## Top 100+ Intuit Aptitude Interview Questions And Answers

Question 1. Two Cars Cover The Same Distance At The Speed Of 60 And sixty four Kmps Respectively. Find The Distance Traveled By Them If The Slower Car Takes 1 Hour More Than The Faster Car?

Answer :

60(x + 1) = 64x

X = 15

60 * sixteen = 960 km.

Question 2. Two Trains Each 250 M In Length Are Running On The Same Parallel Lines In Opposite Directions With The Speed Of 80 Kmph And 70 Kmph Respectively. In What Time Will They Cross Each Other Completely?

Answer :

D = 250 m + 250 m = 500 m

RS = 80 + 70 = 150 * 5/18 = a hundred twenty five/3

T = 500 * three/a hundred twenty five = 12 sec.

Adv Java Interview Questions

Question 3. Two Trains Of Equal Length, Running With The Speeds Of 60 And forty Kmph, Take 50 Seconds To Cross Each Other While They Are Running In The Same Direction. What Time Will They Take To Cross Each Other If They Are Running In Opposite Directions?

Answer :

RS = 60 -40 = 20 * five/18 = one hundred/18

T = 50

D = 50 * a hundred/18 = 2500/nine

RS = 60 + 40 = a hundred * 5/18

T = 2500/nine * 18/500 = 10 sec.

Question 4. How Many Seconds Will A Train 100 Meters Long Take To Cross A Bridge 150 Meters Long If The Speed Of The Train Is 36 Kmph?

Answer :

D = 100 + one hundred fifty = 250

S = 36 * 5/18 = 10 mps

T = 250/10 = 25 sec.

Adv Java Tutorial

Question five. A Can Do A Piece Of Work In 30 Days. He Works At It For five Days And Then B Finishes It In 20 Days. In What Time Can A And B Together It?

Answer :

5/30 + 20/x = 1

x = 24

1/30 + 1/24 = 3/40

40/3 = thirteen 1/three days.

Aptitude Interview Questions

Question 6. A Is Thrice As Efficient As B And Is, Therefore, Able To Finish A Piece Of Work 10 Days Earlier Than B. In How Many Days A And B Will Finish It Together?

Answer :

WC = three:1

WT = 1:3

x ............3x

1/x – 1/3x = 1/10

x = 20/3

three/20 + 1/20 = 1/five => 5 days.

Question 7. 9 Men And 12 Boys Finish A Job In 12 Days, 12 Men And 12 Boys Finish It In 10 Days. 10 Men And 10 Boys Shall Finish It In How Many Days?

Answer :

9M + 12B ----- 12 days

12M + 12B ------- 10 days

10M + 10B -------?

108M + 144B = 120M +120B

24B = 12M => 1M = 2B

18B + 12B = 30B ---- 12 days

20B + 10B = 30B -----? => 12 days.

Advanced Dot Net Interview Questions

Question 8. A Can Do A Piece Of Work In 12 Days. He Worked For 15 Days And Then B Completed The Remaining Work In 10 Days. Both Of Them Together Will Finish It In?

Answer :

15/25 + 10/x = 1 => x = 25

1/25 + 1/25 = 2/25

25/2 = 12 1/2 days.

Question 9. A And B Can Do A Work In 5 Days And 10 Days Respectively. A Starts The Work And B Joins Him After 2 Days. In How Many Days Can They Complete The Remaining Work?

Answer :

Work performed by means of A in 2 days = 2/5

Remaining work = three/5

Work executed by means of both A and B in in the future = 1/5 + 1/10 = 3/10

Remaining paintings = three/five * 10/3 = 2 days.

C and C++ Interview Questions

Question 10. The Ratio Of Two Numbers Is 2:three And The Sum Of Their Cubes Is 945. The Difference Of Number Is?

Answer :

2x.....3x

8x dice + 27x dice = 945

35x dice = 945

x dice = 27 => x = three.

Question 11. The Ratio Of The Number Of Ladies To Gents At A Party Was 1:2 But When 2 Ladies And 2 Gents Left, The Ratio Became 1:three. How Many People Were At The Party Originally?

Answer :

x, 2x

(x-2):(2x-2) = 1:three

3x-6 = 2x-2

x = 4

x+2x = 3x

=> 3*four = 12.

HR Interview Questions

Question 12. Rs.1170 Is Divided So That four Times The First Share, Thrice The 2d Share And Twice The Third Share Amount To The Same. What Is The Value Of The Third Share?

