1. Two numbers are in the ratio 8:3. If the sum of the numbers is 143, find the numbers.
2
2.
of a number is 20 less than the original number. Find the number.
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3. Four-fifths of a number is 10 more than two-thirds of the number. Find the number.
4. Twenty-four is divided into two parts such that 7 times the first part added to 5 times the
second part makes 146. Find each part.
5. Find the number whose fifth part increased by 5 is equal to its fourth part diminished by 5.
6. Three numbers are in the ratio of 4:5:6. If the sum of the largest and the smallest equals
the sum of the third and 55, find the numbers.
7. If 10 be added to four times a certain number, the result is 5 less than five times the
number. Find the number.
8. Two numbers are such that the ratio between them is 3:5. If each is increased by 10, the
ratio between the new numbers so formed is 5 : 7. Find the original numbers.
9. Find three consecutive odd numbers whose sum is 147.
Hint. Let the required numbers be (2x+1), (2x +3) and (2x + 5).
10. Find three consecutive even numbers whose sum is 234.
Hint. Let the required numbers be 2x, (2x +2 ) and (2x + 4).
*1.
11. The sum of the digits of a two-digit number is 12. If the new number formed by reversing
the digits is greater than the original number by 54, find the original number. Check your
solution.
12. The digit in the tens place of a two-digit number is three times that in the units place. If the
digits are reversed, the new number will be 36 less than the original number. Find the
original number. Check your solution.
Answers
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Answer:
1.Given two numbers are in the ratio 8:3
so let the number be 8x and 3x
According to the question we can write as 8x+3x=143
Which implies 11x=143
Again by transpoing x=143/11
Therfore x=13
So the number are 8x=8(13)=104 and 3x=3(3)=39
2.Let the original number be x.
According to the question we can write as (2/3)x+20x
On rearranging x−(2/3)x=20
Now taking the L.C.M od 1 and 3 is 3
(3x−2x)/3=20
x/3=20
Again by transposing x=60
So the original number is 60
3.Let the number required is x
According to the question we can write as
(4/5)x−10=(2/3)x
On rearranging,](4/5)x−(2/3)x=10
Now taking the L.C.M of 3 and 5 is 15
(12x−10x)/15=10
On cross multiplication,
2x=150
x=150/2=75
So the required number is 75
4.Let the two parts of the number be x and (24-x)
According to the question we can write as7x+5(24−x)=146
7x + 120 - 5x = 146
2x = 146 - 120
2x = 26
x = 26/2
⇒x=13
And also (24−x)=24−13=11
So the parts are 13 and 11
5.Let the number be x.
According to the question we can write as (1/5)x+5=(1/4)x−5
On rearranging
(1/5)x−(1/4)x=−5−5
(1/5)x−(1/4)x=−10
Now taking the L.C.M we get
(4x−5x)/20=−10
Again by transposing
X=200
6.Let the numbers be 4x, 5x and 6x
According to the question we can write as
6x+4x=5+55
On rearring we can write as
10x−5x=55
5x=55
x=55/5=11
So the numbers are 44, 55 and 66
7.Let the numbers be x
so , 10 + 4x = 5x - 5
⇒10+5=5x−4x
⇒15=x
∴ the numbers is 15.
8.Let the numbers be 3x and 5x. According to the question we can write as (3x+10) / (5x+10) = (5/7) On cross multiplying we get 7(3x+10) = 5(5x+10) 21x+70=25x+50 On rearranging or transposing 70-50=25x-21x 4x=20 x=20/4=5 So the numbers are 15 and 25
9.let 3 consecutive add numbers be
x-2,x, x+2
As, x-2+x+x+2=147
⇒ 3x=147
⇒x=49
∴ 3 numbers are 47,49,51
10.Let the three numbers be x-2, x, x+2
so, x - 2 + x + x + 2 = 234
3x = 234
x = 78
numbers are = 76, 78, 80
11.Let the digits be x and y, so the number will be = (10x+y), on reversing the digits, the new number will be = (10y+x) According to the question we can write as x + y=12 and also we can write as 10y+x-10x-y=54 Which implies 9y-9x=54 y-x=54/9 y-x=6 y=6+x Now on substituting this in x + y=12 we get x+6+x=12 2x+6=12 2x=12-6 x=6/2=3 Now y=6+x=6+3=9 So the number is 39 To check: digit sum=3+9=12 Reversing the digit numbers becomes 93 and 93-39=54
12. Let the one's digit be y and tens digit be x,
Number = 10x + y
Then,x=3y⋯(i)
Reversed number = 10y + x
A.t.Q :- (10x+y)−(10y+x)=36 Put x = 3y in eq. (i)
⇒9x−9y=36
⇒x−y=4⋯(ii)
⇒3y−y=4
∴2y=4 x=3y ∴x=6
y=2
∴ Number = 62
Step-by-step explanation:
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