Research Aptitude Test

Result

Research Aptitude Test - 9

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Weekly Quiz Competition

Question 1
2 / -0.5

Deepak is brother of Ravi, Rekha is sister of Atul, Ravi Is son of Rekha. How is Deepak related to Rekha?

A
A. Son

B
B. Brother

C
C. Nephew

D
D. Father
Question 2
2 / -0.5

E is the son of A, D is the son of B, E is married to C, C is B’s daughter. How is D related to E?

A
A. Brother

B
B. Uncle

C
C. Father-in-law

D
D. Brother-in-law

Solution

D is brother in law of E.
Question 3
2 / -0.5

Direction : Read the following Information carefully and answer the question below it.

Rahul's roll number in lES Exam is a number consisting of three non-zero distinct digits, such that the sum of the digits at hundreds and unit’s place is equal to that half of the digit at ten’s place. Also the sum of all possible three digit numbers obtained using these three digits without repetition is 2664.

Q. The digits in the unit is

A
A. 2

B
B. 1

C
C. 3

D
D. can't say

Solution

Let the digits be a. y. z such that the number is 106r + 10y + z

Thus, the omer rive 3 digit number which we may obtained using these 3 digits will be

100x + 10z + y

100y + 10x + z

100y + 10z + x

100z + 10x + y

100z + 10y + x

Now when we add ail the six three digit numbers possible to be formed by these three digits:

222 (x + y + z) = 2664 or x + y + z = 12

Also x + z = y/2 thus y = 8

Thus, x + z = 4

As the digits are non zero and distinct thus x and z have to be 1 and 3 but not necessarily in the same order. Thus we cannot say whether the number is 183 or 381.
Question 4
2 / -0.5

Direction : Read the following Information carefully and answer the question below it.

Rahul's roll number in lES Exam is a number consisting of three non-zero distinct digits, such that the sum of the digits at hundreds and unit’s place is equal to that half of the digit at ten’s place. Also the sum of all possible three digit numbers obtained using these three digits without repetition is 2664.

Q. The digits in the tens given is

A
A. 6

B
B. 8

C
C. 9

D
D. 2

Solution

Let the digits be a. y. z such that the number is 106r + 10y + z

Thus, the omer rive 3 digit number which we may obtained using these 3 digits will be

100x + 10z + y

100y + 10x + z

100y + 10z + x

100z + 10x + y

100z + 10y + x

Now when we add ail the six three digit numbers possible to be formed by these three digits:

222 (x + y + z) = 2664 or x + y + z = 12

Also x + z = y/2 thus y = 8

Thus, x + z = 4

As the digits are non zero and distinct thus x and z have to be 1 and 3 but not necessarily in the same order. Thus we cannot say whether the number is 183 or 381.
Question 5
2 / -0.5

Direction : Read the following Information carefully and answer the question below it.

Rahul's roll number in lES Exam is a number consisting of three non-zero distinct digits, such that the sum of the digits at hundreds and unit’s place is equal to that half of the digit at ten’s place. Also the sum of all possible three digit numbers obtained using these three digits without repetition is 2664.

Q. The three digit is number is always divisible by:

A
A. 3

B
B. 18

C
C. 29

D
D. None of these

Solution

Let the digits be a. y. z such that the number is 106r + 10y + z

Thus, the omer rive 3 digit number which we may obtained using these 3 digits will be

100x + 10z + y

100y + 10x + z

100y + 10z + x

100z + 10x + y

100z + 10y + x

Now when we add ail the six three digit numbers possible to be formed by these three digits:

222 (x + y + z) = 2664 or x + y + z = 12

Also x + z = y/2 thus y = 8

Thus, x + z = 4

As the digits are non zero and distinct thus x and z have to be 1 and 3 but not necessarily in the same order. Thus we cannot say whether the number is 183 or 381.
Question 6
2 / -0.5

Direction : A cuboid is divided into 192 identical cube lets. This is done by making minimum number of cuts possible. All cuts are parallel to some of the faces. But before doing so the cube is painted with green colour on one set of opposite faces. But on the other set of opposite faces and red on remaining their part of opposite faces.

What is the maximum number of cubelets possible which are coloured with green colour only ?

A
48

B
64

C
96

D
72

Solution

As we want to maximize the number of cubelets with green in color, the only cuboid has to be painted with green color on the set of opposite faces of 6 x 8. Hence no. of green colored only cubelets will (6 - 2) x (8 - 4) = 4 x 6 = 24, from face. There are two such faces hence maximum total no. of only green painted cube lets will be = 24 x 2 = 48

Hence option (a)
Question 7
2 / -0.5

Direction : A cuboid is divided into 192 identical cubelets This Is done by making minimum no. ot cuts possible. At cuts are parallel to some of the face. But before that, The cube is-painted with green color on one set of opposite faces. Blue on other set of opposite faces and tea on their pair of annosit faces.

Q. What is the number of cubelets possible which are painted with none of the color?

A
A. 16

B
B. 24

C
C. 36

D
D. 48

Solution

Number of unpainted cubelets will be (4 - 2) x (6 - 2) x (8 - 2) = 2 x 4 x 6 = 48
Question 8
2 / -0.5

How many cuboids of dimensions 4 x 5 x 6 are required to form a cube of least size if cuboids have to the placed adjacent, above or below each other?

A
A. 1600

B
B. 1800

C
C. 1200

D
D. None of these

Solution

Least possible dimension of cube is ICM of (4, 5, 6) = 60

The number of cuboids required =
Question 9
2 / -0.5

A cube has been cut into cuboids of size 2 x 3 x 4. What is least possible integer length of the edge of the cube?

A
10

B
12

C
16

D
18

Solution

The least possible dimension of cube LCM of 2,3,4 = 12
Question 10
2 / -0.5

A man walks 1 km towards East and then he turns to South and walks 5 km. Again he turns to East and walks 2 km, after this he turns to North and walks 9 km. Now, how far is he from his starting point?

A
A. 3 km

B
B. 4 km

C
C. 5 km

D
D. 7 km

Solution

The movements of the man are as shown in Fig. (A to B, B to C, C to D, D to E).

Clearly, DF = BC = 5 km.

EF = (DE - DF) = (9 - 5) km = 4 km.

BF = CD = 2 km

AF = AB + BF = AB + CD = (1 + 2) km = 3 km

∴ Man's distance from starting point A