Question

MATHEMATICS
1. Divide 300 into parts so that half of one part is less than the other part by 48. A. 160, 140. B. 168, 132. C. 150, 150. D. 120, 180
2. Find the area of the triangle whose vertices are (x, x), (0, 2x), (2x, 0). A. 4x2 units B. 0 units C. 2x2 units. D. X2 units
3. A person has 3 shares in a lottery containing 2 prizes and 8 blanks. The chance of getting a prize is A. 8/15. B. 12/15. C. 7/15. D. 6/15
4. A student's marks were wrongly entered as 83 instead of 63. As a result, the average marks of the class increased by 0.5. The number of students in the class is. A. 10. B. 20. C. 40. D. 73
5. Water tax is increased by 20% but its consumption is decreased by 20%, then the increase or decrease in the expenditure is A. 5% B. 4% C. 6% D. 8%

Answer 1

1. Let one number be x, and another be y. Create a system of equations:

$$\begin{cases} x+y=300\\ \cfrac{x}{2}=y-48 \end{cases}$$ $$\begin{cases} x=2y-96\\ 3y=396 \end{cases}$$ $$\begin{cases} x=168\\ y=132 \end{cases}$$

Answer: B. 168, 132.

2.Use formula for points A(x₁;y₁) B(x₂;y₂) and C(x₃;y₃):

$$S=\cfrac{1}{2}*|(x₁-x₃)(y₂-y₃)-(x₂-x₃)(y₁-y₃)|$$

A(x;x) B(0;2x) and C(2x;0)

$$S=\cfrac{1}{2}*|(x-2x)(2x-0)-(0-2x)(x-0)|$$

$$S=\cfrac{1}{2}*|-x*2x-(-2x)*x|$$

$$S=\cfrac{1}{2}*|-2x^{2}+2x^{2}|$$

$$S=\cfrac{1}{2}*0 = 0$$

Answer: B. 0 units.

3. Calculate the number of possible options when using three attempts.

$$C_{10}^{3}=\cfrac{10!}{(10-3)!*3!}=\cfrac{10!}{7!*3!}=\cfrac{8*9*10}{2*3}=120$$

Сalculate the number of possible options for which all 3 shares are losing.

$$C_{8}^{3}=\cfrac{8!}{(8-3)!*3!}=\cfrac{8!}{5!*3!}=\cfrac{6*7*8}{2*3}=56$$

Consider the probability (A) at which all 3 shares are losing.

$$A=\cfrac{56}{120}=\cfrac{7}{15}$$

Consider the chance (B) of getting a prize. 

$$B=1-A=1-\cfrac{7}{15}=\cfrac{8}{15}$$

Answer: A. 8/15.

4.  Let the average score be C, the number of students be N and the sum of the scores of students be S. Then:

$$С=\cfrac{S}{N}$$

From the condition of the question we know that:

$$\cfrac{S}{N}+\cfrac{20}{N}=C+0.5;$$

$$C+\cfrac{20}{N}=C+0.5;$$

$$\cfrac{20}{N}=0.5; N=\cfrac{20}{0.5}=40$$

Answer: C. 40.

5. Let the amount of water consumed be x.

After raising the tax the expenditure will be

$$x_1=x+0.2x=1.2x.$$

After reducing the new water consumption the expenditure will be

$$x_2=x_1-0.2x_1=0.8x_1=0.8 \cdot 1.2x=0.96x.$$

So an increase or decrease in expenses is the difference between $x_2$ and $x$, divided by $x$ :

$$\frac{x_2-x}{x}=\frac{0.96x-x}{x}=-0.04$$

$$0.04*100\%=4\%$$

Answer: B. 4%.