30 + Interview Questions in Sutherland Aptitude Interview Questions in Sutherland Aptitude Tutorial Page 1 InterviewSolution

1.	A Dealer Buys Dry Fruits At The Rate Of ` 100, ` 80 And ` 60 Per Kg. He Bought Them In The Ratio 12 : 15 : 20 By Weight. He In Total Gets 20% Profit By Selling The First Two And At Last He Finds He Has No Gain Or No Loss In Selling The Whole Quantity Which He Had. What Was The Percentage Loss He Suffered For The Third Quantity?
Answer» Total quantity RATE =(12 * 100 + 15 * 80 + 20 * 60) = 3600 For first 2 quantity, (12 * 100) + (15 * 80) = 2400 But he GETS 20% profit = 2400 * 1.2 = 2880 So the third quantity = 3600 – 2880 = 720 Actual third quantity rate = 20 * 60 = 1200 Loss suffered = (1200 - 720) / 1200 = 480/1200 = 40 %. Total quantity rate =(12 * 100 + 15 * 80 + 20 * 60) = 3600 For first 2 quantity, (12 * 100) + (15 * 80) = 2400 But he gets 20% profit = 2400 * 1.2 = 2880 So the third quantity = 3600 – 2880 = 720 Actual third quantity rate = 20 * 60 = 1200 Loss suffered = (1200 - 720) / 1200 = 480/1200 = 40 %.

1.

A Dealer Buys Dry Fruits At The Rate Of ` 100, ` 80 And ` 60 Per Kg. He Bought Them In The Ratio 12 : 15 : 20 By Weight. He In Total Gets 20% Profit By Selling The First Two And At Last He Finds He Has No Gain Or No Loss In Selling The Whole Quantity Which He Had. What Was The Percentage Loss He Suffered For The Third Quantity?

Answer»

Total quantity RATE =(12 * 100 + 15 * 80 + 20 * 60) = 3600

For first 2 quantity, (12 * 100) + (15 * 80) = 2400

But he GETS 20% profit = 2400 * 1.2 = 2880

So the third quantity = 3600 – 2880 = 720

Actual third quantity rate = 20 * 60 = 1200

Loss suffered = (1200 - 720) / 1200

= 480/1200 = 40 %.

Total quantity rate =(12 * 100 + 15 * 80 + 20 * 60) = 3600

For first 2 quantity, (12 * 100) + (15 * 80) = 2400

But he gets 20% profit = 2400 * 1.2 = 2880

So the third quantity = 3600 – 2880 = 720

Actual third quantity rate = 20 * 60 = 1200

Loss suffered = (1200 - 720) / 1200

= 480/1200 = 40 %.

Discussion

2.	A Reservoir Is Provided By Two Pipes A And B. A Can Fill The Reservoir 5 Hours Faster Than B. If Both Together Fill The Reservoir In 6 Hours, The Reservoir Will Be Filled By A Alone In?
Answer» If x is the speed then speed of A= x + 5 and B = x Time taken by A and B will be x and x + 5 RESP. 1/x + 1/x + 5 = 1/6 ; X2 – 7x - 30 = 0 x = -3 or x = 10. Since time can’t be negative, x =10. If x is the speed then speed of A= x + 5 and B = x Time taken by A and B will be x and x + 5 resp. 1/x + 1/x + 5 = 1/6 ; x2 – 7x - 30 = 0 x = -3 or x = 10. Since time can’t be negative, x =10.

2.

A Reservoir Is Provided By Two Pipes A And B. A Can Fill The Reservoir 5 Hours Faster Than B. If Both Together Fill The Reservoir In 6 Hours, The Reservoir Will Be Filled By A Alone In?

Answer»

If x is the speed then speed of A= x + 5 and B = x

Time taken by A and B will be x and x + 5 RESP.

1/x + 1/x + 5 = 1/6 ; X2 – 7x - 30 = 0 x = -3 or x = 10.

Since time can’t be negative, x =10.

If x is the speed then speed of A= x + 5 and B = x

Time taken by A and B will be x and x + 5 resp.

1/x + 1/x + 5 = 1/6 ; x2 – 7x - 30 = 0 x = -3 or x = 10.

Since time can’t be negative, x =10.

Discussion

3.	Six Pipes Are Fitted To A Water Tank. Some Of These Are Inlet Pipes And The Others Outlet Pipes. Each Inlet Pipe Can Fill The Tank In 9 Hours And Each Outlet Pipe Can Empty The Tank In 6 Hours. On Opening All The Pipes, An Empty Tank Is Filled In 9 Hours. How Many Inlet Pipes Are There?
Answer» x / 9 – y / 6 = 1/9; 2X – 3y = 2 We can compute that x = 4 and y = 2. Thus, Inlet PIPE = x = 4. x / 9 – y / 6 = 1/9; 2x – 3y = 2 We can compute that x = 4 and y = 2. Thus, Inlet pipe = x = 4.

3.

Six Pipes Are Fitted To A Water Tank. Some Of These Are Inlet Pipes And The Others Outlet Pipes. Each Inlet Pipe Can Fill The Tank In 9 Hours And Each Outlet Pipe Can Empty The Tank In 6 Hours. On Opening All The Pipes, An Empty Tank Is Filled In 9 Hours. How Many Inlet Pipes Are There?

Answer»

x / 9 – y / 6 = 1/9;

2X – 3y = 2

We can compute that x = 4 and y = 2.

Thus, Inlet PIPE = x = 4.

x / 9 – y / 6 = 1/9;

2x – 3y = 2

We can compute that x = 4 and y = 2.

Thus, Inlet pipe = x = 4.

