Basic Level Examination (BLE) - 2080
Kathmandu Metropolitan City, Education Department
Attempt all questions. Figures in the margin indicate full marks.
Two sub-sets of the universal set $U = \{1, 2, 3, 4, 5, 6\}$ are $A = \{1, 3, 4, 5\}$ and $B = \{2, 3, 5\}$.
(a) Are sets $A$ and $B$ overlapping or disjoint sets? Write with reasons. [1]
(b) Write any two sub-sets of set $A$ having a single element. [1]
(c) Show sets $U$, $A$, and $B$ in a Venn diagram. [1]
(a)
Given, $A = \{1, 3, 4, 5\}$, $B = \{2, 3, 5\}$
$$A \cap B = \{3, 5\} \neq \emptyset$$
Since sets $A$ and $B$ have common elements ($3$ and $5$), they are overlapping sets.
(b)
Subsets of set $A$ having a single element are: $\{1\}$ and $\{3\}$
(c)
Venn diagram representing sets $U$, $A$, and $B$ is shown in the figure above.
Gopal went to market to buy a watch. The marked price of the watch is $\text{Rs. } 3500$.
(a) If marked price is represented by $MP$, discount amount by $D$, and selling price by $SP$, write the formula of $SP$. [1]
(b) How much discount did Gopal get while buying the watch on a discount of $12\%$? [1]
(c) The shopkeeper got $25\%$ profit after selling it at $12\%$ discount. What was the cost price ($CP$) of the watch? [2]
(a)
Formula: $$SP = MP - D$$
(b)
Discount Amount $D = 12\% \text{ of } 3500 = \frac{12}{100} \times 3500 = \text{Rs. } 420$
(c)
$$SP = 3500 - 420 = \text{Rs. } 3080$$
$$CP = \frac{SP \times 100}{100 + P\%} = \frac{3080 \times 100}{125} = \text{Rs. } 2464$$
Aashika deposited money in bank A and bank B in the ratio of $3:2$. She deposited $\text{Rs. } 60,000$ in bank A.
(a) How much did Aashika deposit in bank B? [1]
(b) How much simple interest will Aashika obtain in $2$ years at the rate of $5\%$ per annum from bank A? [2]
(c) For how many years will Aashika have to deposit the amount in bank B at the same interest rate to obtain the same amount of interest as from bank A? [2]
(a)
Let deposits be $3x$ and $2x$. $3x = 60000 \implies x = 20000$.
Bank B Deposit = $2 \times 20000 = \text{Rs. } 40,000$
(b)
$$I_A = \frac{P \times T \times R}{100} = \frac{60000 \times 2 \times 5}{100} = \text{Rs. } 6000$$
(c)
$$T_B = \frac{I \times 100}{P_B \times R} = \frac{6000 \times 100}{40000 \times 5} = 3 \text{ years}$$
The number $8657$ is written on the number plate of a taxi.
(a) Write the taxi number $8657$ in scientific notation. [1]
(b) Convert the taxi number $8657$ into the quinary (base-5) number system. [1]
(c) Change the repeating decimal $0.\bar{8}$ into a common fraction. [1]
(d) Simplify: $\frac{7.8 \times 10^{-12}}{1.3 \times 10^{-12}}$. [2]
(a)
$$8657 = 8.657 \times 10^3$$
(b)
$$8657_{10} = (234112)_5$$
(c)
$$0.\bar{8} = \frac{8}{9}$$
(d)
$$\frac{7.8 \times 10^{-12}}{1.3 \times 10^{-12}} = 6$$
The rectangular compound of a $35\text{ m} \times 28\text{ m}$ house belongs to Sagun. He made a circular swimming pool with radius $r = 7\text{ m}$ inside it.
(a) Write the formula to calculate the area of a circular swimming pool. [1]
(b) Calculate the area of the swimming pool. [1]
(c) What is the area of the remaining part of the compound excluding the swimming pool? [2]
(d) Sagun wants to fence the outer compound with one round of wire. Find the total cost of fencing if the cost per meter wire is $\text{Rs. } 40$. [1]
(a)
Area $A = \pi r^2$
(b)
$$A = \frac{22}{7} \times 7^2 = 154 \text{ m}^2$$
(c)
Remaining Area = $(35 \times 28) - 154 = 980 - 154 = 826 \text{ m}^2$
(d)
Perimeter $P = 2(35 + 28) = 126\text{ m}$. Cost = $126 \times 40 = \text{Rs. } 5040$
Answer the following questions:
(a) Express $x^a \times x^b$ as a single power of $x$. [1]
(b) Simplify: $\frac{a}{a-b} - \frac{a}{a+b}$. [2]
(a)
$$x^a \times x^b = x^{a+b}$$
(b)
$$\frac{a(a+b) - a(a-b)}{(a-b)(a+b)} = \frac{2ab}{a^2 - b^2}$$
Simultaneous linear equations $2x + y = 8$ and $x + y = 5$ are given.
