29 + Interview Questions in Esko Aptitude Interview Questions in Esko Aptitude Tutorial

1.	Alvin, Ben And Clinton Run A Race, With Alvin Finishing 48 Meters Ahead Of Ben And 72 Meters Ahead Of Clinton, While Runner Ben Finishes 32 Meters Ahead Of Runner Clinton. Each Runner Travels The Entire Distance At Constant Speed. What Is The Length Of The Race ?
Answer» Let us assume the SPEEDS of alvin,BEN,clinton = A,B,C resp. and the length of race = d let in TIME t alvin finishes the race At=d Bt=d-48 ------- (1) Ct=d-72 --------(2) So,B:C=(d-48):(d-72) ------- (3) Now Ben covers 48 mt more to complete the race. In the same time Clinton covers 40mt (as it is given that he is beaten by 32mt by ben) So,B:C=48:40....(4) (speed proportional to DISTANCE covered if time is same) By, COMPARING equation (3) and (4) d=192 mt Let us assume the speeds of alvin,ben,clinton = A,B,C resp. and the length of race = d let in time t alvin finishes the race At=d Bt=d-48 ------- (1) Ct=d-72 --------(2) So,B:C=(d-48):(d-72) ------- (3) Now Ben covers 48 mt more to complete the race. In the same time Clinton covers 40mt (as it is given that he is beaten by 32mt by ben) So,B:C=48:40....(4) (speed proportional to distance covered if time is same) By, comparing equation (3) and (4) d=192 mt

1.

Alvin, Ben And Clinton Run A Race, With Alvin Finishing 48 Meters Ahead Of Ben And 72 Meters Ahead Of Clinton, While Runner Ben Finishes 32 Meters Ahead Of Runner Clinton. Each Runner Travels The Entire Distance At Constant Speed. What Is The Length Of The Race ?

Answer»

Let us assume the SPEEDS of alvin,BEN,clinton = A,B,C resp.

and the length of race = d

let in TIME t alvin finishes the race

A*t=d

B*t=d-48 ------- (1)

C*t=d-72 --------(2)

So,B:C=(d-48):(d-72) ------- (3)

Now Ben covers 48 mt more to complete the race.

In the same time Clinton covers 40mt (as it is given that he is beaten by 32mt by ben)

So,B:C=48:40....(4) (speed proportional to DISTANCE covered if time is same)

By, COMPARING equation (3) and (4)

d=192 mt

Let us assume the speeds of alvin,ben,clinton = A,B,C resp.

and the length of race = d

let in time t alvin finishes the race

A*t=d

B*t=d-48 ------- (1)

C*t=d-72 --------(2)

So,B:C=(d-48):(d-72) ------- (3)

Now Ben covers 48 mt more to complete the race.

In the same time Clinton covers 40mt (as it is given that he is beaten by 32mt by ben)

So,B:C=48:40....(4) (speed proportional to distance covered if time is same)

By, comparing equation (3) and (4)

d=192 mt

Discussion

2.	Anusha, Banu And Esha Run A Running Race Of 100 Meters. Anusha Is The Fastest Followed By Banu And Then Esha. Anusha, Banu And Esha Maintain Constant Speeds During The Entire Race. When Anusha Reached The Goal Post, Banu Was 10m Behind. When Banu Reached The Goal Post Esha Was 10m Behind. How Far Was Behind Anusha When The Latter Reached The Goal Post.
Answer» By that time Anusha COVERED 100m, Bhanu covered 90M. So ratio of their speeds = 10 : 9 By that time Bhanu reached 100m, Esha covered 90m. So ratio of their speeds = 10 : 9 Ratio of the speed of all the three = 100 : 90 : 81 By that time Anusha covered 100m, Esha Covers only 81. By that time Anusha covered 100m, Bhanu covered 90m. So ratio of their speeds = 10 : 9 By that time Bhanu reached 100m, Esha covered 90m. So ratio of their speeds = 10 : 9 Ratio of the speed of all the three = 100 : 90 : 81 By that time Anusha covered 100m, Esha Covers only 81.

2.

Anusha, Banu And Esha Run A Running Race Of 100 Meters. Anusha Is The Fastest Followed By Banu And Then Esha. Anusha, Banu And Esha Maintain Constant Speeds During The Entire Race. When Anusha Reached The Goal Post, Banu Was 10m Behind. When Banu Reached The Goal Post Esha Was 10m Behind. How Far Was Behind Anusha When The Latter Reached The Goal Post.

Answer»

By that time Anusha COVERED 100m, Bhanu covered 90M.

So ratio of their speeds = 10 : 9 By that time Bhanu reached 100m, Esha covered 90m.

So ratio of their speeds = 10 : 9 Ratio of the speed of all the three = 100 : 90 : 81

By that time Anusha covered 100m, Esha Covers only 81.

By that time Anusha covered 100m, Bhanu covered 90m.

So ratio of their speeds = 10 : 9 By that time Bhanu reached 100m, Esha covered 90m.