Discussion

4.	Sanjay Invested An Amount Of Rs 16,000 For Two Years On Compound Interest And Received An Amount Of Rs 17,640 On Maturity. What Is The Rate Of Interest Per Annum?
Answer» <P>AMOUNT =P(1+r/100)2 17640/16000= (1+r/100)2 441/400= (1+r/100)2 =&GT; (21/20)2= (1+r/100)2 R=5%. Amount =P(1+r/100)2 17640/16000= (1+r/100)2 441/400= (1+r/100)2 => (21/20)2= (1+r/100)2 R=5%.

4.

Sanjay Invested An Amount Of Rs 16,000 For Two Years On Compound Interest And Received An Amount Of Rs 17,640 On Maturity. What Is The Rate Of Interest Per Annum?

Answer»

<P>AMOUNT =P(1+r/100)2

17640/16000= (1+r/100)2

441/400= (1+r/100)2

=&GT; (21/20)2= (1+r/100)2

R=5%.

Amount =P(1+r/100)2

17640/16000= (1+r/100)2

441/400= (1+r/100)2

=> (21/20)2= (1+r/100)2

R=5%.

Discussion

5.	A Certain Function F Satisfies The Equation F(x)+2*f(6-x)=x For All Real Numbers X. The Value Of F(1) Is?
Answer» PUT x =1 ⇒ F(1)+2f(6-1) = 1 ⇒ f(1) + 2f(5) = 1 Put x = 5 ⇒ f(5)+2f(6-5) = 5 ⇒ f(5) + 2f(1) = 5 Put f(5) = 5 - 2f(1) in the first equation ⇒ f(1) + 2(5 - 2f(1)) = 1 ⇒ f(1) + 10 - 4f(1) = 1 ⇒ f(1) = 3. Put x =1 ⇒ f(1)+2f(6-1) = 1 ⇒ f(1) + 2f(5) = 1 Put x = 5 ⇒ f(5)+2f(6-5) = 5 ⇒ f(5) + 2f(1) = 5 Put f(5) = 5 - 2f(1) in the first equation ⇒ f(1) + 2(5 - 2f(1)) = 1 ⇒ f(1) + 10 - 4f(1) = 1 ⇒ f(1) = 3.

5.

A Certain Function F Satisfies The Equation F(x)+2*f(6-x)=x For All Real Numbers X. The Value Of F(1) Is?

Answer»

PUT x =1 ⇒ F(1)+2*f(6-1) = 1 ⇒ f(1) + 2*f(5) = 1

Put x = 5 ⇒ f(5)+2*f(6-5) = 5 ⇒ f(5) + 2*f(1) = 5

Put f(5) = 5 - 2*f(1) in the first equation

⇒ f(1) + 2*(5 - 2*f(1)) = 1

⇒ f(1) + 10 - 4f(1) = 1

⇒ f(1) = 3.

Put x =1 ⇒ f(1)+2*f(6-1) = 1 ⇒ f(1) + 2*f(5) = 1

Put x = 5 ⇒ f(5)+2*f(6-5) = 5 ⇒ f(5) + 2*f(1) = 5

Put f(5) = 5 - 2*f(1) in the first equation

⇒ f(1) + 2*(5 - 2*f(1)) = 1

⇒ f(1) + 10 - 4f(1) = 1

⇒ f(1) = 3.

Discussion

6.	Find The Probability That A Leap Year Chosen At Random Will Have 53 Sundays?
Answer» A leap year has 366 day which is 52 FULL WEEKS + 2 odd days. Now these two odd days may be (sun + mon), (mon + tue), .... (Sat + sun). Now there are TOTAL 7 ways. Of which Sunday appeared two TIMES. So answer 2/7. A leap year has 366 day which is 52 full weeks + 2 odd days. Now these two odd days may be (sun + mon), (mon + tue), .... (Sat + sun). Now there are total 7 ways. Of which Sunday appeared two times. So answer 2/7.

6.

Find The Probability That A Leap Year Chosen At Random Will Have 53 Sundays?

Answer»

A leap year has 366 day which is 52 FULL WEEKS + 2 odd days.

Now these two odd days may be (sun + mon), (mon + tue), .... (Sat + sun).

Now there are TOTAL 7 ways. Of which Sunday appeared two TIMES. So answer 2/7.

A leap year has 366 day which is 52 full weeks + 2 odd days.

Now these two odd days may be (sun + mon), (mon + tue), .... (Sat + sun).

Now there are total 7 ways. Of which Sunday appeared two times. So answer 2/7.

Discussion

7.	A Man Has A Job, Which Requires Him To Work 8 Straight Days And Rest On The Ninth Day. If He Started Work On Monday, Find The Day Of The Week On Which He Gets His 12th Rest Day?
Answer» He works for 8 days and takes REST on the 9th day. So On the 12th rest day, there are 9 x 12 = 108 days PASSED. Number of odd days = (108 - 1) / 7 = 107 / 7 = 2. So the 12th rest day is wednesday. He works for 8 days and takes rest on the 9th day. So On the 12th rest day, there are 9 x 12 = 108 days passed. Number of odd days = (108 - 1) / 7 = 107 / 7 = 2. So the 12th rest day is wednesday.

7.

A Man Has A Job, Which Requires Him To Work 8 Straight Days And Rest On The Ninth Day. If He Started Work On Monday, Find The Day Of The Week On Which He Gets His 12th Rest Day?

Answer»

He works for 8 days and takes REST on the 9th day.

So On the 12th rest day, there are 9 x 12 = 108 days PASSED.

Number of odd days = (108 - 1) / 7 = 107 / 7 = 2. So the 12th rest day is wednesday.

