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Top 100+ Indecomm Aptitude Interview Questions And Answers - May 31, 2020

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Top 100+ Indecomm Aptitude Interview Questions And Answers

Question 1. 12.Five + 10 1/4 Of a hundred ÷ 10% Of 10 = ?

Answer :

12.5 + 10 1/4 of a hundred ÷ 10% of 10

= 12.5 + forty one/four of 100 ÷ 10/one hundred of 10

= 12.5 + 41(25) ÷ 1 = 12.5 + 1025 = 1037.Five.

Question 2. Sixty six% Of 33.33% Of 66.Sixty six% Of ? = 30

Answer :

16.Sixty six% of 33.33% of 66.66% of x = 30

 = 1/6 of 1/three of two/three of x = 30 

=> (30 * 6 * 3 * three)/2  => x = 810

Aptitude Interview Questions
Question 3. A Train Passes A Station Platform In 36 Sec And A Man Standing On The Platform In 20 Sec. If The Speed Of The Train Is 54 Km/hr. What Is The Length Of The Platform?

Answer :

Speed = fifty four * five/18 = 15 m/sec.

Length of the teach = 15 * 20 = three hundred m.

Let the period of the platform be x m . Then, 

(x + 300)/36 = 15 => x = 240 m.

Question four. A 300 M Long Train Crosses A Platform In 39 Sec While It Crosses A Signal Pole In 18 Sec. What Is The Length Of The Platform?

Answer :

Speed = 300/18 = 50/3 m/sec.

Let the duration of the platform be x meters.

Then, (x + 300)/39 = 50/three

3x + 900 = 1950 => x = 350 m.

Question 5. A Train Speeds Past A Pole In 15 Sec And A Platform 100 M Long In 25 Sec, Its Length Is?

Answer :

Let the period of the educate be x m and its velocity be y m/sec.

Then, x/y = 15 => y = x/15

(x + a hundred)/25 = x/15 => x = one hundred fifty m.

Infosys Technical Interview Questions
Question 6. A Train Moves Fast A Telegraph Post And A Bridge 264 M Long In 8 Sec And 20 Sec Respectively. What Is The Speed Of The Train?

Answer :

Let the duration of the educate be x m and its velocity be y m/sec.

Then, x/y = 8 => x = 8y

(x + 264)/20 = y

y = 22

Speed = 22 m/sec = 22 * 18/5 = 79.2 km/hr.

Question 7. How Many Seconds Will A 500 M Long Train Take To Cross A Man Walking With A Speed Of three Km/hr In The Direction Of The Moving Train If The Speed Of The Train Is 63 Km/hr?

Answer :

Speed of educate relative to man = sixty three - three = 60 km/hr.

= 60 * five/18 = 50/three m/sec.

Time taken to skip the person = 500 * three/50 = 30 sec.

HR Interview Questions
Question eight. A Jogger Running At nine Km/hr Along Side A Railway Track Is 240 M Ahead Of The Engine Of A a hundred and twenty M Long Train Running At forty five Km/hr In The Same Direction. In How Much Time Will The Train Pass The Jogger?

Answer :

Speed of train relative to jogger = forty five - nine = 36 km/hr.

= 36 * five/18 = 10 m/sec.

Distance to be protected = 240 + 120 = 360 m.

Time taken = 360/10 = 36 sec.

Question 9. A Train a hundred and ten M Long Is Running With A Speed Of 60 Km/hr. In What Time Will It Pass A Man Who Is Running At 6 Km/hr In The Direction Opposite To That In Which The Train Is Going?

Answer :

Speed of teach relative to guy = 60 + 6 = 66 km/hr.

= 66 * 5/18 = 55/three m/sec.

Time taken to pass the men = 110 * three/fifty five = 6 sec.

Yahoo Aptitude Interview Questions
Question 10. Two Trains a hundred and forty M And 160 M Long Run At The Speed Of 60 Km/hr And 40 Km/hr Respectively In Opposite Directions On Parallel Tracks. The Time Which They Take To Cross Each Other Is?

Answer :

Relative velocity = 60 + 40 = 100 km/hr.

