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1. A 600m Long Train Is Running At 73 Kmph. How Much Time Train Will Take To Cross An Electric Pole?
Formula Used: Time = ( Distance / Speed)
As all the option given in sec., so convert the train speed (Kmph) in to mps multiply by 5/18
speed (mps) = 73 * 5/18
Time = 600 / (73 * 5/18)
= (600 * 18 )/(73 * 5) sec
= (10800 / 365)
Time take by Train = 29.58Sec.
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2. A 120 M Long Train Is Running At 72 Kmph. How Much Time Will It Take To Cross A Man Standing On The Platform?
Formula Used:
Time = ( Distance / Speed)As all the option given in sec., so convert the train speed (Kmph) in to mps multiply by 5/18
speed (mps) = 72 * 5/18 = 20 mps
Time = (120 / 20) sec = 6 sec
Time take by Train = 6 Sec.
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3. Two Trains 400m And 300m Long Run At The Speeds Of 50 Kmph And 40kmph Respectively In Opposite Directions On Parallel Tracks. The Time Taken To Cross Each Other?
Trains are running in opposite Direction:
So need to find Length of two Trains = 300m + 400m = 700m
and Total Speed = 40 Kmph + 50 Kmph (Opposite Direction)
= 90 Kmph
so speed (m/sec) = 90 * 5/18 m/sec = 25 m/sec
Formula Used:
Time = Distance/SpeedTime = 700/ 25 sec
Time = 28 Sec.
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4. Two Stations A And B Are 110 Km Apart On A Straight Line. One Train Starts From A At 7 A.m. And Travels Towards B At 20 Kmph. Another Train Starts From B At 8 A.m. And Travels Towards A At A Speed Of 25 Kmph. At What Time Will They Meet?
Let they meet x hours after 7 a.m.
Distance covered by A in x hours = 20x km.
Distance covered by B in (x – 1) hours = 25(x – 1) km.
So Total Distance
=> 20x + 25(x – 1) = 110
=> 45x – 25 = 110 => 45x = 135
=> x = 3.
As They meet x hrs after 7 a.m. so they meet at 10 a.m.
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5. Two Trains Are Running At 40 Km/hr And 20 Km/hr Respectively In The Same Direction. Fast Train Completely Passes A Man Sitting In The Slower Train In 5 Seconds. What Is The Length Of The Fast Train?
As train are running in same direction
so Relative speed = (40 – 20) km/hr = 20 km/hr
= ( 20 x 5/18 ) m/sec = 50/9 m/sec.
Formula Used:
Distance = Speed * TimeNow Length of Faster Train = ( 50/9 x 5 ) m = 250/9 m
= 27 7/9 m.
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6. A Train Moves Past A Telegraph Post And A Bridge 264 M Long In 8 Seconds And 20 Seconds Respectively. What Is The Speed Of The Train?
Let the length of the train be x meters and its speed by y m/sec.
Then, x/y = 8 => x = 8y ———– (1)
As per the question total distance = (x + 264) meters.
(x + 264)/20 = y
Put the value of x from equation 1.
=> 8y + 264 = 20y
=> y = 22.
Therefore Speed = 22 m/sec = ( 22 x 18 /5) km/hr = 79.2 km/hr.
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7. An Athlete Runs 200 Metres Race In 24 Seconds. His Speed Is?
Speed = (200/24) m/sec = 25/3 m/sec
convert m/sec to km/hr
(25/3 * 18/5) km/hr = 30 km/hr.
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8. Albert Is Travelling On His Cycle And Has Calculated To Reach Point A At 2 P.m. If He Travels At 10 Kmph, He Will Reach There At 12 Noon If He Travels At 15 Kmph. At What Speed Must He Travel To Reach A At 1 P.m.?
Let’s Assume the distance travelled Albert = x km.
Formula Used: Time = Distance/Speed
so (x/10 – x/15) = 2 hrs
=> (3x – 2x)/30 = 2 hrs.
=> (3x – 2x) = 60
x = 60 km.
Time taken to travel 60 km at 10 km/hr = 60/10 hrs = 6 hrs.
Formula Used: Speed = Distance/Time
So, Albert started 6 hours before 2 P.M. i.e @ 8 A.M.
Required speed = 60/5 kmph. = 12 kmph.