Answer :

A+B+C = 1170

4A = 3B = 2C = x

A:B:C = 1/4:1/3:half of = 3:4:6

6/13 * 1170 = Rs.540.

Adv Java Interview Questions

Question thirteen. 4000 Was Divided Into Two Parts Such A Way That When First Part Was Invested At 3% And The Second At five%, The Whole Annual Interest From Both The Investments Is?

Answer :

(x*three*1)/100 + [(4000 - x)*5*1]/100 = 144

3x/a hundred + 200 – 5x/100 = one hundred forty four

2x/one hundred = fifty six è x = 2800.

Question 14. The Equal Amounts Of Money Are Deposited In Two Banks Each At 15% Per Annum For three.5 Years And five Years Respectively. If The Difference Between Their Interests Is Rs.One hundred forty four, Find The Each Sum?

Answer :

(P*5*15)/one hundred - (P*3.Five*15)/100 = a hundred and forty four

75P/100 – 52.5P/one hundred = one hundred forty four

22.5P = 144 * one hundred

=> P = Rs.640.

Question 15. The Average Of 11 Numbers Is 10.Nine. If The Average Of First Six Is 10.5 And That Of The Last Six Is eleven.4 The Sixth Number Is?

Answer :

1 to eleven = 11 * 10.Nine = 119.9

1 to 6 = 6 * 10.Five = sixty three

6 to eleven = 6 * eleven.Four = 68.Four

63 + sixty eight.Four = 131.Four – 119.Nine = eleven.Five

6th range = 11.Five.

VMware Aptitude Interview Questions

Question sixteen. Two Pipes A And B Can Fill A Tank In 4 And 5 Hours Respectively. If They Are Turned Up Alternately For One Hour Each, The Time Taken To Fill The Tank Is?

Answer :

1/4 + 1/five = nine/20

20/nine = 2 2/nine

9/20 * 2 = nine/10 ---- four hours

WR = 1 - nine/10 = 1/10

1 h ---- 1/four

? ----- 1/10

2/five * 60 = 24 = four hrs 24 min.

Question 17. Two Pipes Can Separately Fill A Tank In 20 And 30 Hours Respectively. Both The Pipes Are Opened To Fill The Tank But When The Tank Is Full, A Leak Develops In The Tank Through Which One-0.33 Of Water Supplied By Both The Pipes Goes Out. What Is The Total Time Taken To Fill The Tank?

Answer :

1/20 + 1/30 = 1/12

1 + 1/three = four/three

1 --- 12

four/3 --- ?

4/three * 12 = 16 hrs.

SAP Aptitude Interview Questions

Question 18. A Property Decreases In Value Every Year At The Rate Of 6 1/four% Of Its Value At The Beginning Of The Year Its Value At The End Of 3 Years Was Rs.21093. Find Its Value At The Beginning Of The First Year?

Answer :

6 1/four% = 1/16

x *15/sixteen * 15/16 * 15/sixteen = 21093

x = 25600.24.

Aptitude Interview Questions

Question 19. Find The Compound Interest Accrued On An Amount Of Rs.14,800 At 13.5% P.A At The End Of Two Years. (spherical Off Your Answer To Nearest Integer)

Answer :

CI = 14800 [ 1 + 13.5/100]2 - 1

= 14800 [1 + 27/200]2 - 1

= 14800 2 + 27/20027/2 hundred

= (seventy four)[2 + 27/200](27) =

1998[2 + 27/200] = 3996 + 269.Seventy three = Rs. 4266.

Question 20. A Sum Of Rs.4800 Is Invested At A Compound Interest For Three Years, The Rate Of Interest Being 10% P.A., 20% P.A. And 25% P.A. For The 1st, second And The 3rd Years Respectively. Find The Interest Received At The End Of The Three Years?

Answer :

Let A be the quantity acquired at the stop of the 3 years.

A = 4800[1 + 10/100][1 + 20/100][1 + 25/100]

A = (4800 * eleven * 6 * 5)/(10 * 5 * 4)

A = Rs.7920

So the hobby = 7920 - 4800 = Rs.3120

Oracle Aptitude Interview Questions

Question 21. A Bank Offers five% C.I. Calculated On Half-yearly Basis . A Customer Deposits Rs. 1600 Each On 1st January And 1st July Of A Year. At The End Of The Year, The Amount He Would Have Gained By Way Of Interest Is?

