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- Question 1 of 10
##### 1. Question

1 pointsCategory: Quantitative Aptitude**Directions (1-5): Following questions have two quantities as Quantity I and Quantity II. You have to determine the relationship between them and give answer as,****Quantity 1:**Car A, Car B, and Car C travelled around a circular track and they completed their revolutions in 12 seconds, 20 seconds and 25 seconds respectively. In how many minutes will they come together at the starting point?

**Quantity 2:**Average speed of car A is 20kmph and it reaches the destination in 4 hours and car B covers the same distance in 5 hours. If car A increases the average speed by 8kmph and car B increases the average speed by 4kmph, what would be the difference in its time taken to reach the destination?Correct**Answer: 4) Quantity I < Quantity II**

**Explanation:**

**Q1**

LCM of (12, 20, 25) = 300 seconds

Time = 5 minutes**Q2**

Distance = 20*4 = 80km

Speed of car B = 80/5 = 16km

After increasing speed

Time taken by Car A = 80/28 = 20/7 hours

Time taken by Car B = 80/20 = 4 hours

Difference in time taken = 4 – 20/7 = 8/7 hoursIncorrect**Answer: 4) Quantity I < Quantity II**

**Explanation:**

**Q1**

LCM of (12, 20, 25) = 300 seconds

Time = 5 minutes**Q2**

Distance = 20*4 = 80km

Speed of car B = 80/5 = 16km

After increasing speed

Time taken by Car A = 80/28 = 20/7 hours

Time taken by Car B = 80/20 = 4 hours

Difference in time taken = 4 – 20/7 = 8/7 hours - Question 2 of 10
##### 2. Question

1 pointsCategory: Quantitative Aptitude**Directions (1-5): Following questions have two quantities as Quantity I and Quantity II. You have to determine the relationship between them and give answer as,****Quantity 1:**A container contains a mixture of milk and water in the ratio of 3:5. If 12 litres of mixture is taken out and replaced by the same amount of milk then the new mixture contains milk and water in the ratio 2:3. Find the initial amount of milk?

**Quantity 2:**A container is full of 56 litres of water. If 8 litre of water is taken out and replaced by milk. Further, 7 litres of mixture is taken out and replaced again with milk then find the amount of water at the end?Correct**Answer: 1) Quantity I > Quantity II**

**Explanation:**

**Q1:**

Let the initial quantity of milk = 3x litres and initial quantity of water = 5x litres

After 12 litres of mixture is taken out

Milk left = 3x-4.5 and water left = 5x-7.5

ATQ,

(3x-4.5+12)/(5x-7.5) = 2/3

x = 37.5

Therefore, initial quantity of milk = 3x = 3*37.5 = 112.5 litres**Q2:**

Final quantity:

= 56*(1-8/56)*(1-7/56)

= 56*(6/7)*(7/8)

= 42 litresIncorrect**Answer: 1) Quantity I > Quantity II**

**Explanation:**

**Q1:**

Let the initial quantity of milk = 3x litres and initial quantity of water = 5x litres

After 12 litres of mixture is taken out

Milk left = 3x-4.5 and water left = 5x-7.5

ATQ,

(3x-4.5+12)/(5x-7.5) = 2/3

x = 37.5

Therefore, initial quantity of milk = 3x = 3*37.5 = 112.5 litres**Q2:**

Final quantity:

= 56*(1-8/56)*(1-7/56)

= 56*(6/7)*(7/8)

= 42 litres - Question 3 of 10
##### 3. Question

1 pointsCategory: Quantitative Aptitude**Directions (1-5): Following questions have two quantities as Quantity I and Quantity II. You have to determine the relationship between them and give answer as,****Quantity 1:**3 men can complete a piece of work in 20 days and 12 women can complete the same work in 8 days. In how many days 6 men and 8 women together can complete the same work?

