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Let the three integers be 2x + 1, 2x + 3 , 2x + 5.
Therefore, 6x + 9 = 117,
x = 18
The three integers are 37, 39, 41.
New speed = 40* 3/2 kmph= 60kmph.
C.P. of the article = Rs. 100
So Marked price = (100*100)/80 = Rs. 125
SP after the discount = Rs.(125*88)/100 = 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 4 Years Younger To Raju Whose Age Is 12 Years, Then Find The Age Of Ritu?
Lets assume Ritu present age = x years
so sheema age = (x-6)/6 years
As per question:
sheema age [(x-6)/6 + 4] = 12
(x-6)/6 = 12 - 4
=> (x-6)/6 = 8
=> x - 6 = 48 => x = 54
So Ritu age = 54 years.
Question 5. 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?
Simply appreciates variety 1 by 26% and depreciates variety 2 by 26% as:
For n = 2 we get both values equal.
Variety 2 Variety1
3969.00 1369 Initially
2937.06 1724.94 after I year
2173.42 2173.42 after II year
So the price become same after 2 years.
Lets assume additional hours = x hrs.
So total No. hours in journey = (4+x)
Therefore, Total No. of hrs. in Journey = (x+4) = 6 hrs.
First time distance= H
Second time = 80H/100 = 4H/5
similarly third time 80% of 4H/5 = H(4^2)/(5^2)
and so on..
This will lead to infinite terms of geometric progression
So, Sum = H+ 2*4H/(5(1-4/5)) = 9H.
Question 8. 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?
Lets assume the speed of slower train = x miles/hrs.
So, speed of faster train is = (x+6) miles/hrs.
Given, passed each other at 1:30 PM i.e after 3 1/2 hrs.
Both train travelling towards each other so total relative speed = 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.
Lets assume distance between P and Q = d km.
Time taken by A in both side = d/10 + d/9 = 19d/90 hrs.
Time taken by B in both side = 2d/12 = d/6 hrs.
B tool 10 min. or 1/6 hrs. less than A.
So, 19d/90 - d/6 = 1/6
=> (19d-15d)/90 = 1/6
=> 4d/90 = 1/6
=> d = 15/4 km = 3.75 km.
Lets assume quantity of zinc to added = x kg.
zinc quantity in alloy = 5/8*16=10 Kg.
and copper quantity = 3/8*16=6 kg.
Alloy new ratio of zinc and copper = 3:1
Zinc quantity in alloy => (10+x)/6=3/1 => 10 + x = 18 => x = 8 kg.
Lets assume sum of money = Rs. x ;
A gets Rs. 33 when sum distributed in the ratio of 3:2:5
Given Two No, ratio = 3:4 and their HCF = 4
So, No. = 3*4 =12 and 4*4=16
LCM of 12,16 = 48.
Consider a solution = 1 ltr
X is the certain quantity which has to be replaced
40%of (1-x)+(25%of x)=35/100
=> 40/100 * (1-x) + 25x/100 = 35/100
=> 40/100 - 40x/100 + 25x/100 = 35/100
=> 15x/100 = 5/100
=> 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 8 Meters Ahead Of Runner C. Each Runner Travels The Entire Distance At A Constant Speed. What Was The Length Of The Race?
Lets assume distance of race = x mtrs.
Then when A finishes x m , B has run (x- 12)mtrs and C has run (x-18) mtrs.
=> at this point B is 6 m ahead of C. Now to finish race b needs to run another 12 m,
=> he runs another 12 m. when B finishes race he is 8 m ahead of C.
so last 12 m B has run, C has run 10 m.as speeds are constant,
=> x-12/ x-18 = 12/10 => x = 48 mtrs.
Question 16. 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?
Amount of work done by 20 men = 24 women = 40 boys
i.e 5man = 6 woman = 10 boys. -----------------(i)
According to the 1st condition, 5 men can do a job in 12 days working for 8 hours a day.
Required :- how many more men are required to work with 6 women and 2 boys.
Let the required number of men be M.
