JAMB UTME - Mathematics - 2013

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5x${}^{\circ }$ + (16x - 24)${}^{\circ }$ + 5x${}^{\circ }$ + (4x + 12)${}^{\circ }$ + (6x + 12)${}^{\circ }$ = 360${}^{\circ }$

360x${}^{\circ }$ - 24 + 12 + 12 = 360${}^{\circ }$

36x${}^{\circ }$ = 360${}^{\circ }$

x${}^{\circ }$ = $\frac{{360}^0}{36}$

= 10${}^{\circ }$

Thus, the angle of sector representing Mathematics is 5 x 10${}^{\circ }$ = 50${}^{\circ }$. Hence the number of students who offer mathematics is

$\frac{{55}^o}{36}\times 80\approx 11$

Akọwa Nkọwa

The graph represented by the given equation is y = x² - x - 2. To understand this, we first need to understand what a quadratic equation is. A quadratic equation is a polynomial equation of the second degree, meaning it has an exponent of 2. In other words, it's an equation where the highest exponent of the variable is 2. The equation y = x² - x - 2 is a quadratic equation, where y is the dependent variable and x is the independent variable. The graph of a quadratic equation is a parabola, which is a symmetrical U-shaped curve. Looking at the given equation, we can see that the coefficient of x² is positive, which means that the parabola will open upwards. The coefficient of x is negative, which means that the vertex of the parabola will be to the right of the y-axis. The constant term is -2, which means that the vertex will be two units below the y-axis. So, the graph represented by the given equation is a parabola that opens upwards, with its vertex at (0.5, -2.25) and its axis of symmetry being the line x = 0.5. Therefore, the correct option is the one that represents the equation y = x² - x - 2.

Akọwa Nkọwa

To find the mean of the given values, we need to add them up and divide the result by the total number of values. So, the sum of the given values is: (t + 2) + (2t - 4) + (3t + 2) + 2t = 8t There are four values, so to find the mean, we divide the sum by 4: mean = (8t) / 4 = 2t Therefore, the correct option is 2t, which is the mean of the given values. To summarize, the mean of the values t + 2, 2t - 4, 3t + 2, and 2t is 2t. We added up the values and divided by the total number of values to find the mean.

Akọwa Nkọwa

The midpoint formula states that the midpoint of a line segment with endpoints (x1,y1) and (x2,y2) is: ((x1+x2)/2, (y1+y2)/2) In this problem, we are given the midpoint of the line segment PQ, which is (2,3), and one endpoint, P, which is (-2,1). Let Q have coordinates (x,y), then we can use the midpoint formula to find the coordinates of Q: ((x + (-2))/2, (y + 1)/2) = (2,3) Simplifying this expression, we get: (x-2, y+1) = (4,6) Now we can solve for x and y by equating the x-coordinates and y-coordinates separately: x - 2 = 4 => x = 6 y + 1 = 6 => y = 5 Therefore, the coordinates of point Q are (6,5). So, the correct answer is (6,5).

Akọwa Nkọwa

To find the mode of the seven numbers, we need to determine the number that appears most frequently among the seven numbers. We are given that the mean (average) of the seven numbers is 10. To find the sum of the seven numbers, we can multiply the mean by 7: mean = (sum of seven numbers) / 7 Rearranging this formula, we get: (sum of seven numbers) = mean x 7 Substituting the given mean of 10, we get: (sum of seven numbers) = 10 x 7 = 70 We are also given that six of the seven numbers are 2, 4, 8, 14, 16, and 18. To find the seventh number, we can subtract the sum of these six numbers from the sum of the seven numbers: (seventh number) = (sum of seven numbers) - (sum of six numbers) Substituting the values we know, we get: (seventh number) = 70 - (2 + 4 + 8 + 14 + 16 + 18) (seventh number) = 8 So the seven numbers are: 2, 4, 8, 14, 16, 18, and 8. To find the mode, we need to determine which number appears most frequently among these seven numbers. We can see that the number 8 appears twice, while all the other numbers appear only once. Therefore, the mode of these seven numbers is 8.

Akọwa Nkọwa

y = (2x +2)3
Then $\frac{\delta y}{\delta x}$ = 3(2x +2)22
=6(2x +2)2

Akọwa Nkọwa

The range of a distribution is the difference between the highest and lowest values in the distribution. In this case, the highest score is 10 and the lowest score is 3. Therefore, the range of the distribution is: 10 - 3 = 7 So the correct option is 7.

