Top 100 CAT 2024 Quant Questions (Most Expected)

The region enclosed by the lines is a triangle in the third quadrant formed by the points (0,-7), (0,0) and (9,0). The number of coordinates in the region with x coordinate are as follows
x=0 => 8 points,
x=1, 7points,
x=2, 6points,
x=3, 5points,
x=4, 4points,
x=5, 4points and so on.
Total no. of points =41

The total number of values the expression 4A+9B will take is 10*15 = 150
To find the number of duplicates, suppose the value is same for two pairs (A2,B2) and (A1,B1)

So, 4A2+9B2 = 4A1+9B1

 or, 4(A2-A1) = 9(B1-B2).
This is true only if A2-A1=9 and B1-B2=4
So, A=10 and A1=1. Only one case possible for A's.

B1-B2=4 then  (B1 can be 5,6,7.............15)  

Thus the cases which will repeat are

1. A1=1 B1=5 A2= 10 B2= 1

2. A1=1 B1=6 A2= 10 B2= 2

3. A1=1 B1=7 A2= 10 B2= 3

.

..

...

11. A1=1 B1=15 A2= 10 B2= 11

The number of values B and B1 can take is 11.  (B1=B+4, $$B,B1 \leq 15$$ , B $$\epsilon$$ (1,11) and B1 $$\epsilon$$ (5,15) )
So, number of unique values for 4A+9B = 150-11 = 139

Let A be the amount invested in Bank A. B is the amount invested in Bank B. A+ B = 10,000
A*6*2/100 + B(1.08)*(1.08) - B = 1432
.12A + .1664 B =1432
.12A+ .12B = 1200
So, 0.0464B = 232 or B = 5000
A = 5000.

The given number can be broken into 7*3*7429 = 21*7429
Now if we take the value of $$20^4$$ = 160000 which is close to the given number.
Divide 7429 with 19, we find that it is perfectly divisible by 19.
7429 = 19*17*23
156009 = 17*19*21*23
Sum = 17+19+21+23 = 80

$$6561^2$$ can be written as $$3^{16}$$.

Let a, b and c be the three factors => abc = $$3^{16}$$

If x, y and z are the powers of 3 in a, b and c respectively, then x + y + z = 16 => $$^{18}C_2$$ solutions => 153 solutions.

But, we have to find the unordered solutions => remove cases (k,k,k) and (k,k,m), divide the resulting number by 6 and then add the removed cases again.

Case (k, k, k): As 16 is not divisible by 3, there will be no (k,k,k) cases.

Case (k, k, m): Here 2k + m = 16. k can go from 0 to 8 and correspondingly m will go from 16 to 0. Hence, there will be a total of 9 cases.

So, there are 0 (k,k,k) cases and 9 (k,k,m) cases => 9*3 = 27 solutions have to be removed => 153 - 27 = 126 cases are of the form (k,m,n).

126/6 = 21 => 21 + 9 = 30 unordered solutions.

Note: In the question we have not been asked to find the number of ordered solutions. Every solution of the form (k,m,n) can be arranged in 3! = 6 ways. But only one of them should be counted when counting the number of unordered solutions. So they have been divided by 6. So 126/6 = 21. Also every solution of the form (k,k,m) can be arranged in 3 ways. So they have been divided by 3 to get 27/3 = 9. Thus, the total number of solutions = 21+9 = 30.

f(a)f(b) = f(a) + f(b) + f(ab) - 2
Put a = b = 1.
$$[f(1)]^2 = 3f(1) - 2$$ => f(1) = 1 (or) 2.
Let's assume f(1) = 1
Now, put b = 1.
f(a) = 2f(a) - 1
=> f(a) = 1 => For all values of a, f(a) = 1.
This is false because f(4) = 17.
=> f(1) = 2 is the correct value.
Now put b = 1/a
f(a)f(1/a) = f(a) + f(1/a) + 2 - 2
=> f(a)f(1/a) = f(a) + f(1/a)

So taking RHS terms to LHS and adding 1 to both sides we get
f(a)f(1/a) - f(a) - f(1/a) +1 = 1
(f(a) - 1) (f(1/a)-1) = 1
Let g(x) = f(x)-1
So g(x)*g(1/x) = 1
So g(x) is of the form $$\pm x^n$$
So f(x) is of the form $$\pm x^n + 1$$.

f(a) = $$\pm a^n + 1$$ satisfies the above condition.
$$-4^n + 1$$ = 17 => $$-4^n$$ = 16, which is not possible.
$$4^a + 1$$ = 17 => n = 2
=> f(a) = $$a^2 + 1$$
=> f(7) = $$7^2 + 1$$ = 50

Let the percentage of students who like all three sports be c, the percentage of students who like exactly two sports be b and the percentage of students who like exactly one sport be a.
So, a + b + c = 100
a + 2b + 3c = 65 + 86 + 57 = 208
=> b + 2c = 108
To get the maximum percentage of students who like exactly two sports, we have to maximize b, so the c has to minimum
b + 2c = 108
b + c = 100
So, we get C = 8
Thus the minimum value of c which satisfies these equations is 8
Thus the maximum value of b = 92
Hence the correct answer is 92

3x + 4y + z = a --> (1)

2x + 6y + 4z = b --> (2)

x - y - 2z = c --> (3)

Eliminating z from (1) and (2) , (1)x4 - (2) : 12x + 16y + 4z - (2x + 6y + 4z) = 4a - b

10x + 10 y = 4a - b
=> x + y = (4a - b)/10 - (4)

Eliminating z from (1) and (3), (1)x2 + 3 = 6x + 8y + 2z + (x - y - 2z) = 2a + c
=> x + y = (2a + c)/7 - (5)

Equating equations (4) and (5), we get,

(4a - b)/10 = (2a + c)/7
28a - 7b = 20a + 10c
Thus, 8a = 7b + 10c

Total number of man-hours required to complete the job if the first person works alone = 240
Number of man-hours put in by the first person in a day = 8
Number of man-hours put in by the second person in a day = 16
Number of man-hours put in by the third person in a day = 24 and so on
Let the work be completed in ‘n’ days
8*n + 16*(n-1) + 24*(n-2) + … + 8n*(n-(n-1)) = 240

=> 8n + 16(n-1) + 24(n-2) + … + 8n*1 = 240

If n = 6, the expression on the left hand side becomes 8*6 + 16*5 + 24*4 + 32*3 + 40*2 + 48 = 48*2 + 80*2 + 96*2 = 448
If n = 5, the expression becomes 8*5 + 16*4 + 24*3 + 32*2 + 40 = 280
If n = 4, the expression becomes 8*4 + 16*3 + 24*2 + 32*1 = 160
So, n = 5 is the correct answer

Let the cost of a pen be x, the cost of a pencil be y and the cost of an eraser be z.
4x + 10y + 14z = 120
=> 2x + 5y + 7z = 60 → (1)
6x + 10z = 60 + 12y
=> 6x - 12y + 10z = 60
=> 3x - 6y + 5z = 30 → (2)
We need the value of 78y + 2z - 12x = 2*(-6x + 39y + z)
3*(1) - 4*(2) gives 6x + 15y + 21z - 12x + 24y - 20z = 180 - 120
=> -6x + 39y + z = 60
So, the required value is 2*60 = 120

Let the cost price of 1 kg of rice be C.
For Rani he sells at C but only sells 0.9 kg instead of 1 kg.
For Vani he sells at 1.2C but sells 1.1 kg instead of 1 kg.
Overall he sold 2 kg rice together and the cost price for it is 2C.
His net revenue after both the transactions is C+1.2C = 2.2C
His profit = 0.2C
Profit % = $$ (\frac{0.2C}{2C})\times100$$ = 10% profit.

Let the number of persons who do not take any juice and the number of persons who take all 4 types of juices be ‘x’.
As there are ‘x’ people who take atleast 4 types of juices there should be ‘2x’ people who take atleast 3 types of juices.
This means the number of people who take exactly 3 types of juices = 2x - x = x.
As there are ‘2x’ people who take atleast 3 types of juices there should be ‘4x’ people who take atleast 2 types of juices.
This means the number of people who take exactly 2 types of juices = 4x - x - x = 2x.
As there are ‘4x’ people who take atleast 2 types of juices there should be ‘8x’ people who take atleast 1 type of juices.
This means the number of people who take exactly 1 type of juices = 8x - 4x = 4x.
Thus the total number of people in the Gym = x + 4x + 2x + x + x = 9x.
9x = 207 => x = 23.
The number of people who take exactly 2 types of juices = 2x = 46.

Let the number of large boxes, in which Jay places medium boxes, be m. Hence, m<10.

Let the number of medium boxes, in which Jay places small boxes, be n. Hence, n<m

After adding 7*m medium boxes, the number of empty boxes = (10-m) large boxes + 7m medium boxes = 10+6m. Once a large box gets 7 medium boxes, it is no longer empty.

After adding the 7*n small boxes, the number of empty boxes = (10-m) large boxes + (7m-n) medium boxes + 7n small boxes = 10+6m+6n = 10+6(m+n)

We know that 10+6(m+n) = 82

6(m+n) = 72 => m+n = 12

Now the total number of boxes used = 10 + 7m + 7n = 10+7(m+n) = 10+ 7*12 = 94

Let the present age of Ramesh's son be x
When Ramesh was x, his son was x/4. Thus, the no. of years before from the present during which Ramesh's son's age was x = x - x/4 = 3x/4
Thus, the present age of Ramesh = x + 3x/4 = 7x/4
Thus, the ratio of ages of Ramesh and his son = 7x/4x = 7/4
Thus, since, Ramesh age is at-least 45, his ages could be 49, 56 etc(only integers are possible as we need the ratio on their birth days). When Ramesh is 49 his son is 4/7 x 49 = 28. If Ramesh's age is greater than 49, his son's age would be greater than 30 which is not possible .
Thus, the age of Ramesh is 49 and his son's age is 28
Thus, the required sum = 49 + 28 = 77

Here, let us assume that each man will do 1 unit of work everyday.
Total work = 30*(15) = 450 units.
According to the question 28 men were there on second day,26 on fourth and so on
Work done on first 20 days = 30+28+30+26+30+24………… = (30+28)+(30+26)+(30+24)…………
=58+56……..+42 =$$\frac{10}{2}*(2(58)+(10-1)(-2))$$ = 490
But, 30units is done on 19th day and 10units on 20th day.
The total work of 450 units will be done in first 18 days.

The first equation can be written as
a + b = 12 (when b > 0) I
and a - b = 12 (when b < 0) II {a $$\leq 12$$, for these relations to be valid}
Similarly, for second equation
a + b = 9 (a > 0) III
b - a = 9 (a < 0) IV
In this case, b will always be less than or equal to 9.
On checking the slopes, we can see that lines I and III are parallel and similarly, lines II and IV are also parallel.
I and IV will intersect at b = 10.5 and a = 1.5. This is not possible since < 0 in equation IV. Hence this pair won’t given any solution.
II and III will intersect at a = 10.5 and b = -1.5
This satisfies both the equations and is also true as per the given conditions.
Hence there is the only common solution for both the equations.

From Wilson’s theorem, we know that
(p - 1)! mod p = p-1, where p is a prime number.
So when (N+1) will be a prime number than N! will not be divisible by (N+1).There are 26 prime numbers from 1 to 101. Hence for these 26 values, (N+1) will not be a factor of N!. Moreover, 3! is also not divisible by 4. Hence when N = 3, then also the given statement does not hold true. For all other values of N, N! will be divisible by (N+1). There are 100 numbers from 1 to 100. From these, for 27 numbers can be excluded. Thus, for 73 values of ‘N’, the given relation will hold true.

The speed of the dog is 7 times the speed of Aman. So in the time that Aman completes his journey from P to Q once, the dog will cover the distance 7 times. Both dog and Aman start at the same point. Now when Aman has covered some distance, the dog will be returning back and meeting Aman. When dog is moving from P to Q, it will again meet Aman at a different point. Hence each time the dog crosses Aman, he will do so at a different point. The dog will cross Aman 6 times. Out of these at 3 points both will be moving in the same direction and at the other three points both will be moving in different directions.