= a hundred * 5/18 = 250/nine m/sec.

Distance covered in crossing every different = 140 + a hundred and sixty = three hundred m.

Required time = three hundred * nine/250 = 54/five  = 10.8 sec.

Question 11. Two Trains Are Moving In Opposite Directions At 60 Km/hr And 90 Km/hr. Their Lengths Are 1.10 Km And 0.Nine Km Respectively. The Time Taken By The Slower Train To Cross The Faster Train In Seconds Is?

Answer :

Relative pace = 60 + 90 = one hundred fifty km/hr.

= a hundred and fifty * five/18 = a hundred twenty five/3 m/sec.

Distance included = 1.10 + zero.Nine = 2 km = 2000 m.

Required time = 2000 * three/125 = 48 sec.

SAP Aptitude Interview Questions
Question 12. A Train one hundred twenty five M Long Passes A Man, Running At five Km/hr In The Same Direction In Which The Train Is Going, In 10 Sec. The Speed Of The Train Is?

Answer :

Speed of the teach relative to guy = a hundred twenty five/10 = 25/2 m/sec.

= 25/2 * 18/five = forty five km/hr

Let the speed of the educate be x km/hr. Then, relative velocity = (x - 5) km/hr.

X - five = forty five => x = 50 km/hr.

Aptitude Interview Questions
Question thirteen. A And B Starts A Business With Rs.8000 Each, And After four Months, B Withdraws Half Of His Capital How Should They Share The Profits At The End Of The 18 Months?

Answer :

A invests Rs.8000 for 18 months, but B invests Rs.8000 for the first four months and then withdraws Rs.4000.

So, the investment of B for ultimate 14 months is Rs.4000 handiest.

A             :                 B

8000*18 : (8000*4) + (4000*14)

14400     : 88000

A:B = 18:11.

Question 14. A And B Invests Rs.10000 Each, A Investing For eight Months And B Investing For All The 12 Months In The Year. If The Total Profit At The End Of The Year Is Rs.25000, Find Their Shares?

Answer :

The ratio of their profits A:B = eight:12 = 2:3

Share of A within the general profit = 2/five * 25000 = Rs.10000 Share of A within the total earnings = three/5 * 25000 = Rs.15000.

Question 15. A Is A Working Partner And B Is A Sleeping Partner In The Business. A Puts In Rs.15000 And B Rs.25000, A Receives 10% Of The Profit For Managing The Business The Rest Being Divided In Proportion Of Their Capitals. Out Of A Total Profit Of Rs.9600, Money Received By A Is?

Answer :

15:25 => three:5

9600*10/100 = 960

9600 - 960 = 8640

8640*3/eight = 3240 + 960

                 = 4200.

Oracle Aptitude Interview Questions
Question sixteen. P And Q Started A Business With Respective Investments Of Rs. 4 Lakhs And Rs. 10 Lakhs. As P Runs The Business, His Salary Is Rs. 5000 Per Month. If They Earned A Profit Of Rs. 2 Lakhs At The End Of The Year, Then Find The Ratio Of Their Earnings?

Answer :

Ratio of investments of P and Q is 2 : five 

Total earnings claimed with the aid of P = 12 * 5000 = Rs. 60000 

Total earnings = Rs. 2 lakhs. 

Profit is to be shared = Rs. 140000 

Share of P = (2/7) * 140000 = Rs. 400000

Share of Q = Rs. A hundred thousand

Total earnings of P = (60000 + 40000) = Rs. A hundred thousand

Ratio in their income = 1 : 1.

Question 17. Four Car Rental Agencies A, B, C And D Rented A Plot For Parking Their Cars During The Night. A Parked 15 Cars For 12 Days, B Parked 12 Cars For 20 Days, C Parked 18 Cars For 18 Days And D Parked sixteen Cars For 15 Days. If A Paid Rs. 1125 As Rent For Parking His Cars, What Is The Total Rent Paid By All The Four Agencies?

