Interview Questions.

Top 100+ Elico Aptitude Interview Questions And Answers

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Top 100+ Elico Aptitude Interview Questions And Answers

Question 1. If In A Long Division Sum,the Dividend Is 380606 And The Successive Remainders From The First To The Last Are 434, a hundred twenty five And 413, Then The Divisor Is?

Answer :

Let d = divisor and q = quotient 

First do away with the ultimate the rest: 

d x q + 413 = 380606 

d x q = 380193.

Question 2. The Sum Of The Two Position Number P And Q Is 2.5 Time Their Difference .If The Product Of The Number eighty four Then The What Is The Sum Of Those Two Number?

Answer :

Let the number be P and Q

According to the query

(P + Q) = 2.Five (P - Q)

P + Q = 2.5P - 25Q

3.5Q = 1.5P

P / Q = 7/3 .....(i)

Now PQ = 84.

Electrical and Electronics Engineering Interview Questions
Question three. Out Of Three Given Number The First Number Is Twice The Second And Thrice Number Is 154 Then What Is The Difference Between The First And Third Number ?

Answer :

Let the 1/3 quantity = N

Then the first range = 3N

And second wide variety = 3N/2

According to the question [N + 3N + 3N/2] / three=154.

Question 4. The Difference Between Two Number Is 18 If Four Times The Second Number Is Less Then Three What Is The Sum These Two Number?

Answer :

Let first wide variety = P

And 2nd variety = Q

According to the query 

P - Q = 10 ....(i)

And 3P - 4Q = 18...(ii)

on multiply eq (i) by using 3 and then subtract eq (ii) from it

3P - 3Q = 54

3P - 4Q = 18 

Q = 36

On putting the price of Q within the eq (i) 

P = 18 + Q = 18 +36 

P = fifty four.

Question 5. If The Number 653 Xy Is Divisible By 90, Then (x + Y) = ?

Answer :

ninety = 10 x 9

Clearly, 653xy is divisible via 10, so y = zero

Now, 653x0 is divisible by means of 9.

So, (6 + five + 3 + x + 0) = (14 + x) is divisible by means of nine. So, x = 4.

Hence, (x + y) = (four + zero) = four.

Aptitude Interview Questions
Question 6. A Boy Multiplied 987 By A Certain Number And Obtained 559981 As His Answer. If In The Answer Both ninety eight Are Wrong And The Other Digits Are Correct, Then The Correct Answer Would Be?

Answer :

987 = 3 x 7 x forty seven

So, the desired variety need to be divisible by way of each considered one of three, 7, 47

==> 553681 (Sum of digits = 28, not divisible by 3)

==> 553181 (Sum of digits = 25, not divisible with the aid of 3)

==> 555681 are divisible by using 3, 7, forty seven.

Question 7. The Difference Between Two Number S 1365. When The Larger Number Is Divided By The Smaller One The Quotient Is 6 And The Remainder Is 15. The Smaller Number Is?

Answer :

Let the smaller wide variety be x, then larger range = 1365 + x

Therefore 1365 + x = 6x + 15

5x = 1350 ==> x = 270

Required Number is 270.

Electrical Engineering Interview Questions
Question eight. In Doing A Division Of A Question With Zero Remainder, A Candidate Took 12 As Divisor Instead Of 21. The Quotient Obtained By Him Was 35. The Correct Quotient Is?

Answer :

Dividend = 12 * 35 =420

Now dividend = 420 and divisor = 21

Therefore accurate quotient = 420/21 = 20.

Question nine. Find The Sum Of All Odd Number Up To a hundred?

Answer :

The given number is 1, 3, 5………….Ninety nine

This is an A.P with a = 1, d = 2

Let it incorporate n term 1 + (n-1)2 = ninety nine

==> n = 50

Then required sum = n/2 (first time period + last time period)

==> 50/2(1 + ninety nine) = 2500.

Instrumentational Engineering Interview Questions
Question 10. On Dividing A Number By 357, We Get 39 As Remainder. On Dividing The Same Number 17, What Will Be The Remainder?

Answer :

Let x be the range and y be the quotient. Then,

x = 357 x y + 39

= (17 x 21 x y) + (17 x 2) + five

= 17 x (21y + 2) + 5)

Required the rest = 5.

Question 11. 2056 X 987 =?

Answer :

2056 x 987 = 2056 x (1000 - 13)

= 2056 x 1000 - 2056 x thirteen

= 2056000 - 26728

= 2029272.