Answer :

Amount = [1600 * (1 + 5/(2 * 100)2 + 1600 * (1 + 5/(2 * 100)]

= [1600 * forty one/40(forty one/forty + 1)

= [(1600 * 41 * 81)/(40 * 40)] = Rs. 3321.

C.I. = 3321 - 3200 = Rs. 121.

Question 22. By Selling 50 Meters Of Cloth. I Gain The Selling Price Of 15 Meters. Find The Gain Percent?

Answer :

SP = CP + g

50 SP = 50 CP + 15 SP

35 SP = 50 CP

35 --- 15 CP benefit

100 --- ? => 42 6/7%.

Question 23. If Goods Be Purchased For Rs.840 And One-fourth Be Sold At A Loss Of 20% At What Gain Percent Should The Remainder Be Sold So As To Gain 20% On The Whole Transaction?

Answer :

1/4 CP = 210 SP = 21*(80/one hundred) = 168

SP = 840*(120/100) = 1008

1008 - 168 = 840

three/four SP = 630

Gain = 210

630 --- 210

a hundred --- ? => 33 1/three%.

TCS Aptitude Interview Questions

Question 24. A Can Give B one hundred Meters Start And C two hundred Meters Start In A Kilometer Race. How Much Start Can B Give C In A Kilometer Race?

Answer :

A runs 1000 m at the same time as B runs 900 m and C runs 800 m.

The variety of meters that C runs while B runs a thousand m,

= (a thousand * 800)/900 = 8000/nine = 888.88 m.

B can deliver C = 1000 - 888.88 = 111.12 m.

Advanced Dot Net Interview Questions

Question 25. A Can Run A Kilometer Race In four half Min While B Can Run Same Race In 5 Min. How Many Meters Start Can A Give B In A Kilometer Race, So That The Race Mat End In A Dead Heat?

Answer :

A can deliver B (5 min - four 1/2 min) = 30 sec start.

The distance covered via B in five min = one thousand m.

Distance covered in 30 sec = (a thousand * 30)/three hundred = one hundred m.

A can provide B 100m start.

Question 26. In A Game Of Billiards, A Can Give B 20 Points In 60 And He Can Give C 30 Points In 60. How Many Points Can B Give C In A Game Of 100?

Answer :

A ratings 60 whilst B rating 40 and C scores 30.

The variety of factors that C rankings when B scores a hundred = (a hundred * 30)/forty = 25 * 3 = 75.

In a game of one hundred points, B offers (one hundred - 75) = 25 points to C.

Infosys Aptitude Interview Questions

Question 27. A Mixture Of 150 Liters Of Wine And Water Contains 20% Water. How Much More Water Should Be Added So That Water Becomes 25% Of The New Mixture?

Answer :

Number of liters of water in150 liters of the combination = 20% of 150 = 20/one hundred * one hundred fifty = 30 liters.

P liters of water added to the mixture to make water 25% of the brand new aggregate.

Total amount of water will become (30 + P) and overall extent of combination is (one hundred fifty + P).

(30 + P) = 25/100 * (150 + P)

a hundred and twenty + 4P = 150 + P => P = 10 liters.

C and C++ Interview Questions

Question 28. Two Varieties Of Wheat - A And B Costing Rs. Nine Per Kg And Rs. 15 Per Kg Were Mixed In The Ratio three : 7. If five Kg Of The Mixture Is Sold At 25% Profit, Find The Profit Made?

Answer :

Let the portions of A and B blended be 3x kg and 7x kg.

Cost of 3x kg of A = nine(3x) = Rs. 27x

Cost of 7x kg of B = 15(7x) = Rs. 105x

Cost of 10x kg of the combination = 27x + 105x = Rs. 132x

Cost of five kg of the combination = 132x/10x (five) = Rs. Sixty six

Profit made in promoting 5 kg of the aggregate = 25/a hundred (price of 5 kg of the mixture) = 25/one hundred * 66 = Rs. Sixteen.50.

Question 29. The Cost Of 2 Chairs And 3 Tables Is Rs.1300. The Cost Of 3 Chairs And 2 Tables Is Rs.1200. The Cost Of Each Table Is More Than That Of Each Chair By?

Answer :

2C + 3T = 1300 --- (1)

3C + 3T = 1200 --- (2)

Subtracting second from 1st, we get

-C + T = one hundred => T - C = a hundred.

