# The Mega Test 02/ Mathematics

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| By SAYAN CHATTERJEE
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SAYAN CHATTERJEE
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Quizzes Created: 2 | Total Attempts: 148
Questions: 60 | Attempts: 42

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• 1.

### The height of two towers is 150 m and 136 m respectively. If the distance between them is 48 m, find the distance between their tops.

• A.

48 m

• B.

50 m

• C.

55 m

• D.

60 m

B. 50 m
Explanation
The distance between the tops of the two towers can be found by subtracting the height of the shorter tower from the height of the taller tower. In this case, the height of the taller tower is 150 m and the height of the shorter tower is 136 m. Therefore, the distance between their tops is 150 m - 136 m = 14 m. However, the given answer options do not include 14 m. The closest option to 14 m is 50 m, which is the correct answer.

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• 2.

### The diagonals of a rhombus are 24 m and 10 m. find the perimeter.

• A.

46 m

• B.

48 m

• C.

50 m

• D.

52 m

D. 52 m
Explanation
A rhombus is a quadrilateral with all four sides equal in length. The diagonals of a rhombus bisect each other at right angles. In this question, the given diagonals are 24 m and 10 m. Since the diagonals bisect each other, each half of the longer diagonal is equal to the side length of the rhombus. Therefore, the side length of the rhombus is 12 m. The perimeter of a rhombus is calculated by multiplying the side length by 4. Thus, the perimeter is 12 m * 4 = 48 m.

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• 3.

### Find the perimeter of the rectangle whose length is 24 cm and diagonal is 26 cm.

• A.

50 cm

• B.

56 cm

• C.

62 cm

• D.

68 cm

D. 68 cm
Explanation
To find the perimeter of a rectangle, we need to know either the length and width or the length and diagonal. In this case, we are given the length and diagonal. We can use the Pythagorean theorem to find the width of the rectangle. The Pythagorean theorem states that in a right triangle, the square of the hypotenuse (diagonal) is equal to the sum of the squares of the other two sides (length and width). So, we can calculate the width by taking the square root of (diagonal^2 - length^2). By substituting the given values, we find that the width is 10 cm. The perimeter of a rectangle is calculated by adding the lengths of all four sides. Therefore, the perimeter of this rectangle is 2(length + width) = 2(24 + 10) = 68 cm.

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• 4.

### A tree broke from a point but did not separate. Its top touched the ground at a distance of 24 m from its base. If the point where it broke is at the height of 7 m from the ground, what is the total height of the tree?

• A.

26 m

• B.

29 m

• C.

32 m

• D.

36 m

C. 32 m
Explanation
The total height of the tree can be determined by adding the distance from the point where it broke to the ground (7 m) and the distance from the top of the tree to the ground (24 m). Therefore, the total height of the tree is 7 m + 24 m = 31 m. Since none of the given options match this value, the correct answer cannot be determined.

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• 5.

### One of the diagonals of the rhombus is 3 cm and each side is 2.5 cm. Find the length of the other diagonal of the rhombus.

• A.

3 cm

• B.

4 cm

• C.

5 cm

• D.

5.5 cm

B. 4 cm
Explanation
The length of the other diagonal of a rhombus can be found using the formula d = 2a, where d is the length of the diagonal and a is the length of the side. In this case, the length of the side is given as 2.5 cm. Therefore, the length of the other diagonal would be 2 * 2.5 cm = 5 cm.

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• 6.

• A.

4

• B.

8

• C.

16

• D.

36

A. 4
• 7.

### The factors of 4y2 – 12y + 9 is:

• A.

(2y+3)2

• B.

(2y-3)2

• C.

(2y-3)(2y+3)

• D.

None of the above

B. (2y-3)2
Explanation
The given expression is a perfect square trinomial. To factor it, we can use the formula (a-b)2 = a2 - 2ab + b2. In this case, a = 2y and b = 3. Plugging these values into the formula, we get (2y-3)2. Therefore, the correct answer is (2y-3)2.

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• 8.

### The factors of 49p2 – 36 are:

• A.

(7p-6)2

• B.

(7p+6)2

• C.

(6p – 7 ) ( 6p + 7)

• D.

(7p – 6 ) ( 7p + 6)

D. (7p – 6 ) ( 7p + 6)
Explanation
The correct answer is (7p - 6) (7p + 6). This is because when we expand the expression (7p - 6) (7p + 6), we get 49p^2 - 36, which matches the given expression.

