ALLAMA IQBAL OPEN UNIVERSITY
| Level: Matric | Semester: Spring 2017 | ||||||||||||
| Course Code: General Mathematics-1 (247) | Maximum Marks: 100 | ||||||||||||
| Time Allowed: 03 Hours | Pass Marks: 40 |
Note: Attempt any Five Questions, Question No. 1 is compulsory.
Q1. Fill in the blanks.
i. 0.28 as percentage is 28%
ii. Zakat on an amount of Rs. 4,00,000 is 10,000.
iii. When SP < CP then Loss% is positive value.
iv. The company undertaking the act of insurance is called the insurer or insurance company.
v. If the annual value of a flat is Rs. 8,00,000. Then the tax payable at a rate of 20% is 160,000.
vi. In 46, 6 is called exponent or power.
vii. If an=3n+5 then a1 is 8.
viii. If R2={2,1}, {3,2}, {4,3}, then range of R2 is {1,2,3}
ix. The pair of numbers (4,5) is called an ordered pair.
x. Σ(yi-Y)=0 is one of the properties of arithmetic mean.
Q2.(a) Sadia scored 40 out of 70 in a English test, 70 out of 80 in a Mathematics test and 67 out of 75 in a Chemistry test. In which subject did she perform best?
\[ \text{English: } \frac{40}{70} \times 100 = 57.14\% \] \[ \text{Mathematics: } \frac{70}{80} \times 100 = 87.5\% \] \[ \text{Chemistry: } \frac{67}{75} \times 100 \approx 89.33\% \]
Sadia performed best in Chemistry 89.33%.
Q2.(b) 5 persons can do a job in 15 days. If 2 more persons are employed, how many days will they take to complete the job?
\[ \text{Total work} = 5 \times 15 = 75 \text{ man-days} \]
If 2 more persons are added, total persons = 5 + 2 = 7
\[ \text{Time required} = \frac{75}{7} \approx 10.71 \text{ days} \]
They will complete the job in about 10.7 days.
Q3.(a) Aslam left a property of worth Rs. 48,00,000. Calculate the amount of share of his wife, two sons and two daughters.
\[ \text{Total property} = 48,00,000 \] \[ \text{Wife's share} = \tfrac{1}{8} \times 48,00,000 = 6,00,000 \] \[ \text{Remaining property} = 48,00,000 - 6,00,000 = 42,00,000 \] \[ \text{Children's shares: Sons : Daughters = 2 : 1} \] \[ \text{For 2 sons and 2 daughters, ratio } = 2 + 2 + 1 + 1 = 6 \text{ parts} \] \[ \text{One part} = \tfrac{42,00,000}{6} = 7,00,000 \] \[ \text{Each son} = 2 \times 7,00,000 = 14,00,000 \] \[ \text{Each daughter} = 1 \times 7,00,000 = 7,00,000 \] \[ \therefore \text{Wife } = 6,00,000, \quad \text{Each Son } = 14,00,000, \quad \text{Each Daughter } = 7,00,000 \]
Q3.(b) Three chairs are purchased at Rs. 500 each. One of these is sold at a loss of 10%, at what price should the other two be sold so as to gain 20% on the whole transaction?
\[ \text{Total cost of 3 chairs} = 3 \times 500 = 1500 \] \[ \text{Selling price of 1 chair at 10\% loss} = 500 - \tfrac{10}{100}\times 500 = 450 \] \[ \text{Required total selling price for 20\% gain} = 1500 + \tfrac{20}{100}\times 1500 = 1800 \] \[ \text{Amount to be obtained from 2 chairs} = 1800 - 450 = 1350 \] \[ \text{Selling price of each of the 2 chairs} = \tfrac{1350}{2} = 675 \] \[ \therefore \text{Each of the other two chairs should be sold at Rs. 675.} \]
Q4.(a) Farhan purchases a car in Saudi Arabia for 20,000 Riyals. Delivery was to be made after two months and payment is also to be made at the time of delivery. At the time of contract, the rate was 1 Riyal = Rs. 21.5, while at the time of delivery the rate was 1 Riyal = Rs. 21 02. Determine the loss in rupees due to change in the rate.