IBM Aptitude Interview Questions

Question 30. The Denominator Of A Fraction Is 1 Less Than Twice The Numerator. If The Numerator And Denominator Are Both Increased By 1, The Fraction Becomes 3/5. Find The Fraction?

Answer :

Let the numerator and denominator of the fraction be 'n' and 'd' respectively.

D = 2n - 1

(n + 1)/(d + 1) = 3/5

5n + 5 = 3d + 3

5n + 5 = 3(2n - 1) + 3 => n = 5

d = 2n - 1 => d = 9

Hence the fraction is : 5/nine.

Question 31. The Sum Of Four Consecutive Even Numbers Is 292. What Would Be The Smallest Number?

Answer :

Let the 4 consecutive even numbers be 2(x - 2), 2(x - 1), 2x, 2(x + 1)

Their sum = 8x - four = 292 => x = 37

Smallest range is: 2(x - 2) = 70.

Question 32. P, Q And R Have Rs.6000 Among Themselves. R Has Two-thirds Of The Total Amount With P And Q. Find The Amount With R?

Answer :

Let the quantity with R be Rs.R

r = 2/3 (overall amount with P and Q)

r = 2/3(6000 - r) => 3r = 12000 - 2r

=> 5r = 12000 => r = 2400.

Question 33. In A Class There Are 20 Boys And 25 Girls. In How Many Ways Can A Boy And A Girl Be Selected?

Answer :

We can pick out one boy from 20 boys in 20 methods.

We pick out one female from 25 women in 25 ways

We pick out a boy and girl in 20 * 25 ways i.E., = 500 methods.

HR Interview Questions

Question 34. 1860 + four/7 Of 21.21 - forty one.Four = ?

Answer :

1860 + 4/7 of 21.21 - forty one.Four = 1860 + four(3.03) - forty one.4

= 1860 + 12.12 - 41.4 = 1872.12 - forty one.Four

= 1830.72.

Question 35. [(523 + 27) * (187 - 35) / (424 + 16)] / [110/22 Of (2 * 38)] = ?

Answer :

[(523 + 27) * (187 - 35) / (424 + 16)] / [110/22 of (2 * 38)]

= [(550 * 152) / 440] / [(110/44) * 38]

= (5 * 38)/(5/2 * 38) = 2.

Question 36. 20 - [7 - (3 -2)] + 1/3 (4.2) / 1.Four = ?

Answer :

20 - [7 - (3 -2)] + 1/three (4.2) / 1.Four

=> [20 - (7 - 1) + 1.4] / 1.4

=> (20 - 6 + 1.Four)/1.Four = 10 + 1 = 11.

VMware Aptitude Interview Questions

Question 37. Which Of The Following Groups Of Fractions Is In Descending Order?

Answer :

The fractions taken into consideration are eight/15 9/13 6/eleven

To compare them we make the denominators the same. So the fractions are

(8 * thirteen * 11)/2145, (nine * 15 * 11)/2145 and (6 * 15 * thirteen)/2145

1144/2145, 1485/2145 and 1170/2145

so in descending order the fractions can be

1485/2145, 1170/2145 and 1144/2145 i.E., 9/13 , 6/11 , 8/15.

Question 38. A Train a 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 Seconds. The Speed Of The Train Is?

Answer :

Speed of the train relative to guy = (a hundred twenty five/10) m/sec = (25/2) m/sec. [(25/2) * (18/5)] km/hr = 45 km/hr.

Let the velocity of the train be x km/hr.

Then, relative velocity = (x - 5) km/hr.

X - five = 45 ==> x = 50 km/hr.

Question 39. Two Trains Running In Opposite Directions Cross A Man Standing On The Platform In 27 Seconds And 17 Seconds Respectively And They Cross Each Other In 23 Seconds. The Ratio Of Their Speeds Is?

Answer :

Let the speeds of the 2 trains be x m/sec and y m/sec respectively.

Then, period of the primary train = 27 x meters, and duration of the second one teach = 17 y meters.

(27 x + 17 y) / (x + y) = 23 ==> 27 x + 17 y = 23 x + 23 y ==> 4 x = 6 y ==> x/y = 3/2.

Question forty. A three hundred Meter Long Train Crosses A Platform In 39 Seconds While It Crosses A Signal Pole In 18 Seconds. What Is The Length Of The Platform?

Answer :

Speed = [300 / 18] m/sec = 50/three m/sec.

Let the duration of the platform be x meters.

