25 + Interview Questions in Fidelity Aptitude Interview Questions in Fidelity Aptitude Tutorial

1.	A Bath Tub Can Be Filled By A Cold Water Pipe In 20 Minutes And By A Hot Water Pipe In 30 Minutes. A Person Leaves The Bathroom After Turning On Both Pipes Simultaneously And Returns At The Moment When The Bath Tub Should Be Full. Finding However, That The Waste Pipe Has Been Open, He Now Closes It. In 3 Minutes More The Bath Tub Is Full. In What Time Would The Waste Pipe Empty It ?
Answer» TOTAL work = 1/20 + 1/30 (3+2)/60 = 5/60 = 1/12 For both of them needed = 12 min. But due to leakage it take = 15 min. So we can say for filling pipe it takes 3 minutes work for the waste pipe will be = 12 min. 3 min. of both filling pipe = 12 min. of waste pipe 12 min. of filling pipe will be = 48 min. of waste pipe i.e. Total work = 1/20 + 1/30 (3+2)/60 = 5/60 = 1/12 For both of them needed = 12 min. But due to leakage it take = 15 min. So we can say for filling pipe it takes 3 minutes work for the waste pipe will be = 12 min. 3 min. of both filling pipe = 12 min. of waste pipe 12 min. of filling pipe will be = 48 min. of waste pipe i.e.

1.

A Bath Tub Can Be Filled By A Cold Water Pipe In 20 Minutes And By A Hot Water Pipe In 30 Minutes. A Person Leaves The Bathroom After Turning On Both Pipes Simultaneously And Returns At The Moment When The Bath Tub Should Be Full. Finding However, That The Waste Pipe Has Been Open, He Now Closes It. In 3 Minutes More The Bath Tub Is Full. In What Time Would The Waste Pipe Empty It ?

Answer»

TOTAL work = 1/20 + 1/30

(3+2)/60 = 5/60 = 1/12

For both of them needed = 12 min.

But due to leakage it take = 15 min.

So we can say for filling pipe it takes 3 minutes

work for the waste pipe will be = 12 min.

3 min. of both filling pipe = 12 min. of waste pipe

12 min. of filling pipe will be = 48 min. of waste pipe i.e.

Total work = 1/20 + 1/30

(3+2)/60 = 5/60 = 1/12

For both of them needed = 12 min.

But due to leakage it take = 15 min.

So we can say for filling pipe it takes 3 minutes

work for the waste pipe will be = 12 min.

3 min. of both filling pipe = 12 min. of waste pipe

12 min. of filling pipe will be = 48 min. of waste pipe i.e.

Discussion

2.	A Jar Was Containing 60 Liters Of Mixture Of Milk And Water In The Ratio 7:5. From This Jar 12 Litres Of Mixtures Was Taken Out And 8 Litres Of Pure Milk Was Added. What Is The Respective Ratio Of Milk And Water In The Mixture After The Final Operation ?
Answer» Let n be the MIXTURE quantity of milk = 7/12 × 60 Ltr = 35 Ltr Quantity of water = 5/12 × 60 Ltr. = 25 Ltr 12 Ltr. of mixture removed contains milk = 7/12 × 12 Ltr. = 7 Ltr and water = 5/12 × 12 Ltr = 5 Ltr After adding 8 Ltr of PURE milk, Net milk in the mixture = 35 - 7 + 8 = 36 Ltr. Net water in the mixture = 25 - 5 = 20 Ltr. So the required RATIO of milk and water now = 36/20 = 9/5 Let n be the mixture quantity of milk = 7/12 × 60 Ltr = 35 Ltr Quantity of water = 5/12 × 60 Ltr. = 25 Ltr 12 Ltr. of mixture removed contains milk = 7/12 × 12 Ltr. = 7 Ltr and water = 5/12 × 12 Ltr = 5 Ltr After adding 8 Ltr of pure milk, Net milk in the mixture = 35 - 7 + 8 = 36 Ltr. Net water in the mixture = 25 - 5 = 20 Ltr. So the required ratio of milk and water now = 36/20 = 9/5

2.

A Jar Was Containing 60 Liters Of Mixture Of Milk And Water In The Ratio 7:5. From This Jar 12 Litres Of Mixtures Was Taken Out And 8 Litres Of Pure Milk Was Added. What Is The Respective Ratio Of Milk And Water In The Mixture After The Final Operation ?

