29 + Interview Questions in Sap Aptitude Interview Questions in Sap Aptitude Tutorial Page 1 InterviewSolution

1.	A Man Has Some Hens And Cows. If The Number Of Heads Be 48 And The Number Of Feet Equals 140, Then The Number Of Hens Will Be?
Answer» LET the number of HENS be x and the number of cows be y. Then, x + y = 48 .... (i) and 2x + 4Y = 140 x + 2y = 70 .... (II) Solving (i) and (ii) we get: x = 26, y = 22. The required answer = 26. Let the number of hens be x and the number of cows be y. Then, x + y = 48 .... (i) and 2x + 4y = 140 x + 2y = 70 .... (ii) Solving (i) and (ii) we get: x = 26, y = 22. The required answer = 26.

1.

A Man Has Some Hens And Cows. If The Number Of Heads Be 48 And The Number Of Feet Equals 140, Then The Number Of Hens Will Be?

Answer»

LET the number of HENS be x and the number of cows be y.

Then, x + y = 48 .... (i)

and 2x + 4Y = 140 x + 2y = 70 .... (II)

Solving (i) and (ii) we get: x = 26, y = 22.

The required answer = 26.

Let the number of hens be x and the number of cows be y.

Then, x + y = 48 .... (i)

and 2x + 4y = 140 x + 2y = 70 .... (ii)

Solving (i) and (ii) we get: x = 26, y = 22.

The required answer = 26.

Discussion

2.	There Are Two Examinations Rooms A And B. If 10 Students Are Sent From A To B, Then The Number Of Students In Each Room Is The Same. If 20 Candidates Are Sent From B To A, Then The Number Of Students In A Is Double The Number Of Students In B. The Number Of Students In Room A Is?
Answer» Let the number of STUDENTS in ROOMS A and B be x and y respectively. Then, x - 10 = y + 10 x - y = 20 .... (i) and x + 20 = 2(y - 20) x - 2y = -60 .... (ii) Solving (i) and (ii) we get: x = 100 , y = 80. The required answer A = 100. Let the number of students in rooms A and B be x and y respectively. Then, x - 10 = y + 10 x - y = 20 .... (i) and x + 20 = 2(y - 20) x - 2y = -60 .... (ii) Solving (i) and (ii) we get: x = 100 , y = 80. The required answer A = 100.

2.

There Are Two Examinations Rooms A And B. If 10 Students Are Sent From A To B, Then The Number Of Students In Each Room Is The Same. If 20 Candidates Are Sent From B To A, Then The Number Of Students In A Is Double The Number Of Students In B. The Number Of Students In Room A Is?

Answer»

Let the number of STUDENTS in ROOMS A and B be x and y respectively.

Then, x - 10 = y + 10 x - y = 20 .... (i)

and x + 20 = 2(y - 20) x - 2y = -60 .... (ii)

Solving (i) and (ii) we get: x = 100 , y = 80.

The required answer A = 100.

Let the number of students in rooms A and B be x and y respectively.

Then, x - 10 = y + 10 x - y = 20 .... (i)

and x + 20 = 2(y - 20) x - 2y = -60 .... (ii)

Solving (i) and (ii) we get: x = 100 , y = 80.

The required answer A = 100.

Discussion

3.	What Was The Aggregate Of Marks Obtained By Sajal In All The Six Subjects?
Answer» AGGREGATE marks OBTAINED by Sajal = [ (90% of 150) + (60% of 130) + (70% of 120) + (70% of 100) + (90% of 60) + (70% of 40) ] = [ 135 + 78 + 84 + 70 + 54 + 28 ] = 449. Aggregate marks obtained by Sajal = [ (90% of 150) + (60% of 130) + (70% of 120) + (70% of 100) + (90% of 60) + (70% of 40) ] = [ 135 + 78 + 84 + 70 + 54 + 28 ] = 449.

3.

What Was The Aggregate Of Marks Obtained By Sajal In All The Six Subjects?

Answer»

AGGREGATE marks OBTAINED by Sajal

= [ (90% of 150) + (60% of 130) + (70% of 120) + (70% of 100) + (90% of 60) + (70% of 40) ]

= [ 135 + 78 + 84 + 70 + 54 + 28 ]

= 449.

Aggregate marks obtained by Sajal

= [ (90% of 150) + (60% of 130) + (70% of 120) + (70% of 100) + (90% of 60) + (70% of 40) ]

= [ 135 + 78 + 84 + 70 + 54 + 28 ]

= 449.

Discussion

4.	What Are The Average Marks Obtained By All The Seven Students In Physics?
Answer» Average marks obtained in PHYSICS by all the seven STUDENTS =1/7x [ (90% of 120) + (80% of 120) + (70% of 120) + (80% of 120) + (85% of 120) + (65% of 120) + (50% of 120) ] =1/7x [ (90 + 80 + 70 + 80 + 85 + 65 + 50)% of 120 ] =1/7x [ 520% of 120 ] =624/7 = 89.14. Average marks obtained in Physics by all the seven students =1/7x [ (90% of 120) + (80% of 120) + (70% of 120) + (80% of 120) + (85% of 120) + (65% of 120) + (50% of 120) ] =1/7x [ (90 + 80 + 70 + 80 + 85 + 65 + 50)% of 120 ] =1/7x [ 520% of 120 ] =624/7 = 89.14.