Electronics Interview Questions
Question 12. In A Division Sum, The Divisor Is 10 Times The Quotient And 5 Times The Remainder. If The Remainder Is 46, What Is The Dividend?

Answer :

Divisor = (five x forty six) = 230

10 x Quotient = 230=230= 23/10

Dividend = (Divisor x Quotient) + Remainder

= (230 x 23) + 46

= 5290 + 46

= 5336.

Electrical and Electronics Engineering Interview Questions
Question 13. What Was The Day Of The Week On twelfth January, 1979?

Answer :

Number of peculiar days in (1600 + three hundred) years = (zero + 1) = 1 abnormal day.

Seventy eight years = (19 jump years + 59 normal years)

==> (38 + fifty nine) unusual days = 6 ordinary days

12 days of January have 5 extraordinary days.

Therefore, total range of bizarre days= (1 + 6 + 5) = five ordinary days.

Question 14. A Number When Divide By 6 Leaves A Remainder 3. When The Square Of The Number Is Divided By 6, The Remainder Is?

Answer :

Let x = 6q + three.

Then, x2 = (6q + three)2

= 36q2 + 36q + nine

= 6(6q2 + 6q + 1) + 3

Thus, while x2 is divided by way of 6, then the rest = three.

Question 15. 310 X 999 =?

Answer :

310 x 999 = 310 x (one thousand-1)

= 310 x a thousand -310 x 1

= 310000 – 310

= 309690.

BHEL Interview Questions
Question 16. Persons Can Repair A Road In 12 Days, Working 5 Hours A Day. In How Many Days Will 30 Persons, Working 6 Hours A Day, Complete The Work?

Answer :

Let the specified range of days be x.

Less persons, More days (Indirect Proportion)

More working hours in step with day, Less days (Indirect Proportion)

Persons30:39Working hours/day6:5?? 12:x

=> 30 * 6 * x = 39 * 5 * 12 

=> x= 13.

Question 17. Men Can Complete A Piece Of Work In 18 Days. In How Many Days Will 27 Men Complete The Same Work ?

Answer :

Less Men, method extra Days Indirect Proportion

Let the wide variety of days be x then,

27 : 36 :: 18 : x

[Please pay attention, we have written 27 : 36 rather than 36 : 27, in indirect proportion, if you get it then chain rule is clear to you :)]          

=>27x = 36 * 18

=> x = 24

 So 24 days will be required to get paintings accomplished by using 27 guys.

Manufacturing Industries Interview Questions
Question 18. Some Persons Can Do A Piece Of Work In 12 Days. Two Times The Number Of Such Persons Will Do Half Of That Work In?

Answer :

Let x men can do the in 12 days and the desired quantity of days be z

More men, Less days     [Indirect Proportion]

Less work, Less days     [Direct Proportion  ]

men2x:xwork1:12 :: 12 : z

(2x×1×z)=(x×12×12)

z=3.

Aptitude Interview Questions
Question 19. If The Cost Price Of A Certain Object Doubles, Then The Loss Gets Tripled Of What It Was Initially. The Initial Loss % Was ?

Answer :

Let the value rate be x and selling fee be y.

Loss = x – y

When the cost price doubles, the loss gets tripled.

So it will become like this, 2x – y = 3(x-y)

=> x = 2y

Loss % = (loss/ C.P) x 100 = [(2y-y)/2y] x one hundred= 50 %.

Question 20. A Quantity Of Tea Is Sold At Rs. 5.75 Per Kilogram. The Total Gain By Selling The Tea At This Rate Is Rs. 60. Find The Quantity Of Tea Being Sold If A Profit Of 15% Is Made On The Deal ?

Answer :

Say general fee price of tea is x.

Then general income at a fee of 15% is = (15x/one hundred)

According to question,

15x/100 = 60

so x = four hundred

C.P of the tea is Rs. Four hundred.

So total promoting fee may be = (four hundred+60) = Rs.460

so the amount of the tea may be = (460/5.Seventy five) = 80kg.

ELECTRONICS & INSTRUMENTATION Engineering Interview Questions
Question 21. A Trader Marked The Price Of The T.V. 30% Above The Cost Price Of The T.V. And Gave The Purchaser 10% Discount On The Marked Price, Thereby Gaining Rs.340. Find The Cost Price Of The T.V ?

Answer :

Let 'x' be the fee rate.