We know that $$8a^3-12a^2-6a-1=0$$
Therefore, $$-8a^3 + 12a^2 + 6a + 1 = 0$$
Or, $$8a^3 + 12a^2+ 6a + 1 = 16a^3$$
Therefore, $$(2a+1)^3 = 16a^3$$
Or, $$(2+1/a)^3 = 16$$
Therefore, $$1/a = \sqrt[3]{16}-2$$
Hence, $$a = \frac{1}{\sqrt[3]{16}-2} = \frac{\sqrt[3]{256}+\sqrt[3]{128}+4}{16-8}$$
Therefore, $$a = \frac{\sqrt[3]{4}+\sqrt[3]{2}+1}{2}$$
Hence, $$p=4,q=2,r=2$$ and the required sum equals 8

We have,
$$x(x+1)(x+2)(x+3) = 168$$
$$(x)(x+3)*(x+1)(x+2) = 168$$
$$ (x^2 + 3x)*(x^2 + 3x + 2) = 168 $$
$$ (x^2 + 3x)^2 + 2(x^2 + 3x) = 168 $$
$$ (x^2 + 3x)^2 + 2(x^2 + 3x) + 1 = 169 $$
Let $$ y = x^2 + 3x $$
So, $$ y^2 + 2y + 1 = 169 $$
$$ (y+1)^2 = 13^2 $$

So, $$y + 1 = 13$$ and $$y + 1 = -13$$
$$y = 12$$ or $$y = -14$$
So, $$ x^2 + 3x = 12 $$ or $$ x^2 + 3x = -14 $$
Considering,
$$ x^2 + 3x = -14 $$
$$ x^2 + 3x + \frac{9}{4} = -14 + \frac{9}{4} $$
$$ (x + \frac{3}{2})^2 = - \frac{47}{4} $$
This will give imaginary roots. Hence can be ignored.
Considering,
$$ x^2 + 3x = 12 $$
$$ x^2 + 3x + \frac{9}{4} = 12 + \frac{9}{4} $$
$$ (x + \frac{3}{2})^2 = \frac{57}{4} $$
$$ (x + \frac{3}{2}) = \pm \frac{\sqrt{57}}{2} $$
Thus, $$ x = \frac{\sqrt{57} - 3}{2}$$ or $$x = \frac{-\sqrt{57} - 3}{2} $$
Thus, 2 real roots are possible.
Required sum = $$ \frac{\sqrt{57} - 3}{2} + \frac{-\sqrt{57} - 3}{2} $$ = $$ \frac{-6}{2} = -3 $$

We know that X! has 3 more trailing zeroes than (X-2)!. Also, we know that the number of zeroes is determined by the powers of 5 present in the number. Therefore, for one number to have 3 more zeroes than the other, the larger number must contain 3 additional powers of 5.

This is possible only in 2 cases - X is a multiple of 125 or X-1 is not a multiple of 125.
Eg) 125! has 3 more trailing zeroes than 123! and 126! has 3 more trailing zeroes than 124!.
Therefore, for each multiple of 125 present in the given range, 2 numbers are possible.
From 1000 to 2000, there are 9 multiples of 125. However, X cannot take the value 2001 as it falls outside the given bracket. Therefore, total number of possible values for which the number of zeroes increases at least by 3 = 2*9 - 1 = 17.

Among these 17 numbers, there are 2 multiples of 625. Since 625 is the fourth power of 5, the number of zeroes will increase by 4 in these cases. Therefore, we have to subtract these 4 cases (2 for each number).

Therefore, the total number number of values for which the given condition is satisfied = 17-4 = 13.
Therefore, option C is the right answer.

T = $$\frac{3}{5} + \frac{6}{5^2} + \frac{11}{5^3} + \frac{18}{5^4} + \frac{27}{5^5} + \frac{38}{5^6} + … $$

Substitute $$\frac{1}{5} = x$$

T = $$3x + 6x^2 + 11x^3 + 18x^4 + 27x^5 + 38x^6 + …$$         (1)

T*x = $$3x^2 + 6x^3 + 11x^4 + 18x^5 + 27x^6 + 38x^7 + …$$         (2)

Subtract equation (2) from equation (1)

T*(1 - x) = $$3x + 3x^2 + 5x^3 + 7x^4 + 9x^5 + 11x^6 + …$$        (3)

T*(1 - x)x = $$3x^2 + 3x^3 + 5x^4 + 7x^5 + 9x^6 + 11x^7 + …$$       (4)

Subtract equation (4) from equation (3)

T*(1-x)*(1-x) = $$3x + 0 +2x^3 + 2x^4 + 2x^5 + 2x^6 + …$$

T*(1-x)*(1-x) = $$3x +2(x^3 + x^4 + x^5 + x^6 + …)$$

T*(1-x)*(1-x) = $$3x +2(\frac{x^3}{1-x})$$

Resubstitute $$x = \frac{1}{5} $$

T*$$\frac{4*4}{5*5} = \frac{3}{5} + 2*\frac{1/5^3}{4/5}$$

T = $$\frac{15}{16} + 2*\frac{1}{64}$$

T = $$\frac{15}{16} + \frac{1}{32}$$

T = $$\frac{15}{16} + \frac{1}{32}$$

T = $$\frac{31}{32}$$

Hence, option D is the right choice.

Roots of the equation $$ax^2 + bx + c = 0$$ are $$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$.
We know that the sign of the roots are different. One of the roots must be positive and the other must be negative. Let us evaluate the options one by one.

Option A:
$$a$$ is positive, $$b$$ is positive and $$c$$ is negative

Since $$c$$ is negative, $$\sqrt{b^2-4ac}$$ will be greater than $$b$$. Therefore, the root $$\frac{-b - \sqrt{b^2-4ac}}{2a}$$ will be negative and $$\frac{-b + \sqrt{b^2-4ac}}{2a}$$ will be positive.

Option B:
$$a$$ is negative, $$b$$ is positive and $$c$$ is positive

Since $$a$$ is negative, the term $$\sqrt{b^2-4ac}$$ will be greater than $$b$$. Therefore, the root $$\frac{-b - \sqrt{b^2-4ac}}{2a}$$ will be positive (Since $$a$$ is negative) and the root $$\frac{-b + \sqrt{b^2-4ac}}{2a}$$ will be negative.

Option D:
$$a$$ is negative, $$b$$ is negative and $$c$$ is positive

Since $$a$$ is negative, the term $$\sqrt{b^2-4ac}$$ will be greater than $$b$$. Also, $$b$$ is negative. Therefore the term $$-b$$ will be positive.

The root $$\frac{-b - \sqrt{b^2-4ac}}{2a}$$ will be positive (Since $$a$$ is negative) and the root $$\frac{-b + \sqrt{b^2-4ac}}{2a}$$ will be negative.

Option C:

$$a$$ is positive, $$b$$ is negative and $$c$$ is positive

Since both $$a$$ and $$c$$ are positive, the term $$\sqrt{b^2-4ac}$$ will be less than $$b$$. Also, $$b$$ is negative.

Therefore, both the roots $$\frac{-b - \sqrt{b^2-4ac}}{2a}$$ and $$\frac{-b + \sqrt{b^2-4ac}}{2a}$$ will be positive.
Option C is definitely false and hence, option C is the right answer.

Let the amount at which he sells earphones in good condition be Rs n.
Therefore, he sells $$(50 - x - y )$$ earphones at Rs n each i.e. he earns Rs $$(50 - x - y)n$$ after selling earphones in good condition also it is given that y number of earphones are not properly packed and could only be sold after giving a 50% discount. Therefore, he sells these y earphones at Rs n/2 each earning Rs yn/2. His total profit is 20%, i.e. he sells all the items at 50*100*1.2 = Rs 6000.
Therefore, $$(50 - x - y)n + yn/2 = 6000$$
$$50n - xn - yn + yn/2 = 6000$$
$$50n - xn - yn/2 = 6000 $$
$$100n - 2xn - yn = 12000$$
$$n = \frac{12000}{100-2x-y} $$
When, x = 9 and y = 2 we get n = 150
When, x = 7 and y = 8 we get n = 153.84 which is not an integer and hence x = 7 and y = 8 cannot be the answer.
When, x = 8 and y = 24 we get n = 200
When, x = 13 and y = 24 we get n = 240
Therefore, option B is the correct answer.

Given that $$T_{n} = \dfrac{(n+7)(n+8)}{8}$$ 

$$\Rightarrow$$ $$\dfrac{1}{T_{n}}$$ = $$\dfrac{8}{(n+7)(n+8)}$$

$$\Rightarrow$$ $$\dfrac{1}{T_{n}}$$ = $$8(\dfrac{1}{(n+7)}-\dfrac{1}{(n+8)})$$

Therefore, $$\dfrac{1}{T_{1}}$$ = $$8(\dfrac{1}{8}-\dfrac{1}{9})$$  ... (1)

$$\dfrac{1}{T_{2}}$$ = 8($$\dfrac{1}{9}-\dfrac{1}{10})$$   ... (2)

$$\dfrac{1}{T_{3}}$$ = 8($$\dfrac{1}{10}-\dfrac{1}{11})$$ and so on  ... (3)

$$\dfrac{1}{T_{k}}$$ = 8($$\dfrac{1}{k+7}-\dfrac{1}{k+8})$$

We can see that negative part of each term is same as positive part of subsequent term. Hence we can cancel out these terms. Therefore, 

$$\dfrac{1}{T_{1}}$$+$$\dfrac{1}{T_{2}}$$+$$\dfrac{1}{T_{3}}$$+ ... + $$\dfrac{1}{T_{k}}$$ = 8($$\dfrac{1}{8}-\dfrac{1}{k+8}$$)

$$\Rightarrow$$ 8($$\dfrac{1}{8}-\dfrac{1}{k+8}$$) = $$\dfrac{1000}{1004}$$

$$\Rightarrow$$ 8($$\dfrac{k+8-8}{8(k+8)}$$) = $$\dfrac{1000}{1004}$$

$$\Rightarrow$$ $$\dfrac{k}{(k+8)}$$ = $$\dfrac{1000}{1004}$$

$$\Rightarrow$$ $$k=2000$$

Let '100x' be the seeing price of 1 pen. It is given that selling price of 16 pens is the same the cost price of 25 pens.

Hence, the cost price of 1 pen = $$\dfrac{16*100x}{25} = 64x$$ 

Total discount offered on 50 pens is the same as the selling price of 14 pens. 

Hence, the discount offered on 1 pen = $$\dfrac{14*100x}{50} = 28x$$

Therefore, the marked up price of 1 pen = 100x + 28x = 128x 

Hence, markup percentage = $$\dfrac{128x - 64x}{64x}\times 100$$ = 100%

The milkman buys milk at 100 Rs per liter.
He sells 20L of pure milk at 80 Rs per liter.
Thus, he faces of loss of 400 Rs I.e. -400.
After adding water he has 80L of milk and 20L of water out of which he sells 20L.
In this 20L there will be 16L milk and 4L water
Thus, CP of 16L of milk = 1600
SP of 20L of mixture = 20*85 = 1700
Thus, he makes a profit of 100 Rs here.
Overall loss/profit till now = -400+100 = -300 Rs
He now has 64L of milk and 36L of water.
He sells 25L (at 90 Rs per liter) of this mixture I.e. 16L of milk and 9L of water.
Thus, CP of 16L of milk = 1600 Rs
SP of 25L of mixture = 25*90= 2250 Rs
Thus, he makes a profit of 2250-1600 = 650 Rs here.
Overall loss/profit till now = -300+650 = 350 Rs
He now has 48L of milk and 52L of water.
He sells 25L (at 95 Rs per liter) of this mixture I.e. 12L of milk and 13L of water.
Thus, CP of 12L of milk = 1200 Rs
SP of 25L of mixture = 25*95= 2375 Rs
Thus, he makes a profit of 2375 -1200 = 1175 Rs here.
Thus, Overall Profit/Loss in these transactions = 1175+350 = 1525 Rs.

It is given that a, b, and c are three real roots of the equation $$x^3 - 9x^2 - 10x + 168 = 0$$.
Therefore, (x - a)(x - b)(x - c) = 0.  

