18 + Interview Questions in Value Labs Aptitude Interview Questions in Value Labs Aptitude Tutorial Page 1 InterviewSolution

1.	Two Pipes A And B Can Fill A Tank In 20 And 30 Minutes Respectively. If Both The Pipes Are Used Together, Then How Long It Will Take To Fill The Tank ?
Answer» we first GET the FILLING in 1 minute for both pipes then we will add them to get the RESULT, as PART filled by A in 1 MIN = 1/20 Part filled by B in 1 min = 1/30 Part filled by (A+B) in 1 min = 1/20 + 1/30= 1/12 So both pipes can fill the tank in 12 mins. we first get the filling in 1 minute for both pipes then we will add them to get the result, as Part filled by A in 1 min = 1/20 Part filled by B in 1 min = 1/30 Part filled by (A+B) in 1 min = 1/20 + 1/30= 1/12 So both pipes can fill the tank in 12 mins.

1.

Two Pipes A And B Can Fill A Tank In 20 And 30 Minutes Respectively. If Both The Pipes Are Used Together, Then How Long It Will Take To Fill The Tank ?

Answer»

we first GET the FILLING in 1 minute for both pipes then we will add them to get the RESULT, as

PART filled by A in 1 MIN = 1/20

Part filled by B in 1 min = 1/30

Part filled by (A+B) in 1 min = 1/20 + 1/30= 1/12

So both pipes can fill the tank in 12 mins.

we first get the filling in 1 minute for both pipes then we will add them to get the result, as

Part filled by A in 1 min = 1/20

Part filled by B in 1 min = 1/30

Part filled by (A+B) in 1 min = 1/20 + 1/30= 1/12

So both pipes can fill the tank in 12 mins.

Discussion

2.	The Greatest Number Which On Dividing 1657 And 2037 Leaves Remainders 6 And 5 Respectively, Is:
Answer» REQUIRED NUMBER = H.C.F. of (1657 - 6) and (2037 - 5) = H.C.F. of 1651 and 2032 = 127. Required number = H.C.F. of (1657 - 6) and (2037 - 5) = H.C.F. of 1651 and 2032 = 127.

2.

The Greatest Number Which On Dividing 1657 And 2037 Leaves Remainders 6 And 5 Respectively, Is:

Answer»

REQUIRED NUMBER = H.C.F. of (1657 - 6) and (2037 - 5)

= H.C.F. of 1651 and 2032 = 127.

Required number = H.C.F. of (1657 - 6) and (2037 - 5)

= H.C.F. of 1651 and 2032 = 127.

Discussion

3.	The Average Of Runs Of A Cricket Player Of 10 Innings Was 32. How Many Runs Must He Make In His Next Innings So As To Increase His Average Of Runs By 4?
Answer» Average = total runs / no.of innings = 32 So, total = Average x no.of innings = 32 x 10 = 320. Now INCREASE in avg = 4runs. So, new avg = 32+4 = 36runs Total runs = new avg x new no. of innings = 36 x 11 = 396 Runs made in the 11th inning = 396 - 320 = 76 Average = total runs / no.of innings = 32 So, total = Average x no.of innings = 32 x 10 = 320. Now increase in avg = 4runs. So, new avg = 32+4 = 36runs Total runs = new avg x new no. of innings = 36 x 11 = 396 Runs made in the 11th inning = 396 - 320 = 76

3.

The Average Of Runs Of A Cricket Player Of 10 Innings Was 32. How Many Runs Must He Make In His Next Innings So As To Increase His Average Of Runs By 4?

Answer»

Average = total runs / no.of innings = 32

So, total = Average x no.of innings = 32 x 10 = 320.

Now INCREASE in avg = 4runs. So, new avg = 32+4 = 36runs

Total runs = new avg x new no. of innings = 36 x 11 = 396

Runs made in the 11th inning = 396 - 320 = 76

Average = total runs / no.of innings = 32

So, total = Average x no.of innings = 32 x 10 = 320.

