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.
Q 1 - The entire surface range of a cuboid 24 cm long, 14 cm wide and 7.5 cm high is:
A - 2520 cm2
B - 1260 cm2
C - 1242 cm2
D - 621 cm2
Zone of the entire surface = 2(Lb+ bh +Lh)
= 2 (24*14 + 14*15/2 + 24* 15/2) cm2
= 2(336+105+180) cm2= (621*2) cm2
= 1242 cm2
Q 2 - The length, expansiveness and tallness of a cuboid are in the proportion 6:5:4 and its entire surface region is 33300 cm2. Its volume is:
A - 45000 cm3
B - 405000 cm3
C - 455000 cm3
D - information deficient
Let length = 6x cm, breadth = 5x cm and height = 4x cm
Whole surface area = 2(lb+ bh + lh)
=2 (6x*5x + 5x *4x + 6x *4x) cm2 = (148x2) cm2
∴148x2= 33300 ⇒x2 = 225 ⇒x = √225 = 15 cm
∴L= 90 cm   , B= 75 cm and h= 60 cm
∴Volume = (L*b*h) = (90*75*60) =405000cm3
Q 3 - The length of the askew of a cuboid 30 cm long, 24 cm wide and 18 cm is:
A - 30 cm
B - 15 √2cm
C - 30 √2 cm
D - 60 cm
Length of the diagonal =√ (L 2+ b2+ h2)=  √[ (30)2 +( 24)2+(18)2]
= √ (900+576+324)   =√1800= √900*2
= 30√2 cm
Q 4 - The most extreme length of a pencil that can be kept in a rectangular pencil box of measurements 8cm *6cm* 2 cm is:
A - 2√13 cm
B - 2√14 cm
C - 2√26 cm
D - 10√2 cm
Required  length =  √[(8)2 +(6)2 +(2)2]=√104 cm =√4*26
=2√26 cm.
Q 5 - The length of the longest shaft that can be kept in a room 5 m long, 4 m wide and 3 m high, is:
A - 5√2 m
B - 6√2 m
C - 7√2 m
D - none of these
Required length = √[(5) 2+(4) 2+(3) 2]=√(25+16)+(9)=√50m =√25*2 m =5√2m
Q 6 - The region of the base of a rectangular tank is 6500 cm2 and the volume of the water contained in it in 2.6 cubic meter. The profundity of the water in the tank is:
A - 2.5 m
B - 3 m
C - 5.5 m
D - 4 m
L*b= 6500cm2 , L*b*d=2.6m3=(2.6*100*100*100) cm3
∴ d = (2.6*100*100*100)/6500 cm = (2.6*100*100*100)/6500*100 = 4m
∴ Depth = 4m
Q 7 - What number of blocks will be expected to develop a divider 4 m long, 3 m high and 13 cm expansive , if every block measures 20 cm *12cm*6.5 cm?
A - 500
B - 1000
C - 1500
D - 2000
Volume of wall = (400*300*13) cm3
Volume of each bricks = (20*12*6.5) cm3
No. of bricks = (400*300*13 /20*12*13)*2 = 1000
Q 8 - A divider 24m long , 8 m high and 60cm thick is comprised of blocks, every measuring 24cm * 12cm *8cm , it being given that 10% of the divider comprises of mortar. What number of blocks will be required?
A - 50000
B - 45000
C - 40000
D - 20000
Volume of wall = (24*8*60/100)m3 =576/5m3
Volume of bricks = (90% of 576/5) m3= (90/100*576/5) m3= (144*18/25) m3
Volume if 1 bricks = (24/100*12/100*8/100) m3
Number of bricks = [(144*18/25)*100/24*100/12*100/8) = 45000
Q 9 - The velocities of A and B are in the proportion 3:4. A takes 20 min. more than B to achieve a destination. In what time does A achieve the destination?
A - 4/3 hrs
B - 2 hrs
C - 5/3 hrs
D - 8/3 hrs
Let the time taken by A be x hrs.
Then, time taken by B = (x-20/60) hrs = (x-1/3) hrs
Ratio of speeds = inverse ratio of time taken
∴3:4 =(x- 1/3): x ⇒3x-1/3x = 3/4
⇒12x- 4 = 9x
⇒3x= 4 ⇒x= 4/3 hrs
Required time = 4/3 hrs.