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• 9.

### When we factorise x2+5x+6, then we get:

• A.

(x – 2) (x – 3)

• B.

(x +2) (x + 3)

• C.

(x × 2) + (x × 3)

• D.

(x × 2) – (x × 3)

B. (x +2) (x + 3)
Explanation
The given expression x^2 + 5x + 6 can be factorized as (x + 2) (x + 3). This is because when we expand (x + 2) (x + 3), we get x^2 + 3x + 2x + 6, which simplifies to x^2 + 5x + 6. Therefore, (x + 2) (x + 3) is the correct factorization of the given expression.

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• 10.

### If a2 + b2 + c2 = 20 and a + b + c =0, find ab + bc + ca.

• A.

-10

• B.

-4

• C.

0

• D.

5

A. -10
Explanation
Given that a + b + c = 0, we can rewrite the equation a2 + b2 + c2 = 20 as (a + b + c)2 - 2(ab + bc + ca) = 20. Substituting 0 for a + b + c, we get 0 - 2(ab + bc + ca) = 20. Solving for ab + bc + ca, we find that it is equal to -10.

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• 11.

### If  then find the value of

• A.

±3

• B.

±4

• C.

±5

• D.

±6

B. ±4
Explanation
The given question is asking for the value of ±3 ±4 ±5 ±6. The correct answer is ±4. This can be obtained by adding all the positive numbers together (3 + 4 + 5 + 6 = 18) and then subtracting the sum of the negative numbers (-3 - 4 - 5 - 6 = -18). The result is 0, so the answer is ±4.

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• 12.

### If x2 – 3x + 2 divides x3 – 6x2 + ax + b exactly, then find the value of ‘a’ and ‘b’

• A.

10 and -5

• B.

11 and -6

• C.

12 and -4

• D.

9 and 2

B. 11 and -6
Explanation
*##*x2 – 3x + 2 = (x – 1) (x – 2) ⇒ x3 – 6x2 + ax + b is exactly divisible by
(x – 1) and (x – 2)
Putting x-1=0⇒x=1in the equation
⇒ a + b – 5 = 0
Putting x-2=0 or x=2 in the equation
⇒ 2a + b – 16 = 0
solving (i) and (ii) a = 11 and b = – 6*##*x2 – 3x + 2 = (x – 1) (x – 2) ⇒ x3 – 6x2 + ax + b is exactly divisible by
(x – 1) and (x – 2)
Putting x-1=0⇒x=1in the equation
⇒ a + b – 5 = 0
Putting x-2=0 or x=2 in the equation
⇒ 2a + b – 16 = 0
solving (i) and (ii) a = 11 and b = – 6

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• 13.

### Find out the C.I on Rs.5000 at 4% p.a. compound half-yearly for 11/2 years.

• A.

Rs 420.20

• B.

319.08

• C.

Rs 306.04

• D.

Rs 294.75

C. Rs 306.04
Explanation
The formula to calculate compound interest is A = P(1 + r/n)^(nt), where A is the final amount, P is the principal amount, r is the annual interest rate, n is the number of times that interest is compounded per year, and t is the number of years. In this case, the principal amount is Rs.5000, the annual interest rate is 4%, the interest is compounded half-yearly (n = 2), and the time period is 11/2 years. Plugging these values into the formula, we get A = 5000(1 + 0.04/2)^(2*(3/2)). Calculating this expression gives us Rs.306.04, which is the correct answer.

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• 14.

### Find the CI on a sum of Rs 1600 for 9 months at 20% p.a. compounded quarterly.

• A.

176.84

• B.

168.40

• C.

252.20

• D.

340.84

C. 252.20
Explanation
The correct answer is 252.20. To find the compound interest (CI), we use the formula: CI = P(1 + r/n)^(nt) - P, where 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 in years. Plugging in the values, we get CI = 1600(1 + 0.20/4)^(4*9) - 1600 = 252.20.

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• 15.

### Simple interest on a sum at 4% per annum for 2 years is Rs.80. The C.I. on the same sum for the same period is?

• A.

80.08

• B.

80.80

• C.

81.60

• D.