Akọwa Nkọwa

Angle of major sector = 360° - 120° = 240°

Area of major sector : θ360×πr2 $\frac{\theta }{360}\times \pi r^2$

r = 43√2=23–√cm $\frac{4\sqrt{3}}{2}=2\sqrt{3}cm$

Area : 240360×π×(23–√)2 $\frac{240}{360}\times \pi \times (2\sqrt{3}{)}^2$

= 8πcm2

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Akọwa Nkọwa

y = x sin x
Where u = x and v = sin x
Then $\frac{\delta u}{\delta x}$ = 1 and $\frac{\delta v}{\delta x}$ = cos x
By the chain rule, $\frac{\delta y}{\delta x}=v\frac{\delta u}{\delta x}+u\frac{\delta v}{\delta x}$
= (sin x)1 + x cos x
= sin x + x cos x

Akọwa Nkọwa

Using the cosine rule, we have

p2=q2+r2−2qrcosP $p^2=q^2+r^2-2qr\cos P$

p2=82+62−2(8)(6)(112) $p^2=8^2+6^2-2(8)(6)(\frac{1}{12})$

= 64+36−8 $64+36-8$

p2=92∴p=92−−√cm

Akọwa Nkọwa

In any quadrilateral, the sum of its interior angles is always 360 degrees. Therefore, we can write an equation using the given angles as follows: (3y + 10)° + (2y + 30)° + (y + 20)° + 4y° = 360° Simplifying the equation: 10y + 60 = 360 10y = 300 y = 30 Hence, the value of y is 30. Therefore, option C, "30°", is the correct answer.

Akọwa Nkọwa

We know that the formula for the area of a circle is A = πr^2, where A is the area and r is the radius. We are given that the radius of the circle is increasing at the rate of 0.02 cm/s. This means that dr/dt = 0.02 cm/s. We need to find the rate at which the area is increasing when the radius of the circle is 7 cm. This means we need to find dA/dt when r = 7 cm. To do this, we first differentiate the formula for the area of a circle with respect to time: dA/dt = d/dt(πr^2) Using the chain rule, we get: dA/dt = 2πr (dr/dt) Substituting the given values, we get: dA/dt = 2π(7) (0.02) = 0.88π cm^2/s So, the rate at which the area is increasing when the radius of the circle is 7 cm is 0.88π cm^2/s, which is approximately 2.76 cm^2/s. Therefore, the answer is option (D) 0.88cm^2S^-1.

Akọwa Nkọwa

The probability of picking a fruit that is neither an apple nor an orange is the same as the probability of picking a banana, since bananas are the only remaining fruit in the basket. To calculate the probability, we need to divide the number of bananas in the basket by the total number of fruits in the basket. The total number of fruits in the basket is: 9 (apples) + 8 (bananas) + 7 (oranges) = 24 So the probability of picking a banana is: 8 (bananas) / 24 (total fruits) = 1/3 Therefore, the answer is the third option: 3.

Akọwa Nkọwa

Given that P = |5321| and Q = |4235|, we need to find 2P + Q. First, we need to evaluate P and Q. The vertical bars indicate the absolute value of the number, which means the distance of the number from zero on the number line. Therefore, we have: P = |5321| = 5321 Q = |4235| = 4235 Now, we can substitute these values in 2P + Q and simplify: 2P + Q = 2(5321) + 4235 = 10642 + 4235 = 14877 Therefore, the value of 2P + Q is 14877. Hence, the answer is option B, i.e., |14877|.

Akọwa Nkọwa

Let f(p) = 6p3 - p2 - 47p + 30
Then by the remainder theorem,
(p - 3): f(3) = remainder R,
i.e. f(3) = 6(3)3 - (3)2 - 47(3) + 30 = R
162 - 9 - 141 + 30 = R
192 - 150 = R
R = 42

Akọwa Nkọwa

5 people can take 3 places in;
5P3 ways, = $\frac{5!}{(5-3)!}$ = $\frac{5!}{2!}$
= $\frac{5\times 4\times 3\times 2!}{2!}$
= 5 x 4 x 3
= 60 ways