So ratio of their speeds = 10 : 9 Ratio of the speed of all the three = 100 : 90 : 81

By that time Anusha covered 100m, Esha Covers only 81.

Discussion

3.	Three Runners A, B And C Run A Race, With Runner A Finishing 12 Meters Ahead Of Runner B And 18 Meters Ahead Of Runner C, While Runner B Finishes 8 Meters Ahead Of Runner C. Each Runner Travels The Entire Distance At A Constant Speed. What Was The Length Of The Race?
Answer» LETS assume distance of race = x MTRS. Then when A finishes x m , B has run (x- 12)mtrs and C has run (x-18) mtrs. => at this point B is 6 m ahead of C. Now to finish race b needs to run another 12 m, => he RUNS another 12 m. when B finishes race he is 8 m ahead of C. so last 12 m B has run, C has run 10 m.as speeds are constant, => x-12/ x-18 = 12/10 => x = 48 mtrs. Lets assume distance of race = x mtrs. Then when A finishes x m , B has run (x- 12)mtrs and C has run (x-18) mtrs. => at this point B is 6 m ahead of C. Now to finish race b needs to run another 12 m, => he runs another 12 m. when B finishes race he is 8 m ahead of C. so last 12 m B has run, C has run 10 m.as speeds are constant, => x-12/ x-18 = 12/10 => x = 48 mtrs.

3.

Three Runners A, B And C Run A Race, With Runner A Finishing 12 Meters Ahead Of Runner B And 18 Meters Ahead Of Runner C, While Runner B Finishes 8 Meters Ahead Of Runner C. Each Runner Travels The Entire Distance At A Constant Speed. What Was The Length Of The Race?

Answer»

LETS assume distance of race = x MTRS.

Then when A finishes x m , B has run (x- 12)mtrs and C has run (x-18) mtrs.

=> at this point B is 6 m ahead of C. Now to finish race b needs to run another 12 m,

=> he RUNS another 12 m. when B finishes race he is 8 m ahead of C.

so last 12 m B has run, C has run 10 m.as speeds are constant,

=> x-12/ x-18 = 12/10 => x = 48 mtrs.

Lets assume distance of race = x mtrs.

Then when A finishes x m , B has run (x- 12)mtrs and C has run (x-18) mtrs.

=> at this point B is 6 m ahead of C. Now to finish race b needs to run another 12 m,

=> he runs another 12 m. when B finishes race he is 8 m ahead of C.

so last 12 m B has run, C has run 10 m.as speeds are constant,

=> x-12/ x-18 = 12/10 => x = 48 mtrs.

Discussion

4.	A Team Of Workers Was Employed By A Contractor Who Undertook To Finish 360 Pieces Of An Article In A Certain Number Of Days. Making Four More Pieces Per Day Than Was Planned, They Could Complete The Job A Day Ahead Of Schedule. How Many Days Did They Take To Complete The Job ?
Answer» DAYS taken in the general scenario = 360/N; Days taken when 4 articles are prepared extra per day = 360/N + 4; The difference in the day is one, therefore; 360/n - 360/n+4 =1 N2 + 4N – 1440 = 0; N = 36, i.e. number of item prepared in general scenario is 36, and where 4 articles prepared extra is 40. Therefore no of days taken to complete the job = 360/40 = 9. Days taken in the general scenario = 360/N; Days taken when 4 articles are prepared extra per day = 360/N + 4; The difference in the day is one, therefore; 360/n - 360/n+4 =1 N2 + 4N – 1440 = 0; N = 36, i.e. number of item prepared in general scenario is 36, and where 4 articles prepared extra is 40. Therefore no of days taken to complete the job = 360/40 = 9.

4.

A Team Of Workers Was Employed By A Contractor Who Undertook To Finish 360 Pieces Of An Article In A Certain Number Of Days. Making Four More Pieces Per Day Than Was Planned, They Could Complete The Job A Day Ahead Of Schedule. How Many Days Did They Take To Complete The Job ?

Answer»

DAYS taken in the general scenario = 360/N;

Days taken when 4 articles are prepared extra per day = 360/N + 4;

The difference in the day is one, therefore;

360/n - 360/n+4 =1

N2 + 4N – 1440 = 0;

N = 36,

i.e. number of item prepared in general scenario is 36, and where 4 articles prepared extra is 40.

Therefore no of days taken to complete the job = 360/40 = 9.

Days taken in the general scenario = 360/N;

Days taken when 4 articles are prepared extra per day = 360/N + 4;

The difference in the day is one, therefore;

360/n - 360/n+4 =1

N2 + 4N – 1440 = 0;

N = 36,

i.e. number of item prepared in general scenario is 36, and where 4 articles prepared extra is 40.

Therefore no of days taken to complete the job = 360/40 = 9.

Discussion

5.	In A 200 M Race, A Beats B By 20 M And C By 38m. In A Race Of 300 M B Will Beat C By
Answer» If a RUNS 200M then b runs 200-20=180m then C runs 200-38=162m if b runs 300M then c runs 162/180300=270 therefore b BEATS c by 300-270=30 If a runs 200m then b runs 200-20=180m then c runs 200-38=162m if b runs 300m then c runs 162/180300=270 therefore b beats c by 300-270=30

5.

