CrowdforGeeks | Build Skills with Online Courses from Top Institutions

Top 31 Agilent Technologies Aptitude Interview Questions

Q1. How Much Is eighty% Of forty Is Greater Than 4/5 Of 25?

Zero% of 40 = forty * (eighty/one hundred) and 4/five of 25 = 25 * four/5

[40* (80/100) - 25*(4/5)] = [3200/100 - 100/5]

= 32 - 20

= 12.

Q2. The Average Salary Of 3 Workers Is ninety five Rs. Per Week. If One Earns Rs.A hundred and fifteen And Second Earns Rs.65 How Much Is The Salary Of The 3rd Worker?

Let’s anticipate three workers a a,b and c

Their common salary is (a+b+c)/three = 95

=> a+b+c=285

a=a hundred and fifteen,

b=sixty five

So c=285-a hundred and fifteen-sixty five=one zero five

Q3. A And B Together Can Do A Piece Of Work In 12 Days, Which B And C Together Can Do In 16 Days. After A Has Been Working At It For five Days And B For 7 Days, C Finishes It In thirteen Days. In How Many Days C A

(A and B) 1 day work = 1/12 -----(1)

(B and C) 1 day paintings = 1/16 -----(2)

Given A's 5 days' work + B's 7 days' work + C's thirteen days' work = 1

Simplify the above...

=> (A + B)'s five days' paintings + (B + C)'s 2 days' work + C's 11 days' paintings = 1

Put the values from equation (1) & (2)

=> (5*1/12)+(2*1/16) + C's eleven days' paintings = 1

=> C's eleven days' work = (1-five/12+2/16)=11/24

=> C's 1 day's work = (11/24? 1/11) =1/24

So C by myself can finish the work = 24 days.

Q4. A, B And C Together Earn Rs. Three hundred Per Day, While A And C Together Earn Rs. 188 And B And C Together Earn Rs. 15

B's day by day earning = Rs. (300 - 188) = Rs. 112.

A's every day incomes = Rs. (300 - 152) = Rs. 148.

C's each day earning = Rs. [300 - (112 + 148)] = Rs. Forty.

Q5. How Many Litres Of Water Should Be Added To A 30 Liter Mixture Of Milk And Water Containing Milk And Water In The Ratio Of 7: 3 Such That The Resulting Mixture Has 40% Water In It?

30 litres of the aggregate has milk and water within the ratio 7: @I.E. The solution has 21 litres of milk and nine litres of water.

When you add more water, the quantity of milk within the aggregate remains constant at 21 litres. In the primary case, earlier than addition of in addition water, 21 litres of milk debts for 70% via volume. After water is delivered, the new mixture carries 60% milk and forty% water.

Therefore, the 21 litres of milk bills for 60% through extent.

Hence, one hundred% extent =21/0.6 = 35 litres.

We started out with 30 litres and ended up with 35 litres.

So, 5 litres of water turned into brought.

Q6. Two Lots Of Mango With Equal Quantity, One Costing Rs. 10 Per Kg And The Other Costing Rs. 15 Per Kg, Are Mixed Together And Whole Lot Is Sold At Rs. 15 Per Kg. What Is The Profit Or Loss?

Cost Price of 1 kg mango @ Rs. 10 in line with kg

and Cost of one kg Mango @ Rs. 15 in keeping with kg = 10 + 15 = Rs. 25 (Cost Price for two Kg)

Selling Price for 2 kg = 2 x Rs. 15 = Rs. 30

Here CP < SP

So Profit = (Rs. 30 - Rs. 25) = Rs. 5

Formula Used: Profit% = (profit/CP)*100

Profit % = (5/25) * 100% = 20%

Q7. Virat Travelled 75 Kms In 7 Hours. He Went Some Distance At The Rate Of 12 Km/hr And The Rest At 10 Km/hr. How Far Did He Travel At The Rate Of 12 Km/hr?

