Question 1. Find Three Consecutive Odd Integers Whose Sum Is 117?

Answer :

Let the three integers be 2x + 1, 2x + 3 , 2x + 5.

Therefore, 6x + 9 = 117,

x = 18

The three integers are 37, 39, forty one.

Question 2. A Train Covers A Distance In 50 Min ,if It Runs At A Speed Of forty eight Kmph On An Average.The Speed At Which The Train Must Run To Reduce The Time Of Journey To 40 Min Will Be?

Answer :

Time=50/60 hr=five/6hr

Speed=48mph

distance=S*T=48*5/6=40km

time=40/60hr=2/3hr

New pace = 40* three/2 kmph= 60kmph.

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Question three. The Cost Price Of An Article Ls eighty% Of Its Marked Price For Sale. How Much Percent Does The Trades-guy Gain After Allowing A Discount Of 12%?

Answer :

C.P. Of the item = Rs. A hundred

So Marked fee = (a hundred*a hundred)/80 = Rs. One hundred twenty five

SP after the discount = Rs.(125*88)/a hundred = Rs. 110

therefore Gain percent = 10.

Question 4. If 6 Is Subtracted From The Present Age Of Ritu And The Remainder Is Divided By 6, Then The Present Age Of Sheema Is Obtained. If Sheema Is four Years Younger To Raju Whose Age Is 12 Years, Then Find The Age Of Ritu?

Answer :

Lets assume Ritu present age = x years

so sheema age = (x-6)/6 years

As per query:

sheema age [(x-6)/6 + 4] = 12

(x-6)/6 = 12 - 4

=> (x-6)/6 = 8

=> x - 6 = forty eight => x = fifty four

So Ritu age = fifty four years.

Question five. The Cost Of Two Varieties Of Paint Is Rs.3969 Per 2 Kg And Rs.1369 Per 2 Kg Respectively. After How Many Years, The Value Of Both Paint Will Be The Same, If Variety 1 Appreciates At 26% Per Annum And Variety 2 Depreciates At 26% Per Annum?

Answer :

Simply appreciates variety 1 via 26% and depreciates variety 2 through 26% as:

3969(1-(1/26))^n=1369(1+)1/26))^n

For n = 2 we get both values same.

Variety 2 Variety1

3969.00 1369 Initially

2937.06 1724.94 when I year

2173.Forty two 2173.Forty two after II yr

So the price become same after 2 years.

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Question 6. A Car Averages 55 Mph For The First 4 Hours Of A Trip And Averages 70 Mph For Each Additional Hour. The Average Speed For The Entire Trip Was 60 Mph. How Many Hours Long Is The Trip?

Answer :

Lets expect additional hours = x hrs.

So total No. Hours in adventure = (4+x)

[(55*4)+(70*x)]/(four+x)=60 =>

=>x=2

Therefore, Total No. Of hrs. In Journey = (x+4) = 6 hrs.

Question 7. A Ball Dropped From H Height And Moves 80% Of Height Each Time. Total Distance Covered Is?

Answer :

First time distance= H

Second time = 80H/one hundred = 4H/5

further third time 80% of 4H/five = H(four^2)/(five^2)

and so forth..

This will cause limitless phrases of geometric development

i.E H+2*4H/five+2*16H/25..

So, Sum = H+ 2*4H/(5(1-four/five)) = 9H.

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Question eight. At 10:00 Am 2 Trains Started Travelling Towards Each Other From Station 287 Miles Apart They Passed Each Other At 1:30 Pm The Same Dayy .If Average Speed Of The Faster Train Exceeded By 6 Miles /hr What Is Speed Of Faster Train In Miles/hr?

Answer :

Lets expect the velocity of slower teach = x miles/hrs.

So, pace of faster educate is = (x+6) miles/hrs.

Given, passed each other at 1:30 PM i.E after three 1/2 hrs.

Both educate travelling closer to every other so general relative velocity = x+(x+6) = (2x+6)

So, 287/(2x+6) = 7/2 => 574 = 14x + 42

=> 14x = 542 => x = ~38

So spee of faster train = (x+6) miles/hrs. = 44 miles/hrs.

