Top 100 CAT 2025 Quant Questions PDF with Video Solutions

Let the capacity of the container be 160x liters.
It was half filled by A having liquids p and q in the ratio 3:5.
Thus, the amount of liquid p and q in the container is 30x liters and 50x liters.
They evaporate at the rate of 2 liters per minute and 3 liters per minute.
Thus, amount of p and q left after 30 minutes = 30x-2*30 = 30x-60 liters and 50x-90 liters respectively.
Volume of the portion of the container that is empty = 160x - 30x+60 - 50x+90 = 80x+150 liters
This is filled by liquid B having q and p in the ratio 2:3.
Thus, the amount of q added = (80x+150)*0.4 = 32x+60 liters.
The amount of p added = (80x+150)*0.6 = 48x+90 liters.
The ratio of p and q in the final mixture is 2:3.
Thus, $$\dfrac{30x-60+48x+90}{50x-90+32x+60} = \dfrac{11}{9}$$
=> $$\dfrac{78x+30}{82x-30} = \dfrac{11}{9}$$
Solving, we get, x = 3.
Thus, the capacity of the container is 160*3 = 480 liters.

Total score of the 10 batsmen = 600
Average of bottom five batsmen = 55
Total score of bottom five batsmen = 275
Total score of top 5 batsmen = 325
To obtain the maximum score, the scores of others are to be minimised
The minimum score of the top five batsmen can be 56 when all the bottom five batsman have 55 as their score.
Hence, 56,57,58,59,95 are the scores of the top five batsmen.

Quantity of milk left after the 1st iteration = $$50 * \frac{40}{50}$$
Similarly, after the 2nd iteration, quantity left = $$50 * \frac{40}{50}$$ * $$\frac{42}{50}$$
This will be repeated 5 times with quantity withdrawn decreasing by 2 litres each time.
So, Quantity of milk in final mixture = $$50 * \frac{40}{50} * \frac{42}{50} * \frac{44}{50} * \frac{46}{50} * \frac{48}{50}$$
= $$2^5 (20 * 21 * 22 * 23 * 24) * \frac{1}{50^{4} }$$
= $$\frac{2 * 24!}{19!} * \frac{1}{25^{4} }$$
Hence, option A is the correct answer.

Let us assume that the total number of people at the party be ‘n’. Let the total apple juice consumed at the party be ‘k’ ml.
We have been given that
$$\frac{k}{5} + \frac{1300*n - k}{7} = 1300$$
=> 7k + 6500n - 5k = 45500
=> 2k + 6500n = 45500
=> k + 3250n = 22750
=> We know that ‘k’ is an integer and
0 < k < 1300n
=> 0 < 22750 - 3250n < 1300n
=> 0 < 22750 < 4550n
=> 5 < n < 7
Hence, n must be ‘6’.
Therefore ‘k’ will be 3250
Hence, the required ratio will be
3250/5 : 4550/7 = 1:1

Let the geometric progression be $$a,ar,ar^2,ar^3,….$$
It is given that,
$$a+ar+ar^2=26(\frac{ar^3}{1-r})$$
$$=>1-r^3=26r^3$$
$$=>27r^3=1$$
$$=>r=\frac{1}{3}$$

The required ratio is,
$$\frac{a}{\frac{ar}{1-r}}=\frac{1-r}{r}=2$$

Final volume of A and B does not change because the volume is being replaced.

Suppose final % of milk in A and B is 100a% and 100b%.

Total quantity of milk in solution A = Volume of A*(100a)/100 = 100*(100a)/100 = 100a

Total quantity of milk in solution B = Volume of B*(100b)/100 = 200*(100b)/100 = 200b

So, the total quantity of milk = 100a+200b = 100
After removing 40 litre from A now, the amount of milk in A = 60a

After removing 40 litre from A now, the amount of milk in B = 200b+40a

Now, after adding 40 litre from B, the amount of milk in A = 60a + $$\ \frac{\ 40}{240}$$*(200*b+40*a)

Since, the total quantity of milk will remain unchanged even after mixing,

60*a + $$\ \frac{\ 40}{240}$$*(200*b+40*a) = 100a.

On solving we get 200a=200b i.e. a=b. Hence, 300a =100. Thus, a=b=33.33%

Let the numbers be 2a, 2b, 2c, 2d.

and $$a\le\ b\le\ c\le\ d$$

and at first, 2a + 2b + 2c + 2d = 162

which means a + b + c + d =81 --------equation (1)

and we have to find the value of a + b + 3c + 3d

which is (a + b + c + d) + 2c + 2d ; putting values from equation (1)

= 81 + 2c +2d

= 81 + 2 ( c + d ) ---------equation(2)

also if$$a\le\ b\le\ c\le\ d$$

$$then\ a+1\ \le\ \ b$$

$$\ b+1\ \le\ \ c$$

$$\ c+1\ \le\ \ d$$

from these equations :

$$\ a+2\ \le\ \ c$$  ---------------equation(3)

$$\ b+2\ \le\ \ d$$ ----------------equation(4)

adding equation (3) and (4)

$$a+\ b\le\ c+\ d-4$$ ----------- equation(5)

putting values of a+b from equation (5) to equation (1)

$$\ c+\ d-4\ +c\ +d\ \ge\ 81$$

$$2\left(c+d\right)\ge\ 85$$

$$c+d\ge\ 42.5$$

as c and d are natural numbers so $$c+d\ge\ 43$$

putting this value in equation (2)

a+b+3c+3d $$\ge\ $$ 81 + 2*(43)

a+b+3c+3d $$\ge\ $$ 167

which means the minimum value of the required equation is 167.

We only have one option greater than 167 which is 173.

Hence 173 is the answer.

Let us assume the total amount of rice present is 18x. That is we have 5x kg of Rs 20 rice, 6x kg of Rs 30 rice, and 7x kg of Rs 40 rice.

CP of rice = 20*5x + 30*6x + 40*7x = 560x.
SP of rice = 560x*110/100 = 616x.

To get a profit of 10% the shopkeeper should sell the whole quantity at 616x.

Therefore, selling price of the mixture per kg = 616x/(5x+6x+7x) = $$34 \frac{2}{9}$$

In the amount of time taken by A to run one side of the square, D runs all four sides and reaches where he was initially standing. Hence, A and D can only meet at D's initial position.

Once A, B, C, D reach D's starting position, they can meet again only at point D (since the speeds are in the ratio 1:2:3:4 - by the time A, covers 1 round, B will cover 2 rounds, C will cover 3 rounds and D will cover 4 rounds). Therefore, after meeting at D, all 4 of them will meet when A completes a round.

Let us assume that A completes a round in 12 seconds. B will take 6 seconds, C will take 4 seconds, and D will take 3 seconds to complete a round. A covers a distance of 4s in 12 seconds. => A takes 3 seconds to cover a distance of 's'.

It takes A 9 seconds to reach D (initially). A will complete a round in 12 seconds. Therefore, when they meet for the third time, 12+12+9 = 33 seconds would have elapsed. In 33 seconds, A would have covered 33/3 = 11s.