**Quantity 2:**A and B can do a piece of work in 20 days and 30 days respectively. They started working together but after some days A left the work. B alone completed the remaining work in 12 days. Find after how many days A left the work?Correct**Answer: 4) Quantity I < Quantity II**

**Explanation:**

**Q1**

3 men can complete work in 20 days

1 man can complete the work in 3*20 = 60 days

Therefore, 6 men can complete work in 60/6 = 10 days

Work done by 6 men in one day = 1/10

12 women can complete the work in 8 days

1 woman can complete the work in 12*8 = 96 days

8 women can complete the work in 96/8 = 12 days

Work done by 8 women in one day= 1/12

Let the number of days in which 6 men and 8 women together can complete the same work = x

x*(1/10 + 1/12) = 1

x*(11/60) = 1

x = 60/11 days**Q2**

If A worked for x days then B worked for (x+12) days.

(x / 20) + {(x+12)/30} = 1

(x / 2) + {(x+12)/3} = 10

(3x+2x+24) = 60

5x = 36

x = 36/5 daysIncorrect**Answer: 4) Quantity I < Quantity II**

**Explanation:**

**Q1**

3 men can complete work in 20 days

1 man can complete the work in 3*20 = 60 days

Therefore, 6 men can complete work in 60/6 = 10 days

Work done by 6 men in one day = 1/10

12 women can complete the work in 8 days

1 woman can complete the work in 12*8 = 96 days

8 women can complete the work in 96/8 = 12 days

Work done by 8 women in one day= 1/12

Let the number of days in which 6 men and 8 women together can complete the same work = x

x*(1/10 + 1/12) = 1

x*(11/60) = 1

x = 60/11 days**Q2**

If A worked for x days then B worked for (x+12) days.

(x / 20) + {(x+12)/30} = 1

(x / 2) + {(x+12)/3} = 10

(3x+2x+24) = 60

5x = 36

x = 36/5 days - Question 4 of 10
##### 4. Question

1 pointsCategory: Quantitative Aptitude**Quantity 1:**If the S.I. on a sum of money for 2 years at 6.25% per annum is Rs. 250, what is the C.I. on the same at the same rate and for the same time?

**Quantity 2:**What principal will be amount to Rs. 5350 at compound interest in 2 years, the rate of interest for 1st and 2nd year is 4% and 8% respectively?Correct**Answer: 4) Quantity I < Quantity II**

**Explanation:**

**Q1**

SI = PRT/100

P = 100*SI/RT

P = 100*250/2*6.25

P = 2000

Amount = P*(1+R/100)^{n}

= 2000*(1+6.25/100)^{2}

= 2000*(106.25/100)^{2}

= 2257.8 (approx)**Q2**

5350 = P*(1+4/100)*(1+8/100)

5350 = P*(104/100)*(108/100)

P = (5350*100*100/104*108)

P = 4760 (approx)Incorrect**Answer: 4) Quantity I < Quantity II**

**Explanation:**

**Q1**

SI = PRT/100

P = 100*SI/RT

P = 100*250/2*6.25

P = 2000

Amount = P*(1+R/100)^{n}

= 2000*(1+6.25/100)^{2}

= 2000*(106.25/100)^{2}

= 2257.8 (approx)**Q2**

5350 = P*(1+4/100)*(1+8/100)

5350 = P*(104/100)*(108/100)

P = (5350*100*100/104*108)

P = 4760 (approx) - Question 5 of 10
##### 5. Question

1 pointsCategory: Quantitative AptitudeThere are two bags A and B. Bag A contains 2 red, 4 blue and 3 black pens while Bag B contains 1 red, 5 blue, and 2 black pens.

**Quantity 1:**Probability of getting at most 1 black pen if 3 pens are drawn at random from bag B.

**Quantity 2:**Probability of getting at least 2 blue pens if 3 pens are drawn at random from Bag A.Correct**Answer: 1) Quantity I > Quantity II**

**Explanation:**

**Q1:**

Probability (at most 1 black pen from bag B)

= Probability (no black pen + 1 black pen)

= (6C_{3}+ 6C_{2}*2C_{1})/8C_{3}

= (20+30)/56

= 25/28**Q2:**

Probability (at least 2 blue pens from bag A) =

Probability (2 blue pens + 3 blue pens)

= (4C_{2}*5C_{1}+ 4C_{3})/9C_{3}

= 34/84

= 17/42Incorrect**Answer: 1) Quantity I > Quantity II**

**Explanation:**

**Q1:**

Probability (at most 1 black pen from bag B)