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9. A Person Travelled By Train For 1 Hour At A Speed Of 50 Kmph. He Then Travelled By A Taxi For 30 Minutes At A Speed Of 32 Kmph To Complete His Journey. What Is The Average Speed At Which He Travelled During The Journey?
Total distance travelled in 1 hour
= 50 + (&frac; x 32) km
= 66 km
? Average speed = Total distance/Total time
= 66/(3/2) = 44 kmph.
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10. A Car Covers Four Successive 6 Km Stretches At Speeds Of 25 Kmph, 50 Kmph, 75 Kmph And 150 Kmph Respectively. Its Average Speed Over This Distance Is?
Time = Distance/Speed
Time taken for each 6 km can be given by
6/25, 6/50, 6/75 and 6/150
Total time = (6/25) + (6/50) + (6/75) + (6/150) = (36 + 18 + 12 + 6)/150 = 72/150
Average speed = Distance/time = (24/72) x 150 = 50 kmph.
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11. A’ And ‘b’ Complete A Work Togather In 8 Days.if ‘a’ Alone Can Do It In 12 Days.then How Many Day ‘b’ Will Take To Complete The Work?
A & B one day work = 1/8
A alone one day work = 1/12
B alone one day work = (1/8 – 1/12) = ( 3/24 – 2/24)
=> B one day work = 1/24
so B can complete the work in 24 days.
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12. If A Alone Can Do A Piece Of Work In 8 Days And B Alone Can Do The Same Work In 12 Days. How Many Days A And B Required To Finish The Same Work If They Work Togather?
A alone one day work = 1/8
B alone one day work = 1/12
Both A and B one day work = (1/8 + 1/12) = (3/24 + 2/24)
= 5/24
so A and B together finish the work in 24/5 day
or 4 4/5 days.
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13. A Can Finish A Piece Work In 18 Days And B Can Do The Same Work In Half The Time Taken By A. So If They Working Together, What Part Of The Same Work Can Finished In A Day?
First find the 1 day work of both (A & B)
A’s 1 day’s work = 1/18
and
B’s 1 day’s work = 1/9 (B can do work in half time)
(A + B)’s 1 day’s work = (1/18+1/9)
= (1+2)/18 = 3/18 = 1/6
so A & B together can do 1/6 of work in 1 day.
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14. A And B Can Together Finish A Work In 30 Days. They Worked Together For 20 Days And Then B Left. After Another 20 Days, A Finished The Remaining Work. In How Many Days A Alone Can Finish The Job ?
(A + B)’s 1 day’s work = 1/30
so (A&B) 20 days work = (20*1/30) = 2/3
so left work = (1?2/3)=1/3
1/3 work is done by A = 20 days.
So whole work will be done by A = (20 x 3) = 60 days.
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15. A And B Together Can Do A Piece Of Work In 30 Days. A Having Worked For 16 Days, B Finishes The Remaining Work Alone In 44 Days. In How Many Days Shall B Finish The Whole Work Alone ?
A’s 1 day’s work = x
and B’s 1 day’s work = y
So (A & B) 1 day work = 1/30 => x+y =1/30
=> 30x + 30y = 1 ——– (1)
So 16x + 44y = 1 ——– (2)
By Solving above two equations,
x = 1/60 and y = 1/60
B’s 1 day’s work = 1/60
Hence, B alone shall finish the whole work in 60 days.
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16. Vikas Can Cover A Distance In 1hr 24min By Covering 2/3 Of The Distance At 4 Kmph And The Rest At 5kmph.the Total Distance Is?
Let total distance be S
total time=1hr24min
A to T :: speed=4kmph
diistance=2/3S
T to S :: speed=5km
distance=1-2/3S=1/3S
21/15 hr=2/3 S/4 + 1/3s /5
84=14/3S*3
S=84*3/14*3
= 6km.
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17. A Cistern Can Be Filled By A Tap In 4 Hours While It Can Be Emptied By Another Tap In 9 Hours. If Both Taps Are Opened Simultaneously, Then After How Much Time Will The Cistern Get Filled ?