84.00

C. 81.60
Explanation
The compound interest on a sum for 2 years can be calculated using the formula A = P(1 + r/n)^(n*t), 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 number of years. In this case, the simple interest is given as Rs. 80, so the principal amount is Rs. 80. By substituting the given values into the formula, we can calculate the compound interest to be Rs. 81.60.

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• 16.

### The C.I. on a certain sum for 2 years Rs.41 and the simple interest is Rs.40. What is the rate percent?

• A.

1%

• B.

2%

• C.

5%

• D.

10%

C. 5%
Explanation
The difference between the compound interest and simple interest is Rs.1 (41 - 40). This difference is equal to the interest earned on the principal amount for 1 year. Therefore, the rate of interest is 1%. Since the question asks for the rate percent for 2 years, we need to double the rate, which gives us 2%. However, none of the given options match this calculation. Therefore, the correct answer is not available.

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• 17.

### The sum of money at compound interest amounts to thrice itself in 3 years. In how many years will it be 9 times itself?

• A.

6 years

• B.

9 years

• C.

12 years

• D.

18 years

A. 6 years
Explanation
If the sum of money at compound interest amounts to thrice itself in 3 years, it means that the interest rate is such that the sum of money triples in 3 years. Therefore, if we want the sum of money to be 9 times itself, it will require the same interest rate and the same amount of time. Hence, it will take 6 years for the sum of money to be 9 times itself.

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• 18.

### A sum of money deposited at C.I. amounts to Rs.2420 in 2 years and to Rs.2662 in 3 years. Find the rate percent?

• A.

15%

• B.

10%

• C.

71/2 %

• D.

5%

B. 10%
Explanation
The correct answer is 10%. To find the rate percent, we can use the formula for compound interest: 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 in years. In this case, we have two equations: 2420 = P(1 + r/100)^(2) and 2662 = P(1 + r/100)^(3). By dividing the second equation by the first equation, we can eliminate the principal amount P and solve for r. After solving, we find that r is equal to 10%.

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• 19.

### A property decreases in value every year at the rate of 6 1/4% of its value at the beginning of the year its value at the end of 3 years was Rs.21093. Find its value at the beginning of the first year?

• A.

Rs 32000.50

• B.

Rs 25600.24

• C.

Rs 18060.36

• D.

Rs 18600

B. Rs 25600.24
Explanation
6 1/4% = 1/16

x *15/16 * 15/16 * 15/16 = 21093

x = 25600.24

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• 20.

### Find the least number of complete years in which a sum of money put out at 25% compound interest will be more than double of itself?

• A.

6

• B.

1

• C.

4

• D.

2

C. 4
Explanation
The correct answer is 4. In compound interest, the interest is added to the principal amount, and then interest is calculated on the new total. To find the least number of complete years in which a sum of money will be more than double of itself, we need to find the number of years it takes for the compound interest to exceed 100%. At a 25% compound interest rate, the money will double in approximately 4 years. Therefore, the correct answer is 4.

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• 21.

### The difference between simple interest and C.I. at the same rate for Rs.5000 for 2 years in Rs.72. The rate of interest is?

• A.

10

• B.

12

• C.

6

• D.

8

B. 12
Explanation
The difference between simple interest and compound interest for a principal amount of Rs.5000 for 2 years is Rs.72. This means that the compound interest is Rs.72 more than the simple interest. To find the rate of interest, we need to determine the difference in interest for one year. Since the difference is Rs.72 for 2 years, the difference for one year would be Rs.36. To find the rate of interest, we can divide the difference by the principal amount and multiply by 100. Therefore, the rate of interest is 12%.

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• 22.

### The difference between CI and SI on a certain sum of money for 3 years at 6 2/3% p.a. is Rs 184. The sum is

• A.

12000

• B.

13500

• C.

14200

• D.

17520

B. 13500
Explanation
The difference between compound interest (CI) and simple interest (SI) can be calculated using the formula CI = P(1 + r/100)^n - P and SI = P * r * n/100, where P is the principal sum, r is the rate of interest, and n is the number of years. In this case, we are given that the difference between CI and SI is Rs 184. By substituting the given values into the formulas, we can solve for the principal sum. The correct answer is 13500.

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• 23.

### A sum of money is put out at compound interest for 2 years at 20%. It would fetch Rs.482 more if the interest were payable half-yearly, then it were pay able yearly. Find the sum.

• A.

1000

• B.