Answer»

Let n be the MIXTURE quantity of milk = 7/12 × 60 Ltr = 35 Ltr

Quantity of water = 5/12 × 60 Ltr. = 25 Ltr

12 Ltr. of mixture removed contains milk = 7/12 × 12 Ltr. = 7 Ltr and water = 5/12 × 12 Ltr = 5 Ltr

After adding 8 Ltr of PURE milk,

Net milk in the mixture = 35 - 7 + 8 = 36 Ltr.

Net water in the mixture = 25 - 5 = 20 Ltr.

So the required RATIO of milk and water now = 36/20 = 9/5

Let n be the mixture quantity of milk = 7/12 × 60 Ltr = 35 Ltr

Quantity of water = 5/12 × 60 Ltr. = 25 Ltr

12 Ltr. of mixture removed contains milk = 7/12 × 12 Ltr. = 7 Ltr and water = 5/12 × 12 Ltr = 5 Ltr

After adding 8 Ltr of pure milk,

Net milk in the mixture = 35 - 7 + 8 = 36 Ltr.

Net water in the mixture = 25 - 5 = 20 Ltr.

So the required ratio of milk and water now = 36/20 = 9/5

Discussion

3.	A Clock Showed 5 Min Past 3'0 Clock On Sunday Evening When The Correct Time Was 3'0 Clock It Looses Uniformly And Was Observed To Be 10 Min Slow On The Subsequent Tuesday At 9pm . When Did The Clock Show The Correct Time?
Answer» Clock showed 5 min past 3'0 clock on Sunday evening when the correct time was 3'0 clock. It looses uniformly and was observed to be 10 min SLOW on the subsequent Tuesday at 9pm . In 54 hrs , it lost 15 mins. so 5 mins are lost in 554/15 = 18 hrs so watch showed correct time at 1500+1800-2400= 0900 hrs on Monday i.e 9AM Monday Clock showed 5 min past 3'0 clock on Sunday evening when the correct time was 3'0 clock. It looses uniformly and was observed to be 10 min slow on the subsequent Tuesday at 9pm . In 54 hrs , it lost 15 mins. so 5 mins are lost in 554/15 = 18 hrs so watch showed correct time at 1500+1800-2400= 0900 hrs on Monday i.e 9AM Monday

3.

A Clock Showed 5 Min Past 3'0 Clock On Sunday Evening When The Correct Time Was 3'0 Clock It Looses Uniformly And Was Observed To Be 10 Min Slow On The Subsequent Tuesday At 9pm . When Did The Clock Show The Correct Time?

Answer»

Clock showed 5 min past 3'0 clock on Sunday evening when the correct time was 3'0 clock.

It looses uniformly and was observed to be 10 min SLOW on the subsequent Tuesday at 9pm .

In 54 hrs , it lost 15 mins.

so 5 mins are lost in 5*54/15 = 18 hrs

so watch showed correct time at 1500+1800-2400= 0900 hrs on Monday i.e 9AM Monday

Clock showed 5 min past 3'0 clock on Sunday evening when the correct time was 3'0 clock.

It looses uniformly and was observed to be 10 min slow on the subsequent Tuesday at 9pm .

In 54 hrs , it lost 15 mins.

so 5 mins are lost in 5*54/15 = 18 hrs

so watch showed correct time at 1500+1800-2400= 0900 hrs on Monday i.e 9AM Monday

Discussion

4.	The Average Age Of All The Student Of A Class Is 18 Years. The Average Age Of Boys Of The Class Is 20 Years And That Of The Girls Is 15 Years. If The Number Of Girls In The Class Is 20, Then Find The Number Of Boys In The Class ?
Answer» LET Boys in CLASS = B Girls in class = 20 Now, (20B+1520)/(B+20) = 18 ⇒ B = 30 Let Boys in class = B Girls in class = 20 Now, (20B+1520)/(B+20) = 18 ⇒ B = 30

4.

The Average Age Of All The Student Of A Class Is 18 Years. The Average Age Of Boys Of The Class Is 20 Years And That Of The Girls Is 15 Years. If The Number Of Girls In The Class Is 20, Then Find The Number Of Boys In The Class ?