Time taken by tap A to fill the cistern=4 hrs
so work done by tap A in 1 hour = 1/4th
Time taken by tap B to empty the full cistern = 9 hours
so work done by tap B in 1 hour = 1/9th
=> Work done by (A + B) in 1 hour=(1/4 – 1/9)=5/36
Therefore, the tank will fill the cistern = 36/5 hours=7.2 hours.
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18. In A Class, There Are 15 Boys And 10 Girls. Three Students Are Selected At Random. The Probability That 1 Girl And 2 Boys Are Selected, Is?
Let’s assume the sample space = S
and Event of selecting 1 girl and 2 boys = E
So, n(S) = Number ways of selecting 3 students out of 25 = 25C3
=> (25 * 24 * 23)/(3 * 2 * 1) = 2300.
n(E) = (10C1* 15C2)
= 10 * [(15 * 14)/(2 * 1)] = 1050.
P(E) = n(E)/n(S) = 1050/2300 = 21/46.
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19. A Man Swims Downstream 72 Km And Upstream 45 Km Taking 9 Hours Each Time. What Is The Speed Of The Current?
Man’s downstream speed = 72/ 9 kmph => 8kmph
Man’s up stream speed = 45/ 9 => 5 kmph
So speed of current = (8 – 5)/2 = 1.5 kmph.
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20. A Boat Can Row Upstream At 25 Kmph And Downstream At 35 Kmph, Then The Speed Of The Current Is?
man’s upstream speed = 25 kmph
Man’s downstream speed = 35 kmph
so Speed of current = (35 – 25)/2 = 5 kmph.
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21. A Man Can Row His Boat With The Stream At 6 Km/h And Against The Stream In 4 Km/h. The Man’s Rate Is?
Man’s row in downstream by speed = 6 kmph
and upstream by speed = 4 kmph
so man rate = (6 – 4)/2 = 1 kmph.
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22. Two Pipes A And B Can Fill A Tank In 9 Hours And 3 Hours Respectively. If They Are Opened On Alternate Hours And If Pipe A Is Opened First, In How Many Hours Will The Tank Be Full?
Tank part filled by pipe A in 1 hour =1/9
Tank part filled by pipe B in 1 hour =1/3
Given Pipe A and B are opened alternatively.
So Part filled in every 2 hours =(1/9+1/3)=4/9
Tank Part will be filled in 4 hour =2*4/9=8/9
Remaining part = (1-8/9)=1/9
So next is A turn.
So Pipe A will fill remaining 1/9 part in next 1 hour.
Total Time = (4 hrs + 1 hrs) = 5 hrs.
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23. Pipe A Can Fill A Cistern In 6 Hours Less Than Pipe B. Both The Pipes Together Can Fill The Cistern In 4 Hours. How Much Time Would A Take To Fill The Cistern All By Itself?
Let’s assume time required by Pipe A to fill the cistern = X hours
So Time required by Pipe B to fill the cistern = (X + 6) hours
? Both Pipes (A+B) can fill cistern in 1 hour = [1/X + 1/(X + 6)]
Given Both pipe fill the cistern in 4 hours
=> [1/X + 1/(X + 6)] = 1/4 => [(X+6) + X]/(X+6)*x = 1/4
4X + 24 + 4X = X2 + 6x
X2 – 2X – 24 = 0
(X-6)(X+4) = 0
=> A can fill cistern in 6 hours.
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24. One Pipe Can Fill A Tank Three Times As Fast As Another Pipe. If Together The Two Pipes Can Fill The Tank In 36 Min, Then The Slower Alone Will Be Able To Fill The Tank In?
Lets assume time required by slower pipe alone to fill the tank = x minutes.
Then, faster pipe will fill it in x/3 minutes.
=> 1/x+3/x = 1/36
=>4/x = 1/36 => x = 144 min.
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25. 13 Sheeps And 9 Pigs Were Bought For Rs. 1291.85.if The Average Price Of A Sheep Be Rs. 74. What Is The Average Price Of A Pig?
Average price of a sheep = Rs. 74
:Total price of 13 sheeps = (74*13) = Rs. 962
But, total price of 13 sheeps and 9 pigs
= Rs. 1291.85
Total price of 9 pigs
= Rs. (1291.85-962) = Rs. 329.85
Hence, average price of a pig
= (329.85/9) = Rs. 36.65.