So, total distance covered by A = 11s.

Alternate Solution:
Using diagram,

In each step, A is moving s distance. After 4 steps the configuration is repeated. In the third step, A has travelled 3s, when all four meet. After 4 steps they will meet again for the second time. After 4 more steps they will meet again for the third time.
So total distance travelled by A = 3s+4s+4s = 11s

Let the total work to be done be 480 units (LCM of 60,20,30,8,32)
So the respective efficiencies of different pipes are-
A- 8 units/min (480/60)
B- 24 units/min
C- 16 units/min
D- 60 units/min
E- 15 units/min
Now, one combination of emptying and filling pipe can fill the tank in 60 minutes, i.e. together they can do 480/60=8 units of work per minute.
Out of the combinations possible, only A-C and B-C have a difference of 8 units/min.
So, if A is an emptying the tank and C is a filling it, they will do together 8 units of work per min. When C is an emptying the tank and B is filling it, the work done per minute will be the same 8 units/min.

According to the question, the second combination takes 8/45th of time of first one, i.e 32/3 minutes. Only combination that can do so is when D is the emptying pipe and E is the filling pipe. So among A, B, and C, we cannot know for sure which pipe is the filling one and which one is the emptying one.

Thus, we can’t determine which pipes have been converted to emptying pipes.

When they wave for the first time, the total distance travelled by them is d, where d is the distance between A and B.

For every subsequent point where they wave, they have to travel 2d each.

The first time, they wave, the bus from A has travelled 25/(25+35) x 18 = 7.5 km

For the 10th time that they wave, total distance = d + (9x2)d = 19d = 19 x 18.

Distance travelled by the bus from point A = 25/(25+35) x 19 x 18 = 142.5 km

This means that the bus is 144 - 142.5 = 1.5 km from A. Difference in distance = 7.5 - 1.5 = 6 km.

There is a certain amount of work to be done for the man to reach the top of the escalator. A part of this work is done by the man and the remaining part is done by the escalator.
Let the number of steps on the escalator be L.
Number of steps covered by the man = 100
So, number of steps covered by the escalator = L - 100
Let the time taken for this be t.
So, speed of the man = 100/t
Speed of the escalator = (L-100)/t

In the second case, number of steps covered by the escalator = (L-100)/t * (t - 20 + 18) = (L-100)/t * (t-2)

Number of steps covered by the man = 100/t * (t + 18) = 100 * (t + 18)/t

The sum of the numbers of steps covered by the man and escalator is L.
So, (L-100)/t * (t-2) + 100 * (t+18)/t = L

=> (Lt - 2L - 100t + 200 + 100t + 1800)/t = L
=> 2000 - 2L + Lt = Lt
=> 2L = 2000
=> L = 1000

So, the number of steps on the escalator is 1000.

Two boats A and B are travelling in opposite directions in still water and covers the distance in ratio 3:1 respectively.

Let the two boats travel at 3v and 1v speed.

Let the speed of the stream is 'r'.

The boats will travel in stream water:

Distance covered by A= (3v-r) t

Distance covered by B= (v+r) t

$$\frac{\left(3v-r\right)t}{\left(v+r\right)\ t}=\ \frac{1}{3}$$

r=2v

The ratio of the speed of A and speed of the water =3v/2v

=3/2

The number of points of intersection of two racers running around a circular track in the same direction is |a-b| where a:b is the ratio of their speeds.

The number of points of intersections must be a factor of 6( 1,2,3 or 6) so that they meet only at the corners.

A:|7-6|= 1

B:|3-1|= 2

C:|5-2|= 3

D:|5-1|= 4

4 does not belong to 1, 2, 3 or 6. Hence, D is the answer.

Assuming each man completes 1 unit of work each and each boy completes x unit of work each day.
The work done in 15 days = (5x+3)*15 = Total work
Also, total work = work done by 5 men in 6 days + work done by 5 men and 10 boys in 4 days
=> (3+5x)*15 = 6*5+5*4+10x*4
=> 45+75x =50 + 40x
=> y = 1/7 units
Hence, the efficiency of 1 man is equal to 7 boys.

The first time the two people meet, the total distance travelled by them will be ‘d’ where ‘d’ is the distance between P and Q. The second time they meet the total distance covered will be 3d and for third meeting it will be 5d. In the time that they cover a distance of ‘d’, Virat covers 540 m. So by the time that they cover 3d, Virat would have covered 540*3 = 1620 m. Now, they never cross each other while travelling in same direction. Hence, 1620 = d + 600
=> d = 1020
Thus, the distance between PQ must be 1020.
Now by the third meeting Virat would have covered 540*5 = 2700 m
Thus, required distance from P will be 2700 - 1020*2 = 660 m
Thus, the distance from Q will be 1020 - 660 = 360

Let m be the number of units of work done by a man on one day.
w be the number of units of work done by a woman in one day.

10m*8 = 15w*6

$$\frac{m}{w}$$ = $$\frac{9v}{8v}$$

(9w+xm)*8 = 10m*8

9*8v+x*9v = 10*9v

8+x=10

x = 2

C is the correct answer.

Pipes A, B and C are fitted to a tank, such that A is an inlet pipe, B and C are outlet pipes.
Pipe B is situated at half the height of the tank while C is situated at one fourth the height of the tank from the bottom.

All the pipes are opened simultaneously for an empty tank.

Let the flow in pipe A = y

The ratio of the outlet flow of pipes B & C is 1 : 5.

The flow in pipe B and C is x and 5x.

.

The time taken to fill T/4 tank = t

The net rate of flow of water= y

Thus, yt=T/4.......(1)

.

The time taken to next fill T/4 tank = 6-t

Now, the net rate of flow of water= y-5x

(y-5x)(6-t)= T/4 .......(2)

.

The time taken to next fill T/2 tank = 16-6=10h

The net rate of flow of water= y-5x-x=y-6x.

10(y-6x)=T/2 .......(3)

From (1) and (3)

2*yt= 10(y-6x)

t=5(y-6x)/y ......(4)

From (3) and (2)

(y-5x)(6-t) =5 (y-6x) ........(5)

Thus,

$$\frac{\left(y-5x\right)\left(y+30x\right)}{y}=5\left(y-6x\right)$$  (Subsituting the value of t from (4) into (5))

$$4y^2-55xy+150x^2=0$$

$$4y^2-40xy-15xy+150x^2=0$$

$$\left(4y-15x\right)\left(y-10x\right)=0$$

y/x= 15/4 or y/x=10

Now, 15/4 will be rejected because the flow rate y-6x will become negative.
Hence, y:x = 10:1

Option A

The ratio of their cost price is 3:4 and the ratio of profit is 1:2.
So, the ratio of the selling price will be in between 1:2 and 3:4
Of the given options, only 37:49 is not in between the two ratios

Let the CP be 100. Profit = 25%

=> The SP = MP = 125.

