Intuit Aptitude Interview Questions & Answers

Question 1. Two Cars Cover The Same Distance At The Speed Of 60 And 64 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 * 16 = 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 = 125/3

T = 500 * 3/125 = 12 sec.

Question 3. Two Trains Of Equal Length, Running With The Speeds Of 60 And 40 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 * 5/18 = 100/18

T = 50

D = 50 * 100/18 = 2500/9

RS = 60 + 40 = 100 * 5/18

T = 2500/9 * 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 + 150 = 250

S = 36 * 5/18 = 10 mps

T = 250/10 = 25 sec.

Question 5. A Can Do A Piece Of Work In 30 Days. He Works At It For 5 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 = 13 1/3 days.

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 = 3:1

WT = 1:3

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

1/x – 1/3x = 1/10

x = 20/3

3/20 + 1/20 = 1/5 => 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.

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 done by A in 2 days = 2/5

Remaining work = 3/5

Work done by both A and B in one day = 1/5 + 1/10 = 3/10

Remaining work = 3/5 * 10/3 = 2 days.

Question 10. The Ratio Of Two Numbers Is 2:3 And The Sum Of Their Cubes Is 945. The Difference Of Number Is?

Answer :

2x.....3x

8x cube + 27x cube = 945

35x cube = 945

x cube = 27 => x = 3.

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 1:3. How Many People Were At The Party Originally?

Answer :

x, 2x

(x-2):(2x-2) = 1:3

3x-6 = 2x-2

x = 4

x+2x = 3x

=> 3*4 = 12.

Question 12. Rs.1170 Is Divided So That 4 Times The First Share, Thrice The 2nd 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 = 1/4:1/3:1/2 = 3:4:6

6/13 * 1170 = Rs.540.

Question 13. 4000 Was Divided Into Two Parts Such A Way That When First Part Was Invested At 3% And The Second At 5%, The Whole Annual Interest From Both The Investments Is?

Answer :

(x*3*1)/100 + [(4000 - x)*5*1]/100 = 144

3x/100 + 200 – 5x/100 = 144

2x/100 = 56 è x = 2800.

Question 14. The Equal Amounts Of Money Are Deposited In Two Banks Each At 15% Per Annum For 3.5 Years And 5 Years Respectively. If The Difference Between Their Interests Is Rs.144, Find The Each Sum?

Answer :

(P*5*15)/100 - (P*3.5*15)/100 = 144

75P/100 –  52.5P/100 = 144

22.5P = 144 * 100 

=> P = Rs.640.

Question 15. The Average Of 11 Numbers Is 10.9. If The Average Of First Six Is 10.5 And That Of The Last Six Is 11.4 The Sixth Number Is?

Answer :

1 to 11 = 11 * 10.9 = 119.9

1 to 6 = 6 * 10.5 = 63

6 to 11 = 6 * 11.4 = 68.4

63 + 68.4 = 131.4 – 119.9 = 11.5

6th number = 11.5.

Question 16. 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 + 1/5 = 9/20

 20/9 = 2 2/9

 9/20 * 2 = 9/10 ---- 4 hours

 WR = 1 - 9/10 = 1/10

 1 h ---- 1/4

 ? ----- 1/10

 2/5 * 60 = 24 = 4 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-third 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 + 1/3 = 4/3

1 --- 12

4/3 --- ?

4/3 * 12 = 16 hrs.

Question 18. A Property Decreases In Value Every Year At The Rate Of 6 1/4% 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 1/4% = 1/16

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

x = 25600.24.

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. (round Off Your Answer To Nearest Integer)

Answer :

CI = 14800{ [ 1 + 13.5/100]2 - 1 } 

= 14800 { [1 + 27/200]2 - 1 

= 14800 { 2 + 27/200}{27/200} 

= (74)[2 + 27/200](27) = 

1998[2 + 27/200] = 3996 + 269.73 = 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, 2nd And The 3rd Years Respectively. Find The Interest Received At The End Of The Three Years?

Answer :

Let A be the amount received at the end of the three years. 

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

A = (4800 * 11 * 6 * 5)/(10 * 5 * 4) 

A = Rs.7920 

So the interest = 7920 - 4800 = Rs.3120

Question 21. A Bank Offers 5% 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 * 41/40(41/40 + 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 gain

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*(80/100) = 168

SP = 840*(120/100) = 1008

1008 - 168 = 840

3/4 SP = 630

Gain = 210

630 --- 210

100 --- ? => 33 1/3%.

Question 24. A Can Give B 100 Meters Start And C 200 Meters Start In A Kilometer Race. How Much Start Can B Give C In A Kilometer Race?

Answer :

A runs 1000 m while B runs 900 m and C runs 800 m.

The number of meters that C runs when B runs 1000 m,

= (1000 * 800)/900 = 8000/9 = 888.88 m.

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

Question 25. A Can Run A Kilometer Race In 4 1/2 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 give B (5 min - 4 1/2 min) = 30 sec start.

