3, 7, 15, 39, 63, 127, 255, 511
C. 39
3*2+1 = 7
15*2+1 = *31*
31*2+1 = 63
63*2+1 = 127
127*2+1 = 255
255*2+1 = 511
Balls- B1, B2, B3, B4, B5, B6, B7, B8, B9.
Group1 – (B1, B2, B3), Group2 – (B4, B5, B6), Group3 – (B7, B8, B9)
Now weigh any two groups. Group1 on left side of the scale and Group2 on the right side.
When weighing scale tilts left – Group1 has a heavy ball or right – Group2 has a heavy ball or balanced – Group3 has a heavy ball.
Lets assume Group 1 has a heavy ball.
Now weigh any two balls from Group1. B1 on left side of the scale and B2 on right side.
When weighing scale tilts left – B1 is the heavy or tilts right – B2 is the heavy or balanced – B3 is the heavy.
16
In the example first we have to find the area of the field
we have given the following values ,
cost of fencing per meter = Rs 1.50
Diameter of circular field = 28
We have to find the area , a = ?
We know that,
Area = 2 π r
= 2 x 22/7 x 14
= 2 x 22 x 2
= 44 x 2
= 88 sq. m
So, area of circle is 88 sq.m
Now just multiply it with 1.50
cost of fencing = 88 x 1.50
= Rs. 132
So, the cost of fencing the circular field is Rs. 132
3 miutes
(x**2 – 6* x + 5) = (x-1)*(x-5)
(x**2 + 2 * x + 1) = (x + 1) * (x+1) = (x+1)**2
For what x is (x-1)*(x-5)/( (x+1)**2) a minimum?
One way to answer this question is by using calculus.
Take the derivative, and set to zero.
Since this is a fraction of polynomials, and a fraction is
zero only if it’s numerator is zero, we need calculate only
the numerator of the derivative and set it to zero.
The numerator of the
Derivative of (x-1)*(x-5)/( (x+1)**2) is
( (x-1) + (x-5) ) ( x+1)**2 – (x-1)(x-5)( 2 (x+1) )
= (2 x – 6) (x+1)**2 – (2) (x-1)(x-5) (x+1)
= 0
Divide through by 2 (x+1)
(x-3)(x+1) – (x-1)(x-5) = 0
(x**2 – 2 x – 3 ) – (x**2 – 6 x + 5) = 0
x**2 – x**2 – 2 x + 6 x – 3 – 5 = 0
4 x – 8 = 0
x = 2
Plugging in x = 2 into the original
(x**2-6*x+5)/(x**2+2*x+1)
gives us (2**2 – 6 * 2 + 5)/(2**2 + 2*2 + 1)
= (4 – 12 + 5) / (4 + 4 + 1) = -3/9 = -1/3
Least value is -1/3
24 peacocks and 36 rabbits
Ratio of diameters=1:2
Ratio of radius=1:2
Ratio of volumes=(4/3*3.14*r1^3)/(4/3*3.14*r2^3)
=r1^3/r2^3=1^3/2^3=1/8
30/100*142.85 = approx. 100
so ans is 42.85
7:8
E. Copper
1/3
Solution:
As given, we have,
The cost of one pen = 36 Rs.
So, the cost of 15 pens = 36 × 15 = 540 Rs.
The cost of one book = 45 Rs.
So, the cost of 12 books = 45 × 12 = 540 Rs.
The cost of one pencil = 8 Rs.
So, the cost of 10 pencils = 8 × 10 = 80 Rs.
Now,
the cost of each eraser is 40 Rs. less than the combined costs of pen and pencil.
So,
Combined costs of pen and pencil = 36 + 8 = 44 Rs.
Cost of one eraser = 44 – 40 = 4 Rs.
So, the cost of 5 erasers = 4 × 5 = 20 Rs.
Hence,
The total amount spent is
Hence, the total amount spent is 1180 Rs.
Answer is 16
1,2,4,7,11,….
1+(1) =2
2+(2) =4
4+(3) =7
7+(4) =11
Therefore
11+(5) =16
610 × 717 × 1127
= (2 × 3)10 × 717 × 1127
= 210 × 310 × 717 × 1127
Number of prime factors in the given expression
= (10 + 10 + 17 + 27)
= 64