Now increase in avg = 4runs. So, new avg = 32+4 = 36runs

Total runs = new avg x new no. of innings = 36 x 11 = 396

Runs made in the 11th inning = 396 - 320 = 76

Discussion

4.	The Ratio Between The Length And The Breadth Of A Rectangular Park Is 3: 2. If A Man Cycling Along The Boundary Of The Park At The Speed Of 12 Km/hr Completes One Round In 8 Minutes, Then The Area Of The Park (in Sq. M) Is:
Answer» Perimeter = Distance covered in 8 MIN. =12000x 8m = 1600 m. Let length = 3x METRES and breadth = 2X metres. Then, 2(3x + 2x) = 1600 or x = 160. Length = 480 m and Breadth = 320 m. Area = (480 x 320) m2 = 153600 m2. Perimeter = Distance covered in 8 min. =12000x 8m = 1600 m. Let length = 3x metres and breadth = 2x metres. Then, 2(3x + 2x) = 1600 or x = 160. Length = 480 m and Breadth = 320 m. Area = (480 x 320) m2 = 153600 m2.

4.

The Ratio Between The Length And The Breadth Of A Rectangular Park Is 3: 2. If A Man Cycling Along The Boundary Of The Park At The Speed Of 12 Km/hr Completes One Round In 8 Minutes, Then The Area Of The Park (in Sq. M) Is:

Answer»

Perimeter = Distance covered in 8 MIN. =12000x 8m = 1600 m.

Let length = 3x METRES and breadth = 2X metres.

Then, 2(3x + 2x) = 1600 or x = 160.

Length = 480 m and Breadth = 320 m.

Area = (480 x 320) m2 = 153600 m2.

Perimeter = Distance covered in 8 min. =12000x 8m = 1600 m.

Let length = 3x metres and breadth = 2x metres.

Then, 2(3x + 2x) = 1600 or x = 160.

Length = 480 m and Breadth = 320 m.

Area = (480 x 320) m2 = 153600 m2.

Discussion

5.	Sam And Joan Are Playing A Tennis Match. If The Probability Of Sam's Win Is 0.59, Then Find The Probability Of Joan's Win.
Answer» <P>LET event A = Sam wins and event B = JOAN wins. Then, P(A) = 0.59 Since if Sam wins, Joan cannot win and if Joan wins, Sam cannot win, so we can SAY that the events A and B are mutually exclusive. Other than these two events, there are no any other possible outcomes. So, P(A)+P(B) = 1 0.59+P(B) = 1 P(B) = 1-0.59 = 0.41 Let event A = Sam wins and event B = Joan wins. Then, P(A) = 0.59 Since if Sam wins, Joan cannot win and if Joan wins, Sam cannot win, so we can say that the events A and B are mutually exclusive. Other than these two events, there are no any other possible outcomes. So, P(A)+P(B) = 1 0.59+P(B) = 1 P(B) = 1-0.59 = 0.41

5.

Sam And Joan Are Playing A Tennis Match. If The Probability Of Sam's Win Is 0.59, Then Find The Probability Of Joan's Win.

Answer»

<P>LET event A = Sam wins and event B = JOAN wins. Then,

P(A) = 0.59

Since if Sam wins, Joan cannot win and if Joan wins, Sam cannot win, so we can SAY that the events A and B are mutually exclusive. Other than these two events, there are

no any other possible outcomes. So,

P(A)+P(B) = 1

0.59+P(B) = 1

P(B) = 1-0.59 = 0.41

Let event A = Sam wins and event B = Joan wins. Then,

P(A) = 0.59

Since if Sam wins, Joan cannot win and if Joan wins, Sam cannot win, so we can say that the events A and B are mutually exclusive. Other than these two events, there are

no any other possible outcomes. So,

P(A)+P(B) = 1

0.59+P(B) = 1

P(B) = 1-0.59 = 0.41

Discussion

6.	A Mixture Contains Alcohol And Water In The Ratio 4 : 3. If 5 Liters Of Water Is Added To The Mixture, The Ratio Becomes 4: 5. Find The Quantity Of Alcohol In The Given Mixture.
Answer» Let the quantity of ALCOHOL and water be 4X litres and 3x litres respectively 4x/(3x+5) = 4/5 20x = 4(3x+5) 8X = 20 x = 2.5 Quantity of alcohol = (4 x 2.5) litres = 10 litres. Let the quantity of alcohol and water be 4x litres and 3x litres respectively 4x/(3x+5) = 4/5 20x = 4(3x+5) 8x = 20 x = 2.5 Quantity of alcohol = (4 x 2.5) litres = 10 litres.

6.

A Mixture Contains Alcohol And Water In The Ratio 4 : 3. If 5 Liters Of Water Is Added To The Mixture, The Ratio Becomes 4: 5. Find The Quantity Of Alcohol In The Given Mixture.