Q 10 - 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 11 - The proportion between the paces of two trains is 7:8. On the off chance that the second prepare keeps running in 5 hours 400 km, the pace of the first prepare is :
A - 70 km/hr
B - 200 km/hr
C - 250 km/hr
D - 350 km/hr 
Let the speed of first train be 7x km/hr.
Then the speed of the second train is 8x km/hr.
But speed of the second train=400/5km/hr=80 km/hr
∴8x=80⇒x=10.
Hence the speed of first train is (7*10) km/hr=70 km/hr.
Q 12 - A auto covers a separation of 715 km at a steady speed. On the off chance that the pace of the auto would have been 10 km/hr all the more, then it would have taken 2 hours less to cover the same separation. What is the first speed of the auto?
A - 45km/hr
B - 50 km/hr
C - 55 km/hr
D - 65 km/hr
Let the constant speed be x km/hr. Then,
715/x-715/(x+10) =2⇒1/x-1/(x+10) =2/715
⇒(x+10)-x/x(x+10) =2/715⇒x(x+10) =3575
⇒x2+10x-3575=0⇒x2+65x-55x-3575=0
⇒x(x+65)-55(x+65)=0
⇒(x+65)(x-55)=0
⇒x=55.
∴Original speed of the car is 55km/hr.
Q 13 - Two train approach one another at 30 km/hr and 27 km/hr from two spot 342 km separated. After how long will they meet?
A - 5 hrs.
B - 6 hrs.
C - 7 hrs.
D - 12 hrs.
Suppose the two trains meet after x hours. Then,
30x+27 x= 342 ⇒ 57 x = 342 ⇒ x = 342/57 = 6.
So the two trains will meet after 6 hours.
Q 14 - A understudy strolls from his home at 5/2 km/hr and achieves his school late by 6 min. Following day, he builds his pace by 1 km/hr and achieves a 6 min. before educational time. How far is the school from his home?
A - 5/4 km
B - 7/4 km
C - 9/4 km
D - 11/4 km
Let the required distance be x km. then,
x/ (5/2) - x/ (7/2) = 12/60 (  ∵difference between two times is 12 min.)
⇒ 2x/5 - 2x/7 = 1/5   ⇒ 14 x-10 x = 7 ⇒ 4x= 7 ⇒ x= 7/4
Required distance = 7/4 km
Q 15 - If an understudy strolls from his home to class at 5km/hrs, he is late by 30 min. However, on the off chance that he strolls at 6 km/hr. he is late by 5 min. just. The separation of his school from his home is:
A - 2.5 km
B - 3.6 km
C - 5.5 km
D - 12.5 km
Let the required distance be x km. then,
x/5 - x/6 = 25/7   (difference between two times is 25 min.)
⇒ 12x- 10 x = 25 ⇒2x = 25   ⇒ x= 25/2 km = 12.5 km
Q 16 - 176 + √2401  = ?
A - 14
B - 15
C - 18
D - 24
√176 +√2401 =√(176+49) =√225 =15
Q 17 - 248 + √ 52 +√144
A - 14
B - 16
C - 16.6
D - 18.8
Given exp.= √248  +√52+12  = √248+ √64 = √248+8   =√256 = 16
Q 18 - √(25/16) =?
A - 7/4
B - 5/4
C - 1.125
D - none of these
√25/16  = 5/4
Q 19 - √32+√48/ √8+√12 = ?
A - √2
B - 2
D - 8
(√32+√48)/( √8+√12) =(√16*2 +√16*3)/ √4*2 +√4*3)
= 4√2+4√3/ 2√2+2√3  = 4(√2+√3)/ 2(√2+√3) =2
Q 20 - (√24+√216)/√96= ?
A - 2√6
B - 6√2
C - 2
D - 1/3 √6
(√24+√216)/√96 = (√4*6 +√36*6)/ √16*6 = (2√6+6√6)/4√6
= 8√6 /4√6 = 2
Q 21 - if √24 is approximately equal to 4.898 , then √8/3 is nearly equal
A - 0.544
B - 1.333
C - 1.633
D - 2.666
√8/3 = √8*3/3*3 = √24/3 =4.898/3 = 1.633
Q 22 - given that √3= 1.732, the value of   3 +√6 /(5√3-2√12 -√32+√50)is:
A - 4.899
B - 2.551
C - 1.414
D - 1.732
Given exp = 3+√6/(5√3-2√4*3-√16*2+ √25*2)
= 3+√6/(5√3-4√3-4√2+5√2)
= (3+√6)/( √3+√2)*( √3-√2)/( √3-√2)
= 3√3-3√2+√18-√12/(3-2)
=(3√3-3√2+√9*2 -√4*3 )= (3√3-3√2+3√2-2√3) =√3= 1.732
Q 23 - A certain no. of men finishes a bit of work in 60 days. On the off chance that there were 8 men more work could be done in 10 days less. What number of men arrived initially?