Car cost = 20,000 Riyals
Rate at contract time: 1 Riyal = 21.5 Rs
Rate at delivery time: 1 Riyal = 21.02 Rs
Cost in Rs at contract time: \[ 20,000 \times 21.5 = 430,000 \text{ Rs} \]
Cost in Rs at delivery time: \[ 20,000 \times 21.02 = 420,400 \text{ Rs} \]
Loss due to exchange rate change: \[ 430,000 - 420,400 = 9,600 \text{ Rs} \]
\[ \therefore \text{Loss} = 9,600 \text{ Rs} \]
Q4.(b) If gross pay of a person is Rs. 50,000, then calculate his net take home salary, after deductions of Rs. 500 as income tax. Rs. 1400 as benevolent fund, Rs 1800 as G.P fund and Rs. 300 as group insurance.
Gross pay = Rs. 50,000
Deductions:
Income tax = Rs. 500
Benevolent fund = Rs. 1,400
G.P. fund = Rs. 1,800
Group insurance = Rs. 300
Total deductions: \[ 500 + 1400 + 1800 + 300 = 4000 \text{ Rs} \]
Net take-home salary: \[ \text{Net salary} = \text{Gross pay} - \text{Total deductions} = 50,000 - 4,000 = 46,000 \text{ Rs} \]
Q5.(a) Prove that loga ab = logaa + logab
\[ \text{LHS} = \log_a(ab) \] \[ \text{Using the logarithm property } \log_a(xy) = \log_a x + \log_a y \] \[ \log_a(ab) = \log_a a + \log_a b \] \[ \text{RHS} = \log_a a + \log_a b \] \[ \therefore \log_a(ab) = \log_a a + \log_a b \]
Q5.(b) Using logarithmic table evaluate.
\[ \dfrac{(0.0537)^{\tfrac{1}{2}} \times (1.5)^3}{(0.0014)^{\tfrac{1}{3}} \times (1.435)^{\tfrac{1}{6}}} \]
\[ N \;=\; \dfrac{(0.0537)^{\tfrac{1}{2}} \times (1.5)^3}{(0.0014)^{\tfrac{1}{3}} \times (1.435)^{\tfrac{1}{6}}} \]
Take base ten logarithms
\[ \log N \;=\; \tfrac{1}{2}\log(0.0537) \;+\; 3\log(1.5) \;-\; \tfrac{1}{3}\log(0.0014) \;-\; \tfrac{1}{6}\log(1.435) \]
Using logarithm table values rounded to four decimal places
| Number | Logarithm |
|---|---|
| \(0.0537\) | \(\log(0.0537)= -1.2700\) |
| \(1.5\) | \(\log(1.5)= 0.1761\) |
| \(0.0014\) | \(\log(0.0014)= -2.8539\) |
| \(1.435\) | \(\log(1.435)= 0.1569\) |
Multiply by the powers
\[ \tfrac{1}{2}\log(0.0537) = \tfrac{1}{2}\times(-1.2700) = -0.6350 \]
\[ 3\log(1.5) = 3\times 0.1761 = 0.5283 \]
\[ \tfrac{1}{3}\log(0.0014) = \tfrac{1}{3}\times(-2.8539) = -0.9513 \]
\[ \tfrac{1}{6}\log(1.435) = \tfrac{1}{6}\times 0.1569 \approx 0.0262 \]
Now assemble the terms
\[ \log N = -0.6350 + 0.5283 - (-0.9513) - 0.0262 \]
\[ \log N = -0.6350 + 0.5283 + 0.9513 - 0.0262 = 0.8184 \]
Antilog gives the final result
\[ N = 10^{0.8184} \approx 6.5828 \]
Q6.(a) Find six Arithmetic Means between 5 and 12.
