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**Question 1**
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The shadow of a pole 5√3m high is 5m. Find the angle of elevation of the sun.

**Answer Details**

The height of the pole is 5√3m and its shadow on the ground is 5m. Let x be the angle of elevation of the sun. From the geometry of the situation, we can see that tan(x) = opposite/adjacent = (5√3)/5 = √3. Taking the inverse tangent of both sides, we get: x = tan^(-1)(√3) ≈ 60° Therefore, the angle of elevation of the sun is 60°. So the answer is (D) 60°.

**Question 2**
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The length L of a simple pendulum varies directly as the square of its period T. If a pendulum with period 4 sec. is 64 cm long, find the length of pendulum whose period is 9 sec

**Answer Details**

The problem describes a direct variation between the length of a simple pendulum and the square of its period, which means that we can write the relationship as: L = kT^2 where L is the length of the pendulum, T is its period, and k is a constant of proportionality. To find the value of k, we can use the information given in the problem that a pendulum with a period of 4 seconds has a length of 64 cm. Substituting these values into the equation above, we get: 64 = k(4^2) 64 = 16k k = 4 Now that we know the value of k, we can use the same equation to find the length of a pendulum with a period of 9 seconds: L = 4(9^2) L = 324 Therefore, the length of the pendulum whose period is 9 seconds is 324 cm.

**Question 5**
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The locus of a point which is 5 cm from the line LM is a

**Answer Details**

The locus of a point which is 5cn from the line LM is a pair of lines on opposite sides of LM and parallel to it, each distance 5cm from LM

**Question 6**
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The shaded area in the diagram above is represented by

**Answer Details**

Coordinates of the line are (x1 y1)(0,2) respectively and (x2 y2)(2,0) respectively

Gradient (m) of the line = (y2 - y1) / (x2 - x1) = (0-6) / (2-0) = -6/2

= -3

Equation of the line y-y1 = m(x-x1)

y - 6 = -3(x - 0)

y - 6 = -3x

y + 3x = 6

inequality; from the y axis, the shaded region is below the line

∴ y + 3x ∠ 6 and line is a broken line

**Question 7**
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The mean age of a group of students is 15 years. When the age of a teacher, 45 years old, is added to the age of the students, the mean of their ages becomes 18 years. Find the number of the students in the group

**Answer Details**

Let's start by defining some variables: - n: the number of students in the group - sum_age_students: the sum of the ages of the students We know that the mean age of the group of students is 15, so we can write: mean_age_students = sum_age_students / n = 15 From the problem, we also know that when the age of the teacher is added to the age of the students, the mean of their ages becomes 18. So we can write another equation: (mean_age_students + 45) = (sum_age_students + 45) / (n + 1) = 18 Now we have two equations with two variables (n and sum_age_students), and we can solve for them simultaneously. Let's start by simplifying the second equation: sum_age_students + 45 = 18(n + 1) sum_age_students = 18n + 27 Now we can substitute this expression for sum_age_students in the first equation: sum_age_students / n = 15 (18n + 27) / n = 15 18n + 27 = 15n 3n = 27 n = 9 Therefore, there are 9 students in the group.

**Question 8**
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If y = 3 cos(x/3), find dy/dx when x = (3π/2)

**Answer Details**

y = 3cos(x/3)

dy/dx = 3x(1/3)x - sin (x/3)

= - sin x/3

But x = 3π/2 ∴ -sin(x/3) = -sin(3π/6)

= -sin (3 * 180)/6

= - sin 90

= -1

**Question 9**
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An arc of length 22cm subtends an angle of θ at the center of the circle. What is the value of θ if the radius of the circle is 15cm?[Take π = 22/7]

**Answer Details**

In a circle, the ratio of the arc length to the circumference of the circle is equal to the ratio of the angle subtended by the arc at the center of the circle to the angle of one full revolution (360 degrees). Using this property, we can find the value of the angle θ in the following way: The circumference of the circle is given by: C = 2πr = 2 x (22/7) x 15 = 94.29 cm The given arc length is 22 cm. Therefore, the ratio of the arc length to the circumference of the circle is: 22/94.29 = 0.2332 Let the angle of one full revolution be 360 degrees. Then, the ratio of the angle subtended by the arc at the center of the circle to the angle of one full revolution is also 0.2332. Thus, we can write: θ/360 = 0.2332 Multiplying both sides by 360, we get: θ = 360 x 0.2332 = 83.952 degrees (approximately) Therefore, the value of θ is approximately 84 degrees. The correct option is: - 84o

