Equivalent Expressions MCQ Quiz in हिन्दी - Objective Question with Answer for Equivalent Expressions - मुफ्त [PDF] डाउनलोड करें

To simplify \(10x + 25\) by factoring, we look for the greatest common factor of the terms. The GCF of 10 and 25 is 5. We factor out 5 from the expression: \(10x + 25 = 5(2x + 5)\). To verify, distribute the 5: \(5 \times 2x = 10x\) and \(5 \times 5 = 25\). The expression \(5(2x + 5)\) is correct. Therefore, the correct option is 4.

To find the number of students not in the math club, we first calculate the number of students in the club. \(40\%\) of 150 is \(0.4 \times 150 = 60\). Therefore, the number of students in the math club is 60. The number of students not in the club is \(150 - 60 = 90\). Thus, option 2 is correct. The other options do not correctly subtract the number in the club from the total.

Let the width of the garden be \(w\) meters. Then, the length is \(w + 3\) meters. The area \(A\) is given by \(A = \text{length} \times \text{width} = (w + 3)w = 70\). Expanding gives \(w^2 + 3w = 70\). Rearranging, we get \(w^2 + 3w - 70 = 0\). Factoring, we need numbers that multiply to -70 and add to 3, which are 10 and -7. So, \(w^2 + 3w - 70 = (w - 7)(w + 10) = 0\). Solving \(w - 7 = 0\) or \(w + 10 = 0\), we find \(w = 7\) or \(w = -10\). Since width cannot be negative, the width is 7 meters. Therefore, option 2 is correct.

To factor \(x^2 - 7x + 12\), we need two numbers whose product is 12 and sum is -7. The numbers -3 and -4 satisfy these conditions because \(-3 \times -4 = 12\) and \(-3 + -4 = -7\). Therefore, the factored form is \((x - 3)(x - 4)\). Option 2 is correct. Options 1, 3, and 4 do not satisfy both conditions simultaneously.

To solve \(21y^2 - 9y^2\), subtract the coefficients of the like terms, \(y^2\). This results in \((21 - 9)y^2 = 12y^2\). Thus, the correct answer is option 3. Option 1 incorrectly combines exponents. Option 2 incorrectly treats the exponents. Option 4 incorrectly adds coefficients.

Adding the polynomials \(7p^3 + 15p^3\) involves adding the coefficients of the like terms, \(p^3\): \(7 + 15 = 22\). Therefore, the resulting polynomial is \(22p^3\), making option 1 correct. Option 2 multiplies the coefficients. Option 3 incorrectly subtracts. Option 4 incorrectly adds exponents.

To factor \(x^2 - 5x - 14\), we need two numbers that multiply to -14 and add to -5. The numbers -7 and +2 work because \(-7 \times 2 = -14\) and \(-7 + 2 = -5\). Thus, the expression factors to \((x - 7)(x + 2)\). Therefore, option 3 is correct. Options 1, 2, and 4 do not meet the criteria for both multiplication and addition.

For the expression \(x^2 + 6x + 8\), we seek two numbers whose product is 8 and sum is 6. These numbers are 2 and 4 because \(2 \times 4 = 8\) and \(2 + 4 = 6\). Therefore, the factored form is \((x + 2)(x + 4)\). Hence, option 1 is correct. Options 2, 3, and 4 do not accurately reflect the necessary factor pairs.

To find the total production, add the expressions \(25x^3 + 18x^3\). The coefficients of the like terms, \(x^3\), are added: \(25 + 18 = 43\). Therefore, the total production is \(43x^3\), making option 1 correct. Option 2 incorrectly adds exponents. Option 3 is incorrect due to subtraction. Option 4 incorrectly treats the exponents.

Let \(p\) represent the number of pens and \(n\) represent the number of notebooks. We have two equations: \(p + n = 10\) (total items) and \(2p + 5n = 34\) (total cost). Solving the first equation for \(n\) gives \(n = 10 - p\). Substituting in the second equation: \(2p + 5(10 - p) = 34\). Simplifying: \(2p + 50 - 5p = 34\) leads to \(-3p + 50 = 34\). Solving for \(p\), we get \(-3p = -16\) and \(p = \dfrac{16}{3}\). Since \(p\) must be an integer, we re-evaluate our approach. Checking logical integer values, we find if \(p = 6\), then \(n = 4\) fits both \(p + n = 10\) and \(2p + 5n = 34\) as \(12 + 20 = 34\). Thus, option 3 is correct.