Answer»

LET Boys in CLASS = B

Girls in class = 20

Now, (20B+15*20)/(B+20) = 18

⇒ B = 30

Let Boys in class = B

Girls in class = 20

Now, (20B+15*20)/(B+20) = 18

⇒ B = 30

Discussion

5.	In A Kilometer Race, A Can Give B A 100 M Start And C A 150 M Start. How Many Meters Start Can B Give To C ?
Answer» A can give B a 100 m start and C a 150m. Start MEANS when A runs 1000m, B runs 900m and C runs 850m. When B runs 1000m, C will RUN 1000 X (850/900) m (i.e. 8500/9 m) Thus, B can give C a start of - 1000 - (8500/9), i.e. 500/9 m. A can give B a 100 m start and C a 150m. Start means when A runs 1000m, B runs 900m and C runs 850m. When B runs 1000m, C will run 1000 x (850/900) m (i.e. 8500/9 m) Thus, B can give C a start of - 1000 - (8500/9), i.e. 500/9 m.

5.

In A Kilometer Race, A Can Give B A 100 M Start And C A 150 M Start. How Many Meters Start Can B Give To C ?

Answer»

A can give B a 100 m start and C a 150m. Start MEANS when A runs 1000m, B runs 900m and C runs 850m.

When B runs 1000m, C will RUN 1000 X (850/900) m (i.e. 8500/9 m)

Thus, B can give C a start of - 1000 - (8500/9), i.e. 500/9 m.

A can give B a 100 m start and C a 150m. Start means when A runs 1000m, B runs 900m and C runs 850m.

When B runs 1000m, C will run 1000 x (850/900) m (i.e. 8500/9 m)

Thus, B can give C a start of - 1000 - (8500/9), i.e. 500/9 m.

Discussion

6.	The Fuel Indicator In A Car Shows 1/5th Of The Fuel Tank As Full. When 22 More Liters Of Fuel Are Poured In To The Tank, The Indicator Rests At The 3/4of The Full Mark. Find The Capacity Of The Tank ?
Answer» X/5 + 22 = 3x/4 ⇒ x = 40 LITRES x/5 + 22 = 3x/4 ⇒ x = 40 litres

6.

The Fuel Indicator In A Car Shows 1/5th Of The Fuel Tank As Full. When 22 More Liters Of Fuel Are Poured In To The Tank, The Indicator Rests At The 3/4of The Full Mark. Find The Capacity Of The Tank ?

Answer»

X/5 + 22 = 3x/4 ⇒ x = 40 LITRES

x/5 + 22 = 3x/4 ⇒ x = 40 litres

Discussion

7.	A Boat Travels Upstream From B To A And Downstream From A To B In 3 Hours. If The Speed Of The Boat In Still Water Is 9 Km/hour And The Speed Of The Current Is 3 Km/hour, The Distance Between A And B Is ?
Answer» LET d be the distance between A and B So, d/12 + d/6 = 3 d = 12 KM Let d be the distance between A and B So, d/12 + d/6 = 3 d = 12 km

7.

A Boat Travels Upstream From B To A And Downstream From A To B In 3 Hours. If The Speed Of The Boat In Still Water Is 9 Km/hour And The Speed Of The Current Is 3 Km/hour, The Distance Between A And B Is ?

Answer»

LET d be the distance between A and B

So, d/12 + d/6 = 3 d = 12 KM

Let d be the distance between A and B

So, d/12 + d/6 = 3 d = 12 km

Discussion

8.	The Ratio Between The Number Of Passengers Travelling By I And Ii Class Between The Two Railway Stations Is 1 : 50, Whereas The Ratio Of I And Ii Class Fares Between The Same Stations Is 3 : 1. If On A Particular Day Rs. 1,325 Were Collected From The Passengers Travelling Between These Stations, Then What Was The Amount Collected From The Ii Class Passengers ?
Answer» Let x be the number of passengers and y be the fare taken from passengers. 3XY + 50xy = 1325 => XY = 25 AMOUNT collected from II class passengers = 25 × 50 = RS. 1250. Let x be the number of passengers and y be the fare taken from passengers. 3xy + 50xy = 1325 => xy = 25 Amount collected from II class passengers = 25 × 50 = Rs. 1250.

8.

The Ratio Between The Number Of Passengers Travelling By I And Ii Class Between The Two Railway Stations Is 1 : 50, Whereas The Ratio Of I And Ii Class Fares Between The Same Stations Is 3 : 1. If On A Particular Day Rs. 1,325 Were Collected From The Passengers Travelling Between These Stations, Then What Was The Amount Collected From The Ii Class Passengers ?

