**Directions : Study the following pie and table chart carefully and answer the questions given below.**

**A person travels daily for 8 hours for 5 days to cover a certain distance. The following pie chart shows the percentage of total distance travelled by him in 5 different modes on day1 (M1, M2, M3, M4, and M5) and the percentage of distance travelled by him with the same modes remained the same as each day of the journey.**

The table shows speed of M5 each day and the time it took to travel using M5 out of total travelling time that day.

Days |
Speed of M5 (kmph) |
Time taken by M5 each day as % of total travel time |

1 | 40 | 6.25 |

2 | 60 | 12.5 |

3 | 68 | 3.125 |

4 | 72 | 8.33 |

5 | 120 | 16.67 |

**1 What is the sum of the total distance travelled by the person during the given five days?**

**» Explain it**

**Correct Option: C**

**From common explanation we have**

**Total distance = A + B + C + D + E**

**= 30500/15 = 2033 1/3 km**

**Hence, option C is correct.**

**Common explanation : **

Days |
Time taken by M5 each day as % of total travel time |
8 hour travelling each day |

1 |
6.25 |
6.25% of 8 = 30 min |

2 |
12.5 |
1 hour |

3 |
3.125 |
15 min |

4 |
8.33 |
40 min |

5 |
16.67 |
1 h 20 min |

**From the table given in the question and above,**

**Distance = speed × time = 40 × 1/2 = 20 km**

**From pie chart, for M5, this is 15% of total distance, so**

**Total distance, A = 100 × 20/15 km**

**Similarly, we calculate for each day**

**B = 100 × 60 /15 km**

**C = 100 × 17/15km**

**D = 100 × 48 /15 km**

**E = 100 × 160/15 km**

**2 What is difference between the total distance travelled by Mode 2 (M2) in the five days and the total distance travelled by Mode 3 (M3) in the five days?**

**Correct Option: A**

**From common explanation, we have**

**Distance travelled by M3 in five days**

**25% of A + 25% of B + 25% of C + 25% of D + 25% of E**

**25% of (A + B + C + D + E)**

**= 25% of 2033 1/3 km**

**Similarly, distance travelled by M2 in five days**

**= 35% of 2033 1/3 km**

**Difference = 10% of 2033 1/3 km**

**= 610/3 km = 203 1/3 km**

**Hence, option A is correct.**

**Common explanation : **

Days |
Time taken by M5 each day as % of total travel time |
8 hour travelling each day |

1 | 6.25 | 6.25% of 8 = 30 min |

2 | 12.5 | 1 hour |

3 | 3.125 | 15 min |

4 | 8.33 | 40 min |

5 | 16.67 | 1 h 20 min |

**Let from day 1 to day 5 he travels A, B, C, D, and E km respectively,**

**From the table given in the question and above,**

**Distance = speed × time = 40 × 1/2 = 20 km**

**From pie chart, for M5, this is 15% of total distance, so**

**Total distance, A = 100 × 20/15 km**

**Similarly, we calculate for each day**

**B = 100 × 60 /15 km**

**C = 100 × 17/15 km**

**D = 100 × 48/15 km**

**E = 100 × 160/15 km**

**3.The average speed of the person during the first two days is approximately what percent of the average speed of the person during the last three days?**

A 13.33%

B 33.33%

C 56.67%

D 53.33%

E 43.33%

**Correct Option: D**

**Total distance in first two days from common explanation**

**= A + B = 1600/3 km**

**Total time = 2 × 8 = 16 hours**

**Average speed = (1600/3)/16 = 100/3 kmph**

**Similarly, average speed for last three days**

**= 125/2 kmph**

**Percentage = (100/3) /(125/2) × 100 = 53.33%**

**Hence, option D is correct.**

**Common explanation : **

Days |
Time taken by M5 each day as % of total travel time |
8 hour travelling each day |

1 | 6.25 | 6.25% of 8 = 30 min |

2 | 12.5 | 1 hour |

3 | 3.125 | 15 min |

4 | 8.33 | 40 min |

5 | 16.67 | 1 h 20 min |

Let from day 1 to day 5 he travels A, B, C, D, and E km respectively,

From the table given in the question and above,

Distance = speed × time = 40 × 1/2 = 20 km

From pie chart, for M5, this is 15% of total distance, so

Total distance, A = 100 × 20/15 km

Similarly, we calculate for each day

B = 100 × 60/15 km

C = 100 × 17 /15km

D = 100 × 48/15 km

E = 100 × 160/15 km

**4 Suppose, the person spends 25% of the total time on each day to travel by M1 then the average speed of M1 during the five days is approximately what percent less than the average speed of M5 during the five days?**

A 60%

B 75%

C 80%

D 120%

E 100%

**Correct Option: B**

**From common explanation, we have**

**Total time to travel by M1 in five days = 25% of (5 × 8) = 10 hours**

**Total distance travelled by M1 in five days = 10% of (A + B + C + D + E)**

**= 10% of 2033 1/3 = 610/3 km**

**Average speed of M1 during the five days**

**= (610/3)/10 = 61/3 kmph**

**Now, distance travelled using M5 in five days = 15% of of (A + B + C + D + E)**

**= 15% of 2033 1/3 = 305 km**

**Time of M5 (from table in common explanation)**

**= 1/2 hr + 1 hr + 1/4 hr + 2/3 hr + 4 /3hr = 15/4 hr**

**Average speed = 305/(15/4) = 244/3 kmph**

**Percentage difference = (244/3 – 61/3)/(244/3) × 100 = 75%**

**Hence, option B is correct.**

**Common explanation : **

Days |
Time taken by M5 each day as % of total travel time |
8 hour travelling each day |

1 |
6.25 |
6.25% of 8 = 30 min |

2 |
12.5 |
1 hour |

3 |
3.125 |
15 min |

4 |
8.33 |
40 min |

5 |
16.67 |
1 h 20 min |

**Let from day 1 to day 5 he travels A, B, C, D, and E km respectively,**

**From the table given in the question and above,**

**Distance = speed × time = 40 × 1 /2= 20 km**

**From pie chart, for M5, this is 15% of total distance, so**

**Total distance, A = 100 × 20/15 km**

**Similarly, we calculate for each day**

**B = 100 × 60/15 km**

**C = 100 × 17/15 km**

**D = 100 × 48/15 km**

**E = 100 × 160/15 km**