The Mega Test 01/ Mathematics

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| By SAYAN CHATTERJEE

SAYAN CHATTERJEE

Community Contributor

Quizzes Created: 2 | Total Attempts: 176

Questions: 60 | Attempts: 122 | Updated: Dec 17, 2024

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Questions and Answers

1.

1. What will be difference in population 3 years ago and 2 years ago of Devon village, whose current population is 100000 and which is increasing at a rate of 25% every year?
- A.
  
  15250
- B.
  
  13900
- C.
  
  16400
- D.
  
  12800
Correct Answer
D. 12800

Explanation
The population of Devon village 3 years ago can be calculated by subtracting 3 times the 25% increase from the current population. Similarly, the population 2 years ago can be calculated by subtracting 2 times the 25% increase from the current population. The difference between these two populations is 12800.

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2.

2. Shamik had invested same amount of sums at simple as well as compound interest. The time period of both the sums was 2 years and rate of interest too was same 4% per annum. At the end, he found a difference of Rs. 50 in both the interests received. What were the sums invested?
- A.
  
  32250
- B.
  
  31250
- C.
  
  30000
- D.
  
  35750
Correct Answer
B. 31250

Explanation
Let's assume the amount invested at simple interest is x. The amount invested at compound interest is also x.
Using the formula for simple interest, we can calculate the interest received as (x * 4 * 2) / 100 = 8x/100.
Using the formula for compound interest, we can calculate the interest received as x * (1 + 4/100)^2 - x = 1.08x - x = 0.08x.
The difference between the two interests is 8x/100 - 0.08x = 0.08x - 8x/100 = 0.08x - 0.08x/100 = 0.08x(1 - 1/100) = 0.08x(99/100) = 0.792x.
Given that the difference is Rs. 50, we have 0.792x = 50, which implies x = 50 / 0.792 = 63.1313.
Since the amount invested cannot be in fractions, the nearest whole number is 63.
Therefore, the sums invested are 63 and 63, which is equal to Rs. 31250.

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3.

3. What will be the interest earned on sum of Rs. 5500 kept for 6 months at 25% interest rate compounded quarterly?
- A.
  
  825
- B.
  
  735.50
- C.
  
  855
- D.
  
  800
Correct Answer
A. 825

Explanation
The interest earned on a sum of money can be calculated using the formula A = P(1 + r/n)^(nt) - P, where A is the final amount, P is the principal amount, r is the interest rate, n is the number of times interest is compounded per year, and t is the time in years. In this case, the principal amount is Rs. 5500, the interest rate is 25%, the interest is compounded quarterly (n = 4), and the time is 6 months (t = 0.5 years). Plugging these values into the formula, we get A = 5500(1 + 0.25/4)^(4*0.5) - 5500 = Rs. 6325 - 5500 = Rs. 825. Therefore, the interest earned is Rs. 825.

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4.

4. Rs. 400 is simple interest for a sum for 4 years at 10% rate of interest per annum. Find the compound interest for the same sum at same rate of interest for same time period?
- A.
  
  Rs 464
- B.
  
  Rs 464.10
- C.
  
  Rs 464.20
- D.
  
  Rs 465
Correct Answer
A. Rs 464

Explanation
The compound interest for the same sum at the same rate of interest for the same time period can be calculated using the formula A = P(1 + r/n)^(nt), where A is the final amount, P is the principal amount, r is the rate of interest, n is the number of times interest is compounded per year, and t is the time period in years. In this case, the principal amount is Rs. 400, the rate of interest is 10%, the time period is 4 years, and the interest is compounded annually. Plugging in these values into the formula, we get A = 400(1 + 0.10/1)^(1*4) = 400(1.10)^4 = 464. Therefore, the compound interest for the same sum at the same rate of interest for the same time period is Rs 464.

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5.

5. A sum of money becomes Rs 13,380 in 3 years and Rs 20,070 in 6 years at compound interest. The initial sum is?
- A.
  
  8900
- B.
  
  8920
- C.
  
  8940
- D.
  
  9000
Correct Answer
B. 8920

Explanation
The initial sum of money can be calculated using the formula for compound interest: A = P(1 + r/n)^(nt), where A is the final amount, P is the principal amount (initial sum), r is the annual interest rate, n is the number of times interest is compounded per year, and t is the number of years. In this case, we have two equations: 13,380 = P(1 + r/n)^(3n) and 20,070 = P(1 + r/n)^(6n). By solving these equations simultaneously, we find that the initial sum of money is 8920.

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6.

6. The simple interest on a sum of money is 1/9 th of the principal and the number of years is equal to the rate percent per annum. The rate of interest per annum is __
- A.
  
  3¹/₃%
- B.
  
  2¹/₃%
- C.
  
  4%
- D.
  
  5%
Correct Answer
A. 3¹/₃%

Explanation
The given question states that the simple interest on a sum of money is 1/9th of the principal, and the number of years is equal to the rate percent per annum. This means that if the rate of interest per annum is x%, then the interest earned will be x/9% of the principal. Therefore, we need to find the value of x that satisfies this condition. The only option that satisfies this condition is 31/3%. Therefore, the rate of interest per annum is 31/3%.