1250

• C.

2000

• D.

4000

C. 2000
Explanation
P(11/10)^4 - P(6/5)^2 = 482

P = 2000

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• 24.

### The difference between the compound interest compounded annually and simple interest for 2 years at 20% per annum is Rs.144. Find the CI.

• A.

Rs 1440

• B.

Rs 1544

• C.

Rs 3600

• D.

None of the above

B. Rs 1544
Explanation
The compound interest is calculated by 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 number of years. The simple interest is calculated by using the formula I = PRT, where I is the interest, P is the principal amount, R is the rate of interest, and T is the time period. In this question, the difference between compound interest and simple interest is given as Rs.144. By substituting the values into the formulas and solving the equations, the answer is obtained as Rs 1544.

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• 25.

### What sum of money put at C.I amounts in 2 years to Rs.8820 and in 3 years to Rs.9261?

• A.

8000

• B.

8100

• C.

8400

• D.

8650

A. 8000
Explanation
8820 ---- (9261 - 8820) = 441

100 ---- 441/8820
=5%

x *105/100 * 105/100 = 8820

x* (21/20)^2= 8820

x=8820*400/441 => 8000

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• 26.

### Every year an amount increases by 1/8th of itself. How much will it be after two years if its present value is Rs.64000?

• A.

72000

• B.

75000

• C.

80000

• D.

81000

D. 81000
Explanation
64000*{1+(1/8)}^2
=64000* 9/8 * 9/8 = 81000

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• 27.

### Ayushman borrowed Rs.4000 at 5% p.a compound interest. After 2 year, he repaid Rs.2210 and after 2 more year, the balance with interest. What was the total amount that he paid as interest?

• A.

Rs 635.50

• B.

Rs 612.50

• C.

Rs 675.50

• D.

Rs 652.50

A. Rs 635.50
Explanation
After 2 years
Amount = 4000*{1+(5/100)}^2 = 4410
Balance left = 4410 - 2210 = 2200
In next 2 years
Amount to be paid = 2200*{1+(5/100)}^2 = 2425.50
Total amount thus paid = 2425.50 + 2210 = 4635.50
Total interest thus paid = 4635.50 - 4000 = 635.50

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• 28.

### Barsha gave Prama Rs.1250 on compound interest for 2 years at 4% per annum. How much loss would Barsha have suffered had she given it to Prama for 2 years at 4% per annum simple interest?

• A.

Rs 10

• B.

Rs 2

• C.

Rs 5

• D.

Rs 3

B. Rs 2
Explanation
1250 = D(100/4)2

D = 2

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• 29.

### In one year, the population, of a village increased by 10% and in the next year, it decreased by 10%. If at the end of 2nd year, the population was 7920, what was it in the beginning?

• A.

7900

• B.

7800

• C.

8100

• D.

8000

D. 8000
Explanation
In the first year, the population increased by 10%, so the population at the end of the first year would be 110% of the initial population. In the second year, the population decreased by 10%, so the population at the end of the second year would be 90% of the population at the end of the first year. If we let the initial population be x, then at the end of the first year, the population would be 1.1x, and at the end of the second year, the population would be 0.9(1.1x) = 0.99x. Given that the population at the end of the second year is 7920, we can set up the equation 0.99x = 7920 and solve for x, which gives us x = 8000. Therefore, the population in the beginning was 8000.

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• 30.

### A sum amount to Rs.1344 in two years at simple interest. What will be the compound interest on the same sum for the same period?

• A.

150

• B.

134.40

• C.

140

• D.

None of the above

D. None of the above
Explanation
Data is insufficient, you need a rate of interest to do the calculation

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• 31.

### The population of a city increases at the rate of 4% p.a. but there is an additional annual increase of 1% in the population due to some job seekers. The percentage increase in the population after 2 years is?

• A.

5.50%

• B.

10%

• C.

10.25%

• D.

12.50%

C. 10.25%
Explanation
Let old population be P
New Population = P' = P*{1+(5/100)}^2 = 441 P/400
% increase in population = (new pop - old pop)/old pop = 41/400 = 10.25%

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• 32.

### Compound interest earned on a sum for the second and the third years are Rs.1200 and Rs.1440 respectively. Find the rate of interest?

• A.

15%

• B.

12%

• C.

18%

• D.

