11 - 20 of 62 Questions
# | Question | Ans |
---|---|---|
11. |
Solve the equation \( 3x^2 − 4x − 5 = 0 \) A. x = 1.75 or − 0.15 B. x = 2.12 or − 0.79 C. x = 1.5 or − 0.34 D. x = 2.35 or −1.23 Detailed SolutionUsing the quadratic formula, we have\(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\) From the equation \(3x^{2} - 4x - 5 = 0\), a = 3, b = -4 and c = -5. \(\therefore x = \frac{-(-4) \pm \sqrt{(-4)^{2} - 4(3)(-5)}}{2(3)}\) = \(\frac{4 \pm \sqrt{16 + 60}}{6}\) = \(\frac{4 \pm \sqrt{76}}{6}\) = \(\frac{4 \pm 8.72}{6}\) = \(\frac{4 + 8.72}{6} \text{ or } \frac{4 - 8.72}{6}\) = \(\frac{12.72}{6} \text{ or } \frac{-4.72}{6}\) \(x = \text{2.12 or -0.79}\) There is an explanation video available below. |
|
12. |
The first and last term of a linear sequence (AP) are 6 and 10 respectively. If the sum of the sequence is 40. Find the number of terms A. nth = 3 B. nth = 4 C. nth = 5 D. nth = 6 Detailed Solutionnth term of a linear sequence (AP) = a+(n − 1)dfirst term = 6, last term = 10 sum − 40 i.e. a = 6, l = 10, S = 40 S\(_{n}\)= n/2(2a + (n − 1)d or Sn = ÷2 (a + l) S\(_{n}\) = n/2(a + l) 40 = n/2(6 + 10) 40 = 8n 8n = 40 8n = 40 n = 40/8 = 5 The number of terms = 5 There is an explanation video available below. |
|
13. |
Find the equation of a line which is form origin and passes through the point (−3, −4) A. y = \( \frac{3x}{4} \) B. y = \( \frac{4x}{3} \) C. y = \( \frac{2x}{3} \) D. y = \( \frac{x}{2} \) Detailed SolutionThe slope of the line from (0, 0) passing through (-3, -4) = \(\frac{-4 - 0}{-3 - 0}\)= \(\frac{4}{3}\) Equation of a line is given as \(y = mx + b\), where m = slope and b = intercept. To get the value of b, we use a point on the line, say (0, 0). \(y = \frac{4}{3} x + b\) \(0 = \frac{4}{3}(0) + b\) \(b = 0\) The equation of the line is \(y = \frac{4}{3} x\) There is an explanation video available below. |
|
14. |
The amount A to which a principal P amounts at r% compound interest for n years is given by the formula A = P(1 + (r ÷ 100)\(^n\). Find A, if P = 126, r = 4 and n = 2. A. N132.50K B. N136.30K C. N125.40K D. N257.42K Detailed Solution\( A = P \left(1 + \frac{r}{100}\right)^n \)Where P = 126, r = 4,n = 2 A=126 \( \left(1 + \frac{4}{100}\right)^2 \text{Using LCM} \) =126 \( \left(\frac{100+4}{100}\right)^2 = 126 \left(\frac{104}{100}\right)^2 \) =126 \( \left(1.04^2 \right) \) = 126 * 1.04 * 1.04 =136.28 A = 136.30 (approx.) The Amount A, = N136.30k There is an explanation video available below. |
|
15. |
Make x the subject of the equation A. x = 5[(s − 2) ÷ t] - 3/5y B. x = 25[(s − 2) ÷ t] − 3ty C. x = [1 ÷ (s − 2)3ty] D. x = [5(s − 2) 2 ÷ 3ty] × t |
|
16. |
If log520 = x, find x A. 1.761 B. 1.354 C. 1.861 D. 2.549 |
|
17. |
Find at which rate per annum simple interest N525 will amount to N588 in 3 years. A. 3% B. 2% C. 5% D. 4% |
|
18. |
Given that A = {1, 5, 7} A. {1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 15} B. {1, 2, 3, 5, 6, 8, 12, 15} C. {2, 4, 5, 9, 12, 15} D. {1, 5, 6, 7, 8, 9, 12, 15} |
|
19. |
The extension of a stretched string is directly proportional to its tension. If the extension produced by a tension of 8 Newton’s is 2cm, find the extension produced by a tension of 12 newton’s. A. 2 B. 1 D. 3 Detailed SolutionLet the extension be E and the tension be T.Then \(E \propto T\) \(E = kT\) when T = 8N, E = 2cm \(2 = k \times 8\) \(k = \frac{2}{8} = 0.25\) \(\therefore E = 0.25T\) when T = 12N, \(E = 0.25 \times 12 = 3cm\) There is an explanation video available below. |
|
20. |
Factorize \( x^2 − 2x − 15 \) A. (x + 3)2 B. (x + 5)(x − 3) C. (x − 5)2 D. (x − 5)(x + 3) |
