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Question 1 of 10
1. Question
1 pointsCategory: Quantitative AptitudeDirections for questions 15: Study the following graph carefully to answer the given questions.
A boatman rows up stream at 16 kmph and 2 kmph on Tuesday and Friday resectively. Then find the difference between the downstream speed of boat on Tuesday and Friday together and the speed of boat in still water on the same days together?
Correct
Solution: downstream speed = (2*6(speed of current)) + 16(upstream speed) = 28 km/hr. Since, (downstream speed + Upstream speed)/2 = Speed of boat in still water, therefore 2*speed of boat in still water upstream speed = downstream speed i.e. u = 6 2 = 4 km/hr. Downstream speed of boat on Tuesday and Friday together = 28+ 4= 32 kmph and Still water speed on Tuesday = (Upstream speed+ downstream speed)/2. Putting the values i.e we get 22 kmph on Tuesday and 3 kmph on Friday. Hence, Still water speed on both days together = 25 kmph. Therefore, Difference = 32 – 25 = 7 kmph
Incorrect
Solution: downstream speed = (2*6(speed of current)) + 16(upstream speed) = 28 km/hr. Since, (downstream speed + Upstream speed)/2 = Speed of boat in still water, therefore 2*speed of boat in still water upstream speed = downstream speed i.e. u = 6 2 = 4 km/hr. Downstream speed of boat on Tuesday and Friday together = 28+ 4= 32 kmph and Still water speed on Tuesday = (Upstream speed+ downstream speed)/2. Putting the values i.e we get 22 kmph on Tuesday and 3 kmph on Friday. Hence, Still water speed on both days together = 25 kmph. Therefore, Difference = 32 – 25 = 7 kmph

Question 2 of 10
2. Question
1 pointsCategory: Quantitative AptitudeDirections for questions 15: Study the following graph carefully to answer the given questions.
Find the ratio between the downstream distance travelled by boat on Monday, Wednesday, Thursday and Friday together and the upstream distance travelled by the boat on the same days together?
Correct
Solution: Downstream distance travelled by boat on Monday, Wednesday, Thursday and Friday together = 48 + 30 + 40 + 52 = 170.Upstream distance travelled by boat on Monday, Wednesday, Thursday and Friday together = 90 + 72 + 42 + 36 = 240. Hence ratio =17 : 24.
Incorrect
Solution: Downstream distance travelled by boat on Monday, Wednesday, Thursday and Friday together = 48 + 30 + 40 + 52 = 170.Upstream distance travelled by boat on Monday, Wednesday, Thursday and Friday together = 90 + 72 + 42 + 36 = 240. Hence ratio =17 : 24.

Question 3 of 10
3. Question
1 pointsCategory: Quantitative AptitudeDirections for questions 15: Study the following graph carefully to answer the given questions.
Find the difference between the upstream distance travelled by boat on Tuesday and Friday together and the downstream distance travelled by boat on the same days together?
Correct
Incorrect

Question 4 of 10
4. Question
1 pointsCategory: Quantitative AptitudeDirections for questions 15: Study the following graph carefully to answer the given questions.
Total downstream distance travelled by boat on Monday, Tuesday and Wednesday together is what percentage of total upstream distance travelled by boat on the same days together?
Correct
Solution: Total downstream distance travelled by boat on Monday, Tuesday and Wednesday = 108
Total up stream distance travelled by boat on Monday, Tuesday and Wednesday = 222, hence required % = 108/222 * 100 = 48.6%Incorrect
Solution: Total downstream distance travelled by boat on Monday, Tuesday and Wednesday = 108
Total up stream distance travelled by boat on Monday, Tuesday and Wednesday = 222, hence required % = 108/222 * 100 = 48.6% 
Question 5 of 10
5. Question
1 pointsCategory: Quantitative AptitudeDirections for questions 15: Study the following graph carefully to answer the given questions.
A boatman rows downstream at 6kmph on Wednesday. Find the time taken by the boat to cover upstream distance on the same day?
Correct
Solution: Downstream Speed = 6kmph, Upstream Speed = 8kmph – 6kmph = 2kmph, Time taken by the boat to cover upstream distance = 72/2 = 36 hours
Incorrect
Solution: Downstream Speed = 6kmph, Upstream Speed = 8kmph – 6kmph = 2kmph, Time taken by the boat to cover upstream distance = 72/2 = 36 hours