A - 30
B - 40
C - 32
D - 36
Let there be x men originally.
X man finish the work in 60 days and (x+8) finish it in 50 days.
X man finish the job in 60 days.
⇒ 1 man can finish it in 50 (x+8) days.
∴ 60 x = 50(X+80 ⇒10 x= 400 ⇒x = 40
Hence, there were 40 men originally.



Q 24 - The rates of working of A and B are in the proportion 3:4. The no. of days taken by them to complete the work is in the proportion
A - 3:4
B - 9:16
C - 4:3
D - none of these
Ratio of time taken =  1/3 : 1/4 = 4:3
Q 25 - A man and a kid finish a work together in 24 days. On the off chance that throughout the previous 6 days man alone takes every necessary step then it is finished in 26 days. To what extent the kid will take to complete the work done?
A - 72 days
B - 20 days
C - 24 days
D - 36 days
Work done by  (man + boy ) in 20 days = (1/24*20)= 5/6
              Remaining work = (1-5/6) = 1/6
1/6 work is done by boy in 6 days.
Whole work will be done by the boy in (6*6) = 36 days
Q 26 - 8 men can delve a pit in 20 days. On the off chance that a man works half double as a kid then 4 men and 9 kids can delve a comparable pit in:
A - 10 days
B - 12 days
C - 15 days
D - 16 days
1 man = 3/2 boys , 8 men = (8*3/2) boys = 12 boys
(4men + 9 boys) = (4*3/2 +9) boys = 15 boys
Now, 12 boys dig it in 20 days.
1 boy digs it in (20*12) days.
15 boys will dig it in (20*12)/15 days = 16 days
Q 27 - 8 men can do a bit of work in 12 days. 4 ladies can do it in 48 days and 10 kids can do it in 24 days. In how long can 10 men, 4 ladies and 10 kids together finish the bit of work?
A - 5 days
B - 15 days
C - 28 days
D - 6 days
8 men 1 day work = 1/12  ⇒1 man 1 day work = 1/96
4 women 1 day work = 1/48 ⇒1 women 1 day work = 1/192
10 children 1 day work = 1/24 ⇒1 child 1 day work = 1/240
(10 men + 4 women +10 children)'s 1 day work = (10/96 + 4/192 +10/240)
= (5/48+ 1/48+ 1/24) = (5+1+2)/48 = 8/48 = 1/6
So, they can finish the work in 6 days.
Q 28 - A men, a ladies and a kid can together finish a bit of work in 3 days. In the event that a men alone can do it in 6 days and a kid alone in 18 days, to what extent will a ladies take to finish the work? 
A - 9 days
B - 21 days
C - 24 days
D - 27 days
(1 man+ 1 women +1 boy ) 1 day work = 1/3
Suppose the women alone can do it in x days. Then
1/6 +1/x+1/18= 1/3 ⇒4/18 +1/x = 1/3 ⇒1/x = (1/3- 2/9) = 1/9
So, 1 woman can do the work in 9 days
Q 29 - 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 30 - 9 youngsters can finish a bit of work in 360 days. 18 men can finish the same work of piece in 72 days and 12 ladies can finish it in 162 days. In how long can 4 men, 12 ladies and 10 kids together finish the bit of work?
A - 68 days
B - 81 days
C - 96 days
D - 124 days
9 children 1 day work = 1/360 ⇒1 child 1 day work = 1/3240
18 man 1 day work = 1/72 ⇒1 man 1 day work = 1/1296
12 women 1 day work = 1/162 ⇒1 women 1 day work = 1/1944
(4men +12 women +10 children) 1 day work = (4/1296+ 12/1944+10/3240)
= (1/324+1/162+1/324) = 4/324 = 1/81
Hence they can finish the work in 81 days.
Q 31 - 10 ladies can finish a work in 8 days and 10 youngsters take 12 days to finish the work. How long will 6 ladies and 3 kids together take to finish the work?