Question nine. Find The Approximate Value Of The Following Equation 6.23% Of 258.Forty three - ? + three.11% Of 127 = 13.87?

Answer :

[(6.23/100)*258.43]-'X'+[(3.11/100)*127]=13.87

sixteen.100189-X+three.9497=13.87

X=6.179889.

SAP Aptitude Interview Questions

Question 10. A Cycled From P To Q At 10 Kmph And Returned At The Rate Of nine Kmph. B Cycled Both Ways At 12 Kmph. In The Whole Journey B Took 10 Minutes Less Than A. Find The Distance Between P And Q?

Answer :

Lets assume distance between P and Q = d km.

Time taken via A in each aspect = d/10 + d/9 = 19d/ninety hrs.

Time taken through B in both aspect = 2d/12 = d/6 hrs.

B tool 10 min. Or 1/6 hrs. Much less than A.

So, 19d/90 - d/6 = 1/6

=> (19d-15d)/ninety = 1/6

=> 4d/90 = 1/6

=> d = 15/four km = three.Seventy five km.

Question 11. An Alloy Of Zinc And Copper Contains The Metals In The Ratio 5 : 3. The Quantity Of Zinc To Be Added To 16 Kg Of The Alloy So That The Ratio Of The Metal May Be 3 : 1 Is?

Answer :

Lets expect amount of zinc to introduced = x kg.

Zinc quantity in alloy = 5/eight*16=10 Kg.

And copper quantity = 3/eight*16=6 kg.

Alloy new ratio of zinc and copper = three:1

Zinc quantity in alloy => (10+x)/6=3/1 => 10 + x = 18 => x = eight kg.

Oracle Aptitude Interview Questions

Question 12. A Gets Rs.33 When A Sum Of Money Was Distributed Among A, B And C In The Ratio three:2:five, What Will Be The Sum Of Money?

Answer :

Lets count on sum of money = Rs. X ;

A receives Rs. 33 whilst sum dispensed within the ratio of 3:2:five

so, 33=x*3/10

=> x=110.

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Question 13. The Ratio Of Two Numbers Is 3:four And Their Hcf Is 4.Their Lcm Is?

Answer :

Given Two No, ratio = three:4 and their HCF = 4

So, No. = 3*four =12 and 4*4=16

LCM of 12,16 = 48.

Question 14. A Certain Quantity Of forty% Solution Is Replaced With 25% Solution Such That The New Concentration Is 35%. What Is The Fraction Of The Solution That Was Replaced?

Answer :

Consider an answer = 1 ltr

X is the positive quantity which has to get replaced

Now,

40%of (1-x)+(25p.Cof x)=35/100

=> forty/100 * (1-x) + 25x/100 = 35/one hundred

=> forty/a hundred - 40x/a hundred + 25x/a hundred = 35/a hundred

=> 15x/a hundred = five/100

=> x = 1/3.

Question 15. Three Runners A, B And C Run A Race, With Runner A Finishing 12 Meters Ahead Of Runner B And 18 Meters Ahead Of Runner C, While Runner B Finishes eight Meters Ahead Of Runner C. Each Runner Travels The Entire Distance At A Constant Speed. What Was The Length Of The Race?

Answer :

Lets assume distance of race = x mtrs.

Then while A finishes x m , B has run (x- 12)mtrs and C has run (x-18) mtrs.

=> at this factor B is 6 m in advance of C. Now to complete race b needs to run some other 12 m,

=> he runs some other 12 m. Whilst B finishes race he is eight m in advance of C.

So final 12 m B has run, C has run 10 m.As speeds are constant,

=> x-12/ x-18 = 12/10 => x = 48 mtrs.

TCS Aptitude Interview Questions

Question sixteen. If 20 Men Or 24 Women Or 40 Boys Can Do A Job In 12 Days Working For 8 Hours A Day, How Many Men Working With 6 Women And 2 Boys Take To Do A Job Four Times As Big Working For 5 Hours A Day For 12 Days?