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26. Three Pipes A, B And C Can Fill A Tank In 6 Hours. After Working At It Together For 2 Hours, C Is Closed And A And B Can Fill The Remaining Part In 7 Hours. The Number Of Hours Taken By C Alone To Fill The Tank Is?
Tank part filled by pipes (A+B+C) in 1 hrs = 1/6 ——- (1)
so tank part filled by (A+B+C) in 2 hrs = 2*1/6 = 1/3
Now find the remaining part = (1-1/3) = 2/3
=> (A+B) 7 hs work = 2/3
so (A+B) 1 hrs work = 2/21 —— (2)
To find the C 1 hrs work use eq. 1 & 2
=> 1/6-2/21 = 1/14
so C alone can fill the tank in 14 hrs.
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27. A Tap Can Fill A Tank In 6 Hours. After Half The Tank Is Filled, Three More Similar Taps Are Opened. What Is The Total Time Taken To Fill The Tank Completely?
Time taken by one tap to fill half tank = 3hrs.
Part filled by the four taps in 1 hrs = (4*1/6) = 2/3.
Remaining part = (1-1/2) = 1/2.
so 2/3 : 1/2 :: 1:x => x = 1/2*1*3/2) = 3/4 hrs. => 45 min.
=> Total time taken = 3 hrs 45 min.
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28. If (x + 1/x) = 2, Then The Value Of (x^100 + 1/x^100) Is?
Quick Approach
for x =1 given eq. will be satisfy. (1+1/1)=2
so (x^100 + 1/x^100) = (1^100 + 1/1^100) = 2.
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29. If (2x + 2/x) = 1, Then The Value Of (x^3 + 1/x^3) Is?
(2x + 2/x) = 1
=> (x + 1/x) = 1/2
(x^3+1/x^3) = (x+1/x)^3-3x*1/x(x+1/x)
= (1/2)^3-1/2 = (1/8 – 3/2) = (1-12)/8= -11/8.
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30. 7 Years Ago, The Ages (in Years) Of A And B Were In The Ratio 4:5 And 7 Years Hence They Will Be In The Ratio 5:6. The Present Age Of B Is?
7 years ago, A’s age=4x years
and B’s age=5xyears
so (4x+14)/(5x+l4)=5/6
=> 25x + 70 = 24x + 84
x = (84 – 70) = 14
B’s present age = 5x + 7 = 5*14 + 7 = 77 years.
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31. The Sum Of Ages Of Family Members (both Children And Parents) Is 360 Years.the Total Ages Of Children And Parents Are In The Ratio 2:1 And The Ages Of Wife And Husband Are In The Ratio 5:7.what Will Be The Age Of Husband?
Given sum of ages is 360 years.
The ratio of children and parents ages is 2:1.
So total age of parents = 360 x 1 / 3 = 120 years
Given Ratio of wife and husband age is 5:7.
So the age of husband = 120 x 7
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32. The Sum Of The Ages Of A Mother And Her Son Is 45 Years. Five Years Ago, The Product Of Their Ages Was 3 Times The Mother Age At That Time, Then The Present Age Of The Son?
Let’s assume mother age = x years. —– (1)
sum of mother and her son age = 45
so son age will be = (45-x) years. ——- (2)
Five year ago:
mother age will be = (x-5) years
son age will be = (45-x-x) years = (40-x) year.
As per question
(x-5) * (40-x) = 3*(x-5)
=> (40-x) = 3
=> x = 37 year.
so son age will be (45-37) = 8 years.
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33. Father Is Aged Three Times More Than His Son Mohit. After 8 Years, He Would Be Two And A Half Times Of Mohit’s Age. After Further 8 Years, How Many Times Would He Be Of Mohit’s Age?
Let’s assume Monit’s present age = X years.
So father’s present age = (X + 3X) years = 4X years.
After 8 years.
(4X + 8) =5/2 * (X + 8)
=> 8X + 16 = 5X + 40
=> 3X = 24 so, X=8
Hence, required ratio = (4X + 16) / (X + 16) = 48/24 = 2.
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34. If X^3 + 3x^2 + 3x = 7, Then X Is Equal To?
x^3 + 3x^2 + 3x = 7
Adding both side 1
=> x^3 + 3x^2 + 3x + 1= 7 +1
=> (x+1)^3=2^3
=> x+1 = 2 => x =1.