Akọwa Nkọwa

$r\propto \frac{1}{\sqrt{s}},r\propto \frac{1}{\sqrt{t}}$
$r\propto \frac{1}{\sqrt{s}}$ ..... (1)
$r\propto \frac{1}{\sqrt{t}}$ ..... (2)
Combining (1) and (2), we get
$r=\frac{k}{\sqrt{s}\times \sqrt{t}}=\frac{k}{\sqrt{st}}$
This gives $\sqrt{st}=\frac{k}{r}$
By taking the square of both sides, we get
st = $\frac{k^2}{r^2}$
s = $\frac{k^2}{r^{2}t}$

Akọwa Nkọwa

To calculate the number of ways a student can select 2 subjects from 5 subjects, we can use the formula for combinations: C(n, k) = n! / (k! (n - k)!) where: n = number of subjects (5) k = number of subjects to be selected (2) So, substituting the values in the formula, we get: C(5, 2) = 5! / (2! (5 - 2)!) C(5, 2) = 120 / (2 * 3!) C(5, 2) = 120 / (2 * 6) C(5, 2) = 120 / 12 C(5, 2) = 10 So, a student can select 2 subjects from 5 subjects in 10 different ways.

Akọwa Nkọwa

To find the number of square tiles required to cover the rectangular floor, we need to find out how many tiles can fit into the floor without overlapping. We can start by finding the area of the rectangular floor: Area = Length x Width Area = 7.2m x 4.2m Area = 30.24 square meters Next, we need to find the area of one square tile: Area of tile = Side x Side Area of tile = 30cm x 30cm Area of tile = 900 square centimeters (We need to convert this to square meters to match the area of the floor) Area of tile = 0.09 square meters Now, we can divide the total area of the floor by the area of one tile to get the number of tiles required: Number of tiles = Total area / Area of one tile Number of tiles = 30.24 square meters / 0.09 square meters Number of tiles = 336 Therefore, we need 336 square tiles to cover the rectangular floor. is the correct answer.

Akọwa Nkọwa

Given P(2, -3) and Q(-5, 1)

Midpoint = (2+(−5)2,−3+12) $(\frac{2+(-5)}{2},\frac{-3+1}{2})$

= (−32,−1) $(\frac{-3}{2},-1)$

Slope of the line PQ = 1−(−3)−5−2 $\frac{1-(-3)}{-5-2}$

= −47 $-\frac{4}{7}$

The slope of the perpendicular line to PQ = −1−47 $\frac{-1}{-\frac{4}{7}}$

= 74 $\frac{7}{4}$

The equation of the perpendicular line: y=74x+b $y=\frac{7}{4}x+b$

Using a point on the line (in this case, the midpoint) to find the value of b (the intercept).

−1=(74)(−32)+b $-1=(\frac{7}{4})(\frac{-3}{2})+b$

−1+218=138=b $-1+\frac{21}{8}=\frac{13}{8}=b$

∴ $\therefore$ The equation of the perpendicular bisector of the line PQ is y=74x+138 $y=\frac{7}{4}x+\frac{13}{8}$

≡8y=14x+13⟹8y−14x−13=0

Akọwa Nkọwa

Let's call the shares of the three girls "5x", "3x", and "2x", where "x" is a constant factor that will allow us to determine the actual number of apples in each share. We are told that the highest share is 40 apples, which corresponds to the "5x" share. Therefore, we can set up an equation: 5x = 40 Solving for "x", we get: x = 8 Now we can determine the actual number of apples in each share by multiplying the corresponding factor by "x": - The largest share ("5x") has 5 x 8 = 40 apples - The second-largest share ("3x") has 3 x 8 = 24 apples - The smallest share ("2x") has 2 x 8 = 16 apples Therefore, the smallest share has 16 apples, which is option (C).

Akọwa Nkọwa

$\frac{3^{-5n}}{9^{1-n}}\times {27}^{n+1}$
$\frac{3^{-5n}}{3^{2(1-n)}}\times 3^{3(n+1)}$
$3^{-5n}÷3^{2(1-n)}\times 3^{3(n+1)}$

$3^{-5n-2(1-n)+3(n+1)}$

$3^{-5n-2+2n+3n+3}$
$3^{-5n+5n+3-2}$
$3^1$
= 3

Akọwa Nkọwa

log1041/3 = 1/3 log104
= 1/3 x 0.6021
= 0.2007

Akọwa Nkọwa

2 x s = 280o(Angle at centre = 2 x < at circum)

