Loading....

Press & Hold to Drag Around |
|||

Click Here to Close |

**Question 1**
**Report**

Simplify 3\(\sqrt{45} - 12\sqrt{5} + 16\sqrt{20}\), leaving your answer in surd form.

**Answer Details**

First, we need to simplify each of the surds in the expression. \[\begin{aligned} 3\sqrt{45} &= 3\sqrt{9\times 5} = 3\times 3\sqrt{5} = 9\sqrt{5} \\ 16\sqrt{20} &= 16\sqrt{4\times 5} = 16\times 2\sqrt{5} = 32\sqrt{5} \end{aligned}\] Now we can substitute these simplified surds back into the original expression and simplify it further: \[\begin{aligned} 3\sqrt{45} - 12\sqrt{5} + 16\sqrt{20} &= 9\sqrt{5} - 12\sqrt{5} + 32\sqrt{5} \\ &= (9-12+32)\sqrt{5} \\ &= 29\sqrt{5} \end{aligned}\] Therefore, the simplified expression is 29\(\sqrt{5}\). The correct option is (a).

**Question 2**
**Report**

Which of the following is true for the set \(P = \{-3.2\leq x< 5\}\) where x is an integer

**Question 3**
**Report**

N140,000 is shared between ABU, Kayode and Uche. Abu has twice as much as Kayode, and Kayode has twice as much as Uche. What is Kayode's share?

**Answer Details**

Let's start by using variables to represent the amount of money each person gets. Let Uche's share be x. Then, we know that Kayode's share is twice that amount, so his share is 2x. And we also know that Abu's share is twice Kayode's share, so his share is 2(2x) = 4x. The sum of their shares is given as N140,000, so we can set up an equation: x + 2x + 4x = 140,000 Simplifying the left side of the equation, we get: 7x = 140,000 Dividing both sides by 7, we get: x = 20,000 So Uche's share is N20,000. We can then find Kayode's share, which is twice that amount: 2x = 2(20,000) = 40,000 Therefore, Kayode's share is N40,000. So the answer is, N40,000.

**Question 4**
**Report**

Find the LCM of 2^{3} x 3 x 5^{2}, 2 x 3^{2} x 5 and 2^{3} x 3^{3} x 5

**Question 5**
**Report**

Find the value of x in the diagram

**Question 6**
**Report**

In the diagram, O is the centre of the circle and < PQR = 106^{o}, find the value of y

**Question 7**
**Report**

The table shows the ages(in years) of twenty children chosen at random from a community. What is the mean age? \(\begin{array}{c|c} Age(years) & 1 & 2 & 3 & 4 & 5 \\ \hline {\text {Number of children}} & 2 & 3 & 5 & 6 & 4 \end{array}\)

**Answer Details**

To find the mean age of the twenty children, we need to calculate the sum of the ages and divide by the total number of children. Sum of the ages = (1 x 2) + (2 x 3) + (3 x 5) + (4 x 6) + (5 x 4) = 2 + 6 + 15 + 24 + 20 = 67 Total number of children = 2 + 3 + 5 + 6 + 4 = 20 Mean age = sum of the ages/total number of children = 67/20 = 3.35 years Therefore, the mean age of the twenty children is 3.35 years. The correct option is (b) 3.35 years.

**Question 8**
**Report**

Find the smallest value of k such that 2\(^2\) x 3\(^3\) x 5 x k is a perfect square.

**Question 9**
**Report**

A sector of a circle of radius 14cm containing an angle 60^{o} is folded to form a cone. Calculate the radius of the base of the cone.

**Answer Details**

The sector of a circle can be folded to form a cone if and only if the arc length of the sector is equal to the circumference of the base of the cone. Given that the radius of the sector is 14 cm and the angle is 60^{o}, the arc length of the sector is \(\frac{60}{360}\times2\pi(14)=\frac{14\pi}{3}\) cm. Let's denote the radius of the base of the cone by r. Then, the circumference of the base of the cone is \(2\pi r\) cm. Since the arc length of the sector is equal to the circumference of the base of the cone, we have: \[\frac{14\pi}{3}=2\pi r\] Dividing both sides by 2π gives: \[r=\frac{14}{3}\div2=\frac{7}{3}=2\frac{1}{3}\,\text{cm}\] Therefore, the radius of the base of the cone is \(2\frac{1}{3}\,\text{cm}\). Answer:.

