Indian Navy Tradesman Mate Mock Test - 8 - Airforce X Y / Indian Navy SSR MCQ

'P of Q + R' means Q is the father of P, R is the daughter of Q.

Hence, R is the sister of P.

'L() M ÷ N' means M is mother of L, N is sister of M.

Hence, M and N are both sisters.

(15 x 2 - 3) = 27,

(31 x 2 - 6) = 56

and (45 x 2 - 9) = 81

(5 x 6 x 8) + (7 x 4 x 9) = 492

and (7 x 5 x 4) + (6 x 8 x 9) = 572

Therefore (4 x 3 x 5) + (7 x 2 x 5) = 130.

21 + 31 = 52

and 39 + 46 = 85

Therefore, 16 + 83 = 99.

(18 x 12)/3 = 72

and (32 x 16)/4 = 128

Therefore, (24 x 14)/? = 112

(336/?) = 112

? = (336/112)

? = 3.

According to the question,
B > D
A and C > E
∴ B > D > A and C > E
Clearly, shortest = E

So Option A is correct

According to the question,
N > M
Y > M
N > X > Y
On arranging the above data, we get
&becaul; N > X > Y > M
So, N is the most intelligent among them.

So Option A is correct

The information given in the question may be arranged as follows
Barun is taller than Sanjay.
Barun > Sanjay
Krishna is also not as tall as Bipul but is taller than Barun.
Bipul > Krishna > Barun
So according to above explanation the final arrangement is
Bipul > Krishna > Barun > Sanjay
So, Bipul is the tallest.

So Option C is correct

Let Age of Mahima = x.
Age of Anita = 2x.
Age of Rajan = x/2.
Age of Deepti = 4x.
Age of Lata = x/4.
Now we have the following arrangements in descending order
Deepti > Anita > Mahima > Rajan > Lata

Lata is the youngest.

So Option C is correct

Total holidays are 6 and the number of days in January is 31. Number of working days = 31 - 6 = 25. Thus, this option is correct.

In 30 years from 1971 to 2001, number of odd days = 30 + 8 (from leap years) = 38 odd days
It gives 3 odd days after being divided by 7.
So, December 9, 1971 is Thursday (three days before Sunday).

1990 was not a leap year and had only 1 odd day.
Therefore, January 1, 1991 will be Tuesday.

Local time varies only with longitudes.
The flight travelled from 10E to 70W longitudes i.e. 80° longitudes.
Time change because of longitude change = 80 × 4 = 320 minutes = 5 hours and 20 minutes
Time at 18N 10E after 10 hour journey = 2 am + 10 hours = 12 pm
Because the flight travelled towards West from East, the local time at 36N 70W
= 12 pm - 5 hrs 20 m = 6:40 am

P K D E J H I

Vowels that are neither at the beginning nor at the end are codded as 6, i.e. E will be coded as 6.
Vowels that are either at the beginning or at the end are codded as 9, i.e. I will be coded as 9.
From the above table, the code for P is 7, K is 1, D is 2, J is 5, H is 3.
So, the code for PKDEJHI will be 7 1 2 6 5 3 9.

We can draw the above diagram based on the information given in the question.
If the gender of the person is known, it is denoted by the suffix m or f. If it is not known, then it is denoted by x or y. We know, for example, that G is of the same sex as C, so we can label both as x; B and therefore, F are of the opposite sex from C, so we can label both y, and so on.
Horizontal lines indicate marriage, vertical or diagonal lines indicate children.  
This question orients you, in case you made an unwarranted assumption about the sexes. Since we do not know B's sex for sure, we cannot determine F's as well; this rules out all the wrong choices - including E, since we know H's sex.
Being the child of D's sibling, B, F is D's niece or nephew. 
Therefore, option 2 is the correct answer.

Both (1) and (2) follow because the statement "some papers are pens" can be converted into "some pens are papers" and the statement "all pencils are pens" can be converted into "some pens are pencils". Therefore, both (1) and (2) conclusions follows.

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Given: W > X, X > Y and Y > Z

On combining: W > X > Y > Z

I) X > Z → True

II) W > Z → True

III) Z > W → False (as W > Z)

IV) W > Y → True

Hence, Z > W is definitely wrong.

Statements: P ≥ R, R > T, T ≤ A

Conclusions:

I. P > A → False (As P ≥ R > T ≤ A → Thus, there is no clear relationship given between P and A)

II. T < P → True (As P ≥ R > T → T < P)

Hence, Only conclusion II follows.

Given statements: P < Q < R; R = S ≤ T

On combining: P < Q < R = S ≤ T

Conclusions:

I. Q > T → False (as Q < R and R = S ≤ T → Q < R = S ≤ T → Q ≤ T)

II. S > P → True (as P < Q < R; R = S → P < Q < R = S → P < S)

Hence, only conclusion II is true.

I. M > Q → False (as Q ≤ C ≤ B = M → M gives either M = Q or M > Q).

II. D ≤ Q → False (as Q ≤ C ≤ B = M > D)

III. M = Q → False (as Q ≤ C ≤ B = M → M gives either M = Q or M > Q).

IV. C < D → False (as C ≤ B = M > D).

P > Q ≤ C ≤ B = M > D is given in statement. Therefore but Conclusion I and III forms a complementary pair for either.

Thus, the correct answer is option 1.

A tap can fill a tank in 4 hours.
Therefore the tap can fill half the tank in 2 hours.
Remaining part = 1/2
After half the tank is filled, two more similar taps are opened.
Hence, total number of taps becomes 3.
Part filled by one tap in 1 hour = 1/4
Part filled by three taps in 1 hour 
= 3 × 1/4 = 3/4
i.e., 4 taps can fill remaining half in 40 minutes.
Total time taken = 2 hour + 40 minute 
= 2 hour 40 minutes

part filled in 2 hours = 
Remaining  part =  
∴ (A + B)'s 7 hour's work = 2/3
(A + B)'s 1 hour's work = 2/21
∴ C's 1 hour's work = {(A + B + C)'s 1 hour's  work} - {(A + B)'s 1 hour's work}

∴ C alone can fill the tank in 14 hours.

Gain = 10%
Let the C.P. be x.

Let the original price be Rs. 100.
Marked down price = Rs. 80
Now, we want to raise 80 to 100.

Simple Interest = PRT / 100
Here P = Principal, R = Rate of interest and T = time period
Hence, Simple Interest ∝ T
Required Ratio = (Simple Interest for 5 years) / (Simple Interest for 15 years)
⇒ T₁/ T₂ = 5 / 15
⇒ 1 / 3 = 1 : 3