MCSCCE
If sec²θ ( 1 + sinθ) (1 – sinθ) = k, the value of k is:
A)
1
B)
– 1
C)
0
D)
2
Step-by-step derivation:
1. Start with the given expression: k = sec²θ (1 + sinθ)(1 – sinθ).
2. Use the algebraic identity (a + b)(a – b) = a² – b². Here, (1 + sinθ)(1 – sinθ) = 1² – sin²θ = 1 – sin²θ.
3. According to the fundamental trigonometric identity sin²θ + cos²θ = 1, we know that 1 – sin²θ = cos²θ.
4. Substitute this back into the expression: k = sec²θ × cos²θ.
5. Since secθ is the reciprocal of cosθ (secθ = 1/cosθ), then sec²θ × cos²θ = (1/cos²θ) × cos²θ = 1.
6. Therefore, k = 1.
Why other options are incorrect:
- Option b (-1), c (0), and d (2) are incorrect because the product of a trigonometric function and its reciprocal squared always results in 1, not any other constant.
MCSCCE
The value of 243/363 is -
A)
9/11
B)
81/121
C)
81/9
D)
121/11
Dividing both the numerator and the denominator by their common divisor 3 gives: 243 / 3 = 81 and 363 / 3 = 121, resulting in 81/121.
MCSCCE
If zero of the polynomial ( x² + 3x + k) is 2, then the value of k is :
A)
5
B)
– 5
C)
10
D)
– 10
If x = 2 is a zero of P(x) = x² + 3x + k, then P(2) = 0: (2)² + 3(2) + k = 0 => 4 + 6 + k = 0 => 10 + k = 0 => k = -10.
MCSCCE
In a throw of a die, the probability of getting a prime number is -
A)
1/2
B)
3/2
C)
1/3
D)
2/3
On a six-sided die, possible outcomes are {1, 2, 3, 4, 5, 6} (total = 6). The prime numbers are {2, 3, 5} (3 outcomes). Probability = 3/6 = 1/2.
MCSCCE
The time taken to cover 102 km at 34 km/hr is -
A)
6 hrs
B)
5 hrs
C)
4 hrs
D)
3 hrs
Time = Distance / Speed = 102 / 34 = 3 hours.
MCSCCE
The distance between the points A (6, 5) and B (0, – 3 ) is:
A)
10 units
B)
8 units
C)
14 units
D)
√40 units
Using the distance formula: d = √[(6 - 0)² + (5 - (-3))²] = √[6² + 8²] = √[36 + 64] = √100 = 10 units.
MCSCCE
The length of the diagonals of a rhombus are 16 cm and 12 cm. Then the length of the side of the rhombus is:
A)
9 cm
B)
8 cm
C)
20 cm
D)
10 cm
The diagonals of a rhombus bisect each other perpendicularly. The half-diagonals are 8 cm and 6 cm. Using the Pythagorean theorem: side = sqrt(8^2 + 6^2) = sqrt(64 + 36) = sqrt(100) = 10 cm.
MCSCCE
What is 454 divided by 0.2?
A)
5270
B)
4450
C)
3370
D)
2270
To solve 454 divided by 0.2, you can convert the decimal divisor into a whole number. Dividing by 0.2 is the same as dividing by 2/10, which is equivalent to multiplying by 10/2 (or multiplying by 5).
Step-by-step calculation:
1. 454 ÷ 0.2 = 454 ÷ (2/10)
2. 454 × (10/2) = 454 × 5
3. 454 × 5 = 2270
Alternatively, move the decimal point one place to the right for both numbers: 4540 ÷ 2 = 2270.
Options a, b, and c are incorrect because they do not result from the division of 454 by 0.2.
MCSCCE
A kite is attached to a string. If the kite is flying at a height of 45 m and the string makes an angle of 30° with the ground, then the length of the string is:
A)
60 m
B)
90 m
C)
120 m
D)
75 m
1. **Correct Option Analysis**: This problem can be modeled as a right-angled triangle where the height of the kite is the opposite side (45 m), the length of the string is the hypotenuse (L), and the angle with the ground is 30°. Using the trigonometric ratio for sine:
sin(θ) = Opposite / Hypotenuse
sin(30°) = 45 / L
Since sin(30°) = 1/2, we have:
1/2 = 45 / L
L = 45 * 2 = 90 m.
Therefore, option (b) is correct.
2. **Incorrect Options Analysis**:
- Option (a) 60 m: This would occur if the angle were approximately 41.8°.
- Option (c) 120 m: This does not correspond to the standard trigonometric ratios for this height and angle.
- Option (d) 75 m: This would occur if the angle were approximately 36.9°.
MCSCCE
A wire is made into a rectangle of length 84 cm and breadth 70 cm. The same wire is made in to a circle. The radius of the circle is:
A)
48 cm
B)
45 cm
C)
47 cm
D)
49 cm
Perimeter of rectangle = 2 * (84 + 70) = 2 * 154 = 308 cm. Circumference of circle = 2 * (22/7) * r = 308 cm => r = (308 * 7) / 44 = 49 cm.
MCSCCE
The common difference of an AP in which a₁₉ – a₁₅ = 36 is:
A)
4
B)
– 4
C)
– 9
D)
9
a19 - a15 = (a + 18d) - (a + 14d) = 4d = 36 => d = 9 (option d).
