To evaluate the given expression, we must follow the order of operations, commonly remembered by the acronym PEMDAS/BODMAS (Parentheses/Brackets, Exponents/Orders, Multiplication and Division (from left to right), Addition and Subtraction (from left to right)).
Thus, the evaluated value of the expression is \(-29\).
D) -29 is the correct answer, as derived by following the order of operations.
The value of \( \frac{5}{2} \) + \(\left(\frac{1}{1+\frac{3}{8}}\right)\) - \( \frac{5}{4} \) is
\( \frac{106}{59} \)
\( \frac{87}{44} \)
\( \frac{95}{69} \)
\( \frac{113}{72} \)
To solve this problem, we need to simplify the given expression by performing the arithmetic operations in the correct order. We will first simplify the fraction within the parenthesis, then perform the addition and subtraction.
The term inside the parenthesis is \[\left(\frac{1}{1+\frac{3}{8}}\right)\]. First, simplify the denominator \( 1+\frac{3}{8} \).
\[ 1 + \frac{3}{8} = \frac{8}{8} + \frac{3}{8} = \frac{8+3}{8} = \frac{11}{8} \]Now, the term becomes:
\[ \frac{1}{1+\frac{3}{8}} = \frac{1}{\frac{11}{8}} \]Dividing by a fraction is equivalent to multiplying by its reciprocal:
\[ \frac{1}{\frac{11}{8}} = 1 \times \frac{8}{11} = \frac{8}{11} \]The original expression was \[\frac{5}{2} + \left(\frac{1}{1+\frac{3}{8}}\right) - \frac{5}{4}\]. After simplification, it becomes:
\[ \frac{5}{2} + \frac{8}{11} - \frac{5}{4} \]The denominators are 2, 11, and 4. The least common multiple (LCM) of 2, 11, and 4 is 44.
Now, substitute these equivalent fractions back into the expression:
\[ \frac{110}{44} + \frac{32}{44} - \frac{55}{44} \]\[ = \frac{110 + 32 - 55}{44} \]\[ = \frac{142 - 55}{44} \]\[ = \frac{87}{44} \]B) \( \frac{87}{44} \) is the correct value of the given expression.
To determine which of the given numbers divides 54,10,96,943, we can use the divisibility rules or perform long division for each option. Since the numbers are large, long division can be tedious. A more efficient approach is to perform modular arithmetic or long division with the given options.
Divide 541096943 by 103.
\[ 541096943 \div 103 \]\[ 541096943 = 103 \times 5253368 + 99 \]Since the remainder is 99 (not 0), 103 does not divide 541096943.Divide 541096943 by 97.
\[ 541096943 \div 97 \]\[ 541096943 = 97 \times 5578319 \]Since the remainder is 0, 97 divides 541096943.Divide 541096943 by 87.
\[ 541096943 \div 87 \]\[ 541096943 = 87 \times 6219505 + 68 \]Since the remainder is 68 (not 0), 87 does not divide 541096943.Divide 541096943 by 107.
\[ 541096943 \div 107 \]\[ 541096943 = 107 \times 5056980 + 83 \]Since the remainder is 83 (not 0), 107 does not divide 541096943.B) 97 is the correct option because when 541096943 is divided by 97, the remainder is 0.
If cot²θ = \(\frac{1}{1-a^2}\), |a| < 1, and θ is an acute angle, then find the value of (secθ + cosecθ tan³θ).
(2 - a²)\( \frac{3}{2} \)
(4 - a²)
(a² - 8)
(2 - a\( \frac{3}{2} \))
The problem asks us to find the value of an expression involving trigonometric functions, given a relationship between \(\cot^2\theta\) and a constant \(a\). We will use fundamental trigonometric identities to simplify the given expression and substitute the value of \(\cot^2\theta\).
We are given that \(\cot^2\theta = \frac{1}{1-a^2}\). Since \(\theta\) is an acute angle, all trigonometric ratios are positive.
First, let's find \(\tan^2\theta\):
\[ \tan^2\theta = \frac{1}{\cot^2\theta} = \frac{1}{\frac{1}{1-a^2}} = 1-a^2 \]Next, let's find \(\sec^2\theta\) using the identity \(\sec^2\theta = 1 + \tan^2\theta\):
\[ \sec^2\theta = 1 + (1-a^2) = 2-a^2 \]Since \(\theta\) is acute, \(\sec\theta > 0\). So, \(\sec\theta = \sqrt{2-a^2}\).
Now, let's find \(\cosec^2\theta\) using the identity \(\cosec^2\theta = 1 + \cot^2\theta\):
\[ \cosec^2\theta = 1 + \frac{1}{1-a^2} = \frac{1-a^2+1}{1-a^2} = \frac{2-a^2}{1-a^2} \]Since \(\theta\) is acute, \(\cosec\theta > 0\). So, \(\cosec\theta = \sqrt{\frac{2-a^2}{1-a^2}}\).
