# Top 31 Agilent Technologies Aptitude Interview Questions You Must Prepare 29.Nov.2023

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.

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

(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.

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.

30 litres of the mixture has milk and water in the ratio 7: @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.

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%

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.

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

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: @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.

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

CP* (76/100) = 912 => CP = 912 * 100/76

CP= 12 * 100

=> CP = 1200

Cost price of article = Rs. 1200

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.

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.

Try option & get wer as fourth option:

(200/(48 - 32)) - (200/ (48 + 32)) (12(1/2) min - 2(1/2) min)

= 10 min

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

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

= (182)

= 324

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.

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.

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

The 20 litre mixture contains milk and water in the ratio of 3 : @Therefore, there will be 12 litres of milk in the mixture and 8 litres of water in the mixture.

Step 1: When 10 litres of the mixture is removed, 6 litres of milk is removed and 4 litres of water is removed. Therefore, there will be 6 litres of milk and 4 litres of water left in the container. It is then replaced with pure milk of 10 litres. Now the container will have 16 litres of milk and 4 litres of water.

Step 2: When 10 litres of the new mixture is removed, 8 litres of milk and 2 litres of water is removed. The container will have 8 litres of milk and 2 litres of water in it. Now 10 litres of pure milk is added. Therefore, the container will have 18 litres of milk and 2 litres of water in it at the end of the second step.

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

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

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%

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

Required average

=> (510 * 5 + 240 * 25)/30.

=> 8550/30

=> 285.

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.

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.

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.

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

Area of Pond = 64

Area of field = 64 × 8 = 512

Area of field ⇒ x. 2x = 512

⇒ 2x2 = 512 ⇒ x2 = 256 ⇒ x = 16

Length = 2 × 16 = 32.

SP = Rs. 1080 and CP = Rs. 900

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

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

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.

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