According to the given information,
=>(5 x 12 x 8 )/1=[(M + 5 + 1) x 5 x 12 ]/4. ( 6 women= 5men & 2 boys =1 men)
=>32= M + 6.
Hence, 26 men are working with 6 women and 2 boys.
Question 17. Nitish Sold His Watch And Sunglasses At A Loss Of 4% And Gain Of 4% Respectively For 2600 To Kamal. Kamal Sold The Same Sun Glasses And Watch At A Loss Of 4% And Gain Of 4% Respectively For 2700. The Price Of Watch And Sun Glasses To Nitish Were?
Lets assume the CP of watch = Rs x
and sunglasses = Rs y.
2600=96x/100 + 104y/100
2700= 104x/100 + 96y/100
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?
Lets assume second grandson = x year.
So, first grandson age = (x+2) year ,
third grandson = (x-2) year
i.e grandfather age is 3 times as older as his second grandson.
Question 19. Two Urns Contain Respectively 2 Red 3 White And 3 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?
According to baye's theorem:
p(no of way to drawn a ball from 2nd urns is red and ball transferred from 1 urns)=(3c1/5c1)*(3c1/9c1)=9/45
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?
Lets assume marks of Ajith, Sachu, Karna, Saheep and Ramesh respectively u, v, w, x, y
So, as per the question:
u- 9=z ---(ii)
x+ z=v ---(iii)
Given, u=32 ---- (v)
By solving eq. (i), (ii), (iii), (iv) and (v)
Question 21. The Average Number Of Visitors Of A Library In The First 4 Days Of A Week Was 58. The Average For The 2nd, 3rd, 4th And 5th Days Was 60.if The Number Of Visitors On The 1st And 5th Days Were In The Ratio 7:8 Then What Is The Number Of Visitors On The 5th Day Of The Library?
If number of visitors on 1st, 2nd, 3rd, 4th & 5th day are a, b, c, d & e respectively, then
Subtracting eq. (i) from (ii)
Given, a/e=7/8 ---(iv)
So from eq. (iii) & (iv)
LCM of (26,39,52,65)=780
So smallest number = (780-4) = 776.
Lets assume day required to complete the job = x day
x/4 = 3/4
=> x = 3 days.
Lets assume the radius of 1st circle = x cm.
so, the radius of 2nd circle = (14-x)cm.
By solving above eq.
x = 11 or 3.
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)?
Lets assume the capacity of the vessel = x litres.
x (1- 10/x)^2 = 32
x^2 +100-20x = 32x
x^2 +100 - 52x= 0
x^2 - 50x - 2x + 100 = 0
=> (x-2) (x-50) = 0
=> x=2 & x=50
As 10 litres is withdrawn so vessel capacity will be 50 litres.
Tax reduced from 3 1/2 or 7/2% to 3 1/3% or 10/3%.
So, the difference in tax = (7/2 - 10/3)% = 1/6%.
=> 1/6 % of Rs. 8400 = 8400*1/6% = 8400* (1/6 * 1/100) = Rs. 14.
Lets assume speed of boat in still water = x km/hr
So boat downstream speed = (x+2) km/hr.
boat upstream speed = (x-2) km/hr.
6/(x-2)+6/(x+2)=33/60 => 6/(x-2)+6/(x+2)=11/20
=> 120*[(x-2) + (x+2)] = 11*x^2 -44
=> 240 x = 11*x^2-44
=> 11* x^2 -240x -44 = 0 ---------(1)
By solving eq. (1)
x = 22 km/hr.
Total late in three day = 16*3 = 48 min.
Total late in one day =16 min.
so late in 1 min. = 16/(24*60) = 0.01111;
Hence in 5 hour (5:00 AM to 10:00 AM on 4th day) =.0111*5*60 = 3.333;
So, Exact time will be = 10:52 AM.
If the difference of the sum of digits at odd places and the sum of its digits at even places, is either 0 or divisible by 11, then clearly the number is divisible by 11.
[(Sum of digits at odd places) - (Sum of digits at even places)]/11
=> [(3+4+6+2) - (2+5+8)]/11 => (15 - 15)/11 = 0/11 = 0
similarly others number are divisible by 11.
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