Answer»

Let the quantity of ALCOHOL and water be 4X litres and 3x litres respectively

4x/(3x+5) = 4/5

20x = 4(3x+5)

8X = 20

x = 2.5

Quantity of alcohol = (4 x 2.5) litres = 10 litres.

Let the quantity of alcohol and water be 4x litres and 3x litres respectively

4x/(3x+5) = 4/5

20x = 4(3x+5)

8x = 20

x = 2.5

Quantity of alcohol = (4 x 2.5) litres = 10 litres.

Discussion

7.	A And B Started A Business By Investing Rs.4000/- And Rs.5000/- Respectively. Find The A’s Share Out Of A Total Profit Of Rs.1800:
Answer» A = Rs.4000/- B = Rs.5000/- A share 4 PARTS & B share 5 parts TOTAL 9 parts -----> Rs.1800/- ----> 1 part -------> Rs.200/- A share = 4 parts -----> Rs.800/- A = Rs.4000/- B = Rs.5000/- A share 4 parts & B share 5 parts Total 9 parts -----> Rs.1800/- ----> 1 part -------> Rs.200/- A share = 4 parts -----> Rs.800/-

7.

A And B Started A Business By Investing Rs.4000/- And Rs.5000/- Respectively. Find The A’s Share Out Of A Total Profit Of Rs.1800:

Answer»

A = Rs.4000/-

B = Rs.5000/-

A share 4 PARTS & B share 5 parts

TOTAL 9 parts -----> Rs.1800/-

----> 1 part -------> Rs.200/-

A share = 4 parts -----> Rs.800/-

A = Rs.4000/-

B = Rs.5000/-

A share 4 parts & B share 5 parts

Total 9 parts -----> Rs.1800/-

----> 1 part -------> Rs.200/-

A share = 4 parts -----> Rs.800/-

Discussion

8.	Worker A Takes 8 Hours To Do A Job. Worker B Takes 10 Hours To Do A Job. How Long Should It Take Both A And B, Working Together To Do The Same Job ?
Answer» FIRST we need to calculate 1 HOURS work, then their COLLECTIVE work as A's 1-hour work is 1/8 B's 1-hour work is 1/10 (A+B)'s 1 hour work = 1/8 + 1/10= 9/40 So both will finish the work in 40/9 hours=4 4/9 first we need to calculate 1 hours work, then their collective work as A's 1-hour work is 1/8 B's 1-hour work is 1/10 (A+B)'s 1 hour work = 1/8 + 1/10= 9/40 So both will finish the work in 40/9 hours=4 4/9

8.

Worker A Takes 8 Hours To Do A Job. Worker B Takes 10 Hours To Do A Job. How Long Should It Take Both A And B, Working Together To Do The Same Job ?

Answer»

FIRST we need to calculate 1 HOURS work, then their COLLECTIVE work as A's 1-hour work is 1/8

B's 1-hour work is 1/10

(A+B)'s 1 hour work = 1/8 + 1/10= 9/40

So both will finish the work in 40/9 hours=4 4/9

first we need to calculate 1 hours work, then their collective work as A's 1-hour work is 1/8

B's 1-hour work is 1/10

(A+B)'s 1 hour work = 1/8 + 1/10= 9/40

So both will finish the work in 40/9 hours=4 4/9

Discussion

9.	A Person Crosses A 600 M Long Street In 5 Minutes. What Is His Speed In Km Per Hour ?
Answer» Speed = 600/ 5 x 60 m/sec. = 2 m/sec. = 2 x 18/5km/hr =7.2 km/hr Speed = 600/ 5 x 60 m/sec. = 2 m/sec. = 2 x 18/5km/hr =7.2 km/hr

9.

A Person Crosses A 600 M Long Street In 5 Minutes. What Is His Speed In Km Per Hour ?

Answer»

Speed = 600/ 5 x 60 m/sec. = 2 m/sec. = 2 x 18/5km/hr =7.2 km/hr

Discussion

10.	The Product Of Two Numbers Is 4107. If The H.c.f. Of These Numbers Is 37, Then The Greater Number Is:
Answer» Let the numbers be 37a and 37b. Then, 37a x 37b = 4107 AB = 3. Now, co-primes with PRODUCT 3 are (1, 3). So, the REQUIRED numbers are (37 x 1, 37 x 3) i.e., (37, 111). Greater number = 111. Let the numbers be 37a and 37b. Then, 37a x 37b = 4107 ab = 3. Now, co-primes with product 3 are (1, 3). So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111). Greater number = 111.

