TOEFL Questions

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Here the given words, sullen and moody are synonyms to each other which means that gloomy, resentful. Similarly, repetitious and monotonous are synonyms to each other which means dull, tedious.

Hence, Sullen : moody :: monotonous : repetitious.

Since he is telling about the thing that had happened in past.

FRIEND : FPIBNZ

Follows a pattern that,

F - F

R- second letter to left of R i.e, P

I - I

E - Third letter to left of E i.e, B

N - N

D - Fourth letter to left of D i.e, Z

Similarly, for PATRON 

P - P

A -  second letter to left of A i.e, Y

T - T

R - Third letter to left of R i.e, O

O - O

N - Fourth letter to left of N i.e, J

PATRON : PYTOOJ

Therefore, FRIEND : FPIBNZ :: PATRON : PYTOOJ

ALPHABET is the only has the most letters in it.

Alphabet has all the 26 letters from A - Z. 

Hence, ALPHABET is one word that has the most letters in it.

As Microphone makes sound louder similarly Microscope makes the object magnified.

Hence, Microphone : Loud :: Microscope : Magnify.

The series rtus/rtus/rtus/rtus. Thus, the pattern 'rtus' is repeated.

Here the letters follow the rule (AxZ)+(A+Z)

The number corresponding to the letters are used in the rule

A=1, Z=26 => 26x1 + 1 + 26 = 26+27 = 53 =  053

(LxP)+(L+P) = (12 x16) + 12 + 16 = 192+28 = 220

Similarly (11x12) + 11 + 12 => 132 + 23 = 155 

This question concerns a committee’s decision about which five of eight areas of expenditure to reduce. The question requires you to suppose that K and N are among the areas that are to be reduced, and then to determine which pair of areas could not also be among the five areas that are reduced.

The fourth condition given in the passage on which this question is based requires that exactly two of K, N, and J are reduced. Since the question asks us to suppose that both K and N are reduced, we know that J must not be reduced:

Reduced         ::      K, N
Not reduced   ::      J

The second condition requires that if L is reduced, neither N nor O is reduced. So L and N cannot both be reduced. Here, since N is reduced, we know that L cannot be. Thus, adding this to what we’ve determined so far, we know that J and L are a pair of areas that cannot both be reduced if both K and N are reduced:

Reduced        ::      K, N
Not reduced  ::      J, L

Answer choice (B) is therefore the correct answer.