In A 200 M Race, A Beats B By 20 M And C By 38m. In A Race Of 300 M B Will Beat C By

Answer»

If a RUNS 200M then b runs 200-20=180m

then C runs 200-38=162m

if b runs 300M then c runs 162/180*300=270

therefore b BEATS c by 300-270=30

If a runs 200m then b runs 200-20=180m

then c runs 200-38=162m

if b runs 300m then c runs 162/180*300=270

therefore b beats c by 300-270=30

Discussion

6.	In A Race Of 4 Kms A Beats B By 100 M Or 25 Seconds, Then Time Taken By A Is
Answer» B covers 100M in 25 SECONDS B take time =(400025)/100=1000 sec=16 min 40 sec. A takes time =1000 sec-25sec=975 sec= 16 min 25 sec. B covers 100m in 25 seconds B take time =(400025)/100=1000 sec=16 min 40 sec. A takes time =1000 sec-25sec=975 sec= 16 min 25 sec.

6.

In A Race Of 4 Kms A Beats B By 100 M Or 25 Seconds, Then Time Taken By A Is

Answer»

B covers 100M in 25 SECONDS B take time =(4000*25)/100=1000 sec=16 min 40 sec.

A takes time =1000 sec-25sec=975 sec= 16 min 25 sec.

B covers 100m in 25 seconds B take time =(4000*25)/100=1000 sec=16 min 40 sec.

A takes time =1000 sec-25sec=975 sec= 16 min 25 sec.

Discussion

7.	A Can Run 1.5 Km Distance In 2 Min 20 Seconds, While B Can Run This Distance In 2 Min 30 Sec. By How Much Distance Can A Beat B ?
Answer» A takes time 2.20 MIN. = 140 SEC. B takes time 2.30 min. = 150 Sec. Diffrence = (150-140) = 10 Sec. Now we are to find distance covered in 10 sec by B 150 Sec. = 1500 m. 1 Sec=10 m. 10 Sec = 1010=100 m A takes time 2.20 min. = 140 Sec. B takes time 2.30 min. = 150 Sec. Diffrence = (150-140) = 10 Sec. Now we are to find distance covered in 10 sec by B 150 Sec. = 1500 m. 1 Sec=10 m. 10 Sec = 1010=100 m

7.

A Can Run 1.5 Km Distance In 2 Min 20 Seconds, While B Can Run This Distance In 2 Min 30 Sec. By How Much Distance Can A Beat B ?

Answer»

A takes time 2.20 MIN. = 140 SEC. B takes time 2.30 min. = 150 Sec.

Diffrence = (150-140) = 10 Sec.

Now we are to find distance covered in 10 sec by B

150 Sec. = 1500 m.

1 Sec=10 m.

10 Sec = 10*10=100 m

A takes time 2.20 min. = 140 Sec. B takes time 2.30 min. = 150 Sec.

Diffrence = (150-140) = 10 Sec.

Now we are to find distance covered in 10 sec by B

150 Sec. = 1500 m.

1 Sec=10 m.

10 Sec = 10*10=100 m

Discussion

8.	In A 200 M Race A Beats B By 35 M Or 7 Seconds. A's Time Over The Cause Is ?
Answer» B covers 35M in 7 SECONDS B TAKE time = (2007)/35=40 A TAKES time = (40-7)= 33 Sec. B covers 35m in 7 seconds B take time = (2007)/35=40 A takes time = (40-7)= 33 Sec.

8.

In A 200 M Race A Beats B By 35 M Or 7 Seconds. A's Time Over The Cause Is ?

Answer»

B covers 35M in 7 SECONDS B TAKE time = (200*7)/35=40

A TAKES time = (40-7)= 33 Sec.

B covers 35m in 7 seconds B take time = (200*7)/35=40

A takes time = (40-7)= 33 Sec.

Discussion

9.	A Can Run 3 Kms In 3 Min 18 Sec And B Can Run Same Distance In 3 Min 40 Sec, Then By How Much Distance A Can Beat B ?
Answer» GIVEN, 3KM or 3000m Difference is 22 second 3 min 40sec = 220 SEC so, 300022/220 = 300m Given, 3km or 3000m Difference is 22 second 3 min 40sec = 220 sec so, 300022/220 = 300m

9.

A Can Run 3 Kms In 3 Min 18 Sec And B Can Run Same Distance In 3 Min 40 Sec, Then By How Much Distance A Can Beat B ?