**Question 10**
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If 6logx2 - 3logx3 = 3log50.2, find x.

**Answer Details**

6logx2 - 3logx3 = 3log50.2

= logx26 - 3logx33 = log5(0.2)3

= logx(64/27) = log5(1/5)3

logx(64/27) = log5(1/125)

let logx(64/27) = y

∴xy = 64/27

and log5(1/125) = y

∴ 5y = 1/125

5y = 125-1

5y = 5-3

∴ y = -3

substitute y = -3 in xy = 64/27

implies x-3 = 64/27

1/x3 = 64/27

64x3 = 27

x3 = 27/64

x3 = 3√27/64

x = 3/4

**Question 11**
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Given that 3√42x = 16, find the value of x

**Answer Details**

3√42x = 16

this implies that (3√42x)3 = (16)3

42x = 42*3

42x = 46

∴ 2x = 6

x = 3

**Question 12**
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Determine the locus of a point inside a square PQRS which is eqidistant from PQ and QR

**Answer Details**

The diagonal QS bisects the angle formed by PQ and QR

∴ [A]

**Question 13**
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I. Rectangular bars of equal width

II. The height of each rectangular bar is proportional to the frequency of the corresponding class interval.

III. Rectangular bars have common sides with no gaps in between

A histogram is described completely by

**Answer Details**

A histogram is a graphical representation of a frequency distribution of a dataset, where the data is divided into intervals or bins and the frequency of each interval is represented by the area of a rectangle or bar. To construct a histogram, we need to follow certain rules, which are described by the statements: I. Rectangular bars of equal width II. The height of each rectangular bar is proportional to the frequency of the corresponding class interval. III. Rectangular bars have common sides with no gaps in between. Therefore, a histogram is described completely by statements I, II, and III. Statement I ensures that each bar is of equal width, which is necessary to represent the intervals or bins in a consistent way. Statement II ensures that the height of each bar accurately represents the frequency of the corresponding class interval. This is important because the area of the bar represents the frequency of the interval, which is the product of the width and height of the bar. Statement III ensures that the bars are adjacent to each other without any gaps, which is necessary to accurately represent the continuity of the data. Hence, the correct answer is option (A) I, II, and III.

**Question 14**
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Find the midpoint of the line joining P(-3, 5) and Q(5, -3).

**Answer Details**

The midpoint of a line segment is the point that is exactly halfway between the endpoints of the line segment. The formula for finding the midpoint of a line segment with endpoints (x1, y1) and (x2, y2) is: Midpoint = ((x1 + x2) / 2, (y1 + y2) / 2) In this case, we are given the endpoints of the line segment PQ as P(-3, 5) and Q(5, -3). Using the formula for the midpoint, we can find the coordinates of the midpoint as: Midpoint = ((-3 + 5) / 2, (5 - 3) / 2) = (2 / 2, 2 / 2) = (1, 1) Therefore, the midpoint of the line joining P(-3, 5) and Q(5, -3) is (1, 1). Hence, the correct answer is option (A) (1, 1).

**Question 15**
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In how many ways can 2 students be selected from a group of 5 students in a debating competition?

**Answer Details**

**Question 16**
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Find the remainder when 3x3 + 5x2 - 11x + 4 is divided by x + 3

**Answer Details**

x = -3

substitute x = -3 in 3x3 + 5x2 - 11x + 4

3(-3)3 + 5(-3)2 - 11(-3) + 4

-81 + 45 + 33 + 4

-81 + 82

= 1

**Question 17**
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If the operation * on the set of integers is defined by p * Q = √pq $\sqrt{pq}$, find the value of 4 * ( 8 * 32).