Answer :

The ratio in which the four agencies will be paying the rents = 15 * 12 : 12 * 20 : 18 * 18 : 16 * 15

= one hundred eighty : 240 : 324 : 240 = forty five : 60 : eighty one : 60

Let us recollect the four amounts to be 45k, 60k, 81k and 60k respectively.

The general rent paid by way of the four groups = 45k + 60k + 81k + 60k= 246k

It is given that A paid Rs. 1125

45k = 1125 => ok = 25

246k = 246(25) = Rs. 6150

Thus the whole hire paid by means of all the 4 companies is Rs. 6150.

Apple Technical Interview Questions
Question 18. A Started A Business With An Investment Of Rs. 70000 And After 6 Months B Joined Him Investing Rs. 120000. If The Profit At The End Of A Year Is Rs. 52000, Then The Share Of B Is?

Answer :

Ratio of investments of A and B is (70000 * 12) : (120000 * 6) = 7 : 6

Total earnings = Rs. 52000

Share of B = 6/13 (52000) = Rs. 24000.

Infosys Technical Interview Questions
Question 19. A, B And C Started A Business With Capitals Of Rs. 8000, Rs. 10000 And Rs. 12000 Respectively. At The End Of The Year, The Profit Share Of B Is Rs. 1500. The Difference Between The Profit Shares Of A And C Is?

Answer :

Ratio of investments of A, B and C is 8000 : ten thousand : 12000 = four : five : 6

And also for the reason that, income percentage of B is Rs. 1500

=> five components out of 15 components is Rs. 1500

Now, required distinction is 6 - four = 2 parts

Required difference = 2/five (1500) = Rs. Six hundred.

Question 20. A And B Start A Business, With A Investing The Total Capital Of Rs.50000, On The Condition That B Pays A Interest @ 10% Per Annum On His Half Of The Capital. A Is A Working Partner And Receives Rs.1500 Per Month From The Total Profit And Any Profit Remaining Is Equally Shared By Both Of Them. At The End Of The Year, It Was Found That The Income Of A Is Twice That Of B. Find The Total Profit For The Year?

Answer :

Interest obtained through A from B = 10% of half of of Rs.50000 = 10% * 25000 = 2500.

Amount received by using A according to annum for being a working partner = 1500 * 12 = Rs.18000.

Let 'P' be the a part of the ultimate profit that A gets as his share. Total profits of A = (2500 + 18000 + P)

Total income of B = only his share from the ultimate income = 'P', as A and B proportion the last earnings equally.

Income of A = Twice the income of B

(2500 + 18000 + P) = 2(P)

P = 20500

Total profit = 2P + 18000

= 2*20500 + 18000 = 59000.

TCS Aptitude Interview Questions
Question 21. A Boat Goes 24 Km Downstream In 10 Hours, It Takes 2 Hours More To Cover The Same Distance Against The Stream. What Is The Speed Of The Boat In Still Water?

Answer :

Speed downstream = 24/10 km/hr = 2.Four km/hr 

Speed upstream = 24/12 km/hr = 2 km/hr 

Speed of the boat in still water = ½ (2.Four +2)km/hr = 2.2 km/hr.

Question 22. A Steamer Goes Downstream From One Port To Another In 4 Hours, It Covers The Same Distance Upstream In five Hours. If The Speed Of The Stream Is 2km/hr,the Distance Between The Two Parts Is?

Answer :

Let the distance among the two parts be x km. 

Then velocity downstream = x/four km/hr. 

Speed Upstream = x/5 km/hr 

Speed of the circulate = ½ (x/four –x/five) 

Therefore, ½(x/4 –x/five) = 2. => x/4 –x/5 = 4 => x = eighty. 

Hence, the gap between the two ports is 80km.

Question 23. A Boat Covers A Distance Of 30 Km In 2 ½ Hours Running Downstream While Returning It Covers The Same Distance In 3 ¾ Hours. What Is The Speed Of The Boat In Still Water?

Answer :

Speed downstream = (30 x 2/5)km/hr = 12km/hr

Speed upstream = (30 x four/15) km/hr = eight km/hr 

Speed of boat in still water = ½ (12 + eight)km/hr = 10 km/hr.