New CP = 110, new MP = 137.5.

New profit = 20% => New SP = 132.

Therefore, required discount % on MP = (5.5/137.5)*100 = 4%.

Let cost price of product be C

Selling price (S) = 0.95C (as sold at 5% loss)

Given, if he had sold it for 90 Rs more, he would get 1% profit

S+90 = 1.01C

0.95C + 90 = 1.01C

0.06C = 90

C = 1500

Therefore, initial cost price is Rs 1500

Let the number of bags bought by the shopkeeper be N.
Total cost incurred = 85N
Total selling price of 40% of the bags = 60 * (40/100)*N = 60*2N/5 = 24N
For a 10% profit, the total selling price of all the bags put together should be (110/100)*85N = 93.5N
So, the remaining bags should be sold for a cumulative price of 93.5N - 24N = 69.5N
So, price of each bag = 69.5N/(3N/5) = 69.5*5/3 = Rs 115.83

Myra invested Rs 6000 per month in business for 12 months, hence her share = 12 $$\times\ $$ 6000 + 11 $$\times\ $$ 6000 + 10 $$\times\ $$ 6000 + ...... + 1 $$\times\ $$ 6000 = 78 $$\times\ $$ 6000

Ayra invested Rs 6000 per month in business for 8 months, hence her share = 8 $$\times\ $$ 6000 + 7 $$\times\ $$ 6000 + ....... + 1 $$\times\ $$ 6000 = 36 $$\times\ $$ 6000

Let us assume Kiara invested in business for ‘t’ months, hence her share = $$12000\times\ \ \frac{\ t\left(t+1\right)}{2}$$ = t(t + 1) $$\times\ $$ 6000

Myra received 50% of the profit which means she owns 50% share

78 $$\times\ $$ 6000 + 36 $$\times\ $$ 6000 + t(t+1) $$\times\ $$ 6000 = 156 $$\times\ $$ 6000

t(t+1) = 42

t = 6

This means Kiara invested after 6 months from starting of the business. Therefore, she invested 2 months after Ayra invested.

Answer is option D.

After the first time, the ratio of milk and water is 20-x:x.

Hence, the amount of milk  = 20-x

Assuming the price of 20 litres milk is y.

Hence, the cost price of 1 litres milk  = y/20

Similarly the selling price of 20-x litre milk = y

Hence the selling price of 20-x litres milk = y/(20-x)

The profit percentage = $$\ \dfrac{\ \ \frac{y\ }{20-x}-\ \frac{y\ }{20}}{\ \frac{y\ }{20}}$$  = $$\ \frac{\ 20}{20-x}-1$$ = $$\ \frac{\ 25}{100}$$

=> $$\ \frac{\ 20}{20-x}$$ = 1.25

=> x=4 litres

The ratio of the amounts at the end of the 2nd and 4th years is $$(1+R)^2 = 1.27$$
After 6 years, the ratio is $$(1+R)^6 = 2.05$$
It can be seen that $$(1+R)^5 < 2$$
So, the amount doubles every 6 years

Assuming the annual rate of interest = r

Hence, compound interest paid by the person in 4 years = $$P\left(1+r\right)^4-P$$
Also, compound interest paid by the person in 8 years = $$P\left(1+r\right)^8-P$$
Hence, the ratio = $$\ \frac{\ \left(1+r\right)^8-1}{\left(1+r\right)^4-1}=5$$
=> $$\ \ \left(1+r\right)^8-1=5\left(1+r\right)^4-5$$
Assuming, $$\left(1+r\right)^4$$ =x,
$$x^2-5x+4=0$$
=> (x-1)(x-4)=0
=> x=1 or x=4, x cannot be equal to 1 as the ratio will not be defined. Hence x=4 => $$\left(1+r\right)^4$$ = 4  =>1+r = $$\sqrt{\ 2}$$
=> r=$$\sqrt{\ 2}-1$$ = 0.414
Hence, the rate of interest will be 41% approximately.

Let x+y = $$a$$, x-y =$$ b$$

$$|a| + |b| = 8$$

x = $$\frac{a+b}{2}$$, y = $$\frac{a-b}{2}$$

$$x^2+y^2+3xy$$

= $$\frac{a^2+b^2+2ab}{4}$$+ $$\frac{a^2+b^2-2ab}{4}$$ +3($$\frac{a^2-b^2}{4})$$

=$$\frac{5a^2-b^2}{4}$$

The value of the above expression will be maximum when a = 8, b= 0

= $$\frac{5*8*8}{4}$$ = 80

Alternate Approach:

If x+y $$\geq$$ 0, then |x+y| = x+y
x+y < 0, then |x+y| = -(x+y)
x-y $$\geq$$ 0, then |x-y| = x-y
x+y < 0, then |x-y| = -(x-y)
Applying the above conditions, we get
Case 1:
x+y $$\geq$$ 0, x-y $$\geq$$ 0, then x+y+x-y = 8
x = 4
Case 2:
x+y $$\geq$$ 0, x-y < 0, then x+y-x+y = 8
y = 4
Case 1:
x+y < 0, x-y $$\geq$$ 0, then -x-y+x-y = 8
y = -4
Case 1:
x+y < 0, x-y < 0, then -x-y-x+y = 8
x = -4

The values of both x and y are maximum at (4,4) in order to maximze the equation  $$x^2+y^2+3xy$$.

Similarly in the third quadrant, both the values will be negative. The terms are of even power, thus we can maximize the equation at (-4,-4) also.

The maximum value of $$x^2+y^2+3xy$$ is at (4,4) & (-4,-4)
= $$4^2+4^2+3*4*4$$
=80
80 is the correct answer.

4x-3y+2=0

4x-3y=-2

Since (a,b) is a solution of the equation, we have 4a-3b=-2.

$$-8^b\left(\dfrac{\left(492a-369b\right)}{16^a}\right)=\dfrac{-123\left(4a-3b\right)}{2^{4a-3b}}=\dfrac{\left(123\times\ 2\right)}{2^{-2}}=123\times\ 8=984$$

Hence, the answer is 984.

Let the polynomial be
f(x) = k(x - 1)(x - 2)(x - 3)(x - 4) + 5x
Since, it is given that the coefficient of $$\text{x}^4$$ is 1, k = 1
Putting x = 5, we get
f(5) = 4 * 3 * 2 * 1 + 25 = 49
Hence, option C is the correct answer.

$$\log_9\left(x-1\right)\ =\ \log_3\left(x-3\right)$$

$$\ \frac{\log\left(x-1\right)\ \ }{\log\ 9}=\ \ \frac{\ \log\ \left(x-3\right)\ }{\log\ 3}$$

$$\ \frac{\log\left(x-1\right)\ \ }{2\log\ 3}=\ \ \frac{\ \log\ \left(x-3\right)\ }{\log\ 3}$$

$$(x-1) = (x-3)^2$$

$$x^2-6x+9-x+1$$ = 0

x = 5, 2

Let's see if the values satifies the equation

If x = 2, the expression inside the log will become negative, which is invalid.