**Question 10**
**Report**

Simplify \((0.3\times 10^{5})\div (0.4\times 10^{7})\)leaving your answer in standard form

**Question 11**
**Report**

The area of a square field is 110.25m^{2}. Find the cost of fencing it round at N75.00 per metre

**Answer Details**

The area of the square field is given as 110.25m^{2}. We know that the area of a square is given by the formula A = s^{2}, where A is the area and s is the side length. Therefore, we can find the side length of the square as follows: s^{2} = A s^{2} = 110.25 s = √110.25 = 10.5m The perimeter of the square is given by the formula P = 4s, where P is the perimeter and s is the side length. Therefore, we can find the perimeter of the square as follows: P = 4s = 4(10.5m) = 42m To find the cost of fencing it round at N75.00 per metre, we need to multiply the perimeter by N75.00: Cost = P × N75.00 = 42m × N75.00 = N3,150.00 Therefore, the cost of fencing the square field at N75.00 per metre is N3,150.00. So the correct option is (B) N3,150.00.

**Question 12**
**Report**

Expand the expression(3a - xy)(3a + xy)

**Answer Details**

To expand the expression (3a - xy)(3a + xy), we can use the formula for multiplying two binomials: (a + b)(a - b) = a^{2} - b^{2} If we let a = 3a and b = xy, then we have: (3a - xy)(3a + xy) = (3a)^{2} - (xy)^{2} Expanding the terms on the right side, we get: (3a)^{2} - (xy)^{2} = 9a^{2} - x^{2}y^{2} Therefore, the answer is 9a^{2} - x^{2}y^{2}.

**Question 13**
**Report**

In the diagram, O is the centre of the circle and PQ is a diameter. Triangle RSO is an equilateral triangle of side 4cm. Find the area of the shaded region

**Question 14**
**Report**

If p \(\alpha \frac{I}{Q}\) which of the following is true?

**Answer Details**

Given that p \(\alpha \frac{I}{Q}\), where \(\alpha\) means 'is proportional to'. To determine the relationship between q and p, we need to manipulate the equation so that q is isolated on one side. We can write p = k\(\frac{I}{Q}\), where k is the constant of proportionality. Multiplying both sides by Q, we get pQ = kI. Dividing both sides by p, we get Q = \(\frac{k}{p}\)I. Since k is a constant of proportionality, we can write it as k = cp for some other constant c. Therefore, Q = \(\frac{c}{p}\)I. This means that q is inversely proportional to p. So, the correct option is q \(\alpha \frac{1}{p}\).

**Question 15**
**Report**

The table shows the ages(in years) of twenty children chosen at random from a community. What is the median of the distribution? \(\begin{array}{c|c} Age(years) & 1 & 2 & 3 & 4 & 5 \\ \hline {\text {Number of children}} & 2 & 3 & 5 & 6 & 4 \end{array}\)

**Question 16**
**Report**

If x = 3, Y = 2 and z = 4, what is the value of 3x^{2} - 2y + z?

**Answer Details**

Substituting x = 3, y = 2, and z = 4 into the expression 3x^{2} - 2y + z gives: 3x^{2} - 2y + z = 3(3)^{2} - 2(2) + 4 = 27 - 4 + 4 = 27 Therefore, the value of 3x^{2} - 2y + z is 27.

**Question 17**
**Report**

PQR is an equilateral triangle with sides 2\(\sqrt{3cm}\). calculate its height

**Question 18**
**Report**

A sector of a circle of radius 14cm containing an angle 60^{o} is folded to form a cone. Calculate the radius of the base of the cone

**Question 19**
**Report**

Simplify \(\frac{4}{2x} - \frac{2x + x}{x}\)

**Question 20**
**Report**

For what value of x is the expression \(\frac{2x-1}{x+3}\)not defined?