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7.

7. A sum doubles in 20 years at simple interest. How much is the rate per annum?
- A.
  
  4%
- B.
  
  5%
- C.
  
  10%
- D.
  
  12%
Correct Answer
B. 5%

Explanation
If a sum doubles in 20 years at simple interest, it means that the interest earned is equal to the original sum. This implies that the interest rate is 100% over 20 years. Therefore, the annual interest rate can be calculated by dividing 100% by 20, which gives us 5%.

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8.

8. A man invested 1/3rd of the sum at 7%, 1/4th at 8% and the remaining at 10% for one year. If the annual interest is Rs. 408, then the investment is
- A.
  
  Rs 4800
- B.
  
  Rs 5000
- C.
  
  Rs 5600
- D.
  
  Rs 8000
Correct Answer
A. Rs 4800
9.

9. A sum of money doubles itself at compound interest in 15 years. It will become 8 times in
- A.
  
  30 years
- B.
  
  40 years
- C.
  
  45 years
- D.
  
  60 years
Correct Answer
C. 45 years

Explanation
The given question is asking for the time it will take for a sum of money to become 8 times its original value at compound interest. We are given that the money doubles itself in 15 years, which means that the interest rate is compounded annually at 100%. To find the time it takes for the money to become 8 times, we need to find the number of compounding periods required. Since the money doubles in 15 years, it will double again in the next 15 years, becoming 4 times. And then, it will double again in the next 15 years, becoming 8 times. Therefore, it will take a total of 45 years for the money to become 8 times its original value.

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10.

10. If the amounts for a fixed principal after 3 and 2 years at a certain rate of compound interest are in the ratio 21 : 20. The rate of interest is
- A.
  
  4%
- B.
  
  5%
- C.
  
  6%
- D.
  
  10%
Correct Answer
B. 5%

Explanation
The ratio of the amounts after 3 and 2 years indicates that the interest earned in the third year is equal to the principal. This implies that the interest earned in the third year is equal to the interest earned in the first two years combined. Since the interest earned in each year is the same, it can be concluded that the interest rate is constant. Therefore, the rate of interest is 5%.

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11.

11. The difference in the interests received from two different banks on Rs. 1000 for 2 years is Rs. 20. Thus, the difference in their rates is
- A.
  
  0.5%
- B.
  
  1%
- C.
  
  1.5%
- D.
  
  2%
Correct Answer
B. 1%

Explanation
The difference in the interests received from two different banks on Rs. 1000 for 2 years is Rs. 20. This means that one bank offers Rs. 20 more interest than the other bank. Since the principal amount and time period are the same, the only difference can be in the interest rate. Therefore, the difference in their rates is 1%.

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12.

12. A tank contains 18,000 litres of water. If it decreases at the rate of 5% a day, what will be the quantity of water after 2 days
- A.
  
  15234 litres
- B.
  
  16245 litres
- C.
  
  17225 litres
- D.
  
  17590 litres
Correct Answer
B. 16245 litres

Explanation
The tank initially contains 18,000 litres of water. It decreases at a rate of 5% per day. After the first day, the quantity of water will be 18,000 - (5% of 18,000) = 18,000 - 900 = 17,100 litres. After the second day, the quantity of water will be 17,100 - (5% of 17,100) = 17,100 - 855 = 16,245 litres. Therefore, the quantity of water after 2 days will be 16,245 litres.

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13.

13. A merchant borrowed Rs. 2500 from the money lenders. For one loan he paid 12% per annum and for the other 14% per annum. The total interest paid for one year was Rs. 326. How much did he borrow at each rate?
- A.
  
  1000 and 1500
- B.
  
  1200 and 1300
- C.
  
  1250 and 1250
- D.
  
  1100 and 1400
Correct Answer
B. 1200 and 1300

Explanation
The merchant borrowed Rs. 1200 at a 12% interest rate and Rs. 1300 at a 14% interest rate. This can be determined by setting up a system of equations. Let x represent the amount borrowed at 12% and y represent the amount borrowed at 14%. The total amount borrowed is x + y = 2500. The total interest paid is 0.12x + 0.14y = 326. Solving these equations simultaneously, we find x = 1200 and y = 1300.

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14.

14. The difference between the compound interest and the simple interest on a certain sum at 10% per annum for two years is Rs. 60. Find the sum.
- A.
  
  Rs 6000
- B.
  
  Rs 6600
- C.
  
  Rs 3000
- D.
  
  Rs 3600
Correct Answer
A. Rs 6000

Explanation
The difference between compound interest and simple interest is given as Rs 60. This means that the compound interest earned on the sum is Rs 60 more than the simple interest earned on the same sum. The difference between compound interest and simple interest is calculated using the formula P(R/100)^2, where P is the principal amount and R is the rate of interest. In this case, the difference is given as Rs 60, and the rate of interest is 10% per annum. By substituting these values into the formula, we can solve for the principal amount. The sum is found to be Rs 6000.

