The Mega Test 02/ Mathematics

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.

48 m

50 m

55 m

60 m

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.

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.

2) 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?

10%

10.5%

11%

12%

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%.

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%.

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

Rs 2.40

Re 1

Rs 3

Rs 3.60

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

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

Difference = (123 - 120) = 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.

4) 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?

Rs 24.00

Rs 25.20

Rs 21.60

Rs 20.80

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

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

5) Simple Interest on a sum at 4% per annum for 2 years is 80. The compound interest on the same sum for the same period is?

81.60

80.80

84.00

88.80

not-available-via-ai

Explanation

not-available-via-ai

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

Rs 100

Rs 108

Rs 120

Rs 144

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

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

7) 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.

Rs 2520

Rs 2760

Rs 3120

Rs 3320

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.

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.

8) 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?

15%

12%

18%

20%

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%.

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%.

9) 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?

5.50%

10%

10.25%

12.50%

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%

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%

10) 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?

150

134.40

140

None of the above

Data is insufficient, you need a rate of interest to do the calculation

Explanation

Data is insufficient, you need a rate of interest to do the calculation

11) 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?

7900

7800

8100

8000

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.

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.

12) 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?

Rs 10

Rs 2

Rs 5

Rs 3

1250 = D(100/4)2

D = 2

Explanation

1250 = D(100/4)2

D = 2

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

27°

36°

24°

30°

Given, tan 2A = cot (A – 18°)

⇒ tan 2A = tan (90 – (A – 18°)

⇒ tan 2A = tan (108° – A)

⇒ 2A = 108° – A

⇒ 3A = 108°

⇒ A = 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°

14) 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.

1000

1250

2000

4000

P(11/10)^4 - P(6/5)^2 = 482

P = 2000

Explanation

P(11/10)^4 - P(6/5)^2 = 482

P = 2000

15) 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

12000

13500

14200

17520

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.

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.

16) If the height of a tower and the distance of the point of observation from its foot,both, are increased by 10%, then the angle of elevation of its top

Increases by 20%

Decreases by 20%

Remains unchanged

Have no relation

tan θ = h/x

Where h is height and x is distance from tower,

If both are increased by 10%, then the angle will remain unchanged.

Explanation

tan θ = h/x

Where h is height and x is distance from tower,

If both are increased by 10%, then the angle will remain unchanged.

17) 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?

26 m

29 m

32 m

36 m

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.

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.

18) An observer √3m tall is 3 m away from the pole 2√3 high. What is the angle of elevation of the top?

60°

30°

45°

90°

Height of the pole that is above man’s height = 2 √3 – √3 = √3m

Hence, AB = √3m

BC = 3m

Hence, tan C = AB/BC= 1/√3

Explanation

Height of the pole that is above man’s height = 2 √3 – √3 = √3m

Hence, AB = √3m

BC = 3m

Hence, tan C = AB/BC= 1/√3

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

176.84

168.40

252.20

340.84

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.

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.

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

Rs 420.20

319.08

Rs 306.04

Rs 294.75

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.

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.

21) The factors of 49p² – 36 are:

(7p-6)2

(7p+6)2

(6p – 7 ) ( 6p + 7)

(7p – 6 ) ( 7p + 6)

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.

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.

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

-10

-4

0

5

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.

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.

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

6 years

9 years

12 years

18 years

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.

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.

24) If x² – 3x + 2 divides x³ – 6x² + ax + b exactly, then find the value of 'a' and 'b'

10 and -5

11 and -6

12 and -4

9 and 2

*##*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

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

25) A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole if the angle made by the rope with the ground level is 30°.

10 m

15 m

20 m

30 m

In ΔABC, we have

sin 30°=h/20

⇒1/2=h/20

⇒h=10 m

Hence, the height of the pole is 10 m.

Explanation

In ΔABC, we have

sin 30°=h/20

⇒1/2=h/20

⇒h=10 m

Hence, the height of the pole is 10 m.

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

40 years

20 years

15 years

10 years

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.

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.

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

50 cm

56 cm

62 cm

68 cm

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.

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.

28) In a circle with center O, two diameters AB and CD are drawn perpendicular to each other. Find the length of chord AC

2 AB

AB √2

AB/2

AB/√2

In a circle, the length of a chord is given by the formula 2r*sin(θ/2), where r is the radius of the circle and θ is the angle subtended by the chord at the center of the circle. In this case, since AB and CD are diameters, they divide the circle into four quadrants, each with a central angle of 90 degrees. Therefore, the angle subtended by chord AC is 90 degrees. Since the radius of the circle is AB/2, substituting these values into the formula gives 2*(AB/2)*sin(90/2) = AB/√2. Thus, the length of chord AC is AB/√2.