If x = 5, $$\log_94\ =\log_32$$

The sum of the values of x which satisfies the equation is 5

f(f(x)) = 9

Let f(y) = 9

$$y^2-4y+4=9\ ->y^2-4y-5=0->\left(y-5\right)\left(y+1\right)=0$$

y = 5 or y = -1

Hence, f(x) = 5 or f(x) = -1

$$x^2-4x+4=5\ or\ x^2-4x+4=-1$$

$$x^2-4x-1=0\ or\ x^2-4x+5=0$$

$$x^2-4x+5=0\ $$ has no roots.

The other equation has the roots

$$2\pm\sqrt{5}$$

Sum of roots = 4.

$$a+b+c=P$$ and $$ab+bc+ac=108$$ and $$abc=216$$
Given that $$a^3+b^3+c^3=648$$, it implies that $$a^3+b^3+c^3-3abc=0$$
So, $$a+b+c=0$$ or $$a^2+b^2+c^2-ab-bc-ac=0$$.
As, a, b and c are positive, it implies that $$a+b+c > 0$$ and $$a^2+b^2+c^2-ab-bc-ac=0$$
$$a^2+b^2+c^2-ab-bc-ac=0$$ only if $$a=b=c=6$$
So, $$P=a+b+c=18$$

$$\log_{\ \frac{\ 3}{4}}\left(\ \frac{\ x-4}{x-1}\right)\ >\ 0$$

The term "$$\ \frac{\ x-4}{x-1} $$" should be positive.

$$\ \frac{\ x-4}{x-1}\ >\ 0$$

x $$\in\ $$ $$\left(\ -\ \infty\ ,\ 1\right)\ U\ \left(4,\ \infty\ \right)$$

To satisfy the condition that $$\log_{\ \frac{\ 3}{4}}\left(\frac{x-4}{x-1}\right)\ >\ 0$$:

$$\ \frac{\ x-4}{x-1}\ <\ 1$$

$$\ \frac{\ x-4}{x-1} - 1\ <\ 0$$

x > 1

Solution set = (4, $$\infty$$)

C is the correct answer.

We have,$$\log_2\log_3\left(\ \dfrac{\ x^2+2}{x+14}\right)<0$$
=> $$\log_3\left(\ \dfrac{\ x^2+2}{x+14}\right)<1$$
=> $$\dfrac{\ x^2+2}{x+14}<3$$
=> $$\dfrac{\ x^2+2}{x+14}-3<0$$
=> $$\dfrac{\ x^2+2-3x-42}{x+14}<0$$
=> $$\dfrac{\ x^2-3x-40}{x+14}<0$$
=> $$\dfrac{\ \left(x+5\right)\left(x-8\right)}{x+14}<0$$
=> x $$\in\ \ \left(-\infty\ ,\ -14\right)\ U\ \left(-5,8\right)$$
The argument of log in the base 3 should be positive.
$$\ \dfrac{\ x^2+2}{x+14}>0$$
=> x>-14  ($$\ \ x^2+2>0\ $$ for all real values of x)
Also, the argument in the log base 2 should also be positive.
Hence, $$\log_3\left(\dfrac{x^2+2}{x+14}\right)$$ > 0
=> $$\dfrac{x^2+2}{x+14}>1$$
=> $$\dfrac{x^2+2}{x+14}-1>0$$
=> $$\dfrac{x^2+2-x-14}{x+14}>0$$
=> $$\dfrac{x^2-x-12}{x+14}>0$$
=> $$\dfrac{\left(x-4\right)\left(x+3\right)}{x+14}>0$$
=> x $$\in\ \left(-14,-3\right)\ U\ \left(4,\infty\right)$$
Hence, the final solution set = (-5,-3) U (4,8)
Total number of integers satisfying = 4
4 integers satisfying the inequality are -4,5,6,7.

We have, $$x^2-5x+\frac{1}{x^2}-\frac{5}{x}-4\le0$$ which can be rearranged as:
$$x^2+\frac{1}{x^2}+2-2-5x-\frac{5}{x}-4\le0$$
=> $$x^2+\frac{1}{x^2}+2-5x-\frac{5}{x}-6\le0$$
=> $$\left(x+\frac{1}{x}\right)^2-5\left(x+\frac{1}{x}\right)-6\le0$$
Assuming $$x+\frac{1}{x}=t$$
=> $$t^2-5t-6\le\ 0$$
=> (t-6)(t+1)$$\le\ 0$$
=> $$-1\le\ t\le\ 6$$
=> $$-1\le\ x+\frac{1}{x}\le\ 6$$
Using AM-GM, the minimum value of $$x+\frac{1}{x}\ $$ for x>0 will be 2.
=> $$x+\frac{1}{x}\ \ge2$$
Hence, in $$2 \le\ x+\frac{1}{x}\le\ 6$$
$$x+\frac{1}{x}\ \ge 2,$$ will be true for all positive real values of x. =>x>0
The smallest integral value satisfying will be x=1
Also, $$x+\frac{1}{x}\ \le6$$, the largest value satisfying the equation will be 5 because x=6, then 6+1/6(a positive value) will be greater than 6.
Hence the smallest value integral value satisfying is 1 and the largest integral value satisying is 5.
Difference = 5-1 = 4

We have, (x+1)$$\log_{10}\left(3-x^2\right)$$ < 0
Now, for $$\log_{10}\left(3-x^2\right)$$ to be a real value $$\left(3-x^2\right)$$ should be greater than 0.
=> $$3-x^2\ >\ 0$$  => $$x^2-3<0$$   => x$$\in\ \left(-\sqrt{3},\sqrt{3}\right)$$
Now, if the product of two terms is less than 0, then both are of opposite sign.
So, x+1<0, then x<-1 and x+1>0, then x>-1
Since the base of the log is greater than 1, $$\log_{10}\left(3-x^2\right)$$ will be less than 0 if 3-$$x^2<1$$
=> 2-$$x^2<0$$   => x $$\in\ \left(-\infty\ ,-\sqrt{2}\right)\ \cup\left(\sqrt{\ 2},\infty\ \right)$$
Plotting x+1 on a number line, we get

Plotting $$\log_{10}\left(3-x^2\right)$$ on a number line, we get

Clearly, the signs are opposite between when x $$\in\ \left(-\sqrt{2},-1\right)\ \cup\left(\sqrt{2},\sqrt{3}\right)$$
Approximate range of x = (-1.414,-1) U (1.414,1.732)

Option A: $$\frac{-7}{5}$$ = -1.4 lies in the range
Option B: $$\frac{8}{5}$$ = 1.6 lies in the range.
Option C: $$\frac{29}{20}$$ = 1.45 lies in the range
Option D: $$\frac{-43}{30}$$ = -1.43 does not lie in the range

$$1^3-2^3+3^3-4^3......................-100^3$$

=$$1^3+2^3+3^3.......100^3-2\left(2^3+4^3+6^3.........100^3\right)$$

=$$1^3+2^3+3^3.......100^3-2\times\ 2^3\left(1^3+2^3+3^3.........50^3\right)$$

=$$\left(\ \frac{\ 100\times\ 101}{2}\right)^2\ -\ 16\left(\ \frac{\ 50\times\ 51}{2}\right)^2$$

=$$50^2\times\ 101^{2\ }-4\times\ 50^2\times\ 51^2$$

=$$50^2\left(101^{2\ }-2^2\times\ 51^2\right)$$

= $$50^2\left(101-102\right)\left(101+102\right)$$

=-507500

A is the correct answer.