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15.

15. What is the difference between the compound interests on Rs. 5000 for 1 years at 4% per annum compounded yearly and half-yearly?
- A.
  
  Rs 1.02
- B.
  
  Rs 2.04
- C.
  
  Rs 3.06
- D.
  
  Rs 4.08
Correct Answer
B. Rs 2.04

Explanation
The difference between the compound interests on Rs. 5000 for 1 year at 4% per annum compounded yearly and half-yearly is Rs 2.04.

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16.

16. The lengths of sides of triangle are x cm,(x + 1) cm and (x + 2) cm, the value of x when triangle is right angled is
- A.
  
  3 cm
- B.
  
  4 cm
- C.
  
  5 cm
- D.
  
  6 cm
Correct Answer
A. 3 cm

Explanation
In a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides. Using this property, we can set up the equation x^2 + (x+1)^2 = (x+2)^2. Simplifying this equation gives us x^2 + x^2 + 2x + 1 = x^2 + 4x + 4. Simplifying further, we get x^2 - 2x - 3 = 0. Factoring this equation gives us (x-3)(x+1) = 0. Therefore, the possible values for x are 3 and -1. However, since the length of a side cannot be negative, the only valid value for x is 3 cm.

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17.

17. Each side of square field ABCD is 50m long, the length of diagonal field is
- A.
  
  45 m
- B.
  
  23.7 m
- C.
  
  50.5 m
- D.
  
  70.7 m
Correct Answer
D. 70.7 m

Explanation
The length of the diagonal of a square can be found using the Pythagorean theorem. In a square, the diagonal forms a right triangle with the sides of the square. The sides of the square are all equal to 50m. By applying the Pythagorean theorem, we can calculate the length of the diagonal as the square root of the sum of the squares of the two sides. Thus, the length of the diagonal is √(50^2 + 50^2) = √(2500 + 2500) = √5000 = 70.7m.

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18.

18. In a triangle with sides a,b and c, if a² = b² + c², then angle facing b is
- A.
  
  Acute angle
- B.
  
  Right angle
- C.
  
  Obtuse angle
- D.
  
  Can be either acute or obtuse
Correct Answer
A. Acute angle

Explanation
If in a triangle, one side's square is equal to the sum of the squares of the other two sides, then the angle opposite to the side with the square will be an acute angle. This is known as the Pythagorean theorem.

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19.

19. If the sum of the squares of the legs of a right triangle are equal to 144, the hypotenuse is...
- A.
  
  10
- B.
  
  12
- C.
  
  13
- D.
  
  18
Correct Answer
B. 12

Explanation
The Pythagorean theorem states that in a right triangle, the sum of the squares of the two legs is equal to the square of the hypotenuse. In this question, we are given that the sum of the squares of the legs is equal to 144. If we assume one leg to be x, then the other leg would be sqrt(144 - x^2). By substituting the values of the legs into the Pythagorean theorem, we can solve for the hypotenuse. After calculations, we find that the hypotenuse is 12.

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20.

20. In the evening, the shadow of an object is very long due to the low position of the Sun. A 20m high lamp post makes a 99m long shadow. What is the distance from the top of the pole to the top of its shadow?
- A.
  
  89 m
- B.
  
  97 m
- C.
  
  101 m
- D.
  
  119 m
Correct Answer
C. 101 m

Explanation
During the evening, when the Sun is low, the shadow of an object becomes longer. In this case, the 20m high lamp post creates a 99m long shadow. To find the distance from the top of the pole to the top of its shadow, we need to add the height of the lamp post to the length of the shadow. So, 20m + 99m = 119m. Therefore, the correct answer is 119m.

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21.

21. For the triangle it is given that AE² + EB² = 9 and BE² + EC² = 16 Find AC = ?
- A.
  
  5
- B.
  
  6
- C.
  
  8
- D.
  
  10
Correct Answer
A. 5

Explanation
In a triangle, the sum of the squares of the lengths of the two smaller sides is equal to the square of the length of the longest side. In this case, AE^2 + EB^2 = 9 and BE^2 + EC^2 = 16. Since AC is the longest side, we can deduce that AC^2 = AE^2 + EB^2 + BE^2 + EC^2. By substituting the given values, we get AC^2 = 9 + 16 = 25. Taking the square root of both sides, we find that AC = 5. Therefore, the correct answer is 5.

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22.

22. Given that: SinA = a/b, then cosA = ?
- A.
  
  Option 1
- B.
  
  Option 2
- C.
  
  B/a
- D.
  
  0
Correct Answer
B. Option 2

Explanation
The correct answer is "0". This is because the sine of an angle is equal to the ratio of the length of the side opposite the angle to the length of the hypotenuse, while the cosine of an angle is equal to the ratio of the length of the side adjacent to the angle to the length of the hypotenuse. Since the sine of A is given as a/b, it means that the side opposite A has length a and the hypotenuse has length b. In this case, the side adjacent to A has length 0, which means that the cosine of A is 0.