**Answer Details**

P×Q=√PQ4×(8×32)=4×√8×32=4×√256=4×16=√4×16=√64=8

**Question 18**
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The weight of 10 pupils in a class are 15 kg, 16 kg, 17 kg, 18 kg, 16 kg, 17 kg, 17 kg, 17 kg, 18 kg, and 16 kg. What is the range of this distribution?

**Answer Details**

The range of a distribution is the difference between the maximum and minimum values. In this case, the maximum weight is 18 kg and the minimum weight is 15 kg, so the range is 18 kg - 15 kg = 3 kg. Therefore, the answer is option (B) 3.

**Question 19**
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Simplify 1√3+2 $\frac{1}{\sqrt{3}+2}$ in the form a+b√3

**Answer Details**

1√3+2=1√3+2×√3−2√3−2=√3−2(√3)2−22=√3−23−4=√3−21=−√3+2=2−√3

**Question 20**
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Find the value of x in the figure above

**Answer Details**

Xsin45=15sin60X=15sin45sin60X=15×(1√2)√32X=15√2×2√3X=30√6X=30√6×√6√6=30√66=5√6

**Question 21**
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Find *P*, if 4516 - *P*7 = 3056

**Answer Details**

4516 - P7 = 3056

P7 = 4516 - 3056

P7 = 1426

convert 1426 = 1 * 62 + 4 * 61 + 2 * 60

= 36 + 24 + 2

= 62

Convert 6210 to base 7

62/7 = 8 R 6

8/7 = 1 R 1

1/7 = 0 R 1

∴P7 = 1167

P7 = 4516 - 3056

P7 = 1426

convert 1426 = 1 * 62 + 4 * 61 + 2 * 60

= 36 + 24 + 2

= 62

Convert 6210 to base 7

62/7 = 8 R 6

8/7 = 1 R 1

1/7 = 0 R 1

∴P7 = 1167

**Question 22**
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Some white balls put in a basket containing twelve red balls and sixteen black balls. If the probability of picking a white balls from the baskets is 3/7, how many white balls were introduced?

**Answer Details**

Let's say the number of white balls introduced is x. The total number of balls in the basket before introducing the white balls was 12 + 16 = 28. After introducing the x white balls, the total number of balls becomes 12 + 16 + x. Now, the probability of picking a white ball is given as 3/7. We can express this probability as: Number of white balls / Total number of balls = 3/7 Substituting the values, we get: x / (12 + 16 + x) = 3/7 Solving for x: 7x = 3(28 + x) 7x = 84 + 3x 4x = 84 x = 21 Therefore, the number of white balls introduced is 21. So, option (B) is the correct answer.

**Question 23**
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Find the value of α2 + β2 if α + β = 2 and the distance between points (1, α) and (β, 1)is 3 units

**Answer Details**

PQ=√(β−1)2+(1−α)23=√(β2−2β2+1+1−2α+α2)3=√(α2+β2−2α+2β+2)3=√(α2+β2−2(α+β)+2)3=√(α2+β2−2∗2+2)3=√(α2+β2−2)9=(α2+β2−2)α2+β2=9+2α2+β2=11

**Question 24**
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The graph above shows the cumulative frequency curve of the distribution of marks in a class test. What percentage of the students scored more than 20 marks?

**Answer Details**

class-marks(x)(freq. (f)cum. freq.93.....3146....3+6=9199.....9+9=18243.....18+3=21292.....21+2=23342.....23+2=2525
$\begin{array}{ccc}\text{class-marks(x)}& \text{(freq. (f)}& \text{cum. freq.}\\ 9& & \mathrm{3.....3}\\ 14& & \mathrm{6....3}+6=9\\ 19& & \mathrm{9.....9}+9=18\\ 24& & \mathrm{3.....18}+3=21\\ 29& & \mathrm{2.....21}+2=23\\ 34& & \mathrm{2.....23}+2=25\\ & & 25\end{array}$

% of students scoring more than 20 marks

= 725 $\frac{7}{25}$ x 100% = 28%

**Question 25**
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A farmer planted 5000 grains of maize and harvested 5000 cobs, each bearing 500 grains. What is the ratio of the number of grains sowed to the number harvested?