Question 6 of 10
6. Question
1 pointsCategory: Quantitative AptitudeDirection for Question 68: In the following question, two statements are numbered as I, II and III. On solving these statements you get Quantity I, Quantity II and Quantity III respectively. Solve for the statements individually and mark the correct answer.
Quantity I: If Anil walks at 12kmph instead of 10kmph, he would have walked 18km more. The actual distance travelled by him is?
Quantity II: John travelled from his town to city. John went to city by bicycle at the speed of 25 km/h and came back at the speed of 4 km/h. If John took 5 hours and 48 min to complete his journey, what is the distance between town and city?
Quantity III: A farmer travelled a distance of 61 km in 9 hours. He travelled partly on foot @ 4 km/hr and partly on bicycle @ 9 km/hr. The distance travelled on foot is:Correct
Solution: Quantity I: x/10 = (x+18)/12, therefore 12x = 10x + 180 i.e.
2x = 180, X = 90km. Quantity II: Average speed of John = 2xy/x+y = 2 × 25 × 4 / (25 + 4) = 200/29 km/h. Since, Distance travelled = Speed × Time = 200/29 × 29/5 = 40 Km, Distance between city and town = 40/2 = 20 km. Quantity III Let the distance travelled on foot be x km. Then, distance travelled on bicycle = (61 x) km. So, x/4 + (61x)/9 = 9, 9x + 4(61 x) = 9 x 36. So, 5x = 80, x = 16 km Hence, Quantity I > Quantity II > Quantity IIIIncorrect
Solution: Quantity I: x/10 = (x+18)/12, therefore 12x = 10x + 180 i.e.
2x = 180, X = 90km. Quantity II: Average speed of John = 2xy/x+y = 2 × 25 × 4 / (25 + 4) = 200/29 km/h. Since, Distance travelled = Speed × Time = 200/29 × 29/5 = 40 Km, Distance between city and town = 40/2 = 20 km. Quantity III Let the distance travelled on foot be x km. Then, distance travelled on bicycle = (61 x) km. So, x/4 + (61x)/9 = 9, 9x + 4(61 x) = 9 x 36. So, 5x = 80, x = 16 km Hence, Quantity I > Quantity II > Quantity III 
Question 7 of 10
7. Question
1 pointsCategory: Quantitative AptitudeDirection for Question 68: In the following question, two statements are numbered as I, II and III. On solving these statements you get Quantity I, Quantity II and Quantity III respectively. Solve for the statements individually and mark the correct answer.
The total surface area of a cube, sphere and cylinder is the same. The height of the cylinder is twice its radius.
Quantity I: Volume of the cube.
Quantity II: Volume of the sphere.
Quantity III: Volume of the cylinder.Correct
Solution: Let the cube’s edge length is a, the surface area of cube = 6a^2 and its volume = a^3. Let the radius of the sphere is r, then its volume = (4/3) πr^3 and its surface area will be 4πr^2. Let the height of the cylinder be 2h therefore, its radius will be h. Surface area of the cylinder = 2πh(2h)+2πh^2 = 6πh^2. Now since the surface area are same, 6a^2 = 4πr^2 = 6πh^2, therefore, r= (√3/√(2π))a. Volume of sphere = (√6/√π)a^3, this will be always greater than a^3. Hence Quantity II > Quantity I, Also 6a^2 = 4πr^2 = 6πh^2, a = (√π)/h. Volume of the cylinder = πh^2(2h) = 2πh^3 = 2π(a/√π)^3 = 2a^3/(√π) > a^3. Hence Quantity III > Quantity I. Volume of sphere =(√6/√π)a^3 is > volume of the cylinder = 2a^3/(√π). Hence, Quantity II > Quantity III > Quantity I
Incorrect
Solution: Let the cube’s edge length is a, the surface area of cube = 6a^2 and its volume = a^3. Let the radius of the sphere is r, then its volume = (4/3) πr^3 and its surface area will be 4πr^2. Let the height of the cylinder be 2h therefore, its radius will be h. Surface area of the cylinder = 2πh(2h)+2πh^2 = 6πh^2. Now since the surface area are same, 6a^2 = 4πr^2 = 6πh^2, therefore, r= (√3/√(2π))a. Volume of sphere = (√6/√π)a^3, this will be always greater than a^3. Hence Quantity II > Quantity I, Also 6a^2 = 4πr^2 = 6πh^2, a = (√π)/h. Volume of the cylinder = πh^2(2h) = 2πh^3 = 2π(a/√π)^3 = 2a^3/(√π) > a^3. Hence Quantity III > Quantity I. Volume of sphere =(√6/√π)a^3 is > volume of the cylinder = 2a^3/(√π). Hence, Quantity II > Quantity III > Quantity I