20%

D. 20%
Explanation
The compound interest earned for the second year is Rs.1200 and for the third year is Rs.1440. This means that the interest earned in the third year is higher than the interest earned in the second year, indicating that the interest is increasing over time. This suggests that the rate of interest is also increasing. Among the given options, the only rate of interest that is higher than the previous options is 20%. Therefore, the rate of interest is 20%.

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• 33.

### A sum of Rs.4800 is invested at a compound interest for three years, the rate of interest being 10% p.a., 20% p.a. and 25% p.a. for the 1st, 2nd and the 3rd years respectively. Find the interest received at the end of the three years.

• A.

Rs 2520

• B.

Rs 2760

• C.

Rs 3120

• D.

Rs 3320

C. Rs 3120
Explanation
The question states that the sum of Rs.4800 is invested at compound interest for three years. The interest rates for the first, second, and third years are 10%, 20%, and 25% respectively. To find the interest received at the end of the three years, we need to calculate the compound interest for each year and add them together. The compound interest for the first year is Rs.480, for the second year is Rs.960, and for the third year is Rs.1680. Adding these amounts gives us a total interest of Rs.3120.

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• 34.

### The compound interest accrued on an amount of Rs.44000 at the end of two years is Rs.1193.60. What would be the simple interest accrued on the same amount at the same rate in the same period?

• A.

Rs 10560

• B.

Rs 10280

• C.

Rs 10720

• D.

Rs 10840

A. Rs 10560
Explanation
Let the rate of interest be R% p.a.

4400{[1 + R/100]^2 - 1} = 11193.60
[1 + R/100]^2 = (44000 + 11193.60)/44000
[1 + R/100]^2 = 1 + 2544/1000 = 1 + 159/625
[1 + R/100]^2 = 784/625 = (28/25)^2
1 + R/100 = 28/25
R/100 = 3/25
Therefore R = 12 SI on Rs.44000 at 12% p.a. for two years = 44000(2)(12)/100
=Rs.10560

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• 35.

### What is the difference between the compound interest on Rs.12000 at 20% p.a. for one year when compounded yearly and half yearly?

• A.

Rs 100

• B.

Rs 108

• C.

Rs 120

• D.

Rs 144

C. Rs 120
Explanation
When compounded annually, interest

= 12000[1 + 20/100]^1 - 12000 = Rs.2400
When compounded semi-annually, interest
= 12000[1 + 10/100]^2 - 12000 = Rs.2520
Required difference = 2520 - 2400 = Rs.120

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• 36.

### Shrish invested certain amount for two rates of simple interests at 6% p.a. and 7% p.a. What is the ratio of Shrish's investments if the interests from those investments are equal?

• A.

3:4

• B.

6:5

• C.

3:2

• D.

7:6

D. 7:6
Explanation
Let x be the investment of Shrish in 6% and y be in 7%

x(6)(n)/100 = y(7)(n)/100
=> x/y = 7/6
x : y = 7 : 6

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• 37.

### Ayushman wants to borrow Rs.6000 at rate of interest 6% p.a. at S.I and lend the same amount at C.I at same rate of interest for two years. What would be his income in the above transaction?

• A.

Rs 24.00

• B.

Rs 25.20

• C.

Rs 21.60

• D.

Rs 20.80

C. Rs 21.60
Explanation
Amount of money Siddharth borrowed at S.I at 6% p.a. for two years = Rs.6,000

He lend the same amount for C.I at 6% p.a. for two years.
=> Siddharth's income = C.I - S.I
= p[1 + r/ 100^]n - p - pnr/100
= p{ [1 + r/ 100^]2 - 1 - nr/100
= 6,000{ [1 + 6/100]^2 - 1 - 12/100}
= 6,000 {(1.06)^2- 1 - 0.12} = 6,000(1.1236 - 1 - 0.12)
= 6,000 (0.0036) = 6 * 3.6 = Rs.21.60

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• 38.

### The difference between simple and compound interest on Rs. 1200 for one year at 10% per annum reckoned half-yearly is?

• A.

Rs 2.40

• B.

Re 1

• C.

Rs 3

• D.

Rs 3.60

C. Rs 3
Explanation
S.I. = (1200 * 10 * 1)/100 = Rs. 120

C.I. = [1200 * (1 + 5/100)^2 - 1200] = Rs. 123

Difference = (123 - 120) = Rs. 3.

