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The age of a father is equal to the sum of the ages of his four children. After 20 years, the sum of the ages of the children will be twice the age of the father. Find the age of the father.
Let the father's present age be F and the sum of the ages of his four children be S, so F = S. In 20 years, the father's age becomes F + 20, while each child ages 20 years, increasing their sum by 4 × 20 = 80 to S + 80. The condition gives S + 80 = 2(F + 20). Substituting S = F yields F + 80 = 2F +Read more
Let the father’s present age be F and the sum of the ages of his four children be S, so F = S. In 20 years, the father’s age becomes F + 20, while each child ages 20 years, increasing their sum by 4 × 20 = 80 to S + 80. The condition gives S + 80 = 2(F + 20). Substituting S = F yields F + 80 = 2F + 40, resulting in F = 40 years.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 13 Two Variables, One Line Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-13/
See lessA train moving with uniform speed for a certain distance takes 6 hours less if its speed is increased by 6 km/hour. It would have taken 6 hours more had its speed been decreased by 4 km/hour. Find the distance travelled and the speed of the train.
Let uniform speed be s km/h and scheduled time be t hours, with distance d = st. From (s + 6)(t − 6) = st, expansion gives −6s + 6t − 36 = 0 or t − s = 6. From (s − 4)(t + 6) = st, expansion gives 6s − 4t − 24 = 0 or 3s − 2t = 12. Substituting t = s + 6 into the second equation yields s = 24 km/h, tRead more
Let uniform speed be s km/h and scheduled time be t hours, with distance d = st. From (s + 6)(t − 6) = st, expansion gives −6s + 6t − 36 = 0 or t − s = 6. From (s − 4)(t + 6) = st, expansion gives 6s − 4t − 24 = 0 or 3s − 2t = 12. Substituting t = s + 6 into the second equation yields s = 24 km/h, t = 30 hours and distance d = 24 × 30 = 720 km.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 13 Two Variables, One Line Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-13/
See lessIn a cyclic quadrilateral ABCD, ∠A = (x + 7)°, ∠B = (y + 8)°, ∠C = (3y + 23)° and ∠D = (4x + 12)°. Find all four angles of the cyclic quadrilateral.
In a cyclic quadrilateral, opposite angles are supplementary. Thus, ∠A + ∠C = (x + 7) + (3y + 23) = 180°, which simplifies to x + 3y = 150. Also, ∠B + ∠D = (y + 8) + (4x + 12) = 180°, giving 4x + y = 160. Multiplying the first equation by 4 and subtracting the second eliminates x, yielding 11y = 440Read more
In a cyclic quadrilateral, opposite angles are supplementary. Thus, ∠A + ∠C = (x + 7) + (3y + 23) = 180°, which simplifies to x + 3y = 150. Also, ∠B + ∠D = (y + 8) + (4x + 12) = 180°, giving 4x + y = 160. Multiplying the first equation by 4 and subtracting the second eliminates x, yielding 11y = 440, so y = 40 and x = 30. The angles are 37°, 48°, 143° and 132°.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 13 Two Variables, One Line Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-13/
See lessThe sum of the digits of a two-digit number is 15. The number obtained by interchanging the digits exceeds the given number by 9. Find the number.
Let the tens digit be x and the units digit be y, making the original number 10x + y. The sum of digits gives x + y = 15. Reversing the digits gives 10y + x. The difference condition (10y + x) − (10x + y) = 9 reduces to 9y − 9x = 9 or y − x = 1. Adding x + y = 15 and y − x = 1 yields 2y = 16, givingRead more
Let the tens digit be x and the units digit be y, making the original number 10x + y. The sum of digits gives x + y = 15. Reversing the digits gives 10y + x. The difference condition (10y + x) − (10x + y) = 9 reduces to 9y − 9x = 9 or y − x = 1. Adding x + y = 15 and y − x = 1 yields 2y = 16, giving y = 8 and x = 7. The number is 78.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 13 Two Variables, One Line Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-13/
See lessAt a certain time, Jacob notices that his digital watch reads a minutes after two o’clock. Fifteen minutes later, it reads b minutes after three o’clock. He noticed that a is six times greater than b. What time was it when he looked at his watch for the second time?
The time elapsed from a minutes past two to b minutes past three is 15 minutes. Expressing both in minutes after two o'clock gives the relation (60 + b) − a = 15, which simplifies to a − b = 45. Given that a = 6b, substituting gives 6b − b = 45, leading to 5b = 45 or b = 9. Therefore, Jacob's watchRead more
The time elapsed from a minutes past two to b minutes past three is 15 minutes. Expressing both in minutes after two o’clock gives the relation (60 + b) − a = 15, which simplifies to a − b = 45. Given that a = 6b, substituting gives 6b − b = 45, leading to 5b = 45 or b = 9. Therefore, Jacob’s watch read 3:09 at the second viewing.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 13 Two Variables, One Line Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-13/
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