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Question 1 of 10
1. Question
1 pointsCategory: Quantitative AptitudeA man gives 50% of his money to his son and 30% to his daughter. 80% of the rest is donated to a trust. If he is left with 16,000 now, how much money did he gave in the beginning?
Correct
Answer – 1) Rs 4,00,000
Explanation:
Let he gave in the beginning= Rs x
X * (20/100) * (20/100) = 16000
X = Rs 4,00,000Incorrect
Answer – 1) Rs 4,00,000
Explanation:
Let he gave in the beginning= Rs x
X * (20/100) * (20/100) = 16000
X = Rs 4,00,000 
Question 2 of 10
2. Question
1 pointsCategory: Quantitative AptitudeA man sells an article at a gain of 15%. If he had bought it at 10% less and solid if for Rs 4 less, he would have gained 25%. What is the cost price of the article?
Correct
Answer – 1) Rs 160
Explanation:
Incorrect
Answer – 1) Rs 160
Explanation:

Question 3 of 10
3. Question
1 pointsCategory: Quantitative AptitudeA boy has a few coins of 50 paise , 25 paise and 10 paise in the ratio of 1: 32: 3. If the total amount of the coins is Rs 8.80. What is the number of 10 paise coins?
Correct
Answer – 4) 30
Explanation:
Incorrect
Answer – 4) 30
Explanation:

Question 4 of 10
4. Question
1 pointsCategory: Quantitative AptitudeA man swims downstream a distance of 15km in 1 hour. If the speed of the current is 5km/hour, what is the time taken by the man to swim the same distance upstream?
Correct
Answer – 5) 3 hrs
Explanation:
Let the speed of the boat in still water = X Km/hr
The speed of current, Y = 5 Km/hr
Downstream speed = 15 Km/hr
X+5 = 15
X = 10 Km/hr
Upstream speed, U = X Y
10 – 5 = 5 Km/hr
Upstream Time= Distance/ Upstream Speed
15/5 = 3 hoursIncorrect
Answer – 5) 3 hrs
Explanation:
Let the speed of the boat in still water = X Km/hr
The speed of current, Y = 5 Km/hr
Downstream speed = 15 Km/hr
X+5 = 15
X = 10 Km/hr
Upstream speed, U = X Y
10 – 5 = 5 Km/hr
Upstream Time= Distance/ Upstream Speed
15/5 = 3 hours 
Question 5 of 10
5. Question
1 pointsCategory: Quantitative AptitudeVessels A and B contain mixture of milk and water in the ratio 4:5 and 5:1 respectively. In what ratio should quantities of mixture be Taken from A and B to form a mixture in which milk to water is in the ratio 5:4?
Correct
Answer – 2) 5:2
Explanation:
Incorrect
Answer – 2) 5:2
Explanation:

Question 6 of 10
6. Question
1 pointsCategory: Quantitative AptitudeX^{2 }– 32x +240 = 0
10Y^{2} – 23Y 42 = 0Correct
Answer 2) X>Y
Explanation
X^{2 }– 32x +240
(X12)(X20)
X = 12, 2010Y^{2} – 23Y 42
(5Y + 6)(2Y 7)
Y = 6/5, 7/2Incorrect
Answer 2) X>Y
Explanation
X^{2 }– 32x +240
(X12)(X20)
X = 12, 2010Y^{2} – 23Y 42
(5Y + 6)(2Y 7)
Y = 6/5, 7/2 
Question 7 of 10
7. Question
1 pointsCategory: Quantitative AptitudeX^{2} 4X 32 = 0
Y^{2} +27Y +72 = 0Correct
Answer –5) X=Y or relationship cannot be established
Explanation
X^{2} 4X 32
(X + 4)(X 8)
= 4, 8Y^{2} +27Y +72
(Y + 3)(Y+24)
= 3, 24Incorrect
Answer –5) X=Y or relationship cannot be established
Explanation
X^{2} 4X 32
(X + 4)(X 8)
= 4, 8Y^{2} +27Y +72
(Y + 3)(Y+24)
= 3, 24 
Question 8 of 10
8. Question
1 pointsCategory: Quantitative Aptitude15X^{2} +7X 36 = 0
Y^{2} 31Y +184 = 0Correct
Answer –1) X<Y
Explanation
15X^{2} +7X 36
(3X4)(5X+9)
X = 4/3, 9/5Y^{2} 31Y +184
(Y23)(Y8)
Y= 23, 8Incorrect
Answer –1) X<Y
Explanation
15X^{2} +7X 36
(3X4)(5X+9)
X = 4/3, 9/5Y^{2} 31Y +184
(Y23)(Y8)
Y= 23, 8 
Question 9 of 10
9. Question
1 pointsCategory: Quantitative AptitudeQuantity I :What is the perimeter of an equilateral triangle whose area is 64√3 cm2.
Quantity II: What is the perimeter of a regular hexagonal whose area is 96√3 cm2.Correct
Answer –5) Quantity I= Quantity II
Explanation
Quantity I
We should know that area of the equilateral triangle = √3/4(Side)^{2}
Let’s assume each side of the equilateral triangle = a cm
√3/4 a^{2} = 64√3
A^{2 }= 256
A^{ 2 }= 16
Hence perimeter of the triangle = 3a = 48Quantity II
We should know that area of the regular hexagonal = (3√3)/2 (Side)^{2}
Let assume each side of the regular hexagon = a cm
(3√3)/2 (a)^{2} = 96√3
A^{2} = 64
A = 8
Hence perimeter of the hexagonal 6a cm = 48 cmIncorrect
Answer –5) Quantity I= Quantity II
Explanation
Quantity I
We should know that area of the equilateral triangle = √3/4(Side)^{2}
Let’s assume each side of the equilateral triangle = a cm
√3/4 a^{2} = 64√3
A^{2 }= 256
A^{ 2 }= 16
Hence perimeter of the triangle = 3a = 48Quantity II
We should know that area of the regular hexagonal = (3√3)/2 (Side)^{2}
Let assume each side of the regular hexagon = a cm
(3√3)/2 (a)^{2} = 96√3
A^{2} = 64
A = 8
Hence perimeter of the hexagonal 6a cm = 48 cm 
Question 10 of 10
10. Question
1 pointsCategory: Quantitative AptitudeManoj appeared for IBPS PO exam. What are the minimum marks that manoj needs in order to qualify the exam.
Statement I: When Manoj got 36% of the marks then he just fell short by 9 marks.
Statement II: When Manoj scored 44% of the marks then he scored 3 marks more than what was required to qualify.Correct
Answer –4) Question can be answered by using Statement I and Statement II together
Explanation
Let us assume maximum marks in IBPS PO = 100a
Statement I
Minimum qualifying marks in IBPS exam = 36a + 9 (1)
Statement II
Minimum qualifying marks in the IBPS exam = 44a – 3 (2)
Now using (1) and (2) together
36a + 9 = 44a – 3
A = 1.5
Maximum marks in the IBPS PO = 100 x 1.5 = 150Incorrect
Answer –4) Question can be answered by using Statement I and Statement II together
Explanation
Let us assume maximum marks in IBPS PO = 100a
Statement I
Minimum qualifying marks in IBPS exam = 36a + 9 (1)
Statement II
Minimum qualifying marks in the IBPS exam = 44a – 3 (2)
Now using (1) and (2) together
36a + 9 = 44a – 3
A = 1.5
Maximum marks in the IBPS PO = 100 x 1.5 = 150
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