(a) Write the degree of the given linear equations. [1]
(b) Solve the given system of linear equations using a graph. [2]
(a)
Degree = 1
(b)
Intersection point from graph $P(3, 2) \implies x = 3, y = 2$
Two algebraic expressions $x^2 + 6x + 8$ and $x^2 - 4$ are given.
(a) Find the HCF of the given expressions. [2]
(b) For what values of $x$ does the value of $x^2 + 6x + 8$ become zero? [2]
(a)
$$x^2+6x+8 = (x+2)(x+4)$$
$$x^2-4 = (x+2)(x-2)$$
$$\text{HCF} = x+2$$
(b)
$$(x+2)(x+4) = 0 \implies x = -2 \text{ or } -4$$
In the figure given below, straight lines $AB$ and $CD$ are intersected by transversal line $MN$ at points $E$ and $F$ respectively. Observe the figure carefully and answer the following questions.
(a) Write a pair of alternate angles from the figure. [1]
(b) Find the value of $x$. [2]
(c) At what value of $\angle BEG$ will line segments $AB$ and $CD$ be parallel? [1]
(a)
Alternate angles: $\angle AEF$ and $\angle EFD$
(b)
$$x + 60^\circ + 2x = 180^\circ \implies 3x = 120^\circ \implies x = 40^\circ$$
(c)
$$\angle BEG + 2(40^\circ) = 180^\circ \implies \angle BEG = 100^\circ$$
Answer the following geometry construction and proof questions:
(a) Construct a parallelogram $ABCD$ with side $AB = 7\text{ cm}$, $BC = 5\text{ cm}$, and angle $\angle ABC = 60^\circ$. [3]
(b) Sketch two triangles $\Delta ABC$ and $\Delta ACD$ formed from parallelogram $ABCD$. Prove that $\Delta ABC \cong \Delta CDA$. [2]
Construction Animation of Parallelogram ABCD
(a) Steps of Construction:
1. Draw base $AB = 7\text{ cm}$.
2. At $B$, construct $\angle ABX = 60^\circ$. Cut $BC = 5\text{ cm}$.
3. From $A$ draw arc $5\text{ cm}$ and from $C$ draw arc $7\text{ cm}$ to intersect at $D$. Join $AD$ & $CD$.
(b) Proof:
In $\Delta ABC$ and $\Delta CDA$: $AB=CD$, $BC=DA$, $AC=CA$ (Common). $\therefore \Delta ABC \cong \Delta CDA$ [S.S.S Axiom].
(a) Find the distance between two points $(-a, 0)$ and $(0, -b)$. [1]
(b) If the bearing of point $B$ from point $A$ is $060^\circ$, what is the bearing of $A$ from $B$? [2]
(c) Find the vertices of the image $\Delta D'E'F'$ of $\Delta DEF$ with vertices $D(4, 5)$, $E(-6, 1)$, and $F(-2, 7)$ under rotation through $180^\circ$ about origin $(0,0)$. Show both in a graph. [3]
(a)
Distance $d = \sqrt{a^2 + b^2}$ units
(b)
Bearing of $A$ from $B = 060^\circ + 180^\circ = 240^\circ$
(c)
Image vertices: $D'(-4, -5)$, $E'(6, -1)$, $F'(2, -7)$
In the table given below, the expenses done by a family in a month on food, education, health, and miscellaneous are given:
| Titles | Amount of Expenditure (in Rs.) |
|---|---|
| Food | 12000 |
| Education | 8000 |
| Health | 6000 |
| Miscellaneous | 10000 |
(a) Represent the above information in a pie-chart. [2]
(b) The average expenditure of the family according to the table is $\text{Rs. } 9000$. How much money should be reduced from miscellaneous expenses to make average expenses $\text{Rs. } 8500$? [1]
(a)
Total = Rs. 36000
Angles: Food = $120^\circ$, Education = $80^\circ$, Health = $60^\circ$, Misc = $100^\circ$
(b)
Target Total = $8500 \times 4 = \text{Rs. } 34000$. Reduction = $36000 - 34000 = \text{Rs. } 2000$.