We see that 2 is repeated twice. So use of 2 ends in 2nd term.
Then 4 is repeated four times. So use of 4 ends in 2+4 i.e 6th term
Then 6 is repeated six times. So use of 6 ends in 2+4+6 i.e 12th term
And so on
.
.
Then 170 is repeated 170 times. So use of 170 ends in 2+4+6+…+170
Now, 2+4+6+…+170 = 7310
7311th term would be 172 and will be repeated 172 times.
So 7330th term must be 172.

$$4x^{\sqrt{\ x}^{\sqrt{\ x}^{\sqrt{\ x}^{.^{.^{.^{.^{.\infty\ }}}}}}}}\ =\ 0.0625$$

$$4x^{\sqrt{\ x}^{\sqrt{\ x}^{\sqrt{\ x}^{.^{.^{.^{.^{.\infty\ }}}}}}}}\ =\ \ \frac{\ 1}{16}$$
$$x^{\sqrt{\ x}^{\sqrt{\ x}^{\sqrt{\ x}^{.^{.^{.^{.^{.\infty\ }}}}}}}}\ =\ \ \frac{\ 1}{64}$$

$$\sqrt{\ x}^{\sqrt{\ x}^{\sqrt{\ x}^{.^{.^{.^{.^{\infty\ }}}}}}}=\ \sqrt{\ \ \frac{\ 1}{64}}$$

$$\sqrt{\ x}^{\sqrt{\ x}^{\sqrt{\ x}^{.^{.^{.^{.^{\infty\ }}}}}}}=\ {\ \ \frac{\ 1}{8}}$$

$$\sqrt{\ x}^{\ \frac{\ 1}{8}}\ =\ \ \frac{\ 1}{8}$$

$$x^{\ \frac{\ 1}{8}}\ =\ \ \frac{\ 1}{64}$$

$$x^{\ \frac{\ 1}{8}}\ =\ \ \frac{\ 1}{2^6}$$

$$x^{ }\ =\ \ \frac{\ 1}{2^{48}}$$

B is the correct answer.

Consider average of given numbers = $$a$$

Total sum = 33$$a$$

Assuming x,y and z are hundreds digit, tens digit and units digit of 30th term of the series respectively.

After reversing the digits, new average = $$\frac{{33{a}-(100{x}+10{y}+{z})+((100{z}+10{y}+{z}))}}{33}$$ = $$a+6$$

⇒ $$a$$ + $$\frac{99{(z-x)}}{33}$$ = $$a+6$$

⇒ $$z-x$$ = 2

x cannot be 0 as all are 3 digit numbers

Case 1:   z= 8  x= 6    y=7    The digits are in strictly increasing or strictly decreasing order.

30th number = 678 =a+2*29      Hence 1st term of series = 620, Last term = a+32*d = 684

Hundreds digits of all numbers are the same.

Case 2:   z= 6  x= 4    y=5    The digits are in strictly increasing or strictly decreasing order.

30th number = 456 =a+2*29      Hence 1st term of series = 398, Last term = a+32*d = 462

Hundreds digits of all numbers are not the same.

Case 3:   z= 4  x= 2    y=3    The digits are in strictly increasing or strictly decreasing order.

30th number = 234 =a+2*29      Hence 1st term of series = 176, Last term = a+32*d = 240

Hundreds digits of all numbers are not the same.

Only 1st case is possible.

3rd term = a+2*2 = 624

Sum of digits of 624 = 6+2+4 = 12

We have, $$x^2+2x+33$$ = $$x^2+3x-x-3+3+33$$ = x(x+3)-1(x+3)+3+33
= (x+3)(x-1)+36
Now, first term (x+3)(x-1) is always divisible by (x+3)
Hence, if 36 is divisible by x+3, then the whole value will be divisible by x+3
So, factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18 and 36  (9 values)
x+3 will be equal to these values.
x+3 = 1, x+3 = 2 and x+3 = 3 will be rejected because the values are not positive.
Hence, the total number of remaining values = 9-3 = 6

145 can be written as $$64\ +\ 81\ =\ 2^6+3^4$$

=> a+3 = 6 => a= 3

and b-2 = 4 => b= 6

Also $$2^{2a+1}+3^{2b-7}=2^7+3^{5\ }=371$$

Hence $$\ 2^{3a-5}+3^{b-1}\ =\ 2^4+3^5=259$$

$$(a^{\log_{b}{x}})^2 - 5x^{\log_{b}{a}} + 6$$ = 0
or, $$(x^{\log_{b}{a}})^2 - 5x^{\log_{b}{a}} + 6$$ = 0  (wkt $$(a^{\log_{b}{x}}) = (x^{\log_{b}{a}})$$)
Let $$x^{\log_{b}{a}}$$ be $$y$$.
Then, $$y^2 - 5y + 6$$ = 0
On solving, we get $$y$$ = 2 or $$y$$ = 3
=>$$x^{\log_{b}{a}}$$ = 2 or $$x^{\log_{b}{a}}$$ = 3
=> $$x = 2^{\log_{a}{b}}$$ or $$x = 3^{\log_{a}{b}}$$
Hence, option B is the correct answer.

As per the given condition all the individual elements are in the powers of 2

An even numbered for a real number is always greater than 0 or equal to zero.

Since the sum of the three individual elements is zero.

This is only possible when all three of them are equal to 0.

Hence 3x+y+z+2 = 0

= 3x+y+z = -2  (1)

2x+3y+3z -22 = 0

= 2x+3y+3z = 22    (2)

x-4y+2z+14 = 0

= x-4y+ 2z = -14  (3)

Multiplying (1) with 2 and subtracting (2)- (1)

= (2x+3y+3z = 22) - (9x+3y+3z = -6)

= -7x = 28.

x = -4.

Substituting this in equation (1) = 3(-4)+y+z = -2.

y+z = 10. (4)

Substituting in equation 3 :

-4-4y+2z = -14

2z-4y = -10.

z-2y = -5. (5)

= 2y-z = 5

y+z = 10.

