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Question 1 of 10
1. Question
1 pointsCategory: Quantitative AptitudeDirections(Q.No: 1 – 5). Study the following graph carefully and answer accordingly.
Number of students playing Chess, Volley Ball, Badminton from different Schools
The total number of students playing Chess and Badminton together from school A is what % of the total number of students playing these two games together from School D?Correct
Explanation :
Total number of students playing Chess and Badminton together from school A = 220+140 = 360
Total number of students playing Chess and Badminton together from School D = 260+160 = 420
Percentage = (360/420)*100 = 600/7 %Incorrect
Explanation :
Total number of students playing Chess and Badminton together from school A = 220+140 = 360
Total number of students playing Chess and Badminton together from School D = 260+160 = 420
Percentage = (360/420)*100 = 600/7 % 
Question 2 of 10
2. Question
1 pointsCategory: Quantitative AptitudeDirections(Q.No: 1 – 5). Study the following graph carefully and answer accordingly.
Number of students playing Chess, Volley Ball, Badminton from different Schools
If the number of students playing each game in school B is increased by 20% and the number of students playing each game in school C is decreased by 30% then what will be the difference between the number of students in School B and C?Correct
Explanation :
Number of students in school B = 120% of 680 = 816
Number of students in school C = 70% of 760 = 532
Difference = 816 – 532 = 284Incorrect
Explanation :
Number of students in school B = 120% of 680 = 816
Number of students in school C = 70% of 760 = 532
Difference = 816 – 532 = 284 
Question 3 of 10
3. Question
1 pointsCategory: Quantitative AptitudeDirections(Q.No: 1 – 5). Study the following graph carefully and answer accordingly.
Number of students playing Chess, Volley Ball, Badminton from different Schools
If out of the students playing Volleyball from schools B, D and E 30%,20% and 10% respectively got selected for state level competition, what was the total number of students got selected for state level competition from these three schools together?Correct
Explanation :
Total number of students who got selected for state level competition from schools B D and E
= 30% of 180 + 20% of 320 + 10% of 240
= 54 + 64 + 24 = 142Incorrect
Explanation :
Total number of students who got selected for state level competition from schools B D and E
= 30% of 180 + 20% of 320 + 10% of 240
= 54 + 64 + 24 = 142 
Question 4 of 10
4. Question
1 pointsCategory: Quantitative AptitudeDirections(Q.No: 1 – 5). Study the following graph carefully and answer accordingly.
Number of students playing Chess, Volley Ball, Badminton from different Schools
Total number of students playing Badminton from all schools together is approximately, what percent of the total number of students playing Volleyball from all schools together?Correct
Explanation :
Total number of students playing Badminton from all the schools together = 140 + 260 + 320 + 160 + 180 = 1060
Total number of students playing Volleyball from all the schools together = 360 + 180 + 240 + 320 + 240 = 1340
% = (1060/1340)*100 = 79%(approximately)Incorrect
Explanation :
Total number of students playing Badminton from all the schools together = 140 + 260 + 320 + 160 + 180 = 1060
Total number of students playing Volleyball from all the schools together = 360 + 180 + 240 + 320 + 240 = 1340
% = (1060/1340)*100 = 79%(approximately) 
Question 5 of 10
5. Question
1 pointsCategory: Quantitative AptitudeDirections(Q.No: 1 – 5). Study the following graph carefully and answer accordingly.
Number of students playing Chess, Volley Ball, Badminton from different Schools
From School A, out of the students playing Chess, 80% got selected for state level competition out of which 25% further got selected for national level competition.From School E, out of the students playing Chess, 45% got selected for state level competition out of which twothird further got selected for national level competition.What is the total number of students playing Chess from these two schools who got selected for national level competition?Correct
Explanation :
Number of students who got selected for national level competition from School A = 25% of 80% of 220 = 44
Number of students who got selected for national level competition from School E = (2/3) of 45% of 280 = 84
Total number of students playing Chess from these two schools who got selected for national level competition = 44 + 84 = 128Incorrect
Explanation :
Number of students who got selected for national level competition from School A = 25% of 80% of 220 = 44
Number of students who got selected for national level competition from School E = (2/3) of 45% of 280 = 84
Total number of students playing Chess from these two schools who got selected for national level competition = 44 + 84 = 128 
Question 6 of 10
6. Question
1 pointsCategory: Quantitative AptitudeDirections: Q(610) Solve
Mark A If Quantity 1 > Quantity 2
Mark B If Quantity 1 < Quantity 2
Mark C If Quantity 1 ≥ Quantity 2
Mark D If Quantity 1 ≤ Quantity 2
Mark E If Quantity 1 is not related to Quantity 2Quantity 1:
The present age of Ramesh is one – fourth that of his father. After six years, the father’s age will be twice the age of Kumar. If Kumar celebrated the sixth birthday 8 year ago, What is Ramesh’s Present age?