Answer»

Let x be the number of passengers and y be the fare taken from passengers.

3XY + 50xy = 1325 => XY = 25

AMOUNT collected from II class passengers = 25 × 50 = RS. 1250.

Let x be the number of passengers and y be the fare taken from passengers.

3xy + 50xy = 1325 => xy = 25

Amount collected from II class passengers = 25 × 50 = Rs. 1250.

Discussion

9.	The Mean Daily Profit Made By A Shopkeeper In A Month Of 30 Days Was Rs. 350. If The Mean Profit For The First Fifteen Days Was Rs. 275, Then The Mean Profit For The Last 15 Days Would Be ?
Answer» AVERAGE would be : 350 = (275 + x)/2 On SOLVING, x = 425. Average would be : 350 = (275 + x)/2 On solving, x = 425.

9.

The Mean Daily Profit Made By A Shopkeeper In A Month Of 30 Days Was Rs. 350. If The Mean Profit For The First Fifteen Days Was Rs. 275, Then The Mean Profit For The Last 15 Days Would Be ?

Answer»

AVERAGE would be : 350 = (275 + x)/2

On SOLVING, x = 425.

Average would be : 350 = (275 + x)/2

On solving, x = 425.

Discussion

10.	Two Trains 246 M And 304 M Long Are Travelling Towards One Another At 114 Km/hr And 66 Km/hr Respectively. How Long Do The Trains Take To Pass One Another ?
Answer» 114 km/h => 1145/18 m/s => 95/3 m/s 66km/h => 665/18 m/s => 55/3 m/s Time taken = (246+304)/(55/3 + 95/3) => 11 sec 114 km/h => 1145/18 m/s => 95/3 m/s 66km/h => 665/18 m/s => 55/3 m/s Time taken = (246+304)/(55/3 + 95/3) => 11 sec

10.

Two Trains 246 M And 304 M Long Are Travelling Towards One Another At 114 Km/hr And 66 Km/hr Respectively. How Long Do The Trains Take To Pass One Another ?

Answer»

114 km/h => 114*5/18 m/s => 95/3 m/s

66km/h => 66*5/18 m/s => 55/3 m/s

Time taken = (246+304)/(55/3 + 95/3) => 11 sec

114 km/h => 114*5/18 m/s => 95/3 m/s

66km/h => 66*5/18 m/s => 55/3 m/s

Time taken = (246+304)/(55/3 + 95/3) => 11 sec

Discussion

11.	Five Years Ago The Average Age Of A Family Of Six Members Was 20 Years. A Baby Is Born And Now The Average Age Is 22 Years. How Old Is The Baby At Present ?
Answer» Lets assume the age of all the 6 people = 20 YEARS. After 5 years i.e. now their ages will be 25 years. TOTAL age of all the 6 will be = 256 = 150 Including baby's age, AVG is 22 years LET x be the age of baby. (150+x)/7 = 22 => x = 4 years. Lets assume the age of all the 6 people = 20 years. After 5 years i.e. now their ages will be 25 years. Total age of all the 6 will be = 256 = 150 Including baby's age, avg is 22 years Let x be the age of baby. (150+x)/7 = 22 => x = 4 years.

11.

Five Years Ago The Average Age Of A Family Of Six Members Was 20 Years. A Baby Is Born And Now The Average Age Is 22 Years. How Old Is The Baby At Present ?

Answer»

Lets assume the age of all the 6 people = 20 YEARS.

After 5 years i.e. now their ages will be 25 years.

TOTAL age of all the 6 will be = 25*6 = 150

Including baby's age, AVG is 22 years

LET x be the age of baby.

(150+x)/7 = 22

=> x = 4 years.

Lets assume the age of all the 6 people = 20 years.

After 5 years i.e. now their ages will be 25 years.

Total age of all the 6 will be = 25*6 = 150

Including baby's age, avg is 22 years

Let x be the age of baby.

(150+x)/7 = 22

=> x = 4 years.

Discussion

12.	Sand And Cement Were Mixed In The Ratio 3 : 1 To Make Up 1 Tonne. How Much Sand Must Be Added To Make The Ratio 6 : 1 ?
Answer» Let amount of sand and cement as 3X and X 3x+x = 1000kg x = 250 Sand : cement =&GT; 750 : 250 (750+y)/250 = 6/1 => y = 750 kg. Let amount of sand and cement as 3x and x 3x+x = 1000kg x = 250 Sand : cement => 750 : 250 (750+y)/250 = 6/1 => y = 750 kg.