Let us assume that F(x) = (x - a)(x - b)(x - c) = $$x^3 - 9x^2 - 10x + 168$$. 

Also, the sum of the roots = (a + b+ c) = 9. 

It is given that (a+b), (b+c), and (c+a) are the roots of the equation $$x^3+px^2+qx+r = 0$$.

Therefore, the sum of the roots = (a+b) + (b+c) + (c+a) = -p.

Hence, p = -18. 

Also, $$r = -(a+b)*(b+c)*(c+a)$$ 

$$\Rightarrow$$ $$r = -(9-c)*(9-a)*(9-b)$$              {We know that (a + b+ c) = 9}. 

$$\Rightarrow$$ $$r = -F(9)$$

$$\Rightarrow$$ $$r = -(9^3 - 9*9^2 - 10*9 + 168)$$

$$\Rightarrow$$ $$r = -78$$

Therefore, the value of |r - p|. = |-78+18| = 60.

It is given that, both the root of the quadratic equation, $$ax^2+(a-4)x+a+1=0$$ are greater than 0. Therefore, the sum of roots should be greater than 0.

$$\Rightarrow$$ $$-\dfrac{B}{A}$$ > 0
$$\Rightarrow$$ $$-\dfrac{(a-4)}{a}$$ > 0
$$\Rightarrow$$ $$\dfrac{a-4}{a}$$ < 0
$$\Rightarrow$$ $$0 < a < 4$$   ... (1)

It is given that the quadratic equation has both the roots greater than 0. Therefore D $$\geq$$ 0 
$$\Rightarrow$$ $$(a-4)^2-4*(a)*(a+1) \geq 0$$

$$\Rightarrow$$ $$a^2-8a+16-4a^2-4a \geq 0$$

$$\Rightarrow$$ $$-3a^2-12a+16 \geq 0$$

$$\Rightarrow$$ $$3a^2+12a-16 \leq 0$$

$$\Rightarrow$$ $$\dfrac{-6-2\sqrt{21}}{3} \leq a \leq \dfrac{-6+2\sqrt{21}}{3}$$

$$\Rightarrow$$ $$-5.05 \leq a \leq 1.05$$  ... (2)

From equation (1) and (2), we can see that 'a' can be equal to 1. Let's figure out the value of both the roots explicitly at a = 1. 
At a = 1, we can see that the equation reduces to $$x^2-3x+2=0$$. Therefore, x = 1 or 2. We can say that at a = 1, both the roots of the given equation are greater than 0.

Hence, we can say that there is only one integer value (a=1) integer value of 'a' for which both the root of the quadratic equation $$ax^2+(a-4)x+a+1=0$$ are greater than 0.

Let us try to simplify the expression : 

$$ \log_{3}({\log_{81}{x})}=\log_{81}({\log_{3}{x})} $$

$$ \Rightarrow \dfrac{\log({\log_{81}{x})}}{\log3}=\dfrac{\log({\log_{3}{x})}}{\log{81}} $$

$$ \Rightarrow \dfrac{\log({\log_{81}{x})}}{\log3}=\dfrac{\log({\log_{3}{x})}}{4 \times \log{3}} $$

$$ \Rightarrow {\log({\log_{81}{x})}}=\dfrac{\log({\log_{3}{x})}}{4} $$

$$ \Rightarrow 4 \times {\log({\log_{81}{x})}}=\log({\log_{3}{x})} $$ --------(1)

Let $$\log_{3}{x} = t$$, then $$\log_{81}{x} = \dfrac{logx}{log81} = \dfrac{logx}{4log3} = \dfrac{t}{4} $$

Putting the values in equation (1),

$$ 4 \times {\log(\dfrac{t}{4})} = \log(t) $$

$$ \Rightarrow {\log(\dfrac{t}{4})^{4}} = \log(t) $$

Since all the terms are over 0, we can remove the log. 

$$ \Rightarrow (\dfrac{t}{4})^{4} = \log(t) $$

$$ \Rightarrow \dfrac{t^{4}}{2^{8}} = t $$

$$ \Rightarrow t^{4} - 2^{8}t = 0 $$

$$ \Rightarrow (t)(t^{3} - 2^{8}) = 0 $$

$$ \Rightarrow t = 0$$ and $$t=2^{8/3} $$

$$ \Rightarrow \log_{3}{x} = 0$$ and $$\log_{3}{x}=2^{8/3} $$

$$ \Rightarrow x = 3^{0}=1$$ and $$x =3^{2^{8/3}} $$

1 will not be an acceptable answer in this equation because if we put 1 in the first log, its value will become 0. As we know that the log function cannot have 0, 1 becomes an unacceptable value. 

Thus, $$x =3^{2^{8/3}} $$

Let us draw the diagram according to the information given in the question. 

Let us draw a perpendicular from R to line segment PQ which meets at point U that we get after expanding QP towards P.
 

In right-angle triangle RUP,
$$RP^2=RU^2+UP^2$$
$$\Rightarrow$$ $$x^2+y^2=169$$   ...(1)
In right-angle triangle RUQ,
$$RQ^2=RU^2+UQ^2$$
$$\Rightarrow$$ $$x^2+(y+30)^2=37^2$$ 
$$\Rightarrow$$ $$x^2+y^2+60y+900=1369$$ 
$$\Rightarrow$$ $$x^2+y^2+60y=469$$   ...(2)

From equation (1) and (2), we can say that 60y = 300 $$\Rightarrow$$ y = 5 cm.
By substituting value of y = 5 in equation (1), we get x = 12 cm.
By symmetry we can say that RS = PQ + 2*PU = 30 + 2*5 = 40 cm. 
Since both $$\triangle$$ RPQ and $$\triangle$$ SQP are congruent we can say that $$\triangle$$ PTQ $$\sim$$ $$\triangle$$ STR.

Therefore, the height of both the triangles will be the same as the ratio of the corresponding sides. 
We can see that $$\dfrac{PQ}{RS}$$ = $$\dfrac{30}{40}$$ = $$\dfrac{3}{4}$$

Hence, we can say that height of triangle PTQ = $$\dfrac{3}{4}$$* Height of triangle STR
Also, the sum of the heights of both the triangles = RU = 12 cm
Therefore, height of triangle PTQ = $$\dfrac{3}{7}$$*12 = $$\dfrac{36}{7}$$ cm
Hence, the area of triangle PTQ = $$\dfrac{1}{2}$$*PQ*Height of the triangle = $$\dfrac{1}{2}*30*\dfrac{36}{7}$$ = $$77\dfrac{1}{7}$$ cm$$^2$$. Hence, option B is the correct answer.

For the equation to be defined, 0<18-$$3^x$$ 
So, the maximum value of x can be 2.
Also, 18-$$3^x$$ =$$3^{3-x}$$
18=$$3^x$$+$$3^{3-x}$$
18=$$3^x$$ +$$\frac{3^3}{3^x}$$
Taking $$3^x$$ as t, then:
18=t+$$\frac{3^3}{t}$$
18=t+$$\frac{27}{t}$$
$$t^2-18t+27=0$$

 Let us take the two roots to be A and B. A and B would be values of $$3^x$$ that satisfy the equation.

However, we need to find the product of $$3^x+2$$ that satisfy the equation. So, we must find the value of (A+2)*(B+2) = AB + 2*(A+B) + 4.

From the quadratic equation t^2-18t+27=0, we know that A+B = 18 and AB=27
So, the required product will be 27+36+4 = 67

So the answer is Option C

Let the length of the race be $$x$$ m. Let us denote the speeds of Akram, Bikram, and Chandan with $$a, b$$ and $$c$$ respectively.
Akram can give Bikram a head start of 64 feet.

=> $$\frac{x}{a} = \frac{x-64}{b}$$
=> $$\frac{a}{b} = \frac{x}{x-64}$$ -------------(1)

The head start that Chandan can give Bikram is 40 feet longer than the head start that he can give to Akram.

Let us assume that Chandan gives Akram a head start of $$y$$ feet.
Now, we can form 2 equations.

$$\frac{x}{c} = \frac{x-y}{a}$$
=> $$\frac{a}{c} = \frac{x-y}{x}$$ ----------(2)

$$\frac{x}{c} = \frac{x-y-40}{b}$$
$$\frac{b}{c} = \frac{x-y-40}{x}$$ ------------(3)

Dividing (2) by (3), we get,
$$\frac{a}{b} = \frac{x-y}{x-y-40}$$ -----------------------(4)

Equating (1) and (4), we get, 

$$\frac{x}{x-64} = \frac{x-y}{x-y-40}$$ ----------(5)

We know that the ratio of speeds of 2 of the 3 persons is 2:1. 

If the value of (1) is 2, then $$x$$ will be equal to $$128$$.
Substituting the value of $$x$$ in (5), we get,

$$2 = \frac{128-y}{88-y}$$
$$176-2y = 128-y$$
$$y = 48$$

Let us assume the value of (2) to be $$\frac{1}{2}$$.

$$\frac{1}{2} = \frac{x-y}{x}$$
=> $$y = \frac{x}{2}$$
Substituting the value of $$x$$ in (5), we get,

$$\dfrac{x}{x-64} = \dfrac{\frac{x}{2}}{\frac{x}{2}-40}$$
$$\dfrac{x}{x-64} = \frac{x}{x-80}$$

As we can see, the equation is inconsistent and hence, we can eliminate this case. 

Let us assume the value of (3) to be $$\frac{1}{2}$$.

$$\frac{1}{2} = \frac{x-y-40}{x}$$
=> $$(y+40) = \frac{x}{2}$$
$$x = 2y+80$$ 

Substituting this relationship in (5), we get,
$$\frac{2y+80}{2y+16} = \frac{y+80}{y+40}$$
$$(2y+80)(y+40) = (2y+16)(y+80)$$
$$2y^2+160y+3200 = 2y^2+176y+1280$$
$$16y = 1920$$
=> $$y = 120$$
=> $$x = 320$$ feet. 

As we can see, the length of the track can be $$128$$ feet or $$320$$ feet. Therefore, option D is the right answer.

$$(p+q+r)^2$$ = $$p^2+q^2+r^2+2(pq+qr+pr)$$
$$25=p^2+q^2+r^2+2*3$$
$$p^2+q^2+r^2=19$$
$$(q+r)^2 =q^2+r^2+2qr$$
So $$(q+r)^2 \leq 2*(q^2+r^2)$$
$$(5-p)^2 \leq 2*(19-p^2)$$
$$25+p^2-10p \leq 2*(19-p^2)$$
$$3p^2-10p-13 \leq 0$$
$$(p+1)(3p-13) \leq 0$$
p $$\in$$ [-1,13/3]

So the maximum value p can take is 13/3

Hence B is the correct answer.

To make the calculation easier, let us take the amount the investor wants to invest as $$x$$.
Scheme A :
Amount of the investment after 2 years, $$a_2 = x \times (1.1)^2 = 1.21x $$
Amount of investment after 5 years, $$a_5 = 1.21x \times (1+(0.2 \times 3)) = 1.936x $$
Since the increase is more than 60%, the additional 10% won’t be added to it.
Scheme B :
Amount of investment after 2 years, $$b_2 = x \times (1+(0.1 \times 2)) = 1.2x $$
Amount of investment after 5 years, $$b_5 = 1.2x \times (1.2)^3 = 2.0736x $$
Since the increase is more than 50%, the additional 15% won’t be added to it.
After looking at the interest earned, we can say that the investor should choose scheme B.
Amount earned by the investor = $$10^6 \times (2.0736-1) = 1073600 $$

Let the people be A,B,C.

Let the amount of work done per day by A,B,C be $$4x$$,$$5x$$ and $$6x$$ units 

Thus, total work done in 1 day = $$4x+5x+6x = 15x$$ units 

Total work of the wall = $$15x \times 5 = 75x$$ units 

Total amount of money for the construction of wall = 150000 

The most amount of money will be earned by C as his efficiency is the most and therefore will also do the most amount of work. The least amount of money will be earned by A as his efficiency is the least and therefore will also do the least amount of work. 