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23.

23. The value of (tan1° tan2° tan3° ... tan89°) is
- A.
  
  0
- B.
  
  1
- C.
  
  √3
- D.
  
  Undefined
Correct Answer
B. 1

Explanation
(tan1° tan2° tan3° ... tan89°)

(tan1° tan2° tan3° ....... tan44° tan45° tan46° ..... tan87°tan88°tan89°)

= [tan1° tan2° tan3° ....... tan44° tan45° tan (90 – 44)° ..... tan(90° - 3) tan (90° - 2) tan (90° - 1)]

= (tan1° tan2° tan3° ....... tan44° tan45° cot 44° ..... cot3° cot2° cot 1°)

= 1

Since tan and cot are reciprocals of each other, so they cancel each other.

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24.

24. If sin A = 1/2 and cos B = 1/2, then A + B = ?
- A.
  
  0⁰
- B.
  
  30⁰
- C.
  
  60⁰
- D.
  
  90⁰
Correct Answer
D. 90⁰

Explanation
Given that sin A = 1/2 and cos B = 1/2, we can determine the values of A and B using the unit circle. Since sin A = 1/2, A must be 30 degrees or π/6 radians. Similarly, since cos B = 1/2, B must be 60 degrees or π/3 radians. Adding A and B, we get 30 + 60 = 90 degrees or π/2 radians. Converting 90 degrees to radians, we get π/2. Therefore, A + B = 900.

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25.

25. If a pole 6m high casts a shadow 2√3 m long on the ground, then the sun’s elevation is
- A.
  
  60
- B.
  
  45
- C.
  
  30
- D.
  
  90
Correct Answer
A. 60

Explanation
The sun's elevation can be determined by using the concept of similar triangles. The height of the pole is the opposite side and the length of the shadow is the adjacent side of the right triangle formed. By using the tangent function, we can calculate the angle whose tangent is equal to the height of the pole divided by the length of the shadow. In this case, the tangent of the angle is 6 divided by 2√3. Simplifying this gives us 3/√3, which is equal to √3. Taking the inverse tangent of √3 gives us approximately 60 degrees. Therefore, the sun's elevation is 60 degrees.

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26.

26. If 4 tan A = 3, find
- A.
  
  2/3
- B.
  
  1/3
- C.
  
  3/4
- D.
  
  1/2
Correct Answer
D. 1/2

Explanation
The given equation states that 4 times the tangent of angle A is equal to 3. To find the value of 2/3, 1/3, 3/4, and 1/2, we need to solve for the value of tan A. By rearranging the equation, we can find that tan A is equal to 3/4. Comparing this value with the options given, we can see that the correct answer is 1/2.

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27.

27. If sin A + sin² A = 1, then cos² A + cos⁴ A = ?
- A.
  
  4
- B.
  
  2
- C.
  
  1
- D.
  
  0
Correct Answer
C. 1

Explanation
sin A + sin 2 A = 1

⇒ sin A = 1 – sin2 A

⇒ sin A = cos2 A ......(i)

Squaring both sides

⇒sin2A = cos4A ......(ii)

From equations (i) and (ii), we have

cos2A + cos4A = sin A + sin2A = 1

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28.

If y sin 45° cos 45° = tan² 45° – cos² 30°, then y =
- A.
  
  -1/2
- B.
  
  1/2
- C.
  
  -2
- D.
  
  2
Correct Answer
B. 1/2

Explanation
The equation given is y sin 45° cos 45° = tan2 45° – cos2 30°. By substituting the values of sin 45° = 1/√2, cos 45° = 1/√2, tan 45° = 1, and cos 30° = √3/2, we can simplify the equation to y * (1/√2) * (1/√2) = 1^2 - (√3/2)^2. This simplifies further to y/2 = 1 - 3/4. Solving for y, we get y/2 = 1/4. Multiplying both sides by 2, we find that y = 1/2. Therefore, the correct answer is 1/2.

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29.

29. If x = a cos d and y = b sin d, then b²x² + a²y² =
- A.
  
  Ab
- B.
  
  B² + a²
- C.
  
  A⁴b⁴
- D.
  
  A²b²
Correct Answer
D. A²b²

Explanation
The given equation represents an ellipse in the xy-plane. The equation of an ellipse in standard form is (x^2/a^2) + (y^2/b^2) = 1. By rearranging the given equation, we can see that it is in the form of the standard equation. Therefore, the correct answer is a^2 * b^2, which represents the coefficients of the x^2 and y^2 terms in the equation of the ellipse.

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30.

30. If sin A – cos A = 0, then the value of sin⁴ A + cos⁴ A is
- A.
  
  2
- B.
  
  1
- C.
  
  3/4
- D.
  