Let the distance travelled at the rate of 12 kmph be x km

Time taken to travel this distance = x/12....... (1)

Then the distance travelled at the rate of 10 km/hr will be (75 - x)

Time taken to travel this distance = (75 - x)/10....... (2)

Total time taken to cover the total distance = 7 hrs

=>(x/12) + (seventy five - x)/10 = 7 [sum of (1) and (2)]

=> (10x + 900 - 12x) = 840

=> -2x = -60

=> x = 30.

Therefore, the space travelled on the price of 12 km/hr = 30km.

Q8. P, Q And R Are Three Typists Who Working Simultaneously Can Type 216 Pages In four Hours. In One Hour, R Can Type As Many Pages More Than Q As Q Can Type More Than P. During A Period Of Five Hours, R Can

Let's the variety of pages typed in a single hour by P, Q and R be p, q and r respectively. Then,

P,Q and R typed page in 1 hrs = 216/four

=> p + q + r = 216/4

=> p + q + r = fifty four ... (i)

r - q = q - p => 2p = q + r ...(ii)

5r = 7p => p = 5/7 r ... (iii)

By Solving above (i), (ii) and (iii) equations

=> p = 15, q = 18, q = 21

Q9. A And B Run A five Km Race On A Round Course Of 400 M. If Their Speed Be In The Ratio five : 4, How Often Does The Winner Pass The Other On Circular Track?

Total no. Of round could be 5000/400 = 12.5

No. Of rounds A completes to finish the race could be 12.Five and by the point B can complete most effective 10 rounds.

(As ratio of speed of A & B is five: @So they meet for the first time after A has completed 5 rounds and B has finished 4 rounds.)

So distinction in no. Of spherical might be = 21⁄2

The winner meets the alternative 2 instances because the winner meets the other after every 2000 meters.

Q10. It Takes 8 Hours For A six hundred Km Journey, If 120 Km Is Done By Train And The Rest By Car. It Takes 20 Minutes More, If two hundred Km Is Done By Train And The Rest By Car. The Ratio Of The Speed Of The Train To

Assume Here the speed of the train = x km/hr

and speed of vehicle = y km/hr.

Given First journey records:

(one hundred twenty/x + 480/y) = 8 => (a hundred and twenty/x + 480/y)/eight = 1

(15/x + 60/y) = 1 => (1/x + four/y) = 1/15 ---------- (1)

Given Second journey records:

And, (2 hundred/x + four hundred/y) = 25 => (1/x + 2/y) = 25/2 hundred ------- (2)

By Solving (1) and (2)

x = 60 and y = eighty.

So Ratio of speeds = 60: 80

= three: 4

Q11. After Decreasing 24% In The Cost Price Of An Article, Its Costs Rs.9

CP* (seventy six/100) = 912 => CP = 912 * 100/seventy six

CP= 12 * a hundred

=> CP = 1200

Cost charge of article = Rs. 1200

Q12. Two Trains one hundred thirty And a hundred and ten Meters Long Are Going In The Same Direction. The Faster Train Takes One Minute To Pass The Other Completely. If They Are Moving In Opposite Directions, They Pass Each Other Comp

Total Distance to be travelled via each the trains = a hundred thirty + 110 = 240m

Let ‘F ’and ‘S’ be the speeds of fast and gradual trains in m/sec. 240=60(F-S), 240= (F+S)

In the same direction, F – S = four m/sec …… (1)

In the other course, F + S =eighty m/sec…… (2)

Solving them we get F = 42 m/sec.

Q13. A Man Can Do A Job In 15 Days. His Father Takes 20 Days And His Son Finishes It In 25 Days. How Long Will They Take To Complete The Job If They All Work Together?

Find the 1 day paintings for all three

1 day's work for all three = (1/15 + 1/20 + 1/25)

= (20/three hundred + 15/300 + 12/300) = 47/300

So all together can do the work in 300/47 days. => 6.4 days.

