Answer:
(a) Let Santa's present age = y yr (i) Sarita's age after 5 yr from now \[=(y\text{ }+5)\text{ }yr\] (ii) Santa's age 3 yr back \[=(y-3)\text{ }yr\] (iii) Given, Sarita's grandfather is 6 times of her age So, Sarita's grandfather age \[=6\times \] Sarita's present age \[=6\times y=6yyr\] (iv) Given, grandmother is 2 yr younger than grandfather. Age of grandfather = 6y yr [from (iii)] \[\therefore \] Age of grandmother = [Grandfather's age \[-2\]] yr \[=(6y-2)yr\] (v) Given, Sarita's father's age is 5 yr more than 3 times Sarita's age. Sarita's present age = y yr 3 times of Sarita's age = 3y yr \[\therefore \] Sarita's father age \[=3\times \] Sarita's age + 5 \[=(3y+5)\,yr\] (b) Given, the breadth of the hall = b m Now, according to the question, Length of hall = 4 m less than 3 times the breadth of the hall = (3 \[\times \] breadth of hall) ? 4 \[=(3\times b-4)m\] \[=(3b-4)\,m\] Hence, if breadth is b m, then length of hall is \[(3b-4)\,m\]. (c) Given, the height of a rectangular box = h cm According to the question, Length of the box = 5 times the height \[=5h\,cm\] and breadth of the box = 10 cm less than the length = 10 cm less than \[5h\] \[=(5h-10)\,cm\] (d) Given, the Meena is at step = s Also, Beena is 8 steps ahead. \[\therefore \] Beena is at step = s + 8 and Leena is 7 steps behind \[\therefore \] Leena is at step \[=s-7\] Now, total number of steps = 10 less than 4 times Meena's steps \[=4\times \] Meena's steps \[-10=4s-10\] Hence, total number of steps using scan be expressed as \[(4s-10)\]. (e) Given, speed of the bus \[=\upsilon \,k/h\] i.e. Distance travelled by bus in \[1h=\upsilon \,km\] \[\therefore \] Distance travelled by bus is 5 h from Daspur \[=5\upsilon \,km\] After travelling 5 hours, Beespur is still 20 km away. So, total distance from Daspur to Beespur \[=(5\upsilon +20)\,km\]. Hence, distance from Daspur to Beespur in terms of v can be expressed as \[(5\upsilon +20)km\].
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