If a=4, the square-circle-square….. ratio will be 4:$\pi$:2:$\frac{\pi }{2}$: 1:$\frac{\pi }{4}$

Answer is 4:($\frac{\pi }{2}$)= 16:$\pi$

Option (d)

m${}^2$ -1 =24n. This means that when m${}^2$ is divided by 24 it leaves a remainder -1.

m is of the form (6p+1)${}^2$ or (6p-1)${}^2$.. This is equal 12p(3p+/-1)+1..

We can see that the term 12p(3p+/-1) Is always divisible by 24..,Therefore we have to find the possible values of (6p+1) and ( 6p-1) less than 100.

Answer is 33..

Option(c)
Let the excess of speed over 30 kmph=S

Receipts= R and Expenses=E

Then

R=K${}_1$ S

And E= K${}_2$S${}^2$

Also, at 60kmph, R=E

Thus, K${}_1$=30 K${}_2$

We need to maximize R-E= K${}_1$S- K${}_2$S${}^2$

=SK${}_2$(30-S)

S+30-S, the sum is constant, the product will be maximum when they are equal

Thus S=30-S $\to$ S=15. Speed = 30+15= 45

Option(d)

3 * (1ABCDE)  = ABCDE1 

The only possibility for the last digit for the number is 7, which when multiplied by 3 gives 1 with a carry over 2

Proceeding this way,the number can be determined as

E = 7, D = 5, C = 8, B = 2, A = 4. The sum of the digits = 27

Option (b)

Let a, b and c denote the number of -1s, 1s and 2s in the sequence respectively. So,

-a + b + 2c = 19 and a + b + 4c = 99 (neglecting the zeroes)

So, a = 40 – c and b = 59 – 3c where 0 < c < 19 (as b > 0)

So, $x_1^3+x_2^3+x_3^3+...+x_n^3$ = -a + b + 8c = 19 + 6c

When c = 0 (a = 40, b = 59), the minimum value is 19; when c = 19 (a = 21, b =2), maximum value is 133.

$\angle$ BCA = 30${}^{\circ }$

$\angle$ BAC = 60${}^{\circ }$

$\angle$ ODC = 90${}^{\circ }$ (radius is perpendicular to the tangent)

$\angle$ COD = 60${}^{\circ }$

$\angle$ BOD = 120${}^{\circ }$

Now the sides OB = OD (Radii of same circle)

AB = AD (tangents from the same external point)

so, triangle ABO is congruent to triangle ADO

$\angle$ BOA = $\angle$ DOA = 60${}^{\circ }$

Now AB = 30, so OB = $\frac{30}{\sqrt{3}}$, so area ( AOB) = area (AOD) = $\frac{1}{2}$ * 30 * 30 $\sqrt{3}$ = 150 $\sqrt{3}$

So area (ABOD) = 300 $\sqrt{3}$ = 519.6.

Option(d)
A number divisible by both 3 as well as 5 has to be divisible by 15. In the first 100 natural numbers, there are 6 such numbers-> 15, 30, 45, 60, 75, 90

Favourable number of ways = ${}^6{C}_4$

Option (d)

Go from unitary method, try for smaller numbers. Consider 10x9${}^{10}$ = (10.9)9${}^9$

(10x9)+1

Consider the remainder when $\frac{9^9}{91}$. go by frequency method.

$\frac{9^1}{91}$${|}_R$=9

$\frac{9^1}{91}$${|}_R$=-10

$\frac{9^3}{91}$${|}_R$= -90/91|_R=+1

Therefore $\frac{9^9}{91}$${|}_R$=1

Remainder will be only the first part of the numerator=10x9, in the case we have considered = 90

And in the question = 100x99= 9900

Option (a)

Digit sum is nothing but the remainder obtained when the number is divided by 9. Thus from 6! To 10! the number is completely divisible by 9. Hence, digit sum is 0 1!+2!+3!+4!+5! = 1+2+6+24+120 = 153, which is completely divisible by 9.
Hence, digit sum is 0

Option (d)

All numbers of the form 4k+2 cannot be expressed as difference of squares.

Thus, those 4 digit numbers which can be expressed as difference of squares in atleast one way= number of four digit numbers - 4 digit numbers of the form 4k+2

= 9000 -2250= 6750.