**Answer Details**

The expression \(\frac{2x-1}{x+3}\) is not defined when the denominator, \(x+3\), is equal to zero. Therefore, we can solve the equation \(x+3=0\) to find the value of x that makes the expression undefined. \(x+3=0\) \(x=-3\) So, the value of x that makes the expression undefined is -3. Therefore, the correct answer is (-3).

**Question 21**
**Report**

If the probability of an event occurring is x, what is the probability of the event not occurring?

**Answer Details**

The probability of an event occurring is the measure of the likelihood that the event will happen, given as a number between 0 and 1, inclusive. The probability of an event not occurring is the measure of the likelihood that the event will not happen. Since the sum of the probabilities of an event occurring and not occurring is 1, we can find the probability of an event not occurring by subtracting the probability of the event occurring from 1. Therefore, the probability of an event not occurring is equal to 1 minus the probability of the event occurring. Mathematically, if the probability of an event occurring is x, then the probability of the event not occurring is 1 - x. So, option A, 1 - x, is the correct answer.

**Question 22**
**Report**

In a , < PQR = < PRQ = 45^{o}. which of the following statements is/are correct? i. \(\bigtriangleup\)PQR is an equalateral triangle ii. \(\bigtriangleup\)PQR is an isosceles triangle iii. \(\bigtriangleup\)PQR is a right-angled triangle

**Question 23**
**Report**

The area of a square field is 110.25m^{2}. Find the cost of fencing it round at N75.00 per meter square

**Question 24**
**Report**

Solve the simultaneous equations x + y = 2 and 3x - 2y = 1

**Answer Details**

To solve the simultaneous equations x + y = 2 and 3x - 2y = 1, we can use the method of substitution or elimination. Method of Substitution: We can solve for one variable in terms of the other from the first equation and substitute it into the second equation, then solve for the other variable. Let's solve for y in terms of x from the first equation: x + y = 2 y = 2 - x Now substitute y = 2 - x into the second equation: 3x - 2y = 1 3x - 2(2 - x) = 1 3x - 4 + 2x = 1 5x = 5 x = 1 Now substitute x = 1 into either equation to find y: x + y = 2 1 + y = 2 y = 1 Therefore, the solution is x = 1 and y = 1. So, the correct answer is: x = 1 and y = 1.

**Question 25**
**Report**

Convert 42_{5} to base three numeral

**Answer Details**

To convert a number from base 5 to base 3, we need to first convert it to base 10, and then convert it to base 3. 42_{5} can be written as: $42_{5} = 4\cdot5^{1} + 2\cdot5^{0} = 20 + 2 = 22_{10}$ Now we convert 22_{10} to base 3. We can do this by repeatedly dividing 22 by 3 and taking the remainders, until the quotient is zero. The remainders, read from bottom to top, give us the base 3 numeral. \begin{array}{c|c} \text{Dividend} & \text{Remainder}\\ \hline 22 & 1 \\ 7 & 2 \\ 2 & 2 \\ 0 & \\ \end{array} So, 22_{10} is equal to 212_{3}. Therefore, the correct option is (3) 211_{3}.

**Question 26**
**Report**

A messenger was paid N2.50 an hour during the normal working hours and 4.00 n hour during overtime. If he received N31.00 for 10 hours work, how many hours are for overtime?

**Answer Details**

Let's call the number of normal working hours the messenger did as "x" and the number of overtime hours as "y". We can create two equations based on the given information: Equation 1: The messenger worked a total of 10 hours x + y = 10 Equation 2: The total amount paid is N31.00 2.5x + 4y = 31 We can use the first equation to solve for x in terms of y: x = 10 - y Now we can substitute this expression for x into the second equation: 2.5(10 - y) + 4y = 31 Simplifying and solving for y, we get: 25 - 2.5y + 4y = 31 1.5y = 6 y = 4 Therefore, the messenger worked 4 hours of overtime. Answer: (c) 4.

**Question 27**
**Report**

Q is 32 km away from P on a bearing 042^{o} and R is 25km from P on a bearing of 132^{o}. Calculate the bearing of R from Q.

