Exercise 2.2
1. Two-thirds of a number is 5 more than half of it. What is the number?
2. Fifteen years from now Ravi's age will be four times his present age. What is the present age of Ravi?
3. Ram has 3 times as many 2 coins as he has 5 coins. If he has in all a sum of 77, how many coins
of 2 does he have?
4. The sum of three consecutive numbers is 156. Find the number which is a multiple of 13 out of these
numbers
5. Divide 63 into two parts such that one part is
of the other
6. Find a number whose fifth part increased by 30 is equal to its fourth part decreased by 30.
Answers
Answer:
Which class is this questions
Step-by-step explanation:
Solutions :-
1)
Let the number be X
Half of the number = X/2
Two-third of the number = 2X/3
According to the given problem,
Two-thirds of a number is 5 more than half of it.
=> 2X/3 = (X/2)+5
=> (2X/3)-(X/2) = 5
LCM of 3 and 2 = 6
=> (4X-3X)/6 = 5
=> X/6 = 5
=> X = 5×6
=> X = 30
The required number = 30
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2)
Let the present age of Ravi be X years
Fifteen years from now Ravi's age will be (X+15) years
Four times his present age = 4X years
According to the given problem,
Fifteen years from now Ravi's age will be four times his present age.
=> X+15 = 4X
=> 15 = 4X-X
=> 15 = 3X
=> 3X = 15
=> X = 15/3
=> X = 5
The present age of Ravi is 5 years
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3)
Let the number of 5 coins Ram has be X
The value of 5 coins = 5X
Then the number of 2 coins = 3X
The value of 2 coins = 2×3X = 6X
Their sum = 5X+6X = 11X
According to the given problem,
The sum of all = 77
=> 11X = 77
=> X = 77/11
=> X = 7
Number of 5 coins= 7
Number of 2 coins = 3×7 = 21
The number of 2 coins = 21
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4)
Let the three consecutive numbers be X,X+1,X+2
Their sum = X+X+1+X+2 = 3X+3
According to the given problem,
The sum of three consecutive numbers is 156.
=> 3X+3 = 156
=> 3X = 156-3
=> 3X = 153
=> X = 153/3
=>X = 51
Now, X+1 = 51+1=52
and X+2 = 51+2 = 53
The three consecutive numbers = 51,52,53
52= 13×4
the number which is a multiple of 13 out of these numbers = 52
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5)
Question is insufficient
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6)
Let the number be X
It's fifth part = X/5
It is increased by 30 = (X/5)+30
It's fourth part = X/4
It's fourth part is decreased by 30=
(X/4)-30
According to the given problem
A number whose fifth part increased by 30 is equal to its fourth part decreased by 30.
=> (X/5)+30= (X/4)-30
=> (X/5) - (X/4) = -30 -30
=> (X/5)-(X/4) = -60
LCM of 5 and 4 = 20
=> (4X-5X)/20 = -60
=> -X/20 = -60
=> X/20 = 60
=> X = 60×20
=> X = 1200
The required number = 1200
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