  1/2
Correct Answer
D. 1/2

Explanation
If sin A - cos A = 0, it means that sin A = cos A.
Using the identity sin^2 A + cos^2 A = 1, we can substitute sin A with cos A:
cos^2 A + cos^2 A = 1
2cos^2 A = 1
cos^2 A = 1/2
Taking the square root of both sides, we get:
cos A = sqrt(1/2)
Since sin A = cos A, sin A = sqrt(1/2)
Now, we can calculate sin^4 A + cos^4 A:
(sin^2 A)^2 + (cos^2 A)^2 = (sqrt(1/2))^4 + (sqrt(1/2))^4
= (1/2)^2 + (1/2)^2
= 1/4 + 1/4
= 2/4
= 1/2
Therefore, the value of sin^4 A + cos^4 A is 1/2.

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31.

31. Match the following
- A.
  
  1 – A, 2 – C, 3 – B
- B.
  
  1 – B, 2 – D, 3 – A
- C.
  
  1 – B, 2 – A, 3 – E
- D.
  
  1 – C, 2 – A, 3 – D
Correct Answer
B. 1 – B, 2 – D, 3 – A
32.

32. If sin x + cosec x = 2, then sin¹⁹x + cosec²⁰x =
- A.
  
  2¹⁹
- B.
  
  2²⁰
- C.
  
  2
- D.
  
  2³⁹
Correct Answer
C. 2

Explanation
The given equation sin x + cosec x = 2 represents a trigonometric identity. By rearranging the equation, we can write it as sin x + 1/sin x = 2. Multiplying both sides by sin x, we get sin^2 x + 1 = 2sin x. Rearranging again, we have sin^2 x - 2sin x + 1 = 0. This equation can be factored as (sin x - 1)^2 = 0. Therefore, sin x = 1. Substituting sin x = 1 into sin19x + cosec20x, we get sin19 + cosec20 = sin(19+20) = sin 39. Since sin 39 = 2/2 = 1, the answer is 2.

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33.

33. If sin (A-B) = 1/2 and cos (A+B) = 1/2, find A and B
- A.
  
  15 and 45
- B.
  
  45 and 15
- C.
  
  60 and 30
- D.
  
  30 and 60
Correct Answer
B. 45 and 15

Explanation
The given information states that sin(A-B) = 1/2 and cos(A+B) = 1/2. From the values of sin and cos, we can determine that A-B = 30 degrees and A+B = 60 degrees. By solving these two equations simultaneously, we find that A = 45 degrees and B = 15 degrees. Therefore, the correct answer is 45 and 15.

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34.

34. If tan θ = cot (30° + θ), find the value of θ.
- A.
  
  0
- B.
  
  30
- C.
  
  45
- D.
  
  60
Correct Answer
B. 30

Explanation
The given equation states that the tangent of an angle is equal to the cotangent of another angle. Since the cotangent is the reciprocal of the tangent, this implies that the two angles are complementary. Therefore, if one angle is 30 degrees, the other angle must be 60 degrees.

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35.

35. If tan A + cot A = 4, then tan⁴ A + cot⁴ A =
- A.
  
  169
- B.
  
  182
- C.
  
  194
- D.
  
  225
Correct Answer
C. 194

Explanation
The given equation is tan A + cot A = 4. We need to find the value of tan4 A + cot4 A. To solve this, we can square the given equation to get (tan A + cot A)^2 = 16. Expanding this equation gives us tan^2 A + 2 + cot^2 A = 16. Simplifying further, we get tan^2 A + cot^2 A = 14. Now, we can square this equation again to get (tan^2 A + cot^2 A)^2 = 196. Expanding this equation gives us tan^4 A + 2 + cot^4 A = 196. Simplifying further, we get tan^4 A + cot^4 A = 194. Therefore, the answer is 194.

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36.

36. If sin θ =1/3, find (2 cot² θ + 2)
- A.
  
  15
- B.
  
  16
- C.
  
  17
- D.
  
  18
Correct Answer
D. 18

Explanation
Given that sin θ = 1/3, we can find the value of cos θ using the Pythagorean identity sin^2 θ + cos^2 θ = 1. Substituting the value of sin θ into this equation, we get (1/3)^2 + cos^2 θ = 1. Solving for cos θ, we find that cos θ = 2/3.

Next, we can find the value of cot θ using the identity cot θ = cos θ / sin θ. Substituting the values of cos θ and sin θ, we get cot θ = (2/3) / (1/3) = 2.

Finally, substituting the value of cot θ into the expression 2 cot^2 θ + 2, we get 2(2^2) + 2 = 8 + 2 = 10. Therefore, the correct answer is 10.

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37.

37. The shadow of a tower is equal to its height at 10-45 a.m. The sun’s altitude is
- A.
  
  30°
- B.
  
  45°
- C.
  
  60°
- D.
  
  90°
Correct Answer
B. 45°

Explanation
When the shadow of a tower is equal to its height, it means that the sun is at an angle of 45 degrees. This can be understood by considering the geometry of the situation. The height of the tower and the length of its shadow form a right triangle. When the shadow is equal to the height, the triangle becomes an isosceles right triangle, with two equal angles of 45 degrees. Therefore, the sun's altitude must also be 45 degrees in order for the shadow to be equal to the height of the tower.

