Research Aptitude Test

Result

Research Aptitude Test - 4

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Weekly Quiz Competition

Question 1
2 / -0.5

Two chairs (A and B) are in an empty room overnight. Chair A is made of steel, while chair B is made of wood. In the night and in the morning, the room ’s temperature is 290 K. In the morning, a person chooses between Chair A and Chair B as the seat by feeling (touching) the chair and choosing one which feels warmer. The person chooses__ (BLANK-1) _ as it feels warmer because _ (BLANK-2)___has__ (BLANK-3)______ (BLANK-4)____ than ___(BLANK-5)___.

Find the correct Answer for Blank 1

A
Chair B

B
Chair A

C
neither chair

D
Both the chairs

Solution

Chair B
Explanation: The person would choose Chair B (the wooden chair) as it feels warmer because wood is a better insulator than steel. This means that wood has a lower thermal conductivity than steel, allowing it to retain heat better and feel warmer to the touch.
Question 2
2 / -0.5

If log 2 = 0.3010 and log 3 = 0.4771 then the value of log 4.5 is:

A
0.6532

B
0.7727

C
0.3266

D
None of these

Solution

log 4.5 = log(9/2) = log9 - log2

= log(3² ) - log 2 = 2log3 - log2

= (2 x 0.4771) - (0.3010) = 0.6532.
Question 3
2 / -0.5

The value of is:

A
3

B
2

C
1

D
0

Solution

Given Expression

=
Question 4
2 / -0.5

If log x + log y = log(x+y), then:

A
x = y

B
xy = 1

C
y = (x-1)/x

D
y = x/(x-1)

Solution

logx+logy = log(x+y)

⇒log(x+y) = log(xy)

∴x + y = xy or x = y (x - 1)
Question 5
2 / -0.5

If a be the first term of a G.P., L the n^th term and P the product of first n terms then P =

A
(al)^n/2

B
(a-l)^n/2

C
(a+l)^n/2

D
None of these

Solution

If r is the common ratio of G.P, then

L = a r^n-1 …(i)

The first n terms of the G.P. are

a, ar, ar² , ar³ ........ar^n-1

P = a x ar x ar² x ... x ar^n-1

= aⁿ x r ^{1+2+3 …+(n-1)}
Question 6
2 / -0.5

If 7 times the 7^thterm of an A.P. is equal to 11 times its, 11^thterm, then the 18^thterm of the A.P. is

A
1

B
2

C
0

D
3

Solution

Let a be the first term and d, the common difference of an AP.

∴ a₇ = a + 6d

A₁₁ = a+10d ∴ 7a₇ = 11a₁₁

7 (a + 6d) = 11 (a + 10d)

⇒ 7a + 42d = 11a + 110d

⇒ 4a + 68d = 0

⇒ a + I7d = 0

⇒a₁₈ = 0
Question 7
2 / -0.5

How many numbers of 3 digits can be formed with the digits 1, 2, 3, 4, 5 (repetition of digits not allowed)?

A
125

B
120

C
60

D
150

Solution

A three digit number is to be formed from given 5 digits 1,2,3,4,5.
Now, there are 5 ways to fill ones place.
Since, repetition is not allowed , so tens place can be filled by remaining 4 digits.
So, tens place can be filled in 4 ways.
Similarly, to fill hundreds place, we have 3 digits remaining.
So,hundreds can be filled by 3 ways.
So, required number of ways in which three digit numbers can be formed from the given digits is 5 ×4 ×3=60
Question 8
2 / -0.5

Flow many numbers between 2000 and 3000 can be formed with the digits 0, 1, 2, 3, 4, 5, 6, 7 (repetition of digits not allowed)?

A
42

B
210

C
336

D
440

Solution

Number between 2000 and 3000 are four digit numbers in which 2 is the first digit So, 2 must be taken in box d₁

Now rest box can be filled in ⁷ P₃ ways.

⇒
Question 9
2 / -0.5

In how many ways can a person send Invitation cards to 6 of his friends if he has four servants to distribute the cards?

A
6⁴

B
4⁶

C
24

D
None of these

Solution

Person has six invitation cards 1 card can be given to any of 4 servants so six cards can be distributed in 4 x 4 x 4 x 4 x 4 x 4 = 4⁶ ways
Question 10
2 / -0.5

In how many ways can 7 Indians, 5 Pakistanis and 6 Dutch be seated in a row so that all persons of the same nationality sit together?

A
3!

B
7! 5! 6!

C
3! 7! 5! 6!

D
18!

Solution

We have

→7 Indians

→5 Pakistanis and

→6 Dutch

7 Indians can sit in 7! ways

5 Pakistanis can sit in 5! ways

6 Dutch can sit in 6! ways

Also they can sit together in 3! different ways 7! x 5! x 6! x 3!