He works for 8 days and takes rest on the 9th day.

So On the 12th rest day, there are 9 x 12 = 108 days passed.

Number of odd days = (108 - 1) / 7 = 107 / 7 = 2. So the 12th rest day is wednesday.

Discussion

8.	Rajiv Can Do A Piece Of Work In 10 Days , Venky In 12 Days And Ravi In 15 Days. They All Start The Work Together, But Rajiv Leaves After 2 Days And Venky Leaves 3 Days Before The Work Is Completed. In How Many Days Is The Work Completed ?
Answer» Let the work be 60 units. If venky worked for 3 days, and the REMAINING days of work be x days, total days to COMPLETE the work be x + 3 days. Now Capacities of RAJIV is 60/10 = 6, Venky is 5, Ravi is 4. (6 + 5 + 4) 2 + (5 + 4) (x - 3) + 5 x 3 = 60. Solving we get x = 4. So total days to complete the work is 7 days. Let the work be 60 units. If venky worked for 3 days, and the remaining days of work be x days, total days to complete the work be x + 3 days. Now Capacities of Rajiv is 60/10 = 6, Venky is 5, Ravi is 4. (6 + 5 + 4) 2 + (5 + 4) (x - 3) + 5 x 3 = 60. Solving we get x = 4. So total days to complete the work is 7 days.

8.

Rajiv Can Do A Piece Of Work In 10 Days , Venky In 12 Days And Ravi In 15 Days. They All Start The Work Together, But Rajiv Leaves After 2 Days And Venky Leaves 3 Days Before The Work Is Completed. In How Many Days Is The Work Completed ?

Answer»

Let the work be 60 units. If venky worked for 3 days, and the REMAINING days of work be x days, total days to COMPLETE the work be x + 3 days.

Now Capacities of RAJIV is 60/10 = 6, Venky is 5, Ravi is 4.

(6 + 5 + 4) 2 + (5 + 4) (x - 3) + 5 x 3 = 60.

Solving we get x = 4. So total days to complete the work is 7 days.

Let the work be 60 units. If venky worked for 3 days, and the remaining days of work be x days, total days to complete the work be x + 3 days.

Now Capacities of Rajiv is 60/10 = 6, Venky is 5, Ravi is 4.

(6 + 5 + 4) 2 + (5 + 4) (x - 3) + 5 x 3 = 60.

Solving we get x = 4. So total days to complete the work is 7 days.

Discussion

9.	Mark Told John "if You Give Me Half Your Money I Will Have Rs.75. John Said, "if You Give Me One Third Of Your Money, I Will Have Rs.75/- How Much Money Did John Have ?
Answer» Let the MONEY with MARK and John are M and J respectively. Now M + J/2 = 75 M/3 + J = 75 Solving we get M = 45, and J = 60. Let the money with Mark and John are M and J respectively. Now M + J/2 = 75 M/3 + J = 75 Solving we get M = 45, and J = 60.

9.

Mark Told John "if You Give Me Half Your Money I Will Have Rs.75. John Said, "if You Give Me One Third Of Your Money, I Will Have Rs.75/- How Much Money Did John Have ?

Answer»

Let the MONEY with MARK and John are M and J respectively.

Now

M + J/2 = 75

M/3 + J = 75

Solving we get M = 45, and J = 60.

Let the money with Mark and John are M and J respectively.

Now

M + J/2 = 75

M/3 + J = 75

Solving we get M = 45, and J = 60.

Discussion

10.	Jake Can Dig A Well In 16 Days. Paul Can Dig The Same Well In 24 Days. Jake, Paul And Hari Together Dig The Well In 8 Days. Hari Alone Can Dig The Well In?
Answer» SIMPLE one. LET the total WORK to be done is 48 meters. Now Jake can dig 3 mts, Paul can dig 2 mts a day. Now all of them combined dug in 8 days so per day they dug 48/8 = 6 mts. So Of these 8 mts, Hari capacity is 1 mt. So he takes 48 /1 = 48 days to complete the digging JOB. Simple one. Let the total work to be done is 48 meters. Now Jake can dig 3 mts, Paul can dig 2 mts a day. Now all of them combined dug in 8 days so per day they dug 48/8 = 6 mts. So Of these 8 mts, Hari capacity is 1 mt. So he takes 48 /1 = 48 days to complete the digging job.

10.

Jake Can Dig A Well In 16 Days. Paul Can Dig The Same Well In 24 Days. Jake, Paul And Hari Together Dig The Well In 8 Days. Hari Alone Can Dig The Well In?

Answer»

SIMPLE one. LET the total WORK to be done is 48 meters.

Now Jake can dig 3 mts, Paul can dig 2 mts a day.

Now all of them combined dug in 8 days so per day they dug 48/8 = 6 mts.

So Of these 8 mts, Hari capacity is 1 mt.

So he takes 48 /1 = 48 days to complete the digging JOB.

Simple one. Let the total work to be done is 48 meters.

Now Jake can dig 3 mts, Paul can dig 2 mts a day.

Now all of them combined dug in 8 days so per day they dug 48/8 = 6 mts.

So Of these 8 mts, Hari capacity is 1 mt.

So he takes 48 /1 = 48 days to complete the digging job.

Discussion

11.	Four People Each Roll A Four Die Once. Find The Probability That At Least Two People Will Roll The Same Number ?
Answer» The number of ways of rolling a DICE where no two numbers probability that no ONE rolls the same number = 6 x 5 x 4 x 3 Now total possibilities of rolling a dice = 64 The probability that a no one gets the same number = 6×5×4×364=518 So the probability that at least two people gets same number = 1?518=1318. The number of ways of rolling a dice where no two numbers probability that no one rolls the same number = 6 x 5 x 4 x 3 Now total possibilities of rolling a dice = 64 The probability that a no one gets the same number = 6×5×4×364=518 So the probability that at least two people gets same number = 1?518=1318.