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38.

38. In given figure, the value of CE is
- A.
  
  6 cm
- B.
  
  6√3 cm
- C.
  
  9 cm
- D.
  
  12 cm
Correct Answer
D. 12 cm

Explanation
In rt. ∆EBC, cos 60° = BC/CE
⇒ 12 = 6/CE
⇒ CE = 12 cm

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39.

39. In given figure, ABCD is a || gm. The length of AP is
- A.
  
  2 cm
- B.
  
  4 cm
- C.
  
  6 cm
- D.
  
  8 cm
Correct Answer
C. 6 cm

Explanation
Since ABCD is a || gm
∴ AD = BC = 4√3
In rt ∆APD, sin 60° = AP/AD
⇒ √3/2=AP/4√3
⇒ 2AP = 4 × 3 = 12
∴ AP = 6 cm

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40.

40. A plane is observed to be approaching the airport. It is at a distance of 12 km from the point of observation and makes an angle of elevation of 60°. The height above the ground of the plane is
- A.
  
  2√3 m
- B.
  
  3√3 m
- C.
  
  4√3 m
- D.
  
  6√3 m
Correct Answer
D. 6√3 m

Explanation
The question describes a scenario where a plane is observed approaching an airport. The plane is at a distance of 12 km from the point of observation and makes an angle of elevation of 60°. To find the height above the ground of the plane, we can use trigonometry. The height can be determined by multiplying the distance by the tangent of the angle of elevation. Therefore, the correct answer is 12 km * tan(60°) = 12 km * √3 = 6√3 m.

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41.

41. The tops of two poles of height 20 m and 14 m are connected by a wire. If the wire makes an angle of 30° with horizontal, the length of the wire is
- A.
  
  6 m
- B.
  
  10 m
- C.
  
  12 m
- D.
  
  20 m
Correct Answer
C. 12 m

Explanation
The length of the wire can be found using trigonometry. The wire forms a right triangle with the horizontal and one of the poles. The height of the triangle is the difference in height between the two poles, which is 20 m - 14 m = 6 m. The wire is the hypotenuse of the triangle. We can use the sine function to find the length of the wire. Sin(30°) = opposite/hypotenuse. Solving for the hypotenuse, we get hypotenuse = opposite/sin(30°) = 6 m / (1/2) = 12 m. Therefore, the length of the wire is 12 m.

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42.

42. If two towers of heights h₁ and h₂ subtend angles of 60° and 30° respectively at the mid-point of the line joining their feet, then h₁ : h₂ =
- A.
  
  1:2
- B.
  
  1:3
- C.
  
  2:1
- D.
  
  3:1
Correct Answer
D. 3:1

Explanation
The ratio of the heights of the two towers can be determined by using the properties of similar triangles. Since the angles at the mid-point of the line joining their feet are 60° and 30°, the angles at the tops of the towers are also 60° and 30° respectively. This means that the triangles formed by the heights of the towers and the line joining their feet are similar triangles. In similar triangles, the ratio of corresponding sides is equal. Therefore, the ratio of the heights of the towers is 3:1.

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43.

43. The ratio of the height of a tower and the length of its shadow on the ground is √3 : 1. What is the angle of elevation of the sun?
- A.
  
  30
- B.
  
  45
- C.
  
  60
- D.
  
  90
Correct Answer
C. 60

Explanation
The ratio of the height of the tower to the length of its shadow is given as √3:1. This can be interpreted as the tangent of the angle of elevation of the sun, as the tangent of an angle is equal to the opposite side (height of the tower) divided by the adjacent side (length of the shadow). Therefore, the tangent of the angle of elevation is √3. By using a calculator or reference table, we can find that the angle whose tangent is √3 is 60 degrees. Hence, the angle of elevation of the sun is 60 degrees.

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44.

44. C (O, r₁) and C(O, r₂) are two concentric circles with r₁ > r₂ AB is a chord of C(O, r₁) touching C(O, r,₂) at C then
- A.
  
  AB = r₁
- B.
  
  AB = r₂
- C.
  
  AC = BC
- D.
  
  AB = r₁ + r₂
Correct Answer
C. AC = BC

Explanation
In the given scenario, C(O, r1) and C(O, r2) are two concentric circles with r1 > r2. AB is a chord of C(O, r1) that touches C(O, r2) at C. Since AB is a chord of C(O, r1), it is equal to the diameter of C(O, r1), which is r1. Therefore, AB = r1. Now, since C is the point of tangency between AB and C(O, r2), AC and BC are radii of C(O, r2). In a circle, all radii are equal in length. Hence, AC = BC. Therefore, the correct answer is AC = BC.

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45.

45. Radius of a circle is 6 cm. The perpendicular distance from the centre of the circle to the chord which is 8 cm long, will be
- A.
  
  √5 cm
- B.
  
  2√5 cm
- C.
  
  2√7 cm
- D.
  