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35. A Man Owns 2/3 Of The Market Research Beauro Business And Sells 3/4 Of His Shares For Rs. 75000. What Is The Value Of Business?
3/4 of his share = 75000
so his share = 100000.
2/3 of business value = 100000
so total value = 150000.
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36. From Its Total Income, A Sales Company Spent Rs.20,000 For Advertising, Half Of The Remainder On Commissions And Had Rs.6000 Left. What Was Its Total Income?
Let total income is X
X=20,000+(X-20,000/2)+6000
X-X/2=20,000-10,000+6000
X/2=16,000
X=32,000.
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37. The Sum Of Three Numbers Is 98. If The Ratio Of The First To Second Is 2 :3 And That Of The Second To The Third Is 5 : 8, Then The Second Number Is?
Let the three parts be A, B, C. Then,
A : B = 2 : 3 and B : C = 5 : 8 = 5 * 3/5 : 8 * 3/5 = 3 : 24/5
=> A : B : C = 2 : 3 : 24/5 = 10 : 15 : 24
=> B = 98 x 15/49 = 30.
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38. A Sum Of Money Becomes 2.5 Times Itself At 12.5% Simple Interest P.a. The Period Of Investment Is?
Let the period is ‘T’ and Sum= ‘P’.
As given money become 2.5.
=> 2.5 * P = P + S.I
=> S.I = 1.5 * P ————– (1)
=> S.I = (P * T * 12.5)/100 ———–(2)
By eq. (1) and (2)
=> 1.5 * P = (P * T * 12.5)/100
=> T=12 years.
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39. Three Partners A, B, C Start A Business. Twice A’s Capital Is Equal To Thrice B’s Capital And B’s Capital Is Four Times C’s Capital. Out Of A Total Profit Of Rs. 16,500 At The End Of The Year, B’s Share Is?
Let C capital = x. so B capital = 4x
2 *(A capital) = 3 * 4x
=> A’s capital = 6x
So A : B : C = 6x : 4x : x = 6 : 4 : 1
So, B’s capital = Rs. [16500 * 4/11] = Rs. 6000.
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40. If A Man Walks At The Rate Of 5kmph, He Misses A Train By Only 7min. However If He Walks At The Rate Of 6 Kmph He Reaches The Station 5 Minutes Before The Arrival Of The Train.find The Distance Covered By Him To Reach The Station?
Lets assume the required distance = x km.
Difference in the times taken at two speeds=12mins=1/5 hr.
Therefore (x/5-x/6)=1/5 or (6x-5x) = 6 or x = 6km.
So required distance = 6 km.
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41. A Batsman In His 18th Innings Makes A Score Of 150 Runs And There By Increasing His Average By 6. Find His Average After 18th Innings?
Let the average for 17 innings is x runs
Total runs in 17 innings = 17x
Total runs in 18 innings = 17x + 150
Average of 18 innings = 17x + 150/18
17x + 150/18 = x + 6 — > x = 42
Thus, average after 18 innings = 42.
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42. Sum Of Squares Of Two Numbers Is 2754, Their Hcf Is 9, Lcm Is 135, Find The Numbers?
Product of two no. = H.C.F*L.C.M
So,x*y=135*9=1215 —–(1)
and x^2+y^2=2754
So,(x+y)^2=x^2+y^2+2*x*y = 2754+2*1215=5184
So,x+y=72 ———– (2)
By solving eq. (1) & (2)
nos. are 45 and 27.
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43. Find The Largest 4-digit Number, Which Gives The Remainder 7 And 13 When Divided By 11 And 17?
LCM of 11 and 17 = 187.
When divided by 11 remainder 7,so difference 4.
When devided by 17 remainder 13,so difference 4.
Largest no exactly devide by 11 & 17=9911
The no’s = 9911-4 = 9907.
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44. What Comes Next In The Sequence?
4, 2, 5, 9, 5, 11, 13, 7, 16, 17, 9Split the series as below
4,2,5 | 9,5,11 | 13,7,16 | 17,9,?4,2,5 -> diff b/w 4,5 = 1
9,5,11 -> diff b/w 9,11 = 2
13,7,16 -> diff b/w 13,16 = 3
17,9,21 -> diff b/w 17,21 = 4.
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