12.

Sand And Cement Were Mixed In The Ratio 3 : 1 To Make Up 1 Tonne. How Much Sand Must Be Added To Make The Ratio 6 : 1 ?

Answer»

Let amount of sand and cement as 3X and X

3x+x = 1000kg

x = 250

Sand : cement =&GT; 750 : 250

(750+y)/250 = 6/1

=> y = 750 kg.

Let amount of sand and cement as 3x and x

3x+x = 1000kg

x = 250

Sand : cement => 750 : 250

(750+y)/250 = 6/1

=> y = 750 kg.

Discussion

13.	B's Monthly Salary Is Less By 10% Than That Of A's.c's Monthly Salary Is Less By 25% Than That Of A's.by What Percent Of B's Salary More Than That Of C's ?
Answer» Lets consider C's SALARY = RS. 100 B's will be 90 C's will be 100(1-25/100)=100(75/100)=75 [(90-75)/75]100=(15/75)100=20% Lets consider c's salary = Rs. 100 B's will be 90 C's will be 100(1-25/100)=100(75/100)=75 [(90-75)/75]100=(15/75)100=20%

13.

B's Monthly Salary Is Less By 10% Than That Of A's.c's Monthly Salary Is Less By 25% Than That Of A's.by What Percent Of B's Salary More Than That Of C's ?

Answer»

Lets consider C's SALARY = RS. 100

B's will be 90

C's will be 100(1-25/100)=100*(75/100)=75

[(90-75)/75]*100=(15/75)*100=20%

Lets consider c's salary = Rs. 100

B's will be 90

C's will be 100(1-25/100)=100*(75/100)=75

[(90-75)/75]*100=(15/75)*100=20%

Discussion

14.	A Started Business Investing Rs. 60,000.exactly Six Months Later,b Joined Him Investing Rs. 90,000.if They Make A Profit Of Rs. 1,26,000 At The End Of The Year. How Much Should B's Share In It ?
Answer» A INVESTMENT for 12 months = 60,00012 = 720000 B investment for 6 months = 90000 6 = 540000 A & B investment ratio = 72/54 => 4/3 So, B' SHARE = 126000 * 3/7= 54000 A investment for 12 months = 60,00012 = 720000 B investment for 6 months = 90000 6 = 540000 A & B investment ratio = 72/54 => 4/3 So, B' share = 126000 * 3/7= 54000

14.

A Started Business Investing Rs. 60,000.exactly Six Months Later,b Joined Him Investing Rs. 90,000.if They Make A Profit Of Rs. 1,26,000 At The End Of The Year. How Much Should B's Share In It ?

Answer»

A INVESTMENT for 12 months = 60,000*12 = 720000

B investment for 6 months = 90000 * 6 = 540000

A & B investment ratio = 72/54 => 4/3

So, B' SHARE = 126000 * 3/7= 54000

A investment for 12 months = 60,000*12 = 720000

B investment for 6 months = 90000 * 6 = 540000

A & B investment ratio = 72/54 => 4/3

So, B' share = 126000 * 3/7= 54000

Discussion

15.	If The Value Of The Numerator Is Increased By 20% And That Of The Denominator Is Decreased By 25% The Fraction Becomes Unity The Numerator Of Original Fraction Is ?
Answer» 120n/75d=1 n/d=75/120=5/8 120n/75d=1 n/d=75/120=5/8

15.

If The Value Of The Numerator Is Increased By 20% And That Of The Denominator Is Decreased By 25% The Fraction Becomes Unity The Numerator Of Original Fraction Is ?

Answer»

120n/75d=1

n/d=75/120=5/8

120n/75d=1

n/d=75/120=5/8

Discussion

16.	What Is The Next Number Of The Following Sequence 125102550(?).
Answer» Consider the series as 1 2 5 10 25 50 ____ 12 = 2 22.5 = 5 52 = 10 102.5 = 25 252 = 50 502.5 = 125 Consider the series as 1 2 5 10 25 50 ____ 12 = 2 22.5 = 5 52 = 10 102.5 = 25 252 = 50 502.5 = 125

16.

What Is The Next Number Of The Following Sequence 125102550(?).