4.

What Are The Average Marks Obtained By All The Seven Students In Physics?

Answer»

Average marks obtained in PHYSICS by all the seven STUDENTS

=1/7x [ (90% of 120) + (80% of 120) + (70% of 120) + (80% of 120) + (85% of 120) + (65% of 120) + (50% of 120) ]

=1/7x [ (90 + 80 + 70 + 80 + 85 + 65 + 50)% of 120 ]

=1/7x [ 520% of 120 ]

=624/7

= 89.14.

Average marks obtained in Physics by all the seven students

=1/7x [ (90% of 120) + (80% of 120) + (70% of 120) + (80% of 120) + (85% of 120) + (65% of 120) + (50% of 120) ]

=1/7x [ (90 + 80 + 70 + 80 + 85 + 65 + 50)% of 120 ]

=1/7x [ 520% of 120 ]

=624/7

= 89.14.

Discussion

5.	What Is The Average Candidates Who Appeared From State Q During The Given Years?
Answer» REQUIRED average=8100 + 9500 + 8700 + 9700 + 8950/5 =44950/5 = 8990. Required average=8100 + 9500 + 8700 + 9700 + 8950/5 =44950/5 = 8990.

5.

What Is The Average Candidates Who Appeared From State Q During The Given Years?

Answer»

REQUIRED average=8100 + 9500 + 8700 + 9700 + 8950/5

=44950/5

= 8990.

Required average=8100 + 9500 + 8700 + 9700 + 8950/5

=44950/5

= 8990.

Discussion

6.	A Bag Contains 6 Black And 8 White Balls. One Ball Is Drawn At Random. What Is The Probability That The Ball Drawn Is White?
Answer» LET number of balls = (6 + 8) = 14. Number of WHITE balls = 8. P (drawing a white BALL) =8/14=4/7. Let number of balls = (6 + 8) = 14. Number of white balls = 8. P (drawing a white ball) =8/14=4/7.

6.

A Bag Contains 6 Black And 8 White Balls. One Ball Is Drawn At Random. What Is The Probability That The Ball Drawn Is White?

Answer»

LET number of balls = (6 + 8) = 14.

Number of WHITE balls = 8.

P (drawing a white BALL) =8/14=4/7.

Let number of balls = (6 + 8) = 14.

Number of white balls = 8.

P (drawing a white ball) =8/14=4/7.

Discussion

7.	What Is The Probability Of Getting A Sum 9 From Two Throws Of A Dice?
Answer» In two throws of a dice, N(S) = (6 X 6) = 36. Let E = EVENT of GETTING a sum ={(3, 6), (4, 5), (5, 4), (6, 3)}. P(E) = n(E)/ n(S)=4/36 =1/9. In two throws of a dice, n(S) = (6 x 6) = 36. Let E = event of getting a sum ={(3, 6), (4, 5), (5, 4), (6, 3)}. P(E) = n(E)/ n(S)=4/36 =1/9.

7.

What Is The Probability Of Getting A Sum 9 From Two Throws Of A Dice?

Answer»

In two throws of a dice, N(S) = (6 X 6) = 36.

Let E = EVENT of GETTING a sum ={(3, 6), (4, 5), (5, 4), (6, 3)}.

P(E) = n(E)/ n(S)=4/36 =1/9.

In two throws of a dice, n(S) = (6 x 6) = 36.

Let E = event of getting a sum ={(3, 6), (4, 5), (5, 4), (6, 3)}.

P(E) = n(E)/ n(S)=4/36 =1/9.

Discussion

8.	Tickets Numbered 1 To 20 Are Mixed Up And Then A Ticket Is Drawn At Random. What Is The Probability That The Ticket Drawn Has A Number Which Is A Multiple Of 3 Or 5?
Answer» Here, S = {1, 2, 3, 4, ...., 19, 20}. Let E = EVENT of GETTING a MULTIPLE of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}. P(E) = n(E)/ n(S)=9/20. Here, S = {1, 2, 3, 4, ...., 19, 20}. Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}. P(E) = n(E)/ n(S)=9/20.

8.

Tickets Numbered 1 To 20 Are Mixed Up And Then A Ticket Is Drawn At Random. What Is The Probability That The Ticket Drawn Has A Number Which Is A Multiple Of 3 Or 5?

Answer»

Here, S = {1, 2, 3, 4, ...., 19, 20}.

Let E = EVENT of GETTING a MULTIPLE of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.

P(E) = n(E)/ n(S)=9/20.

Here, S = {1, 2, 3, 4, ...., 19, 20}.

Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.

P(E) = n(E)/ n(S)=9/20.

