Loading....

Press & Hold to Drag Around |
|||

Click Here to Close |

**Question 1**
**Report**

Marks12345Frequency22844 $\begin{array}{cccccc}\text{Marks}& 1& 2& 3& 4& 5\\ \text{Frequency}& 2& 2& 8& 4& 4\end{array}$

The table above shows the marks obtained in a given test. How many students took the test?

**Answer Details**

To determine how many students took the test, we need to sum up the frequencies in the table, since each frequency represents the number of students who obtained the corresponding mark. Adding up the frequencies, we get: 2 + 2 + 8 + 4 + 4 = 20 Therefore, 20 students took the test. The answer is (B) 20.

**Question 2**
**Report**

If the area of △
$\u25b3$PQR above is 12√3
$3$cm2, find the value of q?

**Answer Details**

Let A denote the area of △ $\u25b3$PQR, then A = 12bh $\frac{1}{2}bh$

Using Sin 60∘ $\circ $ = hq $\frac{h}{q}$

h = q sin 60∘ $\circ $

So A = 12b(qsin60o) $\frac{1}{2}b(q\mathrm{sin}{60}^{o})$

12√3=12×8×q×√33 $\sqrt{3}=\frac{1}{2}\times 8\times q\times \frac{\sqrt{3}}{3}$

12√3 $3$ - 2q√3 $3$

q = 122=6 $\frac{12}{2}=6$cm

**Question 3**
**Report**

If Cos θ $\theta $ = 1213 $\frac{12}{13}$. Find θ $\theta $ + cos2θ

**Answer Details**

Cos θ
$\theta $ = 1213
$\frac{12}{13}$

x2 + 122 = 132

x2 = 169- 144 = 25

x = 25

= 5

Hence, tanθ
$\theta $ = 512
$\frac{5}{12}$ and cosθ
$\theta $ = 1213
$\frac{12}{13}$

If cos2θ
$\theta $ = 1 + 1tan2θ
$\frac{1}{ta{n}^{2}\theta}$

= 1 + 1(5)212
$\frac{1}{\frac{(5{)}^{2}}{12}}$

= 1 + 125144
$\frac{1}{\frac{25}{144}}$

= 1 + 14425
$\frac{144}{25}$

= 25+14425
$\frac{25+144}{25}$

= 16925

**Question 4**
**Report**

If [x327] $\left[\begin{array}{cc}x& 3\\ 2& 7\end{array}\right]$= 15, find the value of x.

**Answer Details**

[x327]
$\left[\begin{array}{cc}x& 3\\ 2& 7\end{array}\right]$= 15
*x* x 7 - 2 x 3 = 15

7*x* - 6 = 15

7*x* = 15+6

7*x* = 21
*x* = 21/7*x* = 3

**Question 5**
**Report**

Marks | 1 | 2 | 3 | 4 | 5 |

Frequency | 2 | 2 | 8 | 4 | 4 |

The table above show the marks obtained in a given test.

How many student too the test

**Answer Details**

To find the total number of students who took the test, we need to add up the frequency of all the marks. 2 + 2 + 8 + 4 + 4 = 20 Therefore, 20 students took the test. The answer is (C) 20.

**Question 6**
**Report**

Evaluate ∣∣ ∣∣205463891∣∣ ∣∣

**Answer Details**

∣∣ ∣∣205463891∣∣ ∣∣
$\left|\begin{array}{ccc}2& 0& 5\\ 4& 6& 3\\ 8& 9& 1\end{array}\right|$

= 2(6 - 27) - 0(4 - 24) + 5(36 - 48)

= 2(-21) - 0 + 5(-12)

= -42 + 5(-12)

= -42 - 60

= -102

**Question 7**
**Report**

A student measures a piece of rope and found that it was 1.26m long. If the actual length of the rope is 1.25m, what was the percentage error in the measurement?

**Answer Details**

The percentage error in measurement is the difference between the measured value and the actual value, divided by the actual value, multiplied by 100. In this case, the measured value is 1.26m, and the actual value is 1.25m. So the difference between the measured value and actual value is: 1.26m - 1.25m = 0.01m The percentage error can be calculated as: (0.01m ÷ 1.25m) × 100% = 0.8% Therefore, the percentage error in the measurement is 0.8%, which corresponds to option E.

**Question 8**
**Report**

If p and q are two non zero numbers and 18(p+q) = 918+p)q, which of the following must be true?

**Answer Details**

Using substitution If 18(p+18) = (18 + p)q

let q = 18

imply 18(p+18) = (18+p)18

i.e 18p + 18 x 18 = 18 x 18 + 18p

Since the left hand side = the right hand side

imply that q = 18

**Question 9**
**Report**

Find the gradient of the line passing through the points (-2, 0) and (0, -4).