10.

The Product Of Two Numbers Is 4107. If The H.c.f. Of These Numbers Is 37, Then The Greater Number Is:

Answer»

Let the numbers be 37a and 37b.

Then, 37a x 37b = 4107

AB = 3.

Now, co-primes with PRODUCT 3 are (1, 3).

So, the REQUIRED numbers are (37 x 1, 37 x 3) i.e., (37, 111).

Greater number = 111.

Let the numbers be 37a and 37b.

Then, 37a x 37b = 4107

ab = 3.

Now, co-primes with product 3 are (1, 3).

So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111).

Greater number = 111.

Discussion

11.	The Greatest Number Of Four Digits Which Is Divisible By 15, 25, 40 And 75 Is:
Answer» GREATEST NUMBER of 4-digits is 9999. L.C.M. of 15, 25, 40 and 75 is 600. On dividing 9999 by 600, the REMAINDER is 399. Required number (9999 – 399) = 9600. Greatest number of 4-digits is 9999. L.C.M. of 15, 25, 40 and 75 is 600. On dividing 9999 by 600, the remainder is 399. Required number (9999 – 399) = 9600.

11.

The Greatest Number Of Four Digits Which Is Divisible By 15, 25, 40 And 75 Is:

Answer»

GREATEST NUMBER of 4-digits is 9999.

L.C.M. of 15, 25, 40 and 75 is 600.

On dividing 9999 by 600, the REMAINDER is 399.

Required number (9999 – 399) = 9600.

Greatest number of 4-digits is 9999.

L.C.M. of 15, 25, 40 and 75 is 600.

On dividing 9999 by 600, the remainder is 399.

Required number (9999 – 399) = 9600.

Discussion

12.	In Digging A Pond 20 M * 10 M * 5 M The Volumes Of The Soil Extracted Will Be ?
Answer» 20 * 10 * 5 = 1000 20 * 10 * 5 = 1000

12.

In Digging A Pond 20 M * 10 M * 5 M The Volumes Of The Soil Extracted Will Be ?

Answer»

20 * 10 * 5 = 1000

Discussion

13.	A Alone Can Do A Piece Of Work In 6 Days And B Alone In 8 Days. A And B Undertook To Do It For Rs. 3200. With The Help Of C, They Completed The Work In 3 Days. How Much Is To Be Paid To C ?
Answer» C’s 1 day’s WORK = 1/3 – (1/6 + 1/8) = 1/3 – 7/24 = 1/24 A’s wages : B’s wages : C’s wages 1/6 : 1/8 : 1/24 = 4:3:1 C’s share = 1/8 * 3200 = Rs. 400 C’s 1 day’s work = 1/3 – (1/6 + 1/8) = 1/3 – 7/24 = 1/24 A’s wages : B’s wages : C’s wages 1/6 : 1/8 : 1/24 = 4:3:1 C’s share = 1/8 * 3200 = Rs. 400

13.

A Alone Can Do A Piece Of Work In 6 Days And B Alone In 8 Days. A And B Undertook To Do It For Rs. 3200. With The Help Of C, They Completed The Work In 3 Days. How Much Is To Be Paid To C ?

Answer»

C’s 1 day’s WORK = 1/3 – (1/6 + 1/8) = 1/3 – 7/24 = 1/24

A’s wages : B’s wages : C’s wages

1/6 : 1/8 : 1/24 = 4:3:1

C’s share = 1/8 * 3200 = Rs. 400

C’s 1 day’s work = 1/3 – (1/6 + 1/8) = 1/3 – 7/24 = 1/24

A’s wages : B’s wages : C’s wages

1/6 : 1/8 : 1/24 = 4:3:1

C’s share = 1/8 * 3200 = Rs. 400

Discussion

14.	22004 – 29303 + 101299 – 59532 = ?
Answer» 22004 – 29303 + 101299 – 59532 123303 – 888.35 = 34468 = 34500 22004 – 29303 + 101299 – 59532 123303 – 888.35 = 34468 = 34500

14.

22004 – 29303 + 101299 – 59532 = ?