**Question 28**
**Report**

Evaluate(111_{two})^{2} and leave your answer in base 2.

**Answer Details**

To evaluate (111_{two})^{2}, we need to convert the binary number 111_{two} to its decimal equivalent, then square the result, and finally convert the answer back to binary. 111_{two} = 1 × 2^{2} + 1 × 2^{1} + 1 × 2^{0} = 7_{ten} (111_{two})^{2} = (7_{ten})^{2} = 49_{ten} To convert 49_{ten} back to binary, we repeatedly divide by 2 and write down the remainders in reverse order: 49 ÷ 2 = 24 remainder 1 24 ÷ 2 = 12 remainder 0 12 ÷ 2 = 6 remainder 0 6 ÷ 2 = 3 remainder 0 3 ÷ 2 = 1 remainder 1 1 ÷ 2 = 0 remainder 1 Reading the remainders in reverse order gives us: 49_{ten} = 110001_{two} Therefore, the answer is (B) 110001_{two}.

**Question 29**
**Report**

The venn diagram shows the choice of food of a number of visitors to a canteen. How many people took at least two kinds of food? If there were 35 *visitors* in all

**Question 30**
**Report**

For what value of x is the expression \(\frac{2x - 1}{x + 3}\) not defined?

**Answer Details**

The expression \(\frac{2x-1}{x+3}\) is undefined if the denominator is equal to zero, since division by zero is not allowed in mathematics. Therefore, we need to solve the equation \(x+3=0\) to find the value of x for which the expression is undefined. Solving for x, we have: \begin{align*} x+3 &= 0 \\ x &= -3 \end{align*} Therefore, the expression is not defined for x = -3. Thus, the correct answer is (-3).

**Question 31**
**Report**

A chord of length 6cm is drawn in a circle of radius 5cm. Find the distance of the chord from the center of the circle

**Question 32**
**Report**

Two circles have radii 16cm and 23cm. What is the difference between their circumference? take \(\pi = \frac{22}{7}\)

**Answer Details**

The circumference of a circle is given by the formula 2πr, where r is the radius of the circle and π is a constant value approximately equal to 22/7 or 3.14. Using the given formula, the circumference of the circle with radius 16cm is 2 x (22/7) x 16cm = 100.57cm (to 2 decimal places). Similarly, the circumference of the circle with radius 23cm is 2 x (22/7) x 23cm = 144.57cm (to 2 decimal places). Therefore, the difference between the circumferences of the two circles is 144.57cm - 100.57cm = 44.00cm (to 2 decimal places). Hence, the correct option is (C) 44.00cm.

**Question 33**
**Report**

Find the LCM of \(2^{3}\times 3\times 5^{2}, 2\times 3^{2}\times 5 \hspace{1mm}and \hspace{1mm}2^{2}\times 3^{2}\times 5\)

**Question 34**
**Report**

IN the diagram, |LN| = 4cm, LNM = 90^{o} and tan y = 2/3. What is the area of the ?LMN?

**Question 35**
**Report**

Simplify \(3\sqrt{45}-12\sqrt{5}+16\sqrt{20}\)leaving your answer in surd form

**Answer Details**

First, we need to simplify each term under the square root sign. We can write: \begin{align*} 3\sqrt{45} &= 3\sqrt{9\times 5} = 3\times 3\sqrt{5} = 9\sqrt{5}\\ 12\sqrt{5} &= 2\times 2\times 3\sqrt{5} = 4\sqrt{5}\times 3\\ 16\sqrt{20} &= 16\sqrt{4\times 5} = 16\times 2\sqrt{5} = 32\sqrt{5} \end{align*} Now we can substitute these simplified expressions back into the original expression and simplify: \begin{align*} 3\sqrt{45}-12\sqrt{5}+16\sqrt{20} &= 9\sqrt{5}-4\sqrt{5}\times 3+32\sqrt{5}\\ &= (9-12+32)\sqrt{5}\\ &= 29\sqrt{5} \end{align*} Therefore, the answer is \(29\sqrt{5}\).