Answer»

GIVEN, 3KM or 3000m Difference is 22 second

3 min 40sec = 220 SEC

so, 3000*22/220 = 300m

Given, 3km or 3000m Difference is 22 second

3 min 40sec = 220 sec

so, 3000*22/220 = 300m

Discussion

10.	In A Game Of 80 Points. A Can Give B 5 Points And C 15 Points. Then How Many Points B Can Give C In A Game Of 60 ?
Answer» A can give B 5 points A / B = 80/75 A can give C 15 points A / C = 80/ 65 B can give C a points of ( B / A ) * ( A / C) (75 / 80) * ( 80 / 65) (75 / 65) = 15/13 Need to be FIND in a game of 60 , so MULTIPLY NUMERATOR and DENOMINATOR with 4 such that the ratio BECOMES 60/52 Therefore B can give C of (60-52) = 8 points A can give B 5 points A / B = 80/75 A can give C 15 points A / C = 80/ 65 B can give C a points of ( B / A ) * ( A / C) (75 / 80) * ( 80 / 65) (75 / 65) = 15/13 Need to be find in a game of 60 , so multiply numerator and denominator with 4 such that the ratio becomes 60/52 Therefore B can give C of (60-52) = 8 points

10.

In A Game Of 80 Points. A Can Give B 5 Points And C 15 Points. Then How Many Points B Can Give C In A Game Of 60 ?

Answer»

A can give B 5 points A / B = 80/75

A can give C 15 points A / C = 80/ 65

B can give C a points of ( B / A ) * ( A / C)

(75 / 80) * ( 80 / 65)

(75 / 65) = 15/13

Need to be FIND in a game of 60 , so MULTIPLY NUMERATOR and DENOMINATOR with 4 such that the ratio BECOMES 60/52

Therefore B can give C of (60-52) = 8 points

A can give B 5 points A / B = 80/75

A can give C 15 points A / C = 80/ 65

B can give C a points of ( B / A ) * ( A / C)

(75 / 80) * ( 80 / 65)

(75 / 65) = 15/13

Need to be find in a game of 60 , so multiply numerator and denominator with 4 such that the ratio becomes 60/52

Therefore B can give C of (60-52) = 8 points

Discussion

11.	A Takes 3 Min 45 Seconds To Complete A Kilometre. B Takes 4 Minutes To Complete The Same 1 Km Track. If A And B Were To Participate In A Race Of 2 Kms, How Much Start Can A Give B In Terms Of Distance ?
Answer» A can GIVE B a start of 15 seconds in a km race. B takes 4 minutes to run a km. i.e 1000/4= 250 m/min = 250/60 m/sec THEREFORE, B will cover a distance of = 62.5 meters in 15 seconds. The start that A can give B in a km race therefore, is 62.5 meters, the distance that B run in 15 seconds. Hence in a 2 km race, A can give B a start of 62.5 * 2 = 125 m or 30 seconds. A can give B a start of 15 seconds in a km race. B takes 4 minutes to run a km. i.e 1000/4= 250 m/min = 250/60 m/sec Therefore, B will cover a distance of = 62.5 meters in 15 seconds. The start that A can give B in a km race therefore, is 62.5 meters, the distance that B run in 15 seconds. Hence in a 2 km race, A can give B a start of 62.5 * 2 = 125 m or 30 seconds.

11.

A Takes 3 Min 45 Seconds To Complete A Kilometre. B Takes 4 Minutes To Complete The Same 1 Km Track. If A And B Were To Participate In A Race Of 2 Kms, How Much Start Can A Give B In Terms Of Distance ?

Answer»

A can GIVE B a start of 15 seconds in a km race.

B takes 4 minutes to run a km. i.e 1000/4= 250 m/min = 250/60 m/sec

THEREFORE, B will cover a distance of = 62.5 meters in 15 seconds.

The start that A can give B in a km race therefore, is 62.5 meters, the distance that B run in 15 seconds.

Hence in a 2 km race, A can give B a start of 62.5 * 2 = 125 m or 30 seconds.

A can give B a start of 15 seconds in a km race.

B takes 4 minutes to run a km. i.e 1000/4= 250 m/min = 250/60 m/sec

Therefore, B will cover a distance of = 62.5 meters in 15 seconds.

The start that A can give B in a km race therefore, is 62.5 meters, the distance that B run in 15 seconds.

Hence in a 2 km race, A can give B a start of 62.5 * 2 = 125 m or 30 seconds.

Discussion

12.	In A 100 M Race, A Can Beat B By 25 M And B Can Beat C By 4 M. In The Same Race, A Can Beat C By ?
Answer» we can do a by percentage method A can beat B by 10075/100=75 25 km 7596/100=72 100-72=28 we can do a by percentage method A can beat b by 10075/100=75 25 km 7596/100=72 100-72=28

12.

In A 100 M Race, A Can Beat B By 25 M And B Can Beat C By 4 M. In The Same Race, A Can Beat C By ?

Answer»

we can do a by percentage method

A can beat B by

100*75/100=75

25 km

75*96/100=72

100-72=28

we can do a by percentage method

A can beat b by

100*75/100=75

25 km

75*96/100=72

100-72=28

Discussion

13.	Look At This Series: 1000, 200, 40, …. What Number Should Come Next?
Answer» 8 is the number that comes next. This is a SIMPLE division series. Each number is DIVIDED by 5. 8 is the number that comes next. This is a simple division series. Each number is divided by 5.

13.

Look At This Series: 1000, 200, 40, …. What Number Should Come Next?

Answer»

8 is the number that comes next.