Hence y, z = 5

x-y+z = -4-5+5 = -4

We have, $$\log_{0.5}x+\log_4y\ =-1$$ and xy=4
Substituting  y = $$\dfrac{4}{x}$$ in the first equation.
$$\log_{0.5}x+\log_4\left(\dfrac{4}{x}\right)\ =-1$$
=> $$-\log_2x+1-\log_4x\ =-1$$
=> $$-\log_2x-\frac{1}{2}\log_2x\ =-2$$
=> $$\frac{3}{2}\log_2x=2$$
=> x =$$2^{\frac{4}{3}}$$ = $$\sqrt[\ 3]{16}$$

The diagonal of the first square is the diameter of the first circle = 10. The side of the first square is the diameter of the second circle = $$10/\sqrt2$$. The area of the first circle is $$25\pi$$. The area of the second circle = $$25\pi$$/2. Therefore, the sum of areas = $$25\pi$$ + $$25\pi$$/2 + $$25\pi$$/4 +…. = $$50\pi$$.

In the triangle BOA, tan $$\theta = \frac{5\sqrt{3}}{5} = \sqrt{3}$$
Therefore, $$\theta = 60$$ degrees.

The shaded area = Area of two triangles - area of section of circle.
As the radius is perpendicular to a tangent, triangle OAB is a right angled triangle, with right angle at B, with area=$$1/2*5*5\sqrt{3}$$.
Hence area of both triangles= $$2*1/2*5*5\sqrt{3}=25 \sqrt{3}$$.
The angle BOA=$$tan^{-1} \frac{5 \sqrt{3}}{5} = 60^{\circ}$$.

Hence, angle BOC = 2*BOA = 120°
Hence the area of the section = area of circle/3 = $$\frac{\pi (5^2)}{3}$$.
Hence area of shaded portion= $$25 (\sqrt{3} -\frac{\pi}{3})$$

Let MR=NQ=x and height=y

So, area of the parallelogram=xy=74.

Consider triangle RTS and NSQ. These triangles are congruent since angle RST=angle NSQ (opposite angles), angle TRS=angle NQS (alternate angles), and RS=SQ (s is the midpoint). Thus, area triangle RTS = NSQ.

=> MR=RT=x

=> Area of parallelogram= area of triangle MTN.

Now the figure MNQT is a trapezium whose area is 1/2 X (x+2x) X y= 3/2 xy= 3 X 37.

Area of triangle NTQ= area of MNQT-area of MTN= 37X3-37X2=37 sq units.

Area of the triangle is $$1/2 * x * \sqrt{36-x^2}$$.

The maximum area of the triangle will be if it is an isosceles right triangle with hypotenuse equal to 6 metres. In this case, the other two sides will be 6/√2 and 6/√2 m respectively.

Hence, area of triangle = 1/2 * 6/√2 * 6/√2 = 36/4 = 9 sq m

The figure will be as shown below.

The area covered by the two circles = Sum of the areas of the two circles - Area of the region overlapped by the two circles
Area of each of the circle = $$\pi\times r^2=4\pi$$
Area of the segment enclosed by the line CD and the arc CD = Area of sector ACD - Area of triangle ACD = $$\pi-2$$
Area common to the two circles is $$2\times (\pi - 2)$$.
So, total area covered by the two circles is $$4\pi + 4\pi - (2\pi - 4) = 6\pi + 4$$

Using symmetry, $$\angle\ BOA\ =\angle\ COA$$
Since, $$\angle\ BOC\ \ =120^{\circ\ }$$ = $$\angle\ BOA+\angle\ COA$$ = 2$$\angle\ BOA$$
=> $$\angle\ BOA$$ = 60$$^{\circ\ }$$
Since OB is tangent to the smaller circle, $$\angle OBA\ \ =90^{\circ}$$
In right triangle BOA, $$\angle\ BAO\ =90-\angle\ BOA\ =\ 90-60=30^0$$
Now, OA cos$$\angle\ BAO$$ = AB = r   =>$$\frac{OA\sqrt{3}}{2}=r$$
=> OA = $$\frac{2r}{\sqrt{3}}$$
Now, OD = $$6+4\sqrt{3}$$ = r + $$\frac{2r}{\sqrt{3}}$$
=> r = $$\frac{\left(6+4\sqrt{3}\right)\sqrt{3}}{2+\sqrt{3}}$$
=> r = $$\frac{12+6\sqrt{3}}{2+\sqrt{3}}$$ = $$\frac{6\left(2+\sqrt{3}\right)}{2+\sqrt{3}}$$ = 6 cm

As given that pole subtends equal angle at the ends of the diameter AB. Then it will be at perpendicular from the center as given in the figure below.

Here suppose the pole PQ subtends 30$$\degree$$ on the point R.

Then PQ/PR= tan 30$$\degree$$

PR=6$$\sqrt3$$

Using Pythagoras theorem:

$$PR^2=PO^2+OR^2$$

$$ 36*3=100+X^2 $$

OR = 2$$\sqrt2$$

Just the top surface of the ground is exposed initially. Therefore, the area of sand exposed will be equal to the area of the base circle of the cylinder (well).

It has been given that the diameter of the well is equal to the diameter of the heap.
Let the radius of the cylinder and cone be $$r$$, depth of the cylinder be $$h$$ and the height of the heap (cone) be $$c$$.
Let the slant height of the cone be $$s$$.

It has been given that the amount of sand exposed has increased by 25%.

Area of the ground (sand) exposed initially = base area of the cylinder = $$\pi*r^2$$.
Area of sand exposed in the heap = Curved surface area of the cone = $$\pi*r*s$$ = $$\pi*r*\sqrt{c^2+r^2}$$

We know that the area exposed has increased by 25%.

=> $$1.25*\pi*r^2$$ = $$\pi*r*\sqrt{c^2+r^2}$$
=> $$1.25*r = \sqrt{c^2+r^2}$$
=> $$\frac{5}{4}*r = \sqrt{c^2+r^2}$$

Squaring on both sides, we get,
$$\frac{25}{16}*r^2 = c^2+r^2$$
$$\frac{9}{16}*r^2 = c^2$$
Taking square root on both sides, we get,
$$\frac{3}{4}*r = c$$ -----------------(1)

Now, we know that the volume of sand excavated will be equal to the volume of the heap.
=> $$\pi*r^2*h = \frac{1}{3}*\pi*r^2*c$$
=> $$c = 3h$$-------------------------(2)

Substituting (2) in (1), we get,

$$\frac{3}{4}*r = 3h$$
$$r = 4h$$
=> Diameter, $$d=8h$$
Diameter of the well is $$n$$ times the depth of the well.

=> n = 8.

Therefore, 8 is the right answer.

Area of the smaller discs = 3$$\pi\ r^2$$

The radius of the bigger disc= 2r/$$\sqrt{\ 3}$$+r

Area of the bigger disc=  $$\frac{\pi\ r^2\left(2+\sqrt{\ 3}\right)^2}{3}$$

Area of the shaded region= $$\frac{\pi\ r^2\left(2+\sqrt{\ 3}\right)^2}{3}$$ - 3$$\pi\ r^2$$

=$$\frac{\pi\ r^2\left(4\sqrt{\ 3}\ -2\right)}{3}$$

Since the radii of the small circles are maximum, the centres of the small circles should lie on the lines joining the vertices of the big triangle with the center of the big circle. Let the radius of the small circle be r.
OA = 8
OP = 8*sin 30 + r = 4+r
OD = 4+2r = radius of the big circle = 8 => r = 2
So, OP = 6 => side of triangle PQR = OP cos 30 * 2 = 6 * $$\sqrt3$$/2 * 2 = 6 * $$\sqrt3$$

Triangle PQR is also an equilateral triangle.

