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GATE Numerical Reasoning Aptitude (Analytical) Questions - 2


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GATE Numerical Reasoning Aptitude (Analytical) Questions - 2 - Question 1

Fill in the missing value

Detailed Solution for GATE Numerical Reasoning Aptitude (Analytical) Questions - 2 - Question 1

Middle number is the average of the numbers on both sides.(Left and Right)
Average of 6 and 4 is 5
Average of (7 + 4) and (2 + 1) is 7
Average of (1 + 9 + 2) and (1 + 2 + 1) is 8
Average of (4 + 1) and (2 + 3) is 5
Therefore, Average of (3) and (3) is 3

The number appearing in the centre line is average of the sum of numbers appearing on left  and right of numbers.

Hence, the unknown number is given by 3 + 3 / 2 = 3

GATE Numerical Reasoning Aptitude (Analytical) Questions - 2 - Question 2

Given (9 inches)1/2 = (0.25 yards) 1/2, which one of the following statements is TRUE?

Detailed Solution for GATE Numerical Reasoning Aptitude (Analytical) Questions - 2 - Question 2

Take square on both side, we get

9 inches = 0.25 yards

Why not option (A) ?

Let me give you an example

(3m)2=9m2

It means that if number gets squared then the units also get squared. Above is an example of a square having side 3m, the area of this square would be 9m2.

Similarly,

(9 m)½ = 3 m½

Hence, if I have

(9 inches)½ = (0.25 yards)½

If I take the square root, then we get

3 inches½ = 0.5 yards½

and not 3 inches = 0.5 yards

So option (A) is wrong

GATE Numerical Reasoning Aptitude (Analytical) Questions - 2 - Question 3

M and N start from the same location. M travels 10 km East and then 10 km North-East. N travels 5 km South and then 4 km South-East. What is the shortest distance (in km) between M and N at the end of their travel?

Detailed Solution for GATE Numerical Reasoning Aptitude (Analytical) Questions - 2 - Question 3

So the adjoining figure for solution

GATE Numerical Reasoning Aptitude (Analytical) Questions - 2 - Question 4

The number that least fits this set: (324, 441, 97 and 64) is _____.

Detailed Solution for GATE Numerical Reasoning Aptitude (Analytical) Questions - 2 - Question 4

18= 324
21= 441
8= 64
X2 ! =  97
All the above are perfect squares but 97 is a prime number.

GATE Numerical Reasoning Aptitude (Analytical) Questions - 2 - Question 5

Find the area bounded by the lines 3x + 2y = 14, 2x - 3y = 5 in the first quadrant.

Detailed Solution for GATE Numerical Reasoning Aptitude (Analytical) Questions - 2 - Question 5

The total area of the triangle which is bounded by the 2 given lines in the first quadrant

= 1/2 x 14/3 x 7

=  98 / 6

= 16.33 sq units

And area bounded with x axis and the lines in the 1st quadrant   

= 1/2 x ( 4.67 - 2.5)

= 1.08 sq units

So area bounded with y axis  =  total area in 1st quadrant - area in 1st quadrant bounded by the lines and the   x axis

= 16.33 - 1.08      

= 15.25 sq units

GATE Numerical Reasoning Aptitude (Analytical) Questions - 2 - Question 6

A straight line is fit to a data set (ln x, y). This line intercepts the abscissa at ln x = 0.1 and has a slope of −0.02. What is the value of y at x = 5 from the fit?

Detailed Solution for GATE Numerical Reasoning Aptitude (Analytical) Questions - 2 - Question 6

The equation of a line is
y = mx + c where m is the slope & c is the y-intercept
Now, In this question x is replaced with lnx
So, the equation of line becomes,
y = mlnx + c
or, y = -0.02lnx + c
We have given with abscissa which is essentially x-intercept. So, now we have to find ‘c’ the y-intercept.
for, y=0, lnx = 0.1 (given in the question)
Putting the value,
0 = -0.02 × 0.1 + c
or, c = 0.002
So, the equation of line becomes,
y = -0.02lnx + 0.002
putting x = 5 (asked in the question)
y = -0.002ln5 + 0.002 = -0.002 × 1.6 + 0.002 = -0.03
(ln5 = 1.6)

GATE Numerical Reasoning Aptitude (Analytical) Questions - 2 - Question 7

A cube of side 3 units is formed using a set of smaller cubes of side 1 unit. Find the proportion of the number of faces of the smaller cubes visible to those which are NOT visible.

Detailed Solution for GATE Numerical Reasoning Aptitude (Analytical) Questions - 2 - Question 7

Total number of cubes = 9 × 3 = 27
∴Total number of faces = 27 × 6 = 162
∴Total number of non-visible faces = 162 - 54 = 108

∴ Number of visible faces / Number of non visible faces = 54/108 = 1/2

GATE Numerical Reasoning Aptitude (Analytical) Questions - 2 - Question 8

Two and a quarter hours back, when seen in a mirror, the reflection of a wall clock without number markings seemed to show 1:30. What is the actual current time shown by the clock?

Detailed Solution for GATE Numerical Reasoning Aptitude (Analytical) Questions - 2 - Question 8

Present time is given  by

= 10: 30 + 2 : 15

= 12 : 45

GATE Numerical Reasoning Aptitude (Analytical) Questions - 2 - Question 9

A wire of length 340 mm is to be cut into two parts. One of the parts is to be made into a square and the other into a rectangle where sides are in the ratio of 1:2. What is the length of the side of the square (in mm) such that the combined area of the square and the rectangle is a MINIMUM?

Detailed Solution for GATE Numerical Reasoning Aptitude (Analytical) Questions - 2 - Question 9

Let’s assume that the piece from which rectangle is made, has length x mm.

Perimeter of rectangle = x

∴ Breadth of rectangle = x/6 and length of rectangle = 2x/6 = x/3

⇒ Area of rectangle = x/6 × x/3 = x2/18

Perimeter of square = 340 – x

Length of square = (340 – x)/4 = 85 – x/4

⇒ Area of square =(85 − x/4)2

Total area =(85 − x/4)2 + x2/18 = f(x)

Now, f′(x)=2×(85 − x/4)× − 1 + 2x/18 = 0

Solving, we get: x = 180

Length of square = 85 – x/4 = 85 – 45 = 40 mm

GATE Numerical Reasoning Aptitude (Analytical) Questions - 2 - Question 10

It takes 10s and 15s, respectively, for two trains travelling at different constant speeds to completely pass a telegraph post. The length of the first train is 120 m and that of the second train is 150 m. The magnitude of the difference in the speeds of the two trains (in m/s) is _____.

Detailed Solution for GATE Numerical Reasoning Aptitude (Analytical) Questions - 2 - Question 10

Speed = length / time

⇒ length = speed x time

120 = 10 x s1 ⇒ s= 12

150 = 15 x s⇒ s2 = 10

|s1 - s2| = 2

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