Then, x + three hundred / 39 = 50/three

three(x + 300) = 1950 è x = 350m.

SAP Aptitude Interview Questions

Question 41. A Train a hundred and ten Meters Long Is Running With A Speed Of 60 Kmph. In What Time Will It Pass A Man Who Is Running At 6 Kmph In The Direction Opposite To That In Which The Train Is Going?

Answer :

Speed of teach relative to man = (60 + 6) km/hr = 66 km/hr

[66 * 5/18] m/sec = [55/3] m/sec.

Time taken to pass the person = [110 * 3/55] sec = 6 sec.

Question 42. A Train Speeds Past A Pole In 15 Seconds And A Platform one hundred M Long In 25 Seconds. Its Length Is?

Answer :

Let the duration of the teach be x meters and its pace be y m/sec.

They, x / y = 15 => y = x/15

x + a hundred / 25 = x / 15

x = one hundred fifty m.

Oracle Aptitude Interview Questions

Question 43. A Train Crosses A Platform Of one hundred twenty M In 15 Sec, Same Train Crosses Another Platform Of Length one hundred eighty M In 18 Sec. Then Find The Length Of The Train?

Answer :

Length of the train be ‘X’

X + 120/15 = X + a hundred and eighty/18

6X + 720 = 5X + 900

X = 180m .

Question 44. If Rs.3250 Be Divided Among Ram, Shyam And Mohan In The Ratio Of half of:1/3:1/4 Then The Share Of Each Are?

Answer :

half of:1/3:1/4 = 6:four:3

Ram = 6/13 * 3250 = 1500

Shyam = 4/13 * 3250 = a thousand

Mohan = 3/thirteen * 3250 = 750.

Question 45. A, B And C Are Entered Into A Partnership. A Invested Rs.6500 For 6 Months, B Invested Rs.8400 For five Months And C Invested For Rs.10000 For three Months. A Is A Working Partner And Gets 5% Of The Total Profit For The Same. Find The Share Of C In A Total Profit Of Rs.7400?

Answer :

sixty five * 6 : 84 * 5 : 100 * 3

26:28:20

C share = 74000 * 95/one hundred = 7030 * 20/seventy four => 1900.

Question forty six. Two Persons A And B Take A Field On Rent. A Puts On It 21 Horses For 3 Months And 15 Cows For 2 Months; B Puts 15 Cows For 6months And 40 Sheep For 7 half Months. If One Day, 3 Horses Eat As Much As five Cows And 6 Cows As Much As 10 Sheep, What Part Of The Rent Should A Pay?

Answer :

3h = 5c

6c = 10s

A = 21h*three + 15c*2

= 63h + 30c

= 105c + 30c = 135c

B = 15c*6 + 40s*7 half

= 90c + 300s

= 90c + 180c = 270c

A:B = a hundred thirty five:270

27:fifty two

A = 27/79 = 1/three.

Question forty seven. A Man Can Swim In Still Water At four.Five Km/h, But Takes Twice As Long To Swim Upstream Than Downstream. The Speed Of The Stream Is?

Answer :

M = 4.Five

S = x

DS = four.Five + x

US = 4.5 + x

four.Five + x = (four.5 - x)2

4.Five + x = nine -2x

3x = 4.Five

x = 1.5.

Question 48. A Man Rows His Boat eighty five Km Downstream And forty five Km Upstream, Taking 2 1/2 Hours Each Time. Find The Speed Of The Stream?

Answer :

Speed downstream = d/t = eighty five/(2 1/2) = 34 kmph

Speed upstream = d/t = forty five/(2 1/2) = 18 kmph

The speed of the movement = (34 - 18)/2 = eight kmph.

Question 49. The Time Taken By A Man To Row His Boat Upstream Is Twice The Time Taken By Him To Row The Same Distance Downstream. If The Speed Of The Boat In Still Water Is 42 Kmph, Find The Speed Of The Stream?

Answer :

The ratio of the times taken is two:1.

The ratio of the rate of the boat in nevertheless water to the rate of the flow = (2+1)/(2-1) = 3/1 = 3:1

Speed of the flow = 42/three = 14 kmph.

Question 50. A Man Can Row At five Km/hr In Still Water, If The River Is Running At 1 Km/hr It Takes Him 75 Minutes To Row To A Place And Back. How Far Is The Place?