11.

Four People Each Roll A Four Die Once. Find The Probability That At Least Two People Will Roll The Same Number ?

Answer»

The number of ways of rolling a DICE where no two numbers probability that no ONE rolls the same number = 6 x 5 x 4 x 3

Now total possibilities of rolling a dice = 64

The probability that a no one gets the same number = 6×5×4×364=518

So the probability that at least two people gets same number = 1?518=1318.

The number of ways of rolling a dice where no two numbers probability that no one rolls the same number = 6 x 5 x 4 x 3

Now total possibilities of rolling a dice = 64

The probability that a no one gets the same number = 6×5×4×364=518

So the probability that at least two people gets same number = 1?518=1318.

Discussion

12.	The Fourteen Digits Of A Credit Card Are To Be Written In The Boxes Shown Above. If The Sum Of Every Three Consecutive Digits Is 18, Then The Value Of X Is?
Answer» Let US assume right most two squares are a , b Then Sum of all the squares = 18 x 4 + a + b .......... (1) ALSO Sum of the squares before 7 = 18 Sum of the squares between 7, x = 18 and sum of the squares between x , 8 = 18 So Sum of the 14 squares = 18 + 7 + 18 + x + 18 + 8 + a + b...... (2) EQUATING 1 and 2 we get x = 3. Let us assume right most two squares are a , b Then Sum of all the squares = 18 x 4 + a + b .......... (1) Also Sum of the squares before 7 = 18 Sum of the squares between 7, x = 18 and sum of the squares between x , 8 = 18 So Sum of the 14 squares = 18 + 7 + 18 + x + 18 + 8 + a + b...... (2) Equating 1 and 2 we get x = 3.

12.

The Fourteen Digits Of A Credit Card Are To Be Written In The Boxes Shown Above. If The Sum Of Every Three Consecutive Digits Is 18, Then The Value Of X Is?

Answer»

Let US assume right most two squares are a , b

Then Sum of all the squares = 18 x 4 + a + b .......... (1)

ALSO Sum of the squares before 7 = 18

Sum of the squares between 7, x = 18

and sum of the squares between x , 8 = 18

So Sum of the 14 squares = 18 + 7 + 18 + x + 18 + 8 + a + b...... (2)

EQUATING 1 and 2 we get x = 3.

Let us assume right most two squares are a , b

Then Sum of all the squares = 18 x 4 + a + b .......... (1)

Also Sum of the squares before 7 = 18

Sum of the squares between 7, x = 18

and sum of the squares between x , 8 = 18

So Sum of the 14 squares = 18 + 7 + 18 + x + 18 + 8 + a + b...... (2)

Equating 1 and 2 we get x = 3.

Discussion

13.	Eesha Bought 18 Sharpeners For Rs.100. She Paid 1 Rupee More For Each White Sharpener Than For Each Brown Sharpener. What Is The Price Of A White Sharpener And How Many White Sharpener Did She Buy ?
Answer» If she bought 10 white SHARPENERS at Rs.6 PER PIECE, She has SPENT Rs.60 already. And with the remaining Rs.40, she bought 8 brown sharpeners at 40/8 = Rs.5 which is Rs.1 less than White sharpener. If she bought 10 white sharpeners at Rs.6 per piece, She has spent Rs.60 already. And with the remaining Rs.40, she bought 8 brown sharpeners at 40/8 = Rs.5 which is Rs.1 less than White sharpener.

13.

Eesha Bought 18 Sharpeners For Rs.100. She Paid 1 Rupee More For Each White Sharpener Than For Each Brown Sharpener. What Is The Price Of A White Sharpener And How Many White Sharpener Did She Buy ?

Answer»

If she bought 10 white SHARPENERS at Rs.6 PER PIECE,

She has SPENT Rs.60 already.

And with the remaining Rs.40,

she bought 8 brown sharpeners at 40/8 = Rs.5 which is Rs.1 less than White sharpener.

If she bought 10 white sharpeners at Rs.6 per piece,

She has spent Rs.60 already.

And with the remaining Rs.40,

she bought 8 brown sharpeners at 40/8 = Rs.5 which is Rs.1 less than White sharpener.

Discussion

14.	In How Many Ways A Team Of 11 Must Be Selected From 5 Men And 11 Women Such That The Team Must Comprise Of Not More Than 3 Men?
Answer» The team may consist of 0 MEN + 11 women, 1 men + 10 women, 2 men + 9 women, or 3 men + 8 women. So Number of WAYS are = 11C11+5C1×11C10+5C2×11C9+5C3×11C8 = 2256. The team may consist of 0 men + 11 women, 1 men + 10 women, 2 men + 9 women, or 3 men + 8 women. So Number of ways are = 11C11+5C1×11C10+5C2×11C9+5C3×11C8 = 2256.

14.

In How Many Ways A Team Of 11 Must Be Selected From 5 Men And 11 Women Such That The Team Must Comprise Of Not More Than 3 Men?

Answer»

The team may consist of 0 MEN + 11 women,

1 men + 10 women,

2 men + 9 women,

or 3 men + 8 women.

So Number of WAYS are = 11C11+5C1×11C10+5C2×11C9+5C3×11C8 = 2256.

The team may consist of 0 men + 11 women,

1 men + 10 women,

2 men + 9 women,

or 3 men + 8 women.

So Number of ways are = 11C11+5C1×11C10+5C2×11C9+5C3×11C8 = 2256.