Now, let's simplify the expression we need to evaluate: \(\sec\theta + \cosec\theta \tan^3\theta\).
We can rewrite \(\cosec\theta \tan^3\theta\) as \[\frac{1}{\sin\theta} \frac{\sin^3\theta}{\cos^3\theta} = \frac{\sin^2\theta}{\cos^3\theta}\]. This doesn't seem to simplify things easily in terms of \(\sec\theta\) and \(\tan\theta\).
Let's try to express it in terms of \(\sec\theta\) and \(\tan\theta\):
\[ \cosec\theta \tan^3\theta = \frac{1}{\sin\theta} \frac{\sin^3\theta}{\cos^3\theta} = \frac{\sin^2\theta}{\cos^3\theta} = \frac{\tan^2\theta}{\cos\theta} = \tan^2\theta \sec\theta \]So the expression becomes:
\[ \sec\theta + \tan^2\theta \sec\theta \]Factor out \(\sec\theta\):
\[ \sec\theta (1 + \tan^2\theta) \]Using the identity \((1 + \tan^2\theta) = \sec^2\theta\), the expression simplifies to:
\[ \sec\theta (\sec^2\theta) = \sec^3\theta \]Now, substitute the value of \(\sec^2\theta = 2-a^2\):
\[ \sec^3\theta = (\sec^2\theta)^{\frac{3}{2}} = (2-a^2)^{\frac{3}{2}} \]A) \((2 - a^2)^{\frac{3}{2}}\) is the correct answer, as derived in the step-by-step solution.
To find the final selling price after successive discounts, we need to apply each discount one after another. First, calculate the price after the first discount. Then, apply the second discount on this new price.
The marked price of the product is ₹7,500.
The first discount is 8%.
\[ \text{Discount 1 amount} = 8\% \text{ of } \text{₹}7,500 = \frac{8}{100} \times 7500 = \text{₹}600 \]Price after first discount = Marked Price - Discount 1 amount
\[ \text{Price after 1st discount} = 7500 - 600 = \text{₹}6,900 \]The second discount is 5%, which is applied to the price after the first discount (₹6,900).
\[ \text{Discount 2 amount} = 5\% \text{ of } \text{₹}6,900 = \frac{5}{100} \times 6900 = \text{₹}345 \]Final Selling Price = Price after 1st discount - Discount 2 amount
\[ \text{Final Selling Price} = 6900 - 345 = \text{₹}6,555 \]Alternatively, we can calculate the remaining percentage after each discount:
Correct Option: B) ₹6,555
Simplify: 3((\( \frac{4}{3} \))x² - 24x + 18) - 4(x² + 7x - 18)
To simplify the given algebraic expression, we need to distribute the constants into their respective parentheses and then combine like terms. This involves multiplying each term inside the first parenthesis by 3 and each term inside the second parenthesis by -4. Finally, we group terms with \(x^2\), terms with \(x\), and constant terms together and perform the additions/subtractions.
Multiply each term inside the first parenthesis by 3:
\[ 3 \left( \frac{4}{3}x^2 - 24x + 18 \right) = 3 \times \frac{4}{3}x^2 - 3 \times 24x + 3 \times 18 \] \[ = 4x^2 - 72x + 54 \]Multiply each term inside the second parenthesis by -4:
\[ -4(x^2 + 7x - 18) = -4 \times x^2 - 4 \times 7x - 4 \times (-18) \] \[ = -4x^2 - 28x + 72 \]Now, add the two simplified expressions from Step 1 and Step 2:
\[ (4x^2 - 72x + 54) + (-4x^2 - 28x + 72) \] \[ = 4x^2 - 72x + 54 - 4x^2 - 28x + 72 \]Group the \(x^2\) terms, \(x\) terms, and constant terms:
\[ (4x^2 - 4x^2) + (-72x - 28x) + (54 + 72) \]So, the simplified expression is:
\[ 0 - 100x + 126 = -100x + 126 \]Correct Option: B) -100x + 126
The selling price of 50 books is equal to the cost price of 16 books. Find the loss or gain percentage.
To find the loss or gain percentage, we need to establish a relationship between the selling price (SP) and cost price (CP) of a single item based on the given information. We will then use the formula for percentage loss or gain.
Let the Cost Price (CP) of one book be \(CP_1\) and the Selling Price (SP) of one book be \(SP_1\).