Now Marked rate = x + 30x/one hundred = 13x/10

10% cut price = 10/100 x 13x/10 = 13x/one hundred 

Selling rate = 13x/10 - 13x/one hundred = 117x/one hundred

Given gain = 340

Here advantage = 117x/one hundred - x = 17x/one hundred = 340 => x = Rs. 2000.

Question 22. A Driver Of Auto Rickshaw Makes A Profit Of 20% On Every Trip When He Carries 3 Passengers And The Price Of Petrol Is Rs. 30 A Litre. Find The % Profit For The Same Journey If He Goes For four Passengers Per Trip And The Price Of Petrol Reduces To Rs. 24 Litres ?

Answer :

When three passengers profits become 3x

cost= Rs.30 

earnings =20% of 30 = Rs.6 

That way his earning is Rs.36. So that according to passenger fare should be Rs.12.

When 4 passengers

incomes = 12x4=Rs.48.

Price =Rs.24.

Income = Rs.24 = a hundred%.

Question 23. The Cost Price Of An Article Is 54% Of The Marked Price. Calculate The Gain Percent After Allowing A Discount Of 15%?

Answer :

Let marked price = Rs. A hundred. 

Then, C.P. = RS. 54, 

S.P. = Rs. 85

Gain % = 31/64 x a hundred = forty eight.Four%.

Analogue electronics Interview Questions
Question 24. A Shopkeeper Who Deals In Books Sold A Book At 16% Loss. Had She Charged An Additional Rs.60 While Selling It , Her Profit Would Have Been 14%. Find The Cost Price, In Rupees, Of The Book ?

Answer :

Let the C.P be 'x'

Then, the selling fee S.P = x - 16x/one hundred

= 84x/a hundred = 21x/25

Now, if the S.P is 60 greater, then the income is 14% 

=> 21x/25 + 60 = x + 14x/100

=> 114x/one hundred - 21x/25 = 60

=> (fifty seven - 42)x/50 = 60

=> 15x/50 = 60

x = 3000/15 = 2 hundred

Therefore, the Cost fee C.P = x = Rs. 200.

Electrical Engineering Interview Questions
Question 25. Sambhu Buys Rice At Rs. 10/kg And Puts A Price Tag On It So As To Earn A Profit Of 20%. However, His Faulty Balance Shows one thousand Gm When It Is Actually 800 Gm. What Is His Actual Gain Percentage ?

Answer :

CP of 1000gm = Rs. 10

SP of 800gm = Rs. 12

SP of 1000gm =12x1000/800 = Rs. 15

Now take 1000gm as connection with calculate profit.

Profit=SP-CP=15-10=Rs. Five

Profit % = 5x100/10 = 50%.

Question 26. In A Certain Business, The Profit Is 220% Of The Cost. If The Cost Increases By 25% But The Selling Price Remains Constant, Approximately What Percentage Of The Selling Price Is The Profit ?

Answer :

Let C.P.= Rs. A hundred.

Then, Profit = Rs.220, 

S.P. = Rs.320.

New C.P. = 125% of Rs. A hundred = Rs. A hundred twenty five 

New S.P. = Rs.320.

Profit = Rs. (320 - one hundred twenty five) = Rs. 195 

Required percentage = 195/320×100== 60.9 =~ 61%.

BHEL Aptitude Interview Questions
Question 27. Arun Purchased 30 Kg Of Wheat At The Rate Of Rs. Eleven.50 Per Kg And 20 Kg Of Wheat At The Rate Of 14.25 Per Kg. He Mixed The Two And Sold The Mixture. Approximately What Price Per Kg Should Be Sell The Mixture To Make 30% Profit ?

Answer :

C.P. Of fifty kg wheat = (30 * 11.50 + 20 * 14.25) = Rs. 630.

S.P. Of fifty kg wheat = 130% of Rs. 630 = 130/one hundred * 630 = Rs. 819.

S.P. According to kg = 819/50 = Rs. Sixteen.38 = 16.30.

Instrumentational Engineering Interview Questions
Question 28. A Merchant Buys Two Items For Rs. 7500. One Item He Sells At A Profit Of sixteen% And The Other Item At 14% Loss. In The Deal, The Merchant Makes Neither Any Profit Nor Any Loss. What Is The Difference Between The Selling Price Of Both The Items?

Answer :

Let the C.P of one item is Rs. P

and that of different is Rs. (7500 - P)

According to the information given

C.P = S.P 

=> Px(116/one hundred) + (7500-P)x(86/one hundred) = 7500 

=> 30P = 105000

=> P = 3500

Required distinction between selling fees

= Rs. [(3500/100) x 116] - [(4000/100) x 86]

= 4060-3440

= Rs. 620.