= Probability (no black pen + 1 black pen)

= (6C_{3}+ 6C_{2}*2C_{1})/8C_{3}

= (20+30)/56

= 25/28**Q2:**

Probability (at least 2 blue pens from bag A) =

Probability (2 blue pens + 3 blue pens)

= (4C_{2}*5C_{1}+ 4C_{3})/9C_{3}

= 34/84

= 17/42 - Question 6 of 10
##### 6. Question

1 pointsCategory: Quantitative Aptitude**Directions (6-10): In each of the following questions, a question is followed by three statements I, II and III. Read all the statements to find the answer to given question and then answer accordingly that which statement/s can give the answer alone/together.****Find the present age of Rakhi?**

1. Average of the present age of Rakhi, Ankita, Teena, and Suchi is 16 years.

2. Average of the ages of Ankita and Suchi before 5 years was 12 years. Payal is 6 years older than Teena.

3. Average of the present age of Payal and Teena is 15 years.Correct**Answer: 5) All 1, 2 and 3.**

Explanation:

From 1

Rakhi + Ankita + Teena + Suchi = 4*16 = 64 years………..a

From 2

Before 5 years

Ankita + Suchi = 2*12 = 24

Present

Ankita + Suchi = 24 + 2*5 = 34 ………………b

Payal = 6 + Teena………………………………..c

From 3

Payal + Teena = 2*15 = 30 ………………..d

Solving, equation a, b, c and d

Rakhi + 34 + 12 = 64

Rakhi = 18 years

Hence, all 1, 2 and 3.Incorrect**Answer: 5) All 1, 2 and 3.**

Explanation:

From 1

Rakhi + Ankita + Teena + Suchi = 4*16 = 64 years………..a

From 2

Before 5 years

Ankita + Suchi = 2*12 = 24

Present

Ankita + Suchi = 24 + 2*5 = 34 ………………b

Payal = 6 + Teena………………………………..c

From 3

Payal + Teena = 2*15 = 30 ………………..d

Solving, equation a, b, c and d

Rakhi + 34 + 12 = 64

Rakhi = 18 years

Hence, all 1, 2 and 3. - Question 7 of 10
##### 7. Question

1 pointsCategory: Quantitative Aptitude**Directions (6-10): In each of the following questions, a question is followed by three statements I, II and III. Read all the statements to find the answer to given question and then answer accordingly that which statement/s can give the answer alone/together.****Find the age of Kasima after 8 years?**

1. The ratio of the ages of Sumit and Eakta before 6 years was 5:4 respectively. After 3 years, the ratio of their ages will be 6:5 respectively.

2. Sum of the present ages of Kasima and Wasim is equal to the sum of the present ages of Sumit and Eakta.

3. Kasima is 4 years older than Eakta.Correct**Answer: 3) Only 1 and 3.**

Explanation:

**From 1 and 3**

Let the ages of Sumit and Eakta before 6 years be 5x and 4x respectively.

ATQ,

(5x+9)/(4x+9) = 6/5

x = 9

Present age of Eakta = 4x + 6 = 4*9 + 6 = 42 years

Present age of Kasima = 42 + 4 = 46

Age of Kasima after 8 years = 46 + 8 = 54 yearsIncorrect**Answer: 3) Only 1 and 3.**

Explanation:

**From 1 and 3**

Let the ages of Sumit and Eakta before 6 years be 5x and 4x respectively.

ATQ,

(5x+9)/(4x+9) = 6/5

x = 9

Present age of Eakta = 4x + 6 = 4*9 + 6 = 42 years

Present age of Kasima = 42 + 4 = 46

Age of Kasima after 8 years = 46 + 8 = 54 years - Question 8 of 10
##### 8. Question

1 pointsCategory: Quantitative Aptitude**Directions (6-10): In each of the following questions, a question is followed by three statements I, II and III. Read all the statements to find the answer to given question and then answer accordingly that which statement/s can give the answer alone/together.****Pipe S is opened for the first 4 minutes then closed and pipe R is opened. Find the time taken by pipe R to fill the remaining part of the tank.**

