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S.chand books class 8 maths solution chapter playing with numbers

PLAYING WITH NUMBERS

EXERCISE 5 (A)


QUESTION 1

The sum of the digits of a 2 digit number is 9.The numbers is 6 times the unit digit.Find the number.

Sol :

The sum of two digit number is 9. Where u is unit digit and t is ten’s place digit .
t + u = 9
u = ( 9 – t )

i.e. 23 can be written as 20 + 3 = 10*2 + 3

The number is 6 times the unit digit.

Let the original number be tu then can be written as

10t + u = 6u
10t + u – 6u = 0
10t – 5u = 0
10t – 5(9-t) = 0

10t – 45 + 5t = 0
10t + 5t = 45
15t = 45
t = 45/15
t = 3 is the tens digit
then
9-3 = 6 is units digit

Then the original number is 36

 


QUESTION 2

The sum of the digits of a 2 digit number is 7.If the digits are reserved, the new number increased by 3 less than 4 times the original number. Find the original number.

Sol :

The sum of two digit number is 7. Where u is unit digit and t is ten’s place digit of original number
t + u = 7
u = ( 7 – t ) …(1)

When digits are reversed  , new number is 3 less than 4 times original number  .

10t + u = 4(10u + t) – 3

10t + u = 40u + 4t -3

10t – 4t + u – 40u = -3

6t – 39u = -3

Taking common 3 both sides

3(2t – 13u) = 3(-1) [3 cancel both sides]

2t – 13u = -1

2t – 13(7 – t) = -1 [from (1)]

2t – 91 + 13t = -1

15t = -1 + 91

15t = 90

t = 6

Putting value of t in equation (1) , we get

u = ( 7 – t )

u = ( 7 – 6 )

u = 1

Then the new number is 61

and original number is 16

 


QUESTION 3

The sum of a number of 2 digits and of the number formed by reversing the digits is 110; and the difference of the digits is 6. Find the number.

Sol :

Let once place digit= u and tens place digit be t in original number

Hence , the number is (10t + u)

Reverse of this number be (10u + t)

According to the question ,

(10t + u) + (10u + t) = 110

10t + t + u + 10u = 110

11t + 11u = 110

11(t + u) = 110

t+u=\dfrac{110}{11}

t + u = 10 …(1)

It is given that difference of digits

t – u = 6

t = 6 + u …(2)

Putting (2) in (1) , we get

(6 + u) + u = 10

6 + 2u = 10

2u = 10 – 6

u=\dfrac{4}{2}

u = 2

putting value of u in (2) , we get

t = 6 + 2 = 8

Hence , reversed number is 82

Original number is 28

 


QUESTION 4

A certain number between 10 and 100 is 8 times the sum of its digits, and if 45 be subtracted from it the digits will be reserved. Find the number.

Sol :

Let the number be xy

value of number = 10x+y

10x + y = 8(x+y)

10x + y = 8x+8y

2x – 7y = 0………………..(1)

According to the question ,

10x + y – 45 = 10y + x

9x – 9y = 45

x-y = 5……………..(2)

multiply by -2

-2x+ 2y = -10

add to (1)

-5y = -10

y = 2

x – y = 5

x = 7

The original Number = 72

 


QUESTION 5

A number consists of three digits, the right-hand digit being zero. If the left hand and the middle digits be interchanged the number is diminished by 180.If the left hand digit be halved and the middle and right hand digits be interchanged the number is diminished by 454.Find the number.

Hint: let the original number be 100a + 10a + 0, i.e., 100a +10b, then ,by the first condition (100a+10b)-(100b-10a) = a-b = 2

Sol :

Since the number contains three digits, it has a hundreds place, a tens place and a ones place.

It is given that the right most digit is 0. Let the digits in the hundreds place and tens place be x and y respectively. Thus, the number is 100x + 10y + 0 = 100x +10y.

It is also given that if the digits in the hundreds place and tens place are interchanged, the difference between the original number and the new number after interchange is 180.

=> 100x + 10y – (100y + 10x) = 180

=> 100x + 10y – 100y – 10x = 180

=> 90x – 90y = 180

=> 90(x – y) = 180

=> x – y = 2 —> (i)

It is also given that if the digit in the hundreds place is halved and digits in tens place and ones place are interchanged, the difference between the original number and the new number after interchange is 180.

=> 100x + 10y – (100x/2 + 0 + y) = 454

=> 100x + 10y -50x -y = 454

=> 50x + 9y = 464 —>(ii)

Multiplying both sides of Eqn (i) by 50, we get

50x – 50y = 100

Eqn (i) – Eqn (ii) => (50x – 50y) – (50x + 9y) = 100 – 454

=> 50x -50y -50x -9y = -354

=> -59y = -354

=> y = 6

Now, substitute the value of y in the equation  x – y = 2.

=> x – 6 = 2

=> x = 8

Therefore, the three digit number xy0 = 860.

 


 

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