This is a SIMPLE division series. Each number is DIVIDED by 5.

8 is the number that comes next.

This is a simple division series. Each number is divided by 5.

Discussion

14.	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

14.

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

15.	A Merchant Marks His Goods Up By 75% Above His Cost Price. What Is The Maximum % Discount That He Can Offer So That He Ends Up Selling At No Profit Or Loss ?
Answer» Let us assume that the cost PRICE of the article = Rs.100 Therefore, the merchant would have marked it to Rs.100 + 75% of Rs.100 = 100 +75 = 175. Now, if he sells it at no profit or loss, he sells it at the cost price.i.e., he offers a discount of Rs.75 on his SELLING price of Rs.175. Therefore, his % discount = (75/175) X 100= 42.85% Let us assume that the cost price of the article = Rs.100 Therefore, the merchant would have marked it to Rs.100 + 75% of Rs.100 = 100 +75 = 175. Now, if he sells it at no profit or loss, he sells it at the cost price.i.e., he offers a discount of Rs.75 on his selling price of Rs.175. Therefore, his % discount = (75/175) x 100= 42.85%

15.

A Merchant Marks His Goods Up By 75% Above His Cost Price. What Is The Maximum % Discount That He Can Offer So That He Ends Up Selling At No Profit Or Loss ?

Answer»

Let us assume that the cost PRICE of the article = Rs.100

Therefore,

the merchant would have marked it to Rs.100 + 75% of Rs.100 = 100 +75 = 175.

Now, if he sells it at no profit or loss, he sells it at the cost price.i.e., he offers a

discount of Rs.75 on his SELLING price of Rs.175.

Therefore, his % discount = (75/175) X 100= 42.85%

Let us assume that the cost price of the article = Rs.100

Therefore,

the merchant would have marked it to Rs.100 + 75% of Rs.100 = 100 +75 = 175.

Now, if he sells it at no profit or loss, he sells it at the cost price.i.e., he offers a

discount of Rs.75 on his selling price of Rs.175.

Therefore, his % discount = (75/175) x 100= 42.85%

Discussion

16.	A Sum Of Money At Simple Interest Amounts To Rs. 815 In 3 Years And To Rs. 854 In 4 Years. The Sum Is:
Answer» S.I. for 1 YEAR = RS. (854 - 815) = Rs. 39. S.I. for 3 years = Rs.(39 x 3) = Rs. 117. Principal = Rs. (815 - 117) = Rs. 698 S.I. for 1 year = Rs. (854 - 815) = Rs. 39. S.I. for 3 years = Rs.(39 x 3) = Rs. 117. Principal = Rs. (815 - 117) = Rs. 698

16.

A Sum Of Money At Simple Interest Amounts To Rs. 815 In 3 Years And To Rs. 854 In 4 Years. The Sum Is:

Answer»

S.I. for 1 YEAR = RS. (854 - 815) = Rs. 39.

S.I. for 3 years = Rs.(39 x 3) = Rs. 117.

Principal = Rs. (815 - 117) = Rs. 698

S.I. for 1 year = Rs. (854 - 815) = Rs. 39.

S.I. for 3 years = Rs.(39 x 3) = Rs. 117.

Principal = Rs. (815 - 117) = Rs. 698

Discussion

17.	There Are 6 Consecutive Odd Numbers. The Difference Between The Square Of The Average Of The First Three Numbers And The Square Of The Average Of The Last Three Numbers Is 288. What Is The Last Odd Number ?
Answer» LET the 6 CONSECUTIVE odd no.’s are: X, X+2, X+4, X+6, X+8, X+10 Avg. of 1st three no’s is X+2. Avg. of LAST three no’s is X+8. Given that (X+8)2-(X+2)2=288 X=19 Last Odd no. is X+10= 29. Let the 6 consecutive odd no.’s are: X, X+2, X+4, X+6, X+8, X+10 Avg. of 1st three no’s is X+2. Avg. of Last three no’s is X+8. Given that (X+8)2-(X+2)2=288 X=19 Last Odd no. is X+10= 29.

17.

There Are 6 Consecutive Odd Numbers. The Difference Between The Square Of The Average Of The First Three Numbers And The Square Of The Average Of The Last Three Numbers Is 288. What Is The Last Odd Number ?

Answer»

LET the 6 CONSECUTIVE odd no.’s are:

X, X+2, X+4, X+6, X+8, X+10

Avg. of 1st three no’s is X+2.

Avg. of LAST three no’s is X+8.

Given that (X+8)2-(X+2)2=288

X=19

Last Odd no. is X+10= 29.

Let the 6 consecutive odd no.’s are:

X, X+2, X+4, X+6, X+8, X+10

Avg. of 1st three no’s is X+2.

Avg. of Last three no’s is X+8.

Given that (X+8)2-(X+2)2=288

X=19

Last Odd no. is X+10= 29.