Question 29. If Books Bought At Prices Ranging From Rs 2 hundred To Rs 350 Are Sold At Prices Ranging From Rs 300 To Rs 425, What Is The Greatest Possibleprofit That Might Be Made In Selling Eight Books?

Answer :

least fee charge      = two hundred*8 = 1600 

best sold fee = 425 * eight = 3400 

earnings required = 3400- 1600 = 1800.

Sony India Aptitude Interview Questions
Question 30. A Supplier Supplies Cartridges To A News Paper Publishing House. He Earns A Profit Of 25% By Selling Cartridges For Rs. 1540. Find The Cost Price Of The Cartridges ?

Answer :

Let Cost Price(C.P) = P

advantage% = (S.P-C.P)/C.P x 100

25 = (1540-P)/P x one hundred

25/a hundred = (1540-P)/P

=> P = 4(1540)-4P

=> 5P = four(1540)

=> P = 1232

So, Cost Price = Rs. 1232.

Question 31. Selling An Article At A Profit Of 20%, Aman Gets Rs. Four hundred More Than Selling At A Loss Of 20%. The Cost Price Of The Article Is ?

Answer :

Cost price of the item is given by means of

= 400x100/(20+20)

= Rs.1000.

Question 32. Uma Sold An Article For 3400 And Got A Profit Of 25%. If He Had Sold The Article For Rs. 3265. How Much Profit% Would Uma Have Got ?

Answer :

one hundred twenty five% ---- 3400

=> 100% ---- ?

=> ? = 3400x100/a hundred twenty five = 2720

=> Cost charge of the item = Rs. 2720

Profit while article offered at Rs. 3265 = 3265 - 2720 = 545

Hence, Profit% = Gain x one hundred/fee price

=> P% = 545 x a hundred/2720

=> P% = 20%.

ELGI Aptitude Interview Questions
Question 33. If The Simple Interest On A Sum Of Money For 2 Years At 5% Per Annum Is Rs. 50, What Is The Compound Interest On The Same At The Same Rate And For The Same Time?

Answer :

Sum = Rs.(50*a hundred)/2*5=Rs.500

Amount=Rs.[500*(1+5/100)2]

= Rs. 551.25

C.I = Rs.(551.25-500)= Rs.Fifty one.25.

Electronics Interview Questions
Question 34. Albert Invested An Amount Of Rs.8000 In A Fixed Deposit Scheme For 2 Years At Compound Interest Rate five P.C.P.A. How Much Amount Will Albert Get On Maturity Of The Fixed Deposit ?

Answer :

Amount

=Rs.[8000x(1+5/100)²]

= Rs.[8000 x 21/20x21/20]

= Rs.8820. 

Question 35. Simon Deposits $400 In An Account That Pays three% Interest Compounded Annually. What Is The Balance Of Simon’s Account At The End Of 2 Years?

Answer :

I=Prt

I=12

Balance = P +Prt

412

Find the balance at the cease of the second yr.

I = Prt=12.36

Balance =P + Prt

424.36.

Question 36. The Compound Interest On Rs. 30,000 At 7% Per Annum Is Rs. 4347. The Period (in Years) Is?

Answer :

Amount = Rs. (30000 + 4347) = Rs. 34347.

Let the time be n years.

Then, 30000*(1+7/a hundred)^n=34347

n= 2 years.

BHEL Interview Questions
Question 37. An Amount Of five,000 Is Invested At A Fixed Rate Of eight Per Cent Per Annum. What Amount Will Be The Value Of The Investment In Five Years Time, If The Interest Is Compounded Annually?

Answer :

The handiest a part of this kind of calculation that wishes particular

care is that concerning the interest fee. The system assumes that

r is a proportion, and so, in this situation:

r = 0.08

In addition, we've got P = five,000 and n = 5, so:

V = P(1 + r)five = 5,000 x (1 + 0.08)5 = five,000 x 1.469328 = 7,346.Sixty four

Thus the value of the investment could be 7,346.Sixty four.

Question 38. A Sum Is Equally Invested In Two Different Schemes On Ci At The Rate Of 15% And 20% For Two Years. If Interest Gained From The Sum Invested At 20% Is Rs. 528.75 More Than The Sum Invested At 15%, Find The Total Sum?