Answer»

22004 – 29303 + 101299 – 59532

123303 – 888.35 = 34468 = 34500

22004 – 29303 + 101299 – 59532

123303 – 888.35 = 34468 = 34500

Discussion

15.	150 Men Consume 1050 Kg Of Rice In 30 Days. In How Many Days Will 70 Men Consume 980 Kg Of Rice?
Answer» Rate of consumption of each man = 1050/(150 * 30) = 7/30 KG/day Let us say 70 men take x days to consume 150 kg. Quantity CONSUMED by each item in x days = (7x/30) kg Quantity consumed by 70 men in x days = (7/30 x)(70) kg = (7/30 x) * (70) = 960 x = 60 days. Rate of consumption of each man = 1050/(150 * 30) = 7/30 kg/day Let us say 70 men take x days to consume 150 kg. Quantity consumed by each item in x days = (7x/30) kg Quantity consumed by 70 men in x days = (7/30 x)(70) kg = (7/30 x) * (70) = 960 x = 60 days.

15.

150 Men Consume 1050 Kg Of Rice In 30 Days. In How Many Days Will 70 Men Consume 980 Kg Of Rice?

Answer»

Rate of consumption of each man = 1050/(150 * 30) = 7/30 KG/day

Let us say 70 men take x days to consume 150 kg.

Quantity CONSUMED by each item in x days = (7x/30) kg

Quantity consumed by 70 men in x days = (7/30 x)(70) kg

= (7/30 x) * (70) = 960

x = 60 days.

Rate of consumption of each man = 1050/(150 * 30) = 7/30 kg/day

Let us say 70 men take x days to consume 150 kg.

Quantity consumed by each item in x days = (7x/30) kg

Quantity consumed by 70 men in x days = (7/30 x)(70) kg

= (7/30 x) * (70) = 960

x = 60 days.

Discussion

16.	In A Game Of 90 Points A Can Give B 15 Points And C 30 Points. How Many Points Can B Give C In A Game Of 100 Points ?
Answer» While A SCORES 90 points, B scores (90-15)=75 points and C scores (90-30)= 60 points i.e., when B scores 75 points, C scores 60 points => When B scores 100 points, C scores 6075×100 = 80 points i.e., in a game of 100 points, B can give C (100-80)=20 points While A scores 90 points, B scores (90-15)=75 points and C scores (90-30)= 60 points i.e., when B scores 75 points, C scores 60 points => When B scores 100 points, C scores 6075×100 = 80 points i.e., in a game of 100 points, B can give C (100-80)=20 points

16.

In A Game Of 90 Points A Can Give B 15 Points And C 30 Points. How Many Points Can B Give C In A Game Of 100 Points ?

Answer»

While A SCORES 90 points, B scores (90-15)=75 points and C scores (90-30)= 60 points

i.e., when B scores 75 points, C scores 60 points

=> When B scores 100 points, C scores 6075×100 = 80 points

i.e., in a game of 100 points, B can give C (100-80)=20 points

While A scores 90 points, B scores (90-15)=75 points and C scores (90-30)= 60 points

i.e., when B scores 75 points, C scores 60 points

=> When B scores 100 points, C scores 6075×100 = 80 points

i.e., in a game of 100 points, B can give C (100-80)=20 points

Discussion

17.	Ram Bought 4 Dozen Apples At Rs. 12 Per Dozen And 2 Dozen Apples At Rs. 16 Per Dozen. He Sold All Of Them To Earn 20%. At What Price Per Dozen Did He Sell The Apples ?
Answer» 6 dozen APPLES = Rs. (12 × 4 + 16 × 2) = Rs. 80 Gain = 20% S.P = Rs. (120/100 × 80) = Rs. 96 Per dozen = Rs. (96/6) = Rs. 16 6 dozen apples = Rs. (12 × 4 + 16 × 2) = Rs. 80 Gain = 20% S.P = Rs. (120/100 × 80) = Rs. 96 Per dozen = Rs. (96/6) = Rs. 16

17.

Ram Bought 4 Dozen Apples At Rs. 12 Per Dozen And 2 Dozen Apples At Rs. 16 Per Dozen. He Sold All Of Them To Earn 20%. At What Price Per Dozen Did He Sell The Apples ?