**Question 36**
**Report**

Which of the following statement is true for the ste P = {-3.2 \(\leq\) x < 5} where x is an integer?

**Answer Details**

The given set P = {-3.2 ≤ x < 5} contains all the integers that are greater than or equal to -3.2 but less than 5. However, since x has to be an integer, the least possible value of x is -3 (since -3 is the greatest integer less than or equal to -3.2) and the greatest possible value of x is 4 (since 4 is the greatest integer less than 5). Therefore, the statement "least value of x is -3" is true for the set P, and the statement "greatest value of x is 5" is false.

**Question 37**
**Report**

The bearing of a point P from another point Q is 310^{o}. If |PQ| = 200m, how far west of Q is P?

**Question 38**
**Report**

At what rate per cent per annum will N520.00 yield a simple interest of N39.00 in three years?

**Answer Details**

We can use the formula for simple interest: Simple Interest = (P * r * t) / 100 where P is the principal, r is the rate of interest per annum, and t is the time period in years. We are given that P = N520.00 and the simple interest, SI = N39.00. We are also given that the time period is 3 years. Substituting these values into the formula, we get: N39.00 = (N520.00 * r * 3) / 100 Simplifying this equation, we get: r = (N39.00 * 100) / (N520.00 * 3) = 2.5% Therefore, the rate of interest per annum is 2.5%. So, the correct option is: - 2\(\frac{1}{2}\)%

**Question 39**
**Report**

The table above gives the distribution of the marks of a number of students in a test.

\(\begin{array}{c|c} Mark &1 & 2 & 3 & 4 & 5 & 6 \\ \hline Frequency & 5 & 3 & 6 & 4 & 2 & 5\end{array}\), find the mode of the distribution.

**Question 40**
**Report**

In the diagram |LN| = 4cm, LNM = 90^{o} and tan y = \(\frac{2}{3}\). What is the area of the \(\bigtriangleup\)LMN?

**Question 41**
**Report**

For what range of values of x is 4x - 3(2x - 1) > 1?

**Answer Details**

To solve the inequality 4x - 3(2x - 1) > 1, we can simplify it as follows: 4x - 6x + 3 > 1 -2x + 3 > 1 -2x > -2 x < 1 Therefore, the range of values of x that satisfy the inequality is x < 1. So, the correct option is "x < 1".

**Question 42**
**Report**

Find the product of 0.0409 and 0.0021 leaving your answer in the standard form

**Question 43**
**Report**

make w the subject of the relation \(\frac{a + bc}{wd + f}\) = g

**Answer Details**

To make w the subject of the relation \(\frac{a + bc}{wd + f} = g\), we can follow these steps: 1. Multiply both sides by \((wd + f)\) 2. Divide both sides by g 3. Divide both sides by (a + bc) This gives: \begin{align*} \frac{a + bc}{wd + f} &= g \\ (a + bc) &= g(wd + f) \\ a + bc &= gwd + gf \\ gwd &= a + bc - gf \\ w &= \frac{a + bc - gf}{gd} \end{align*} Therefore, the solution is: \begin{equation*} w = \frac{a + bc - gf}{gd} \end{equation*} Hence, the option that matches this answer is: - \(\frac{a + bc - fg}{dg}\)

**Question 44**
**Report**

Divide the sum of 8, 6, 7, 2, 0, 4, 7, 2, 3, by their mean

**Answer Details**

To divide the sum of the given numbers by their mean, we need to first find their sum and mean. Sum = 8 + 6 + 7 + 2 + 0 + 4 + 7 + 2 + 3 = 39 Mean = (8 + 6 + 7 + 2 + 0 + 4 + 7 + 2 + 3) / 9 = 4.33 (rounded to two decimal places) Now, to divide the sum by the mean, we simply divide the sum by the mean: 39 / 4.33 = 9 (rounded to the nearest whole number) Therefore, the answer is 9.

**Question 45**
**Report**

Which of the following is a factor of 2 - x - x^{2}?

**Question 46**
**Report**

I am x years old and my brother is 3 years older how old was my brother last year

**Question 47**
**Report**

If \(p\propto \frac{1}{q}\), which of the following is true?

