Let the total amount of money with him be 300x. It is given that he can buy 60 shirts or 50 trousers with this amount of money.

Hence, the cost price of one shirt is (300x/60) = 5x, and the cost price of one trousers is (300x/50) = 6x.

He uses 10% of his money to buy a cold drink, which implies the cost price of the cold drink is (300x * 0.1) = 30x. He buys 24 shirts, and the cost price of 24 shirts = (24*5x) = 120x.

The amount of money still left with him = (300x - (30x+120x)) = 150x.

For 6x, he can buy 1 trouser => he can buy (150x/6x) = 25 trousers with the rest of the money

Let then number be n = 2622m + 69 = 2620m + 2m + 68 +1 = multiple of 4 + 2m +1
Now if m = 1 then answer is 3;
m= 2, the answer is 1.
Hence, the remainder can be either 1 or 3 depending on m.
So, it can't be determined

Let Arjun’s initial average be ‘a’.
Total marks scored in 95 mocks = 95a
Total marks scored in 96 mocks = 96(a+x)
=>95a+384 = 96(a+x)
=>a = 384-96x
The values possible for x are 0,1,2,3
If x = 4,
a will be 0 (It is given that a is a natural number)
Hence x cannot be 4.
The number of possible values of x is 4.

There are a total of 25 prime numbers below hundred.

Based on this the original probability of picking a prime numbered ball must be 25/100 = 1/4.

But since Ramesh misreads this to 140 he calculates this for 140.

Between 100 and 140  the additional prime numbers are 101, 103, 107, 109, 113, 127, 131, 137, 139.

Hence the probability that Ramesh calculates is 34/140.

The original probability is 1/4 = 35/140.

The probability calculated by Ramesh = 34/140.

The difference between the two probabilities  = 35/140 - 34/140 = 1/140

Hence, Option C is the correct choice.

Let us assume that a man does ‘x’ units of work a day and woman does ‘y’ units of work a day.
Total work = 20(x)(10)+10(y)(10) = 12(x)(10)+16(y)(10)
Hence, 80x = 60y.
4x = 3y.
Total work = 120x+160y = 30(3y)+160y = 250y.
Let the number of days taken by 25 women be ‘d’.
25(y)(d) = 250y
d = 10 days.

There are eight letters of which only 6 are unique. The remaining two are repeated.
=> Number of words that can be formed = $$\frac{8!}{2!*2!}$$ = 10080

Let the capacity of the tank be 180 units. So inlet pipe can fill 12 units in 1 minute and outlet pipe can empty 5 units in a minute. When both pipes are open, 7 units will be filled in a minute. Hence in 10 minutes, 70 units will be filled. Now another outlet pipe is connected. Hence the effective rate of inflow would become 2 units per minute. So it would take 55 minutes to fill 110 units.
Thus the total time which is taken = 55 + 10 = 65 minutes.

One of the roads would be parallel to the length and the other to the breadth of the park.

Let the width be d.

Area of the road parallel to length = 30d

Area of road parallel to width = 40d

Area of overlap = $$d^2$$

Hence, total area of road = $$40d + 30d - d^2$$ = $$70d - d^2$$

Total area of the park = 30 x 40 = 1200

As the total area of the two roads is equal to the remaining area of the park, the total area of the roads should be half the total area of the park. 

$$70d - d^2$$=1200/2 = 600

$$d^2 - 70d + 600 =0$$

d(d-60)-10(d-60)=0

(d-10)(d-60)=0

d=10m or d=60m

As d cannot be greater than the dimensions of the park, d=10m

Let the speed of Abhish be b (in kms/hr) and the speed of Aditi be d (in kms/hr). Then, 5b+5d=660 (from the first statement)

Also, it is given that if both of them head north, they will meet in 55 hours. This means that Abhish will have to travel 660 kms+the distance covered by Aditi in 55 hours and Aditi will travel the distance covered by her in 55 hours.

So, 660+55d=55b

Substituting the value of d in equation 2 we get

660+55(132-b)=55b

Or 660+7260=110b

So, b=72

So, to cover a distance of 660kms, Abhish will take $$\frac{660}{72}$$=9.1667 hours

=9 hours and 10 minutes

The area of a cyclic quadrilateral is given by $$\sqrt{(s-a)(s-b)(s-c)(s-d)}$$, where s is the semi-perimeter.
In this case, s = (14+17+18+23)/2 = 36.
Area = $$\sqrt{(36-14)(36-17)(36-18)(36-23)}$$ = 6$$\sqrt{2717}$$.

Let the distance travelled by the person by bus be D.
So, distance travelled by bike = 250 - D
Speed of the bus = 30 kmph
Speed of the bike = 40 kmph
Time taken for the bus journey = D/30
Time taken for the bike journey = (250-D)/40

So, D/30 + (250-D)/40 = 8

=> 4D + 3*(250-D) = 8 * 120
=> 4D - 3D + 750 = 960
=> D = 960 - 750 = 210 km

So, distance travelled by the person by bike = 250 - 210 = 40 km

It is given that the length of the tangent is equal to the radius of the circle.
Thus, OB = AB = 12 cm.
Triangle OAB is an isosceles right triangle as shown.

