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Question 1 of 10
1. Question
1 pointsCategory: Quantitative AptitudeDirections for Questions (15) Study the following information carefully and answer the questions given below:
Six different companies P, Q, R, S, T and U make two items I and II. The total number of items produced by these six companies is 2,00,000. The cost of production of each item is Rs.20,000. The distribution of the total production by these companies is given in the following piechart and the table shows the ratio of production of Item I to that of Item II and the percentage profit earned by these companies on each of these items.
What is the profit earned by Company R on Item II?
Correct
Solution: Profit earned by company R on Item II, (17)/(100) ×200000× 3/5×20000×12/100=4.8 crore
Incorrect
Solution: Profit earned by company R on Item II, (17)/(100) ×200000× 3/5×20000×12/100=4.8 crore

Question 2 of 10
2. Question
1 pointsCategory: Quantitative AptitudeDirections for Questions (15) Study the following information carefully and answer the questions given below:
Six different companies P, Q, R, S, T and U make two items I and II. The total number of items produced by these six companies is 2,00,000. The cost of production of each item is Rs.20,000. The distribution of the total production by these companies is given in the following piechart and the table shows the ratio of production of Item I to that of Item II and the percentage profit earned by these companies on each of these items.
What is the total cost of production of Item I by companies S and T together?
Correct
Solution: Total cost of production of item I by companies S and T together, ((19)/(100)×200000×1/3×20000)+((11)/(100)×200000×1/4×20000)=25.33 crore+11.11 crore=36.44 crore
Incorrect
Solution: Total cost of production of item I by companies S and T together, ((19)/(100)×200000×1/3×20000)+((11)/(100)×200000×1/4×20000)=25.33 crore+11.11 crore=36.44 crore

Question 3 of 10
3. Question
1 pointsCategory: Quantitative AptitudeDirections for Questions (15) Study the following information carefully and answer the questions given below:
Six different companies P, Q, R, S, T and U make two items I and II. The total number of items produced by these six companies is 2,00,000. The cost of production of each item is Rs.20,000. The distribution of the total production by these companies is given in the following piechart and the table shows the ratio of production of Item I to that of Item II and the percentage profit earned by these companies on each of these items.
What is the ratio of the cost of production of Item II by Company P to the cost of production of Item I by Company Q?
Correct
Solution: P :Q=(1/5×(22)/(100)) :(3/5×(25)/(100))=22 :75
Incorrect
Solution: P :Q=(1/5×(22)/(100)) :(3/5×(25)/(100))=22 :75

Question 4 of 10
4. Question
1 pointsCategory: Quantitative AptitudeDirections for Questions (15) Study the following information carefully and answer the questions given below:
Six different companies P, Q, R, S, T and U make two items I and II. The total number of items produced by these six companies is 2,00,000. The cost of production of each item is Rs.20,000. The distribution of the total production by these companies is given in the following piechart and the table shows the ratio of production of Item I to that of Item II and the percentage profit earned by these companies on each of these items.
What is the total of the profit earned by Company U on production of Item II and the profit of Company S on production of Item I?
Correct
Solution: U+S =((6)/(100)×200000×3/5×20000×15/100) + ((19)/(100) ×200000 ×1/3× 20000× 25/100) =2 crore+6.48 crore =8.48 crore
Incorrect
Solution: U+S =((6)/(100)×200000×3/5×20000×15/100) + ((19)/(100) ×200000 ×1/3× 20000× 25/100) =2 crore+6.48 crore =8.48 crore

Question 5 of 10
5. Question
1 pointsCategory: Quantitative AptitudeDirections for Questions (15) Study the following information carefully and answer the questions given below:
Six different companies P, Q, R, S, T and U make two items I and II. The total number of items produced by these six companies is 2,00,000. The cost of production of each item is Rs.20,000. The distribution of the total production by these companies is given in the following piechart and the table shows the ratio of production of Item I to that of Item II and the percentage profit earned by these companies on each of these items.
What is the total of the profit earned by Company ‘T’ on production of Item I and the profit of Company P on production of Item II?
Correct
Solution: T + P = ((11)/(100)×200000×1/4×20000×15/100) + ((22)/(100)× 200000× 1/5× 20000× 24/100) = 1.67+4.27 =5.94 crore
Incorrect
Solution: T + P = ((11)/(100)×200000×1/4×20000×15/100) + ((22)/(100)× 200000× 1/5× 20000× 24/100) = 1.67+4.27 =5.94 crore

