TSCAB Recruitment Quantitative Aptitude Exam Pattern | ||||
---|---|---|---|---|

S. No. | Name of Tests (objective) | No. of questions | Max. Marks | Time allotted for each test (Separately timed) |

1. | General/Financial Awareness |
50 | 50 | 35 Minutes |

2. | General English |
40 | 40 | 35 Minutes |

3. | Reasoning Ability & Computer Aptitude. |
50 | 50 | 45 Minutes |

4. | Quantitative Aptitude |
50 | 50 | 45 Minutes |

Total |
190 |
190 |
160 Minutes |

There will be penalty for wrong answers marked in the Objective Tests. For each question for which a wrong answer has been given by the candidate one fourth or 0.25 of the marks assigned to that question will be deducted as penalty to arrive at corrected score. If a question is left blank, i.e., no answer is marked by the candidate, there will be no penalty for that question.

TSCAB Recruitment Quantitative Aptitude – Syllabus | |
---|---|

1. | Number system |

2. | Volume and Surface Area |

3. | Simplification |

4. | Permutation and Combination |

5. | Time and Distance |

6. | Partnership |

7. | Logarithm |

8. | Time & Work |

9. | Functions |

10. | Simple Interest |

11. | Problems on H.C.F and L.C.M. |

12. | Mensuration |

- A. Rs. 4462.50

B. Rs. 8032.50

C. Rs. 8900

D. Rs. 8925

E. None of these

**Answer**: Option D

**Explanation**:

Principal

= Rs. \(\frac{100 \times 4016.25}{9 \times 5}\)

= Rs. \(\frac{401625}{45}\)

= Rs. 8925.

- A. 720

B. 900

C. 1200

D. 1800

**Answer**: Option C

**Explanation**:

2(15 + 12) x h = 2(15 x 12)

h = \(\frac{180}{27}\) m = \(\frac{20}{3}\) m.

Volume = (\({15 \times 12 \times \frac {20}{3}}\)) m\(^{3}\) = 1200 m\(^{3}\).

- A. 1

B. 3

C. 5

D. 10

**Answer**: Option B

**Explanation**:

\(log_{10}\) 5 + \(log_{10}\) (5x + 1) = \(log_{10}\) (x + 5) + 1

\(log_{10}\) 5 + \(log_{10}\) (5x + 1) = \(log_{10}\) (x + 5) + \(log_{10}\) 10

\(log_{10}\) [5 (5x + 1)] = \(log_{10}\) [10(x + 5)]

5(5x + 1) = 10(x + 5)

5x + 1 = 2x + 10

3x = 9

x = 3.

- A. 30 birds

B. 60 birds

C. 72 birds

D. 90 birds

**Answer**: Option A

**Explanation**:

Let the total number of shots be x. Then,

Shots fired by A = \(\frac {5}{8}\)x

Shots fired by B = \(\frac {3}{8}\)x

Killing shots by A = \(\frac {1}{3}\) of \(\frac {5}{8}\) x = \(\frac {5}{24}\)x

Shots missed by B = \(\frac {1}{2}\) of \(\frac {3}{8}\) x = \(\frac {3}{16}\)x

\(\frac {3 \times x}{16}\) = 27 or x = \(\frac {27 \times 16}{3}\) = 144.

Birds killed by A = \(\frac {5x}{24}\) = \(\frac {5}{24}\) x 144 = 30.

- A. 14 km

B. 15 km

C. 16 km

D. 17 km

**Answer**: Option C

**Explanation**:

Let the distance traveled on foot be x km.

Then, distance travelled on bicycle = (61 -x) km.

So, \(\frac {x}{4}\) + \(\frac {(61 -x)}{9}\) = 9

9x + 4(61 -x) = 9 x 36

5x = 80

x = 16 km.

I. Single flag of different colors

II. Any two flags in a different sequence of colors.

III. three flags in a different sequence of colors.

The maximum number of codes that can be generated is.

- A. 6

B. 9

C. 15

D. 18

**Answer**: Option C

**Explanation**:

This type of question becomes very easy when we assume three colors are red(R) blue(B) and Green(G).

We can choose any color.

Now according to the statement 1 i.e.., codes can be generated by waving single flag of different colours, the number of ways are three i.e.., R.B.G from statement III three flags in different sequence of colors, the number of ways is six i.e.., RBG, BGR, GBR, RGB, BRG, GRB.

Hence total number of ways by changing flag = 3+ 6 +6 = 15.

- A. Rs. 2736

B. Rs. 3800

C. Rs. 4788

D. Rs. 6840

**Answer**: Option A

**Explanation**:

Ratio of investment = Ratio of Profit

Ratio of Profit of Rohan to Mohan = 5:2

Rohan’s Share = Rs. 6840 = \(\frac {5}{5+2}\) x (90% Total Profit) ———-> 10% given to hospital

∴ 6840 = \(\frac {5}{7}\) x \(\frac {90}{100}\) x Total Profit

∴ Total Profit = Rs. 10,640/-

Vijay’s Share = 10,640 – 6840 – 1064 = Rs. 2736/-

- A. 35

B. 37

C. 42

D. 30

**Answer**: Option B

**Explanation**:

The number of liters in each can = HCF of 80, 144 and 368 = 16 liters.

Number of cans of Maaza = \(\frac{80}{16}\) = 5

Number of cans of Pepsi = \(\frac{11}{16}\) = 9

Number of cans of Sprite = \(\frac{368}{16}\) = 23

The total number of cans required = 5 + 9 + 23 = 37 cans.