So, area of the triangle PQR = $$\sqrt3/4 * side^2$$ = $$\sqrt3/4 * 36 * 3$$ = $$27\sqrt3 cm^2$$

According to the given conditions,

$$\frac{\left(n-2\right)180}{n}:\frac{\left(m-2\right)180}{m}=3:4$$

$$1-\frac{2}{n}:1-\frac{2}{m}=3:4$$

$$\frac{1-\frac{2}{n}}{1-\frac{2}{m}}=\frac{3}{4}-->4-\frac{8}{n}=3-\frac{6}{m}-->\frac{8}{n}-\frac{6}{m}=1$$

$$8m-6n=mn\ -->\ mn-8m+6n-48=-48$$

$$m\left(n-8\right)+6\left(n-8\right)=-48$$

$$\left(n-8\right)\left(m+6\right)=-48$$

$$\left(8-n\right)\left(m+6\right)=48$$

Now, n can take values from 3 to 7.

If n = 3, we get 8 - 3 as 5, that is not divisible by 48.

If n = 4, 8 - 4 = 4, hence m = 6

n = 5, m = 10

n = 6, m = 18

n = 7, m = 42

Hence, for different values, we get different values of m+n. Hence, the answer cannot be determined.

The sum of the areas of quadrilaterals TRCQ and STPA = The sum of the areas of quadrilaterals TRDS and QTPB = $$\frac{1}{2}\times\ \left(area\ of\ rectangle ABCD\right)$$

Therefore, the area of the rectangle = $$2\times\ \left(43+17\right)\ =120\ cm^2$$

Let $$l$$ and $$b$$ be the length and breadth of the rectangle. Then,

$$Area\left(\triangle\ SDR\right)=Area\left(\triangle\ QCR\right)=\frac{1}{2}\times\ \frac{l}{2}\times\ \frac{b}{2}=\frac{lb}{8}$$

The sum of both the triangles = $$\frac{lb}{4}$$

We know that $$\left(l\times\ b\right)\ =\ 120\ cm^2$$

On substituting the value of $$\left(l\times\ b\right)$$, we get the sum of triangles = 30 $$cm^2$$

Area of the triangle, ABC= $$\frac{\sqrt{\ 3}}{4}a^2$$= $$\frac{\sqrt{\ 3}}{4}12^2$$= $$36\sqrt{\ 3}$$

The area of the equilateral triangle DEB= $$\frac{\sqrt{\ 3}}{4}6^2$$= $$9\sqrt{\ 3}$$

A side joining the midpoints of the of two sides will divide the triangle in the ratio 1:3.

The triangles ADF, DEF, EFC are equilateral and congruent. ADEC is divided into three equal parts. So the area of DEF and DEB will be equal.

Thus the triangle DEF=  $$9\sqrt{\ 3}$$ => Side of DEF = 6

IJ divides AD in the ratio 1:1 => It also divides FD in the ratio 1:1. Hence FH = 3

Since DEF and HFG are similar triangles, $$\frac{\left(area\ of\ HFG\right)}{\left(Are\ of\ DEF\right)}=\ \frac{FH^2}{DE^2}$$

$$\left(area\ of\ HFG\right)=\ \frac{9}{36}\cdot9\sqrt{\ 3}$$

The area of the triangle HFG =1/4 * area of the triangle DEF.

=> The area of the triangle HFG = $$\frac{9}{4}\sqrt{\ 3}$$

Option D

Let us draw the diagram according to the information provided in the problem statement.

Let us draw the third median from point Q to side PR. The median will pass through point U (centroid) as shown in the figure below

In triangles PTR and RSP,
PT = RS {Since PQ = QR}
$$\angle$$TPR = $$\angle$$SRP {In triangle PQR, PQ = QR. Hence, $$\angle$$ QPR = $$\angle$$ QRP}
Side PR is common on both the triangles. Hence, we can say that both the triangles are congruent.
Therefore, PS = RT. Consequently, PU = RU.

In triangles PUV and RUV,
PU = RU
PV = RV
Also, side UV is common in both the triangles. Hence, we can say that both triangles PUV and RUV are congruent.

Therefore, we can say that $$\angle$$ UVP = $$\angle$$ UVR = $$\dfrac{180°}{2}$$ = 90°
Also, $$\angle$$ VUP = $$\angle$$ VUR = $$\dfrac{60°}{2}$$ = 30°

In right-angle triangle PUV,
tanPUV = $$\dfrac{PV}{UV}$$

$$\Rightarrow$$ tan(30°) = $$\dfrac{15}{UV}$$
$$\Rightarrow$$ $$\dfrac{1}{\sqrt{3}}=\dfrac{15}{UV}$$
$$\Rightarrow$$ $$UV=15\sqrt{3}$$ cm.

We know that the centroid divides in 2:1.
Hence, we can say that QV = 3*UV = $$3*15\sqrt{3}$$ cm = $$45\sqrt{3}$$ cm

Hence, the area of triangle PQR = $$\dfrac{1}{2}$$*PR*QV = $$\dfrac{1}{2}*30*45\sqrt{3}$$ = $$675\sqrt{3}$$ cm$$^2$$. Therefore, option E is the correct answer.

The two train stops can be selected in $$^{12} C_2$$ ways and the direction can be selected in 2 ways. So, total number of ways is 12*11 = 132

An even digit palindrome number is always divisible by 11, while an odd digit palindrome number is not always divisible by 11. So, P(10000) > P(5000) > P(1000).

Let the term with coefficient $$x^8$$ in the expansion $$(px^2+\frac{1}{qx})^{13}$$ be the $$a^{th}$$ term.
So, 2*a-(13-a) = 8
or a=7
So, the coefficient of $$x^8$$ is $$^{13} C_7 *p^7*q^{-6}$$

Similarly, the term with coefficient $$x^{-8}$$ in the expansion $$(px-\frac{1}{qx^2})^{13}$$ is the sixth term.
So, the coefficient of $$x^{-8}$$ is -$$^{13} C_6 * p^6*q^{-7}$$

As, both of them are equal, pq = -1

When two dices are thrown simultaneously, the total possible outcomes are 6*6 = 36
Instead of calculating the outcomes where the sum is less than 9, we can easily calculate the cases where the sum is greater than or equal to 9. Such cases are
(5,4), (4,5), (6,3), (3,6), (5,5), (6,4), (4,6), (6,5), (5,6), (6,6)
So there are 10 cases. Hence the probability that sum is greater than or equal to 9 is $$\frac{10}{36}$$ = $$\frac{5}{18}$$
So the probability that sum is less than 9 is 1- $$\frac{5}{18}$$ = $$\frac{13}{18}$$

It has been given that there are 16 equally spaced points on the circumference of the circle.
If the diameter is one of the sides of the triangle, then the triangle will be right-angled.
If 2 points are present on one side of the diameter and the third point is present on the other side, the resulting triangle will be acute-angled.
For a triangle to be obtuse-angled, all the 3 points must be present on the same side of the diameter.