1. Pipe P and Q can together fill the tank in 12 minutes.

2. Pipe P, Pipe Q and Pipe S together can fill the tank in 9 minutes.

3. Pipe R and pipe S together can fill the tank in 10 minutes.Correct**Answer: 1) All 1, 2 and 3.**

Explanation:

From 1, 2 and 3

1/P + 1/Q = 1/12 ………………a

1/P + 1/Q + 1/S = 1/9 ………….b

1/R + 1/S = 1/10 …………c

Form equation b – equation a

1/S = 1/9 – 1/12

1/S = 1/36

Putting 1/S = 1/36 in equation c

1/R + 1/36 = 1/10

1/R = 1/10 – 1/36

1/R = 13/180

Let the time taken by R to fill the tank is t

4/36 + 13t/180 = 1

13t/180 = 1 – 1/9

13t/180 = 8/9

13t/20 = 8

t = 160/13Incorrect**Answer: 1) All 1, 2 and 3.**

Explanation:

From 1, 2 and 3

1/P + 1/Q = 1/12 ………………a

1/P + 1/Q + 1/S = 1/9 ………….b

1/R + 1/S = 1/10 …………c

Form equation b – equation a

1/S = 1/9 – 1/12

1/S = 1/36

Putting 1/S = 1/36 in equation c

1/R + 1/36 = 1/10

1/R = 1/10 – 1/36

1/R = 13/180

Let the time taken by R to fill the tank is t

4/36 + 13t/180 = 1

13t/180 = 1 – 1/9

13t/180 = 8/9

13t/20 = 8

t = 160/13 - Question 9 of 10
##### 9. Question

1 pointsCategory: Quantitative Aptitude**In a shop there are 20% of books in Spanish. To find the total number of books in the shop which of the following information is/are required?**

1) There are 30% of books in English.

2) There are 3500 books in a non-English language.

3) There are 3000 books which are neither in English nor in Spanish.Correct**Answer: 4) Either 1 and 2 together or 1 and 3 together are sufficient.**

Explanation:

From 1 and 2

30% books are in English, so rest 70% books are in non-English language

Total books = 3500*(100/70) = 5000

**From 1 and 3**

There are 20% books in Spanish.

There are 30% books in English.

There are 3000 books which are neither in English nor in Spanish.

50% = 3000

100% = 6000Incorrect**Answer: 4) Either 1 and 2 together or 1 and 3 together are sufficient.**

Explanation:

From 1 and 2

30% books are in English, so rest 70% books are in non-English language

Total books = 3500*(100/70) = 5000

**From 1 and 3**

There are 20% books in Spanish.

There are 30% books in English.

There are 3000 books which are neither in English nor in Spanish.

50% = 3000

100% = 6000 - Question 10 of 10
##### 10. Question

1 pointsCategory: Quantitative Aptitude**Puneet buys 20 watches and some belts for Rs. 18000 and sells out only the belts. Then the number of belts purchased by Puneet is –**

1) The ratio of C.P. of each watch and belt is in the ratio 6:5.

2) The cost price of a set of 2 watches and a belt is Rs. 1700.Correct**Answer: 3) Both 1 and 2**

Explanation:

Let the CP of a watch and a belt be x and y respectively.

And let the no. of belts be B

Given: 20x +By = 18000……………….1

From 1 and 2,

x:y = 6:5 then, let x = 6u and y = 5u

2x+y = 1700

2(6u) + 5u = 1700

12u + 5u = 1700

17u = 1700

u = 100

20x +By = 18000

20(6u) +B(5u) = 18000

120u + 5Bu = 18000

120*100+ 5B*100 = 18000

12000 + 500B = 18000

500B = 6000

B = 6000/500

B = 12Incorrect**Answer: 3) Both 1 and 2**

Explanation:

Let the CP of a watch and a belt be x and y respectively.

And let the no. of belts be B

Given: 20x +By = 18000……………….1

From 1 and 2,

x:y = 6:5 then, let x = 6u and y = 5u

2x+y = 1700

2(6u) + 5u = 1700

12u + 5u = 1700

17u = 1700

u = 100

20x +By = 18000

20(6u) +B(5u) = 18000

120u + 5Bu = 18000

120*100+ 5B*100 = 18000

12000 + 500B = 18000

500B = 6000

B = 6000/500

B = 12

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