33+ Agilent Technologies Aptitude Interview Questions And Answers

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Agilent Technologies Aptitude Interview Questions

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 each the trains = 130 + 110 = 240m

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

In the identical path, F – S = Four m/sec …… (1)

In the wrong way, 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, pace of the Boat in nonetheless water (B) =10 km/hr

Let S be the pace of move of river, then

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

91/13 – 91/7 = 20

So, S = Three 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 inhabitants = 2, 94,000

Mahindra & Mahindra = 1, 50,000

So, Non Mahindra vans = 1, 44,000

Since out of each 1000 Mahindra tractors, 98 are crimson, out of 1, 50,000 Mahindra tractors 14,700 are crimson.

5.3% of two, 94,000 = 15,582 are crimson tractors in all.

So non Mahindra tractors that are crimson =15,582 – 14,700 = 882

Hence proportion of non Mahindra tractors which are crimson = 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 misplaced = 7 – 26/5 = 9/5%

Difference in import = 12 – 3 = 9 million as 9/5% % of whole 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 possibility & get reply as fourth possibility:

(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 spherical might be 5000/400 = 12.5

No. of rounds A completes to complete the race might be 12.5 and by the point B can full solely 10 rounds.

(As ratio of pace of A & B is 5: 4. So they meet for the primary time after A has completed 5 rounds and B has completed Four rounds.)

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

The winner meets the opposite 2 occasions as a result of the winner meets the opposite after each 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 whole revenue be = 5

Increased revenue might be = 5.5

Increased expenditure might be = 3.36

Increased saving might be = 5.5 – 3.36 = 2.14

Percentage enhance might 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 two smaller parks = New larger 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 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 subject ⇒ 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 in the future work = 1/8

A alone in the future work = 1/12

B alone in the future work = (1/8 – 1/12) = ( 3/24 – 2/24)

=> B in the future work = 1/24

So B can full 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 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 web 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 end 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 completed by B in 23 days = (140? 23) =2340

Remaining work = (1-23/40) =17/40

Now, 17/360 work was completed by (A + B) = 1 day.

17/40 a part of work was completed 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 each day incomes = Rs. (300 – 188) = Rs. 112.

A’s each day incomes = Rs. (300 – 152) = Rs. 148.

C’s each day incomes = 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 collectively can do the work in 300/47 days. => 6.Four days.

Question 16. 287 X 287 + 269 X 269 – 2 X 287 X 269 =?

Given Exp. = a2 + b2 – 2ab, the place 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 third Worker?

Let’s assume three employees a a,b and c

Their common wage 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 common

=> (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 worth 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 financial institution = (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% revenue, 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 two Kg)

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

Here CP < SP

So Profit = (Rs. 30 – Rs. 25) = Rs. 5

Formula Used: Profit% = (revenue/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 gap travelled on the charge of 12 kmph be x km

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

Then the gap travelled on the charge of 10 km/hr might be (75 – x)

Time taken to journey this distance = (75 – x)/10……. (2)

Total time taken to cowl the full distance = 7 hrs

=>(x/12) + (75 – x)/10 = 7 [sum of (1) and (2)]

=> (10x + 900 – 12x) = 840

=> -2x = -60

=> x = 30.

Therefore, the gap travelled on the charge 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 point = distance/pace

Time taken by the athlete for the primary 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 minutes

Time taken to cowl the full distance is 135 minutes.

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 pace of the practice = x km/hr

and pace of automobile = y km/hr.

Given First journey info:

(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 info:

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 actual time is 11 am

Then 30miles/ph reaches it on 10am

20 miles/ph attain it on 12am

If they begin strolling 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 accommodates 7x and 5 xs of mixtures A and B respectively.

Quantity of A in combination left = (7x-7*9/12) Litres = (7x – 21/4) litres.

Quantity of B in combination 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 combination has milk and water within the ratio 7: 3. I.e. the answer has 21 litres of milk and 9 litres of water.

When you add extra water, the quantity of milk within the combination stays fixed at 21 litres. In the primary case, earlier than addition of additional water, 21 litres of milk accounts for 70% by quantity. After water is added, the brand new combination accommodates 60% milk and 40% water.

Therefore, the 21 litres of milk accounts for 60% by quantity.

Hence, 100% quantity =21/0.6 = 35 litres.

We began 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?

The 20 litre combination accommodates milk and water within the ratio of three : 2. Therefore, there might be 12 litres of milk within the combination and eight litres of water within the combination.

Step 1: When 10 litres of the combination is eliminated, 6 litres of milk is eliminated and Four litres of water is eliminated. Therefore, there might be 6 litres of milk and Four litres of water left within the container. It is then changed with pure milk of 10 litres. Now the container could have 16 litres of milk and Four 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 container could have Eight litres of milk and a pair of litres of water in it. Now 10 litres of pure milk is added. Therefore, the container could have 18 litres of milk and a pair of litres of water in it on the finish of the second step.

So, the ratio of milk and water is 18: 2 or 9: 1.


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