The number of points that will be present on one side of the diameter = (16/2) - 1 = 7.
Let us first select the left-most point of the triangle. This can be done in 16 ways.
The other 2 points that form the triangle should be selected from the 7 points to the right of the triangle.
Number of ways in which 2 out of these 7 points can be selected = 7C2.
Therefore, the number of obtuse-angled triangles that can be formed = 16*7C2
Number of ways in which 3 points can be selected from 16 points = 16C3.
Therefore, the required probability is 16*7C2/16C3 = (16*7*3)/(16C3) = 3/5.
Therefore, option D is the right answer.

The toddler is placed at (0, 0) Hence, we can say that he has to travel 5 units in the x-direction and 5 units in the y-direction.
Possible cases of 5 units that have to be covered by the toddler = 5 steps of 1 unit each (1+1+1+1+1) or 2 steps of 1 unit each and 1 step of 3 units (3+1+1).

Case 1: When the toddler moves by (1+1+1+1+1) in both x and y directions.
The number of ways in which the toddler can reach to the toy = $$\dfrac{10!}{5!*5!}$$ = 252.

Case 2: When toddler moves by (1+1+1+1+1) in any one of x and y-direction and by (3+1+1) in the other direction.
The number of ways in which the toddler can reach to the toy = $$2*\dfrac{8!}{5!*2!}$$ = 336.

Case 3: When the toddler moves by (3+1+1) in both x and y directions.
The number of ways in which the toddler can reach to the toy = $$\dfrac{6!}{2!*2!}$$ = 180.

Hence, the total number of ways in which the toddler can reach the toy = 252+336+180 = 768.

$$17^{128} = (21-4)^{128}$$ So, the reminder when $$17^{128}$$ is divided by 21 equals the reminder when $$4^{128}$$ is divided by 21. $$4^3$$ leaves a reminder of 1 when divided by 21. Hence, the reminder when $$17^{128}$$ is divided by 21 equals the reminder when $$4^{126}*4^2$$ is divided by 21. That is 16.

Let the two digit consecutive even numbers be 2n, 2n+2, 2n+4 and 2n+6

2n$$\geq$$10 and 2n+6 $$\leq$$ 98

n$$\geq$$5 and n$$\leq$$46

$$2n+2n+2+2n+4+2n+6 = p^{2}$$
As p is even, let it be 2k.
So, $$2n+3 = k^{2.}$$
Assume $$k=2l+1$$,

$$2n\ +\ 3\ =\ k^2$$

$$2n+2\ =\ k^2-1$$

$$2\left(n+1\right)=\left(k-1\right)\left(k+1\right)$$

$$2\left(n+1\right)=2l\left(k+1\right)$$

$$n\ +\ 1\ =\ l\left(k+1\right)$$

k = 2l+1

So, $$n+1=2*l*(l+1)$$.
As, 5$$\leq$$n$$\leq$$46, for l = 2,3 and 4, we get n = 11,23 and 39 or the first even digit as 22, 46 and 78.
Hence, number of quartets is 3

Rita has to run the tool once to get the LCM of two numbers.

To find the LCM of two numbers say 2, 4, he has to run the tool once.

To find the LCM of three numbers say 2, 4, 6

Rita will run the tool once to find the LCM of 2, 4 and then run the tool once again with 6 and the LCM of 2, 4

On similar lines

To calculate the LCM of 20 numbers, she has to run the tool 19 times.

$$\frac{1}{X}+\frac{5}{Y}=\frac{1}{10}$$

$$\frac{1}{X}=\frac{1}{10}-\frac{5}{Y}$$

$$\frac{1}{X}=\frac{Y-50}{10Y}$$

$$X=\frac{10Y}{Y-50}$$

Now, since both X and Y are positive, Y has to be > 50.

Y = 50 + K, K > 0, K < 20

$$X=\frac{10Y-500}{Y-50}+\frac{500}{Y-50}$$

$$X=10+\frac{500}{Y-50}$$

X = 10 + 500/K

K has to be a factor of 500.

K = 1, 2, 4, 5, 10

Y = 51, 52, 54, 55, 60

Given that $$N = 882000 = 2^4*3^2*5^3*7^2$$
The number of the factors that will have only one zero when we will have only one pair of 2 and 5.

Case 1: Exactly one 2 and at least one 5.
Then total number of factors = 1*3*3*3 = 27

Case 2: Exactly one 5 and at least one 2.
Then total number of factors = 4*3*1*3 = 36

We can see that we have double counted the cases when we will have exactly one 2 and exactly one 5. Total number of such factors = 1*3*1*3 = 9

Therefore, total number of factors ending with exactly one zero = 27 + 36 - 9 = 54

Let the number of students who play all the three sports be ‘c’. Number of students who play exactly two sports = b
Number of students who play exactly one sport = a
So, a + b + c = 200
a + 2b + 3c = 60 + 80 + 120 = 260
So, b + 2c = 60
c = 20
So, b = 60 - 40 = 20
So, the number of students who play at least two sports = b + c = 20 + 20 = 40

Number of times students passed= 35+45+25+30=135

I+II+III+IV=50

I+2*II+3*III+4*IV=135

so, II+2*III+3*IV=85

to maximize I i.e. the number of students passed in exactly one subject we have to maximize IV, but maximum value of IV can be 25

.'. II+2III=10 now maximize III, the maximum value of III=5.

So, IV=25, III= 5, II=0

.'.I=50-5-25=20

The maximum number of students who have passed in exactly one subject=20 students.

Let the no. of students who chose 1 elective, 2 electives and 3 electives are X, Y, Z respectively.

X= a+b+c

Y=d+e+f

Z = g

X +2Y+3Z = 50+70+90=210  ---- (1)
X+Y + Z =100-------(2)

Subtracting 2nd equation from 1st We get, Y+2Z =110 Now the minimum value of z is not 0, because if z=0, y becomes 110 which is not possible.

Y+2Z=110 -----(3)
X+Y+Z=100 ------(4) Subtract 4 from 3
z-x=10,
z=10+x The minimum value of z=10 when x=0.

Let R, M and C be the sets of students of GRE, GMAT and CAT. Let x be the # who are taking all three exams.
To maximize x, we assume that $$n(R \cap M \cap C') = n(R \cap C \cap M') = n(M \cap C \cap R')=0$$.
Hence, 400=(n(R)-x)+(n(M)-x)+(n(C)-x)+x=300+220+50-2x=400.
Hence x=85. However, x cannot be greater than 50. Hence the maximum # of students is 50.