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Question 1 of 10
1. Question
1 pointsCategory: Quantitative AptitudeDirections (15): 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 minutesQ2
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 minutesQ2
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 AptitudeDirections (15): 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 = 3x4.5 and water left = 5x7.5
ATQ,
(3x4.5+12)/(5x7.5) = 2/3
x = 37.5
Therefore, initial quantity of milk = 3x = 3*37.5 = 112.5 litresQ2:
Final quantity:
= 56*(18/56)*(17/56)
= 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 = 3x4.5 and water left = 5x7.5
ATQ,
(3x4.5+12)/(5x7.5) = 2/3
x = 37.5
Therefore, initial quantity of milk = 3x = 3*37.5 = 112.5 litresQ2:
Final quantity:
= 56*(18/56)*(17/56)
= 56*(6/7)*(7/8)
= 42 litres 
Question 3 of 10
3. Question
1 pointsCategory: Quantitative AptitudeDirections (15): 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 daysQ2
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 daysQ2
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 AptitudeDirections (15): 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: 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 AptitudeDirections (15): Following questions have two quantities as Quantity I and Quantity II. You have to determine the relationship between them and give answer as,
There 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
= 25/28Q2:
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
= 25/28Q2:
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 AptitudeDirections (610): 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 AptitudeDirections (610): 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 AptitudeDirections (610): 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 AptitudeDirections (610): 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.
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 nonEnglish 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 nonEnglish 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 nonEnglish 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 AptitudeDirections (610): 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.
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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