 # MOST IMPORTANT AGILENT TECHNOLOGIES INTERVIEW QUESTIONS

Which are the MOST IMPORTANT AGILENT TECHNOLOGIES INTERVIEW QUESTIONS? Looking for Agilent Technologies interview questions? You can find the latest Agilent Technologies interview questions here. These interview questions are collected from the job seekers/candidates who have attended the Agilent Technologies interview recently. Make sure that you are prepared well on all skills and projects or subjects you have mentioned in the resume. Technical Interview questions will be purely based on the topics you have mentioned in your resume. It is also important that you know every detail about the projects you have mentioned in your resume. To shortlist in first attempt, you need to study your technical subjects well and rehearse the answers to common HR and managerial interview questions. To crack your interview, you need to prepare well for the commonly asked list of Agilent Technologies interview questions and interview tips before the interview.

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Question 1. Two Trains 130 And 110 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 Completely In 3 Seconds. Find The Speed Of The Faster Train?
Total Distance to be travelled by both the trains = 130 + 110 = 240m
Let ‘F ’and ‘S’ be the speeds of fast and slow trains in m/sec. 240=60(F-S), 240= (F+S)
In the same direction, F – S = 4 m/sec …… (1)
In the opposite direction, F + S =80 m/sec…… (2)
Solving them we get F = 42 m/sec.

Question 2. A Motor Boat Can Travel At 10 Km/h In Still Water. It Travelled 91 Km Downstream In A River And Then Returned, Taking Altogether 20 Hours. Find The Rate Of Flow Of The River?
Given, speed of the Boat in still water (B) =10 km/hr
Let S be the speed of flow of river, then
91/(10+S) + 91/(10-S) = 20, Then going by options
91/13 – 91/7 = 20
So, S = 3 km/hr.

Question 3. The Total Tractor Population In A State Is 2, 94,000, 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 5.3% Of The Total Tractor Population Is Red. Find The Percentage Of Non-mahindra Tractors That Are Red?
Total tractor population = 2, 94,000
Mahindra & Mahindra = 1, 50,000
So, Non Mahindra trucks = 1, 44,000
Since out of every 1000 Mahindra tractors, 98 are red, out of 1, 50,000 Mahindra tractors 14,700 are red.
5.3% of 2, 94,000 = 15,582 are red tractors in all.
So non Mahindra tractors which are red =15,582 – 14,700 = 882
Hence percentage of non Mahindra tractors that are red = 882/144,000 * 100 = 0.6125%

Question 4. 7% Of The Total Quantity Of Wheat Is Lost In Grinding When A Country Has To Import 12 Million Tones, But When Only 51⁄5 Is Lost, It Can Import 3 Million Tones. Find The Quantity Of Wheat Grown In The Country?
Difference in % of wheat lost = 7 – 26/5 = 9/5%
Difference in import = 12 – 3 = 9 million as 9/5% % of total qty of wheat = 9 million
⇒ 9x/500 = 9
⇒ x = 500 million

Question 5. A Man Who Can Swim 48 M/min In Still Water Swims 200m Against The Current And 200 M With The Current. If The Difference Between Those 2 Times Is 10 Minutes, Find The Speed Of The Current?
Try option & get answer as fourth option:
(200/(48 – 32)) – (200/ (48 + 32)) (12(1/2) min – 2(1/2) min)
= 10 min

Question 6. A And B Run A 5 Km Race On A Round Course Of 400 M. If Their Speed Be In The Ratio 5 : 4, How Often Does The Winner Pass The Other On Circular Track?
Total no. of round will be 5000/400 = 12.5
No. of rounds A completes to finish the race will be 12.5 and by the time B can complete only 10 rounds.
(As ratio of speed of A & B is 5: 4. So they meet for the first time after A has finished 5 rounds and B has finished 4 rounds.)
So difference in no. of round will be = 21⁄2
The winner meets the other 2 times because the winner meets the other after every 2000 meters.

Question 7. Mira’s Expenditure And Saving Are In The Ratio 3:2. Her Income Increases By 10%. Her Expenditure Also Increases By 12%. By How Much % Do Her Saving Increase?
Let total income be = 5
Increased income will be = 5.5
Increased expenditure will be = 3.36
Increased saving will be = 5.5 – 3.36 = 2.14
Percentage increase will be = 14/2 * 100 = 7%

Question 8. 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 bigger one
π (8)² + π(6)² = πr²
π [64+36] = πr²
π *100 = πr²
r= 10

Question 9. The Length Of A Rectangular Field Is Double Its Width, Inside The Field There Is A Square-shaped Pond 8 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 field = 64 × 8 = 512
Area of field ⇒ x. 2x = 512
⇒ 2×2 = 512 ⇒ x2 = 256 ⇒ x = 16
Length = 2 × 16 = 32.