Infosys Aptitude Interview Questions
Question 24. The Speed Of A Motor Boat Is That Of The Current Of Water As 36:five. The Boat Goes Along With The Current In 5 Hours 10 Min, It Will Come Back In?

Answer :

Let the speed of motor boat be 36x km/hr and that of current of water be 5x km/hr 

Speed downstream = (36x +5x)km/hr = 41x km/hr 

Speed upstream = (36 -5x)km/hr = 31x km/hr 

Distance blanketed downstream = (41x × 31/6)km 

Distance upstream = [1271x/6 × 1/31x]hrs 

= forty one/6 hrs = 6hrs 50 min.

HR Interview Questions
Question 25. The Length Of A Rectangular Plot Is Increased By 25%. To Keep It’s Area Uncharged The Width Of The Plot Should Be?

Answer :

Let the period be x metres and breadth be y metres 

Then it’s vicinity =(xy)m2 

New duration =(one hundred twenty five/one hundred x)m =(5x/4)m. 

Let the brand new breadth be 2 metres 

Then xy 5x/4 x 2 => 2 = 4/five y 

Decreased in width =(y -4/5 y) = y/five metres 

Decreased % in width =(y/5 x 1/y x a hundred)% = 20%.

Question 26. If The Length Of A Rectangle Is Increased By 20% And It’s Breadth Is Decreased By 20% Then It Area?

Answer :

Let period = x metres and breadth = y metres, then region =(xy) 

New length =(a hundred and twenty/100x)m = 6x/5 m 

New breadth =(80/100 y)m 

New location =(6x/five x 4y/five)m2 = (24/25 xy)m2 

Decreased in area =(xy – 24/25 xy)m2= xy/25 m2

Decreased % =(xy/25 x 1/xy x a hundred)% = four%.

IBM Aptitude Interview Questions
Question 27. The Perimeter Of A Rectangle Is 60m. If It’s Length Is Twice Its Breadth Then Its Area Is?

Answer :

Let the breadth be x metres then duration = 2x metres

=> 2(2x+x) = 60 

=> 6x =60 

=> x =10 

l = 20m and b =10m 

=> Area =(20 x 10) Sq.M = 2 hundred Sq.M.

Yahoo Aptitude Interview Questions
Question 28. The Area Of A Rectangle Is 12 Sq.Metres And It’s Length Is 3 Times That Of It’s Breadth. What Is The Perimeter Of The Rectangle?

Answer :

Let the breadth be x metres , then it’s period = 3x metres

=>3x × x =12 

=> x2= 4 

=> x =2 l=6m, b =2m 

=> Perimeter = 2(6+2)m = 16m.

Question 29. The Length Of A Rectangular Increased By 10% And It;s Breadth Is Decreased By 10 %. Then The Area Of The New Rectangle Is?

Answer :

Length of the be allow ‘l’ units and breadth be b devices 

Area =lb sq. Units 

Now period =(one hundred ten/a hundred l) = 11l/10 

New breadth =(90/one hundred b) = 96/10 

Now place (11l/10 x ninety six/10)Sq.Gadgets = (ninety nine/one hundred lb)squareunit 

Area reduced = (lb -99/a hundred lb)sq.Units =lb/a hundred squareUnits 

Percentage reduced =(lb/one hundred x 1/lb x a hundred)% = 1%.

Capgemini Aptitude Interview Questions
Question 30. How Many Meters Of Carpet 63cm Will Be Required To Be A Floor Of A Room 14m By 9cm?

Answer :

Width of the carpet = sixty three/100 m allow it’s period be x metres

Then x × 63/a hundred = 14 x 9 

=> x =(14 x nine x 100/63) = 200m 

Length = 200m.

Question 31. The Wheels Revolve Round A Common Horizontal Axis. They Make 15, 20 And forty eight Revolutions In A Minute Respectively. Starting With A Certain Point On The Circumference Down Wards. After What Interval Of Time Will They Come Together In The Same Position?

Answer :

Time for one revolution = 60/15 = four

60/ 20 = 3

60/forty eight = five/4

LCM of four, three, 5/4

LCM of Numerators/HCF of Denominators = 

60/1 = 60.