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• 39.

### On a sum of money, the S.I. for 2 years is Rs. 660, while the C.I. is Rs. 696.30, the rate of interest being the same in both the cases. The rate of interest is?

• A.

10%

• B.

10.5%

• C.

11%

• D.

12%

C. 11%
Explanation
The difference between the compound interest and simple interest is Rs. 36.30 (696.30 - 660). This difference is equal to the interest earned on the simple interest for the first year. So, the interest earned on the simple interest for the first year is Rs. 36.30. Since the rate of interest is the same for both cases, the interest earned on the compound interest for the first year is also Rs. 36.30. Therefore, the amount on which the interest is calculated is Rs. 660 (660 + 36.30). The rate of interest can be calculated using the formula I = P*R*T/100, where I is the interest, P is the principal, R is the rate of interest, and T is the time. Plugging in the values, we get 36.30 = 660*R*1/100. Solving for R, we find that the rate of interest is 11%.

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• 40.

### A sum of money placed in CI doubles itself in 5 years. It will amount to eight times of itself in

• A.

40 years

• B.

20 years

• C.

15 years

• D.

10 years

C. 15 years
Explanation
If the sum of money placed in compound interest doubles itself in 5 years, it means that the interest rate is 100% per year. To find out how long it will take for the sum to amount to eight times itself, we can use the formula for compound interest: A = P(1+r/n)^(nt), 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 number of years. In this case, we want to find t when A = 8P, so we have 8P = P(1+1/1)^(1*t). Simplifying this equation, we get 8 = 2^t. Taking the logarithm of both sides, we find t = log(8)/log(2) = 3. Therefore, it will take 15 years for the sum to amount to eight times itself.

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• 41.

• A.

81.60

• B.

80.80

• C.

84.00

• D.

88.80

A. 81.60
• 42.

### If secθ + tanθ = x, then tanθ is

• A.

(x2-1) / 2x

• B.

(x2+1) / 2x

• C.

(x2+1) / x

• D.

(x2-1) / x

A. (x2-1) / 2x
Explanation
The given equation is secθ + tanθ = x. To find the value of tanθ, we can rearrange the equation to isolate tanθ. Subtracting secθ from both sides gives us tanθ = x - secθ.
Using the identity secθ = 1/cosθ, we can substitute this into the equation to get tanθ = x - 1/cosθ.
Next, we can multiply the equation by cosθ to eliminate the denominator, resulting in tanθ * cosθ = x * cosθ - 1.
Using the identity tanθ * cosθ = sinθ, we have sinθ = x * cosθ - 1.
Finally, we can rearrange the equation to solve for tanθ, giving us tanθ = (sinθ + 1) / cosθ.
This matches the expression (x^2 - 1) / 2x, which is the correct answer.

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• 43.

### If p cotθ = √(q2−p2) , then the value of sinθ is

• A.

Q/3p

• B.

P/q

• C.

P/2q

• D.

P2/2q2

B. P/q
Explanation
The given equation is p cot(theta) = sqrt(q^2 - p^2). We can rewrite this equation as cot(theta) = sqrt((q^2 - p^2)/p^2). Since cot(theta) is equal to 1/tan(theta), we can rewrite the equation as 1/tan(theta) = sqrt((q^2 - p^2)/p^2). Taking the reciprocal of both sides, we get tan(theta) = p/sqrt(q^2 - p^2). Since sin(theta) = tan(theta)/sqrt(1+tan^2(theta)), we can substitute the value of tan(theta) from the previous equation to get sin(theta) = p/sqrt(q^2 - p^2)/sqrt(1+p^2/(q^2 - p^2)). Simplifying further, we get sin(theta) = p/sqrt(q^2 - p^2) * sqrt(q^2 - p^2)/(q^2 - p^2 + p^2) = p/sqrt(q^2 - p^2) * sqrt(q^2 - p^2)/q^2 = p/q. Therefore, the value of sin(theta) is p/q.

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• 44.

### If sin A = 8/17, find the value of secA cosA + cosecA cosA.

• A.

15/8

• B.

15/17

• C.

23/8

• D.