Question 10. A’ And ‘b’ Complete A Work Together In 8 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 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.

Question 11. P, Q And R Are Three Typists Who Working Simultaneously Can Type 216 Pages In 4 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 Type As Many Pages As P Can During Seven Hours. How Many Pages Does Each Of Them Type Per Hour?
Let’s the number of pages typed in one hour by P, Q and R be p, q and r respectively. Then,
P,Q and R typed page in 1 hrs = 216/4
=> p + q + r = 216/4
=> p + q + r = 54 … (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

Question 12. 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 5 Days And B For 7 Days, C Finishes It In 13 Days. In How Many Days C Alone Will Do The Work?
(A and B) 1 day work = 1/12 —–(1)
(B and C) 1 day work = 1/16 —–(2)
Given A’s 5 days’ work + B’s 7 days’ work + C’s 13 days’ work = 1
Simplify the above…
=> (A + B)’s 5 days’ work + (B + C)’s 2 days’ work + C’s 11 days’ work = 1
Put the values from equation (1) & (2)
=> (5*1/12)+(2*1/16) + C’s 11 days’ work = 1
=> C’s 11 days’ work = (1-5/12+2/16)=11/24
=> C’s 1 day’s work = (11/24? 1/11) =1/24
So C alone can finish the work = 24 days.

Question 13. A And B Can Do A Piece Of Work In 45 Days And 40 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 Days After Which A Left The Work Was?
A’s 1 day work = 1/45
B’s 1 day work = 1/40
so (A + B)’s 1 day’s work = (1/45+1/40)=17/360
Work done by B in 23 days = (140? 23) =2340
Remaining work = (1-23/40) =17/40
Now, 17/360 work was done by (A + B) = 1 day.
17/40 part of work was done by (A + B) = (1? 360/17? 17/40) = 9 days.
A left after 9 days.

Question 14. A, B And C Together Earn Rs. 300 Per Day, While A And C Together Earn Rs. 188 And B And C Together Earn Rs. 152. The Daily Earning Of C Is?
B’s daily earning = Rs. (300 – 188) = Rs. 112.
A’s daily earning = Rs. (300 – 152) = Rs. 148.
C’s daily earning = Rs. [300 – (112 + 148)] = Rs. 40.

Question 15. 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 work for all three
1 day’s work for all three = (1/15 + 1/20 + 1/25)
= (20/300 + 15/300 + 12/300) = 47/300
So all together can do the work in 300/47 days. => 6.4 days.

Question 16. 287 X 287 + 269 X 269 – 2 X 287 X 269 =?
Given Exp. = a2 + b2 – 2ab, where a = 287 and b = 269
= (a – b)2 = (287 – 269)2
= (182)
= 324

Question 17. The Average Salary Of 3 Workers Is 95 Rs. Per Week. If One Earns Rs.115 And Second Earns Rs.65 How Much Is The Salary Of The 3rd Worker?
Let’s assume three workers a a,b and c
Their average salary is (a+b+c)/3 = 95
=> a+b+c=285
a=115,
b=65
So c=285-115-65=105

Question 18. 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 * 5 + 240 * 25)/30.
=> 8550/30
=> 285.

Question 19. How Much Is 80% Of 40 Is Greater Than 4/5 Of 25?
0% of 40 = 40 * (80/100) and 4/5 of 25 = 25 * 4/5
[40* (80/100) – 25*(4/5)] = [3200/100 – 100/5]
= 32 – 20
= 12.

Question 20. After Decreasing 24% In The Cost Price Of An Article, Its Costs Rs.912. Find The Actual Cost Of An Article?
CP* (76/100) = 912 => CP = 912 * 100/76
CP= 12 * 100
=> CP = 1200
Cost price of article = Rs. 1200

Question 21. Ram Donated 4% Of His Income To A Charity And Deposited 10% Of The Rest In A Bank. If Now He Has Rs 8640 Left With Him, Then His Income Is?
Let Ram’s Income = Rs. 100.
Donation to charity = Rs. 4
Amount deposited in bank = (96 * l0)/100 = Rs. 9.6
Savings = (100 – 13.6) = Rs. 86.4
So Rs. 86.4 = 100
=> Rs. 8640 = (100/86.4) * 8640 = Rs. 10000