**Answer Details**

To find the gradient of the line passing through two points, we use the formula: gradient = (change in y) / (change in x) In this case, the two points are (-2, 0) and (0, -4). So the change in y is -4 - 0 = -4, and the change in x is 0 - (-2) = 2. Therefore, the gradient of the line passing through the points (-2, 0) and (0, -4) is: gradient = (change in y) / (change in x) = -4 / 2 = -2 So the correct answer is -2.

**Question 10**
**Report**

If y = x sin x, Find d2yd2x

**Answer Details**

To find the second derivative of the given function, we need to differentiate it twice with respect to x. First, we differentiate y with respect to x using the product rule: y = x sin x y' = x cos x + sin x Then, we differentiate y' with respect to x using the product rule again: y' = x cos x + sin x y'' = cos x - x sin x + cos x Simplifying the expression: y'' = 2cos x - x sin x Therefore, the second derivative of y = x sin x is y'' = 2cos x - x sin x.

**Question 11**
**Report**

Find the sum to infinity of the following series. 0.5 + 0.05 + 0.005 + 0.0005 + .....

**Answer Details**

Using S∞
$\infty $ = a1−r
$\frac{a}{1-r}$

r = 0.050.5
$\frac{0.05}{0.5}$ = 110
$\frac{1}{10}$

S∞
$\infty $ = 0.5110
$\frac{0.5}{\frac{1}{10}}$

= 0.5(910)
$\frac{0.5}{(\frac{9}{10})}$

= 0.5×109
$\frac{0.5\times 10}{9}$

= 59

**Question 12**
**Report**

Find the standard deviation of 2, 3, 5 and 6

**Answer Details**

xx−¯x(x−¯x)22−243−11511624∑x=16∑(x−¯x2)=0
$\begin{array}{ccc}x& x-\stackrel{\xaf}{x}& (x-\stackrel{\xaf}{x}{)}^{2}\\ 2& -2& 4\\ 3& -1& 1\\ 5& 1& 1\\ 6& 2& 4\\ \sum x=16& & \sum (x-{\stackrel{\xaf}{x}}^{2})=0\end{array}$

___________________________________

¯x
$\stackrel{\xaf}{x}$ = ∑xN
$\frac{\sum x}{N}$

= 164
$\frac{16}{4}$

= 4

S = √(x−¯x)2N
$\sqrt{\frac{(x-\stackrel{\xaf}{x}{)}^{2}}{N}}$

= √(10)4
$\sqrt{\frac{(10)}{4}}$

= √(5)2

**Question 13**
**Report**

Find the distance between the points (12 $\frac{1}{2}$, -12 $\frac{1}{2}$).

**Answer Details**

Let D denote the distance between (12
$\frac{1}{2}$, -12
$\frac{1}{2}$) then using

D = √(x2−x1)2+(y2−y1)2
$\sqrt{({x}_{2}-{x}_{1}{)}^{2}+({y}_{2}-{y}_{1}{)}^{2}}$

= √(−12−12)2+(−12−12)2
$\sqrt{(-\frac{1}{2}-\frac{1}{2}{)}^{2}+(-\frac{1}{2}-\frac{1}{2}{)}^{2}}$

= √(−1)2+(−1)2
$\sqrt{(-1{)}^{2}+(-1{)}^{2}}$

= √1+1
$\sqrt{1+1}$

= √2

**Question 14**
**Report**

If ∣∣∣x327∣∣∣ $\left|\begin{array}{cc}x& 3\\ 2& 7\end{array}\right|$ = 15, find the value of x

**Answer Details**

The expression |x327| means the absolute value of x to the power of 327. The given equation |x327| = 15 means that the absolute value of x to the power of 327 is equal to 15. To solve for x, we can take the 327th root of both sides of the equation. Thus, we have: |x327| = 15 Taking the 327th root of both sides: |x| = 15^(1/327) Since x can be positive or negative, we have two solutions: x = 15^(1/327) or x = -15^(1/327) Using a calculator, we can approximate the value of x as approximately 2.905 or -2.905. However, only one of these values is among the answer choices, which is x = 3. Therefore, the correct answer is 3.

**Question 15**
**Report**

Marks2345678No. of students3152423 $\begin{array}{cccccccc}\text{Marks}& 2& 3& 4& 5& 6& 7& 8\\ \text{No. of students}& 3& 1& 5& 2& 4& 2& 3\end{array}$

From the table above, if the pass mark is 5, how many students failed the test?

**Answer Details**

To determine how many students failed the test, we need to add up the frequencies of the students who obtained marks less than 5, since the pass mark is 5. Looking at the table, the marks less than 5 are 2, 3, and 4. Adding up the corresponding frequencies, we get: 3 + 1 + 5 = 9 Therefore, 9 students failed the test. The answer is (C) 9.

**Question 16**
**Report**

W is directly proportional to U. If W = 5 when U = 3, find U when W = 27

**Answer Details**

W α
$\alpha $ U

W = ku

u = wk
$\frac{w}{k}$; 27
$\frac{2}{7}$ x 35
$\frac{3}{5}$

= 635

**Question 17**
**Report**

Marks12345Frequency22844 $\begin{array}{cccccc}\text{Marks}& 1& 2& 3& 4& 5\\ \text{Frequency}& 2& 2& 8& 4& 4\end{array}$

The table above shows the marks obtained in a given test. Find the mean mark.