S = $\frac{{280}^o}{2}$

= 140

< O = 360 - 280 = 80o

60 + 80 + 140 + a = 360o

(< in a quad); 280 = a = 360

a = 360 - 280

a = 80o

Akọwa Nkọwa

Let A = $\left(\begin{array}{cc}5& 3\\ 6& 4\end{array}\right)$
Then |A| = $\left(\begin{array}{cc}5& 3\\ 6& 4\end{array}\right)$ = 20 - 18 = 2
Hence A-1 = $\frac{1}{|A|}\left(\begin{array}{cc}4& -3\\ -6& 5\end{array}\right)$
= $\frac{1}{2}\left(\begin{array}{cc}4& -3\\ -6& 5\end{array}\right)$
= $\left(\begin{array}{cc}4\times 1/2& -3\times 1/2\\ -6\times 1/2& 5\times 1/2\end{array}\right)$
= $\left|\begin{array}{cc}2& -\frac{3}{2}\\ -3& \frac{5}{2}\end{array}\right|$

Akọwa Nkọwa

To calculate the time taken for ₦3000 to earn ₦600 if invested at 8% simple interest, we can use the formula: time = (interest / principal) / rate where: interest = ₦600 principal = ₦3000 rate = 8% So, substituting the values in the formula, we get: time = (600 / 3000) / 0.08 time = (0.2) / 0.08 time = 2.5 years So, it takes approximately 2.5 years for ₦3000 to earn ₦600 if invested at 8% simple interest.

Akọwa Nkọwa

∫1+xx3dx $\int \frac{1+x}{x^3}\mathrm{d}x$

= ∫(1x3+xx3)dx $\int (\frac{1}{x^3}+\frac{x}{x^3})\mathrm{d}x$

= ∫(x−3+x−2)dx $\int (x^{-3}+x^{-2})\mathrm{d}x$

= −12x2−1x+k

Akọwa Nkọwa

Length of chord = 2rsin(θ2) $2r\sin (\frac{\theta }{2})$

= 2×8×sin(902) $2\times 8\times \sin (\frac{90}{2})$

= 16×2√2 $16\times \frac{\sqrt{2}}{2}$

= 82–√cm

Akọwa Nkọwa

To find the median age of the frequency distribution in the table, we need to arrange the ages in ascending order and then find the middle value. However, since we are given the frequency of each age group, we need to use cumulative frequency to find the median. First, we calculate the cumulative frequency by adding up the frequency of each age group. We can add the frequencies from the bottom up or from the top down. In this case, we will add from the bottom up: Age | Frequency | Cumulative Frequency ----- | -----------| --------------------- 20 | 3 | 3 25 | 5 | 8 30 | 1 | 9 35 | 1 | 10 40 | 2 | 12 45 | 3 | 15 Next, we find the median age by finding the age that corresponds to the middle value of the cumulative frequency. Since there are 15 people in total, the middle value is (15 + 1) ÷ 2 = 8. Looking at the cumulative frequency column, we can see that the age group that corresponds to the 8th person is 25-29, which has a cumulative frequency of 8. Therefore, the median age is in the range of 25-29. Since the age group 25-29 has a frequency of 5, we can assume that the median age lies somewhere in the middle of this range. We can take the midpoint of this range as the median age. The midpoint is (25 + 29) ÷ 2 = 27. Therefore, the median age of the frequency distribution in the table is 27. Answer: Option (B) 25.

Akọwa Nkọwa

tanθ=oppadj=34 $\tan \theta =\frac{opp}{adj}=\frac{3}{4}$

hyp2=opp2+adj2 $hy{p}^2=op{p}^2+ad{j}^2$

hyp=32+42−−−−−−√ $hyp=\sqrt{3^2+4^2}$

= 5

sinθ=35;cosθ=45 $\sin \theta =\frac{3}{5};\cos \theta =\frac{4}{5}$

sinθ+cosθ=35+45 $\sin \theta +\cos \theta =\frac{3}{5}+\frac{4}{5}$

= 75=125

Akọwa Nkọwa

To find the probability that an integer between 1 and 25 is divisible by both 2 and 3, we need to count the number of integers that satisfy this condition, and divide by the total number of integers between 1 and 25. A number is divisible by both 2 and 3 if and only if it is divisible by their product, which is 6. Therefore, we need to count the number of integers between 1 and 25 that are divisible by 6. To do this, we can start by finding the smallest multiple of 6 that is greater than or equal to 1, which is 6 itself. The next multiples of 6 are 12, 18, and 24, all of which are less than or equal to 25. Therefore, there are 4 integers between 1 and 25 that are divisible by both 2 and 3. The total number of integers between 1 and 25 is 25, so the probability that an integer chosen at random from this range is divisible by both 2 and 3 is 4/25. Therefore, among the given options, the answer is the one represented by the fraction 4/25.