11. |
Solve the equation \( 3x^2 − 4x − 5 = 0 \) A. x = 1.75 or − 0.15 B. x = 2.12 or − 0.79 C. x = 1.5 or − 0.34 D. x = 2.35 or −1.23 Detailed SolutionUsing the quadratic formula, we have\(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\) From the equation \(3x^{2} - 4x - 5 = 0\), a = 3, b = -4 and c = -5. \(\therefore x = \frac{-(-4) \pm \sqrt{(-4)^{2} - 4(3)(-5)}}{2(3)}\) = \(\frac{4 \pm \sqrt{16 + 60}}{6}\) = \(\frac{4 \pm \sqrt{76}}{6}\) = \(\frac{4 \pm 8.72}{6}\) = \(\frac{4 + 8.72}{6} \text{ or } \frac{4 - 8.72}{6}\) = \(\frac{12.72}{6} \text{ or } \frac{-4.72}{6}\) \(x = \text{2.12 or -0.79}\) There is an explanation video available below. |
|
12. |
The first and last term of a linear sequence (AP) are 6 and 10 respectively. If the sum of the sequence is 40. Find the number of terms A. nth = 3 B. nth = 4 C. nth = 5 D. nth = 6 Detailed Solutionnth term of a linear sequence (AP) = a+(n − 1)dfirst term = 6, last term = 10 sum − 40 i.e. a = 6, l = 10, S = 40 S\(_{n}\)= n/2(2a + (n − 1)d or Sn = ÷2 (a + l) S\(_{n}\) = n/2(a + l) 40 = n/2(6 + 10) 40 = 8n 8n = 40 8n = 40 n = 40/8 = 5 The number of terms = 5 There is an explanation video available below. |
|
13. |
Find the equation of a line which is form origin and passes through the point (−3, −4) A. y = \( \frac{3x}{4} \) B. y = \( \frac{4x}{3} \) C. y = \( \frac{2x}{3} \) D. y = \( \frac{x}{2} \) Detailed SolutionThe slope of the line from (0, 0) passing through (-3, -4) = \(\frac{-4 - 0}{-3 - 0}\)= \(\frac{4}{3}\) Equation of a line is given as \(y = mx + b\), where m = slope and b = intercept. To get the value of b, we use a point on the line, say (0, 0). \(y = \frac{4}{3} x + b\) \(0 = \frac{4}{3}(0) + b\) \(b = 0\) The equation of the line is \(y = \frac{4}{3} x\) There is an explanation video available below. |
|
14. |
The amount A to which a principal P amounts at r% compound interest for n years is given by the formula A = P(1 + (r ÷ 100)\(^n\). Find A, if P = 126, r = 4 and n = 2. A. N132.50K B. N136.30K C. N125.40K D. N257.42K Detailed Solution\( A = P \left(1 + \frac{r}{100}\right)^n \)Where P = 126, r = 4,n = 2 A=126 \( \left(1 + \frac{4}{100}\right)^2 \text{Using LCM} \) =126 \( \left(\frac{100+4}{100}\right)^2 = 126 \left(\frac{104}{100}\right)^2 \) =126 \( \left(1.04^2 \right) \) = 126 * 1.04 * 1.04 =136.28 A = 136.30 (approx.) The Amount A, = N136.30k There is an explanation video available below. |
|
15. |
Make x the subject of the equation A. x = 5[(s − 2) ÷ t] - 3/5y B. x = 25[(s − 2) ÷ t] − 3ty C. x = [1 ÷ (s − 2)3ty] D. x = [5(s − 2) 2 ÷ 3ty] × t |
16. |
If log520 = x, find x A. 1.761 B. 1.354 C. 1.861 D. 2.549 |
|
17. |
Find at which rate per annum simple interest N525 will amount to N588 in 3 years. A. 3% B. 2% C. 5% D. 4% |
|
18. |
Given that A = {1, 5, 7} A. {1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 15} B. {1, 2, 3, 5, 6, 8, 12, 15} C. {2, 4, 5, 9, 12, 15} D. {1, 5, 6, 7, 8, 9, 12, 15} |
|
19. |
The extension of a stretched string is directly proportional to its tension. If the extension produced by a tension of 8 Newton’s is 2cm, find the extension produced by a tension of 12 newton’s. A. 2 B. 1 D. 3 Detailed SolutionLet the extension be E and the tension be T.Then \(E \propto T\) \(E = kT\) when T = 8N, E = 2cm \(2 = k \times 8\) \(k = \frac{2}{8} = 0.25\) \(\therefore E = 0.25T\) when T = 12N, \(E = 0.25 \times 12 = 3cm\) There is an explanation video available below. |
|
20. |
Factorize \( x^2 − 2x − 15 \) A. (x + 3)2 B. (x + 5)(x − 3) C. (x − 5)2 D. (x − 5)(x + 3) |