According to the problem statement, the selling price of 50 books is equal to the cost price of 16 books. We can write this as an equation:
\[ 50 \times SP_1 = 16 \times CP_1 \]From this equation, we can express \(SP_1\) in terms of \(CP_1\):
\[ SP_1 = \frac{16}{50} \times CP_1 \]\[ SP_1 = \frac{8}{25} \times CP_1 \]Since \(SP_1 < CP_1\) (because \(\frac{8}{25} < 1\)), there is a loss. Now, we calculate the loss percentage.
The formula for Loss Percentage is:
\[ \text{Loss Percentage} = \frac{\text{Loss}}{\text{CP}} \times 100 \]First, calculate the Loss:
\[ \text{Loss} = CP_1 - SP_1 \]\[ \text{Loss} = CP_1 - \frac{8}{25} CP_1 \]\[ \text{Loss} = \left(1 - \frac{8}{25}\right) CP_1 \]\[ \text{Loss} = \left(\frac{25 - 8}{25}\right) CP_1 \]\[ \text{Loss} = \frac{17}{25} CP_1 \]Now, substitute the Loss into the Loss Percentage formula:
\[ \text{Loss Percentage} = \frac{\frac{17}{25} CP_1}{CP_1} \times 100 \]\[ \text{Loss Percentage} = \frac{17}{25} \times 100 \]\[ \text{Loss Percentage} = 17 \times 4 \]\[ \text{Loss Percentage} = 68\% \]Since \(SP_1 < CP_1\), it is a loss of 68%.
Correct Option: C) 68% loss
We need to find the measure of \(\angle AEB\) in \(\triangle ABC\). We are given that \(BD \perp AC\), \(\angle DBC = 29°\), and \(\angle CAE = 25°\). We will use the properties of right-angled triangles and the angle sum property of a triangle.
Consider the right-angled triangle \(\triangle BDC\). Since \(BD \perp AC\), we have \(\angle BDC = 90°\).
In \(\triangle BDC\), the sum of angles is \(180°\). So, \(\angle BCD + \angle CBD + \angle BDC = 180°\).
We are given \(\angle CBD = 29°\) and \(\angle BDC = 90°\).
Therefore, \(\angle BCD + 29° + 90° = 180°\).
This simplifies to \(\angle BCD + 119° = 180°\).
So, \(\angle BCD = 180° - 119° = 61°\).
Thus, \(\angle C = 61°\).
Now, consider \(\triangle AEC\). We need to find \(\angle AEC\), which is the same as \(\angle AEB\) since E lies on BC.
In \(\triangle AEC\), the sum of angles is \(180°\). So, \(\angle CAE + \angle ACE + \angle AEC = 180°\).
We are given \(\angle CAE = 25°\) and we found \(\angle ACE = \angle BCD = 61°\).
Therefore, \(25° + 61° + \angle AEC = 180°\).
This simplifies to \(86° + \angle AEC = 180°\).
So, \(\angle AEC = 180° - 86° = 94°\).
The question asks for \(\angle AEB\). \(\angle AEB\) and \(\angle AEC\) are supplementary angles if A, E, B are collinear, but E is on BC, so \(\angle AEB\) is an interior angle of \(\triangle ABE\). The point E is on BC, so \(\angle AEB\) is the same as \(\angle AEC\) if we consider the triangle \(\triangle AEC\) or \(\triangle ABE\).
Let's re-evaluate. \(\angle AEB\) is an exterior angle to \(\triangle AEC\) at E, or an interior angle of \(\triangle ABE\). The question asks for \(\angle AEB\). If E is a point on BC, then \(\angle AEB\) and \(\angle AEC\) form a linear pair. So, \(\angle AEB + \angle AEC = 180°\).
From step 3, we found \(\angle AEC = 94°\).
Therefore, \(\angle AEB + 94° = 180°\).
So, \(\angle AEB = 180° - 94° = 86°\).
B) 86°
To find the weight of the metallic box, we first need to calculate the volume of the metal used. This can be done by finding the difference between the external volume and the internal volume of the box. Once the volume of the metal is known, we can multiply it by the given density (weight per unit volume) to find the total weight in grams, and then convert it to kilograms.
Given external length \((L) = 60 \text{ cm}\)
Given external width \((W) = 40 \text{ cm}\)
Given external height \((H) = 23 \text{ cm}\)
The box is open, meaning it has no lid. The thickness of the metal is \(3 \text{ cm}\).