Question 32. Find The Lowest four-digit Number Which When Divided By 3, 4 Or 5 Leaves A Remainder Of 2 In Each Case?

Answer :

Lowest 4-digit quantity is a thousand.

LCM of 3, 4 and five is 60.

Dividing a thousand by means of 60, we get the remainder 40. Thus, the bottom four-digit variety that exactly divisible via three, 4 and 5 is 1000 + (60 - forty) = 1020.

Now, add the remainder 2 it's required. Thus, the solution is 1022.

Cognizant Aptitude Interview Questions
Question 33. Find The Greatest Number That, While Dividing forty seven, 215 And 365, Gives The Same Remainder In Each Case?

Answer :

Calculate the variations, taking  numbers at a time as follows:

(215-forty seven) = 168

(365-215) = 150

(365-47) = 318

HCF of 168, a hundred and fifty and 318 we get 6, that is the greatest quantity, which at the same time as dividing 47, 215 and 365 gives the equal.

Emainder in every instances is 5.

SAP Aptitude Interview Questions
Question 34. A Room Is 6 Meters 24 Centimeters In Length And 4 Meters 32 Centimeters In Width. Find The Least Number Of Square Tiles Of Equal Size Required To Cover The Entire Floor Of The Room?

Answer :

Length = 6 m 24 cm = 624 cm

Width = four m 32 cm = 432 cm

HCF of 624 and 432 = forty eight

Number of square tiles required = (624 * 432)/(48 * 48) = 13 * nine = 117.

Question 35. A Drink Vendor Has eighty Liters Of Maaza, one hundred forty four Liters Of Pepsi And 368 Liters Of Sprite. He Wants To Pack Them In Cans, So That Each Can Contains The Same Number Of Liters Of A Drink, And Doesn't Want To Mix Any Two Drinks In A Can. What Is The Least Number Of Cans Required?

Answer :

The number of liters in each can = HCF of eighty, 144 and 368 = sixteen liters.

Number of cans of Maaza = eighty/sixteen = 5

Number of cans of Pepsi = 144/16 = 9

Number of cans of Sprite = 368/sixteen = 23

The general variety of cans required = five + nine + 23 = 37 cans.

Question 36. The Sum Of The Two Numbers Is 12 And Their Product Is 35. What Is The Sum Of The Reciprocals Of These Numbers?

Answer :

Let the numbers be a and b. Then,

a + b = 12 and ab = 35.

(a + b)/ab = 12/35

= (1/b + 1/a) = 12/35

Sum of reciprocals of given numbers = 12/35.

Oracle Aptitude Interview Questions
Question 37. How Many Of The Following Numbers Are Divisible By 132?

Answer :

264, 396, 462, 792, 968, 2178, 5184, 6336

132 = eleven * three * four

Clearly, 968 isn't divisible by using three

None of 462 and 2178 is divisible with the aid of 4

And, 5284 is not divisible by 11

Each one of the closing 4 numbers is divisible through each one in every of four, three and eleven.

So, there are 4 such numbers.

Question 38. Which One Of The Following Numbers Is Exactly Divisible By eleven?

Answer :

(four + 5 + 2) - (1 + 6 + 3) = 1, now not divisible with the aid of 11.

(2 + 6 + 4) - (4 + 5 + 2) = 1, now not divisible by way of 11.

(4 + 6 + 1) - (2 + five + 3) = 1, not divisible through eleven.

(four + 6 + 1) - (2 + five + four) = 0. 

So, 415624 is divisible through eleven.

Question 39. Nine 3/4 + 7 2/17 - 9 1/15 = ?

Answer :

 Given sum = nine + three/4 + 7 + 2/17 - (nine + 1/15)

= (9 + 7 - 9) + (three/four + 2/17 - 1/15)

= 7 + (765 + 120 - 68)/1020 = 7 817/1020.

Question forty. The Surface Area Of A Cube Is 726m2, Its Volume Is?

Answer :

6a2= 726 

=> a2 = 121 

=> a = eleven Cm 

Therefore, Volume of the dice 

= (eleven × eleven × eleven) cm3 = 1331 cm3.