Answer :

Amount of work done through 20 men = 24 women = forty boys

i.E 5man = 6 woman = 10 boys. -----------------(i)

According to the 1st condition, five men can do a process in 12 days running for eight hours a day.

Required :- what number of greater men are required to work with 6 women and 2 boys.

Let the desired wide variety of guys be M.

According to the given facts,

=>(5 x 12 x 8 )/1=[(M + 5 + 1) x 5 x 12 ]/four. ( 6 women= 5men & 2 boys =1 guys)

=>32= M + 6.

=>M=26.

Hence, 26 men are running with 6 ladies and 2 boys.

Question 17. Nitish Sold His Watch And Sunglasses At A Loss Of four% And Gain Of 4% Respectively For 2600 To Kamal. Kamal Sold The Same Sun Glasses And Watch At A Loss Of four% And Gain Of four% Respectively For 2700. The Price Of Watch And Sun Glasses To Nitish Were?

Answer :

Lets assume the CP of watch = Rs x

and sunglasses = Rs y.

2600=96x/one hundred + 104y/a hundred

2700= 104x/a hundred + 96y/a hundred

By fixing,

y=seven hundred

x=1960.

Infosys Aptitude Interview Questions

Question 18. The Age Of The Grand Father Is The Sum Of His Three Grandsons.The Second Is 2 Year Younger Than First One And The Third One Is 2 Year Younger Than The Second One. Then What Will Be The Age Of The Grandfather?

Answer :

Lets anticipate 2d grandson = x year.

So, first grandson age = (x+2) yr ,

1/3 grandson = (x-2) year

So, x+2+x+x-2=3x

i.E grandfather age is 3 times as older as his second grandson.

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Question 19. Two Urns Contain Respectively 2 Red 3 White And three Red,5 White Balls.One Ball Is Drawn At Random From The First Urn And Transferred Into The Second.A Ball Is Now Drawn From The Second Urn And It Turns Out To Be Red. What Is The Probability That The Transferred Ball Was White?

Answer :

According to baye's theorem:

p(no of manner to drawn a ball from 2nd urns is red and ball transferred from 1 urns)=(3c1/5c1)*(3c1/9c1)=nine/45

Total=(2c1/5c1)*(4c1/9c1)+(3c1/5c1)*(3c1/9c1)=17/45

so chance=(9/forty five)/(17/45)=nine/17.

Question 20. In An Exam, Ajith, Sachu, Karna, Saheep And Ramesh Scored An Average Of 39 Marks. Saheep Scored 7 Marks More Than Ramesh. Ramesh Scored 9 Fewer Than Ajith. Sachu Scored As Many As Saheep And Ramesh Scored.Sachu And Karna Scored 110 Marks Them. If Ajith Scores 32 Marks Then How Many Marks Did Karna Score?

Answer :

Lets anticipate marks of Ajith, Sachu, Karna, Saheep and Ramesh respectively u, v, w, x, y

So, as in step with the question:

z+7=x ---(i)

u- nine=z ---(ii)

x+ z=v ---(iii)

v+w=110 ---(iv)

Given, u=32 ---- (v)

By solving eq. (i), (ii), (iii), (iv) and (v)

w=fifty seven.

IBM Aptitude Interview Questions

Question 21. The Average Number Of Visitors Of A Library In The First 4 Days Of A Week Was fifty eight. The Average For The 2nd, 3rd, 4th And fifth Days Was 60.If The Number Of Visitors On The 1st And fifth Days Were In The Ratio 7:8 Then What Is The Number Of Visitors On The fifth Day Of The Library?

Answer :

If wide variety of visitors on 1st, second, third, 4th & 5th day are a, b, c, d & e respectively, then

a+b+c+d=fifty eight*four=232 ----(i)

b+c+d+e=60*four=240 ----(ii)

Subtracting eq. (i) from (ii)

e-a=eight ---(iii)

Given, a/e=7/8 ---(iv)

So from eq. (iii) & (iv)

a=fifty six, e=sixty four.

