3
1. Two numbers are in the ratio 8 : 3. If the sum of the numbers is 143, find the numbers.
EXERCISE 8B
of a number is 20 less than the original number. Find the number.
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).
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 th
original number. Check your solution.
Answers
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 be x.
Four-fifths of the number is (4/5)x
Two-thirds of the number is (2/3)x.
10 more than two-thirds of the number is (2/3)x +10.
Hence the given problem will be (4/5)x =(2/3)x +10
I.e., 4x/5 = 2x/3 +10
Multiply through out by 15 which is the LCM of 5and 3 we get
12x=10x+150
12x-10x = 150
2x = 150
x = 75
Hence the required number is 75.
4.
First part be x and the second part y.
we have
7x=5y+3
7x-5y=3….1
x+y=24……2
multiplying equation 2 by 5
5x+5y=120…..3
adding equation 1&3
12x=123
x=123/12=10.25
substituting In 2
10.25+y=24
y=24–10.25=13.75.
first part is 10.25.
second part is 13.75