Question
(a) Sunita is half the age of her mother Geeta. Find their ages.
(i) After 4 years
(ii) Before 3 years
(b) A bus travels at 'v' km per hour. It is going from Daspur to Beespur. After the bus has travelled 5 hours, Beespur is still 20 km away. What is the distance from Daspur to Beespur? Express it using v.
(c) Sarita’s father’s age is 5 years more than 3 times Sarita’s age. What is her father's age?
[4 MARKS]
Answer: Option A
:
(a) Solution: 2 Marks (1 Mark for Geeta and 1 mark for Sunita)
(b) Solution: 1 Mark
(c) Solution: 1 Mark
(a) Let the present age of Sunita’s mother Geeta = 2x years
Present age of Sunita =2x2=x years
(i) After 4 years:
Age of Geeta =(2x+4) years
Age of Sunita =(x+4) years
(ii) Before 3 years
Age of Geeta =(2x−3) years
Age of Sunita =(x−3) years
(b) Speed of the bus = vkmhr i.e. (in 1 hr bus travels v km )
∴ Distance travelled by bus after 5 hrs = 5v km
Distance remaining to Beespur = 20 km
Distance from Daspur to Beespur
= Distance travelled by bus in 5 hrs + Distance remaining
= (5v+20) km
(c) Let Sarita's present age be x years
∴ three times of Sarita's age =3x years
∴ Sarita's father age = 3x + 5 years
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:
(a) Solution: 2 Marks (1 Mark for Geeta and 1 mark for Sunita)
(b) Solution: 1 Mark
(c) Solution: 1 Mark
(a) Let the present age of Sunita’s mother Geeta = 2x years
Present age of Sunita =2x2=x years
(i) After 4 years:
Age of Geeta =(2x+4) years
Age of Sunita =(x+4) years
(ii) Before 3 years
Age of Geeta =(2x−3) years
Age of Sunita =(x−3) years
(b) Speed of the bus = vkmhr i.e. (in 1 hr bus travels v km )
∴ Distance travelled by bus after 5 hrs = 5v km
Distance remaining to Beespur = 20 km
Distance from Daspur to Beespur
= Distance travelled by bus in 5 hrs + Distance remaining
= (5v+20) km
(c) Let Sarita's present age be x years
∴ three times of Sarita's age =3x years
∴ Sarita's father age = 3x + 5 years
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