Question 6 of 10
6. Question
1 pointsCategory: Quantitative AptitudeDirections for question 68: Read the following paragraph and answer the questions that follow:
George has recently acquired shares of four companies, namely Asian Paints(AZ), BMZ, ChaeWoo(CW) and Dataman(DT). The financial results for these companies for the year ended 200203 revealed these interesting facts: Profits of AZ were 10% of its sales, while the profits of BMZ were 20% of its sales. While the profits of CW and DT were the same, the sales of CW were the same as those of BMZ. The total expenses of CW were 400% more than its profits while the sales of DT were 200% more than its profits. The total expenses of CW were Rs. 10 million and the total expenses of CW were 11.11% more than those of AZ.
Which company had the lowest total expense?
Correct
Solution: Let the sales of AZ = x, then its profits = 0.1 x. Sales of BMZ = Sales of CW = y. Profit of BMZ is 20% of its sales = 0.2y. Sales of CW are 400% more than profits of CW. Hence, profit of CW = 0.166y = Profits of DT. Sales of DT are 200% more than profits of DT. Hence, sales of DT = 0.5y. Also expenses of CW = y – 0.166y = 0.833y = 10 million. Hence, y = 12. Also, expenses of AZ = 0.9x = 9 million. Hence, x= 10 milion. Thus, following information can be deduced:
Incorrect
Solution: Let the sales of AZ = x, then its profits = 0.1 x. Sales of BMZ = Sales of CW = y. Profit of BMZ is 20% of its sales = 0.2y. Sales of CW are 400% more than profits of CW. Hence, profit of CW = 0.166y = Profits of DT. Sales of DT are 200% more than profits of DT. Hence, sales of DT = 0.5y. Also expenses of CW = y – 0.166y = 0.833y = 10 million. Hence, y = 12. Also, expenses of AZ = 0.9x = 9 million. Hence, x= 10 milion. Thus, following information can be deduced:

Question 7 of 10
7. Question
1 pointsCategory: Quantitative AptitudeDirections for question 68: Read the following paragraph and answer the questions that follow:
George has recently acquired shares of four companies, namely Asian Paints(AZ), BMZ, ChaeWoo(CW) and Dataman(DT). The financial results for these companies for the year ended 200203 revealed these interesting facts: Profits of AZ were 10% of its sales, while the profits of BMZ were 20% of its sales. While the profits of CW and DT were the same, the sales of CW were the same as those of BMZ. The total expenses of CW were 400% more than its profits while the sales of DT were 200% more than its profits. The total expenses of CW were Rs. 10 million and the total expenses of CW were 11.11% more than those of AZ.
If next year, the profits of AZ were to equal CW’s current profit, then with an increase of 12.5% in sales, what would be the profit of AZ expressed as a percentage of sales next year.
Correct
Solution: Let the sales of AZ = x, then its profits = 0.1 x. Sales of BMZ = Sales of CW = y. Profit of BMZ is 20% of its sales = 0.2y. Sales of CW are 400% more than profits of CW. Hence, profit of CW = 0.166y = Profits of DT. Sales of DT are 200% more than profits of DT. Hence, sales of DT = 0.5y. Also expenses of CW = y – 0.166y = 0.833y = 10 million. Hence, y = 12. Also, expenses of AZ = 0.9x = 9 million. Hence, x= 10 milion. Thus, following information can be deduced:
Incorrect
Solution: Let the sales of AZ = x, then its profits = 0.1 x. Sales of BMZ = Sales of CW = y. Profit of BMZ is 20% of its sales = 0.2y. Sales of CW are 400% more than profits of CW. Hence, profit of CW = 0.166y = Profits of DT. Sales of DT are 200% more than profits of DT. Hence, sales of DT = 0.5y. Also expenses of CW = y – 0.166y = 0.833y = 10 million. Hence, y = 12. Also, expenses of AZ = 0.9x = 9 million. Hence, x= 10 milion. Thus, following information can be deduced:

Question 8 of 10
8. Question
1 pointsCategory: Quantitative AptitudeDirections for question 68: Read the following paragraph and answer the questions that follow:
George has recently acquired shares of four companies, namely Asian Paints(AZ), BMZ, ChaeWoo(CW) and Dataman(DT). The financial results for these companies for the year ended 200203 revealed these interesting facts: Profits of AZ were 10% of its sales, while the profits of BMZ were 20% of its sales. While the profits of CW and DT were the same, the sales of CW were the same as those of BMZ. The total expenses of CW were 400% more than its profits while the sales of DT were 200% more than its profits. The total expenses of CW were Rs. 10 million and the total expenses of CW were 11.11% more than those of AZ.
What is the ratio of the highest and the lowest profit?
Correct
Solution: Let the sales of AZ = x, then its profits = 0.1 x. Sales of BMZ = Sales of CW = y. Profit of BMZ is 20% of its sales = 0.2y. Sales of CW are 400% more than profits of CW. Hence, profit of CW = 0.166y = Profits of DT. Sales of DT are 200% more than profits of DT. Hence, sales of DT = 0.5y. Also expenses of CW = y – 0.166y = 0.833y = 10 million. Hence, y = 12. Also, expenses of AZ = 0.9x = 9 million. Hence, x= 10 milion. Thus, following information can be deduced:
Incorrect
Solution: Let the sales of AZ = x, then its profits = 0.1 x. Sales of BMZ = Sales of CW = y. Profit of BMZ is 20% of its sales = 0.2y. Sales of CW are 400% more than profits of CW. Hence, profit of CW = 0.166y = Profits of DT. Sales of DT are 200% more than profits of DT. Hence, sales of DT = 0.5y. Also expenses of CW = y – 0.166y = 0.833y = 10 million. Hence, y = 12. Also, expenses of AZ = 0.9x = 9 million. Hence, x= 10 milion. Thus, following information can be deduced:

Question 9 of 10
9. Question
1 pointsCategory: Quantitative AptitudeDirection for Question 910: In the following question, two statements are numbered as I and II. On solving these statements you get Quantity I and Quantity II respectively. Solve for both the statements individually and mark the correct answer.
Quantity I: Find the area of the trapezium whose parallel sides are 15cm and 12 cm and the perpendicular distance between them is 19cm.
Quantity II: Find the area of the quadrilateral piece of ground one of whose diagonal is 15 cm long and the perpendicular from the other two vertices are 12cm and 9cm respectively.Correct
Solution: Quantity I, Area of the Trapezium=1/2(Sum of the parallel sides) *(Perpendicular distance between them) =1/2(12+15) ×19 =1/2(27×19) =256.5 cm2. Quantity II, Area of the quadrilateral =1/2×d× (h1+h2), d=15cm, h1=12cm, h2=9cm. Area=1/2×15(12+9) =1/2×15(21) =315/2=157.5 cm2.Hence Quantity I >Quantity II
Incorrect
Solution: Quantity I, Area of the Trapezium=1/2(Sum of the parallel sides) *(Perpendicular distance between them) =1/2(12+15) ×19 =1/2(27×19) =256.5 cm2. Quantity II, Area of the quadrilateral =1/2×d× (h1+h2), d=15cm, h1=12cm, h2=9cm. Area=1/2×15(12+9) =1/2×15(21) =315/2=157.5 cm2.Hence Quantity I >Quantity II

Question 10 of 10
10. Question
1 pointsCategory: Quantitative AptitudeDirection for Question 910: In the following question, two statements are numbered as I and II. On solving these statements you get Quantity I and Quantity II respectively. Solve for both the statements individually and mark the correct answer.
Quantity 1: If side of a square is 5 cm. Then find its diagonal?
Quantity II: What is the diagonal if length & breadth of a rectangle is 4cm & 3cm?Correct
Solution: Quantity I, Diagonal of square = √2 × side = √2 ×5 = 7.07 cm. Quantity II, Diagonal of rectangle = √(l^2+b^2 ) = √(4^2+ 3^2 ) = 5 cm. Hence, Quantity I > Quantity II.
Incorrect
Solution: Quantity I, Diagonal of square = √2 × side = √2 ×5 = 7.07 cm. Quantity II, Diagonal of rectangle = √(l^2+b^2 ) = √(4^2+ 3^2 ) = 5 cm. Hence, Quantity I > Quantity II.
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