Quantity 2:
Ramya’s present age is four times her daughter, Veena’s present age and twothirds of her mother, Rani’s present age. The total of the present ages of all of them is 132 years. What is the difference between Ramya’s and her mother Rani’s present age?Correct
Explanation :
Quantity 1:
Kumar’s present age = 8 + 6 = 14
Kumar’s age after 6 years = 14 + 6 = 20
Father age = 2 * 20 = 40
Father’s present age = 34
Ramesh’s present age = 34 / 4 = 8.5 yearsQuantity 2:
Ramya’s present age – x
x + (x / 4) + (3 / 2) x = 132
11x = 132 * 4
x = 48
Ramya’s mother age = 3/2 * 48 = 72
Difference = 72 – 48 = 24 years.
Quantity 1 < Quantity 2Incorrect
Explanation :
Quantity 1:
Kumar’s present age = 8 + 6 = 14
Kumar’s age after 6 years = 14 + 6 = 20
Father age = 2 * 20 = 40
Father’s present age = 34
Ramesh’s present age = 34 / 4 = 8.5 yearsQuantity 2:
Ramya’s present age – x
x + (x / 4) + (3 / 2) x = 132
11x = 132 * 4
x = 48
Ramya’s mother age = 3/2 * 48 = 72
Difference = 72 – 48 = 24 years.
Quantity 1 < Quantity 2 
Question 7 of 10
7. Question
1 pointsCategory: Quantitative AptitudeDirections: Q(610) Solve
Mark A If Quantity 1 > Quantity 2
Mark B If Quantity 1 < Quantity 2
Mark C If Quantity 1 ≥ Quantity 2
Mark D If Quantity 1 ≤ Quantity 2
Mark E If Quantity 1 is not related to Quantity 2Quantity 1:
Two cars A and B starts at the same time from Delhi and Agra respectively which is 200 km apart, If the two cars travel in opposite directions they meet after 1 hour and if they travel in same direction from Delhi and Agra then the car which starts at Delhi meets the another car after 5 hours, What is the speed of car starting from Delhi?
Quantity 2:
A travel bus normally reaches its destination at 60 kmph in 20 hours. Find the speed of that travel bus at which it travels to reduce the time by 5 hours?Correct
Explanation :
Quantity 1:
relative speed = a+b
total distance travelled = 200 km
time taken = 1 hour
a + b = 200 —(1)
Relative speed of the faster car = a – b
total distance travelled = 200 km
time taken = 5 hour
200 = 5(ab)
40 = a – b —(2)
From (1) and (2) 2a = 240
a = 120
Speed of car starting from Delhi = 120km/hr
Quantity 2:
60 * 20 = x *15
x = 80 kmph
Quantity 1 > Quantity 2Incorrect
Explanation :
Quantity 1:
relative speed = a+b
total distance travelled = 200 km
time taken = 1 hour
a + b = 200 —(1)
Relative speed of the faster car = a – b
total distance travelled = 200 km
time taken = 5 hour
200 = 5(ab)
40 = a – b —(2)
From (1) and (2) 2a = 240
a = 120
Speed of car starting from Delhi = 120km/hr
Quantity 2:
60 * 20 = x *15
x = 80 kmph
Quantity 1 > Quantity 2 
Question 8 of 10
8. Question
1 pointsCategory: Quantitative AptitudeDirections: Q(610) Solve
Mark A If Quantity 1 > Quantity 2
Mark B If Quantity 1 < Quantity 2
Mark C If Quantity 1 ≥ Quantity 2
Mark D If Quantity 1 ≤ Quantity 2
Mark E If Quantity 1 is not related to Quantity 2Quantity 1:
Cost Price of two mobiles is same. One is sold at a profit of 20% and the other for Rs. 5200 more than the first. If the net profit is 40%. Find the cost price of each mobile?