Discussion

18.	A Boat Can Travel With A Speed Of 13 Km / Hr In Still Water. If The Speed Of The Stream Is 4 Km / Hr. Find The Time Taken By The Boat To Go 68 Km Downstream?
Answer» SPEED Downstream= (13 + 4) km/hr = 17 km/hr. Time TAKEN to TRAVEL 68 km downstream =(68 / 17)HRS = 4 hrs. Speed Downstream= (13 + 4) km/hr = 17 km/hr. Time taken to travel 68 km downstream =(68 / 17)hrs = 4 hrs.

18.

A Boat Can Travel With A Speed Of 13 Km / Hr In Still Water. If The Speed Of The Stream Is 4 Km / Hr. Find The Time Taken By The Boat To Go 68 Km Downstream?

Answer»

SPEED Downstream= (13 + 4) km/hr

= 17 km/hr.

Time TAKEN to TRAVEL 68 km downstream =(68 / 17)HRS

= 4 hrs.

Speed Downstream= (13 + 4) km/hr

= 17 km/hr.

Time taken to travel 68 km downstream =(68 / 17)hrs

= 4 hrs.

Discussion

19.	Two Pipes Can Fill The Cistern In 10hr And 12 Hr Respectively, While The Third Empty It In 20hr. If All Pipes Are Opened Simultaneously, Then The Cistern Will Be Filled In
Answer» WORK done by all the TANKS working together in 1 hour. => 1/10 + 1/12 - 1/20 = 2/15 Hence, tank will be FILLED in 15/2 = 7.5 hour Work done by all the tanks working together in 1 hour. => 1/10 + 1/12 - 1/20 = 2/15 Hence, tank will be filled in 15/2 = 7.5 hour

19.

Two Pipes Can Fill The Cistern In 10hr And 12 Hr Respectively, While The Third Empty It In 20hr. If All Pipes Are Opened Simultaneously, Then The Cistern Will Be Filled In

Answer»

WORK done by all the TANKS working together in 1 hour.

=> 1/10 + 1/12 - 1/20 = 2/15

Hence, tank will be FILLED in 15/2 = 7.5 hour

Work done by all the tanks working together in 1 hour.

=> 1/10 + 1/12 - 1/20 = 2/15

Hence, tank will be filled in 15/2 = 7.5 hour

Discussion

20.	Two Students Appeared At An Examination. One Of Them Secured 9 Marks More Than The Other And His Marks Was 56% Of The Sum Of Their Marks. The Marks Obtained By Them Are:
Answer» LET their MARKS be (x + 9) and x. Then, x + 9 = (56/100)(x + 9 + x) 25(x + 9) = 14(2x + 9) 3x = 99 x = 33 So, their marks are 42 and 33. Let their marks be (x + 9) and x. Then, x + 9 = (56/100)(x + 9 + x) 25(x + 9) = 14(2x + 9) 3x = 99 x = 33 So, their marks are 42 and 33.

20.

Two Students Appeared At An Examination. One Of Them Secured 9 Marks More Than The Other And His Marks Was 56% Of The Sum Of Their Marks. The Marks Obtained By Them Are:

Answer»

LET their MARKS be (x + 9) and x.

Then, x + 9 = (56/100)(x + 9 + x)

25(x + 9) = 14(2x + 9)

3x = 99

x = 33

So, their marks are 42 and 33.

Let their marks be (x + 9) and x.

Then, x + 9 = (56/100)(x + 9 + x)

25(x + 9) = 14(2x + 9)

3x = 99

x = 33

So, their marks are 42 and 33.

Discussion

21.	A Merchant Marks His Goods Up By 75% Above His Cost Price. What Is The Maximum % Discount That He Can Offer So That He Ends Up Selling At No Profit Or Loss?
Answer» Let us assume that the COST PRICE of the article = Rs.100 THEREFORE, the merchant would have MARKED it to Rs.100 + 75% of Rs.100 = 100 + 75 = 175. Now, if he sells it at no profit or loss, he sells it at the cost price.i.e. he offers a discount of Rs.75 on his selling price of Rs.175 Therefore, his % discount = (75/175) x 100= 42.85% Let us assume that the cost price of the article = Rs.100 Therefore, the merchant would have marked it to Rs.100 + 75% of Rs.100 = 100 + 75 = 175. Now, if he sells it at no profit or loss, he sells it at the cost price.i.e. he offers a discount of Rs.75 on his selling price of Rs.175 Therefore, his % discount = (75/175) x 100= 42.85%

21.

A Merchant Marks His Goods Up By 75% Above His Cost Price. What Is The Maximum % Discount That He Can Offer So That He Ends Up Selling At No Profit Or Loss?

Answer»

Let us assume that the COST PRICE of the article = Rs.100

THEREFORE, the merchant would have MARKED it to Rs.100 + 75% of Rs.100 = 100 + 75 = 175.

Now, if he sells it at no profit or loss, he sells it at the cost price.i.e. he offers a discount of Rs.75 on his selling price of Rs.175

Therefore, his % discount = (75/175) x 100= 42.85%

Let us assume that the cost price of the article = Rs.100

Therefore, the merchant would have marked it to Rs.100 + 75% of Rs.100 = 100 + 75 = 175.