Internal length \((l) = \text{External length} - 2 \times \text{thickness}\)
\[ l = 60 \text{ cm} - 2 \times 3 \text{ cm} = 60 \text{ cm} - 6 \text{ cm} = 54 \text{ cm} \]Internal width \((w) = \text{External width} - 2 \times \text{thickness}\)
\[ w = 40 \text{ cm} - 2 \times 3 \text{ cm} = 40 \text{ cm} - 6 \text{ cm} = 34 \text{ cm} \]Internal height \((h) = \text{External height} - \text{thickness}\) (since it's open, thickness is only subtracted from the bottom)
\[ h = 23 \text{ cm} - 3 \text{ cm} = 20 \text{ cm} \]Weight of \(1 \text{ cm}^3\) of metal = \(0.5 \text{ grams}\)
Total weight in grams = \(V_{\text{metal}} \times 0.5 \text{ g/cm}^3\)
\[ \text{Weight}_{\text{grams}} = 18480 \text{ cm}^3 \times 0.5 \text{ g/cm}^3 \] \[ \text{Weight}_{\text{grams}} = 9240 \text{ grams} \]Since \(1 \text{ kg} = 1000 \text{ grams}\),
\[ \text{Weight}_{\text{kg}} = \frac{9240}{1000} \text{ kg} \] \[ \text{Weight}_{\text{kg}} = 9.24 \text{ kg} \]Correct Option: B) 9.24
This problem involves profit sharing in a partnership, where profit is distributed in proportion to the product of the capital invested and the time for which it was invested. We will set up equations based on the given information about investments, time periods, and profit shares to find the capital contributed by M.
Let N's investment be \(N_c\) and M's investment be \(M_c\).
Given that M invests ₹39,000 more than N, we have:
\[ M_c = N_c + 39000 \quad (1) \]M invests for 4 months and N invests for 7 months.
The ratio of profits is equal to the ratio of (capital \(\times\) time).
Ratio of M's profit to N's profit \((P_M : P_N)\) is:
\[ \frac{P_M}{P_N} = \frac{M_c \times 4}{N_c \times 7} \]Total profit is ₹4,064.
M's share is ₹508 more than N's share.
So, \(P_M = P_N + 508\).
We also know that \(P_M + P_N = 4064\).
Substitute \(P_M\) in the total profit equation:
\[ (P_N + 508) + P_N = 4064 \]\[ 2P_N + 508 = 4064 \]\[ 2P_N = 4064 - 508 \]\[ 2P_N = 3556 \]\[ P_N = \frac{3556}{2} \]\[ P_N = 1778 \]Now, find M's profit share:
\[ P_M = P_N + 508 \]\[ P_M = 1778 + 508 \]\[ P_M = 2286 \]Now, use the profit ratio equation:
\[ \frac{P_M}{P_N} = \frac{M_c \times 4}{N_c \times 7} \]\[ \frac{2286}{1778} = \frac{4M_c}{7N_c} \]Simplify the fraction \(\frac{2286}{1778}\). Both are divisible by 2:
\[ \frac{1143}{889} = \frac{4M_c}{7N_c} \]Cross-multiply and rearrange to find a relationship between \(M_c\) and \(N_c\):
\[ 1143 \times 7N_c = 889 \times 4M_c \]\[ 8001N_c = 3556M_c \]\[ N_c = \frac{3556}{8001} M_c \]Substitute this expression for \(N_c\) into equation (1):
\[ M_c = \frac{3556}{8001} M_c + 39000 \]\[ M_c - \frac{3556}{8001} M_c = 39000 \]\[ \left(1 - \frac{3556}{8001}\right) M_c = 39000 \]\[ \left(\frac{8001 - 3556}{8001}\right) M_c = 39000 \]\[ \left(\frac{4445}{8001}\right) M_c = 39000 \]\[ M_c = 39000 \times \frac{8001}{4445} \]Perform the calculation:
Notice that \(39000 / 4445\) is approximately 8.77. Let's check if 39000 is divisible by 4445. It is not directly. Let's recheck calculations.
Let's simplify \(\frac{1143}{889}\) further. \(1143 = 3 \times 381 = 3 \times 3 \times 127\). \(889 = 7 \times 127\). So, \[\frac{1143}{889} = \frac{9 \times 127}{7 \times 127} = \frac{9}{7}\].
So, the profit ratio is \[\frac{P_M}{P_N} = \frac{9}{7}\].
Now, back to step 6:
\[ \frac{9}{7} = \frac{4M_c}{7N_c} \]\[ 9 \times 7N_c = 7 \times 4M_c \]\[ 63N_c = 28M_c \]\[ N_c = \frac{28}{63} M_c \]\[ N_c = \frac{4}{9} M_c \]Substitute \(N_c = \frac{4}{9} M_c\) into equation (1):
\[ M_c = \frac{4}{9} M_c + 39000 \]\[ M_c - \frac{4}{9} M_c = 39000 \]\[ \left(1 - \frac{4}{9}\right) M_c = 39000 \]\[ \left(\frac{9 - 4}{9}\right) M_c = 39000 \]\[ \frac{5}{9} M_c = 39000 \]\[ M_c = 39000 \times \frac{9}{5} \]\[ M_c = 7800 \times 9 \]\[ M_c = 70200 \]D) ₹70,200