Explanation

In a circle, the length of a chord is given by the formula 2r*sin(θ/2), where r is the radius of the circle and θ is the angle subtended by the chord at the center of the circle. In this case, since AB and CD are diameters, they divide the circle into four quadrants, each with a central angle of 90 degrees. Therefore, the angle subtended by chord AC is 90 degrees. Since the radius of the circle is AB/2, substituting these values into the formula gives 2*(AB/2)*sin(90/2) = AB/√2. Thus, the length of chord AC is AB/√2.

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

Tan A

Cot A

Sec A

1

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).

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).

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

8000

8100

8400

8650

8820 ---- (9261 - 8820) = 441

100 ---- 441/8820
=5%

x *105/100 * 105/100 = 8820

x* (21/20)^2= 8820

x=8820*400/441 => 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

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

1%

2%

5%

10%

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.

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.

32) 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?

Rs 32000.50

Rs 25600.24

Rs 18060.36

Rs 18600

6 1/4% = 1/16

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

x = 25600.24

Explanation

6 1/4% = 1/16

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

x = 25600.24

33) 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?

15%

10%

71/2 %

5%

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%.

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%.

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

Q/3p

P/q

P/2q

P2/2q2

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.

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.

35) If angle OAB = 40 degree, angle ACB =

50

40

60

30

Since angle OAB is given as 40 degrees, angle ACB can be found by subtracting 40 degrees from 180 degrees (the sum of angles in a triangle). Therefore, angle ACB is 180 - 40 = 140 degrees.

Explanation

Since angle OAB is given as 40 degrees, angle ACB can be found by subtracting 40 degrees from 180 degrees (the sum of angles in a triangle). Therefore, angle ACB is 180 - 40 = 140 degrees.

36) AD is the diameter of a circle and AB is a chord. If AD = 34 cm and AB = 30 cm, the distance of AB from the center is

17 cm

15 cm

4 cm

8 cm

In a circle, the distance from the center to a chord is equal to half the length of the chord. Therefore, the distance of AB from the center is equal to half of AB. Since AB is 30 cm, half of AB is 15 cm. Therefore, the correct answer is 15 cm.

Explanation

In a circle, the distance from the center to a chord is equal to half the length of the chord. Therefore, the distance of AB from the center is equal to half of AB. Since AB is 30 cm, half of AB is 15 cm. Therefore, the correct answer is 15 cm.

37) If secθ + tanθ = x, then tanθ is

(x2-1) / 2x

(x2+1) / 2x

(x2+1) / x

(x2-1) / x

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.

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.

38) If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through A, B and C is

6 cm

10 cm

12 cm

20 cm

In a right triangle, the hypotenuse is twice the length of the shorter leg. Since AB is perpendicular to BC, AB is the shorter leg and BC is the hypotenuse. Therefore, BC is twice the length of AB. Since AB is 12 cm, BC is 24 cm. The radius of the circle passing through A, B, and C is equal to half the length of BC, which is 24/2 = 12 cm. Therefore, the correct answer is 12 cm.

Explanation

In a right triangle, the hypotenuse is twice the length of the shorter leg. Since AB is perpendicular to BC, AB is the shorter leg and BC is the hypotenuse. Therefore, BC is twice the length of AB. Since AB is 12 cm, BC is 24 cm. The radius of the circle passing through A, B, and C is equal to half the length of BC, which is 24/2 = 12 cm. Therefore, the correct answer is 12 cm.

39) The angles of elevation of the top of a tower from two points distant s and t from its foot are complementary. Then the height of the tower is:

St

S2t2

√st

S/t

The angles of elevation of the top of a tower from two points distant s and t from its foot are complementary. This means that the sum of the two angles is 90 degrees. Let's assume that the angle of elevation from point s is x degrees. Then, the angle of elevation from point t would be 90 - x degrees. Using trigonometry, we can set up the following equation: tan(x) = height of the tower / s and tan(90 - x) = height of the tower / t. Since tan(90 - x) is equal to cot(x), we can rewrite the second equation as cot(x) = height of the tower / t. By cross multiplying these equations, we get height of the tower = s * t / sqrt(s^2 + t^2). Therefore, the height of the tower is sqrt(st).

Explanation

The angles of elevation of the top of a tower from two points distant s and t from its foot are complementary. This means that the sum of the two angles is 90 degrees. Let's assume that the angle of elevation from point s is x degrees. Then, the angle of elevation from point t would be 90 - x degrees. Using trigonometry, we can set up the following equation: tan(x) = height of the tower / s and tan(90 - x) = height of the tower / t. Since tan(90 - x) is equal to cot(x), we can rewrite the second equation as cot(x) = height of the tower / t. By cross multiplying these equations, we get height of the tower = s * t / sqrt(s^2 + t^2). Therefore, the height of the tower is sqrt(st).

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

46 m

48 m

50 m

52 m

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.

