(3) The sum of father's age and twice the age of his son is 70. If we double
of the father and add it to the age of his son the sum is 95. Find their presente
(4) The denominator of a fraction is 4 more than twice its numerator. Dersom
becomes 12 times the numerator, if both the numerator and the denominator
reduced by 6. Find the fraction.
(5) Two types of boxes A, B are to be placed in a truck having capacity of 10
When 150 boxes of type A and 100 boxes of type B are loaded in the trad
weighes 10 tons. But when 260 boxes of type A are loaded in the truck, it can
accommodate 40 boxes of type B, so that it is fully loaded. Find the weight of
type of box.
(6) Out of 1900 km, Vishal travelled some distance by bus and some by aerop
Bus travels with average speed 60 km/hr and the average speed of aeropla
700 km/hr. It takes 5 hours to complete the journey. Find the distance,
travelled by bus.
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Answers
Question 3
Father's Present age = 40
Son's Present Age = 15
Let us take the father‘s age as “X” and son’s age as “Y”.
Now as given in question the sum of father’s age and twice the son’s age,
X + 2Y = 70 _______ (1)
As per given data, doubling the age of the father and adding it to the son’s age sum is 95,
2X + Y = 95 _______ (2)
Hence to find their present age of X and Y, we equate
(X + 2Y) = 70 and 2X + Y = 95.
To solve (2) equation, substitute X = 70 – 2Y in X of 2X + Y =95,
2(70 – 2Y) + Y = 95
140 – 4Y + Y = 95
Y = 15
Substitute Y = 15 in X = 70 – 2Y, we get
X = 40
_____________________________________
Question 4
Fraction = 7/18
Let the fraction be x/y.
given y = 2x + 4
2x − y = −4
multiply the equation with 6
12x − 6y = −24 ................... (1)
(y − 6) = 12(x − 6)
12x − y = 66 ........................(2)
subtract (1) and (2)
y = 18 x = 7
Therefore,
Fraction = 7/18
______________________________________
Question 5
weight of each A type box = 30 kg
weight of each B type box = 55 kg
Let us assume,
weight of each A type box = x tons
weight of each B type box = y tons
according to the problem given ,
150 boxes of type A and 10 boxes of type
B weighs = 10 tons
150x + 100y = 10
15x + 10y = 1 ---( 1 )
260 boxes of type A and 40 boxes of
type B weighs = 10tons
260x + 40 y = 10
26x + 4y = 1 ---( 2 )
multiply equation ( 1 ) with 2 and equation
( 2 ) with 5 , we get
30x + 20y = 2 ---( 3 )
130x + 20y = 5 --( 4 )
subtract equation ( 3 ) from equation ( 4 )
we get
x = 3/100
put x in equation ( 3 ). we get
y = 11/200 tons
Therefore ,
weight of each type a box = x = 3/100 tons
x = ( 3/100 ) × 1000 kg
x = 30 kg
weight of each type B box = y = 11/200tons
y = ( 11/200 ) × 1000 kg
= 55 kg
______________________________________
Question 6
Distance traveled by bus = 150 km
Let the distance traveled by bus = x
Time taken traveling by bus = x/60
Distance traveled by plane = (1900-x)
Time taken traveling by plane = (1900-x)/700
Total time = x/60 + (1900-x)/700 = 5
x/60 – x/700 = 16/7
8/525x = 16/7
x = 150
Therefore, the distance traveled by bus = 150km