Akọwa Nkọwa

Let f(x) = x2 - x - k
Then by the factor theorem,
(x - 4): f(4) = (4)2 - (4) - k = 0
16 - 4 - k = 0
12 - k = 0
k = 12

Akọwa Nkọwa

80 + 160 + 200 + 80 = 128 + 72 = 720minutes

Akọwa Nkọwa

Akọwa Nkọwa

Using Tn = $\frac{3n+1}{n+1}$,
T1 = $\frac{3\left(1\right)+1}{\left(1\right)+1}=\frac{4}{2}$
T2 = $\frac{3\left(2\right)+1}{\left(2\right)+1}=\frac{7}{3}$
T3 = $\frac{3\left(3\right)+1}{\left(3\right)+1}=\frac{10}{4}$

Akọwa Nkọwa

In the diagram above, < CDE = < CED = 15o

(base < s of isos. △)

< ECD = 180o - (15 + 15)o

= 180o - 30o = 150o

But x + 110o = 150o

(Sum of opp. interior < s of a△ = opp. exterior < )

x = 150o - 110o = 40o

Akọwa Nkọwa

P $\propto$ mu, p $\propto \frac{1}{q}$
p = muk ................ (1)
p = $\frac{1}{q}k$.... (2)
Combining (1) and (2), we get
P = $\frac{mu}{q}k$
4 = $\frac{m\times u}{1}k$
giving k = $\frac{4}{6}=\frac{2}{3}$
So, P = $\frac{mu}{q}\times \frac{2}{3}=\frac{2mu}{3q}$
Hence, P = $\frac{2\times 6\times 4}{3\times \frac{8}{5}}$
P = $\frac{2\times 6\times 4\times 5}{3\times 8}$
p = 10

Akọwa Nkọwa

The locus of points which are equidistant from the line PQ is a line that is perpendicular to PQ and passes through the midpoint of PQ. This is because a point is equidistant from a line if and only if it is on the perpendicular bisector of the line segment connecting any two points on the line. So, the correct option is "perpendicular line to PQ". The points on this line are equidistant from the line PQ, and the line passes through the midpoint of PQ, which is equidistant from both P and Q. A circle with center P or Q would not include all the points equidistant from PQ, and a pair of parallel lines would not be equidistant from PQ.

Akọwa Nkọwa

S = $\sqrt{t^2-4t+4}$
S2 = t2 - 4t + 4
t2 - 4t + 4 - S2 = 0
Using $t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
Substituting, we have;
Using $t=\frac{-(-4)\pm \sqrt{(-4{)}^2-4(1\left)\right(4-S^2)}}{2\left(1\right)}$
$t=\frac{4\pm \sqrt{16-4(4-S^2)}}{2}$
$t=\frac{4\pm \sqrt{16-16+4S^2}}{2}$
$t=\frac{4\pm \sqrt{4S^2}}{2}$
$t=\frac{2(2\pm S)}{2}$
Hence t = 2 + S or t = 2 - S