The distance covered by B in 5 min = 1000 m.

Distance covered in 30 sec = (1000 * 30)/300 = 100 m.

A can give 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 scores 60 while B score 40 and C scores 30.

The number of points that C scores when B scores 100 = (100 * 30)/40 = 25 * 3 = 75.

In a game of 100 points, B gives (100 - 75) = 25 points to C.

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 mixture = 20% of 150 = 20/100 * 150 = 30 liters.

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

Total amount of water becomes (30 + P) and total volume of mixture is (150 + P).

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

120 + 4P = 150 + P => P = 10 liters.

Question 28. Two Varieties Of Wheat - A And B Costing Rs. 9 Per Kg And Rs. 15 Per Kg Were Mixed In The Ratio 3 : 7. If 5 Kg Of The Mixture Is Sold At 25% Profit, Find The Profit Made?

Answer :

Let the quantities of A and B mixed be 3x kg and 7x kg.

Cost of 3x kg of A = 9(3x) = Rs. 27x

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

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

Cost of 5 kg of the mixture = 132x/10x (5) = Rs. 66

Profit made in selling 5 kg of the mixture = 25/100 (cost of 5 kg of the mixture) = 25/100 * 66 = Rs. 16.50.

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 2nd from 1st, we get

-C + T = 100 => T - C = 100.

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 : 5/9.

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

Answer :

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

Their sum = 8x - 4 = 292 => x = 37

Smallest number 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 amount with R be Rs.r

r = 2/3 (total 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 select one boy from 20 boys in 20 ways.

We select one girl from 25 girls in 25 ways

We select a boy and girl in 20 * 25 ways i.e., = 500 ways.

Question 34. 1860 + 4/7 Of 21.21 - 41.4 = ?

Answer :

1860 + 4/7 of 21.21 - 41.4 = 1860 + 4(3.03) - 41.4

= 1860 + 12.12 - 41.4 = 1872.12 - 41.4

= 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)} / 1.4 = ?

Answer :

{20 - [7 - (3 -2)] + 1/3 (4.2)} / 1.4 

=> [20 - (7 - 1) + 1.4] / 1.4 

=> (20 - 6 + 1.4)/1.4 = 10 + 1 = 11.

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

Answer :

The fractions considered are 8/15 9/13 6/11

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

(8 * 13 * 11)/2145, (9 * 15 * 11)/2145 and (6 * 15 * 13)/2145

1144/2145, 1485/2145 and 1170/2145

so in descending order the fractions will be 

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

Question 38. A Train 125 M Long Passes A Man, Running At 5 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 man = (125/10) m/sec = (25/2) m/sec. [(25/2) * (18/5)] km/hr = 45 km/hr.

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

Then, relative speed = (x - 5) km/hr.

x - 5 = 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 two trains be x m/sec and y m/sec respectively.

Then, length of the first train = 27 x meters, and length of the second train = 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 40. A 300 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 = 50/3 m/sec.

Let the length of the platform be x meters.

Then, x + 300 / 39 = 50/3 

3(x + 300) = 1950 è x = 350m.

Question 41. A Train 110 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 train relative to man = (60 + 6) km/hr = 66 km/hr

[66 * 5/18] m/sec = [55/3] m/sec. 

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

Question 42. A Train Speeds Past A Pole In 15 Seconds And A Platform 100 M Long In 25 Seconds. Its Length Is?

Answer :

Let the length of the train be x meters and its speed be y m/sec.

They, x / y = 15 => y = x/15 

x + 100 / 25 = x / 15 

 x = 150 m. 

Question 43. A Train Crosses A Platform Of 120 M In 15 Sec, Same Train Crosses Another Platform Of Length 180 M In 18 Sec. Then Find The Length Of The Train?

Answer :

Length of the train be ‘X’

X + 120/15 = X + 180/18

6X + 720 = 5X + 900

X = 180m .

Question 44. If Rs.3250 Be Divided Among Ram, Shyam And Mohan In The Ratio Of 1/2:1/3:1/4 Then The Share Of Each Are?

Answer :

1/2:1/3:1/4 = 6:4:3

Ram = 6/13 * 3250 = 1500

Shyam = 4/13 * 3250 = 1000

Mohan = 3/13 * 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 5 Months And C Invested For Rs.10000 For 3 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 :

65 * 6 : 84 * 5 : 100 * 3

26:28:20

C share = 74000 * 95/100 = 7030 * 20/74 => 1900.

Question 46. 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 1/2 Months. If One Day, 3 Horses Eat As Much As 5 Cows And 6 Cows As Much As 10 Sheep, What Part Of The Rent Should A Pay?

Answer :

3h = 5c

6c = 10s

A = 21h*3 + 15c*2

    = 63h + 30c

    = 105c + 30c = 135c

 B = 15c*6 + 40s*7 1/2

     = 90c + 300s

     = 90c + 180c = 270c

 A:B = 135:270

            27:52

 A = 27/79 = 1/3.