Now, if he sells it at no profit or loss, he sells it at the cost price.i.e. he offers a discount of Rs.75 on his selling price of Rs.175

Therefore, his % discount = (75/175) x 100= 42.85%

Discussion

22.	A Train Left Station At A Hour B Minutes. It Reached Station Y At B Hour C Minutes On The Same Day, After Travelling C Hours A Minutes (clock Shows Time From 0 Hours To 24 Hours). Number Of Possible Value Of A Is?
Answer» A hours + C ours = B hours .....(i) A, C and B cannot have VALUES greater than or EQUAL to 24 B minutes + A minutes = C minutes .....(ii) Looking at two equation, we get no value of A satisfies both equation A hours + C ours = B hours .....(i) A, C and B cannot have values greater than or equal to 24 B minutes + A minutes = C minutes .....(ii) Looking at two equation, we get no value of A satisfies both equation

22.

A Train Left Station At A Hour B Minutes. It Reached Station Y At B Hour C Minutes On The Same Day, After Travelling C Hours A Minutes (clock Shows Time From 0 Hours To 24 Hours). Number Of Possible Value Of A Is?

Answer»

A hours + C ours = B hours .....(i)

A, C and B cannot have VALUES greater than or EQUAL to 24

B minutes + A minutes = C minutes .....(ii)

Looking at two equation, we get no value of A satisfies both equation

A hours + C ours = B hours .....(i)

A, C and B cannot have values greater than or equal to 24

B minutes + A minutes = C minutes .....(ii)

Looking at two equation, we get no value of A satisfies both equation

Discussion

23.	Sakshi Can Do A Piece Of Work In 20 Days. Tanya Is 25% More Efficient Than Sakshi. The Number Of Days Taken By Tanya To Do The Same Piece Of Work Is:
Answer» RATIO of times TAKEN by Sakshi and Tanya = 125:100 = 5:4. Suppose Tanya takes x days to do the work. 5 : 4 :: 20 : x ? x =(4 x 20)/5 ? x = 16 days Hence, Tanya takes 16 days to COMPLETE the work. Ratio of times taken by Sakshi and Tanya = 125:100 = 5:4. Suppose Tanya takes x days to do the work. 5 : 4 :: 20 : x ? x =(4 x 20)/5 ? x = 16 days Hence, Tanya takes 16 days to complete the work.

23.

Sakshi Can Do A Piece Of Work In 20 Days. Tanya Is 25% More Efficient Than Sakshi. The Number Of Days Taken By Tanya To Do The Same Piece Of Work Is:

Answer»

RATIO of times TAKEN by Sakshi and Tanya = 125:100 = 5:4.

Suppose Tanya takes x days to do the work.

5 : 4 :: 20 : x ? x =(4 x 20)/5

? x = 16 days

Hence, Tanya takes 16 days to COMPLETE the work.

Ratio of times taken by Sakshi and Tanya = 125:100 = 5:4.

Suppose Tanya takes x days to do the work.

5 : 4 :: 20 : x ? x =(4 x 20)/5

? x = 16 days

Hence, Tanya takes 16 days to complete the work.

Discussion

24.	A Father Tells His Son, “i Was Of Your Present Age When You Were Born”. If The Father Is 36 Now, How Old Was The Boy 5 Years Back ?
Answer» LET the FATHER’s AGE be x and the son’s age be y. Then, x-y = y or x = 2Y Now, x = 36. So, 2y = 36 or y = 18. Therefore son’s present age = 18 years. So, son’s age 5 years ago = 13 years. Let the father’s age be x and the son’s age be y. Then, x-y = y or x = 2y Now, x = 36. So, 2y = 36 or y = 18. Therefore son’s present age = 18 years. So, son’s age 5 years ago = 13 years.

24.

A Father Tells His Son, “i Was Of Your Present Age When You Were Born”. If The Father Is 36 Now, How Old Was The Boy 5 Years Back ?

Answer»

LET the FATHER’s AGE be x and the son’s age be y. Then, x-y = y or x = 2Y

Now, x = 36. So, 2y = 36 or y = 18.

Therefore son’s present age = 18 years.

So, son’s age 5 years ago = 13 years.

Let the father’s age be x and the son’s age be y. Then, x-y = y or x = 2y

Now, x = 36. So, 2y = 36 or y = 18.

Therefore son’s present age = 18 years.

So, son’s age 5 years ago = 13 years.

Discussion

25.	Read The Following Information Carefully And Answer The Questions That Follow : A + B Means A Is The Son Of B; A – B Means A Is The Wife Of B; A*b Means A Is The Brother Of B; A/b Means A Is The Mother Of B And A = B Means A Is The Sister Of B. What Does P + R –q Mean ?
Answer» <P>Clearly , P + R – Q MEANS P is the SON of R who is the wife of Q i.e. Q is the father of P. Clearly , P + R – Q means P is the son of R who is the wife of Q i.e. Q is the father of P.

25.

Read The Following Information Carefully And Answer The Questions That Follow : A + B Means A Is The Son Of B; A – B Means A Is The Wife Of B; A*b Means A Is The Brother Of B; A/b Means A Is The Mother Of B And A = B Means A Is The Sister Of B. What Does P + R –q Mean ?