Angle AOB = 45 degrees.
Area of shaded region = Area of triangle AOB - Area of the sector BOC
Area of shaded region = $$\frac{1}{2}*12*12-\frac{\pi*12*12}{8}$$
Area of shaded region = $$18(4-\pi)$$

Let the cost required for manufacturing the TV be $$100x$$.

Thus, the wholesaler buys the TV from the manufacturer at $$1.1\times 100x = 110x$$.

The wholesaler then sells the TV to a Retailer at $$1.2\times 110x = 132x$$

The Retailer marks up the price by $$25$$%, and thus, the customer buys it at $$1.25\times 132x = 165x$$

The customer pays Rs.$$66330$$ for the TV. Thus,

$$165x = 66330$$

$$x$$ = $$\frac{66330}{165}$$

$$x = 402$$

The cost of manufacturing the TV = $$100\times 402 = 40200$$

Hence, $$40200$$ is the correct answer.

Draw a perpendicular to side AC from B. Now a perpendicular from vertex to the opposite side in an equilateral triangle divides it into two congruent triangles.

FBG = $$\frac{60}{3}$$=20$$^{\circ\ }$$

Let the perpendicular be BX. Then XBF=10$$^{\circ\ }$$. Therefore angle BFG= 180-(90+10)=80$$^{\circ\ }$$.

The incircle of an equilateral triangle will make an angle of $$\frac{360}{3}=120^{\circ\ }$$ with any two vertices.

Therefore, the absolute difference= 120-80=40$$^{\circ\ }$$.

Cost of a pen two years from now = 10 * (1+500%) * (1+500%) = Rs. 360

The given series is a geometric progression with first term equal to $$\frac{-3}{2}$$ and common ratio equal to -2
Hence, the sum of the first n terms of the series equals $$\frac{-3}{2}*\frac{(-2)^n-1}{-2-1}=-1024.5$$
Hence, $$n=11$$
 

Since the amounts are equal.

$$P_1\left(1+0.2\cdot2\right) = P_2\left(1+0.2\right)^2$$

$$1.4P_1 = 1.44P_2$$

$$\frac{P_1}{P_2} = \frac{1.44}{1.4} = \frac{144}{140} = \frac{36}{35}$$

Since the total amount is 71 millions, hence the first principle is 36 millions and the second principle is 35 millions.

Hence, $$P_1$$ is 36 millions.

Let $$\frac{\log_{1/3} a}{x-y} = \frac{\log_{1/3} b}{y-z} = \frac{\log_{1/3} c}{z-x}$$ = k

=> a = $$(1/3)^{k(x-y)}$$, b = $$(1/3)^{k(y-z)}$$, and c = $$(1/3)^{k(z-x)}$$

So, $$a^z * b^x * c^y$$ = $$(1/3)^{kxz - kyz + kyx - kzx + kzy - kxy}$$ = $$(1/3)^0$$ = 1

Total value in all the experiments put together = 131.5
Sum of values of the first 3 experiments = 55.5
Sum of the values of the remaining 4 = 76
Sum of the values of 6th and 7th experiments = 21*2 = 42
=> 4th value + 5th value = 34
=> (x+1)+x = 34 => 2x = 33 => x = 16.5
So, value obtained in the 4th experiment = 16.5 + 1 = 17.5

$$(\frac{1}{a}-1)= \frac{b+c}{a}$$.

=$$(\frac{b}{a}+\frac{c}{a})(\frac{a}{b}+\frac{c}{b})(\frac{a}{c}+\frac{b}{c})$$

=$$(1+\frac{c}{a}+\frac{c}{b}+\frac{c^{2}}{ab})(\frac{a}{c}+\frac{b}{c})$$

=$$\frac{a}{c}+1+\frac{a}{b}+\frac{c}{b}+\frac{b}{c}+\frac{b}{a}+1+\frac{c}{a}$$
So, the product when expanded equals $$2+\frac{a}{b}+\frac{b}{a}+\frac{a}{c}+\frac{c}{a}+\frac{b}{c}+\frac{c}{b}$$
It should be noted that the sum of the six fractions is greater than or equal to 6.
So, the minimum value of $$(\frac{1}{a}-1)(\frac{1}{b}-1)(\frac{1}{c}-1)$$ equals 8

Let a, b and c be the coins used of each denomination.
 Hence, a+4b+16c=75. The maximum value c can take is 4.
 When c is 4, b can at max be 2. Hence, b can vary from 0 to 2: 3 cases.
 When c=3, b can vary from 0 to 6: 7 cases.
when c = 2, b can vary from 0 to 10: 11 cases
when c = 1, b can vary from 0 to 14: 15 cases
when c = 0, b can vary from 0 to 18: 19 cases
Hence the total number of ways = 3+7+11+15+19=55.

A+B = X+Y = a
AB = c, XY = c+3
Since their sum is equal, A + B = X + Y => B -Y = -(A - X)
Case 1 : A - X = 1
=> A = 1 + X and B = Y - 1
Hence AB = (X+1)(Y-1) = XY+Y-X-1 = c
=> c+3+Y-X-1 = c
=> X-Y = 2
Case 2: A -X = -1
=> A = X - 1 and B = Y + 1
Hence AB = (X-1)(Y+1) = XY+X-Y-1 = c
=> c+3+X-Y-1 = c
=> X-Y = -2
Hence absolute value of difference between X and Y is 2