Answer :

Speed downstream = (five+1)km/hr = 6 km/hr

Speed upstream = (five-1)km/hr = 4 km/hr

Let the required distance be x km.

Then x/6 + x/four = 75/60 = 5/four => (2x + 3x) =15 => x =3.

Required distance = 3 km.

Question 51. A Rectangular Field Has Area Equal To 150 Sq M And Perimeter 50 M. Its Length And Breadth Must Be?

Answer :

2(l + b) = 50 => l + b = 25

l – b = 5

l = 15 b = 10.

Question 52. Sides Of A Rectangular Park Are In The Ratio 3: 2 And Its Area Is 3750 Sq M, The Cost Of Fencing It At 50 Ps Per Meter Is?

Answer :

3x * 2x = 3750 => x = 25

2(seventy five + 50) = 250 m

250 * half of = Rs.One hundred twenty five.

Question fifty three. A Room Is four Meters 37 Cm Long And 3 Meters 23cm Broad. It Is Required To Pave The Floor With Minimum Square Slabs. Find The Number Of Slabs Required For This Purpose?

Answer :

HCF of 323, 437 = 19

323 * 437 = 19 * 19 * x

x = 391.

Question fifty four. How Many Minimum Number's Of Whole Square Slabs Are Required For Paving The Floor 12.96 Meters Long And three.Eighty four Meters Side?

Answer :

HCF of 384, 1296 = 48

forty eight * forty eight * x = 384 * 1296

x = 216.

Question 55. How Many Figures Are Required To Number The Pages The Pages Of A Book Containing 365 Pages?

Answer :

1 to 9 = 9 * 1 = 9

10 to ninety nine = ninety * 2 = a hundred and eighty

100 to 365 = 266 * 3 = 798

-----------

987

Question fifty six. A Laborer Is Engaged For 30 Days On The Condition That He Receives Rs.25 For Each Day He Works And Is Fined Rs.7.50 For Each Day Is Absent. He Gets Rs.425 In All. For How Many Days Was He Absent?

Answer :

30 * 25 = 750

425

-----------

325

25 + 7.50 = 32.Five

325/32.Five = 10.

Question fifty seven. The Sum Of Five Consecutive Odd Numbers Of Set P Is 435. What Is The Sum Of Five Consecutive Numbers Of Another Set Q. Whose Largest Number Is 45 More Than The Largest Number Of Set P?

Answer :

Let the 5 consecutive odd numbers of set p be 2n - 3, 2n - 1, 2n + 1, 2n + three, 2n + five.

Sum of these five numbers

= 2n - 3 + 2n - 1 + 2n + 1 + 2n + 3 + 2n + 5

= 10n + 5 = 435 => n = forty three

Largest number of set p = 2(43) + five = 91

The largest quantity of set q = 91 + 45 = 136

=> The five numbers of set q are 132, 133, 134, one hundred thirty five, 136.

Sum of above numbers = 132 + 133 + 134 + a hundred thirty five + 136 = 670.

Question fifty eight. The Radius Of A Cylindrical Vessel Is 7cm And Height Is 3cm. Find The Whole Surface Of The Cylinder?

Answer :

r = 7 h = 3

2πr(h + r) = 2 * 22/7 * 7(10) = 440.

Question fifty nine. A Brick Measures 20 Cm * 10 Cm * 7.5 Cm How Many Bricks Will Be Required For A Wall 25 M * 2 M * zero.75 M?

Answer :

25 * 2 * zero.Seventy five = 20/100 * 10/a hundred * 7.5/100 * x

25 = 1/100 * x => x = 25000.

Question 60. If The Length, Breadth And The Height Of A Cuboid Are In The Ratio 6: 5: 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, seventy five , 60 cm.

Question sixty one. The Sum Of Three Consecutive Integers Is 102. Find The Lowest Of The Three?

Answer :

Three consecutive numbers may be taken as (P - 1), P, (P + 1).

So, (P - 1) + P + (P + 1) = 102

3P = 102 => P = 34.

The lowest of the three = (P - 1) = 34 - 1 = 33.

Question sixty two. The Sum Of The Two Digits Of A Number Is 10. If The Number Is Subtracted From The Number Obtained By Reversing Its Digits, The Result Is 54. Find The Number?

Answer :

Any digit quantity may be written as (10P + Q), wherein P is the digit within the tens location and Q is the digit inside the devices vicinity.