Money paid to C = $$\dfrac{6}{4+5+6} \times 150000 = 60000 $$

Money paid to A = $$\dfrac{4}{4+5+6} \times 150000 = 40000 $$ 

Difference in amount = 60000 - 40000 = 20000

Let the original speed of the car be $$x$$ km/hr and the original time of the journey be $$y$$

Thus, the distance between P and Q, $$d = x \times y = xy$$ kms 

The speed of the car after the fault = $$\dfrac{4}{5} \times x = 0.8x $$ km/hr 

Case 1: Distance covered during first 2 hours = $$x \times 2 = 2x$$ km/hr 

Time travelled by the car at reduced speed = $$y+2-2 = y$$ hours 

Distance covered by the car at reduced speed = $$0.8x \times y = 0.8xy $$ 

We know that the total distance travelled in both the cases is same. 

Total distance travelled in first case = $$d$$

$$\Rightarrow (x \times 2)+(0.8x \times y) = x \times y $$ 

$$ \Rightarrow 2x+0.8xy = xy $$ 

$$ \Rightarrow 2+0.8y = y $$ 

$$ \Rightarrow 0.2y = 2 $$ 

$$ \Rightarrow y = 10 $$ hours 

Let the fault develop at $$a$$ hours after the car starts moving 

Distance travelled before the fault = $$x \times a $$ 

Time travelled by the car at reduced speed = $$10+1-a = 11-a$$ hours 

Distance covered by the car at reduced speed = $$0.8x \times (11-a)$$ 

Total distance travelled = $$d$$ 

$$ \Rightarrow ax+8.8x-0.8ax = x \times 10 $$ 

$$ \Rightarrow 8.8x+0.2ax = 10x $$ 

$$ \Rightarrow 8.8+0.2a = 10 $$ 

$$ \Rightarrow 0.2a = 1.2 $$ 

$$ \Rightarrow a = 6 $$ hours 

Thus the fault should've happened at 6 hours to make the delay equal to 1 hour. 

Let us consider the first term of GP=a

Common ratio=r

$$\frac{a}{1-r}$$=10

r=1-$$\frac{a}{10}$$

-1 < r< 1

-1 < 1-$$\frac{a}{10}$$ < 1

0 < a < 20 but the value of a cannot be 10

Sum of possible integral values of a =Sum of first 19 Natural Numbers - 10

=$$\frac{19*20}{2}$$ - 10

=180

From question, OD:DC = 2:3,  OD=4 cm, DC = 6cm

$$\angle BDC= 105 => \angle ADC = 180-105 =75 $$

In the circle, using sin rule in triangle ADC, $$\frac{AC}{sin \angle ADC} = \frac{DC}{sin \angle DAC}$$

=> $$\frac{11.52}{0.96} = \frac{6}{sin \angle DAC}$$

=> $$sin \angle DAC = 0.5 $$

$$\angle DAC=30^{\tiny 0}$$,

Now $$\angle DAC$$ and $$\angle BOC$$ are subtended from same chord, 2$$\angle DAC= \angle BOC$$

$$\angle BOC=2 \times 30^{\tiny 0}=60^{\tiny 0}$$

Area of $$\triangle BOC= \frac{1}{2} \times 10 \times 10 \times sin 60^{\tiny 0} = 25 \sqrt{3}$$

A is the answer.

To get the value of  [a/3]-[a/5]=2, [a/3] should be$$\geq$$2 because [a/3]$$\geq$$[a/5]

At a=6,  [a/3]=2 but [a/5]= 1. So [a/3]-[a/5] = 1

We will take the next multiple of 3.

At a=9, [a/3]=3 but [a/5]= 1. So [a/3]-[a/5] = 2.............(1st case)

In this question, we will check the multiples of 3 and 5 as these points will change the values of [a/3] and [a/5].
Given ‘a’ is a positive integer.

We can create a following table:

For any value greater than 20 the equation [a/3]-[a/5]=2 will be greater than 3 

Hence total of 8 values.

Answer A

All the internal angles of the hexagon are equal.
Let's calculate the measure of each internal angle.
Sum of the internal angles of a polygon of n sides = (n-2)*180
Each internal angle =$$\frac{(n-2)*180}{n}$$
=$$120^0$$
So the Exterior angle of angle P is $60^0$. APQ forms an equilateral triangle of side 1 cm.
Similarly, RBX forms an equilateral triangle of side 2 cm.
Length of AB =7 cm
ABC is an equilateral triangle of side 7 cm.
Length of CY = length of CB- length of YB
=7-4=3 cm
Length of PZ = Length of AC - (length of AP+ length of CZ)
7-(1+3)
=3 cm
Hence the length of ZY is 3 cm and PZ is 3 cm

A is the correct answer.

The coordinates of the rectangle are (3,2),(9,2),(3,6) and (9,6).

The line dividing it in two congruent parts should pass through the central point of the rectangle (3+9)/2=6 and (2+6)/2=4 i.e (6,4)

The slope of the line passing through (3,0) and (6,4) = 4/3=a/b

a+b=4+3=7

$$(\alpha+1)(\alpha+2)(\alpha+4)(\alpha+5)$$

$$[(\alpha+1)(\alpha+5)][(\alpha+2)(\alpha+4)]$$

$$[(\alpha)^2+6\alpha+5][(\alpha)^2+6\alpha+8]$$

Since $$\alpha$$ is the root of the equation $$\alpha^2+6\alpha+10$$=0   => $$\alpha^2+6\alpha$$ =-10

=> $$[(\alpha)^2+6\alpha+5][(\alpha)^2+6\alpha+8]$$ =[(-10+5)*(-10+8)]

=-5*-2=10

Let p,X,q,Y,r,Z,s be the row of apples, where X, Y, Z be the apples which are selected.

Let's analyse the values of p, q, r, s.

  p $$\geq$$ 0, q $$\geq$$ 3,r $$\geq$$ 3, s $$\geq$$ 0

p+q+r+s=30     (3 apples are already selected, hence 33-3=30)

p+q+3+r+3+s=30

p+q+r+s=24

Number of ways in which 24 distinct objects arranged in a row can be partitioned into 4 parts is $$^{24+4-1}C_{4-1}$$

=2925

B is the correct answer.

The smallest 4 digit number in base 5 = $$(1000)_5 = (125)_{10}$$

The largest 4 digit number in base 5 = $$(4444)_5$$ = $$(624)_{10}$$

The number of perfect cubes between 125 and 624 is 4 i.e 125, 216, 343, 512

4 is the correct answer.,

 Let 2a+1, 2b+1, 2c+1 be the number of pens received by each of the three boys.

2a+1+2b+1+2c+1=33   [0 $$\leq$$ a, b, c $$\leq$$ 15]

2a+2b+2c=30

a+b+c=15

Required number of ways  =$$^{15+3-1}C_{3-1}$$

=$$^{17}C_2$$

=136

A is the correct answer.

$$\mid \log_{(6x+4)}{(3x-2)}\mid$$=1

$$3x-2>0$$

$$x>2/3$$........(1)

$$  \log_{(6x+4)}{(3x-2)} =\pm$$1

Case 1

$$  \log_{(6x+4)}{(3x-2)} =$$1

$$\therefore 6x+4 = 3x-2$$

$$x=-2$$

which is not possible according to the equation 1

Case 2

$$  \log_{(6x+4)}{(3x-2)}=-$$1

$$\therefore \frac {1}{6x+4} = 3x-2$$

$$(6x+4)(3x-2)=1$$

$$18x^2=9$$

$$x=\pm \frac{1}{\sqrt2}$$

$$1/{\sqrt2}> 2/3$$

thus only one value of x i.e. $$1/{\sqrt2}$$ can satisfy the equation.

Initially, the distance traveled by Type 1 ball from the drop till the ground = 12m

The ball will bounce back two-third of the height= $$\frac{2\times12}{3}$$

The distance it will cover after the first bounce = 2$$\times \frac{2\times12}{3}$$

The total distance covered by Type 1 ball= 12+ 2$$\times \frac{2\times12}{3}$$+ 2$$\times\frac{2}{3}\times\frac{2\times12}{3}$$+.....

The total distance covered by Type 1 ball= 12+ 2$$\times$$12{ $$\frac{\frac{2}{3}}{1-\frac{2}{3}}$$}

The total distance covered by Type 1 ball= 60m

Similarly,

The total distance covered by Type 2 ball= 36+ 2$$\times \frac{1\times36}{2}$$+ 2$$\times\frac{1}{2}\times\frac{1\times36}{2}$$+.....

The total distance covered by Type 2 ball=108m

Total distance they will travel= 108+60=168m

From figure, FG is perpendicular of OA and GE is perpendicular to OB, where G is the centre of the semicircle. 

In OFGE, GE=GF=r (assuming radius semicircle = r) and all the angles are 90.

Hence OFGE is a square. 

=> OG = r$$\sqrt{2}$$.....

From symmetry, $$\angle$$ OGC = $$\angle$$ OGD = 90

In right triangle OGC, OG$$^2$$+CG$$^2$$=OC$$^2$$

 => OC$$^2$$ = r$$^2$$+ (r$$\sqrt{2})^2$$    (CG=r, OG = r$$\sqrt{2}$$)

=> R$$^2$$ = 3r$$^2$$.........(1)  (OC=R= radius of the quarter circle)

The area of quarter circle = $$\frac{1}{4} \pi$$R$$^2$$ 

The area of semicircle = $$\frac{1}{2} \pi$$r$$^2$$

Area of the shaded region = $$\frac{1}{4} \pi$$R$$^2$$ - $$\frac{1}{2} \pi$$r$$^2$$ 

= $$\frac{1}{4} \pi$$R$$^2$$ - $$\frac{1}{2*3} \pi$$R$$^2$$ (From 1, R$$^2$$ = 3r$$^2$$)

= $$\frac{1}{12} \pi$$R$$^2$$ = $$\frac{1}{12} \pi(2\sqrt{21})^2$$ = 22 cm$$^2$$

Assuming the number of males is 2x and the number of females is 3x. 

The number of males who make Rs 60000 can be assumed y and the number of females who make Rs 30000 can be assumed 4y. 

The ratio of the average salary of males to the average salary of females = $$\ \frac{\left(\ 60000y+30000\left(2x-y\right)\right)\times\ 3x}{2x\times\ \left(60000\left(3x-4y\right)\right)}$$ = $$\ \frac{\ 15}{16}$$

=> $$\ \frac{\ \left(2y+2x-y\right)\times\ 3}{\left(2\left(3x-4y\right)+4y\right)\times\ 2}=\ \frac{\ 15}{16}$$

=> 8x+4y = 15x-10y

=> 7x = 14y  =>x = 2y ..........(1)

Number of employees earning Rs 60000 = y+3x-4y = 3x-3y = 6y-3y = 3y ........(1)

Number of employees earning Rs 30000 =2x-y+4y = 2x+3y = 4y+3y  = 7y .........(2)

Average salary = $$\ \frac{\ 60000\times\ 3y+30000\times\ 7y}{3y+7y}$$ = 39000

Let a, b be the numbers turned up on blue and green dice respectively.

The number turned up on red dice = a+b

a+b $$\leq$$ 6 , a$$\geq$$ 1, b $$\geq$$ 1

Let a be c+1, b = d+1

c+1+d+1 $$\leq$$ 6

c+d$$\leq$$ 4

c+d+t = 4   ($$t\ge\ 0$$)

Number of non negative solutions = $$^6C_2$$ = 15

Total sample space = 6*6*6 = 216

Required probability = $$\frac{15}{216}$$ = $$\frac{5}{72}$$

A is the correct answer.

f(x) = $$x^2+7x+10$$
-2, -5 are the roots of f(x)
Now for f(f(x)) = 0,
$$x^2+7x+10$$ = -2 or $$x^2+7x+10$$ = -5
$$x^2+7x+12$$ = 0 or $$x^2+7x+15$$ = 0
x = -3, -4 and the second equation has imaginary roots
Now for f(f(x)) = 0,
$$x^2+7x+10$$ = -3 or $$x^2+7x+10$$ = -4
$$x^2+7x+13=0$$ or $$x^2+7x+14$$ = 0
Now both the equations has imaginary roots.
A is the correct answer.