Question 47. A Man Can Swim In Still Water At 4.5 Km/h, But Takes Twice As Long To Swim Upstream Than Downstream. The Speed Of The Stream Is?

Answer :

M = 4.5

S = x

DS = 4.5 + x

US = 4.5 + x

4.5 + x = (4.5 - x)2

4.5 + x = 9 -2x

3x = 4.5

x = 1.5.

Question 48. A Man Rows His Boat 85 Km Downstream And 45 Km Upstream, Taking 2 1/2 Hours Each Time. Find The Speed Of The Stream?

Answer :

Speed downstream = d/t = 85/(2 1/2) = 34 kmph

Speed upstream = d/t = 45/(2 1/2) = 18 kmph

The speed of the stream = (34 - 18)/2 = 8 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 2:1.

The ratio of the speed of the boat in still water to the speed of the stream = (2+1)/(2-1) = 3/1 = 3:1

Speed of the stream = 42/3 = 14 kmph.

Question 50. A Man Can Row At 5 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 = (5+1)km/hr = 6 km/hr

Speed upstream = (5-1)km/hr = 4 km/hr 

Let the required distance be x km. 

Then x/6 + x/4 = 75/60 = 5/4 => (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(75 + 50) = 250 m

250 * 1/2 = Rs.125.

Question 53. A Room Is 4 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 54. How Many Minimum Number's Of Whole Square Slabs Are Required For Paving The Floor 12.96 Meters Long And 3.84 Meters Side?

Answer :

HCF of 384, 1296 = 48

48 * 48 * 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 99 = 90 * 2      = 180

100 to 365 = 266 * 3 = 798

                                  -----------

                                     987

Question 56. 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 = 32.5

325/32.5 = 10.

Question 57. 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 five consecutive odd numbers of set p be 2n - 3, 2n - 1, 2n + 1, 2n + 3, 2n + 5.

Sum of these five numbers

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

= 10n + 5 = 435 => n = 43

Largest number of set p = 2(43) + 5 = 91

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

=> The five numbers of set q are 132, 133, 134, 135, 136.

Sum of above numbers = 132 + 133 + 134 + 135 + 136 = 670.

Question 58. 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 59. A Brick Measures 20 Cm * 10 Cm * 7.5 Cm How Many Bricks Will Be Required For A Wall 25 M * 2 M * 0.75 M?

Answer :

25 * 2 * 0.75 = 20/100 * 10/100 * 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: 4 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 

90, 75 , 60 cm.

Question 61. The Sum Of Three Consecutive Integers Is 102. Find The Lowest Of The Three?

Answer :

Three consecutive numbers can 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 62. 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 two digit number can be written as (10P + Q), where P is the digit in the tens place and Q is the digit in the units place.

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

(10Q + P) - (10P + Q) = 54

9(Q - P) = 54

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

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

The required number 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 = 40.

5 years, their ages will be 45 and 25.

Sum of their ages = 45 + 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 :

100 * 100 = 10000

80 * 110 = 8800

10000------- 1200

100 ------- ? = 12%.

Question 65. 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%

62% - 38% = 24%

24% -------- 288

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

Question 66. Find The 37.5% Of 976 =

Answer :

   37.5 % of 976 

= 37.5/100 * 976 = 375/1000 * 976 = 3/8 * 976

= 3 * 122 = 366.

Question 67. 64 Is What Percent Of 80?

Answer :

Let x percent of 80 be 64.

80 * x/100 = 64 => x = (64 * 100)/80 => x = 80.

80% of 80 is 64.

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

Answer :

Let the required percentage be x%.

3/20 = x% of 12/25 => 3/20 = x/100 * 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 - 9 = 0

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

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

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

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

Answer :

The triangle with sides 26 cm, 24 cm and 10 cm is right angled, where the hypotenuse is 26 cm.

Area of the triangle = 1/2 * 24 * 10 = 120 cm².

Question 71. A Wire In The Form Of A Circle Of Radius 3.5 M Is Bent In The Form Of A Rectangule, Whose Length And Breadth Are In The Ratio Of 6 : 5. 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 * 3.5 

 =>  x = 1 

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

Question 72. 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 * 3 Feet And Three Windows Of Dimensions 4 Feet * 3 Feet Each?

Answer :

Area of the four walls = 2h(l + b)

Since there are doors and windows, area of the walls = 2 * 12 (15 + 25) - (6 * 3) - 3(4 * 3) = 906 sq.ft.

Total cost = 906 * 5 = Rs. 4530.

Question 73. 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 number of exhaustive outcomes is 36.

Let E be the event of getting an even number on one die and an odd number on the other.

let the event of getting either both even or both odd then  = 18/36 = 1/2

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

Question 74. The Probability That A Speaks Truth Is 3/5 And That Of B Speaking Truth Is 4/7. What Is The Probability That They Agree In Stating The Same Fact?

Answer :

If both agree stating the same fact, either both of them speak truth of both speak false.

Probability = 3/5 * 4/7 + 2/5 * 3/7

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