Answer»

6 dozen APPLES = Rs. (12 × 4 + 16 × 2) = Rs. 80

Gain = 20%

S.P = Rs. (120/100 × 80) = Rs. 96

Per dozen = Rs. (96/6) = Rs. 16

6 dozen apples = Rs. (12 × 4 + 16 × 2) = Rs. 80

Gain = 20%

S.P = Rs. (120/100 × 80) = Rs. 96

Per dozen = Rs. (96/6) = Rs. 16

Discussion

18.	A Train 110 M Long Is Running With A Speed Of 60 Km/hr. In What Time Will It Pass A Man Who Is Running At 6 Km/hr In The Direction Opposite To That In Which The Train Is Going ?
Answer» Speed of train RELATIVE to man = 60 + 6 = 66 km/hr. = 66 * 5/18 = 55/3 m/sec. Time taken to pass the men = 110 * 3/55 = 6 sec. Speed of train relative to man = 60 + 6 = 66 km/hr. = 66 * 5/18 = 55/3 m/sec. Time taken to pass the men = 110 * 3/55 = 6 sec.

18.

A Train 110 M Long Is Running With A Speed Of 60 Km/hr. In What Time Will It Pass A Man Who Is Running At 6 Km/hr In The Direction Opposite To That In Which The Train Is Going ?

Answer»

Speed of train RELATIVE to man = 60 + 6 = 66 km/hr.

= 66 * 5/18 = 55/3 m/sec.

Time taken to pass the men = 110 * 3/55 = 6 sec.

Speed of train relative to man = 60 + 6 = 66 km/hr.

= 66 * 5/18 = 55/3 m/sec.

Time taken to pass the men = 110 * 3/55 = 6 sec.

Discussion

Explore topic-wise InterviewSolutions in .

Two Pipes A And B Can Fill A Tank In 20 And 30 Minutes Respectively. If Both The Pipes Are Used Together, Then How Long It Will Take To Fill The Tank ?

The Greatest Number Which On Dividing 1657 And 2037 Leaves Remainders 6 And 5 Respectively, Is:

The Average Of Runs Of A Cricket Player Of 10 Innings Was 32. How Many Runs Must He Make In His Next Innings So As To Increase His Average Of Runs By 4?

The Ratio Between The Length And The Breadth Of A Rectangular Park Is 3: 2. If A Man Cycling Along The Boundary Of The Park At The Speed Of 12 Km/hr Completes One Round In 8 Minutes, Then The Area Of The Park (in Sq. M) Is:

Sam And Joan Are Playing A Tennis Match. If The Probability Of Sam's Win Is 0.59, Then Find The Probability Of Joan's Win.

A Mixture Contains Alcohol And Water In The Ratio 4 : 3. If 5 Liters Of Water Is Added To The Mixture, The Ratio Becomes 4: 5. Find The Quantity Of Alcohol In The Given Mixture.

A And B Started A Business By Investing Rs.4000/- And Rs.5000/- Respectively. Find The A’s Share Out Of A Total Profit Of Rs.1800:

Worker A Takes 8 Hours To Do A Job. Worker B Takes 10 Hours To Do A Job. How Long Should It Take Both A And B, Working Together To Do The Same Job ?

A Person Crosses A 600 M Long Street In 5 Minutes. What Is His Speed In Km Per Hour ?

The Product Of Two Numbers Is 4107. If The H.c.f. Of These Numbers Is 37, Then The Greater Number Is:

The Greatest Number Of Four Digits Which Is Divisible By 15, 25, 40 And 75 Is:

In Digging A Pond 20 M * 10 M * 5 M The Volumes Of The Soil Extracted Will Be ?

A Alone Can Do A Piece Of Work In 6 Days And B Alone In 8 Days. A And B Undertook To Do It For Rs. 3200. With The Help Of C, They Completed The Work In 3 Days. How Much Is To Be Paid To C ?

22004 – 29303 + 101299 – 59532 = ?

150 Men Consume 1050 Kg Of Rice In 30 Days. In How Many Days Will 70 Men Consume 980 Kg Of Rice?

In A Game Of 90 Points A Can Give B 15 Points And C 30 Points. How Many Points Can B Give C In A Game Of 100 Points ?

Ram Bought 4 Dozen Apples At Rs. 12 Per Dozen And 2 Dozen Apples At Rs. 16 Per Dozen. He Sold All Of Them To Earn 20%. At What Price Per Dozen Did He Sell The Apples ?

A Train 110 M Long Is Running With A Speed Of 60 Km/hr. In What Time Will It Pass A Man Who Is Running At 6 Km/hr In The Direction Opposite To That In Which The Train Is Going ?