Quantitative Aptitude Quiz for IBPS CLERK Mains & Syndicate Po – Set 3
This quiz has been specially coined for IBPS CLERK Mains and Syndicate PO Exams. Attempt and analyze your preparation so far !
Quant Clerk - 03
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Question 1 of 10
1. Question
The volume of a cuboid, 3 times as broad as it is high and 6 times as long as it is high, is 1152 m3. Find the breadth of the cuboid.
Correct
Let the height of the cuboid be ‘x’ metres.
Then, breadth = 3x metres and length = 6x metres.
6x × 3x × x = 1152
x = 4
So, breadth of the cuboid = 3x = 12 metres
Hence, option 4.
Incorrect
Let the height of the cuboid be ‘x’ metres.
Then, breadth = 3x metres and length = 6x metres.
6x × 3x × x = 1152
x = 4
So, breadth of the cuboid = 3x = 12 metres
Hence, option 4.
Question 2 of 10
2. Question
From a group of 5 females and 4 males, four people are to be selected. The number of ways in which the 4 people can be selected so that the males are in majority is-
Correct
Incorrect
Question 3 of 10
3. Question
The monthly salary of A and B are in the ratio of 4: 5 and their monthly expenditures are in the ratio 3: 4. If both of them save Rs. 20000 each, then A’s monthly salary is-
Correct
Let the monthly salaries of A and B be Rs. ‘4x’ and Rs. ‘5x’ respectively and their monthly expenditures be Rs. ‘3y’ and Rs. ‘4y’ respectively. Then,
On a journey, a bus travels at an average speed of 20 km/hr for 40% of the total distance, at 60 km/hr for 30% of the total distance and 40 km/hr for the remaining distance. The average speed for the whole journey was-
Correct
Incorrect
Question 5 of 10
5. Question
A radio was sold for a certain price and there was a gain of 30%. If it would have been sold for Rs.324 less; there would have been a gain of 12%. Find the cost price of the radio?
Correct
Let the cost price be Rs. x
130% of x – Rs. 324 = 112% of x
18% of x = 324
0.18 × x = 324
So, x = Rs. 1800
Hence, option 4.
Incorrect
Let the cost price be Rs. x
130% of x – Rs. 324 = 112% of x
18% of x = 324
0.18 × x = 324
So, x = Rs. 1800
Hence, option 4.
Question 6 of 10
6. Question
Directions: Each question is followed by two statements I and II. Indicate your answer on the basis of the following:
What is the average of m, n, o and p?
I. The average of o and p is 35.
II. The average of m and n is 40.
Correct
Clearly both statements independently are not sufficient to answer the question.
Combining both the statements
o + p = 80
m + n = 70
m + n + o + p = 150
So, required average = 150 ÷ 4 = 37.5
So, both the statements together are sufficient.
Incorrect
Clearly both statements independently are not sufficient to answer the question.
Combining both the statements
o + p = 80
m + n = 70
m + n + o + p = 150
So, required average = 150 ÷ 4 = 37.5
So, both the statements together are sufficient.
Question 7 of 10
7. Question
Directions: Each question is followed by two statements I and II. Indicate your answer on the basis of the following:
How many rectangular sheets whose length and breadth are in the ratio of 2: 1 will cover a chart of area 576 m^{2?}
I. The sum of the length and breadth of the rectangular sheet is 36 cm.
II. The perimeter of the chart is 100 cm.
Correct
Area of the chart = 576 m^{2 }= 5760000 cm^{2}
Let the length and breadth of the rectangular sheet be ‘2x’ cm and ‘x’ cm respectively.
From I, 2x + x = 36
So, x = 12
So, length and breadth of the rectangular sheet are 24 cm and 12 cm respectively.
Area of the rectangular sheet = 24 × 12 = 288 cm^{2}
Number of rectangular sheets required = 5760000 ÷ 288 = 20000
From II, we cannot find the length and breadth of the rectangular sheet.
Hence, option 1.
Incorrect
Area of the chart = 576 m^{2 }= 5760000 cm^{2}
Let the length and breadth of the rectangular sheet be ‘2x’ cm and ‘x’ cm respectively.
From I, 2x + x = 36
So, x = 12
So, length and breadth of the rectangular sheet are 24 cm and 12 cm respectively.
Area of the rectangular sheet = 24 × 12 = 288 cm^{2}
Number of rectangular sheets required = 5760000 ÷ 288 = 20000
From II, we cannot find the length and breadth of the rectangular sheet.
Hence, option 1.
Question 8 of 10
8. Question
Directions: In the questions, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the correct option.
I. x2 – 11x + 28 = 0
II. y2 – 16y + 60 = 0
Correct
From I, x^{2} – 11x + 28 = 0
x^{2} – 4x – 7x + 28 = 0
x(x – 4) – 7(x – 4) = 0
(x – 4)(x – 7) = 0
x = 4, 7
From II, y^{2} – 16y + 60 = 0
y^{2} – 6y – 10y + 60 = 0
y(y – 6) – 10(y – 6) = 0
(y – 6)(y – 10) = 0
y = 6, 10
Hence, option 3.
Incorrect
From I, x^{2} – 11x + 28 = 0
x^{2} – 4x – 7x + 28 = 0
x(x – 4) – 7(x – 4) = 0
(x – 4)(x – 7) = 0
x = 4, 7
From II, y^{2} – 16y + 60 = 0
y^{2} – 6y – 10y + 60 = 0
y(y – 6) – 10(y – 6) = 0
(y – 6)(y – 10) = 0
y = 6, 10
Hence, option 3.
Question 9 of 10
9. Question
Directions: In the questions, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the correct option.
I. x2 – 16x + 64 = 0
II. y2 – 11y + 24 = 0
Correct
From I, x^{2} – 16x + 64 = 0
x^{2} – 8x – 8x + 64 = 0
x(x – 8) – 8(x – 8) = 0
(x – 8)(x – 8) = 0
x = 8, 8
From II, y^{2} – 11y + 24 = 0
y^{2} – 8y – 3y + 24 = 0
y(y – 8) – 3(y – 8) = 0
(y – 8)(y – 3) = 0
y = 8, 3
Hence, option 4.
Incorrect
From I, x^{2} – 16x + 64 = 0
x^{2} – 8x – 8x + 64 = 0
x(x – 8) – 8(x – 8) = 0
(x – 8)(x – 8) = 0
x = 8, 8
From II, y^{2} – 11y + 24 = 0
y^{2} – 8y – 3y + 24 = 0
y(y – 8) – 3(y – 8) = 0
(y – 8)(y – 3) = 0
y = 8, 3
Hence, option 4.
Question 10 of 10
10. Question
Directions: In the questions, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the correct option.
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