A - 7
B - 10
C - 9
D - 12
10 women 1 days work =1/8 ⇒1 women 1 day work = 1/80
10 children 1 day work = 1/12 ⇒1 children 1 day work = 1/120
(6 women + 3 children) 1 day work = (6/80 +3/120) = (3/40+ 1/40) = 4/40 = 1/10
So, they can finish the work in 10 days.
Q 32 - If 5 men or 9 ladies can complete a bit of work in 19 days, 3 men and 6 ladies will do likewise work in
A - 10 days
B - 12 days
C - 13 days
D - 15 days 
5 men 1 day work = 1/19 ⇒ 1 man's 1 day work = 1/95
9 women 1 day work = 1/19 ⇒1 women 1 day work = 1/171
(3 men + 6 women) 1 day work = (3/95 + 6/171) = (27+30)/855 = 57/ 855 = 1/15
∴ 3 men and 6 women can finish the work in 15 days.
Q 33 - A man loses Rs 55.50 yearly when the yearly rate of interest tumbles from 11.5% to 10%. His capital is:
A - Rs 3700
B - Rs 7400
C - Rs 8325
D - Rs 11100
Let the capital be Rs. x. then,
(x*23/2*1/100*1) ?(x*10*1/100*1) = 55.50
=> 23x/200 ?x/10 =111/2 => 23x-20x =11100 => 3x= 11100 => x = 3700
Hence, the capital is Rs. 3700.
Q 34 - A cash bank finds that because of a fall in the yearly rate of interest from 8% to 31/4% his yearly salary reduces by Rs 61.50. His capital is:
A - Rs 22400
B - Rs 23800
C - Rs 24600
D - Rs 26000
Let the capital be RS. x. then
(x*8/100*1) ?(x*31/4*1/100*1) = 123/2
=> 2x/25-31x/400 = 123/2 => 32x-31x = 24600 => x= 24600
Hence, the capital is Rs. 24600.
Q 35 - A commission operators permits a refund of 2% to a financial specialist while the organization pays on interest of 15% on the speculation. What rate of premium does the financial specialist really acquire on his venture?
A - 17%
B - 131/8%
C - 17%
D - 750/49%
Instead of Rs.100, the investor invests = Rs. (100-2) = Rs. 98
Instead of Rs. 98 = Rs. 15
Interest on Rs. 100= Rs. (15/98*100) = Rs. 750/49
Q 36 - An aggregate of Rs 10000 is loaned mostly at 8% and the staying at 10% p.a. On the off chance that the year?s interest on the normal is 9.2% the cash loaned at 10% is:
A - Rs 6000
B - Rs 5500
C - Rs 5000
D - Rs 4500
Ratio of investments = 0.8: 1.2: = 2:3
Money at 10% = Rs. (10000*3/5) = Rs. 6000>
Q 37 - An aggregate of Rs 5000 was loaned mostly at 6% and incompletely at 9% basic interest. On the off chance that the aggregate yearly premium be Rs 390, the proportion in which the cash was loaned at given rates is: 
A - 1:1
B - 1:2
C - 2:3
D - 3:2
Let the money invested at the two rates be rs. x and Rs. (5000-x)
Then, (x*6/100*1) + (5000-x) *9/100*1 = 390
=> 3x/50 + 9(5000-x)/100 = 390
=> 6x+45000-9x = 39000 => 3x= 6000 => x= 2000.
Required ratio = 2000:3000= 2:3
Q 38 - The distinction between the premiums got from two unique banks on Rs 5000 for a long time is Rs 25. The distinction between their rates is:
A - 1%
B - 2.5%
C -  0.5%
D - 0.25%
Let the rates be x% p.a. and Y % p.a. Then,
(5000*x/100*2)- (5000*y/ 100 *2) =25
=> 100(x-y) = 25 => x-y =0.25
Required difference in rates = 0.25% p.a.
Q 39 - An acquires Rs 8000 at 12% p.a simple interest and B gets Rs 9100 at 10%p.a. simple interest . In how long will their measures of obligations be equivalent?