Answer»

<P>Clearly , P + R – Q MEANS P is the SON of R who is the wife of Q i.e. Q is the father of P.

Clearly , P + R – Q means P is the son of R who is the wife of Q i.e. Q is the father of P.

Discussion

26.	May 6, 1993 Was Thursday. What Day Of The Week Was On May 6, 1992 ?
Answer» 1992 being a leap year, it has 2 odd DAYS. So, the day on May, 1993 is 2 days beyond the day on May 6, 1992. But, on May 6, 1993 it was THURSDAY. So, on May 6, 1992 it was TUESDAY. 1992 being a leap year, it has 2 odd days. So, the day on May, 1993 is 2 days beyond the day on May 6, 1992. But, on May 6, 1993 it was Thursday. So, on May 6, 1992 it was Tuesday.

26.

May 6, 1993 Was Thursday. What Day Of The Week Was On May 6, 1992 ?

Answer»

1992 being a leap year, it has 2 odd DAYS.

So, the day on May, 1993 is 2 days beyond the day on May 6, 1992.

But, on May 6, 1993 it was THURSDAY.

So, on May 6, 1992 it was TUESDAY.

1992 being a leap year, it has 2 odd days.

So, the day on May, 1993 is 2 days beyond the day on May 6, 1992.

But, on May 6, 1993 it was Thursday.

So, on May 6, 1992 it was Tuesday.

Discussion

27.	Pointing Out To A Lady, Rajan Said, “she Is The Daughter Of The Woman Who Is The Mother Of The Husband Of My Mother. “who Is The Lady To Rajan ?
Answer» Mother’s HUSBAND – FATHER: Father’s mother – Grandmother: Grandmother’s daughter – Father’s sister: Father’s sister – AUNT. So, the LADY is Rajan’s Aunt. Mother’s husband – Father: Father’s mother – Grandmother: Grandmother’s daughter – Father’s sister: Father’s sister – Aunt. So, the lady is Rajan’s Aunt.

27.

Pointing Out To A Lady, Rajan Said, “she Is The Daughter Of The Woman Who Is The Mother Of The Husband Of My Mother. “who Is The Lady To Rajan ?

Answer»

Mother’s HUSBAND – FATHER: Father’s mother – Grandmother: Grandmother’s daughter – Father’s sister: Father’s sister – AUNT.

So, the LADY is Rajan’s Aunt.

Mother’s husband – Father: Father’s mother – Grandmother: Grandmother’s daughter – Father’s sister: Father’s sister – Aunt.

So, the lady is Rajan’s Aunt.

Discussion

28.	Look At This Series: 80, 10, 70, 15, 60, ... What Number Should Come Next ?
Answer» This is an alternating ADDITION and subtraction series. In the first PATTERN, 10 is subtracted from each NUMBER to arrive at the next. In the second, 5 is ADDED to each number to arrive at the next This is an alternating addition and subtraction series. In the first pattern, 10 is subtracted from each number to arrive at the next. In the second, 5 is added to each number to arrive at the next

28.

Look At This Series: 80, 10, 70, 15, 60, ... What Number Should Come Next ?

Answer»

This is an alternating ADDITION and subtraction series. In the first PATTERN, 10 is subtracted from each NUMBER to arrive at the next. In the second, 5 is ADDED to each number to arrive at the next

This is an alternating addition and subtraction series. In the first pattern, 10 is subtracted from each number to arrive at the next. In the second, 5 is added to each number to arrive at the next

Discussion

29.	The Difference Between The Simple Interest And Compound Interest On A Certain Sum Of Money For 2 Years At 15% P. A. Is Rs. 45. Find The Sum.
Answer» Since we know that the interest RATE is 0.15, and KNOWING that the difference between two YEARS of compound interest is nothing but interest on interest, we can find the first year’s interest as – 45/0.15 = 300. Now if the interest is 300 at the end of ONE year, then the principal is 300 / 0.15 = 2,000 Since we know that the interest rate is 0.15, and knowing that the difference between two years of compound interest is nothing but interest on interest, we can find the first year’s interest as – 45/0.15 = 300. Now if the interest is 300 at the end of one year, then the principal is 300 / 0.15 = 2,000

29.

The Difference Between The Simple Interest And Compound Interest On A Certain Sum Of Money For 2 Years At 15% P. A. Is Rs. 45. Find The Sum.

Answer»

Since we know that the interest RATE is 0.15, and KNOWING that the difference between two YEARS of compound interest is nothing but interest on interest, we can find the first year’s interest as –

45/0.15 = 300.

Now if the interest is 300 at the end of ONE year, then the principal is 300 / 0.15 = 2,000

Since we know that the interest rate is 0.15, and knowing that the difference between two years of compound interest is nothing but interest on interest, we can find the first year’s interest as –

45/0.15 = 300.

Now if the interest is 300 at the end of one year, then the principal is 300 / 0.15 = 2,000

Discussion

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