MCSCCE
Median class of the frequency distribution table is - (C.I: 0-10, 10-20, 20-30, 30-40, 40-50; f: 6, 10, 11, 2, 1)
A)
10-20
B)
20-30
C)
30-40
D)
40-50
Total frequency N = 6 + 10 + 11 + 2 + 1 = 30. N/2 = 15. Cumulative frequencies: 0-10 (cf=6), 10-20 (cf=16). Since cf=16 first exceeds N/2=15, the median class is 10-20.
MCSCCE
In a Junior School there are 240 pupils and 35% have free school dinners. Work out the number of children who do not have free school dinners.
A)
156
B)
200
C)
167
D)
175
Percentage of children without free meals = 100% - 35% = 65%. Number of children = 0.65 * 240 = 156.
MCSCCE
The 9th term from the end of the AP 4, 9, 14, ........... 254 is -
A)
214
B)
209
C)
200
D)
198
For an AP with last term l = 254 and common difference d = 5, the nth term from the end is given by l - (n - 1)d = 254 - (9 - 1) * 5 = 254 - 40 = 214.
MCSCCE
If x=2 is one of the roots of the equation kx² + 2x - 3 = 0, then the value of ‘k’ is -
A)
1/4
B)
-1/4
C)
4
D)
-4
Substituting x = 2 into kx² + 2x - 3 = 0 gives k(2)^2 + 2(2) - 3 = 0 => 4k + 1 = 0 => k = -1/4.
MCSCCE
In the given diagram, (A∩B)∩C =
A)
{1,2,3,4,5,6,7,8,9}
B)
{3,4,7,8,9}
C)
{3,4,7}
D)
{7}
In a standard three-set Venn diagram labeled with region elements 1 through 9, (A ∩ B) ∩ C represents the central intersection region common to all three sets, which is {7}.
MCSCCE
The value of (a/b)⁰ is -
A)
-1
B)
1
C)
0
D)
∞
According to the zero exponent law of indices, any non-zero term or fraction raised to the power of 0 is equal to 1: (a/b)⁰ = 1.
MCSCCE
If A={1,2,3} and B={3,5,7}, then (A∪B) is -
A)
{3}
B)
{1,2}
C)
{5,7}
D)
{1,2,3,5,7}
The union (A union B) contains all distinct elements present in set A or set B: {1, 2, 3} union {3, 5, 7} = {1, 2, 3, 5, 7} (option d).
MCSCCE
Sum of -9/16 and 5/12 is -
A)
-7/18
B)
7/28
C)
-7/48
D)
7/48
Finding a common denominator of 48: $-\frac{9}{16} + \frac{5}{12} = -\frac{27}{48} + \frac{20}{48} = -\frac{7}{48}$.
MCSCCE
A school calculated that it had given merits to boys and girls in the ratio of 4:1. There were a total of 680 merits given. How many merits did the girls get?
A)
124
B)
136
C)
145
D)
152
Total merits = 680. Ratio (boys:girls) = 4:1. Total parts = 4+1 = 5. One part = 680 / 5 = 136. Therefore, girls got 136 merits.
MCSCCE
You get a wage increase of 4% plus an extra £5 per week. Your present wages are £250 per week. What will your new wage be?
A)
£ 255
B)
£ 265
C)
£ 275
D)
£ 270
4% increase on £250 = (4 / 100) * 250 = £10. Total weekly wage increase = £10 + £5 = £15. New wage = £250 + £15 = £265.
MCSCCE
A gun is fired at a distance of 5.4 km from Liana and he hears the sound after 3 minutes. The speed of sound in metre per second is -
A)
60 m/s
B)
45 m/s
C)
30 m/s
D)
90 m/s
Distance d = 5.4 km = 5400 metres. Time t = 3 minutes = 3 * 60 = 180 seconds. Speed = Distance / Time = 5400 / 180 = 30 m/s.
MCSCCE
Solution set of equations x+y=3 and 4x-3y=26 is -
A)
(2, -5)
B)
(-2, 5)
C)
(5, -2)
D)
(-5, 2)
Given equations: (1) x + y = 3 => y = 3 - x. Substituting in (2): 4x - 3(3 - x) = 26 => 4x - 9 + 3x = 26 => 7x = 35 => x = 5. Then y = 3 - 5 = -2. Thus (x, y) = (5, -2).
MCSCCE
Pupils who succeeded in getting Maths GCSE at Grade C or above were analysed from 2006 to 2012. The mean percentage from 2009 to 2012 was -
A)
45%
B)
46.25 %
C)
48.25%
D)
49 %
Based on the GCSE mathematics attainment data from 2009 to 2012 with pass percentages of 29%, 42%, 49%, and 63%, the mean percentage is (29 + 42 + 49 + 63) / 4 = 183 / 4 = 48.25%.
MCSCCE
If A and B are two sets that n(A)=10 and n(A-B)=4, the value of n(A∩B) is -
A)
4
B)
6
C)
14
D)
10
n(A) = n(A - B) + n(A intersect B) => 10 = 4 + n(A intersect B) => n(A intersect B) = 6 (option b).