Discussion

9.	Two, Trains, One From Howrah To Patna And The Other From Patna To Howrah, Start Simultaneously. After They Meet, The Trains Reach Their Destinations After 9 Hours And 16 Hours Respectively. The Ratio Of Their Speeds Is?
Answer» Let us NAME the TRAINS as A and B. Then, (A's speed) : (B's speed) = b : a = 16 : 9 = 4 : 3. Let us name the trains as A and B. Then, (A's speed) : (B's speed) = b : a = 16 : 9 = 4 : 3.

9.

Two, Trains, One From Howrah To Patna And The Other From Patna To Howrah, Start Simultaneously. After They Meet, The Trains Reach Their Destinations After 9 Hours And 16 Hours Respectively. The Ratio Of Their Speeds Is?

Answer»

Let us NAME the TRAINS as A and B. Then,

(A's speed) : (B's speed) = b : a = 16 : 9 = 4 : 3.

Let us name the trains as A and B. Then,

(A's speed) : (B's speed) = b : a = 16 : 9 = 4 : 3.

Discussion

10.	Six Years Ago, The Ratio Of The Ages Of Kunal And Sagar Was 6 : 5. Four Years Hence, The Ratio Of Their Ages Will Be 11 : 10. What Is Sagar's Age At Present?
Answer» Let the ages of Kunal and Sagar 6 YEARS ago be 6x and 5x years RESPECTIVELY. Then, (6x + 6) + 4/(5x + 6) + 4 =11/10 10(6x + 10) = 11(5x + 10) 5x = 10 X = 2. Sagar's present age = (5x + 6) = 16 years. Let the ages of Kunal and Sagar 6 years ago be 6x and 5x years respectively. Then, (6x + 6) + 4/(5x + 6) + 4 =11/10 10(6x + 10) = 11(5x + 10) 5x = 10 x = 2. Sagar's present age = (5x + 6) = 16 years.

10.

Six Years Ago, The Ratio Of The Ages Of Kunal And Sagar Was 6 : 5. Four Years Hence, The Ratio Of Their Ages Will Be 11 : 10. What Is Sagar's Age At Present?

Answer»

Let the ages of Kunal and Sagar 6 YEARS ago be 6x and 5x years RESPECTIVELY.

Then, (6x + 6) + 4/(5x + 6) + 4 =11/10

10(6x + 10) = 11(5x + 10)

5x = 10

X = 2.

Sagar's present age = (5x + 6) = 16 years.

Let the ages of Kunal and Sagar 6 years ago be 6x and 5x years respectively.

Then, (6x + 6) + 4/(5x + 6) + 4 =11/10

10(6x + 10) = 11(5x + 10)

5x = 10

x = 2.

Sagar's present age = (5x + 6) = 16 years.

Discussion

11.	A Man Is 24 Years Older Than His Son. In Two Years, His Age Will Be Twice The Age Of His Son. The Present Age Of His Son Is?
Answer» LET the SON's present AGE be x years. Then, man's present age = (x + 24) years. (x + 24) + 2 = 2(x + 2) x + 26 = 2x + 4 x = 22. Let the son's present age be x years. Then, man's present age = (x + 24) years. (x + 24) + 2 = 2(x + 2) x + 26 = 2x + 4 x = 22.

11.

A Man Is 24 Years Older Than His Son. In Two Years, His Age Will Be Twice The Age Of His Son. The Present Age Of His Son Is?

Answer»

LET the SON's present AGE be x years. Then, man's present age = (x + 24) years.

(x + 24) + 2 = 2(x + 2)

x + 26 = 2x + 4

x = 22.

Let the son's present age be x years. Then, man's present age = (x + 24) years.

(x + 24) + 2 = 2(x + 2)

x + 26 = 2x + 4

x = 22.

Discussion

12.	Present Ages Of Sameer And Anand Are In The Ratio Of 5 : 4 Respectively. Three Years Hence, The Ratio Of Their Ages Will Become 11 : 9 Respectively. What Is Anand's Present Age In Years?
Answer» LET the present ages of SAMEER and Anand be 5x years and 4x years respectively. Then, 5x + 3/4x + 3=11/9 9(5x + 3) = 11(4x + 3) 45x + 27 = 44x + 33 45x - 44x = 33 - 27 x = 6. Anand's present AGE = 4x = 24 years. Let the present ages of Sameer and Anand be 5x years and 4x years respectively. Then, 5x + 3/4x + 3=11/9 9(5x + 3) = 11(4x + 3) 45x + 27 = 44x + 33 45x - 44x = 33 - 27 x = 6. Anand's present age = 4x = 24 years.

12.

Present Ages Of Sameer And Anand Are In The Ratio Of 5 : 4 Respectively. Three Years Hence, The Ratio Of Their Ages Will Become 11 : 9 Respectively. What Is Anand's Present Age In Years?

Answer»

LET the present ages of SAMEER and Anand be 5x years and 4x years respectively.