Assume the volume of the solution just before pure alcohol replacement = V  and the volume of water = x
Then, the original volume of mixture = P = V-10 and the original volume of water = x-10
=> the volume of alcohol = V-x
After adding 10 litres of water the ratio reverses, which means that what was initially the proportion of alcohol in the total volume will become the proportion of water in the total volume after 10 litres of water is added.
Hence, $$\frac{V-x}{V-10}$$ = $$\ \frac{\ x}{V}$$
=>$$V^2=\ x\left(2V-10\right)$$
=> $$x\ =\ \frac{V^2}{2V-10}$$  ....(1)
After replacing mixture with 10 litres of pure alcohol, the proportion of water = $$\frac{2}{5}=\frac{x-\frac{10x}{V}}{V}$$
=> $$\frac{2V^2}{5\left(V-10\right)}=x$$  ......(2)
Equating (1) and (2),
$$\ \frac{\ 1}{2V-10}=\frac{2}{5V-50}$$
=> 5V-50 = 4V-20 
=> V = 30
Hence, P = V-10 = 30-10 = 20

The sum of the first 20 terms = $$\frac{20}{2}\left[2\times\ 6+\left(20-1\right)\times\ 2\right]220​[2× 6+(20−1)× 2]$$  = 500
After the sum increases by 56%, the new sum increase by 500*56/100 = 280
The common difference increases by = 12-2 = 10
Here the remaining terms = 20-x
Until xth term both correct and wrong series will be the same.

Assuming xth term = a
Correct series: (x+1)th term, (x+2)th term.........20th term = a+2, a+4, a+6.........
Wrong series: (x+1)th term, (x+2)th term.........20th term = a+12,a+24,a+36.......
Clearly, the difference is 10, 20, 30...........(20-x)*10

Hence, the increase = 10[1+2+3+.....20-x] = 10$$\frac{\left(20-x\right)\left(21-x\right)}{2}$$ = 280
=> (x-21)(x-20) = 56
=> $$x^2-41x+364\ =\ 0$$
=> x= 13, x=28
Rejecting the value greater than 20, we get x=13

Let the amount paid in each instalment be Rs x
The rate of interest is 50% paid semiannually,
The interest is 25%
The amount at the end of the first six months = 7200*1.25 = Rs 9000
(9000-x)1.25 = x
x = 5000

Assume the amount of water per litre mixture of R = x
After Q and R in the ratio 1:2, the ratio of water and alcohol  = 4:11
Assuming the volume of Q = 5 litres, then volume of R will be 10 litres,
Hence, water in 5 litres of Q = $$\frac{5}{1+4}\times\ 1$$ = 1 litre
Water in 10 litres of R = 10x
Hence total water in the mixture = $$\left(10+5\right)\times\ \frac{4}{4+11}$$  = 10x+1
=> x = 0.3
Now, if P, Q and R are combined in 3:5:2, then the volume can be assumed as 30, 50 and 20.
So amount of water in P = $$30\times\ \frac{2}{5}=\ 12$$
Similarly,  amount of water in Q = $$50\times\ \frac{1}{5}$$ = 10
The amount of water in R = $$20\times\ 0.3\ \ =6$$
Hence, the total amount of water in (30+50+20) = 100 litres mixture = 12+10+6 = 28 litres
Hence, the percentage  = 28

Assuming the cost price of 6 apples and marked price of 5 mangoes be 30x.
=> Cost price of 1 apple = 5x and marked price of 1 mango = 6x
Assuming the cost price of 5 mangoes and the marked price of 2 apples is 10y.
=> Cost price of 1 mango = 2y and marked price of 1 apple = 5y
The selling price of 6 apples and 2 free mangoes= 6*5y + 2*0= 30y
The cost price of 6 apples and 2 mangoes = 5*6x = 30x+2*2y = 30x+4y
Now, cost price$$\left(1+\frac{\Pr ofit}{100}\right)$$ = Selling price
=> (30x+4y)(1+$$\left(1+\frac{87.5}{100}\right)$$) = 30y
=> (30x+4y)(1+$$\left(1+\frac{7}{8}\right)$$) = 30y
=> 30x+4y = 16y
=> $$\frac{x}{y}=\frac{2}{5}$$
Hence, cost price of 1 mango = 2y, marked price of 1 mango = 6x = 6*$$\frac{2y}{5}$$ = 12y/5
=> Profit % = $$\frac{\left(\frac{12y}{5}-2y\right)}{2y}\times\ 100\ =\ 20\%$$

Assuming the number of employees in department P = p
Then the number of employees in department R = 2p
Now, the total number of employees in Q and R is 280.
Since the transfer is from R to Q, the total number of employees will still be 280.
Hence, the number of employees of R after transfer = 2p-20
=> The number of employees of Q after transfer = 280-(2p-20) = 300-2p
Now, the ratio of the number of employees in department P to that of Q will be the same as the ratio of the number of employees in department Q to that of department R.
=> $$\frac{p}{300-2p}=\frac{300-2p}{2p-20}$$
=> $$p\left(p-10\right)=\left(150-p\right)\left(300-2p\right)$$
=> $$p^2-10p=45000-600p+2p^2$$
=> $$p^2-590p+45000=0$$
=> (p-90)(p-500)=0
=> p=90  (Rejecting the 500, because the total number of employees in R will become 2p = 1000 but there is a condition that the total number of employees in Q and R is 280)
The number of employees in department P = 90
Hence the number of employees in department R = 180

We have, 

Denoting the length AB by $$a$$ and PB by $$b$$. (It is given that $$b$$ = 6-$$3\sqrt{2}$$ cm)
Now, the area of ABC = $$\frac{\sqrt{3}}{4}a^2\ $$ ....(1)
In the triangle APQ, AP = $$a-b$$
$$\tan\ 60^{\circ\ }=\frac{PQ}{AP}$$
=> PQ = ($$a-b$$)$$\sqrt{3}$$
Now, area of APQ = $$\frac{1}{2}\times\ \sqrt{3}\left(a-b\right)\left(a-b\right)$$ ......(2)
Equating (1) and (2),
$$\frac{1}{2}\times\ \sqrt{3}\left(a-b\right)\left(a-b\right)=\ \frac{\sqrt{3}}{4}a^2$$
=> $$a^2= 2(a-b)^2$$
=> $$a=\sqrt{2}(a-b)$$
=> $$b$$ = $$\frac{a}{\sqrt{2}}\left(\sqrt{2}-1\right)$$
=> $$a$$ = $$\frac{b\sqrt{2}}{\sqrt{2}-1}=b\sqrt{2}(\sqrt{2}+1)$$ = $$\sqrt{2}\left(6-3\sqrt{2}\right)\left(\sqrt{2}+1\right)$$ = $$6(\sqrt{2}-1)(\sqrt{2}+1)$$ = 6 cm

y = $$x^2 - 8x + 13$$
=$$x^2-2*x*4 + 16-3$$
=$$(x-4)^2-3$$
The graph of y at x = 4+p and x = 4-p is the same.
Hence the graph of y = $$x^2 + 8x + 13$$ is symmetric about x = 4
4 is the correct answer.

Two runners are running to and fro from points A and B at the speed 7x and 3x.

Let's consider the distance between A and B=10x

The first point of intersection= 7x

For the second point of intersection, the slower runner will cover the distance =d from the first point of intersection

The faster runner will cover 3x+3x+d distance.

Thus 

d/6x+d=3/7

7d=18x+3d

d=4.5x

4.5x=45

x=10

The distance between A and B=10x=100m

The sum of all two-digit numbers that give a remainder of A when they are divided by 7 is 654. This must be an AP of difference 7.

There are 90, two-digit numbers from 10 to 99. 90/7=12.85.

Thus there must be 12 or 13 terms.

Sum of n terms $$\frac{n}{2}\left(2a+\left(n-1\right)d\right)$$=654 Since, 13 is not a factor of 654

Thus n must be 12.

The first term = 14+A, Last term=  91+A

Sum of n terms=$$\frac{n}{2}\left(2\left(14+A\right)+\left(n-1\right)7\right)$$

$$6\left(2\left(14+A\right)+77\right)$$=654

A=2

Consider a point G such that the figure becomes symmetrical about the vertical line passing through A. 

From symmetry, FGCD is a rectangle.
Now, FG = CD which is the side of the pentagon.
Also, AF = AG = AB which is the side of the pentagon.
Hence, AF =  AG = FG 
=> AFG is an equilateral triangle. 
Hence, $$\angle FAG\ =\ 60$$
Now, $$\angle BAE\ =\ \frac{\left(5-2\right)}{5}\times\ 180\ =\ 108$$
$$\angle BAE\ =\ \angle BAF+\angle BAG\ +\ \angle GAE\ =\ 108$$
=> x+60+x = 108
=> x = 24
Now, $$\angle FAE\ =\ \angle FAG+\angle GAE\ =\ 60+24\ =\ 84$$

Let p1, p2, p3 be distinct prime numbers which are not equal to 5.

From the given information, we can conclude that N can be

Case 1: N= p1*p2*p3

Case 2: N= $$p1^3*p2$$

Case 3: N=$$p1^7$$

Now, 2 can or cannot be one of p1,p2,p3

In case 1:

If $$2\ne\ $$ p1/p2/p3 then number of factors of 2*p1*p2*p3=16.

If 2 is equal to p1 or p2 or p3 then the number of factors of 2*N will be 12.

In case 2:

If $$2\ne\ $$ p1 or p2, then number of factors of 2*N=16.

If 2=p1, the number of factors of 2*N will be 10

If 2=p2, the number of factors of 2*N will be 12.

Case 3:

If $$2\ne\ $$ p, the number of factors of 2*N=16.

If 2=p1, the number of factors of 2*N = 9

.'. The number of factors of 2N can be 9,10,12,16.

f(0) = 1 = $$2\times\ 0^2+1$$

f(1) = 3 = $$2\times\ 1^2+1$$

f(2) = 9 = $$2\times\ 2^2+1$$

f(3) = 19 = $$2\times\ 3^2+1$$

Since f(x) is a 4 degree polynomial,

f(x) - ($$2x^2+1$$) has 4 roots 0,1,2 and 3

thus f(x) = $$a\left(x\right)\left(x-1\right)\left(x-2\right)\left(x-3\right)\ +\ 2x^2+1$$

Coef of $$x^4$$ is 2 in f(x) thus a =2

f(x)= $$2\left(x\right)\left(x-1\right)\left(x-2\right)\left(x-3\right)\ +\ 2x^2+1$$

f(-1) = $$2\left(-1\right)\left(-2\right)\left(-3\right)\left(-4\right)+\ 2\left(-1^2\right)+1$$

f(-1) = 48+2+1 = 51

The problem figure in the question can be drawn as:

If we construct an equilateral triangle with base CD, we will get:

AM and DK are parallel to each other (makes 60 degrees with the base) and AD and MK are parallel to each other. 

.'. AMKD is a parallelogram and hence AM= DK and AD= MK.

But since AM=AB=AD.

So, AMKD is a rhombus. 

Also, since MK= KD and M and D are the points on the circle, K must be centre of the circle and so, the radius= KD= CD= 12 cm.

Rectangle has sides 9 and 12. Thus diagonal AC = $$\sqrt{\ 9^2+12^2}=15$$

3 AE =AC. Thus, AE =5. Also CF =5

Given 2 AG =15, thus AG =7.5

We can say from this that E and F trisect the diagonal AC. F is the midpoint of diagonal AC

Area of $$\bigtriangleup\ABC $$ = $$\frac{1}{2}\left(AB\right)\left(BC\right)=\frac{1}{2}\left(9\right)\left(12\right)\ =\ 54$$

It can also be written as $$\frac{1}{2}\left(AC\right)h\ =54$$

$$7.5h\ =54$$...(I)

Triangle ABC and BGF have base on a common line and share the same vertice B. Thus height will be same

GF = CG-CF = 7.5-5 = 2.5

Area = $$\frac{1}{2}2.5h$$ = $$\frac{1}{2}2.5\left(\frac{54}{7.5}\right)$$ = 9

S= $$1^3+2^3+3^3+...\ 1000^3$$.