Discussion

15.	On A 26 Question Test, Five Points Were Deducted For Each Wrong Answer And Eight Points Were Added For Each Correct Answer. If All The Questions Were Answered, How Many Were Correct, If The Score Was Zero ?
Answer» If 10 are correct, his score is 10 x 8 = 80. But 16 are WRONG. So total negative MARKING is 16 x 5 = 80. So FINAL score is zero. If 10 are correct, his score is 10 x 8 = 80. But 16 are wrong. So total negative marking is 16 x 5 = 80. So final score is zero.

15.

On A 26 Question Test, Five Points Were Deducted For Each Wrong Answer And Eight Points Were Added For Each Correct Answer. If All The Questions Were Answered, How Many Were Correct, If The Score Was Zero ?

Answer»

If 10 are correct, his score is 10 x 8 = 80.

But 16 are WRONG.

So total negative MARKING is 16 x 5 = 80.

So FINAL score is zero.

If 10 are correct, his score is 10 x 8 = 80.

But 16 are wrong.

So total negative marking is 16 x 5 = 80.

So final score is zero.

Discussion

16.	A Man Buys A Book For Rs.29.50 And Sells It For Rs 31.10. Find His Gain Percent?
Answer» So we have C.P. = 29.50 S.P. = 31.10 GAIN = 31.10 - 29.50 = RS. 1.6 Gain%=(Gain/Cost∗100)% =(1.6/29.50∗100)%=5.4%. So we have C.P. = 29.50 S.P. = 31.10 Gain = 31.10 - 29.50 = Rs. 1.6 Gain%=(Gain/Cost∗100)% =(1.6/29.50∗100)%=5.4%.

16.

A Man Buys A Book For Rs.29.50 And Sells It For Rs 31.10. Find His Gain Percent?

Answer»

So we have C.P. = 29.50

S.P. = 31.10

GAIN = 31.10 - 29.50 = RS. 1.6

Gain%=(Gain/Cost∗100)%

=(1.6/29.50∗100)%=5.4%.

So we have C.P. = 29.50

S.P. = 31.10

Gain = 31.10 - 29.50 = Rs. 1.6

Gain%=(Gain/Cost∗100)%

=(1.6/29.50∗100)%=5.4%.

Discussion

17.	Teena Is Younger Than Rani By 6 Years. If The Ratio Of Their Ages Is 6:8, Find The Age Of Teena?
Answer» If Rani AGE is x, then Teena age is x-6, so (x-6)/x = 6/8 => 8x-48 = 6x => 2x = 48 => x = 24 So Teena age is 24- 6 = 18 years . If Rani age is x, then Teena age is x-6, so (x-6)/x = 6/8 => 8x-48 = 6x => 2x = 48 => x = 24 So Teena age is 24- 6 = 18 years .

17.

Teena Is Younger Than Rani By 6 Years. If The Ratio Of Their Ages Is 6:8, Find The Age Of Teena?

Answer»

If Rani AGE is x, then Teena age is x-6,

so (x-6)/x = 6/8

=> 8x-48 = 6x

=> 2x = 48

=> x = 24

So Teena age is 24- 6 = 18 years .

If Rani age is x, then Teena age is x-6,

so (x-6)/x = 6/8

=> 8x-48 = 6x

=> 2x = 48

=> x = 24

So Teena age is 24- 6 = 18 years .

Discussion

18.	A Can Do A Work In 40 Days And B In 28 Days. If A And B Together Do The Work, Then Approximately In How Many Days Will The Same Work Be Completed?
Answer» A's 1DAY's WORK = 1/40 B's 1day's work = 1/28 They can work TOGETHER in = 1/40 + 1/28 = 16 DAYS (approximation). A's 1day's work = 1/40 B's 1day's work = 1/28 They can work together in = 1/40 + 1/28 = 16 days (approximation).

18.

A Can Do A Work In 40 Days And B In 28 Days. If A And B Together Do The Work, Then Approximately In How Many Days Will The Same Work Be Completed?

Answer»

A's 1DAY's WORK = 1/40

B's 1day's work = 1/28

They can work TOGETHER in = 1/40 + 1/28 = 16 DAYS (approximation).

A's 1day's work = 1/40

B's 1day's work = 1/28

They can work together in = 1/40 + 1/28 = 16 days (approximation).

Discussion

19.	A Running Train Passes A Pole In 15 Seconds And A Platform Of 100 M Long In 25 Seconds. Find The Length Of The Train?
Answer» Let's suppose TRAIN is RUNNING with a speed of X m/sec having a length of x METERS, so Speed = length/time s= x/15 (while passing a POLE) ........... (i) s= (x+100)/25(while passing the platform)...........( ii) Equating both equations, we get x/15= (x+100)/25 5x=300+3x 2x=300 x=150 meters. Let's suppose train is running with a speed of X m/sec having a length of x meters, so Speed = length/time s= x/15 (while passing a pole) ........... (i) s= (x+100)/25(while passing the platform)...........( ii) Equating both equations, we get x/15= (x+100)/25 5x=300+3x 2x=300 x=150 meters.

19.

A Running Train Passes A Pole In 15 Seconds And A Platform Of 100 M Long In 25 Seconds. Find The Length Of The Train?

Answer»

Let's suppose TRAIN is RUNNING with a speed of X m/sec having a length of x METERS, so

Speed = length/time

s= x/15 (while passing a POLE) ........... (i)

s= (x+100)/25(while passing the platform)...........( ii)

Equating both equations, we get

x/15= (x+100)/25

5x=300+3x

2x=300

x=150 meters.