Apple Technical Interview Questions
Question forty one. If The Length, Breadth And The Height Of A Cuboid Are In The Ratio 6: five: four And If The Total Surface Area Is 33300 Cm2, Then Length Breadth And Height In Cms, Are Respectively?

Answer :

Length = 6x

Breadth = 5x 

Height = 4x in cm 

Therefore, 2(6x × 5x + 5x × 4x + 6x × 4x) = 33300 

148x2 = 33300 

=> x2 = 33300/148 = 225 

=> x = 15. 

Therefore, Length = 90cm, 

Breadth = 75cm, 

Height = 60cm 

ninety, 75 , 60 cm.

Question 42. Rajan Got Married 8 Years Ago. His Present Age Is 6/5 Times His Age At The Time Of His Marriage. Rajan's Sister Was 10 Years Younger To Him At The Time Of His Marriage. The Age Of Rajan's Sister Is?

Answer :

Let Rajan's present age be x years. 

Then, his age on the time of marriage = (x - 8) years.

X = 6/five (x - 8) 

5x = 6x - forty eight => x = 48

Rajan's sister's age on the time of his marriage = (x - eight) - 10 = 30 years.

Rajan's sister's gift age = (30 + eight) = 38 years.

TCS Aptitude Interview Questions
Question forty three. The Sum Of The Ages Of 5 Children Born At The Intervals Of three Years Each Is 50 Years. What Is The Age Of The Youngest Child?

Answer :

Let the ages of the youngsters be x, (x + three), (x + 6), (x + nine) and (x +12) years. 

Then, x + (x + three) + (x + 6) + (x + nine) + (x + 12) = 50

5x = 20 => x = 4.

Age of youngest child = x = 4 years.

Question forty four. Ayesha's Father Was 38 Years Of Age When She Was Born While Her Mother Was 36 Years Old When Her Brother Four Years Younger To Her Was Born. What Is The Difference Between The Ages Of Her Parents?

Answer :

Mother's age when Ayesha's brother become born = 36 years.

Father's age while Ayesha's brother changed into born = (38 + four) = forty two years.

Required distinction = (forty two - 36) = 6 years.

Question 45. The Sum Of The Ages Of A Son And Father Is 56 Years After Four Years The Age Of The Father Will Be Three Times That Of The Son. Their Ages Respectively Are?

Answer :

Present a long time of son and father be x years (fifty six -x)years

(56- x +four) = three(x + four) or 4x =forty eight or x = 12 

Ages are 12 years, 44 years.

Question forty six. Pushpa Is Twice As Old As Rita Was Two Years Ago If The Difference Between Their Ages Is 2 Years, How Old Is Pushpa Today?

Answer :

Rita’s age 2 years in the past be x years 

Pushpa’s gift age =(2x) years 

2x –(x +2) =2 => x =four 

Therefore pushpa’s gift age = eight years.

Question forty seven. Sameer Spends forty % Of His Salary On Food Article, And 1/three Rd Of The Remaining On Transport. If He Saves Rs. 450 Per Month Which Is Half Of The Balance After Spending On Food Items And Transport. What Is This Monthly Salary?

Answer :

Suppose, salary = Rs. One hundred 

Expenditure on food = Rs. Forty 

Balance = Rs. 60 

Expenditure on delivery = 1/three × Rs. 60 = Rs. 20 

Now, stability = Rs. Forty 

Saving = Rs. 20 

If saving is = Rs. 20 

Salary = Rs. 100 

If saving is Rs. 450, 

revenue = Rs (100/20 x 450) = Rs. 2250.

Question 48. The Price Of An Article Has Been Reduced By 25 %. In Order To Restore The Original Price, The New Price Must Be Increased By?

Answer :

Let unique charge = Rs. 100. 

Reduced Price = Rs. 75. 

Increase on Rs. 75 = Rs. 25 

Increase on 100 = (25/75 x one hundred) % = 33 1/three %.

Question forty nine. The Length Of A Rectangle Is Increased By 60 %. By What Percent Would The Width Have To Be Decreased To Maintain The Same Area?