Then, 5x + 3/4x + 3=11/9

9(5x + 3) = 11(4x + 3)

45x + 27 = 44x + 33

45x - 44x = 33 - 27

x = 6.

Anand's present AGE = 4x = 24 years.

Let the present ages of Sameer and Anand be 5x years and 4x years respectively.

Then, 5x + 3/4x + 3=11/9

9(5x + 3) = 11(4x + 3)

45x + 27 = 44x + 33

45x - 44x = 33 - 27

x = 6.

Anand's present age = 4x = 24 years.

Discussion

13.	A Bike Runs At The Speed Of 50 Km/h When Not Serviced And Runs At 60 Km/h When Serviced. After Servicing The Bike Covers A Certain Distance In 6 H. How Much Time Will The Car Take To Cover The Same Distance When Not Serviced?
Answer» From the given data, after servicing speed of the BIKE = 60 km/h Distance COVERED in 6h. = (60 * 6) km = 360KM When it’s not service,the time taken to COVER 360 km = (360 / 50) = 7.2h From the given data, after servicing speed of the bike = 60 km/h Distance covered in 6h. = (60 * 6) km = 360km When it’s not service,the time taken to cover 360 km = (360 / 50) = 7.2h

13.

A Bike Runs At The Speed Of 50 Km/h When Not Serviced And Runs At 60 Km/h When Serviced. After Servicing The Bike Covers A Certain Distance In 6 H. How Much Time Will The Car Take To Cover The Same Distance When Not Serviced?

Answer»

From the given data, after servicing speed of the BIKE = 60 km/h

Distance COVERED in 6h. = (60 * 6) km = 360KM

When it’s not service,the time taken to COVER 360 km = (360 / 50) = 7.2h

From the given data, after servicing speed of the bike = 60 km/h

Distance covered in 6h. = (60 * 6) km = 360km

When it’s not service,the time taken to cover 360 km = (360 / 50) = 7.2h

Discussion

14.	A Farmer Travelled A Distance Of 61 Km In 9 Hours. He Travelled Partly On Foot @ 4 Km/hr And Partly On Bicycle @ 9 Km/hr. The Distance Travelled On Foot Is?
Answer» Let the distance travelled on foot be X KM. Then, distance travelled on bicycle=(61 -x)km. So,x/4+(61 -x)/9= 9 9x + 4(61 -x) = 9 x 36 5x = 80 x = 16 km. Let the distance travelled on foot be x km. Then, distance travelled on bicycle=(61 -x)km. So,x/4+(61 -x)/9= 9 9x + 4(61 -x) = 9 x 36 5x = 80 x = 16 km.

14.

A Farmer Travelled A Distance Of 61 Km In 9 Hours. He Travelled Partly On Foot @ 4 Km/hr And Partly On Bicycle @ 9 Km/hr. The Distance Travelled On Foot Is?

Answer»

Let the distance travelled on foot be X KM.

Then, distance travelled on bicycle=(61 -x)km.

So,x/4+(61 -x)/9= 9

9x + 4(61 -x) = 9 x 36

5x = 80

x = 16 km.

Let the distance travelled on foot be x km.

Then, distance travelled on bicycle=(61 -x)km.

So,x/4+(61 -x)/9= 9

9x + 4(61 -x) = 9 x 36

5x = 80

x = 16 km.

Discussion

15.	It Takes Eight Hours For A 600 Km Journey, If 120 Km Is Done By Train And The Rest By Car. It Takes 20 Minutes More, If 200 Km Is Done By Train And The Rest By Car. The Ratio Of The Speed Of The Train To That Of The Cars Is?
Answer» LET the speed of the train be x km/hr and that of the car be y km/hr. Then,120/x+480/y= 8 -> 1/x+4/y =1/15...(1) And,200/x+400/y=25/3-> 1/x+2/y =1/24....(2) SOLVING(1)and(2)we get: x = 60 and y = 80. RATIO of speeds=60:80=3:4. Let the speed of the train be x km/hr and that of the car be y km/hr. Then,120/x+480/y= 8 -> 1/x+4/y =1/15...(1) And,200/x+400/y=25/3-> 1/x+2/y =1/24....(2) solving(1)and(2)we get: x = 60 and y = 80. Ratio of speeds=60:80=3:4.

15.

It Takes Eight Hours For A 600 Km Journey, If 120 Km Is Done By Train And The Rest By Car. It Takes 20 Minutes More, If 200 Km Is Done By Train And The Rest By Car. The Ratio Of The Speed Of The Train To That Of The Cars Is?

Answer»

LET the speed of the train be x km/hr and that of the car be y km/hr.

Then,120/x+480/y= 8 -> 1/x+4/y =1/15...(1)

And,200/x+400/y=25/3-> 1/x+2/y =1/24....(2)

SOLVING(1)and(2)we get: x = 60 and y = 80.

RATIO of speeds=60:80=3:4.

Let the speed of the train be x km/hr and that of the car be y km/hr.

