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Fill in the missing value
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
Given (9 inches)^{1/2} = (0.25 yards) ^{1/2}, which one of the following statements is TRUE?
Take square on both side, we get
9 inches = 0.25 yards
Why not option (A) ?
Let me give you an example
(3m)^{2}=9m^{2}
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 9m^{2}.
Similarly,
(9 m)^{½} = 3 m^{½}
Hence, if I have
(9 inches)^{½} = (0.25 yards)^{½}
If I take the square root, then we get
3 inches^{½} = 0.5 yards^{½}
and not 3 inches = 0.5 yards
So option (A) is wrong
M and N start from the same location. M travels 10 km East and then 10 km NorthEast. N travels 5 km South and then 4 km SouthEast. What is the shortest distance (in km) between M and N at the end of their travel?
So the adjoining figure for solution
The number that least fits this set: (324, 441, 97 and 64) is _____.
18^{2 }= 324
21^{2 }= 441
8^{2 }= 64
X^{2} ! = 97
All the above are perfect squares but 97 is a prime number.
Find the area bounded by the lines 3x + 2y = 14, 2x  3y = 5 in the first quadrant.
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
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?
The equation of a line is
y = mx + c where m is the slope & c is the yintercept
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 xintercept. So, now we have to find ‘c’ the yintercept.
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)
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.
Total number of cubes = 9 × 3 = 27
∴Total number of faces = 27 × 6 = 162
∴Total number of nonvisible faces = 162  54 = 108
∴ Number of visible faces / Number of non visible faces = 54/108 = 1/2
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?
Present time is given by
= 10: 30 + 2 : 15
= 12 : 45
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?
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 = x^{2}/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} + x^{2}/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
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 _____.
Speed = length / time
⇒ length = speed x time
120 = 10 x s_{1} ⇒ s_{1 }= 12
150 = 15 x s_{2 }⇒ s_{2} = 10
s_{1}  s_{2} = 2
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