## Numerical Ability

#### Numerical Ability

1. Industrial consumption of power doubled from 2000-2001 to 2010-2011. Find the annual rate of increase in percent assuming it to be uniform over the years.

1. 10 or more than 10 can't be the right answer because of compounding. Hence, 7.2 will be the right answer.

##### Correct Option: B

10 or more than 10 can't be the right answer because of compounding. Hence, 7.2 will be the right answer.

1. A firm producing air purifiers sold 200 units in 2012. The following pie chart presents the share of raw material, labour, energy, plant & machinery, and transportation costs in the total manufacturing cost of the firm in 2012. The expenditure on labour in 2012 is Rs. 4,50,000. In 2013, the raw material expenses increased by 30% and all other expenses increased by 20%. What is the percentage increase in total cost for the company in 2013?

1. Percentage increase in total cost = 22%

##### Correct Option: A

Percentage increase in total cost = 22%

1. A five digit number is formed using the digits 1,3,5,7 and 9 without repeating any of them. What is the sum of all such possible five digit numbers?

1. The digit in unit place is selected in 4! Ways
The digit in tens place is selected in 4! Ways
The digit in hundreds place is selected in 4! Ways
The digit in thousands place is selected in 4! Ways
The digit in ten thousands place is selected in 4! Ways
Sum of all values for 1
4! × 1 × (100 + 101 + 102 + 103 +104) = 4! × 11111 × 1
Similarly for ‘3’ 4! × (11111) × 3
Similarly for ‘5’ 4! × (11111) × 5
Similarly for ‘7’ 4! × (11111) × 7
Similarly for ‘9’ 4! × (11111) × 9
∴ sum of all such numbers
4! × (11111)× (1+ 3 + 5 + 7 + 9)
= 24 × (11111) × 25
= 6666600

##### Correct Option: B

The digit in unit place is selected in 4! Ways
The digit in tens place is selected in 4! Ways
The digit in hundreds place is selected in 4! Ways
The digit in thousands place is selected in 4! Ways
The digit in ten thousands place is selected in 4! Ways
Sum of all values for 1
4! × 1 × (100 + 101 + 102 + 103 +104) = 4! × 11111 × 1
Similarly for ‘3’ 4! × (11111) × 3
Similarly for ‘5’ 4! × (11111) × 5
Similarly for ‘7’ 4! × (11111) × 7
Similarly for ‘9’ 4! × (11111) × 9
∴ sum of all such numbers
4! × (11111)× (1+ 3 + 5 + 7 + 9)
= 24 × (11111) × 25
= 6666600

1. An electric bus has onboard instruments that report the total electricity consumed since the start of the trip as well as the total distance covered. During a single day of operation, the bus travels on stretches M, N, O and P, in that order. The cumulative distances travelled and the corresponding electricity consumption are shown in the table below.

The stretch where the electricity consumption per km is minimum is

1.  For M ⇒ 12 = 0.6 20

 N ⇒ 12 = 0.52 25

 O ⇒ 20 = 0.66 30

 P ⇒ 12 = 0.48 30

##### Correct Option: D

 For M ⇒ 12 = 0.6 20

 N ⇒ 12 = 0.52 25

 O ⇒ 20 = 0.66 30

 P ⇒ 12 = 0.48 30

1. Ram and Ramesh appeared in an interview for two vacancies in the same department. The probability of Ram's selection is 1/6 and that of Ramesh is 1/8. What is the probability that only one of them will be selected?

1. P (Ram) = 1/6; p(Ramesh) = 1/8 p(only at) = p (Ram) × p (not ramesh) + p(Ramesh) × p(n0 × Ram)

 = 1 + 7 + 1 × 5 6 8 8 6

 ⇒ 12 = 1 40 4

##### Correct Option: B

P (Ram) = 1/6; p(Ramesh) = 1/8 p(only at) = p (Ram) × p (not ramesh) + p(Ramesh) × p(n0 × Ram)

 = 1 + 7 + 1 × 5 6 8 8 6

 ⇒ 12 = 1 40 4