Question 22. Find The Smallest Number Which Leaves 22,35, forty eight And 61 As Remainders When Divided By 26, 39, 52 And 65 Respectively?

Answer :

26-22=four

39-35=4

fifty two-48=four

65-61=four

LCM of (26,39,52,65)=780

So smallest variety = (780-4) = 776.

Question 23. A Constructor Estimates That three People Can Paint Mr. Khans House In four Days. If He Uses four People Instead Of three,how Long Will They Take To Complete The Job?

Answer :

Lets expect day required to finish the task = x day

Men Day

---------------------

three(u) four(d)

four x

x/four = three/4

=> x = 3 days.

Capgemini Aptitude Interview Questions

Question 24. Two Circles Touch Each Other Externally. The Distance Between Their Centres Is 14 Cm And The Sum Of Their Areas Is a hundred thirty Cm^2. Find Their Radii?

Answer :

Lets count on the radius of 1st circle = x cm.

So, the radius of 2nd circle = (14-x)cm.

(pi*x*x)+(pi*(14-x)*(14-x))=130

By fixing above eq.

X = eleven or three.

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Question 25. A Vessel Is Filled To Its Capacity With Pure Milk. Ten Litres Are Withdrawn From It And Replaced By Water. This Procedure Is Repeated Again. The Vessel Now Has 32 Litres Of Milk. Find The Capacity Of The Vessel (in Litres)?

Answer :

Lets anticipate the potential of the vessel = x litres.

X (1- 10/x)^2 = 32

x^2 +a hundred-20x = 32x

x^2 +100 - 52x= 0

x^2 - 50x - 2x + 100 = zero

=> (x-2) (x-50) = zero

=> x=2 & x=50

As 10 litres is withdrawn so vessel capacity can be 50 litres.

Question 26. If The Sales Tax Reduced From three half% To 3 1/3%, Then What Difference Does It Make To A Person Who Purchases An Article With Market Price Of Rs. 8400 ?

Answer :

Tax decreased from three 1/2 or 7/2% to 3 1/3% or 10/three%.

So, the distinction in tax = (7/2 - 10/3)% = 1/6%.

=> 1/6 % of Rs. 8400 = 8400*1/6% = 8400* (1/6 * 1/one hundred) = Rs. 14.

Cognizant Aptitude Interview Questions

Question 27. In A Stream Running At 2 Kmph,a Motar Boat Goes 6 Km Upstream And Back Again To The Starting Point In 33 Minutes. Find The Speed Of The Motarboat In Still Water?

Answer :

Lets assume speed of boat in nonetheless water = x km/hr

So boat downstream pace = (x+2) km/hr.

Boat upstream velocity = (x-2) km/hr.

6/(x-2)+6/(x+2)=33/60 => 6/(x-2)+6/(x+2)=11/20

=> a hundred and twenty*[(x-2) + (x+2)] = 11*x^2 -forty four

=> 240 x = 11*x^2-forty four

=> eleven* x^2 -240x -44 = 0 ---------(1)

By fixing eq. (1)

x = 22 km/hr.

SAP Aptitude Interview Questions

Question 28. Exact Time Of A Clock Was Set Right At 5:00 Am Then It Loses sixteen Min. Every Day, On The 4th Day When It Shows 10:00 Am, What Is The Exact Time?

Answer :

Total overdue in 3 day = 16*three = forty eight min.

Total late in sooner or later =sixteen min.

So past due in 1 min. = 16/(24*60) = 0.01111;

Hence in five hour (5:00 AM to ten:00 AM on 4th day) =.0111*5*60 = 3.333;

So, Exact time may be = 10:52 AM.

Question 29. Which Of The Following Numbers Are Completely Divisible By 11?

Answer :

If the distinction of the sum of digits at bizarre places and the sum of its digits at even places, is either 0 or divisible by means of eleven, then truly the variety is divisible by way of 11.

[(Sum of digits at odd places) - (Sum of digits at even places)]/11

=> [(3+4+6+2) - (2+5+8)]/11 => (15 - 15)/eleven = 0/11 = 0

in addition others range are divisible with the aid of eleven.

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