Q14. A Man Who Can Swim forty eight M/min In Still Water Swims 200m Against The Current And 2 hundred M With The Current. If The Difference Between Those 2 Times Is 10 Minutes, Find The Speed Of The Current?

Try option & get wer as fourth option:

(two hundred/(forty eight - 32)) - (2 hundred/ (48 + 32)) (12(half) min - 2(1/2) min)

= 10 min

Q15. 287 X 287 + 269 X 269 - 2 X 287 X 269 =?

Given Exp. = a2 + b2 - 2ab, wherein a = 287 and b = 269

= (a - b)2 = (287 - 269)2

= (182)

= 324

Q16. A Can Contains A Mixture Of Two Liquids A And B Is The Ratio 7:

Suppose the can initially consists of 7x and 5 xs of combinations A and B respectively.

Quantity of A in combination left = (7x-7*nine/12) Litres = (7x - 21/four) litres.

Quantity of B in mixture left = (5x - five*nine/12) Litres = (5x - 15/4) litres.

[(7x - 21/4)/ (5x-15/4) +9] = 7/9

=> (28x - 21)/ (20x + 21) = 7/nine

=> 252x - 189 = 140x + 147 => 112x = 336

=> x = three.

So, can contained 21 litres of A.

Q17. A' And 'b' Complete A Work Together In eight Days. If 'an' 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 by myself someday work = 1/12

B alone sooner or later work = (1/eight - 1/12) = ( 3/24 - 2/24)

=> B in the future work = 1/24

So B can whole the paintings in 24 days.

Q18. A Vendor Bought Toffees At 6 For A Rupee. How Many For A Rupee Must He Sell To Gain 20%?

C.P. Of 6 toffees = Re. 1

S.P. Of 6 toffees = a hundred and twenty% of Re. 1 = Rs. (120/one hundred) = Rs. 6/five

So toffees in Re 1 = 6 * five/6 = five

Q19. A 20 Litre Mixture Of Milk And Water Contains Milk And Water In The Ratio three:

The 20 litre aggregate contains milk and water in the ratio of three : @Therefore, there can be 12 litres of milk in the aggregate and eight litres of water in the mixture.

Step 1: When 10 litres of the aggregate is removed, 6 litres of milk is removed and 4 litres of water is removed. Therefore, there may be 6 litres of milk and 4 litres of water left in the container. It is then changed with natural milk of 10 litres. Now the box will have sixteen litres of milk and 4 litres of water.

Step 2: When 10 litres of the brand new combination is eliminated, eight litres of milk and a pair of litres of water is eliminated. The field could have eight litres of milk and a couple of litres of water in it. Now 10 litres of natural milk is delivered. Therefore, the container can have 18 litres of milk and a pair of litres of water in it on the stop of the second one step.

So, the ratio of milk and water is 18: 2 or nine: 1.

Q20. Two Small Circular Parks Of Diameters 16 M, 12 M Are To Be Replaced By A Bigger Circular Park. What Would Be The Radius Of This New Park, If The New Park Occupies The Same Space As The Two Small Parks

Some of areas of 2 smaller parks = New larger one

π (eight)² + π(6)² = πr²

π [64+36] = πr²

π *one hundred = πr²

r= 10

Q21. The Total Tractor Population In A State Is 2, ninety four,000, Out Of Which 1, 50,000 Are Made By Mahindra & Mahindra. Out Of Every 1,000 Mahindra Tractors, 98 Are Red In Colour, But Only five.Three% Of The Total Tra

Total tractor population = 2, 94,000

Mahindra & Mahindra = 1, 50,000

So, Non Mahindra vans = 1, forty four,000

Since out of every a thousand Mahindra tractors, 98 are purple, out of one, 50,000 Mahindra tractors 14,700 are pink.

Five.Three% of two, 94,000 = 15,582 are red tractors in all.