Akọwa Nkọwa

3(x + 2) > 6(x + 3)
3x + 6 > 6x + 18
3x - 6x > 18 - 6
-3x > 12
x < -4

Akọwa Nkọwa

To find the standard deviation of the given set of numbers, we need to find the variance first, using the formula: Variance = (sum of squares of deviations from the mean) / (number of observations) Let's start by finding the mean of the given set of numbers: Mean = (3 + x + 6 + 4 + 7 - x) / 5 = 20 / 5 = 4 Next, we can substitute this mean into the formula for variance, which gives us: 4 = [(3 - 4)^2 + (6 - 4)^2 + (4 - 4)^2 + (x - 4)^2 + (7 - x - 4)^2] / 5 Simplifying this equation, we get: 20 = (3 - 4)^2 + (6 - 4)^2 + (4 - 4)^2 + (x - 4)^2 + (7 - x - 4)^2 20 = 1 + 4 + 0 + (x - 4)^2 + (3 - x)^2 Simplifying further, we get: 15 = 2(x^2 - 8x + 13) Expanding and simplifying this equation, we get: 2x^2 - 16x + 26 = 15 2x^2 - 16x + 11 = 0 We can solve this quadratic equation using the quadratic formula: x = [16 ± sqrt(16^2 - 4(2)(11))] / (2(2)) x = [16 ± sqrt(176)] / 4 x = [16 ± 4sqrt(11)] / 4 x = 4 ± sqrt(11) Now that we have found the possible values of x, we can calculate the standard deviation using the formula: Standard deviation = sqrt(variance) We already know the variance is 4, so: Standard deviation = sqrt(4) = 2 Therefore, the answer is (B) 2.

Akọwa Nkọwa

We can evaluate the integral using the following steps: 1. Recall that the integral of sin(x) with respect to x is -cos(x) + C, where C is the constant of integration. 2. Using this formula, we can integrate sin(x) between the limits of integration π/2 and 0: ∫π/20 sin(x) dx = [-cos(x)]π/20 = (-cos(π/2)) - (-cos(0)) = (-0) - (-1) = 1 3. Therefore, the value of the integral is 1. So the correct answer is 3.

Akọwa Nkọwa

Using Sn = $a\frac{r^2-1}{r-1}$
we get S2 = 3 = $a\frac{r^2-1}{r-1}$
giving 3(r - 1) = a(r2 - 1)
3r - 3 = ar2 - a
ar2 - 3r - a = -3 ..... (1)
ar + ar2 = -6 ..... (2)
From (2), a = $\frac{-6}{(r+r^2)}$
Substitute $\frac{-6}{(r+r^2)}$ for a in (1)
$\left(\frac{-6}{(r+r^2)}\right)r^2-3r-\frac{-6}{(r+r^2)}=-3$
Multiply through by (r + r2) to get
-6r2 - 3r(r + r2) + 6 = -3(r + r2)
-6r2 - 3r2 - 3r3 + 6 = -3r - 3r2
Equating to zero, we have
3r3 - 3r2 + 3r2 + 6r2 - 3r - 6 = 0
This reduces to;
3r3 + 6r2 - 3r - 6 = 0
3(r3 + 2r2 - r - 2) = 0
By the factor theorem,
(r + 2): f(-2) = (-2)3 + 2(-2)2 - (-2) - 2
-8 + 8 + 2 - 2 = 0
giving r = -2 as the only valid value of r for the G.P.
From (3), = $\frac{-6}{-2+(-2{)}^2}=\frac{-6}{-2+4}$
a = -6/2 = -3
Hence (a + r) = (-3 + -2) = -5

Akọwa Nkọwa

To find 2 * (3 * 4), we need to first evaluate the expression inside the parentheses, which is 3 * 4. According to the definition of the binary operation * that was given, x * y = x + 2y, we can substitute x = 3 and y = 4 to get 3 * 4 = 3 + 2 * 4 = 3 + 8 = 11. Next, we can substitute x = 2 and y = 11 into the expression x * y = x + 2y to find 2 * 11, which is 2 + 2 * 11 = 2 + 22 = 24. So, 2 * (3 * 4) = 24.

Akọwa Nkọwa

To convert the number 27 from base 10 to base 3, we need to divide 27 by 3 repeatedly until the quotient becomes zero. The remainders of each division, read from bottom to top, will be the digits of the number in base 3. 27 divided by 3 is 9 with a remainder of 0. 9 divided by 3 is 3 with a remainder of 0. 3 divided by 3 is 1 with a remainder of 0. 1 divided by 3 is 0 with a remainder of 1. Therefore, the number 27 in base 3 is 1000.

Akọwa Nkọwa

$\frac{\sqrt{5}(\sqrt{147}-\sqrt{12}}{\sqrt{15}}$
$\frac{\sqrt{5}(\sqrt{49\times 3}-\sqrt{4\times 3}}{\sqrt{5\times 3}}$
$\frac{\sqrt{5}(7\sqrt{3}-2\sqrt{3}}{\sqrt{5}\times \sqrt{3}}$
$\frac{\sqrt{3}(7-2}{\sqrt{3}}$
= 5