**Answer Details**

To find the ratio of the number of grains sowed to the number harvested, we need to divide the number of grains sowed by the number of grains harvested. Number of grains sowed = 5000 Number of grains harvested = 5000 cobs × 500 grains per cob = 2,500,000 So the ratio of the number of grains sowed to the number harvested is: 5000 : 2,500,000 Simplifying the ratio by dividing both sides by 5000, we get: 1 : 500 Therefore, the answer is: 1 : 500.

**Question 27**
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An unbiased die is rolled 100 times and the outcome is tabulated above.

What is the probability of obtaining a 5?

**Answer Details**

Number of times 5 was obtained = 20

P(obtaining 5) = 20/100

= 1/5

**Question 28**
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In the diagram above, KLMN is a cyclic quadrilateral. /KL/ = /KN/, ∠NKM = 55o and ∠KML = 40o. Find ∠LKM.

**Answer Details**

**Question 29**
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Factorize completely ac - 2bc - a + 4b2

**Answer Details**

To factorize the expression, we can group the terms as follows: ac - 2bc - a + 4b^2 = a(c - 1) - 2b(c - 2b) Now we can factor out the common factors: a(c - 1) - 2b(c - 2b) = (c - 1)(a - 2b) So the completely factored form of the expression is: ac - 2bc - a + 4b^2 = (c - 1)(a - 2b) Therefore, the correct answer is option A: (a - 2b)(c - a - 2b).

**Question 30**
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Given that the first and forth terms of G.P are 6 and 162 respectively, find the sum of the first three terms of the progression

**Answer Details**

To find the sum of the first three terms of a G.P, we need to determine the common ratio (r) of the progression. Given that the first and fourth terms are 6 and 162 respectively, we can use the formula for the nth term of a G.P to obtain: a1 = 6, a4 = 162 a4 = a1 * r^3 162 = 6 * r^3 r^3 = 27 r = 3 Now, we can find the second and third terms of the progression using the common ratio: a2 = a1 * r = 6 * 3 = 18 a3 = a2 * r = 18 * 3 = 54 The sum of the first three terms of the G.P is: a1 + a2 + a3 = 6 + 18 + 54 = 78 Therefore, the correct answer is option (C) 78.

**Question 31**
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What are the integer values of x which satisfy the inequality -1 < 3 -2x ≤ $\le $ 5?

**Answer Details**

The inequality given is: -1 < 3 - 2x ≤ 5. We need to find the integer values of x that satisfy this inequality. First, let's solve for the left side of the inequality: -1 < 3 - 2x. Add 2x to both sides: 2x - 1 < 3 Add 1 to both sides: 2x < 4 Divide both sides by 2: x < 2 Now, let's solve for the right side of the inequality: 3 - 2x ≤ 5. Subtract 3 from both sides: -2x ≤ 2 Divide both sides by -2, remembering to flip the inequality: x ≥ -1 So, the integer values of x that satisfy the inequality are -1, 0, 1, and 2. Therefore, the correct option is: - -1, 0, 1,

**Question 32**
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Evaluate ∫31(X2−1)dx

**Answer Details**

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

**Question 33**
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y = is inversely proportional to x and y = 4 when x = 1/2. Find x when y = 10.

**Answer Details**

y ∝ 1/x

y = K/x

K = xy

K = (1/2) * 4 ( when y = 4 and x = 1/2)

K = 2

∴y = 2/x

10 = 2/x (when y = 10)

10x = 2

x = 2/10

x = 1/5

**Question 34**
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PQRSTV is a regular polygon of side 7 cm inscribed in a circle. Find the circumference of the circle PQRSTV.

**Answer Details**

Each interior Angle of the polygon = | (n-2) | 180 |

_n |

= | (6-2) | 180 |

_6 |

= 4 x 30

= 120∘ $\circ $

i.e ∠PQR = 120∘ $\circ $ ∴ RQD = 120/2 = 60∘ $\circ $ also

∴ Δ QOR is equilateral Δ ∴ QO = 7cm

∴ PO = QO = RO ...... = 7cm radius of the circumference of the circle = 2πr

= 2 x 22/7 x 7/1

= 44 cm

**Question 35**
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