We know that the sum of cubes of natural number starting from 1 upto n is given by: $$\left[\frac{n\left(n+1\right)}{2}\right]^2$$.

Here, n= 1000. So, S= $$\left[\frac{1000\left(1001\right)}{2}\right]^2$$

=>S= $$\left[\frac{10^3\left(11\times7\times13\right)}{2}\right]^2$$

=> S= $$\left[\frac{2^3\times5^3\left(11\times7\times13\right)}{2}\right]^2$$

=> S= $$\left[2^2\times5^3\left(11\times7\times13\right)\right]^2$$

.'.S= $$2^4\times5^6\times11^2\times7^2\times13^2$$

So, for even factors, we need at least one 2 among the four 2s available with S. We can select either one, two, three or four 2s. For the other prime factors, we also have the option to choose 0 out of the available number of 5s, 11s, 7s and 13s.

So, total even factors= $$4\times7\times3\times3\times3$$= 756.

Let the thickness of the small, medium and large pizzas be $$t_1,\ t_2,\ t_3$$ respectively.

The diameter of small, medium and large pizzas= 8, 11 and 14 inches.

Volume of any pizza= Area$$\times\ $$thickness= $$\pi\ r^2\times t\ $$

Volume of small pizza= $$\pi\ \left(\frac{8}{2}\right)^2\times t_1=16\pi t_1\ \ $$

Volume of the medium pizza= $$\pi\ \left(\frac{11}{2}\right)^2\times t_1=\frac{121}{4}\pi t_2\ \ $$

Volume of the large pizza= $$\pi\ \left(\frac{14}{2}\right)^2\times t_1=49\pi t_3\ \ $$

$$P_1:P_2:P_3=V_1:V_2:V_3\ $$

=> 2:3:4= $$16\pi t_1:\ \frac{121}{4}\pi t_2:49\pi t_3=2:3:4\ \ \ $$

64t1=2x
121t2=3x
196t3=4x
$$\text{}t1:t2:t3\ =\ \frac{1}{32}:\frac{3}{121}:\frac{1}{49}$$

Tap 1 fills alone in 8 hours, tap 2 empties alone in 12 hours and tap 3 fills the entire tank alone in 4 hours. 

Let there be 24 L of water(LCM of 8, 12 and 4) in the tank. 

Tap 1 fills 3L in an hour, tap 2 empties 2L in an hour and tap 3 fills 6L in an hour.

Now irrespective of where taps 1 and 3 are located, they will begin filling the tank from the beginning but tap 2 will start working only when the tank is half filled.

So, let the time taken by tap 1 and 3 to fill the tank by 50% be 't' hours.

(3+6)t= 12

=> t= $$\frac{12}{9}=\frac{4}{3}\ $$ hours.

Now, in 1 hour tap 1 and 3 will fill 9L of water and tap 2 will empty 2L in an hour. In total, all three operating together will fill 7L of water per hour.

.'. Time taken by all the three taps to fill the other half of the tank, 'x' can be found using 7x= 12L

.'.x = $$\frac{12}{7}\ $$ hours.

Total time to fill the tank= t+x= $$\frac{12}{7}+\frac{4}{3}=\ \frac{\ 36+28}{21}=\frac{64}{21}=3\frac{\ 1}{21}\ $$ hours

Selecting squares of side 1 unit each:

We can do so by selecting 1 unit from the 8 squares along the length and 1 unit from 8 squares along the breadth. So, total ways of selecting a square of length 1 unit= $$^8C_1\times^8C_1=\ 8\times8=\ 64\ \ $$

Selecting squares of side 2 units each:

We can do so by selecting 2 continuous units from the 8 squares along the length and 2 continuous units from 8 squares along the breadth. So, total ways of selecting a square of length 1 unit= $$7\times7=49$$

Selecting squares of side 3 units each:

We can do so by selecting 3 continuous units from the 8 squares along the length and 3 continuous units from 8 squares along the breadth. So, total ways of selecting a square of length 1 unit= $$6\times6=36$$

Selecting squares of side 4 units each:

We can do so by selecting 4 continuous units from the 8 squares along the length and 4 continuous units from 8 squares along the breadth. So, total ways of selecting a square of length 1 unit= $$5\times5=25$$

.'. Total ways of selecting a square of side length at most 4 units from a chess board= 64+49+36+25= 174

First, we will deal with the inner modulus. The critical point of the inner modulus is -2.

Case 1: x$$\ge\ $$-2

|(x-2)-|x+2||-5 becomes |(x-2)-x-2|-5=|-4|-5=-1 which is always <0

Therefore for x$$\ge\ $$-2, the inequality does not holds.

Case 2: x<-2

|(x-2)-|x+2||-5 becomes |(x-2)+x+2|-5=|2x|-5.

So, |2x|$$\ge\ $$5 is satisfied under 2 conditions:

(i) 2x$$\le$$-5 => x$$\le$$ $$\frac{-5}{2}$$. We have already asumed that x<-2 and upon solving, we get x $$\le$$ $$\frac{-5}{2}$$. 

.'. x$$\le$$ $$\frac{-5}{2}$$ holds true.

(ii) 2x>5 => x>$$\frac{5}{2}$$. We have already assumed that x<-2 and upon solving, we get x>$$\frac{5}{2}$$ which contradicts our main assumption. So, we reject this.

So, the range of x is $$\left(-\infty\ ,\frac{-5}{2}\right]$$

The company purchased 30 barrels of oil for 6 months. So, it has 180 barrels of oil in total. 

Cost of 1 barrel of oil in March= $25 and the cost per barrel increases by $10 every month.

So, it follows an A.P. with first term, 'a'= $25 and common difference, 'd'= $10 and 'n', number of terms= 6 (March to August).

If someone purchased 1 barrel a month, his total cost= Sum of A.P.= $$\frac{n}{2}\left[2a+\left(n-1\right)d\right]=\frac{6}{2}\left[50+\left(5\times10\right)\right]=3\times100=300\ \ \ $$

.'. Cost of 30 oil barrels every month= $300$$\times\ $$30= $9000

Now, half of the oil barrels, i.e. 90 barrels were sold at $45 each making a sale of $90$$\times\ $$45= $4050.

Remaining barrels= 90.

Half of it were, i.e. 45 barrels were sold at $50 each making a sales of $50$$\times\ $$45= $2250

All the remaining barrels, 45 barrels were sold at $70 each making further sales of $70$$\times\ $$45= $3150

Total Selling Price= $(4050+2250+3150)= $9450.

Profit= $9450-$9000= $450

Profit %= $$\ \frac{\ \text{Profit}}{\text{CP}}\times100\ $$=$$\ \frac{\ 450}{9000}\times100=\ \frac{450}{90}=5\%\ $$

The first function is 

y = $$x^2+2x+1\ =\ \left(x+1\right)^2$$ Plotting the graph we get,

The second function is 

y = |x| + 1. Plotting the graph we get,

They meet at 2 points, first time in the second quadrant, and the second time at (0,1).

Let's find out where they meet in the first quadrant,

$$x^2+2x+1=-x+1$$

$$x^2+3x=0$$

$$x\left(x+3\right)=0$$

x = -3, 0.

At x = -3, y = |x|+1 = 4.

So, they meet at (-3,4).

So, drawing both the graphs together, we get,

Hence, 

f(x) = $$\left(x+1\right)^2$$ for x$$\in\ \left(-\infty\ ,-3\right)U\left(0,\infty\ \right)$$

f(x) = |x| + 1 for x$$\in\ \left[-3\ ,0\right]$$

Hence the minimum value of f(x) is 1 at x = 0.

xy + 5x + 6y < 20

If we add 30 to both sides, we can easily factorize the left side.

xy + 5x + 6y + 30 < 50

x(y+5) + 6(y+5) < 50

(y+5)(x+6) < 50

(x+6)(y+5) < 50

It has been given that x is a whole number and y is a natural number.

Hence, let us assume the above inequality as AB < 50, where A = x+6 and B = y+5

Since, x$$\ge$$0 and y$$\ge$$1, the minimum value of A = 6 and the minimum value of B = 6

Hence, possible ordered pairs of A and B satisfying the inequality is,

6 x 6 < 50

6 x 7 < 50

6 x 8 < 50

7 x 6 < 50

7 x 7 < 50

8 x 6 < 50

Hence, (A,B) can be (6,6), (6,7), (6,8), (7,6), (7,7) or (8,6).

Hence, ordered pair (x,y) can be (0,1), (0,2), (0,3), (1,1), (1,2), (2,1).

Hence, count = 6

Let the total cost be Rs. x

Labour cost= 0.05x

Tyre replacement cost= 0.15x

Accessory replacement cost= 0.25x

Servicing cost= 0.35x

Oil replacement cost= x-(0.05x+0.15x+0.25x+0.35x)= 0.2x

After discount, labour cost= 0.9$$\times$$0.05x= 0.045x

Servicing cost= 0.9$$\times$$0.35x= 0.315x

Oil replacement cost= 0.9$$\times$$0.2x= 0.18x

Accessory replacement cost= 0.8$$\times$$0.25x= 0.2x

'.' No discount is given on replacing tyres, its cost remains the same at 0.15x

Total amount payable by Prasoon= 0.045x+0.315x+0.18x+0.2x+0.15x= 0.89x

Total amount reimbursed by the service center= 0.8$$\times$$0.89x=0.712x

Given, 0.712x=56960

.'.x= Rs. 80000

Original accessory replacement cost=0.25x= 0.25$$\times$$80000= Rs. 20000

The ratio of A, B, C and D in X, Y and Z is given to be

Since 1 unit of M has 3X, 4Y and 6Z, we can show the contribution of A, B, C and D in 1 unit of M or 3 units of X, 4 units of Y and 6 units of Z.

Therefore 1 unit of M has $$\frac{147}{60}$$ units of A, $$\frac{201}{60}$$ units of B, $$\frac{12}{5}$$ units of C and $$\frac{24}{5}$$ units of D.

Also, the total quantity of 1 unit of M= (3+4+6) units= 13 units.

.'. Percentage contribution of A in 1 unit of M= $$\frac{\frac{147}{60}}{13}\times100=49\times\frac{5}{13}\%=18.84\%\ $$

Let the five kids be given A, B, C, D and E notebooks.

As per the demands, A= 1+a; where a$$\ge$$0

B=2+b; where b$$\ge$$0

C= 3+c; where c$$\ge$$0

D= 4+d; where d$$\ge$$0 and

E= 5+e; where e$$\ge$$0

Now, A+B+C+D+E= 22

=>(1+a)+(2+b)+(3+c)+(4+d)+(5+e)= 22

=>15+a+b+c+d+e= 22

=>a+b+c+d+e= 7

This is similar to distributing n similar objects to r different people. This can be done in $$^{\left(n+r-1\right)}C_{\left(r-1\right)}$$

In our case, n= 7, r=5.

Total ways of distribution= $$^{11}C_4=\ \frac{\ 11\times10\times9\times8\ \ \ }{4\times3\times2\times1\ \ \ }$$= 330

Hence,

40 x 25 + 60 x a + 100 x b = 200 x 56

1000 + 60a + 100b = 11200

100 + 6a + 10b = 1120

6a + 10b = 1020 .... (i)

Hence,

100 x 56 + 40 x a + 60 x b = 200 x 59.5

5600 + 40a + 60b = 11900

560 + 4a + 6b = 1190

4a + 6b = 630 .... (ii)

(i) x 2 - (ii) x 3

12a + 20b = 2040

12a + 18b = 1890

2b = 150

b = 75%

Hence, a = 45%

Hence, b - a = 30%

Given, $$2^{\left(1+3+5+...x\right)}=\left(0.0625\right)^{-25}$$

Taking log on both sides, we will get,

$$\left(1+3+5+...x\right)\log2=\ -25\log0.0625\ $$

=>$$\left(1+3+5+...x\right)\log2=\ -25\log\left(\frac{1}{16}\right)\ $$

=>$$\left(1+3+5+...x\right)\log2=\ -25\log\left(2^{-4}\right)\ $$

=>$$\left(1+3+5+...x\right)\log2=\ 100\log2\ $$

=> Sum of odd numbers upto x= 100.

