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Question 1 of 10

1. Question

1 points

Category: Quantitative Aptitude

A 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 4,00,000

Incorrect

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 points

Category: Quantitative Aptitude

A 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 points

Category: Quantitative Aptitude

A 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 points

Category: Quantitative Aptitude

A 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 hours

Incorrect

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 points

Category: Quantitative Aptitude

Vessels 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?

Quantity 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 = 48

Quantity 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

Incorrect

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 = 48

Quantity 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 points

Category: Quantitative Aptitude

Manoj 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 = 150

Incorrect

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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