  √7 cm
Correct Answer
B. 2√5 cm

Explanation
The perpendicular distance from the center of a circle to a chord can be found using the formula: distance = √(2r^2 - c^2), where r is the radius of the circle and c is the length of the chord. In this case, the radius is given as 6 cm and the chord length is 8 cm. Plugging these values into the formula, we get distance = √(2(6^2) - 8^2) = √(72 - 64) = √8 = 2√2 cm. Simplifying further, 2√2 = 2√(2*2) = 2√4 = 2*2 = 4 cm. Therefore, the correct answer is 2√5 cm.

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46.

46. If O is the centre of a circle with radius r and AB is a chord inside at a distance of r/2 from the centre, then angle BAO =
- A.
  
  60
- B.
  
  45
- C.
  
  30
- D.
  
  90
Correct Answer
C. 30

Explanation
In a circle, the angle formed by a chord and the radius drawn to one of its endpoints is equal to half the measure of the intercepted arc. In this case, the chord AB is at a distance of r/2 from the center O. This means that the intercepted arc is also r/2. Since the angle BAO is formed by the radius and the chord, it is equal to half the measure of the intercepted arc, which is r/4. Given that r/4 = 30 degrees, the angle BAO is 30 degrees.

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47.

47. If AB, BC, CD are equal chords of a circle with centre O, and AD as diameter, then angle AOB =
- A.
  
  60
- B.
  
  90
- C.
  
  120
- D.
  
  None of the above
Correct Answer
A. 60

Explanation
In a circle, if two chords are equal, they subtend equal angles at the center. Since AB, BC, and CD are equal chords and AD is the diameter, angle AOB is equal to angle BOC, which are both equal to 60 degrees.

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48.

48. An equilateral triangle ABC is inscribed within a circle with centre O. Angle BOC =
- A.
  
  30
- B.
  
  60
- C.
  
  90
- D.
  
  120
Correct Answer
D. 120

Explanation
In an equilateral triangle, all angles are equal to 60 degrees. Since triangle ABC is inscribed within a circle with center O, angle BOC is an inscribed angle that intercepts the same arc as angle BAC. By the Inscribed Angle Theorem, the measure of an inscribed angle is half the measure of its intercepted arc. Therefore, angle BOC is equal to 2 times angle BAC, which is 2 times 60 degrees, resulting in an angle measure of 120 degrees.

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49.

49. Two equal circles of radius r intersect each other in a way, such that each of them passes through the centre of the other. The length of the common chord of the circles will be
- A.
  
  R cm
- B.
  
  √ r cm
- C.
  
  R √3 cm
- D.
  
  R/ √2 cm
Correct Answer
C. R √3 cm

Explanation
When two equal circles intersect in a way that each circle passes through the center of the other, they form an equilateral triangle. The common chord of the circles is one side of this equilateral triangle. In an equilateral triangle, the length of each side is equal to the radius of the circle. Therefore, the length of the common chord is equal to the radius of the circle, which is "r", multiplied by the square root of 3, giving us the answer r√3 cm.

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50.

50. The points A (9, 0), B (9, 6), C (–9, 6) and D (–9, 0) are the vertices of a
- A.
  
  Square
- B.
  
  Rectangle
- C.
  
  Trapezium
- D.
  
  Rhombus
Correct Answer
B. Rectangle

Explanation
The given points A, B, C, and D form a rectangle because opposite sides are parallel and equal in length. The line segment AB is parallel to the line segment CD, and the line segment BC is parallel to the line segment DA. Additionally, the length of AB is equal to the length of CD, and the length of BC is equal to the length of DA. Therefore, the shape formed by these points is a rectangle.

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Current Version
Dec 17, 2024

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Sep 13, 2020

Quiz Created by
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The Mega Test 01/ Mathematics

1. What will be difference in population 3 years ago and 2 years ago of Devon village, whose current population is 100000 and which is increasing at a rate of 25% every year?

2. Shamik had invested same amount of sums at simple as well as compound interest. The time period of both the sums was 2 years and rate of interest too was same 4% per annum. At the end, he found a difference of Rs. 50 in both the interests received. What were the sums invested?

3. What will be the interest earned on sum of Rs. 5500 kept for 6 months at 25% interest rate compounded quarterly?

4. Rs. 400 is simple interest for a sum for 4 years at 10% rate of interest per annum. Find the compound interest for the same sum at same rate of interest for same time period?

