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Numerical Ability or Quantitative Aptitude Section is very important to get the best score in Banking and other competitive exams. Now a day, the questions asked in this section are calculative and time-consuming. Once dealt with proper strategy, speed, and accuracy, this section will definitely award you the maximum marks in the examination. Here is another set of Questions to improve your speed and practice. Make most of it.
Directions (Q. 1-5): Each of the questions below consists of a question and two statements numbered I and II given below it. You have to decide whether the data provided in the statements are sufficient to answer the question. Read both the statements and give answer
1) if the data in Statement I alone are sufficient to answer the question, while the data in Statement II alone are not sufficient to answer the question.
2) if the data in Statement II alone are sufficient to answer the question, while the data in Statement I alone are not sufficient to answer the question.
3) if the data in either Statement I alone or in Statement II alone are sufficient to answer the question.
4) if the data in both the Statements I and II are not sufficient to answer the question.
5) if the data in both the Statements I and II together are necessary to answer the question.
1. What is the minimum passing percentage in a test?
I. Raman scored 25% marks in the test and Sunil scored 288 marks, which is 128 more than Raman.
II. Raman scored 64 marks less than the minimum passing marks.
2. What is the value of x²+ y + z?
I. 4x + 3y + 5z = 60, and 2x = y, 2y = z
II. 3x + 3y + 2z = 34; 2x + 5y + 6z = 72
3. Whose body weight is the second highest among the five boys Arun, Vinay, Suraj, Raju and Pratap?
I. The average weight of Arun, Suraj and Vinay is 68 kg and the average weight of Raju and Pratap is 72 kg. Also, Suraj is 78 kg, Raju is 68 kg and Vinay is 46 kg.
II. The average weight of Arun, Suraj, Vinay and Raju is 68 kg and also Suraj is 78 kg, Raju is 68 kg and Vinay is 46 kg. All of them have different weights.
4. What is the ratio of the length of a rectangle to the side of a square?
I. The area of the square is 576 sq cm and the area of the rectangle is 600 sq cm.
II. The breadth of the rectangle is half the side of the square.
5. What is the smaller angle of a parallelogram?
I. The ratio of the angles of a triangle is 3 : 5 : 4 and the larger angle of the parallelogram is 34º more than the largest angle of the triangle.
II. The larger angle of the parallelogram is 38° more than its smaller angle.
- 5; From I. Sunil scored 288 marks.
Raman scored 288 – 128 = 160
So, 25 percentage marks = 160
Maximum aggregate marks = 160/25 × 100 = 640
From II. Passing marks = 160 + 64 = 224
From I and II. Now, combining both information, we have required passing percentage = 224/640 × 100 = 35%
- 1; From I. 4x + 3y + 52 = 60, given 2x = y and 2y = 2
2y + 3y + 5 × 2y = 60
15y = 60 ∴ y = 4
x = 2, z = 8
Then, x2 + y + Z = 4 + 4 + 8 = 16
Hence, Statement I alone is sufficient.
From II. 3x + 3y + 22 … (i)
2x + 5y + 62 = 72 … (ii)
Eqn (i) subtracted from eqn (ii) we have,
x – 2y – 42 = –38
We can’t find the value of x, y and z.
Thus, II alone is not sufficient.
- 1; From I. Average weight of Arun, Suraj and Vinay = Arun + Suraj + Vinay/3 = 68
Arun + Suraj + Vinay = 68 × 3 = 204 … (i)
Average weight of Raju and Pratap = Raju + Pratap/2 = 72
Raju + Pratap = 144 … (ii)
According to given information,
Raju’s weight = 68
Now, from eqn (i),
Pratap’s weight = 144 – 68 = 76 kg
Again, Suraj’s weight = 78 kg
and Vinay’s weight = 46
From eqn (i), Arun’s weight = 204 – (46 + 78) = 204 – 124 = 80 kg
Hence, Suraj’s weight is second highest, which is 78 kg.
From II. Average weight of Arun, Suraj, Vinay and Raju = Arun + Suraj + Vinay + Raju/4 = 68
or, Arun + Suraj + Vinay + Raju = 68 × 4 = 272
∴ Arun’s weight = 272 – (78 + 68 + 46) = 272 – 192 = 80 kg
But, we can’t find the weight of Pratap. Because there is no information given about Pratap. Hence, II alone is not sufficient.
- 5; From I. Area of square = 576
∴ Side of square = 24 cm
Area of rectangle = l × b
l × b = 600
From II. Let the breadth of the rectangle be x. Then side of the square = 2x,
Combing both the information, we have
x = 12 cm
Now, l × 12 = 600 l = 50 cm
∴ Reqd ratio = 50/24 = 25/12 = 25:12
Both the statements together are sufficient.
- 5; From I. Let the angles of the triangle be 3x, 5x and 4x.
Sum of angles of the triangle = 180°
∴ 3x + 5x + 4x = 180°
or, x = 15°
Angles of the triangle = 45°, 75° and 60°
Largest angle of the parallelogram = 75 + 34 = 109°
From II. Smaller angle of the parallelogram = larger angle – 38°
Combining both the information, we get smaller angle of the parallelogram. = 109° – 38° = 71°