Answer»

Consider the series as 1 2 5 10 25 50 ____

1*2 = 2

2*2.5 = 5

5*2 = 10

10*2.5 = 25

25*2 = 50

50*2.5 = 125

Consider the series as 1 2 5 10 25 50 ____

1*2 = 2

2*2.5 = 5

5*2 = 10

10*2.5 = 25

25*2 = 50

50*2.5 = 125

Discussion

17.	The Average Of 20 Numbers Is 15 And The Average Of First Five Is 12. The Average Of The Rest Is ?
Answer» If the AVERAGE of remaining numbers be x, then 20 × 15 = 5 × 12 + 15x ⇒ 300 = 60 + 15x ⇒ 15x = 300 – 60 = 240 => x = 240/15 = 16 If the average of remaining numbers be x, then 20 × 15 = 5 × 12 + 15x ⇒ 300 = 60 + 15x ⇒ 15x = 300 – 60 = 240 => x = 240/15 = 16

17.

The Average Of 20 Numbers Is 15 And The Average Of First Five Is 12. The Average Of The Rest Is ?

Answer»

If the AVERAGE of remaining numbers be x, then

20 × 15 = 5 × 12 + 15x

⇒ 300 = 60 + 15x

⇒ 15x = 300 – 60 = 240

=> x = 240/15 = 16

If the average of remaining numbers be x, then

20 × 15 = 5 × 12 + 15x

⇒ 300 = 60 + 15x

⇒ 15x = 300 – 60 = 240

=> x = 240/15 = 16

Discussion

18.	If A, B, C, D, E Are Five Consecutive Odd Numbers, Their Average Is ?
Answer» b = a + 2 c = b + 2 = a + 4 d = c + 2 = a + 6 E = d + 2 = a + 8 Therefore, Required average = (a + a + 2 + a + 4 + a + 6 + a + 8)/5 = a + 4 b = a + 2 c = b + 2 = a + 4 d = c + 2 = a + 6 e = d + 2 = a + 8 Therefore, Required average = (a + a + 2 + a + 4 + a + 6 + a + 8)/5 = a + 4

18.

If A, B, C, D, E Are Five Consecutive Odd Numbers, Their Average Is ?

Answer»

b = a + 2

c = b + 2 = a + 4

d = c + 2 = a + 6

E = d + 2 = a + 8

Therefore, Required average = (a + a + 2 + a + 4 + a + 6 + a + 8)/5

= a + 4

b = a + 2

c = b + 2 = a + 4

d = c + 2 = a + 6

e = d + 2 = a + 8

Therefore, Required average = (a + a + 2 + a + 4 + a + 6 + a + 8)/5

= a + 4

Discussion

19.	The Difference Of Two Numbers Is 11 And One Fifth Of Their Sum Is 9. The Numbers Are ?
Answer» x − y = 11, x + y = 5 × 9 x − y = 11, x + y = 45, y = 17, x = 28 x − y = 11, x + y = 5 × 9 x − y = 11, x + y = 45, y = 17, x = 28

19.

The Difference Of Two Numbers Is 11 And One Fifth Of Their Sum Is 9. The Numbers Are ?

Answer»

x − y = 11, x + y = 5 × 9 x − y = 11, x + y = 45, y = 17, x = 28

Discussion

20.	Simplify (0.001344 / 0.3 X 0.7) = ?
Answer» (0.001344 / 0.3 X 0.7) = 0.0064 (0.001344 / 0.3 x 0.7) = 0.0064

20.

Simplify (0.001344 / 0.3 X 0.7) = ?

Answer»

(0.001344 / 0.3 X 0.7) = 0.0064

(0.001344 / 0.3 x 0.7) = 0.0064

Discussion

21.	If 378 Coins Consist Of Rupee, 50 Paise And 25 Paise Coins, Whose Values Are Proportional To 13 :11 : 7, The Number Of 50 Paise Coins Will Be ?
Answer» If values are proportional to 13 : 11 : 7, then the number of COINS will be proportional to 13/1 : 11/0.50 : 7/0.25 ⇒ 13 : 22 : 28. Now from this the number of coins of 50 paise will be 378 × 22/63 = 132. If values are proportional to 13 : 11 : 7, then the number of coins will be proportional to 13/1 : 11/0.50 : 7/0.25 ⇒ 13 : 22 : 28. Now from this the number of coins of 50 paise will be 378 × 22/63 = 132.

21.