Answer :

Let Rs. K invested in each scheme

Two years C.I on 20% = 20 + 20 + 20x20/one hundred = forty four%

Two years C.I on 15% = 15 + 15 + 15x15/100 = 32.25%

Now,

(P x 44/a hundred) - (P x 32.25/100) = 528.75

=> 11.Seventy five P = 52875

=> P = Rs. 4500

Hence, general invested money = P + P = 4500 + 4500 = Rs. 9000.

Question 39. Find The Compound Interest On Rs. 2680 At 8% Per Annum For 2 Years ?

Answer :

We recognise Compound Interest = C.I. = P1+r100t - 1

Here P = 2680, r = 8 and t = 2

C.I. = 26801 + 81002-1= 268027252-12= 26802725+12725-1= 2680 5225×225

= (2680 x fifty two x 2)/625

= 445.95

Compound Interest = Rs. 445.95.

Question forty. Find The Simple Interest Of Rs.14000 At 10% Per Annum For 3 Months ?

Answer :

The method to calculate easy interest is :

SI=(P x T x R)/a hundred

In the given query Principal(P)=Rs. 14000, Time(T)=(3/12)year. And Rate of interest(R)=10%

So, SI=(14000 x 1/4 x 10)/a hundred

SI= Rs. 350.

Manufacturing Industries Interview Questions
Question forty one. A Train Travels 325 Km In 3.Five Hours And 470 Km In 4 Hours. Find The Average Speed Of Train ?

Answer :

As we know that Speed = Distance / Time

for common velocity = Total Distance / Total Time Taken

Thus, Total Distance = 325 + 470 = 795 km

Thus, Total Speed = 7.5 hrs

Average Speed = 795/7.Five => 106 kmph.

Question 42. If A Student Walks At The Rate Of 4 Mts/min From His Home, He Is 4 Minutes Late For School, If He Walks At The Rate Of 6 Mts/min He Reaches Half An Hour Earlier. How Far Is His School From His Home ?

Answer :

Let the space between domestic and faculty is 'x'.

Let actual time to attain be 't'.

Thus, x/4 = t + four ---- (1)

and x/6 = t - 30 -----(2)

Solving equation (1) and (2)

=> x = 98 mts.

ELECTRONICS & INSTRUMENTATION Engineering Interview Questions
Question forty three. An Express Traveled At An Average Speed Of one hundred Km/hr, Stopping For four Min After Every seventy five Km. How Long Did It Take To Reach Its Destination six hundred Km From The Starting Point ?

Answer :

Time taken to cover 600 km = six hundred/100 = 6 hrs.

Number of stoppages = six hundred/75 - 1 = 7

Total time of stoppages = 4 x 7 = 28 min

Hence, general time taken = 6 hrs 28 min.

Question forty four. 'k' Cycled From A To B At 10 Kmph And Returned At The Rate Of nine Kmph. 'l' Cycled Both Ways At 12 Kmph. In The Whole Journey 'l' Took 10 Minutes Less Than 'ok'. Find The Distance Between A And B ?

Answer :

Let 'd' be the distance among A and B

K -time = d/10 + d/9 = 19d/90 hours

L -time = second/12 = d/6 hours

We recognize that, 10 min = 1/6 hours

Thus, time difference between K and L is given as 10 mins. 

=> 19d/90 - d/6 = 1/6

=> (19d-15d)/ninety = 1/6

=> 4d/ninety = 1/6

Thus,

d= 15/four km = 3.75 km.

Consequently the distance among A and B is 3.75 km.

Question 45. You Drive To The Store At 20 Kmph And Return By The Same Route At 30 Kmph. Discounting The Time Spent At The Store, What Was Your Average Speed ?

Answer :

Average speed=total distance/general time

Let distance to shop be  K

then, general time =(K/20)+(K/30)=K/12

and, overall time =(2K)

so common velocity= 2K / (K/12) = 24kmph.

Question forty six. K Is 50% Faster Than L. If L Starts At nine A.M. And K Starts At 10 A.M. L Travels At A Speed Of 50 Km/hr. If L And K Are 300 Kms Apart, The Time When They Meet When They Travel In Opposite Direction Is ?

Answer :

let 't' be the time and then they met due to the fact that L starts offevolved.

Given K is 50% quicker than L

50 t + 1.Five*50(t-1) = 300

50 t +seventy five t = 300 + 75

t = 375 / one hundred twenty five = three hrs beyond the time that L begins

So they meet at (9 + 3)hrs = 12:00 midday.