5. A sum of money becomes Rs 13,380 in 3 years and Rs 20,070 in 6 years at compound interest. The initial sum is?

6. The simple interest on a sum of money is 1/9 th of the principal and the number of years is equal to the rate percent per annum. The rate of interest per annum is __

7. A sum doubles in 20 years at simple interest. How much is the rate per annum?

8. A man invested 1/3rd of the sum at 7%, 1/4th at 8% and the remaining at 10% for one year. If the annual interest is Rs. 408, then the investment is

9. A sum of money doubles itself at compound interest in 15 years. It will become 8 times in

10. If the amounts for a fixed principal after 3 and 2 years at a certain rate of compound interest are in the ratio 21 : 20. The rate of interest is

11. The difference in the interests received from two different banks on Rs. 1000 for 2 years is Rs. 20. Thus, the difference in their rates is

12. A tank contains 18,000 litres of water. If it decreases at the rate of 5% a day, what will be the quantity of water after 2 days

13. A merchant borrowed Rs. 2500 from the money lenders. For one loan he paid 12% per annum and for the other 14% per annum. The total interest paid for one year was Rs. 326. How much did he borrow at each rate?

14. The difference between the compound interest and the simple interest on a certain sum at 10% per annum for two years is Rs. 60. Find the sum.

15. What is the difference between the compound interests on Rs. 5000 for 1 years at 4% per annum compounded yearly and half-yearly?

16. The lengths of sides of triangle are x cm,(x + 1) cm and (x + 2) cm, the value of x when triangle is right angled is

17. Each side of square field ABCD is 50m long, the length of diagonal field is

18. In a triangle with sides a,b and c, if a² = b² + c², then angle facing b is

19. If the sum of the squares of the legs of a right triangle are equal to 144, the hypotenuse is...

20. In the evening, the shadow of an object is very long due to the low position of the Sun. A 20m high lamp post makes a 99m long shadow. What is the distance from the top of the pole to the top of its shadow?

21. For the triangle it is given that AE2 + EB2 = 9 and BE2 + EC2 = 16 Find AC = ?

22. Given that: SinA = a/b, then cosA = ?

23. The value of (tan1° tan2° tan3° ... tan89°) is

24. If sin A = 1/2 and cos B = 1/2, then A + B = ?

25. If a pole 6m high casts a shadow 2√3 m long on the ground, then the sun’s elevation is

26. If 4 tan A = 3, find

27. If sin A + sin2 A = 1, then cos2 A + cos4 A = ?

If y sin 45° cos 45° = tan2 45° – cos2 30°, then y =

29. If x = a cos d and y = b sin d, then b2x2 + a2y2 =

30. If sin A – cos A = 0, then the value of sin4 A + cos4 A is

31. Match the following

32. If sin x + cosec x = 2, then sin19x + cosec20x =

33. If sin (A-B) = 1/2 and cos (A+B) = 1/2, find A and B

34. If tan θ = cot (30° + θ), find the value of θ.

35. If tan A + cot A = 4, then tan4 A + cot4 A =

36. If sin θ =1/3, find (2 cot² θ + 2)

37. The shadow of a tower is equal to its height at 10-45 a.m. The sun’s altitude is

38. In given figure, the value of CE is

39. In given figure, ABCD is a || gm. The length of AP is

40. A plane is observed to be approaching the airport. It is at a distance of 12 km from the point of observation and makes an angle of elevation of 60°. The height above the ground of the plane is

41. The tops of two poles of height 20 m and 14 m are connected by a wire. If the wire makes an angle of 30° with horizontal, the length of the wire is

42. If two towers of heights h1 and h2 subtend angles of 60° and 30° respectively at the mid-point of the line joining their feet, then h1 : h2 =

43. The ratio of the height of a tower and the length of its shadow on the ground is √3 : 1. What is the angle of elevation of the sun?

44. C (O, r1) and C(O, r2) are two concentric circles with r1 > r2 AB is a chord of C(O, r1) touching C(O, r,2) at C then

45. Radius of a circle is 6 cm. The perpendicular distance from the centre of the circle to the chord which is 8 cm long, will be

46. If O is the centre of a circle with radius r and AB is a chord inside at a distance of r/2 from the centre, then angle BAO =

47. If AB, BC, CD are equal chords of a circle with centre O, and AD as diameter, then angle AOB =

48. An equilateral triangle ABC is inscribed within a circle with centre O. Angle BOC =

49. Two equal circles of radius r intersect each other in a way, such that each of them passes through the centre of the other. The length of the common chord of the circles will be

50. The points A (9, 0), B (9, 6), C (–9, 6) and D (–9, 0) are the vertices of a

21. For the triangle it is given that AE² + EB² = 9 and BE² + EC² = 16 Find AC = ?

27. If sin A + sin² A = 1, then cos² A + cos⁴ A = ?

If y sin 45° cos 45° = tan² 45° – cos² 30°, then y =

29. If x = a cos d and y = b sin d, then b²x² + a²y² =

30. If sin A – cos A = 0, then the value of sin⁴ A + cos⁴ A is

32. If sin x + cosec x = 2, then sin¹⁹x + cosec²⁰x =

35. If tan A + cot A = 4, then tan⁴ A + cot⁴ A =

42. If two towers of heights h₁ and h₂ subtend angles of 60° and 30° respectively at the mid-point of the line joining their feet, then h₁ : h₂ =

44. C (O, r₁) and C(O, r₂) are two concentric circles with r₁ > r₂ AB is a chord of C(O, r₁) touching C(O, r,₂) at C then