Then,120/x+480/y= 8 -> 1/x+4/y =1/15...(1)

And,200/x+400/y=25/3-> 1/x+2/y =1/24....(2)

solving(1)and(2)we get: x = 60 and y = 80.

Ratio of speeds=60:80=3:4.

Discussion

16.	Robert Is Travelling On His Cycle And Has Calculated To Reach Point A At 2 P.m. If He Travels At 10 Kmph, He Will Reach There At 12 Noon If He Travels At 15 Kmph. At What Speed Must He Travel To Reach A At 1 P.m.?
Answer» Let the DISTANCE travelled by x km. Then,x/10-x/15= 2 3x - 2x = 60 x = 60 km. Time taken to travel 60 km at 10 km/hr =[60/10]HRS= 6 hrs. So, ROBERT started 6 hours before 2 P.M. i.e., at 8 A.M. REQUIRED speed =[60/5]kmph.= 12 kmph. Let the distance travelled by x km. Then,x/10-x/15= 2 3x - 2x = 60 x = 60 km. Time taken to travel 60 km at 10 km/hr =[60/10]hrs= 6 hrs. So, Robert started 6 hours before 2 P.M. i.e., at 8 A.M. Required speed =[60/5]kmph.= 12 kmph.

16.

Robert Is Travelling On His Cycle And Has Calculated To Reach Point A At 2 P.m. If He Travels At 10 Kmph, He Will Reach There At 12 Noon If He Travels At 15 Kmph. At What Speed Must He Travel To Reach A At 1 P.m.?

Answer»

Let the DISTANCE travelled by x km.

Then,x/10-x/15= 2

3x - 2x = 60

x = 60 km.

Time taken to travel 60 km at 10 km/hr =[60/10]HRS= 6 hrs.

So, ROBERT started 6 hours before 2 P.M. i.e., at 8 A.M.

REQUIRED speed =[60/5]kmph.= 12 kmph.

Let the distance travelled by x km.

Then,x/10-x/15= 2

3x - 2x = 60

x = 60 km.

Time taken to travel 60 km at 10 km/hr =[60/10]hrs= 6 hrs.

So, Robert started 6 hours before 2 P.M. i.e., at 8 A.M.

Required speed =[60/5]kmph.= 12 kmph.

Discussion

17.	In A Flight Of 600 Km, An Aircraft Was Slowed Down Due To Bad Weather. Its Average Speed For The Trip Was Reduced By 200 Km/hr And The Time Of Flight Increased By 30 Minutes. The Duration Of The Flight Is?
Answer» Let the duration of the flight be x hours. Then,600/x-600/x + (1/2)= 200 600/x-1200/2X + 1= 200 x(2x + 1) = 3 2X2 + x - 3 = 0 (2x + 3)(x - 1) = 0 x= 1 hr. Let the duration of the flight be x hours. Then,600/x-600/x + (1/2)= 200 600/x-1200/2x + 1= 200 x(2x + 1) = 3 2x2 + x - 3 = 0 (2x + 3)(x - 1) = 0 x= 1 hr.

17.

In A Flight Of 600 Km, An Aircraft Was Slowed Down Due To Bad Weather. Its Average Speed For The Trip Was Reduced By 200 Km/hr And The Time Of Flight Increased By 30 Minutes. The Duration Of The Flight Is?

Answer»

Let the duration of the flight be x hours.

Then,600/x-600/x + (1/2)= 200

600/x-1200/2X + 1= 200

x(2x + 1) = 3

2X2 + x - 3 = 0

(2x + 3)(x - 1) = 0

x= 1 hr.

Let the duration of the flight be x hours.

Then,600/x-600/x + (1/2)= 200

600/x-1200/2x + 1= 200

x(2x + 1) = 3

2x2 + x - 3 = 0

(2x + 3)(x - 1) = 0

x= 1 hr.

Discussion

18.	If A Person Walks At 14 Km/hr Instead Of 10 Km/hr, He Would Have Walked 20 Km More. The Actual Distance Travelled By Him Is?
Answer» Let the actual DISTANCE travelled be x KM. Then, x/10=x + 20/14 14X = 10X + 200 4x = 200 x= 50 km. Let the actual distance travelled be x km. Then, x/10=x + 20/14 14x = 10x + 200 4x = 200 x= 50 km.

18.

If A Person Walks At 14 Km/hr Instead Of 10 Km/hr, He Would Have Walked 20 Km More. The Actual Distance Travelled By Him Is?

Answer»

Let the actual DISTANCE travelled be x KM.

Then, x/10=x + 20/14

14X = 10X + 200

4x = 200

x= 50 km.

Let the actual distance travelled be x km.

Then, x/10=x + 20/14

14x = 10x + 200

4x = 200

x= 50 km.

Discussion

19.	How Many Digits Will Be There To The Right Of The Decimal Point In The Product Of 95.75 And .02554 ?
Answer» Sum of decimal PLACES = 7. Since the last DIGIT to the EXTREME right will be zero (since 5 x 4 = 20), so there will be 6 significant digits to the right of the decimal POINT. Sum of decimal places = 7. Since the last digit to the extreme right will be zero (since 5 x 4 = 20), so there will be 6 significant digits to the right of the decimal point.