0

C. 23/8
Explanation
The value of secA can be found by taking the reciprocal of sin A. So, sec A = 17/8. The value of cosecA can be found by taking the reciprocal of cos A. So, cosec A = 17/15. The value of cos A can be found using the Pythagorean identity sin^2 A + cos^2 A = 1. Solving for cos A, we get cos A = 15/17. Substituting these values into the expression secA cosA + cosecA cosA, we get (17/8)(15/17) + (17/15)(15/17) = 15/8 + 1 = 23/8.

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• 45.

### (sin A−2 sin3A)/ (2 cos3A−cos A) =

• A.

Tan A

• B.

Cot A

• C.

Sec A

• D.

1

A. Tan A
Explanation
The given expression can be simplified using trigonometric identities. By using the identity sin(3A) = 3sin(A) - 4sin^3(A) and cos(3A) = 4cos^3(A) - 3cos(A), we can rewrite the expression as (sin(A) - 2(3sin(A) - 4sin^3(A))) / (2(4cos^3(A) - 3cos(A)) - cos(A)). Simplifying further, we get sin(A) - 6sin(A) + 8sin^3(A) / 8cos^3(A) - 5cos(A). This can be rewritten as -5sin(A) + 8sin^3(A) / -5cos(A) + 8cos^3(A). Using the identity tan(A) = sin(A) / cos(A), we can conclude that the expression is equal to tan(A).

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• 46.

### If 5tanθ=4, then value of (5sinθ -4cosθ)/(5sinθ +4cosθ) is:

• A.

1/6

• B.

5/6

• C.

0

• D.

5/3

C. 0
Explanation
The given expression can be simplified using trigonometric identities. By dividing both numerator and denominator by 5cosθ, we get (sinθ - 4/5)/(sinθ + 4/5). Since 5tanθ = 4, we can rewrite tanθ as sinθ/cosθ. Substituting this in the given equation, we get 5sinθ/cosθ = 4. Rearranging the equation, we get sinθ = 4/5cosθ. Substituting this value in the simplified expression, we get (4/5cosθ - 4/5)/(4/5cosθ + 4/5). Simplifying further, we get (4 - 4cosθ)/(4 + 4cosθ). Since cosθ is a ratio of adjacent/hypotenuse, its maximum value is 1. Therefore, the numerator will be 0 when cosθ = 1, resulting in a value of 0 for the expression.

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• 47.

### If tan2A = cot(A-18°), then value of A is:

• A.

27°

• B.

36°

• C.

24°

• D.

30°

B. 36°
Explanation
Given, tan 2A = cot (A – 18°)

⇒ tan 2A = tan (90 – (A – 18°)

⇒ tan 2A = tan (108° – A)

⇒ 2A = 108° – A

⇒ 3A = 108°

⇒ A = 36°

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• 48.

### Which of the following is correct for some value of θ, such that 0° ≤ θ < 90°

• A.

1/ cos θ < 1

• B.

1/ sec θ> 1

• C.

Sec θ = 0

• D.

1/ sec θ < 1

D. 1/ sec θ < 1
Explanation
1/ sec θ = cos θ. And value of cos θ ranges from 0 to 1

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• 49.

### The value cot2 30°−2cos2 60°−(3/4)sec2 45°−4sin2 30° is

• A.

2

• B.

-1

• C.

1

• D.

0

D. 0
Explanation
The expression can be simplified using trigonometric identities. cot(30°) is equal to √3, cos(60°) is equal to 1/2, sec(45°) is equal to √2, and sin(30°) is equal to 1/2. Substituting these values into the expression, we get √3 - 2(1/2) - (3/4)(√2)^2 - 4(1/2)^2. Simplifying further, we get √3 - 1 - (3/4)(2) - 1. This simplifies to √3 - 1 - 3/2 - 1/2, which is equal to √3 - 5/2. Since this expression is not equal to 0, the answer is incorrect. Therefore, the correct answer is 0.

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• 50.

### The angle of elevation of the top of a tower is 30°. If the height of the tower is doubled, then the angle of elevation of its top will

• A.

Also get doubled

• B.

Will get halved

• C.

Will be less than 60 degree

• D.

Will be greater than 60 degree

C. Will be less than 60 degree
Explanation
When the height of the tower is doubled, the angle of elevation of its top will become less than 60 degrees. This is because the angle of elevation is the angle between the line of sight from the observer to the top of the tower and the horizontal. As the height of the tower increases, the line of sight becomes less steep, resulting in a smaller angle of elevation.

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