Quantity 2:
A shopkeeper bought 150 pen drives at the rate of Rs. 500 per pen drive. He spent Rs. 500 on transportation and packing. If the marked price of pen drive is Rs. 520 per pen drive and the shopkeeper gives a discount of 5% on the marked price then what will be the percentage profit gained by the shopkeeper?Correct
Explanation:
Quantity 1 is not related to Quantity 2
Quantity 1 deals with Cost Price of Mobile
Quantity 2 deals with Percentage profit
Solution:
Quantity 1:
CP of each mobile be Rs.x then
(2 x 1.20 x x) + 5200 = 2 x 1.4 x N
⇒ 0.4 x = 5200
∴ N = 13000
Quantity 2:
C.P. of 150 calculators = 150 x 500 = Rs. 75000
∴ Total C.P. = 75000 + 500 = Rs. 75500
Marked price of 150 calculator = 150 x 570 = Rs. 82500
Selling price after discount = 82500 x 95 / 100 = Rs. 78375
∴ percentage profit = [(78375 – 75500) / 75500] x 100 = 3.8%
Quantity 1 is not related to Quantity 2Incorrect
Explanation:
Quantity 1 is not related to Quantity 2
Quantity 1 deals with Cost Price of Mobile
Quantity 2 deals with Percentage profit
Solution:
Quantity 1:
CP of each mobile be Rs.x then
(2 x 1.20 x x) + 5200 = 2 x 1.4 x N
⇒ 0.4 x = 5200
∴ N = 13000
Quantity 2:
C.P. of 150 calculators = 150 x 500 = Rs. 75000
∴ Total C.P. = 75000 + 500 = Rs. 75500
Marked price of 150 calculator = 150 x 570 = Rs. 82500
Selling price after discount = 82500 x 95 / 100 = Rs. 78375
∴ percentage profit = [(78375 – 75500) / 75500] x 100 = 3.8%
Quantity 1 is not related to Quantity 2 
Question 9 of 10
9. Question
1 pointsCategory: Quantitative AptitudeDirections: Q(610) Solve
Mark A If Quantity 1 > Quantity 2
Mark B If Quantity 1 < Quantity 2 Mark C If Quantity 1 ≥ Quantity 2 Mark D If Quantity 1 ≤ Quantity 2 Mark E If Quantity 1 is not related to Quantity 2 Quantity 1 x²  30x + 216 = 0 Quantity 2: y²  23y + 132 = 0Correct
Explanation:
x² – 30x + 216 = 0
x = 12, 18
y² – 23y + 132 = 0
y = 12, 11Incorrect
Explanation:
x² – 30x + 216 = 0
x = 12, 18
y² – 23y + 132 = 0
y = 12, 11 
Question 10 of 10
10. Question
1 pointsCategory: Quantitative AptitudeDirections: Q(610) Solve
Mark A If Quantity 1 > Quantity 2
Mark B If Quantity 1 < Quantity 2
Mark C If Quantity 1 ≥ Quantity 2
Mark D If Quantity 1 ≤ Quantity 2
Mark E If Quantity 1 is not related to Quantity 2Quantity 1:
Vikram borrows a sum of Rs.1500 at the beginning of a year. After four months Rs.2100 more is borrowed at a rate of interest double the previous one. At the end of one year, the sum of interest on both the loans is Rs.416. What is the first rate of interest per annum?
Quantity 2:
Ankita borrows Rs.7000 at simple Interest from a lender. At the end of 3 years, she again borrows Rs.3000 and settled that amount after paying Rs.4615 as interest after 8 years from the time she made the first borrowing. what is the rate of interest?Correct
Explanation:
Quantity 1:
P = 1500
Rate of Interest = x
SI = 1500x/100 = 15x
P = 2100
Rate of Interest = 2x
SI = 4200x/100 = 42x
57x = 416
x = 7.3%
Quantity 2:
SI for Rs.7000 for 8 years= (7000*r*8)/100
Again borrowed=3000
SI = (3000*r*5)/100
Total interest= [(7000*r*8)/100] + [(3000*r*5)/100] = 4615
560r + 150r = 4615
710r = 4615
r = 6.5%Incorrect
Explanation:
Quantity 1:
P = 1500
Rate of Interest = x
SI = 1500x/100 = 15x
P = 2100
Rate of Interest = 2x
SI = 4200x/100 = 42x
57x = 416
x = 7.3%
Quantity 2:
SI for Rs.7000 for 8 years= (7000*r*8)/100
Again borrowed=3000
SI = (3000*r*5)/100
Total interest= [(7000*r*8)/100] + [(3000*r*5)/100] = 4615
560r + 150r = 4615
710r = 4615
r = 6.5%
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