Question 22. 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 = 120% of Re. 1 = Rs. (120/100) = Rs. 6/5
So toffees in Re 1 = 6 * 5/6 = 5

Question 23. 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%

Question 24. A Shopkeeper Loses 15%,if An Article Is Sold For Rs. 102. What Should Be The Selling Price Of The Article To Gain 20%?
Given that SP = Rs. 102 and loss = 15%
CP = (100(SP))/(100 – 15%) = (100 * 102)/85 = Rs. 120
To get 20% profit, New SP = [(100 + p %) CP]/100
= (120 * 120)/100
= Rs. 144

Question 25. 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 per kg
and Cost of 1 kg Mango @ Rs. 15 per kg = 10 + 15 = Rs. 25 (Cost Price for 2 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%

Question 26. A Person Crosses A 600 M Long Street In 5 Minutes. What Is His Speed In Km Per Hour?
Speed = 600/ (5*60) m/sec. = 600/300 m/sec = 2 m/sec
Now convert m/sec to km/hr
= 2 x 18/5 km/hr
= 7.2 km/hr.

Question 27. 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) + (75 – x)/10 = 7 [sum of (1) and (2)]
=> (10x + 900 – 12x) = 840
=> -2x = -60
=> x = 30.
Therefore, the distance travelled at the rate of 12 km/hr = 30km.

Question 28. 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 time = distance/speed
Time taken by the athlete for the first half = 15/10
And time taken by the athlete for the second half = 15/20
Total time = 15/10 + 15/20 = (45/20) x 60 = 135 mins
Time taken to cover the total distance is 135 mins.

Question 29. It Takes 8 Hours For A 600 Km Journey, If 120 Km Is Done By Train And The Rest By Car. It Takes 20 Minutes More, If 200 Km Is Done By Train And The Rest By Car. The Ratio Of The Speed Of The Train To That Of The Cars Is?
Assume Here the speed of the train = x km/hr
and speed of car = y km/hr.
Given First journey information:
(120/x + 480/y) = 8 => (120/x + 480/y)/8 = 1
(15/x + 60/y) = 1 => (1/x + 4/y) = 1/15 ———- (1)
Given Second journey information:
And, (200/x + 400/y) = 25 => (1/x + 2/y) = 25/200 ——- (2)
By Solving (1) and (2)
x = 60 and y = 80.
So Ratio of speeds = 60: 80
= 3: 4

Question 30. 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 The Speed With Which It Walks?
The exact time is 11 am
Then 30miles/ph reaches it on 10am
20 miles/ph reach it on 12am
If they start walking in 6am
Then 30m/ph-(10am-6am) 4h*30=120miles
20m/ph-(12am-6am) 6h*20=120miles

Question 31. A Can Contains A Mixture Of Two Liquids A And B Is The Ratio 7: 5. When 9 Litres Of Mixture Are Drawn Off And The Can Is Filled With B, The Ratio Of A And B Becomes 7: 9. How Many Litres Of Liquid A Were Contained By The Can Initially?
Suppose the can initially contains 7x and 5 xs of mixtures A and B respectively.
Quantity of A in mixture left = (7x-7*9/12) Litres = (7x – 21/4) litres.
Quantity of B in mixture left = (5x – 5*9/12) Litres = (5x – 15/4) litres.
[(7x – 21/4)/ (5x-15/4) +9] = 7/9
=> (28x – 21)/ (20x + 21) = 7/9
=> 252x – 189 = 140x + 147 => 112x = 336
=> x = 3.
So, can contained 21 litres of A.

Question 32. 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 mixture has milk and water in the ratio 7: 3. I.e. the solution has 21 litres of milk and 9 litres of water.
When you add more water, the amount of milk in the mixture remains constant at 21 litres. In the first case, before addition of further water, 21 litres of milk accounts for 70% by volume. After water is added, the new mixture contains 60% milk and 40% water.
Therefore, the 21 litres of milk accounts for 60% by volume.
Hence, 100% volume =21/0.6 = 35 litres.
We started with 30 litres and ended up with 35 litres.
So, 5 litres of water was added.

Question 33. A 20 Litre Mixture Of Milk And Water Contains Milk And Water In The Ratio 3: 2. 10 Litres Of The Mixture Is Removed And Replaced With Pure Milk And The Operation Is Repeated Once More. At The End Of The Two Removals And Replacement, What Is The Ratio Of Milk And Water In The Resultant Mixture?