**Answer Details**

To find the mean, we need to calculate the sum of all the marks obtained, and then divide by the total number of students who took the test. The sum of all marks obtained is: 1 x 2 + 2 x 2 + 3 x 8 + 4 x 4 + 5 x 4 = 2 + 4 + 24 + 16 + 20 = 66 The total number of students who took the test is: 2 + 2 + 8 + 4 + 4 = 20 Therefore, the mean mark is: 66 / 20 = 3.3 So the answer is option (C) 3.3

**Question 18**
**Report**

If P = (2?311) what is P−1

**Answer Details**

P = (2?311)
$\left(\begin{array}{cc}2& ?3\\ 1& 1\end{array}\right)$

|P| = 2 - 1 x -3 = 5

P-1 = 15
$\frac{1}{5}$(13?12)
$\left(\begin{array}{cc}1& 3\\ ?1& 2\end{array}\right)$

= (1535?1525)

**Question 19**
**Report**

A cylindrical pipe 50cm long with radius 7m has one end open. What is the total surface area of the pipe?

**Answer Details**

To calculate the total surface area of the cylindrical pipe, we need to add the surface area of the curved part and the surface area of the two circular ends. The surface area of the curved part can be calculated by multiplying the circumference of the circle (2πr) by the length of the pipe (50cm), which gives us: 2πr x h = 2π x 7m x 50cm = 7π m^2 The surface area of one circular end can be calculated by multiplying the area of the circle (πr^2) by 1, since one end of the pipe is open and has no surface area. Thus, the total surface area of both circular ends is: 2πr^2 = 2π x 7m^2 = 14π m^2 Finally, we add the surface area of the curved part and the surface area of the two circular ends to get the total surface area of the pipe: 7π m^2 + 14π m^2 = 21π m^2 Therefore, the total surface area of the pipe is 21π square meters. The closest option to this answer is 749π, but it is not the correct answer.

**Question 20**
**Report**

actorize completely x3+3x2−10x2x2−8

**Answer Details**

x3+3x2−10x2x2−8
$\frac{{x}^{3}+3{x}^{2}-10x}{2{x}^{2}-8}$ = x(x2+3x−10)2(x2−4)
$\frac{x({x}^{2}+3x-10)}{2({x}^{2}-4)}$

= x(x2+5x−2x−10)2(x+2)(x−2)
$\frac{x({x}^{2}+5x-2x-10)}{2(x+2)(x-2)}$

= x(x−2)(x+5)2(x+2)(x−2)
$\frac{x(x-2)(x+5)}{2(x+2)(x-2)}$

= x(x+5)2(x+2)

**Question 21**
**Report**

Evaluate ∫31(X2−1)dx

**Answer Details**

∫31(x2−1)dx=[13x2−x]31=(9−3)−(13−1)=6−(−23)=6+23=623

**Question 22**
**Report**

Rationalise | 2√3+√5 |

√5-√3 |

**Answer Details**

To rationalize the given expression, we need to eliminate the radical from the denominator. To do that, we can multiply both the numerator and denominator by the conjugate of the denominator. The conjugate of √5-√3 is √5+√3. Therefore, we have: (2√3+√5) / (√5-√3) x (√5+√3) / (√5+√3) Simplifying the numerator and the denominator using FOIL (First, Outer, Inner, Last) method, we get: = [2√3(√5) + 2√3(√3) + √5(√5) + √5(√3)] / [(√5)(√5) - (√3)(√5) + (√5)(√3) - (√3)(√3)] = [2√15 + 6 + 5 + √15] / [5 - 3 + √15 - 3] = [3√15 + 11] / 2 Therefore, the answer is (3√15 + 11) / 2.

**Question 23**
**Report**

Solve for x and y if x - y = 2 and x2 - y2 = 8

**Answer Details**

x - y = 2 ...........(1)

x2 - y2 = 8 ........... (2)

x - 2 = y ............ (3)

Put y = x -2 in (2)

x2 - (x - 2)2 = 8

x2 - (x2 - 4x + 4) = 8

x2 - x2 + 4x - 4 = 8

4x = 8 + 4 = 12

x = 124
$\frac{12}{4}$

= 3

from (3), y = 3 - 2 = 1

therefore, x = 3, y = 1

**Question 24**
**Report**

Evaluate (8116)-14×2-1

**Answer Details**

(8116)−14×2−1=1(8116)14×12=(1681)14×12=(23)4×14=23×12=13

**Question 25**
**Report**

For what range of values of x is 12 $\frac{1}{2}$x + 14 $\frac{1}{4}$ > 13 $\frac{1}{3}$x + 12 $\frac{1}{2}$?

**Answer Details**

12 $\frac{1}{2}$x +