Mettl Profit and Loss Quiz 1

Answer:
Let Cost price of each T-shirt = s
Cost price of each Jeans = p
Selling price of each Tshirt = h
And, Selling price of each Jeans = a
From question
h + a = 1.25 (s + p) …….(i)
h = 1.2s …… (ii)
(h + a – 200) = 1.3 (s + p – 200) …. (iii)
And,p = \frac{h}{2} - 80 ……… (iv)
Solving the above system of equation we get,
s = 800, p = 400, h = 960 and, a = 540
Total cost price of a pair of T-shirt and jeans = 800 + 400 = 1200
Required marked price = 1200 × 1.5 = Rs.1800

Answer:
Cost price of Honda Bike for Raj = \frac{12100}{1.1} = 11000
Since selling price of Honda Bike by Rakesh = Rs. 11000
From question Cost price of Honda Bike for Rakesh
=\frac{11000}{0.9} = Rs. 12222.22
Since selling price of Honda Bike by Sohan = Rs. 12222.22
According to the question, Sohan got a 20% profit
Since Cost Price of Honda Bike for Sohan
=\frac{12222.22}{1.2} = Rs. 10185.2
This cost price includes an Rs. 1500 for repairs.
Hence purchase price for Sohan = 10185.2 - 1500 = Rs. 8685

Answer:
Let the number of balls, wickets, and bats purchased be A, B, and C, respectively.
Thus,
20A + 5B + C = 1000 and A + B + C = 100
Solving the above two equations by eliminating C, we Get
19A + 4B = 900
\Rightarrow B = 225 - \frac{19}{4}A
Now, as B is the number of wickets and 0 < B < 99,
So, putting these limiting values of B in the above equation will provide the value of A as
27 < A < 47. Since A has to be the multiple of 4, possible values of A are 28, 32, 36, 40, and 44. Now, for A = 28 and 32; A + B > 100, so these values of A can be rejected.
For all other values of A, we get the desired solution:
A = 36, B = 54, C = 10
A = 40, B = 35, C = 25
A = 44, B = 16, C = 40
Thus, there are three possible solutions.

Answer:
For profit to be maximum, the derivative of p with reference to x must be 0 and hence
=\frac{d (250x – 5x^{2})}{dx} = 0 = 250 – 10x = 0
So x = 25
Now p for x = 25 is
= 250 (25) – 5 (25)² = Rs 3125

Answer:
Let the price of the ticket be = Rs. (60 + x), where x is greater than zero.
The number of people in the audience would then be (300 - 2x).
The revenue of the theatre owner be = (60 + x) x (300 – 2x)
= (18000 + 180x - 2x²)
This is quadratic expression which achieves a maximum value when x
= \frac{–\text{ coefficient of x }}{2\times \text{ coefficient of }x^{2}}
Quadratic equation has achieved a maximum value when x
= \frac{-b}{2a}
So, x = \frac{-180}{-2\times 2} = 45
Hence, the price of ticket at maximum revenue = (60 + 45) = Rs.105.

Answer:
After a markup of P%, the marked price becomes Rs. (1000 + 10 × P)
After a discount of (0.4 × P) % , the selling price becomes
(1000 + 10 \times P)\times \left[ 1 - \left( 0.4\times \frac{P}{100} \right) \right]
=1000 + 6 \times P – \frac{P^2}{25}......(i)

Given that the final profit = (0.4 x P)%
⇒ Selling Price =
\left [ 1000 + 1000\times \left ( 0.4\times \frac{P}{100} \right ) \right ]
= (1000 + 4 × P)…. (ii)
As, equation (i) is equal to equation (ii)
1000 + 6 \times P - \frac{P^{2}}{25} = (1000 + 4 × P)
\Rightarrow (6 \times P - 4 \times P) = \frac{P^{2}}{25}
⇒ P = 50
As, discount = Market price – Selling price
= (1000 + 10 x P) – (1000 + 4 x P) = 6 × P = (6 x 50) = 300
Hence, amount of discount = Rs. 300.

Answer:

Let the cost price are Rs. 200, Rs. 300 and Rs. 400

Answer:
Let the number of apricots, bananas and Guava bought to be a, b and g.
Given that a + b + g = 80
a ≥ 25, b ≥ 25, g ≥ 25
⇒ 25 ≤ a, b, c ≤ 30
As the increase in cost per guava by Re.1 and the increase in cost per banana by Rs.4 increases the overall bill by Rs.136, g + 4b = 136
In order to satisfy, the above condition, g must be a multiple of 4. Hence, it has to be 28.
Hence, b is 27 and a is 25.
Hence, Pranav purchased 27 bananas

Answer:
Let the capitals of Aman and Bhanu be in the ratio of a : b
i.e \frac{a}{b} = \frac{400}{600} = \frac{2}{3}..........(i)
Further \frac{a - 3000}{b - 1000} = \frac{375}{625} = \frac{3}{5}..........(ii)
Solving (1) and (2)
a = 24000 where b = 36,000
If Aman had invested 3000 more and Bhanu had invest 3000 less,
then a : b = (24000 + 3000) : (36000 – 3000) = 9 : 11
Bhanu's share = Profit\frac{11}{11 + 9}
= 1000\times \frac{11}{20} = 550

Answer:

The cost and the number (actual and 'if) of balls are tabulated below.