Mains exams are very important to clear every

Result | No. of Students | |||
---|---|---|---|---|

Section A | Section B | Section C | Section D | |

Students failed in both Exams | 28 | 23 | 17 | 27 |

Students failed in half-yearly but passed in Annual Exams | 14 | 12 | 8 | 13 |

Students passed in half-yearly but failed in Annual Exams | 6 | 17 | 9 | 15 |

Students passed in both Exams | 64 | 55 | 46 | 76 |

**1. If the number of students passing an examination be considered a criteria for comparision of difficulty level of two examinations, which of the following statements is true in this context**?

**Answer**: Option C

**Explanation**:

Number of students who passed half-yearly exams in the school

= (Number of students passed in half-yearly but failed in annual exams)

+ (Number of students passed in both exams)

= (6 + 17 + 9 + 15) + (64 + 55 + 46 + 76)

= 288.

Also, Number of students who passed annual exams in the school

= (Number of students failed in half-yearly but passed in annual exams)

+ (Number of students passed in both exams)

= (14 + 12 + 8 + 13) + (64 + 55 + 46 + 76)

= 288.

Since, the number of students passed in half-yearly = the number of students passed in annual exams. Therefore, it can be inferred that both the examinations had almost the same difficulty level.

Thus Statements (a), (b) and (d) are false and Statement (c) is true.

**2. How many students are there in Class IX in the school**?

**Answer**: Option D

**Explanation**:

Since the classification of the students on the basis of their results and sections form independent groups, so the total number of students in the class:

= (28 + 23 + 17 + 27 + 14 + 12 + 8 + 13 + 6 + 17 + 9 + 15 + 64 + 55 + 46 + 76)

= 430.

**3. Which section has the maximum pass percentage in at least one of the two examinations**?

**Answer**: Option D

**Explanation**:

Pass percentages in at least one of the two examinations for different sections are:

For Section A[ \(\frac{(14 + 6 + 64)}{(28 + 14 + 6 + 64) × 100}\) ]% = [ \(\frac{(84)}{(112) × 100}\) ] = 75%

For Section B[ \(\frac{12 + 17 + 55)}{(23 + 12 + 17 + 55) × 100}\) ]% = [ \(\frac{(84)}{(107) × 100}\) ] = 78.5%

For Section C[ \(\frac{(14 + 6 + 64)}{(17 + 8 + 9 + 46) × 100}\) ]% = [ \(\frac{(63)}{(80) × 100}\) ] = 78.5%

For Section D[ \(\frac{(13 + 15 + 76)}{(27 + 13 + 15 + 76) × 100}\) ]% = [ \(\frac{(104)}{(31) × 100}\) ] = 79.39%

**4. Which section has the maximum success rate in annual examination**?

**Answer**: Option A

**Explanation**:

Total number of students passed in annual exams in a section

= [ (No. of students failed in half-yearly but passed in annual exams)

+ (No. of students passed in both exams)

Success rate in annual exams in Section A

[ \(\frac{No. of students of Section A passed in annual exams}{Total number of students in Section A × 100 }\) ]%

[ \(\frac{(14 + 64)}{(28 + 14 + 6 + 64) × 100% }\) ]%

[ \(\frac{(78)}{(112) × 100 }\) ]%

= 69.64%.

Similarly, success rate in annual exams in:

Section B [ \(\frac{(13 + 76)}{(27 + 13 + 15 + 76) × 100% }\) ]%

Similarly Section D[ \(\frac{(89)}{(131) × 100 }\) ]% = 67.94%

Clearly, the success rate in annual examination is maximum for Section A

**5. Which section has the minimum failure rate in half-yearly examination**?

**Explanation**: Option D

**Explanation**:

Total number of failures in half-yearly exams in a section

= [ (Number of students failed in both exams)

+ (Number of students failed in half-yearly but passed in Annual exams)

] in that section

Failure rate in half-yearly exams in Section A

[ \(\frac{Number of students of Section A failed in half-yearly}{Total number of students in Section A × 100 }\) ]%

[ \(\frac{(28 + 14)}{(28 + 14 + 6 + 64) × 100% }\) ]%

[ \(\frac{(42)}{(112) × 100 }\) ]%

= 37.5%.

Similarly, Section D [ \(\frac{Number of students of Section A failed in half-yearly}{Total number of students in Section A × 100 }\) ]%

[ \(\frac{(27 + 13)}{(27 + 13 + 15 + 76) × 100% }\) ]%

[ \(\frac{(40)}{(131) × 100 }\) ]%

30.53%

Clearly, the failure rate is minimum for Section D.

**Explanation**: Option A

**Explanation**:

Numbers are alternatively multiplied by 2 and increased by 3.

So, the missing number = 61 x 2 = 122.

**7. 15, 31, 63, 127, 255, (….)**

**Explanation**: Option B

**Explanation**:

Each number is double the preceding one plus 1.

So, the next number is (255 x 2) + 1 = 511.

**8. 1, 8, 27, 64, 125, 216, (….)**

**Explanation**: Option B

**Explanation**:

Numbers are \({1}^{3}\), \({2}^{3}\),\({3}^{3}\), \({4}^{3}\), \({5}^{3}\), \({6}^{3}\).

So, the missing number is 73 = 343.

**9. 3, 7, 6, 5, 9, 3, 12, 1, 15, (….)**

**Explanation**: Option C

**Explanation**:

There are two series, beginning respectively with 3 and 7. In one 3 is added and in another 2 is subtracted.

The next number is 1 – 2 = -1.

**10. 11, 13, 17, 19, 23, 29, 31, 37, 41, (….)**

Numbers are all primes. The next prime is 43.

**Explanation**: Option A

**Explanation**:

\({x}^{2}\)+ 5x -3x- 15 = 0

x(x + 5) – 3(x + 5) = 0

(x – 3)(x + 5) = 0

=> x = 3 or x = -5.

**12. Find the roots of quadratic equation: 2\({x}^{2}\) + 5x + 2 = 0?**

**Explanation**: Option A

**Explanation**:

2\({x}^{2}\) + 4x + x + 2 = 0

2x(x + 2) + 1(x + 2) = 0

(x + 2)(2x + 1) = 0 => x = -2, -1/2

**13. Find the roots of quadratic equation: \({x}^{2}\) + x – 42 = 0?**

**Explanation**: Option D

**Explanation**:

\({x}^{2}\) + 7x – 6x + 42 = 0

x(x + 7) – 6(x + 7) = 0

(x + 7)(x – 6) = 0 => x = -7, 6

**14. Find the roots of quadratic equation: 3\({x}^{2}\) – 7x – 6 = 0?**

**Explanation**: Option B

**Explanation**:

3\({x}^{2}\) – 9x + 2x – 6 = 0

3x(x – 3) + 2(x – 3) = 0

(x – 3)(3x + 2) = 0 => x = 3, -2/3

**15. The sum of the squares of two consecutive positive integers exceeds their product by 91. Find the integers?**

**Explanation**: Option A

**Explanation**:

Let the two consecutive positive integers be x and x + 1

\({x}^{2}\) + \({x + 1}^{2}\) – x(x + 1) = 91

\({x}^{2}\) + x – 90 = 0

(x + 10)(x – 9) = 0 => x = -10 or 9.

As x is positive x = 9

Hence the two consecutive positive integers are 9 and 10.