19.

How Many Digits Will Be There To The Right Of The Decimal Point In The Product Of 95.75 And .02554 ?

Answer»

Sum of decimal PLACES = 7.

Since the last DIGIT to the EXTREME right will be zero (since 5 x 4 = 20),

so there will be 6 significant digits to the right of the decimal POINT.

Sum of decimal places = 7.

Since the last digit to the extreme right will be zero (since 5 x 4 = 20),

so there will be 6 significant digits to the right of the decimal point.

Discussion

20.	A Tailor Has 37.5 Metres Of Cloth And He Has To Make 8 Piecesout Of A Metre Of Cloth. How Many Pieces Can He Make Out Of This Cloth?
Answer» The tailor has to MAKE 8 piece from 1 metre. So for 1 piece ,the measurement will be 1/8=0.125 m from 37.5 m tailor can get 37.5/0.125 = 300 PIECES The tailor has to make 8 piece from 1 metre. So for 1 piece ,the measurement will be 1/8=0.125 m from 37.5 m tailor can get 37.5/0.125 = 300 pieces

20.

A Tailor Has 37.5 Metres Of Cloth And He Has To Make 8 Piecesout Of A Metre Of Cloth. How Many Pieces Can He Make Out Of This Cloth?

Answer»

The tailor has to MAKE 8 piece from 1 metre.

So for 1 piece ,the measurement will be 1/8=0.125 m

from 37.5 m tailor can get 37.5/0.125 = 300 PIECES

The tailor has to make 8 piece from 1 metre.

So for 1 piece ,the measurement will be 1/8=0.125 m

from 37.5 m tailor can get 37.5/0.125 = 300 pieces

Discussion

21.	If Shalu Buys 6 More Apples, His Carton Will Weigh 14.5 Kilograms.if The Weight Of One Apple Is 250 Grams.how Many Apples Did He Initially Had In His Carton?
Answer» WEIGHT of 1 apple = 250 g Weight of 6 APPLES = 250 * 6= 1.5 KG INITIAL weight = 14.5 – 1.5= 13 kg Number of apples in his carton initially : = 13000/250 =52 apples Weight of 1 apple = 250 g Weight of 6 apples = 250 * 6= 1.5 kg Initial weight = 14.5 – 1.5= 13 kg Number of apples in his carton initially : = 13000/250 =52 apples

21.

If Shalu Buys 6 More Apples, His Carton Will Weigh 14.5 Kilograms.if The Weight Of One Apple Is 250 Grams.how Many Apples Did He Initially Had In His Carton?

Answer»

WEIGHT of 1 apple = 250 g

Weight of 6 APPLES = 250 * 6= 1.5 KG

INITIAL weight = 14.5 – 1.5= 13 kg

Number of apples in his carton initially :

= 13000/250

=52 apples

Weight of 1 apple = 250 g

Weight of 6 apples = 250 * 6= 1.5 kg

Initial weight = 14.5 – 1.5= 13 kg

Number of apples in his carton initially :

= 13000/250

=52 apples

Discussion

22.	The Diameter Of A Circle Is 21 Metres. It Will Take How Many Revolutions To Cover A Distance Of 6.6 Km?
Answer» No. of REVOLUTIONS = DISTANCE/circumference. Distance = 6.6 × 1000 = 6600 METRES. Circumference = 2 × 22/7 × 21/2 = 66 metres. No. of revolutions = 6600/66 = 100 revolutions. No. of revolutions = Distance/circumference. Distance = 6.6 × 1000 = 6600 metres. Circumference = 2 × 22/7 × 21/2 = 66 metres. No. of revolutions = 6600/66 = 100 revolutions.

22.

The Diameter Of A Circle Is 21 Metres. It Will Take How Many Revolutions To Cover A Distance Of 6.6 Km?

Answer»

No. of REVOLUTIONS = DISTANCE/circumference.

Distance = 6.6 × 1000 = 6600 METRES.

Circumference = 2 × 22/7 × 21/2 = 66 metres.

No. of revolutions = 6600/66 = 100 revolutions.

No. of revolutions = Distance/circumference.

Distance = 6.6 × 1000 = 6600 metres.

Circumference = 2 × 22/7 × 21/2 = 66 metres.

No. of revolutions = 6600/66 = 100 revolutions.

Discussion

23.	The Length And Breadth Of A Rectangle Are Increased By 20% And 40% Respectively. What Is The Percentage Increase In Its Area?
Answer» APPLY the PERCENTAGE formula. The percentage increase in the area will be P + Q + PQ / 100. So we get the answer as 20 +40 + 20 × 40/100 = 68%. Apply the percentage formula. The percentage increase in the area will be P + Q + PQ / 100. So we get the answer as 20 +40 + 20 × 40/100 = 68%.