**Question 48**
**Report**

If 8^{x-1} = \(\frac{1}{4}\), find x

**Answer Details**

We know that 8^{x-1} = \(\frac{1}{4}\). We can write 8^{x-1} as (2^{3})^{x-1} = 2^{3x-3}. Similarly, we can write 1/4 as 2^{-2}. Therefore, we can write the given equation as 2^{3x-3} = 2^{-2}. Since the bases are equal, we can equate the exponents to get: 3x - 3 = -2 Adding 3 to both sides, we get: 3x = 1 Dividing both sides by 3, we get: x = \(\frac{1}{3}\) Therefore, the value of x is \(\frac{1}{3}\).

**Question 49**
**Report**

In the diagram, |SP| = |SR| and < PRS = 50^{o}. Calculate < PQR

**Question 50**
**Report**

If \(8^{x+1}=\frac{1}{4}\), find x

**Answer Details**

We can solve the given equation by taking logarithms to the base 2 on both sides. This gives: \begin{align*} 8^{x+1} &= \frac{1}{4}\\ \Rightarrow \quad 2^{3(x+1)} &= 2^{-2}\\ \Rightarrow \quad 3(x+1) &= -2\\ \Rightarrow \quad 3x+3 &= -2\\ \Rightarrow \quad 3x &= -5\\ \Rightarrow \quad x &= -\frac{5}{3} \end{align*} Therefore, the value of $x$ is $-\frac{5}{3}$. So, the correct option is \(-\frac{5}{3}\).

**Question 51**
**Report**

Solve the simultaneous equation: x+y=2 and 3x-2y=1

**Answer Details**

To solve this system of simultaneous equations, we need to find the values of x and y that satisfy both equations at the same time. One way to do this is to use the method of substitution. From the first equation, we have x + y = 2, which we can rearrange as x = 2 - y. We can then substitute this expression for x into the second equation, giving 3(2-y) - 2y = 1. Simplifying the left-hand side, we get 6 - 5y = 1, and solving for y, we find y = 1. We can then substitute this value for y into either equation to find x. Using the first equation, we get x + 1 = 2, so x = 1. Therefore, the solution to the simultaneous equations is x = 1 and y = 1.

**Question 52**
**Report**

A chord of length 6cm is drawn in a circle of radius 5cm. Find the distance of the chord from the centre of the circle.

**Answer Details**

When a chord is drawn in a circle, it divides the circle into two equal parts. The line connecting the midpoint of the chord and the center of the circle is perpendicular to the chord. Therefore, to find the distance of the chord from the center of the circle, we need to draw a perpendicular bisector to the chord and measure the distance from the center of the circle to the perpendicular bisector. Let O be the center of the circle, AB be the chord of length 6cm and M be the midpoint of AB. Then OM is the perpendicular bisector of AB, and AM = MB = 3cm. Using the Pythagorean theorem in triangle OAM, we have: $$OA^2 = OM^2 + AM^2 = 5^2 - 3^2 = 16$$ Taking the square root of both sides, we have OA = 4cm. Therefore, the distance of the chord from the center of the circle is 4cm. So the correct option is (d) 4.0cm.

**Question 53**
**Report**

Simplify(0.3 x 10^{5}) \(\div\) (0.4 \(\times\) 10^{7}), leaving you answer in the standard form.

**Answer Details**

To simplify this expression, we need to perform division and then express the answer in standard form. (0.3 x 10^{5}) \(\div\) (0.4 x 10^{7}) = (0.3/0.4) x 10^{5}−7 = 0.75 x 10^{-2} = 7.5 x 10^{-3} Therefore, the answer is 7.5 x 10^{-3}.

**Question 54**
**Report**

From the diagram, find the value of x in the diagram.

**Question 55**
**Report**

N140,000 is shared between Abu, Kayode and Uche. Abu has twice as much as Kayode and Kayode has twice as much as Uche. What is Kayode'sshare?

