InterviewSolution
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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects. |
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Answer» Let the marks scored by Shefali in Maths be ‘x’ and marks scored in English by Shefali will be (30 – x). (x + 2) (30 – x – 3) = 210 (x + 2)(27 – x) = 210 27x – x2 + 54 – 2x = 210 -x2 + 25x + 54 = 210 -x2 + 25x + 54 – 210 = 0 -x2 + 25x – 156 = 0 x2 – 25x + 156 = 0 x2 – 13x – 12x + 156 = 0 x(x – 13) – 12 (x- 13) = 0 (x – 13) (x – 12) = 0 If x – 13 = 0, then x = 13 If x – 12 = 0, then x = 12 ∴ x = 13 OR x = 12 ∴ Shefali scores in Maths, x = 13 Shefali’s marks in English = 30 – x = 30 – 13 = 17. |
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| 2. |
In a class test, the sum of the marks obtained by P in mathematics and science is 28. Had he got 3 more marks in mathematics and 4 marks less in science, the product of marks obtained in the two subjects would have been 180. Find the marks obtained by him in the two subjects separately. |
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Answer» Let the marks obtained by P in mathematics and science be x and (28-x)respectively. According to the given condition, Hence, he obtained 12 marks in mathematics and 16 marks in science or 9 marks in mathematics and 19 marks in science. |
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| 3. |
In a class test, the sum of the marks obtained by P in mathematics and science is 28. Had he got 3 more marks in mathematics and 4 marks less in science, the product of marks obtained in the two subjects would have been 180. Find the marks obtained by him in the two subjects separately. |
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Answer» Let the marks obtained by P in mathematics and science be x and (28 - x) respectively. According to the given condition, (x + 3)(28 - x - 4) = 180 When x = 12, 28 - x = 28 - 12 = 16 When x =9, 28 - x = 28 - 9= 19 Hence, he obtained 12 marks in mathematics and 16 marks in science or 9 marks in mathematics and 19 marks in science. |
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| 4. |
Check whether the equation 6x2 – 7x + 2 = 0 has real roots, and if it has, find them by the method of completing the squares. |
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Answer» The discriminant = b2 – 4ac = 49 – 4 × 6 × 2 = 1 > 0 So, the given equation has two distinct real roots. Now, 6x2 – 7x + 2 = 0 i.e., 36x2 – 42x + 12 = 0 i.e., 6x - (7/2)2+ 12 - 49/4 = 0 i.e., 6x - (7/2)2- (1/2)2= 0 or (6x - 7/2)2= (1/2)2 The roots are given by 6x - 7/2 = ±1/2 i.e., 6x = 4, 3 i.e., x = 2/3, 1/2. |
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| 5. |
State whether the quadratic equation(x + 4)2– 8x = 0have two distinct real roots. Justify your answer. |
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Answer» The equation (x + 4)2– 8x = 0 has no real roots. Simplifying the above equation, x2+ 8x + 16 – 8x = 0 x2+ 16 = 0 D = b2– 4ac =(0) – 4(1) (16) < 0 Hence, the roots are imaginary. |
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| 6. |
A train travels at a certain average speed for a distanced of 54 km and then travels a distance of 63 km at an average speed of 6 km/hr more than the first speed. If it takes 3 hours to complete the total journey, what is its first speed? |
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Answer» Let the first speed of the train be x km/h. Hence, the first speed of the train is 36 km/h. |
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| 7. |
Had Ajita scored 10 more marks in her mathematics test out of 30 marks, 9 times these marks would have been the square of her actual marks. How many marks did she get in the test? |
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Answer» Let her actual marks be x Therefore, 9 (x +10) = x2 i.e., x2 – 9x – 90 = 0 i.e., x2 – 15x + 6x – 90 = 0 i.e., x(x – 15) + 6(x –15) = 0 i.e., (x + 6) (x –15) = 0 Therefore, x = – 6 or x = 15 Since x is the marks obtained, x ≠ – 6. Therefore, x = 15. So, Ajita got 15 marks in her mathematics test. |
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| 8. |
Check whether the equation 8x2+ 2x – 3 = 0 has real roots, and if it has, find them. |
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Answer» To check whether the quadratic equation has real roots or not, we need to check the discriminant value i.e., D = b2- 4ac Given,8x2+ 2x – 3 = 0 ∴D = 22- 4(8) (-3) ⇒D = 4 + 96 > 0 Hence, the roots are real and distinct. To find the roots, use the formula,
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| 9. |
A train travels at a certain average speed for a distance of 63 km and then travels a distance of 72 km at an average speed of 6 km/h more than its original speed. If it takes 3 hours to complete the total journey, what is its original average speed? |
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Answer» Let its original average speed be x km/h. Therefore, 63/x + 72/(x + 6) = 3 7/x + 8/(x + 6) = 3/9 = 1/3 (7(x + 6) + 8x)/(x (x + 6)) = 1/3 i.e., 21 (x + 6) + 24x = x (x + 6) i.e., 21x + 126 + 24x = x2 + 6x i.e., x2 – 39x – 126 = 0 i.e., (x + 3) (x – 42) = 0 i.e., x = – 3 or x = 42 Since x is the average speed of the train, x cannot be negative. Therefore, x = 42. So, the original average speed of the train is 42 km/h. |
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