Question

1) find a number which when multiplied by 7 and then reduced by 3 is equal to 53.

2) sum of two numbers is 95. if one exceeds the other by 3, find the numbers.

3) sum of three consecutive integers is 24. find the integers.
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Answer 1

Answer:

1. Let the no. be X

According to question,

7x-3=53

7x=50

x=50/7

2. Let the first no. be X

hence, the second number= x+3

Their sum=95

Then,

x+x+3=95

2x+3=95

2x=92

x=46

So, the 1st number=x=46

2nd no. =x+3=49

3. Let the 1st integer be x

so,

2nd integer=x+1

3rd integer=x+2

Their sum=24

According to question,

x+x+1+x+2=24

3x+3=24

3x=21

x=7

so, 1st integer=7

2nd integer=8

3rd integer=9

Answer 2

Step-by-step explanation:

3) 7+8+9= 24

1)Let the number be x

and after multiplying with 7 it became 7x

and reduced by 3 which means 7x - 3

7x - 3 = 53

7x = 53 + 3

7x = 56

x = 56/7

x = 8

2) Let one number be x and other number be y

x+ y = 95 ....i

one number exceeds other by 3

x = y + 3

x - y = 3 ......ii

by adding equation 1 and 2

x+ y + x - y = 95 + 3

2x = 98

x = 98 / 2

x = 49