23.

The Length And Breadth Of A Rectangle Are Increased By 20% And 40% Respectively. What Is The Percentage Increase In Its Area?

Answer»

APPLY the PERCENTAGE formula.

The percentage increase in the area will be P + Q + PQ / 100.

So we get the answer as 20 +40 + 20 × 40/100 = 68%.

Apply the percentage formula.

The percentage increase in the area will be P + Q + PQ / 100.

So we get the answer as 20 +40 + 20 × 40/100 = 68%.

Discussion

24.	In A Family, The Average Age Of A Father And A Mother Is 35 Years. The Average Age Of The Father, Mother And Their Only Son Is 27 Years. What Is The Age Of The Son?
Answer» Father+Mother=235=70 YEARS Father+Mother+SON=273=81 years THEREFORE Son's age=81-70=11 years Father+Mother=235=70 years Father+Mother+Son=273=81 years therefore Son's age=81-70=11 years

24.

In A Family, The Average Age Of A Father And A Mother Is 35 Years. The Average Age Of The Father, Mother And Their Only Son Is 27 Years. What Is The Age Of The Son?

Answer»

Father+Mother=2*35=70 YEARS

Father+Mother+SON=27*3=81 years

THEREFORE Son's age=81-70=11 years

Father+Mother=2*35=70 years

Father+Mother+Son=27*3=81 years

therefore Son's age=81-70=11 years

Discussion

25.	The Average Of The First 100 Positive Integers Is?
Answer» N(n+1)/2 1+2+3+....+n THEREFORE AVERAGE of these numbers=n+1/2 therefore REQUIRED average 100+1/2=50.5 n(n+1)/2 1+2+3+....+n therefore average of these numbers=n+1/2 therefore required average 100+1/2=50.5

25.

The Average Of The First 100 Positive Integers Is?

Answer»

N(n+1)/2

1+2+3+....+n

THEREFORE AVERAGE of these numbers=n+1/2

therefore REQUIRED average

100+1/2=50.5

n(n+1)/2

1+2+3+....+n

therefore average of these numbers=n+1/2

therefore required average

100+1/2=50.5

Discussion

26.	What Annual Payment Will Discharge A Debt Of Rs. 6,450 Due In 4 Years At 5% Per Annum Simple Interest?
Answer» Let the annual INSTALLMENT be RS. x therefore (x + x35/100)+(x + x25/100) +(x + x15/100)+x =6450 =&GT;115x/100+110x/100+105x/100+x=6450 =>115x+110x+105x+100x=6450100 =>430x=6450100 x=6450100/430=rs.1500 Let the annual installment be rs. x therefore (x + x35/100)+(x + x25/100) +(x + x15/100)+x =6450 =>115x/100+110x/100+105x/100+x=6450 =>115x+110x+105x+100x=6450100 =>430x=6450100 x=6450100/430=rs.1500

26.

What Annual Payment Will Discharge A Debt Of Rs. 6,450 Due In 4 Years At 5% Per Annum Simple Interest?

Answer»

Let the annual INSTALLMENT be RS. x

therefore (x + x*3*5/100)+(x + x*2*5/100)

+(x + x*1*5/100)+x =6450

=&GT;115x/100+110x/100+105x/100+x=6450

=>115x+110x+105x+100x=6450*100

=>430x=6450*100

x=6450*100/430=rs.1500

Let the annual installment be rs. x

therefore (x + x*3*5/100)+(x + x*2*5/100)

+(x + x*1*5/100)+x =6450

=>115x/100+110x/100+105x/100+x=6450

=>115x+110x+105x+100x=6450*100

=>430x=6450*100

x=6450*100/430=rs.1500

Discussion

27.	A Function F(x) Is Defined As F(x) = F(x – 2) - X(x + 2) For All The Integer Values Of X And F(1) + F(4) = 0. What Is The Value Of F(1) + F(2) + F(3) + F(4) + F(5) + F(6)?
Answer» LET S = f(1) + f(2) + f(3) + f(4) + f(5) + f(6) As f(1) + f(4) = 0, therefore S = f(2) + f(3) + f(5) + f(6) ------ (1) f(2) = f(0) - 8 f(3) = f(1) - 15 f(4) = f(2) - 24 = f(0) - 32 f(5) = f(3) - 35 = f(1) - 50 f(6) = f(4) - 48 = f(0) - 80 Put the above values in EQUATION (1), we get S = f(0) - 8 + f(1) - 15 + f(1) - 50 + f(0) - 80 S = 2(f(0) + f(1)) - 153 ------ (2) As we already know f(1) + f(4) = 0 ⇒f(1) + f(0) - 32 = 0 ⇒f(1) + f(0) = 32 Putting this value in equation 2, we get S = 2(32) - 153 = -89 Let S = f(1) + f(2) + f(3) + f(4) + f(5) + f(6) As f(1) + f(4) = 0, therefore S = f(2) + f(3) + f(5) + f(6) ------ (1) f(2) = f(0) - 8 f(3) = f(1) - 15 f(4) = f(2) - 24 = f(0) - 32 f(5) = f(3) - 35 = f(1) - 50 f(6) = f(4) - 48 = f(0) - 80 Put the above values in equation (1), we get S = f(0) - 8 + f(1) - 15 + f(1) - 50 + f(0) - 80 S = 2(f(0) + f(1)) - 153 ------ (2) As we already know f(1) + f(4) = 0 ⇒f(1) + f(0) - 32 = 0 ⇒f(1) + f(0) = 32 Putting this value in equation 2, we get S = 2(32) - 153 = -89

27.