If 378 Coins Consist Of Rupee, 50 Paise And 25 Paise Coins, Whose Values Are Proportional To 13 :11 : 7, The Number Of 50 Paise Coins Will Be ?

Answer»

If values are proportional to 13 : 11 : 7, then the number of COINS will be proportional to 13/1 : 11/0.50 : 7/0.25 ⇒ 13 : 22 : 28.

Now from this the number of coins of 50 paise will be 378 × 22/63 = 132.

If values are proportional to 13 : 11 : 7, then the number of coins will be proportional to 13/1 : 11/0.50 : 7/0.25 ⇒ 13 : 22 : 28.

Now from this the number of coins of 50 paise will be 378 × 22/63 = 132.

Discussion

22.	Pointing To A Lady A Person Said, "the Son Of Her Only Brother Is The Brother Of My Wife." How Is The Lady Related To The Person ?
Answer» BROTHER of person's WIFE brother-in-law of the person. HENCE, the son of lady's brother is brother-in-law of the person. Therefore, the brother of the lady is the father-in-law of the person. Hence, the lady is the sister of the person's father-in-law. Brother of person's wife brother-in-law of the person. Hence, the son of lady's brother is brother-in-law of the person. Therefore, the brother of the lady is the father-in-law of the person. Hence, the lady is the sister of the person's father-in-law.

22.

Pointing To A Lady A Person Said, "the Son Of Her Only Brother Is The Brother Of My Wife." How Is The Lady Related To The Person ?

Answer»

BROTHER of person's WIFE brother-in-law of the person. HENCE, the son of lady's brother is brother-in-law of the person.

Therefore, the brother of the lady is the father-in-law of the person. Hence, the lady is the sister of the person's father-in-law.

Brother of person's wife brother-in-law of the person. Hence, the son of lady's brother is brother-in-law of the person.

Therefore, the brother of the lady is the father-in-law of the person. Hence, the lady is the sister of the person's father-in-law.

Discussion

23.	A Man Pointing To A Photograph Says, The Lady In The Photograph Is My Nephew's Maternal Grandmother. How Is The Lady In The Photograph Related To The Man's Sister Who Has No Other Sister ?
Answer» Clearly, the lady is the grandmother of man's SISTER's son. i.e., the MOTHER of the mother of man's sister's son. => The mother of man's sister. So, the lady is man's mother. Clearly, the lady is the grandmother of man's sister's son. i.e., the mother of the mother of man's sister's son. => The mother of man's sister. So, the lady is man's mother.

23.

A Man Pointing To A Photograph Says, The Lady In The Photograph Is My Nephew's Maternal Grandmother. How Is The Lady In The Photograph Related To The Man's Sister Who Has No Other Sister ?

Answer»

Clearly, the lady is the grandmother of man's SISTER's son.

i.e., the MOTHER of the mother of man's sister's son.

=> The mother of man's sister.

So, the lady is man's mother.

Clearly, the lady is the grandmother of man's sister's son.

i.e., the mother of the mother of man's sister's son.

=> The mother of man's sister.

So, the lady is man's mother.

Discussion

24.	Pipe A Can Fill A Tank In 8 Hours And Pipe B Can Fill It In 6 Hours. If Both The Pipes Are Opened But After 2 Hours Pipe A Is Closed, Then The Other Pipe Will Fill The Tank In ?
Answer» Tank part filled by Pipe A in 1 HOUR = 1/8 or Time taken to fill 1/8 part of tank filled A = 1 hour ------------- (1) Tank part filled by Pipe B in 1 HOURS = 1/6 or Time taken to fill 1/6 part of tank filled B = 1 hour ---------(2) => Tank Part filled by (A+B) in 1 hour = (1/8 + 1/6) => Tank Part filled by (A+B) in 2 hour = 2(1/8 + 1/6) = 2(3+4)/24 = 27/24 = 7/12 After 2 hours LEFT over portion of tank = (1-7/12) = 5/12 Time taken to fill 1/6 part of tank filled B = 1 hour so time taken to fill 5/12 part of tank by B = 6 5/12 = 5/2 = 2 1/2 => So it TAKES 2 1/2 hrs (2hrs 30 min) to fill 5/12 of tank. Tank part filled by Pipe A in 1 hour = 1/8 or Time taken to fill 1/8 part of tank filled A = 1 hour ------------- (1) Tank part filled by Pipe B in 1 hours = 1/6 or Time taken to fill 1/6 part of tank filled B = 1 hour ---------(2) => Tank Part filled by (A+B) in 1 hour = (1/8 + 1/6) => Tank Part filled by (A+B) in 2 hour = 2(1/8 + 1/6) = 2(3+4)/24 = 27/24 = 7/12 After 2 hours left over portion of tank = (1-7/12) = 5/12 Time taken to fill 1/6 part of tank filled B = 1 hour so time taken to fill 5/12 part of tank by B = 6 5/12 = 5/2 = 2 1/2 => So it takes 2 1/2 hrs (2hrs 30 min) to fill 5/12 of tank.