**Answer Details**

Let Uche's share be x. Since Kayode has twice as much as Uche, Kayode's share will be 2x. Also, since Abu has twice as much as Kayode, Abu's share will be 2(2x) = 4x. The sum of their shares is N140,000, so we have: x + 2x + 4x = N140,000 7x = N140,000 x = N20,000 Therefore, Kayode's share is 2x = 2(N20,000) = N40,000. Hence, the answer is N40,000.

**Question 56**
**Report**

I am x years old and my brother is 3 years older. How old was my brother last year?

**Answer Details**

If you are currently x years old, and your brother is 3 years older than you, then your brother's current age is x + 3. To find out how old your brother was last year, you need to subtract 1 from his current age: (x + 3) - 1 = x + 2 Therefore, your brother was x + 2 years old last year. So, the correct option is (b) (x + 2) years.

**Question 57**
**Report**

Simplify\(\frac{3x^{3}}{(3x)^{3}}\)

**Answer Details**

To simplify the expression, we need to first expand the denominator of the fraction, which is \((3x)^3\). This can be written as \(3^3 x^3\) or \(27x^3\). Substituting this in the original fraction gives: $$\frac{3x^3}{(3x)^3} = \frac{3x^3}{27x^3} = \frac{1}{9}$$ Therefore, the simplified form of the expression is \(\frac{1}{9}\). Hence, the correct option is \(\frac{1}{9}\).

**Question 58**
**Report**

Evaluate \((111_{two})^2\) and leave your answer in base 2

**Answer Details**

To solve this question, we need to convert the number 111 from binary to decimal, then square the result and convert the answer back to binary. Starting with the binary number 111, we can convert it to decimal using the place value system. The rightmost digit represents 1, the second digit from the right represents 2, and the leftmost digit represents 4. Adding these values together, we get: 1 + 2 + 4 = 7 So 111 in binary is equal to 7 in decimal. To square 7, we simply multiply it by itself: 7 x 7 = 49 So the decimal equivalent of (111_{two})^{2} is 49. To convert this back to binary, we use the same place value system but in reverse. Starting with the largest power of 2 that is less than or equal to 49, we subtract that value and place a 1 in the corresponding digit. We then repeat this process with the remainder until we reach 0. 49 is greater than or equal to 32, so we subtract 32 and place a 1 in the 6th digit from the right. The remainder is 17. 17 is greater than or equal to 16, so we subtract 16 and place a 1 in the 5th digit from the right. The remainder is 1. 1 is less than 2, so we place a 1 in the 1st digit from the right. The remainder is 0, so we have our final answer: 49 in binary is equal to 110001_{two} Therefore, the correct answer is (b) 110001_{two}.

**Question 59**
**Report**

Find the volume of a solid cylinder with base radius 10cm and height 14cm.

**Answer Details**

The volume of a cylinder is given by the formula V = πr^{2}h, where r is the radius of the base and h is the height of the cylinder. In this case, the base radius is 10cm and the height is 14cm. Substituting these values into the formula, we have: V = π(10cm)^{2}(14cm) = π(100cm^{2})(14cm) = 1400πcm^{3} Using the value of π ≈ 3.14, we can approximate the answer to: V ≈ 1400(3.14) ≈ 4400cm^{3} Therefore, the volume of the cylinder is approximately 4400cm^{3}. The correct option is (D) 4400cm^{3}.

**Question 60**
**Report**

If the ratio x:y = 3:5 and y:z = 4:7, find the ratio x:y:z

**Question 61**
**Report**

If x =3, y = 2 and z = 4 what is the value of \(3x^{2}-2y+z\)

**Answer Details**

Substituting x = 3, y = 2 and z = 4 into the expression \(3x^{2}-2y+z\), we get: $$3(3)^2 - 2(2) + 4 = 27 - 4 + 4 = 27$$ Therefore, the value of \(3x^{2}-2y+z\) when x = 3, y = 2 and z = 4 is 27. The correct option is (b) 27.

**Question 62**
**Report**

The venn diagram shows the choice of food of a number of visitors to a canteen. If there were 35 visitors in all, find the value of x

**Question 63**
**Report**