Research Aptitude Test

Result

Research Aptitude Test - 22

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Weekly Quiz Competition

Question 1
2 / -0.5

log₃₂x = 0.8, then x is equal to:

A
A. 25.6

B
B. 16

C
C. 10

D
D. 12.8

Solution

log₃₂ x = 0.8 ⇒ x = (32)^0.8

= (2⁵ )^4/5 = 2⁴ = 16.
Question 2
2 / -0.5

A
A. 1

B
B. 2

C
C. 3

D
D. 3/2

Solution

= log_abc a + log_abc b + log_abc c

= log_abc (abc) = 1
Question 3
2 / -0.5

If log₂ [log₃ (log₂ x)] = 1, then x is equal to:

A
512

B
128

C
12

D
0

Solution

log₂ [log₃ (log₂ x)] = 1

⇒log₃ (log₂ x) = 2

⇒log₂ x = 3² = 9

⇒x = 2 ⁹ = 512
Question 4
2 / -0.5

If log₁₀2 = 0.3010, then log₂10 is :

A
A. 0.3322

B
B. 3.2320

C
C. 3.3222

D
D. 5

Solution
Question 5
2 / -0.5

A ball rolling up an incline covers 36 metres during the first second, 32 metres during the second, 28 meters during the next and so on. How much distance will it travel during the 8^thsecond?

A
A. 8 metres

B
B. 6 metres

C
C. 7 metres

D
D. 9 metres

Solution

Distance covered during the 1st second = 36 m

Distance covered during the 2^nd second = 32 m

Distance covered during the 3^rd second = 28 m.

The distance covered form an A.P.

= 36 + 32 + 28 .... in which

a = 38, d = -4

∴ Distance covered in 8th second

= 8th term of the A.P.

= a + 7d = 36 + 7 (-4)

= 38 - 28 = 8 metres.
Question 6
2 / -0.5

The sum of first three terms of a GP. is to the sum of first six terms is 125:152. Find the common ratio of G.P.

A
A. 2/5

B
B. 4/5

C
C. 3/5

D
D. 1/5

Solution

⇒ 125r³ = 27 ⇒ r³ = 27/125

or, r³ = (3/5)³ , ∴ r = 3/5
Question 7
2 / -0.5

A
A.

B
B.

C
C.

D
D. None of these

Solution

(2 + 3¹ ) + (2 + 3² ) +...+ (2 + 3¹¹ )

= 2 + 2 + 2 + …..up to 11 terms + (3 + 3² + 3³ + …...up tp 11 terms)
Question 8
2 / -0.5

The common ratio of a G.P. is and the sum to infinity is 80/9. Find the First term.

A
A. 14

B
B. 16

C
C. 12

D
D. 10

Solution

Hence, the first term is 16.
Question 9
2 / -0.5

In how many ways can Ram choose a vowel and a consonant from the letter ALLAHABAD?

A
A. 4

B
B. 6

C
C. 9

D
D. 5

Solution

In Allahabad we have 4 As that is only one vowel and 4 consonants (B, D, H, L)

So any four combinations.

AB, AD, AH, AL are only 4 possibilities so..
Question 10
2 / -0.5

In how many ways can the letters of the word EQUATION be arranged so that all the vowels come together?

A
A. ⁹ C₄ .⁹ C₆

B
B. 4! . 5!

C
C. 9! / 5!

D
D. 9! - 4!5!

Solution

In EQUATION there are five vowels A, E, I, O and U and three consonants N, Q, T so the possible. arrangements, will b e 4! x 5 ! ways.