24.

Pipe A Can Fill A Tank In 8 Hours And Pipe B Can Fill It In 6 Hours. If Both The Pipes Are Opened But After 2 Hours Pipe A Is Closed, Then The Other Pipe Will Fill The Tank In ?

Answer»

Tank part filled by Pipe A in 1 HOUR = 1/8

or

Time taken to fill 1/8 part of tank filled A = 1 hour ------------- (1)

Tank part filled by Pipe B in 1 HOURS = 1/6

or

Time taken to fill 1/6 part of tank filled B = 1 hour ---------(2)

=> Tank Part filled by (A+B) in 1 hour = (1/8 + 1/6)

=> Tank Part filled by (A+B) in 2 hour = 2*(1/8 + 1/6) = 2*(3+4)/24 = 2*7/24 = 7/12

After 2 hours LEFT over portion of tank = (1-7/12) = 5/12

Time taken to fill 1/6 part of tank filled B = 1 hour

so time taken to fill 5/12 part of tank by B = 6 * 5/12 = 5/2 = 2 1/2

=> So it TAKES 2 1/2 hrs (2hrs 30 min) to fill 5/12 of tank.

Tank part filled by Pipe A in 1 hour = 1/8

or

Time taken to fill 1/8 part of tank filled A = 1 hour ------------- (1)

Tank part filled by Pipe B in 1 hours = 1/6

or

Time taken to fill 1/6 part of tank filled B = 1 hour ---------(2)

=> Tank Part filled by (A+B) in 1 hour = (1/8 + 1/6)

=> Tank Part filled by (A+B) in 2 hour = 2*(1/8 + 1/6) = 2*(3+4)/24 = 2*7/24 = 7/12

After 2 hours left over portion of tank = (1-7/12) = 5/12

Time taken to fill 1/6 part of tank filled B = 1 hour

so time taken to fill 5/12 part of tank by B = 6 * 5/12 = 5/2 = 2 1/2

=> So it takes 2 1/2 hrs (2hrs 30 min) to fill 5/12 of tank.

Discussion

25.	A Project Manager Estimated That He Would Complete The Project In Time If He Hires 42 People For 38 Days. At The End Of 30 Days He Realized That Only 3/5th Of The Work Is Complete. How Many More Men Does He Need To Hire To Complete The Work In Time ?
Answer» 42 people in 30 DAYS did 3/5th of the WORK Let x people are REQUIRED to complete the REMAINING 2/5th of the work in 8 days Applying chain rule we get x = 42 × 30/8 × (2×5)/(3×5) = 105 people Additional no. of people required = ( 105 - 42 ) = 63. 42 people in 30 days did 3/5th of the work Let x people are required to complete the remaining 2/5th of the work in 8 days Applying chain rule we get x = 42 × 30/8 × (2×5)/(3×5) = 105 people Additional no. of people required = ( 105 - 42 ) = 63.

25.

A Project Manager Estimated That He Would Complete The Project In Time If He Hires 42 People For 38 Days. At The End Of 30 Days He Realized That Only 3/5th Of The Work Is Complete. How Many More Men Does He Need To Hire To Complete The Work In Time ?

Answer»

42 people in 30 DAYS did 3/5th of the WORK

Let x people are REQUIRED to complete the REMAINING 2/5th of the work in 8 days

Applying chain rule we get x = 42 × 30/8 × (2×5)/(3×5) = 105 people

Additional no. of people required = ( 105 - 42 ) = 63.

42 people in 30 days did 3/5th of the work

Let x people are required to complete the remaining 2/5th of the work in 8 days

Applying chain rule we get x = 42 × 30/8 × (2×5)/(3×5) = 105 people

Additional no. of people required = ( 105 - 42 ) = 63.

Discussion

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