Question 8 of 10
8. Question
1 pointsCategory: Quantitative AptitudeDirection for Question 68: In the following question, two statements are numbered as I, II and III. On solving these statements you get Quantity I, Quantity II and Quantity III respectively. Solve for the statements individually and mark the correct answer.
Quantity I: If Anil walks at 12kmph instead of 10kmph, he would have walked 18km more. The actual distance travelled by him is?
Quantity II: John travelled from his town to city. John went to city by bicycle at the speed of 25 km/h and came back at the speed of 4 km/h. If John took 5 hours and 48 min to complete his journey, what is the distance between town and city?
Quantity III: A farmer travelled a distance of 61 km in 9 hours. He travelled partly on foot @ 4 km/hr and partly on bicycle @ 9 km/hr. The distance travelled on foot is:There are 5 red balls, 6 green balls, 9 blue balls and remaining yellow balls in a lot of 50 balls.
Quantity I: Probability of picking 2 balls such that one is green and the other is blue.
Quantity II: Probability of picking up three balls such that at least one of them is red.
Quantity III: Probability of picking 3 balls such that at least one is blue.Correct
Solution: Quantity I: 6*9/(50C2) = 54/1225. Quantity II = 1 45C3/(50C3)= 1 – 14190/19600 = 541/1960 > Quantity I . Quantity III = 1 – 41C3/(50C3)= 1 – 10660/19600 = 894/1960 > Quantity II > Quantity I. Hence, Quantity III > Quantity II > Quantity I
Incorrect
Solution: Quantity I: 6*9/(50C2) = 54/1225. Quantity II = 1 45C3/(50C3)= 1 – 14190/19600 = 541/1960 > Quantity I . Quantity III = 1 – 41C3/(50C3)= 1 – 10660/19600 = 894/1960 > Quantity II > Quantity I. Hence, Quantity III > Quantity II > Quantity I

Question 9 of 10
9. Question
1 pointsCategory: Quantitative AptitudeRs. 5887 is divided between Shyam and Ram, such that Shyam’s share at the end of 9 years is equal to Ram’s share at the end of 11 years, compounded annually at the rate of 5%. Find the share of Shyam.
Correct
Solution: Shyam’s share * (1+0.05)^9 = Ram’s share * (1 + 0.05)^11. Shyam’s share / Ram’s share = (1 + 0.05)^11 / (1+ 0.05)^9 = (1+ 0.05)^2 = 441/400. Therefore Shyam’s share = (441/841) * 5887 = 3087
Incorrect
Solution: Shyam’s share * (1+0.05)^9 = Ram’s share * (1 + 0.05)^11. Shyam’s share / Ram’s share = (1 + 0.05)^11 / (1+ 0.05)^9 = (1+ 0.05)^2 = 441/400. Therefore Shyam’s share = (441/841) * 5887 = 3087

Question 10 of 10
10. Question
1 pointsCategory: Quantitative AptitudeA and B entered into a partnership investing Rs. 16,000 and Rs. 12,000 respectively. After 3 months, A withdrew Rs. 5000 while B invested Rs. 5000 more, After 3 more months, C joins the business with a capital of Rs. 21,000, The share of B exceeds that of C, out of a total profit of Rs. 26,400 after one year by?
Correct
Solution: A : B : C = (16000 * 3 + 11000 * 9) : (12000 * 3 + 17000 * 9) : (21000 * 6) = 147 : 189 : 126 = 7 : 9 ; 6. Difference of B and C’s shares = Rs. (26400 * 9/22 – 26400 * 6/22 ) = Rs. 3600.
Incorrect
Solution: A : B : C = (16000 * 3 + 11000 * 9) : (12000 * 3 + 17000 * 9) : (21000 * 6) = 147 : 189 : 126 = 7 : 9 ; 6. Difference of B and C’s shares = Rs. (26400 * 9/22 – 26400 * 6/22 ) = Rs. 3600.
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