P + Q = 10 ----- (1)

(10Q + P) - (10P + Q) = fifty four

nine(Q - P) = 54

(Q - P) = 6 ----- (2)

Solve (1) and (2) P = 2 and Q = 8

The required range is = 28

Question 63. The Sum Of The Present Ages Of Two Persons A And B Is 60. If The Age Of A Is Twice That Of B, Find The Sum Of Their Ages 5 Years Hence?

Answer :

A + B = 60, A = 2B

2B + B = 60 => B = 20 then A = forty.

Five years, their a long time may be 45 and 25.

Sum of their ages = forty five + 25 = 70.

Question 64. The Tax On A Commodity Is Diminished By 20% But Its Consumption Is Increased By 10%. Find The Decrease Percent In The Revenue Derived From It?

Answer :

a hundred * a hundred = 10000

eighty * one hundred ten = 8800

10000------- 1200

a hundred ------- ? = 12%.

Question sixty five. There Were Two Candidates In An Election. Winner Candidate Received 62% Of Votes And Won The Election By 288 Votes. Find The Number Of Votes Casted To The Winning Candidate?

Answer :

W = 62% L = 38%

sixty two% - 38% = 24%

24% -------- 288

62% -------- ? => 744.

Question 66. Find The 37.5% Of 976 =

Answer :

37.Five % of 976

= 37.5/100 * 976 = 375/a thousand * 976 = three/eight * 976

= three * 122 = 366.

Question 67. 64 Is What Percent Of eighty?

Answer :

Let x percentage of 80 be 64.

Eighty * x/100 = sixty four => x = (sixty four * one hundred)/80 => x = eighty.

Eighty% of 80 is 64.

Question 68. 3/20 Is What Percent Of 12/25?

Answer :

Let the specified percentage be x%.

Three/20 = x% of 12/25 => three/20 = x/one hundred * 12/25

=> 12x = (300 * 25)/20 => x = (25 * 25)/20

=> x = 31.25%.

Question 69. Find The Roots Of The Quadratic Equation: 2x2 + 3x - 9 = 0?

Answer :

2x2 + 6x - 3x - nine = 0

2x(x + 3) - three(x + 3) = 0

(x + 3)(2x - 3) = zero

=> x = -3 or x = three/2.

Question 70. If The Sides Of A Triangle Are 26 Cm, 24 Cm And 10 Cm, What Is Its Area?

Answer :

The triangle with aspects 26 cm, 24 cm and 10 cm is right angled, in which the hypotenuse is 26 cm.

Area of the triangle = half * 24 * 10 = a hundred and twenty cm2.

Question seventy one. A Wire In The Form Of A Circle Of Radius three.5 M Is Bent In The Form Of A Rectangule, Whose Length And Breadth Are In The Ratio Of 6 : five. What Is The Area Of The Rectangle?

Answer :

The circumference of the circle is equal to the permeter of the rectangle.

Let l = 6x and b = 5x 2(6x + 5x) = 2 * 22/7 * three.Five

=> x = 1

Therefore l = 6 cm and b = 5 cm Area of the rectangle = 6 * 5 = 30 cm2.

Question seventy two. The Dimensions Of A Room Are 25 Feet * 15 Feet * 12 Feet. What Is The Cost Of White Washing The Four Walls Of The Room At Rs. 5 Per Square Feet If There Is One Door Of Dimensions 6 Feet * three Feet And Three Windows Of Dimensions four Feet * 3 Feet Each?

Answer :

Area of the 4 partitions = 2h(l + b)

Since there are doorways and home windows, place of the partitions = 2 * 12 (15 + 25) - (6 * three) - 3(4 * 3) = 906 sq.Toes.

Total cost = 906 * 5 = Rs. 4530.

Question seventy three. If Two Dice Are Thrown Together, The Probability Of Getting An Even Number On One Die And An Odd Number On The Other Is -.

Answer :

The wide variety of exhaustive outcomes is 36.

Let E be the occasion of having a fair quantity on one die and an peculiar quantity on the opposite.

Allow the event of having both each even or both extraordinary then = 18/36 = half of

P(E) = 1 - 1/2 = half of.

Question 74. The Probability That A Speaks Truth Is 3/five And That Of B Speaking Truth Is four/7. What Is The Probability That They Agree In Stating The Same Fact?

Answer :

If both agree stating the identical reality, either both of them talk fact of both talk false.

Probability = three/five * 4/7 + 2/5 * three/7

= 12/35 + 6/35 = 18/35.