Answer :

Let length = one hundred m. 

Breath = 100 m. 

New duration = one hundred sixty m.

New breath = x meters 

Then, = one hundred sixty × x = 100 × 100 

(or) X = (one hundred ×a hundred)/160 × one hundred twenty five/2 

Decrease in breadth = (100- a hundred twenty five/2) % = 37 ½ %.

Question 50. Water Tax Is Increased By 20% But Its Consumption Is Decreased By 20 %. Then, The Increase Or Decrease In The Expenditure Of The Money Is?

Answer :

Let tax = Rs. 100 and intake = 100 gadgets 

Original Expenditure = Rs. (100 × one hundred) = Rs. 10000 

New Expenditure = Rs. (one hundred twenty × eighty) = Rs. 9600 

Decrease in expenditure = (400/10000 x 100) % = four %.

Question fifty one. On Decreasing The Price Of T.V. Sets By 30 % Its Sale Is Increased By 20 %. What Is The Effect On The Revenue Received By The Shopkeeper?

Answer :

Let price = Rs. A hundred, Sale = a hundred 

Then Sale Value = Rs. (a hundred × 100) = Rs. Ten thousand 

New Sale Value = Rs. (70 × one hundred twenty) = Rs. 8400 

Decrease % = (one thousand/10000 x one hundred) % = 16 %.

Question fifty two. A Man’s Basic Pay For A 40 Hour Week Is Rs. 20 Overtime Is Paid For At 25 % Above The Basic Rate, In A Certain Week He Worked Overtime And His Total Wage Was Rs. 25. He Therefore Worked For A Total Of?

Answer :

Basic price according to hour = Re. (20/forty) = Re. Half of. 

Overtime in line with hour = 125 % of Re. 1/2 = Re.Five/eight. 

He labored x hours beyond regular time. 

Then 20 + 5/eight x = 25 (or) = five/8 x = 5 

X = forty/five = eight hours 

=> he worked in concerned with (forty + 8) = forty eight hours.

Question 53. If 35% Of A Number Is 12 Less Than 50% Of That Number, Then The Number Is?

Answer :

Let the variety be x. Then,

50% of x - 35% of x = 12

50/one hundred x - 35/100 x = 12

x = (12 * a hundred)/15 = eighty.

Question 54. The Price Of An Article Is Cut By 10%. To Restore It To The Former Value, The New Price Must Be Increased By?

Answer :

Let the price of the item = Rs.100

New price = one hundred - 10 = 90

Therefore the brand new fee ought to be elevated by using(a hundred−ninety)×100/ninety=a hundred/9%=11 1/9%.

Question 55. The Income Of A Broker Remains Uncharged Though The Rate Of Commission Is Increased From four % To 5 % The Percentage Of A Slump In Business Is?

Answer :

Let the commercial enterprise price adjustments from x to y. 

Then 4 % of x = five % of y of 4/one hundred × x = 5/a hundred × y 

or y = four/five x. 

Change in enterprise = (x - 4/five x) = four/5 x. 

Percentage slump in business 

= (1/5 x ×1/x × 100)% = 20%.

Question fifty six. If 20% Of A Number, Then a hundred and twenty% Of That Number Will Be?

Answer :

Let the range x. Then,

20% of x = a hundred and twenty

x = (a hundred and twenty * one hundred)/20 = six hundred

one hundred twenty% of x = (one hundred twenty/100 * 600) = 720.

Question 57. Two-fifth Of One-1/3 Of Three-seventh Of A Number Is 15. What Is forty% Of That Number?

Answer :

Let the range be x. Then,

2/5 of 1/three of three/7 of 

x = 15 => x = (15 * 5/3 * 3 * five/2) = 525/2

forty% of 525/2 = (forty/one hundred * 525/2) = one hundred and five.

Question 58. The Difference Between A Number And Its Two-fifth Is 510. What Is 10% Of That Number?

Answer :

Let the variety be x. Then,

x - 2/5 x = 510

x = (510 * 5)/three = 850

10% of 850 = eighty five.




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