Let the total odd numbers be n, then sum of n odd numbers starting from 1= $$n^2$$

So, $$n^2$$=100 or n, Number of odd terms= 10.

.'. x is the 10th odd number from the beginning= 19.

Since the first 4 digits are fixed, all the arrangement and selection is to be done for the next 6 digits.

Now, we can select the 6 digits from all 10 digits, without repetition in $$^{10}C_6$$ ways.

Once, we select any set of 6 numbers, there is only one arrangement in which they appear in ascending order. So, for each possible selection, the number of ways = 1.

Hence, the total number of ways of selecting 10-digit numbers = $$^{10}C_6\times\ 1=^{10}C_6=^{10}C_4=\frac{10\times\ 9\times\ 8\times\ 7}{4\times\ 3\times\ 2\times\ 1}=210$$

Let us assume that the student answers all the questions.

If he has answered all of the questions correctly, he gets a total of 34 x 3 = 102 marks.

Now, for every wrong answer, there is a subtraction of 4 marks, so for n incorrect answers, marks achieved = 102 - 4n

Now, we need to equate this to 67, which is the actual marks achieved. But this is not possible, because 67 is odd, and 102-4n is even

Similarly, we can assume he had attempted 33 questions, with n wrong answers,

33 x 3 - 4n = 67

99 - 4n = 67

4n = 32, n = 8. This is possible.

For 32 attempted questions also,

32 x 3 - 4n = 67 is not possible because left side is even and right side is odd.

For 31 attempted questions,

31 x 3 - 4n = 67

93 - 4n = 67

4n = 26, not possible

Similarly, 30 attempted questions is not possible.

For 29 attempted questions,

29 x 3 - 4n = 67

87 - 4n = 67

n = 5. This is possible.

Again for 28, 27, 26 attempted questions, it is not possible.

For 25 questions attempted,

25 x 3 - 4n = 67

75 - 4n = 67

n = 2

It is not possible for any other combination.

Hence, it is possible in 3 different ways.

It is given that the base has a diameter of 16 units, it implies that the radius is 8 units and hence the circumference is $$16\pi$$ units

If we cut open the cylinder into a 2-D paper then, the ant has to travel a horizontal distance of half the circumference = $$8\pi$$ units. and vertical distance will be $$6\pi$$. 

In order to minimise the distance, he has to travel along the hypotenuse formed which is of length = $$\sqrt{\ \left(8\pi\ \right)^2+\left(6\pi\ \right)^2}=\sqrt{\ 100\pi^2}=10\pi$$

g(x) is formed by shifting f(x) towards the right along the x-axis by 5 units,

g(x) = $$3\left(x-5\right)^2+9\left(x-5\right)+3$$

g(x) = $$3\left(x^2-10x+25\right)+9\left(x-5\right)+3$$

g(x) = $$3x^2-30x+75+9x-45+3$$

g(x) = $$3x^2-21x+33$$

Hence g(x) = 0

$$3x^2-21x+33\ =\ 0$$

$$x^2-7x+11\ =\ 0$$

Hence, sum of the roots = $$\frac{-\left(-7\right)}{1}=7$$

Given, a= $$\log_{125}40$$

=>$$\log_{5^3}\ \left(5\times2^3\right)=a$$

=>$$\frac{1}{3}\log_5\ \left(5\times2^3\right)=a$$

=>$$\log_5\ 5\ +\log_52^3=3a$$

=> $$3\log_52=3a-1$$

.'. $$\log_52=a-\frac{1}{3}$$        .......(1)

b= $$\log_598$$

=> $$\log_5\left(7^2\times2\right)\ =b$$

=> $$\log_5\left(7^2\right)\ +\log_52\ =b$$

=> $$2\log_57\ +a-\frac{1}{3}\ =b$$

=> $$\log_57\ =\frac{b}{2}-\frac{a}{2}+\frac{1}{6}$$         ......(2)

Adding (1) and (2), we get: 

$$\log_57\ +\log_52=\frac{b}{2}-\frac{a}{2}+\frac{1}{6}+a-\frac{1}{3}$$

=> $$\log_514=\frac{b}{2}+\frac{a}{2}-\frac{1}{6}$$

=> $$\log_514=\frac{\left(3b+3a-1\right)}{6}$$

=> $$\log_{14}5\ =\ \frac{6}{3a+3b-1}$$

Let us look at the inequality

$$ |x+3| < 7 $$ or $$ -7 < x+3 < 7 $$ or $$ -10< x < 4 $$ ...(I)

Similarly $$ |y-4| < 8$$ , $$ -8 < y-4 < 8 $$ or -4< y < 12 $$ ...(II)

Coming to the equation,

$$x+ xy+ y $$ = $$x + xy+ y +1 -1 $$

$$x+ xy+ y $$ = $$\left(x+1\right)\left(y+1\right)-1$$

The value will be max when both $$(x+1) $$ and $$(y+1) $$ are either positive or negative. for it to be minimum, one has to be positive and other has to be negative

From (I) and (II) we can say that,

$$ -9 < x+1 < 5$$ and $$ -3 < y+1 < 13 $$

Therefore (x+1) = [-8,-7,-6,-5,......2,3,4] and (y+1) = [-2,-1,0,....11,12] as $$x$$ and $$y$$ are integers

(x+1)(y+1) will have max value of 48 and min value -96

(x+1)(y+1)-1  will have max value of 47 and min value -97

Difference = 47-(-97) = 47+97 = 144

Let ABCD be a kite with the uncommon sides be 7 and 24. And point at which diagonals intersect is O.

It can be shown as

One of the diagonal has a length of 25. It is possible when AC =25 only. If DB = 25 then AD+BD =14 < DB=25 which violates triangle inequality.

Thus we observe that ADB and ABC is a right-triangle as $$7^2+24^2=\ 25^2$$

Area of the kite = $$2\times\ \frac{1}{2}\left(7\right)\left(24\right)\ =168\ $$ sq units.

From symmetry $$\bigtriangleup AOD $$ and  $$\bigtriangleup AOB$$ are congruent and hence $$\angle\ AOD\ =\angle\ AOB=90^0$$

Area of kite = $$\frac{1}{2}\left(AC\right)\left(DO\right)+\frac{1}{2}\left(AC\right)\left(OB\right)=\ \frac{1}{2}\left(AC\right)\left(DB\right)=\ 168$$

$$\ \frac{1}{2}\left(25\right)\left(DB\right)=\ 168$$

$$DB=\ \frac{168}{12.5}=13.44$$ units

We are given with $$S_n=2n^3+n^2+3n+1$$.

We know that any term $$t_n$$ can be found using $$t_n=S_n-S_{n-1}$$.

Here, $$t_x=S_x-S_{x-1}$$

So, $$2x^3+x^2+3x+1-\left[2\left(x-1\right)^3+\left(x-1\right)^2+3(x-1)+1\right]=84$$

=> $$2x^3+x^2+3x+1-\left[2\left(x^3-1-3x^2+3x\right)+x^2+1-2x+3x-3+1\right]=84$$

=> $$2x^3+x^2+3x+1-2x^3+2+6x^2-6x-x^2-1+2x-3x+3-1=84$$

=>$$4+6x^2-6x+2x=84$$

=> $$6x^2-4x=80$$

=> $$3x^2-2x-40=0$$

=> $$\left(x-4\right)\left(x+\frac{10}{3}\right)=0$$

'.' x cannot be negative. 

.'. x=4.

The differences between the terms of the series are in AP.

1 - 3 = -2

-2 - 1 = -3

-6 + 2 = -4 and so on.

We represent such a series as 

T(n) = $$an^2+bn+c$$

Substituting n = 1,2,3, we get

a + b + c = 3 ....(i)

4a + 2b + c = 1 ....(ii)

9a + 3b + c = -2 ....(iii)

(ii) - (i), we get

3a + b = -2

(iii) - (i), we get

8a + 2b = -5

Solving these 2 equations, we get,

a = -0.5, b = -0.5

Putting these values in equation (i), we get c = 4.

Hence, T(n) = $$-0.5n^2-0.5n+4$$

S(n) = $$-0.5\frac{n\left(n+1\right)\left(2n+1\right)}{6}-0.5\frac{n\left(n+1\right)}{2}+4n$$

S(n) = $$-0.5\frac{50\times\ 51\times\ 101}{6}-0.5\frac{50\times\ 51}{2}+4\times\ 50=-21900$$

According to the information,

$$70\le\frac{65x+78y}{x+y}\le72$$

This breaks down to

$$70\le\frac{65x+78y}{x+y}$$ ....(i)

and 

$$\frac{65x+78y}{x+y}\le72$$ ....(ii)

Solving i,

$$\frac{x}{y}\le\frac{8}{5}$$

Solving ii,

$$\frac{x}{y}\ge\frac{6}{7}$$

Hence,

$$\frac{6}{7}\le\frac{x}{y}\le\frac{8}{5}$$

We have been given the range of values y can take, putting the extremes, we get,

When y = 625,

$$\frac{6}{7}\times\ 625\le x\le\frac{8}{5}\times\ 625$$

$$535.7\le x\le1000$$

When y = 630,

$$\frac{6}{7}\times\ 630\le x\le\frac{8}{5}\times\ 630$$

$$540\le x\le1008$$

Hence, the minimum value x can take is 536 and the maximum it can take is 1008.

Hence difference = 472.

Let the newspapers be abbreviated as follows:

The Telegraph - TT

The Economic Times - ET

The Times of India - TOI

Let the number of people in the society be 100 (only for calculation, later we will substitute the correct value)

Therefore, n(TT) = 20

n(ET) = 30

n(TOI) = 40

To maximise the number of people reading no newspaper, we will have to maximise the number of people who read 3 and 2 newspapers as follows.

Hence, maximum number of people who read no newspapers = 60 = 60%.

Total number of people in the society = 600

Hence, the maximum number of people in the society who read none of the newspapers = 0.6 x 600 = 360

For any right-angled triangle with hypotenuse=h and the other two sides= 'a' and 'b', we know that:

Circumradius= $$\frac{h}{2}$$ and inradius= $$\frac{a+b-h}{2}$$

So, h= 2(10) cm= 20 cm.

Also, (a+b-20)=2(4)

=> a+b-20=8

.'. a+b= 28      ....(1)

Also, we have: $$a^2+b^2=h^2$$

From equation (1), we have $$a^2+\left(28-a\right)^2=20^2$$

=> $$a^2+784+a^2-56a=400$$

=> $$2a^2+384-56a=0$$

=>$$a^2-28a+192=0$$.

The two roots of this equation will be a and (28-a) which is nothing but a and b. Also, the product of the roots then become 192.

Area of a right angled triangle with non-hypotenuse sides 'a' and 'b'= $$\ \frac{\ 1}{2}ab$$= $$\frac{192}{2}=96$$ sq. cm

Let us draw a rectangle on the coordinate axis. With (0,0) be one of the edge A. B can be (6,0), C(6,12) and D(0,12).  E and F divide the AC in 3 equal part. E has to be (2,4) and F (4,8)

G is  centroid of DEF, Coordinates of G will be $$\left(\frac{0+2+4}{3},\frac{12+8+4}{3}\right)=\left(\frac{6}{3},\frac{24}{3}\right)\ =\ \left(2,8\right),$$

Upon finding G the required diagram looks like this

Area DGBC = Area GBC + Area DCG

Area DGBC = $$\frac{1}{2}\left(12\right)\left(4\right)$$ + $$\frac{1}{2}\left(6\right)\left(4\right)$$

Area DGBC = 24+12 = 36 sq units

The divisibility rule of 3 states that the sum of digits has to be divisible by 3 in order to be divisible by 3.

Such numbers are 1002,1005,1008.....9999 : 3000 numbers

The divisibility rules of 8 imply that the last 3 digits have to be divisible by 8: Thus "dcba" has to be divisible by 8. Such number are : 1000,1008,..9992 : 1125

We have included the number which are divisible by 24 twice. Thus we have to remove those extra numbers

Such number are : 1008,1032,...9984: 375 numbers

The numbers of 4-digit number  which are divisible by 3 or 8: 3000+1125-375 = 3750

There are a total of 11 alphabets in VENKATESHAN which when arranged in ascending order are A, A, E, E, H K, N, N, S, T and V.