Let's suppose train is running with a speed of X m/sec having a length of x meters, so

Speed = length/time

s= x/15 (while passing a pole) ........... (i)

s= (x+100)/25(while passing the platform)...........( ii)

Equating both equations, we get

x/15= (x+100)/25

5x=300+3x

2x=300

x=150 meters.

Discussion

20.	Find The Smallest Number Which Leaves A Remainder Of 2, When Divided By 3, 4, 5, And 6?
Answer» Taking L.C.M of 3, 4, 5 and 6 = 60. Hence, 60 is the number which is completely DIVISIBLE by these number, but we need that number which leaves the remainder of 2 while dividing by these number, so the REQUIRED number is 62. Taking L.C.M of 3, 4, 5 and 6 = 60. Hence, 60 is the number which is completely divisible by these number, but we need that number which leaves the remainder of 2 while dividing by these number, so the required number is 62.

20.

Find The Smallest Number Which Leaves A Remainder Of 2, When Divided By 3, 4, 5, And 6?

Answer»

Taking L.C.M of 3, 4, 5 and 6 = 60.

Hence, 60 is the number which is completely DIVISIBLE by these number,

but we need that number which leaves the remainder of 2 while dividing by these number,

so the REQUIRED number is 62.

Taking L.C.M of 3, 4, 5 and 6 = 60.

Hence, 60 is the number which is completely divisible by these number,

but we need that number which leaves the remainder of 2 while dividing by these number,

so the required number is 62.

Discussion

21.	In How Many Ways Can Letter Of Word Corporation Can Be Arranged So That All Vowels Always Come Together?
Answer» The word CORPORATION have 5 vowels which are 'O', 'O', 'A', 'I', 'O'. So grouping all vowels together, we can consider all the vowels as 1 letter So it can be ARRANGED as CRPRTN(OOAIO) It has 7 letter so can be arranged as !7 in which R is 2 times so !7/!2 And again we can arrange all the vowels in !5/!3 ways So total number of ways = (!7/!2)(!5/!3) = 50400. The word CORPORATION have 5 vowels which are 'O', 'O', 'A', 'I', 'O'. So grouping all vowels together, we can consider all the vowels as 1 letter So it can be arranged as CRPRTN(OOAIO) It has 7 letter so can be arranged as !7 in which R is 2 times so !7/!2 And again we can arrange all the vowels in !5/!3 ways So total number of ways = (!7/!2)(!5/!3) = 50400.

21.

In How Many Ways Can Letter Of Word Corporation Can Be Arranged So That All Vowels Always Come Together?

Answer»

The word CORPORATION have 5 vowels which are 'O', 'O', 'A', 'I', 'O'. So grouping all vowels together, we can consider all the vowels as 1 letter

So it can be ARRANGED as CRPRTN(OOAIO)

It has 7 letter so can be arranged as !7 in which R is 2 times so

!7/!2

And again we can arrange all the vowels in !5/!3 ways

So total number of ways = (!7/!2)*(!5/!3) = 50400.

The word CORPORATION have 5 vowels which are 'O', 'O', 'A', 'I', 'O'. So grouping all vowels together, we can consider all the vowels as 1 letter

So it can be arranged as CRPRTN(OOAIO)

It has 7 letter so can be arranged as !7 in which R is 2 times so

!7/!2

And again we can arrange all the vowels in !5/!3 ways

So total number of ways = (!7/!2)*(!5/!3) = 50400.

Discussion

22.	Rahul Purchased 12 Dozens Of Toys At The Rate Of Rs.300 Per Dozen. He Sold Each Of Toys At The Rate Of Rs.29. What Is His Percentage Profit?
Answer» COST price of 1 toy = 300/12= Rs.25 SELLING price of 1 toy = Rs.29 Gain = S.P.-C.P. =29-25= Rs.4 Profit %= 4100/25= 16%. Cost price of 1 toy = 300/12= Rs.25 Selling price of 1 toy = Rs.29 Gain = S.P.-C.P. =29-25= Rs.4 Profit %= 4100/25= 16%.

22.

Rahul Purchased 12 Dozens Of Toys At The Rate Of Rs.300 Per Dozen. He Sold Each Of Toys At The Rate Of Rs.29. What Is His Percentage Profit?

Answer»

COST price of 1 toy = 300/12= Rs.25

SELLING price of 1 toy = Rs.29

Gain = S.P.-C.P. =29-25= Rs.4

Profit %= 4*100/25= 16%.

Cost price of 1 toy = 300/12= Rs.25

Selling price of 1 toy = Rs.29

Gain = S.P.-C.P. =29-25= Rs.4

Profit %= 4*100/25= 16%.

Discussion

23.	If Log 2= 0.30103, Then What Will Be The Number Of Digits In 2^64?
Answer» To FIND the NUMBER of digits in 2^64 take the log of the number log(2^64)= 64log&LT;sup>2</sup>=640.30103= 19.26592 Its characteristic is 19 hence to OBTAIN the digit we need to add in characteristic. Hence TOTAL number of digits= 19+1= 20. To find the number of digits in 2^64 take the log of the number log(2^64)= 64log<sup>2</sup>=640.30103= 19.26592 Its characteristic is 19 hence to obtain the digit we need to add in characteristic. Hence total number of digits= 19+1= 20.

23.

If Log 2= 0.30103, Then What Will Be The Number Of Digits In 2^64?

Answer»

To FIND the NUMBER of digits in 2^64 take the log of the number

log(2^64)= 64*log&LT;sup>2</sup>=64*0.30103= 19.26592

Its characteristic is 19 hence to OBTAIN the digit we need to add in characteristic.