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.

41) 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?

3:4

6:5

3:2

7:6

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

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

42) 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?

Rs 635.50

Rs 612.50

Rs 675.50

Rs 652.50

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

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

43) 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.

3 cm

4 cm

5 cm

5.5 cm

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.

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.

44) 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?

72000

75000

80000

81000

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

Explanation

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

45) The factors of 4y²– 12y + 9 is:

(2y+3)2

(2y-3)2

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

None of the above

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.

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.

46) If then find the value of

±3

±4

±5

±6

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.

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.

47) 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?

80.08

80.80

81.60

84.00

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.

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.

48) 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?

10

12

6

8

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%.

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%.

49) 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?

Rs 10560

Rs 10280

Rs 10720

Rs 10840

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

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

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

(x – 2) (x – 3)

(x +2) (x + 3)

(x × 2) + (x × 3)

(x × 2) – (x × 3)

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.

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.

51) Angle CAO =

30

45

60

90

The angle CAO is 60 degrees. This can be determined by looking at the given options and identifying the angle that is closest to the angle formed by line segments CA and AO. In this case, the angle CAO is an acute angle, and the closest option is 60 degrees.

Explanation

The angle CAO is 60 degrees. This can be determined by looking at the given options and identifying the angle that is closest to the angle formed by line segments CA and AO. In this case, the angle CAO is an acute angle, and the closest option is 60 degrees.

52) Find area of the triangle

4

8

16

36

not-available-via-ai

Explanation

not-available-via-ai

53) 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

Also get doubled

Will get halved

Will be less than 60 degree

Will be greater than 60 degree

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.

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.

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

2

-1

1

0

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.

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.

55) 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?

6

1

4

2

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.

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.

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

1/ cos θ < 1

1/ sec θ> 1

Sec θ = 0

1/ sec θ < 1

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

Explanation

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

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

Rs 1440

Rs 1544

Rs 3600

None of the above

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.

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.

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

1/6

5/6

0

5/3

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.

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.

59) Length of a chord is equal to the radius of the circle. Angle subtended by the chord on the circumference is

30

45

60

90

The angle subtended by a chord on the circumference of a circle is half the measure of the angle subtended by the same chord at the center of the circle. Since the length of the chord is equal to the radius of the circle, the angle subtended at the center is 60 degrees. Therefore, the angle subtended by the chord on the circumference is half of 60 degrees, which is 30 degrees.

Explanation

The angle subtended by a chord on the circumference of a circle is half the measure of the angle subtended by the same chord at the center of the circle. Since the length of the chord is equal to the radius of the circle, the angle subtended at the center is 60 degrees. Therefore, the angle subtended by the chord on the circumference is half of 60 degrees, which is 30 degrees.

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

15/8

15/17

23/8

0

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.

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.

The Mega Test 02/ Mathematics

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.

2) 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?

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

4) 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?

5) Simple Interest on a sum at 4% per annum for 2 years is 80. The compound interest on the same sum for the same period is?

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

7) 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.

8) 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?

9) 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?

10) 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?

11) 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?

12) 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?

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

14) 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.

15) 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

16) If the height of a tower and the distance of the point of observation from its foot,both, are increased by 10%, then the angle of elevation of its top

17) 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?

18) An observer √3m tall is 3 m away from the pole 2√3 high. What is the angle of elevation of the top?

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

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

21) The factors of 49p2 – 36 are:

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

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

24) If x2 – 3x + 2 divides x3 – 6x2 + ax + b exactly, then find the value of 'a' and 'b'

25) A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole if the angle made by the rope with the ground level is 30°.

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

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

28) In a circle with center O, two diameters AB and CD are drawn perpendicular to each other. Find the length of chord AC

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

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

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

32) 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?

33) 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?

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

35) If angle OAB = 40 degree, angle ACB =

36) AD is the diameter of a circle and AB is a chord. If AD = 34 cm and AB = 30 cm, the distance of AB from the center is

37) If secθ + tanθ = x, then tanθ is

38) If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through A, B and C is

39) The angles of elevation of the top of a tower from two points distant s and t from its foot are complementary. Then the height of the tower is:

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

41) 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?

42) 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?

43) 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.

44) 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?

45) The factors of 4y2 – 12y + 9 is:

46) If then find the value of

47) 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?

48) 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?

49) 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?

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

51) Angle CAO =

52) Find area of the triangle

53) 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

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

55) 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?

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

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

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

59) Length of a chord is equal to the radius of the circle. Angle subtended by the chord on the circumference is

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

15) 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

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

21) The factors of 49p² – 36 are:

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

24) If x² – 3x + 2 divides x³ – 6x² + ax + b exactly, then find the value of 'a' and 'b'

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

32) 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?

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

45) The factors of 4y²– 12y + 9 is:

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

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