So non Mahindra tractors which can be purple =15,582 – 14,700 = 882

Hence percentage of non Mahindra tractors that are red = 882/a hundred and forty four,000 * a hundred = zero.6125%

Q22. A Shopkeeper Loses 15%,if An Article Is Sold For Rs.

Given that SP = Rs. 102 and loss = 15%

CP = (one hundred(SP))/(100 - 15%) = (100 * 102)/eighty five = Rs. A hundred and twenty

To get 20% income, New SP = [(100 + p %) CP]/a hundred

= (120 * a hundred and twenty)/one hundred

= Rs. 144

Q23. A Library Has An Average Of 510 Visitors On Sundays And 240 On Other Days. The Average Number Of Visitors Per Day In A Month Of 30 Days Beginning With A Sunday Is?

Required average

=> (510 * five + 240 * 25)/30.

=> 8550/30

=> 285.

Q24. An Athlete Runs Half Of The Distance At The Speed Of 10 Km/hr And Remaining 15km At The Speed Of 20 Km/hr, How Much Time Will Be Taken To Cover The Total Distance?

We know that point = distance/pace

Time taken by means of the athlete for the primary half = 15/10

And time taken by using the athlete for the second one half of = 15/20

Total time = 15/10 + 15/20 = (forty five/20) x 60 = 135 mins

Time taken to cover the entire distance is 135 mins.

Q25. A Motor Boat Can Travel At 10 Km/h In Still Water. It Travelled ninety one Km Downstream In A River And Then Returned, Taking Altogether 20 Hours. Find The Rate Of Flow Of The River?

Given, velocity of the Boat in still water (B) =10 km/hr

Let S be the velocity of float of river, then

91/(10+S) + 91/(10-S) = 20, Then going by using alternatives

ninety one/13 - ninety one/7 = 20

So, S = 3 km/hr.

Q26. A And B Can Do A Piece Of Work In 45 Days And forty Days Respectively. They Began To Do The Work Together But A Leaves After Some Days And Then B Completed The Remaining Work In 23 Days. The Number Of Da

A's 1 day paintings = 1/45

B's 1 day work = 1/forty

so (A + B)'s 1 day's work = (1/forty five+1/40)=17/360

Work performed via B in 23 days = (one hundred forty? 23) =2340

Remaining paintings = (1-23/forty) =17/forty

Now, 17/360 paintings became finished by means of (A + B) = 1 day.

17/40 part of paintings became executed by using (A + B) = (1? 360/17? 17/forty) = nine days.

A left after nine days.

Q27. If I Walk With 30 Miles/hr I Reach 1 Hour Before And If I Walk With 20 Miles/hr I Reach 1 Hour Late. Find The Distance Between 2 Points And The Exact Time Of Reaching Destination Is 11 Am Then Find Th

The exact time is eleven am

Then 30miles/ph reaches it on 10am

20 miles/ph attain it on 12am

If they start on foot in 6am

Then 30m/ph-(10am-6am) 4h*30=120miles

20m/ph-(12am-6am) 6h*20=120miles

Q28. The Length Of A Rectangular Field Is Double Its Width, Inside The Field There Is A Square-shaped Pond eight M Long. If The Area Of The Pond Is 1/8 Of The Field, What Is The Length Of The Field?

Area of Pond = 64

Area of subject = 64 × 8 = 512

Area of discipline ⇒ x. 2x = 512

⇒ 2x2 = 512 ⇒ x2 = 256 ⇒ x = sixteen

Length = 2 × sixteen = 32.

Q29. A Cycle Is Bought For Rs.900 And Sold For Rs.1080, Find The Gain Percent?

SP = Rs. 1080 and CP = Rs. 900

Profit = Rs. (1080 - 900) = Rs. 180

Gain% = (180/900) * 100 % = 20%

Q30. A Person Crosses A six hundred M Long Street In five Minutes. What Is His Speed In Km Per Hour?

Speed = six hundred/ (five*60) m/sec. = six hundred/300 m/sec = 2 m/sec

Now convert m/sec to km/hr