Hence TOTAL number of digits= 19+1= 20.

To find the number of digits in 2^64 take the log of the number

log(2^64)= 64*log<sup>2</sup>=64*0.30103= 19.26592

Its characteristic is 19 hence to obtain the digit we need to add in characteristic.

Hence total number of digits= 19+1= 20.

Discussion

24.	To Complete A Piece Of Work A Takes Twice The Time Of B And Thrice The Time Of C. If All Working Together Can Finish The Same Work In 8 Days, Then How Many Days Are Required To Finish The Work By B Alone?
Answer» SUPPOSE, A takes X days to finish a piece of work, so B will take X/2 days, and C will take X/3 days Then, (1/X+2/X+3/X) = 1/8 SOLVING the above EQUATION: 6/x= 1/8 X= 48 days So B alone can finish the work in 48/2= 24 days. Suppose, A takes X days to finish a piece of work, so B will take X/2 days, and C will take X/3 days Then, (1/X+2/X+3/X) = 1/8 Solving the above equation: 6/x= 1/8 X= 48 days So B alone can finish the work in 48/2= 24 days.

24.

To Complete A Piece Of Work A Takes Twice The Time Of B And Thrice The Time Of C. If All Working Together Can Finish The Same Work In 8 Days, Then How Many Days Are Required To Finish The Work By B Alone?

Answer»

SUPPOSE, A takes X days to finish a piece of work, so B will take X/2 days, and C will take X/3 days

Then, (1/X+2/X+3/X) = 1/8

SOLVING the above EQUATION:

6/x= 1/8

X= 48 days

So B alone can finish the work in 48/2= 24 days.

Suppose, A takes X days to finish a piece of work, so B will take X/2 days, and C will take X/3 days

Then, (1/X+2/X+3/X) = 1/8

Solving the above equation:

6/x= 1/8

X= 48 days

So B alone can finish the work in 48/2= 24 days.

Discussion

25.	In How Many Different Ways The Letter Of Words Optical Can Be Arranged So That All The Consonants Come Together?
Answer» The word OPTICAL has 7 letters which have 4 consonants PTCL, so if all consonants ALWAYS come together then, OPTICAL can be arranged as (PTCL) OIA, where all consonants can be CONSIDERED as a single letter Hence we can arrange it in !4 WAYS = 4321= 24 ways Again PTCL also arrange in !4 ways= 4321= 24 ways So total no of ways OPTICAL letters can be arranged = 2424= 576. The word OPTICAL has 7 letters which have 4 consonants PTCL, so if all consonants always come together then, OPTICAL can be arranged as (PTCL) OIA, where all consonants can be considered as a single letter Hence we can arrange it in !4 ways = 4321= 24 ways Again PTCL also arrange in !4 ways= 4321= 24 ways So total no of ways OPTICAL letters can be arranged = 2424= 576.

25.

In How Many Different Ways The Letter Of Words Optical Can Be Arranged So That All The Consonants Come Together?

Answer»

The word OPTICAL has 7 letters which have 4 consonants PTCL, so if all consonants ALWAYS come together then,

OPTICAL can be arranged as (PTCL) OIA, where all consonants can be CONSIDERED as a single letter

Hence we can arrange it in !4 WAYS = 4*3*2*1= 24 ways

Again PTCL also arrange in !4 ways= 4*3*2*1= 24 ways

So total no of ways OPTICAL letters can be arranged = 24*24= 576.

The word OPTICAL has 7 letters which have 4 consonants PTCL, so if all consonants always come together then,

OPTICAL can be arranged as (PTCL) OIA, where all consonants can be considered as a single letter

Hence we can arrange it in !4 ways = 4*3*2*1= 24 ways

Again PTCL also arrange in !4 ways= 4*3*2*1= 24 ways

So total no of ways OPTICAL letters can be arranged = 24*24= 576.

Discussion

26.	The Sum Of Three From The Four Numbers A, B, C, D Are 4024, 4087, 4524 And 4573. What Is The Largest Of The Numbers A, B, C, D?
Answer» a+B+C=4024 b+c+d= 4087 a+c+d=4524 a+b+d=4573 COMBINING all we get 3(a+b+c+d) = 17208 ⇒ a + b + c +d = 3736 Now we find INDIVIDUAL values. a = 1649, b = 1212, c = 1163, d = 1712. So maximum value is 1712. a+b+c=4024 b+c+d= 4087 a+c+d=4524 a+b+d=4573 combining all we get 3(a+b+c+d) = 17208 ⇒ a + b + c +d = 3736 Now we find individual values. a = 1649, b = 1212, c = 1163, d = 1712. So maximum value is 1712.

26.

The Sum Of Three From The Four Numbers A, B, C, D Are 4024, 4087, 4524 And 4573. What Is The Largest Of The Numbers A, B, C, D?

Answer»

a+B+C=4024

b+c+d= 4087

a+c+d=4524

a+b+d=4573

COMBINING all we get 3(a+b+c+d) = 17208

⇒ a + b + c +d = 3736

Now we find INDIVIDUAL values. a = 1649, b = 1212, c = 1163, d = 1712. So maximum value is 1712.

a+b+c=4024

b+c+d= 4087

a+c+d=4524

a+b+d=4573

combining all we get 3(a+b+c+d) = 17208

⇒ a + b + c +d = 3736

Now we find individual values. a = 1649, b = 1212, c = 1163, d = 1712. So maximum value is 1712.