Question 47. Laxmi And Prasanna Set On A Journey. Laxmi Moves Northwards At A Speed Of 20kmph And Prasanna Moves Southward At A Speed Of 30 Kmph. How Far Will Be Prasanna From Laxmi After 60 Minutes ?

Answer :

We recognise 60 min = 1 hr

Total northward Laxmi's distance = 20kmph x 1hr = 20 km

Total southward Prasanna's distance = 30kmph x 1hr = 30 km

Total distance among Prasanna and Laxmi is = 20 + 30 = 50 km.

Question forty eight. Joel Travels The First 3 Hours Of His Journey At 60 Mph Speed And The Remaining five Hours At 24 Mph Speed. What Is The Average Speed Of Joel's Travel In Kmph ?

Answer :

Average speed = Total distance / Total time.

Total distance traveled by means of Joel = Distance included inside the first 3 hours + Distance protected in the subsequent 5 hours.

Distance covered within the first 3 hours = 3 x 60 = one hundred eighty miles

Distance covered in the subsequent 5 hours = 5 x 24 = a hundred and twenty miles

Therefore, overall distance traveled = one hundred eighty + a hundred and twenty = three hundred miles.

Total time taken = three + five = 8 hours.

Average speed = 300/eight = 37.5 mph.

We know that 1 mile = 1.6 kms

=> 37.5 miles = 37.5 x 1.6 = 60 kms

Average pace in Kmph = 60 kmph.

Question 49. A Train-a Passes A Stationary Train B And A Pole In 24 Sec And nine Sec Respectively. If The Speed Of Train A Is 48 Kmph, What Is The Length Of Train B?

Answer :

Length of educate A = forty eight x 9 x five/18 = a hundred and twenty mts

Length of train B = forty eight x 24 x five/18 - 120

=> 320 - one hundred twenty = two hundred mts.

Question 50. Tilak Rides On A Cycle To A Place At Speed Of 22 Kmph And Comes Back At A Speed Of 20 Kmph. If The Time Taken By Him In The Second Case Is 36 Min. More Than That Of The First Case, What Is The Total Distance Travelled By Him (in Km)?

Answer :

Let the distance travelled through Tilak in first case or 2nd case = d kms

Now, from the given statistics,

d/20 = d/22 + 36 min

=> d/20 = d/22 + three/5 hrs

=> d = 132 km.

Hence, the whole distance travelled by him = d + d = 132 + 132 = 264 kms.

Question 51. Two Trains Are Running With Speeds 30 Kmph And 58 Kmph Respectively In The Same Direction. A Man In The Slower Train Passes The Faster Train In 18 Seconds. Find The Length Of The Faster Train?

Answer :

Speeds of  trains = 30 kmph and 58 kmph

=> Relative pace = 58 - 30 = 28 kmph = 28 x five/18 m/s = 70/nine m/s

Given a man takes time to pass duration of quicker teach = 18 sec

Now, required Length of faster train = speed x time = 70/9  x  18 = one hundred forty mts.

Question 52. A Car Covers Its Journey At The Speed Of 80km/hr In 10hours. If The Same Distance Is To Be Covered In 4 Hours, By How Much The Speed Of Car Will Have To Increase?

Answer :

Initial speed = 80km/hr

Total distance = eighty * 10 = 800km

new speed = 800/4 =200km/hr

Increase in velocity = 2 hundred - eighty = 120km/hr.

Question 53. A Monkey Climbs Up A Greased Pole, Ascends 20m And Slip 4m In Alternate Minutes. If The Pole Is 96m High, How Many Minutes Will It Take To Reach The Top?

Answer :

net peak climbed in 2 min = 20m - 4m = 16m

In 10 min the internet top climbed = sixteen * 10/2 = 80m

Remaining height = 96m - 80m = 16m

In the 11th min , the monkey may be ascending up. 

Time taken to ascend the closing 16min

16/20 = four/5min

Total time taken = 10 + 4/five = 54/five min.

Question 54. A Car Is Running At 7/10 Of Its Own Speed Reached A Place In 22 Hours. How Much Time Could Be Saved If The Train Would Run At Its Own Speed ?

Answer :

Since the automobile runs at 7/11 th of its very own velocity, the time it take is eleven/seventh of its common pace.

Allow the usual time taken be t hours

then we are able to write, 11t / 7 = 22

or t  = 22 * 7 / 11 = 14 hours

Time stored  = 22 - 14 = 8hours.




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