Total possible permutation are =  $$\frac{11!}{2!\ 2!\ 2!}$$

Now we need to find the cases where all the "A"s are before "E"s. There are total of 2 A and 2 E. If we pick places for A and E, they can be placed in one way. 4 places can be picked in $$^{11}C_4$$ ways.

Remaining 7 alphabets can be arranged in $$\frac{7!}{2!}$$ ways.

The number of ways possible  = $$^{11}C_4\times\ \frac{7!}{2!}\ =\ \frac{11!}{7!\ 4!}\times\ \frac{7!}{2!}=\frac{11!}{4!\ 2!}$$

Probability = $$\frac{11!}{4!\ 2!}\div\frac{11!}{2!\ 2!\ 2!}$$

Probability = $$\frac{11!}{4!\ 2!}\times\ \frac{2!\ 2!\ 2!}{11!}=\ \frac{1}{6}$$

A and B start on bike and A drops B at R. The time taken is $$T_1$$. Hence, PR = 40$$T_1$$. 

At the same time, C walks from P to Q. Distance covered = 5$$T_1$$.

Distance between A and C = 40$$T_1$$ - 5$$T_1$$ = 35$$T_1$$.

Relative speed between A and C = 40 + 5 = 45.

Time taken to meet = 35/45$$T_1$$ = 7/9$$T_1$$

Now, in the meantime, B also walks from R to T. His speed is same as C, so distance travelled is same, i.e, 5 x 7/9$$T_1$$ = 35/9$$T_1$$.

Hence, the distnce between the bike and B when the bike starts from S is ST, i.e, 35$$T_1$$.

Now, let they meet at Z after time $$T_2$$.

Relative speed = 40 - 5 = 35.

Hence, 35 $$T_2$$ = 35$$T_1$$

$$T_2$$ = $$T_1$$

Hence, time taken by the overall journey = $$T_1$$ + 7/9$$T_1$$ + $$T_1$$ = 25/9$$T_1$$

Time spent by B walking = Toatl time - $$T_1$$ = 25/9$$T_1$$ - $$T_1$$ = 16/9$$T_1$$.

Hence, percentage time = 16/25 x 100 = 64%.

Alternate solution:

Let us divide all the activities into certain time periods for simplification. Let all of them start at time t = 0.

1) From time t = 0 to $$t_1$$. A and B will move towards Z on bike and C will walk towards Z. At exactly $$t_1$$, A will drop B and start moving back to pick up C. Distance covered by A = $$40t_1$$, Distance covered by B = $$40t_1$$ and Distance covered by C = $$5t_1$$

2) From time t = $$t_1$$ to $$t_2$$. B and C will be walking towards Z. A will be alone on bike moving away from city Z as he is going to pick up C. Distance covered by B = $$5(t_2 - t_1)$$, Distance covered by C = $$5(t_2 - t_1)$$ and distance by A in this time period = $$-40\left(t_2-t_1\right)$$ . We have taken"-" as A is moving in opposite direction

3) From time t = $$t_2 $$ to $$t_3$$. A has picked up C and now both are on bike moving towards city Z. 0. B is walking towards city Z. They all reach at city Z at same time $$t_3$$Distance travelled by A = $$40\left(t_3-t_2\right)$$. Distance covered by B = $$5(t_3 - t_2)$$ . Distance travelled by C = $$40\left(t_3-t_2\right)$$

Let the overall distance be $$d$$

Distance travelled by A = $$40t_1$$ $$-40\left(t_2-t_1\right)$$ + $$40\left(t_3-t_2\right)$$ = $$d$$

Thus $$40\left(2t_1+t_3-2t_2\right)=d$$ .....(I)

Distance travelled by B = $$40t_1$$ + $$5(t_2 - t_1)$$ + $$5(t_3 - t_2)$$

$$35t_1+5t_3=d$$ ....(II)

Distance travelled by C = $$5t_1$$ + $$5(t_2 - t_1)$$ + $$40\left(t_3-t_2\right)$$ = $$d$$

$$40t_3-35t_2=d$$ ...(III)

Equating (II) and (III) 

$$35t_1+5t_3=40t_3-35t_2$$

$$t_1=t_3-t_2$$ ....(IV)

Equating (I) and (III)

$$80t_1+40t_3-80t_2=40t_3-35t_2$$

$$ \implies$$ $$80t_{1\ }=45t_2$$

Or $$t_2=\frac{16}{9}t_{1\ }$$....(V)

Putting (V) in(IV)

$$t_1=t_3- \frac{16}{9}t_1$$

$$ \frac{25}{9} t_1 = t_3$$.

From time $$t_1 to t_3$$ B was walking and total journey time was $$t_3$$

% time on walking = $$\frac{t_3-t_1}{t_3}\times\ 100\%$$

% time on walking = $$\left(1-\frac{t_1}{t_3}\right)\times\ 100\%$$ = $$\left(1-\frac{9}{25}\right)\times\ 100\%$$

% time walking = $$\frac{16}{25}\times\ 100\%\ =64\%$$

Let us start by writing the N in empirical form which is $$2^5\times\ 3^8\times\ 2^4\times\ 5^3\times\ 2\times\ 3\times\ 7^2$$

Which equals $$2^{10}\times\ 3^9\times\ 5^3\times\ \ 7^2$$

Let A = number of factors not divisible by 9

B = Number of Factors which are odd

We have to find $$\frac{P\left(A\cap B\right)}{P\left(A\right)}$$

$$ (A \cap B )$$ = Number of factors not divisible by 9 and are odd. It implies it does not have 2 as a factor and the highest power of 3 can be  1. Such type of factors are = (1+1)(3+1)(2+1) = 24

A = Number of factors that are not divisible by 9. Thus the maximum power of 3 they can have is 1. Such number are = (10+1)(1+1)(3+1)(2+1) = 264

Probability = $$\frac{24}{264}=\frac{1}{11}$$

m+n=12

For a logarithm function to be defined, we need the base to be greater than 0. So x+1 > 0 or x > -1.

As the base can't equal 1, x+1 has to be greater than 1 or x>0.

x can be 1,2,3,4.... (natural numbers)

Also, as the argument of a logarithm can't be equal to 0, x can't equal 3 or 4.

So x belongs to {1,2,5,6,7,8,9,...}

Now, for $$\log_{\left(x+1\right)}\left(x-4\right)\left(x-3\right)\ <1$$,

(x-4)(x-3)<(x+1)

=> $$x^2-7x+12<x+1$$

=> $$x^2-8x+11<0$$

=> $$x^2-8x+16<5$$

=> $$\left(x-4\right)^2<5$$

For the above inequality to satisfy, we can only have 3 natural number values in the form of x= 2, 5 or 6.

Alternate Solution:

We have the given inequality: $$\log_{\left(x+1\right)}\left(x-4\right)\left(x-3\right)\ <0$$

For this to be defined,we need to satisfy the basic conditions- For any log function $$\log_ab$$, 

(i) a>0  (ii) a$$\ne$$1  and (iii) b>0

So, in $$\log_{\left(x+1\right)}\left(x-4\right)\left(x-3\right)\ <1$$,

(i) x+1>0 

=> x>-1.

(ii) x+1$$\ne$$1

=> x$$\ne$$0.

(iii) (x-3)(x-4)>0

.'. $$x\in\left(-\infty,3\right)U\left(4,\infty\right)$$.

Combining all three, we have the domain as: $$x\in\left(-1,0\right)U\left(0,3\right)U\left(4,\infty\right)$$.

Now, for $$\log_{\left(x+1\right)}\left(x-4\right)\left(x-3\right)\ <1$$ to be satisfied, we need to have:

$$\frac{\left(x-4\right)\left(x-3\right)}{\left(x+1\right)}<1$$ ....(1)

and 0<(x-4)(x-3) ....(2)

Solving (1), we get:

 $$\frac{x^2-7x+12}{\left(x+1\right)}-1<0$$

=> $$\frac{x^2-7x+12-x-1}{\left(x+1\right)}<0$$

=> $$\frac{x^2-8x+11}{\left(x+1\right)}<0$$

For the numerator, we will use x= $$\frac{-b\pm\sqrt{b^2-4ac\ }\ }{2a}$$ to re-write the expression.

So, $$x^2-8x+11$$ has its roots= $$\frac{8\pm\sqrt{8^2-44\ }\ }{2}=\frac{8\pm\sqrt{20}}{2}=4\pm\sqrt{5}$$

Now, $$\frac{\left(x-4+\sqrt{5}\right)\left(x-4-\sqrt{5}\right)}{\left(x+1\right)}<0$$. 

Critical points of this inequality are: -1, $$4-\sqrt{5},\ 4+\sqrt{5}$$.

Using the wavy curve method, the values of x which satisfy this are- x$$\in\left(-\infty\ ,-1\right)U\left(4-\sqrt{5},\ 4+\sqrt{5}\right)$$.

But we already have the domain of the function as: $$x\in\left(-1,0\right)U\left(0,3\right)U\left(4,\infty\right)$$..

Using these two, the possible values of x becomes: x$$\in\left(4-\sqrt{5},3\right)U\left(4,\ 4+\sqrt{5}\right)$$.

Solving the inequality (2), we get:

(x-4)(x-3)>0

=> x$$\in\left(-\infty\ ,3\right)U\left(4,\infty\ \right)$$

Combining (1) and (2), we get x$$\in\left(4-\sqrt{5},3\right)U\left(4,\ 4+\sqrt{5}\right)$$.

.'. The possible integers in this range are- 2, 5 and 6 only.

Let the cost of the pen be Rs $$x$$ and the marked price be $$y$$

The cost of 5 pen  = $$5x$$

Selling price of 5 pens = $$ y +(y-5)+(y-10)+(y-15)+(y-20)$$ = 5y - 50

As per question

$$\left(1+145\%\right)5x\ =\ 5y-50$$

Or $$2.45x\ =\ y-10$$...(I)

Putting  $$x=200$$  gives, $$y=500$$

The selling price of the $$n^{th} $$ pen is = $$ 500 -(n-1)5$$

To make the sure that profit is obtained on every pen sold

SP > CP

$$ 500 -(n-1)5 >200$$

$$ 500 -5n +5 > 200$$

$$ 305 > 5n $$ or $$ n <61$$

Thus maximum possible pens he can sell is 60

Let us divide the number line into 3 regions

Case 1: x>=32

Case 2: 24<=x<32

Case 3: x<24

Case 1: x >= 32

$$\left|x+5\left|x-32\right|+\left|24-x\right|\right|<5$$

$$\left|x+5\left(x-32\right)-\left(24-x\right)\right|<5$$

$$\left|x+5x-160-24+x\right|<5$$

$$\left|7x-184\right|<5$$

Since x>=32, we get

$$7x<189\ ->\ x<27$$

Hence, there is no intersecting in this case, so no region satisfies case 1.

Case 2: 24<=x<32

$$\left|x+5\left|x-32\right|+\left|24-x\right|\right|<5$$

$$\left|x-5\left(x-32\right)-\left(24-x\right)\right|<5$$

$$\left|x-5x+160-24+x\right|<5$$

$$\left|-3x+136\right|<5$$

Since 24<=x<32,

$$-3x+136<5$$

$$-3x<-131$$

$$3x>131$$

$$x>42.xx$$

Hence, again, we get no overlapping in the number line, hence, this region does not satisfy the equation. 

Case 3: x<24

$$\left|x+5\left|x-32\right|+\left|24-x\right|\right|<5$$

$$\left|x-5\left(x-32\right)+\left(24-x\right)\right|<5$$

$$\left|x-5x+160+24-x\right|<5$$

$$\left|-5x+184\right|<5$$

Since x<24,

$$-5x+184<5$$

$$-5x<-179$$

$$5x>179\ ->x>35.xx$$

Hence, again, we get no overlapping in the number line, hence, this region does not satisfy the equation.

Hence, no point or region satisfies the equation.