Numerical Ability
 Find the missing sequence in the letter series below: A, CD, GHI,?, UVVVXY

 LMN
 MNO
 MNOP
 NOPQ

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Correct Option: C
 If x > y > 1, which of the following must be true?
i. In x > In y
ii. e^{x} > e^{y}
ii. y^{x} > x^{y}
iv. cos x > cosy

 i and ii
 i and iii
 iii and iv
 ii and iv

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For whole numbers, greater the value greater will be its log.
Same logic for power of e.Correct Option: A
For whole numbers, greater the value greater will be its log.
Same logic for power of e.
 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?

 47/48
 1/4
 13/48
 35/48

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P (Ram) = 1/6; p(Ramesh) = 1/8 p(only at) = p (Ram) × p (not ramesh) + p(Ramesh) × p(n_{0} × R_{am})
= 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(n_{0} × R_{am})
= 1 + 7 + 1 × 5 6 8 8 6 ⇒ 12 = 1 40 4
 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

 M
 N
 O
 P

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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
 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?

 6666660
 6666600
 6666666
 6666606

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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 × (10^{0} + 10^{1} + 10^{2} + 10^{3} +10^{4}) = 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
= 6666600Correct 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 × (10^{0} + 10^{1} + 10^{2} + 10^{3} +10^{4}) = 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