A - 18 years
B - 20 years
C - 22 years
D - 24 years
Let the required time be x years. Then,
8000+8000*12/100*x= 9100+9100*10/100*x
=> 50x =1100 => x= 22 years
Q 40 - Rs. 6000 adds up to Rs. 7920 in 4 years at a sure rate of interest. On the off chance that the rate gets to be 1.5 times of itself, the measure of the same rule in 5 years will be:
A - Rs. 9240
B - Rs. 9600
C - Rs. 9000
D - Rs. 9450
P = Rs. 6000, S.I = Rs. (7920-6000) = Rs. 1920, T= 4 years.
R = (100*1920/6000*4) = 8% p.a.
New rate = (1.5*8) % p.a. = 12% p.a.
Now = Rs. 6000, R= 12% p.a. and T = 5 years.
S.I = Rs. (6000*12/100*5) = Rs.3600
Amount = (6000+3600) = Rs. 9600.
Q 41 - {  1+1/1+ 1/(1+1/3) } / 11/7  = ?
A - 1
B - 4/3
C - 5/4
D - 8/7
given exp.   = { 1+  1/ (1+1/(4/3)} /11/7
= { 1+  1/( 1+3/4)} /11/7
= [ 1+ 1/(7/4)]/11/7
= (1+4/7) /11/7
= 11/7 /11/7
= 1
Q 42 - 3/2 /1+1/(1+1/4)= ?
A - 5/4
B - 3/2
C - 5/6
D - 1
given exp. = (3/2) / 1+ 1(/5/4)
=  (3/2)/ (1+4/5)
= (3/2) /(9/5)
= ( 3/2 * 5/9)
= 5/6  
Q 43 - 5/ 3+3 /(1-2/3 )  =?
A - 3/5
B - 5/3
C - 5/12
D - 5
given exp. =  5/ 3+  3/(1/3)
= 5/ (3+9)
= 5/12
Q 44 - 79/14  / 5+  3/  (3+ 1/3 /5  =?
A - 1
B - 3/2
C - 2
D - 5/2
given exp.  =  (79/14)/  5+ 3/(3+5/3)
=  (79/14)/  5+ 3/(14/3)
= (79/14)/ 5+9/14
=  (79/14)/(79/14)
= 1
Q 45 - 4- 5/ 1+ 1 / 3+ 1 /  2+ 1/4 = ?
A - 31/40
B - 1/8
C - 4/9
D - 40
given exp. =   4 - [5/ {1+ 1/ 3+(1/ 9/4)}] 
= 4 -  [ 5/{ 1+ 1/ (3+4/9)}]
= 4 -  [5 / {1+ 1/ (31/9)}]
= 4 -  5/ (1+9/31)
= 4 -  5/ (40/31)
= 4- ( 5*31)/40
= 4- 31/8
= (32-31)/8
= 1/8
Q 46 - 1+ 1/1+ 1/1 + 1/9 = ?
A - 29/10
B - 10/19
C - 29/19
D - 10/9
given exp.  =  1 + 1/(1+ 1/(9/10))
=1 + 1/(19/10)
=1 + 10/19
= 29/19
Q 47 - If 1/ 1+ 1 / 1+ 1/x = 2 then x = ?
A - ? 1/3
B - ? 1
C - 1/3
D - 1
1/(1+ 1/ (1+1/x)) = 2
⇒ 1/(1+x/(x+1)) = 2
⇒ (x+1)/ (2x+1) = 2
⇒ x+1= 4x+2
⇒ 3x= -1
⇒ x= -1/3
Q 48 - What is the selling price of a toy? If the cost of the toy is Rs. 90 and a profit of 15% over selling price is earned?
A - Rs. 103.5
B - Rs. 101.5
C - Rs. 107
D - Rs. 115
Cost price + profit = Selling Price
Cost price is Rs. 90
Profit is 15% of Cost Price=13.5
Selling Price = CP + Profit
=90+13.5 = 103.5
SP is Rs. 103.5
Q 49 - A person incurs 10% loss by selling a refrigerator for Rs 5400. At what price should the refrigerator be sold to earn 10% profit?
A - 5700
B - 6000
C - 6300
D - 6600
SP=5400 X  (110)/90=6600.
Q 50 - By selling a Book for Rs.20, a man loses one eleventh of what it costs him. The cost of the Book is
A - 22
B - 21
C - 20
D - 19
Let us assume cost price is X
i.e., X - 20 = ( 1)/11 X
=>X=22

    87 Lectures 
     22.5 hours 

 Programming Line
 Print
 Add Notes
 Bookmark this page