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We are given only one or two equations, but constraints are there on variables such as the solutions must be integers, positive integers etc.
Eg. ax + by = c
Step 1: Find any single integer solution by trial and error. Let the solution be (x0, y0)
Step 2: For every b units x increases, y must decrease by a units (or vice-versa).
Step 3: Check the constraints
Let a and b be natural numbers. If $$a^2 + ab + a = 14$$ and $$b^2 + ab + b = 28$$, then $$(2a + b)$$ equals
a(a + b + 1) = 14 ...... (1)
b(a + b + 1) = 28 ...... (2)
$$\frac{a}{b}=\frac{1}{2}$$
b = 2a
Substituting in (1), we get
a(3a + 1) = 14
$$3a^2+a-14=0$$
$$3a^2-6a+7a-14=0$$
$$3a\left(a-2\right)+7\left(a-2\right)=0$$
Given, a and b are natural numbers.
Therefore, a = 2 and b = 4
2a + b = 2(2) + 4 = 8
The answer is option A.
For natural numbers x, y, and z, if xy + yz = 19 and yz + xz = 51, then the minimum possible value of xyz is
Correct Answer: 34
It is given, y(x + z) = 19
y cannot be 19.
If y = 19, x + z = 1 which is not possible when both x and z are natural numbers.
Therefore, y = 1 and x + z = 19
It is given, z(x + y) = 51
z can take values 3 and 17
Case 1:
If z = 3, y = 1 and x = 16
xyz = 3*1*16 = 48
Case 2:
If z = 17, y = 1 and x = 2
xyz = 17*1*2 = 34
Minimum value xyz can take is 34.
A donation box can receive only cheques of ₹100, ₹250, and ₹500. On one good day, the donation box was found to contain exactly 100 cheques amounting to a total sum of ₹15250. Then, the maximum possible number of cheques of ₹500 that the donation box may have contained, is
Correct Answer: 12
Let the number of 100 cheques, 250 cheques and 500 cheques be x, y and z respectively.
We need to find the maximum value of z.
x + y + z = 100 ...... (1)
100x + 250y + 500z = 15250
2x + 5y + 10z = 305 ...... (2)
2x + 2y + 2z = 200 ....... (1)
(2) - (1), we get
3y + 8z = 105
At z = 12, x = 3
Therefore, maximum value z can take is 12.
A fruit seller has a stock of mangoes, bananas and apples with at least one fruit of each type. At the beginning of a day, the number of mangoes make up 40% of his stock. That day, he sells half of the mangoes, 96 bananas and 40% of the apples. At the end of the day, he ends up selling 50% of the fruits. The smallest possible total number of fruits in the stock at the beginning of the day is
Correct Answer: 340
Let us assume the initial stock of all the fruits is S.
Let us take we have 'b' and 'a' mangoes initially.
Stock of Mangoes = 40% of S = 2S/5
The total number of fruits sold are Mangoes Sold + Apples Sold + Bananas Sold
= 2S/10 + 96 + 4a/10 = S/2 (Given)
=> S/5 + 96 + 2a/5 = S/2
=> S = $$\dfrac{\left(4a+960\right)}{3}$$
=> $$\dfrac{4a}{3}+320$$
'a' has to be a multiple of 3 for the above term to be an integer.
But 'a' has to be a multiple of 5 for 4a/10 to be an integer.
=> The smallest value of 'a' satisfying both conditions is 15.
=> $$\dfrac{4a}{3}+320=\dfrac{4\left(15\right)}{3}+320$$ = 340
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