Step 1: Formula
The general term of an arithmetic progression (AP) is: \[ a_n = a_1 + (n-1)d \]Step 2: Known values
Here, \[ a_1 = 5, \quad a_8 = 12, \quad n = 8 \]Step 3: Put values into the formula
\[ a_8 = a_1 + (8-1)d \] \[ 12 = 5 + 7d \]Step 4: Solve for \(d\)
\[ 12 - 5 = 7d \] \[ 7 = 7d \] \[ d = 1 \]Step 5: Write all terms
\[ a_1 = 5 \] \[ a_2 = a_1 + d = 5 + 1 = 6 \] \[ a_3 = a_1 + 2d = 5 + 2(1) = 7 \] \[ a_4 = a_1 + 3d = 5 + 3(1) = 8 \] \[ a_5 = a_1 + 4d = 5 + 4(1) = 9 \] \[ a_6 = a_1 + 5d = 5 + 5(1) = 10 \] \[ a_7 = a_1 + 6d = 5 + 6(1) = 11 \] \[ a_8 = a_1 + 7d = 5 + 7(1) = 12 \]Final Answer
The six arithmetic means are: \[ 6, \; 7, \; 8, \; 9, \; 10, \; 11 \]Q6.(b) Find the nth term of a Geometric Progression,
\[ \text{If } \frac{a_5}{a_3} = \frac{16}{9} \quad \text{and} \quad a_2 = \frac{6}{7} \]\[ \text{Given: } \quad \frac{a_5}{a_3} = \frac{16}{9}, \quad a_2 = \frac{6}{7} \] \[ \text{In a G.P., } \quad a_n = a \cdot r^{\,n-1} \] \[ a_5 = a \cdot r^4, \quad a_3 = a \cdot r^2 \] \[ \frac{a_5}{a_3} = \frac{a \cdot r^4}{a \cdot r^2} = r^2 \] \[ r^2 = \frac{16}{9} \quad \Rightarrow \quad r = \pm \frac{4}{3} \] \[ a_2 = a \cdot r \] \[ \frac{6}{7} = a \cdot r \quad \Rightarrow \quad a = \frac{6}{7r} \] \[ \text{Case 1: } r = \frac{4}{3} \] \[ a = \frac{6}{7 \cdot \tfrac{4}{3}} = \frac{6}{\tfrac{28}{3}} = \frac{18}{28} = \frac{9}{14} \] \[ \text{So, } a = \frac{9}{14}, \quad r = \frac{4}{3} \] \[ \text{Case 2: } r = -\frac{4}{3} \] \[ a = \frac{6}{7 \cdot \left(-\tfrac{4}{3}\right)} = \frac{6}{-\tfrac{28}{3}} = \frac{18}{-28} = -\frac{9}{14} \] \[ \text{So, } a = -\frac{9}{14}, \quad r = -\frac{4}{3} \] \[ \text{Final Answer: } \quad (a, r) = \left(\tfrac{9}{14}, \tfrac{4}{3}\right) \quad \text{or} \quad (a, r) = \left(-\tfrac{9}{14}, -\tfrac{4}{3}\right) \]
Q7.(a) \(\text{If } U=\{1,2,3,\dots,20\},\ A=\{2,4,6,\dots,20\} \text{ and } B=\{1,3,5,\dots,19\} \text{ then verify } (A \cup B)^{c} = A^{c} \cap B^{c}\)
\(U = \{1,2,3,\dots,20\}\)
\(A = \{2,4,6,\dots,20\}\)
\(B = \{1,3,5,\dots,19\}\)
\(A^{c} = U - A = \{1,3,5,\dots,19\} = B\)
\(B^{c} = U - B = \{2,4,6,\dots,20\} = A\)
\(A \cup B = U\)
\((A \cup B)^{c} = U^{c} = \varnothing\)
\(A^{c} \cap B^{c} = B \cap A = \varnothing\)
\(\therefore (A \cup B)^{c} = A^{c} \cap B^{c}\)
Q7.(b)
Q8.(a)
Q8.(b)
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