Discussion

27.	When Usha Was Thrice As Old As Nisha, Her Sister Asha Was 25, When Nisha Was Half As Old As Asha, Then Sister Usha Was 34. Their Ages Add To 100. How Old Is Usha?
Answer» Let the age of Usha is 3x then Nisha is x and Asha is 25 Also Usha 34, Nisha y, and Asha 2y. We know that 3x - 34 = x - 2y = 25 - 2y SOLVING above three equations we get x = 9, y = 16 Their AGES are 34, 16, 32.whose sum = 82. So after 18 years their ages will be equal to 100. So Usha age is 34 + 6 = 40. Let the age of Usha is 3x then Nisha is x and Asha is 25 Also Usha 34, Nisha y, and Asha 2y. We know that 3x - 34 = x - 2y = 25 - 2y Solving above three equations we get x = 9, y = 16 Their ages are 34, 16, 32.whose sum = 82. So after 18 years their ages will be equal to 100. So Usha age is 34 + 6 = 40.

27.

When Usha Was Thrice As Old As Nisha, Her Sister Asha Was 25, When Nisha Was Half As Old As Asha, Then Sister Usha Was 34. Their Ages Add To 100. How Old Is Usha?

Answer»

Let the age of Usha is 3x then Nisha is x and Asha is 25

Also Usha 34, Nisha y, and Asha 2y.

We know that 3x - 34 = x - 2y = 25 - 2y

SOLVING above three equations we get x = 9, y = 16

Their AGES are 34, 16, 32.whose sum = 82. So after 18 years their ages will be equal to 100. So Usha age is 34 + 6 = 40.

Let the age of Usha is 3x then Nisha is x and Asha is 25

Also Usha 34, Nisha y, and Asha 2y.

We know that 3x - 34 = x - 2y = 25 - 2y

Solving above three equations we get x = 9, y = 16

Their ages are 34, 16, 32.whose sum = 82. So after 18 years their ages will be equal to 100. So Usha age is 34 + 6 = 40.

Discussion

28.	The Balls In A Bag Are Numbered From 1 To 20 And Mixed Up. If One Ball Is Drawn Out Randomly Then What Is The Probability That The Drawn Ball Has A Number, Which Is A Multiple Of 3 Or 5?
Answer» Total number of multiple of 3 upto 20 = 20/3= 6 (only taking whole number) Total number of multiple of 5 upto 20= 20/5 = 4 Total number of 35 upto 20= 1 So total multiple upto of 3 or 5 upto 20= 6+4-1= 9 So REQUIRED PROBABILITY = 9/20. Total number of multiple of 3 upto 20 = 20/3= 6 (only taking whole number) Total number of multiple of 5 upto 20= 20/5 = 4 Total number of 35 upto 20= 1 So total multiple upto of 3 or 5 upto 20= 6+4-1= 9 So required probability = 9/20.

28.

The Balls In A Bag Are Numbered From 1 To 20 And Mixed Up. If One Ball Is Drawn Out Randomly Then What Is The Probability That The Drawn Ball Has A Number, Which Is A Multiple Of 3 Or 5?

Answer»

Total number of multiple of 3 upto 20 = 20/3= 6 (only taking whole number)

Total number of multiple of 5 upto 20= 20/5 = 4

Total number of 3*5 upto 20= 1

So total multiple upto of 3 or 5 upto 20= 6+4-1= 9

So REQUIRED PROBABILITY = 9/20.

Total number of multiple of 3 upto 20 = 20/3= 6 (only taking whole number)

Total number of multiple of 5 upto 20= 20/5 = 4

Total number of 3*5 upto 20= 1

So total multiple upto of 3 or 5 upto 20= 6+4-1= 9

So required probability = 9/20.

Discussion

29.	If Log 27= 1.431, Then What Will Be The Value Of Log 9?
Answer» LOG(3)3= 1.431 3 LOG3= 1.431 log3= 0.477 LOG9= log(3)2= 2 log3= 20.477= 0.954. log(3)3= 1.431 3 log3= 1.431 log3= 0.477 log9= log(3)2= 2 log3= 20.477= 0.954.

29.

If Log 27= 1.431, Then What Will Be The Value Of Log 9?

Answer»

LOG(3)3= 1.431

3 LOG3= 1.431

log3= 0.477

LOG9= log(3)2= 2 log3= 2*0.477= 0.954.

log(3)3= 1.431

3 log3= 1.431

log3= 0.477

log9= log(3)2= 2 log3= 2*0.477= 0.954.

Discussion

30.	Two Trains Are Running With The Speed Of 60 Km/hr And 40 Km/hr Respectively In The Same Direction. The Fast Moving Train Completely Passes A Man Sitting In Slower Train In 9 Sec. Find The Length Of The Fast Moving Train?
Answer» RELATIVE SPEED of both TRAIN = (60-40) km/hr = 20 km/hr 20km/hr= 20(5/18) m/sec = 100/18m/sec So the length of fast moving train= (100/18)9= 50m. Relative speed of both train = (60-40) km/hr = 20 km/hr 20km/hr= 20(5/18) m/sec = 100/18m/sec So the length of fast moving train= (100/18)9= 50m.

30.

Two Trains Are Running With The Speed Of 60 Km/hr And 40 Km/hr Respectively In The Same Direction. The Fast Moving Train Completely Passes A Man Sitting In Slower Train In 9 Sec. Find The Length Of The Fast Moving Train?

Answer»

RELATIVE SPEED of both TRAIN = (60-40) km/hr = 20 km/hr

20km/hr= 20*(5/18) m/sec = 100/18m/sec

So the length of fast moving train= (100/18)*9= 50m.

Relative speed of both train = (60-40) km/hr = 20 km/hr

20km/hr= 20*(5/18) m/sec = 100/18m/sec

So the length of fast moving train= (100/18)*9= 50m.

Discussion

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