A Function F(x) Is Defined As F(x) = F(x – 2) - X(x + 2) For All The Integer Values Of X And F(1) + F(4) = 0. What Is The Value Of F(1) + F(2) + F(3) + F(4) + F(5) + F(6)?

Answer»

LET S = f(1) + f(2) + f(3) + f(4) + f(5) + f(6)

As f(1) + f(4) = 0, therefore S = f(2) + f(3) + f(5) + f(6) ------ (1)

f(2) = f(0) - 8

f(3) = f(1) - 15

f(4) = f(2) - 24 = f(0) - 32

f(5) = f(3) - 35 = f(1) - 50

f(6) = f(4) - 48 = f(0) - 80

Put the above values in EQUATION (1), we get

S = f(0) - 8 + f(1) - 15 + f(1) - 50 + f(0) - 80

S = 2(f(0) + f(1)) - 153 ------ (2)

As we already know f(1) + f(4) = 0 ⇒f(1) + f(0) - 32 = 0 ⇒f(1) + f(0) = 32

Putting this value in equation 2, we get S = 2(32) - 153 = -89

Let S = f(1) + f(2) + f(3) + f(4) + f(5) + f(6)

As f(1) + f(4) = 0, therefore S = f(2) + f(3) + f(5) + f(6) ------ (1)

f(2) = f(0) - 8

f(3) = f(1) - 15

f(4) = f(2) - 24 = f(0) - 32

f(5) = f(3) - 35 = f(1) - 50

f(6) = f(4) - 48 = f(0) - 80

Put the above values in equation (1), we get

S = f(0) - 8 + f(1) - 15 + f(1) - 50 + f(0) - 80

S = 2(f(0) + f(1)) - 153 ------ (2)

As we already know f(1) + f(4) = 0 ⇒f(1) + f(0) - 32 = 0 ⇒f(1) + f(0) = 32

Putting this value in equation 2, we get S = 2(32) - 153 = -89

Discussion

28.	If The Graphs Of The Equations X + Y = 0 And 5y + 7x = 24 Intersect At (m, N), Then The Value Of M +n Is?
Answer» By SOLVING the given TWO equations, we GET the INTERSECTION point (12, - 12). So, m = 12, n = -12 Hence, m + n = 12 – 12 = 0. By solving the given two equations, we get the intersection point (12, - 12). So, m = 12, n = -12 Hence, m + n = 12 – 12 = 0.

28.

If The Graphs Of The Equations X + Y = 0 And 5y + 7x = 24 Intersect At (m, N), Then The Value Of M +n Is?

Answer»

By SOLVING the given TWO equations, we GET the INTERSECTION point (12, - 12).

So, m = 12, n = -12

Hence, m + n = 12 – 12 = 0.

By solving the given two equations, we get the intersection point (12, - 12).

So, m = 12, n = -12

Hence, m + n = 12 – 12 = 0.

Discussion

29.	The Average Score Of A Cricketer For 13 Matches Is 42 Runs. If His Average Score For The First 5 Matches Is 54, Then What Was His Average Score (in Runs) For Last 8 Matches?
Answer» TOTAL Score = Average * Number of MATCHES Total score of 13 matches = 13 × 42 = 546 Total score of first 5 matches = 5 × 54 = 270 Therefore, total score of LAST 8 matches = 546 - 270 = 276 Average = 276/8 = 34.5 Total Score = Average * Number of matches Total score of 13 matches = 13 × 42 = 546 Total score of first 5 matches = 5 × 54 = 270 Therefore, total score of last 8 matches = 546 - 270 = 276 Average = 276/8 = 34.5

29.

The Average Score Of A Cricketer For 13 Matches Is 42 Runs. If His Average Score For The First 5 Matches Is 54, Then What Was His Average Score (in Runs) For Last 8 Matches?

Answer»

TOTAL Score = Average * Number of MATCHES

Total score of 13 matches = 13 × 42 = 546

Total score of first 5 matches = 5 × 54 = 270

Therefore, total score of LAST 8 matches = 546 - 270 = 276

Average = 276/8 = 34.5

Total Score = Average * Number of matches

Total score of 13 matches = 13 × 42 = 546

Total score of first 5 matches = 5 × 54 = 270

Therefore, total score of last 8 matches = 546 - 270 = 276

Average = 276/8 = 34.5

Discussion

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