37= 1628 ⇒22/7* r= 22 ⇒ r = 7m h =(37-7)= 30 m Volume = πr2h= (22/7* 7* 7* 30) m3= 4620 m3
Q 38 - 0.88 m3. Of iron is softened and shaped as iron poles, each of distance across 2cm and length 7 m. What number of bars is shaped?
Answer : A
Explanation
For each iron rod, r = 1 cm = 1/100 m and h= 7m Volume of 1 iron rod = πr2h = (22/7* 1/100 *1/100 *7) m3 = 11/5000 m3 No. of iron rods= (88/100* 5000/11) = 400
Q 39 - Two barrel have break even with volumes and their statures are in the proportion 1:3. The proportion of their radii is
Answer : D
Explanation
Let their radii be R and r and the heights be h and 3h. Then πR2h = πr2*3h⇒ R2/r2 =3 ⇒(R/r) 2 = (√3)2 ⇒R/r= √3/1 = 3/√3 ∴ R: r = 3:√3
Q 40 - The range of a wire is diminished to 33%. In the event that the volume continues as before, then the length of new wire must be how often the first length?
Answer : D
Explanation
Let the original radius = r and original length = h New radius =⅓ r, let new length =H. Then, πr2h= π (⅓r) 2 *H = πr2H/9 ⇒h= H/9 ⇒ H= 9h
Answer : A
Explanation
√256/√x =2 ⇒ 16/√x =2 ⇒√x= 16/2 = 8 ⇒ x= 82= 64
Answer : B
Explanation
Let x = √2+√2+√2+√2+...∞. Then x= √2+x ∴ 2+x= x2 ⇒ x2-x-2= 0 ⇒ x2-2x+x-2 =0 ⇒ x(x-2)+(x-2)= 0 ⇒ (x-2)(x+1)=0 ⇒ x= 2 (∵x≠ -1)
Answer : D
Explanation
√0.9 = √ 0.94 * 0.94 = 0.94
Answer : C
Explanation
√0.121 = √ 0.347 * 0.347 = 0.347
Answer : D
Explanation
√0.064 = √ 0.252 * 0.252 = 0.252
Q 46 - If in a examination condition of marks is that for the right answer student gets 3 marks but reduce 2 marks for every wrong answer . if 30 questions attempted and 40 marks gets by the student. Find out the number of question correctly done by the student.
Answer : B
Explanation
Let correctly attempted questions be x. Then , 3x-2 (30-x) = 40 ⇒ 5x = 100 ⇒ x = 20. Hence, 20 question correctly done by the student.
Q 47 - 1/4th of Naresh money is equal to 1/6th of Suresh money. If the joint amount of Naresh and Suresh is Rs. 600 then what should be the differences between both amounts?
Answer : C
Explanation
1/4 *x = 1/6 *y = h . then , x= 4h and y = 6h 4h+6h = 600 ⇒ 10h= 600 ⇒ h = 60. Required difference = Rs. (6h-4h) = 2h = Rs. (2*60) = Rs.120.
Q 48 - 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?
Answer : B
Explanation
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.
Q 49 - If the average score of boys and girls is 71 and 73 but combined average of boys and girls is 71.8. How much percentage of girls is present in the class?
Answer : A
Explanation
If the number of girls is x than the number of boys in the class (100 -x). Then 73 x+ 71( 100- x) = 71.8 * 100 ⇒ 73x - 71 x = 7180 - 7100 = 2x = 80 ⇒ x = 40 Hence, it proves that 40% girls present in the class.
Q 50 - If we have two companies A and B . If no. of employess tansfered from A to B , then numbers of employees becomes equal . On the other hand if 20 employees are transferred from B to A Then the number of employess in A becomes doubles in the comparision to B company. How much Employees are working in A's company?
Answer : D
Explanation
We assume that x employee in the A company and y in the B company. Then, (x-10) = ( y+10) ⇒ x- y = 20 ... (i) 2 ( y-20) = (x+20) ⇒ 2y-x = 60 ...(ii) On solving (i) and (ii) , we get x = 100 ∴ There are 100 employees in company A.
Answer Sheet
| Question Number | Answer Key |
|---|---|
| 1 | B |
| 2 | B |
| 3 | C |
| 4 | D |
| 5 | A |
| 6 | B |
| 7 | C |
| 8 | C |
| 9 | A |
| 10 | A |
| 11 | B |
| 12 | C |
| 13 | D |
| 14 | A |
| 15 | B |
| 16 | B |
| 17 | A |
| 18 | C |
| 19 | B |
| 20 | A |
| 21 | B |
| 22 | B |
| 23 | C |
| 24 | D |
| 25 | D |
| 26 | D |
| 27 | C |
| 28 | C |
| 29 | A |
| 30 | C |
| 31 | C |
| 32 | B |
| 33 | D |
| 34 | C |
| 35 | A |
| 36 | A |
| 37 | B |
| 38 | A |
| 39 | D |
| 40 | D |
| 41 | A |
| 42 | B |
| 43 | D |
| 44 | C |
| 45 | D |
| 46 | B |
| 47 | C |
| 48 | B |
| 49 | A |
| 50 | D |