Find the two consecutive positive integers whose sum is 63.?

Question

Anu is four years older than sunil.Eight years ago,Anu was three times sunil's age.Find the ages of Sunil and Anu?

Answer 1

x+(x+1)=63
2x+1=63
x=31
x+1=32

Sunil=x
Anu=x+4
x+4-8=3(x-8)
x-4=3x-24
2x=20
x=10
Sunil=10
Anu=14

Answer 2

1. The square of a positive integer is 98 less than twice the square of the next consecutive positive integer.What are the integers? x^2 + 98 = 2(x + 1)^2 x^2 + 98 = 2(x^2 + 2x + 1) x^2 + 98 = 2x^2 + 4x + 2 0 = x^2 + 4x - 96 (x + 12) ( x - 8) = 0 x = -12 or 8; Since we need positive integers, rule out the -12. Answer: 8 and 9 2. Find the numbers whose sum is 15 and the sum of whose squares is 113. x + y = 15 or x = 15 - y x^2 + y^2 = 113 (15 - y)^2 + y^2 = 113 225 - 30y + y^2 + y^2 = 113 2y^2 - 30y + 112 = 0 2 (y^2 - 15y + 56) = 0 2 (y - 8) (y - 7) = 0 y = 8 or 7 x = 8 or 7 Answer: 8 and 7. 3. Find two numbers whose sum is 11 and the sum of whose squares is 85. x + y = 11 or x = 11 - y x^2 + y^2 = 85 (11 - y)^2 + y^2 = 85 121 - 22y + y^2 + y^2 = 85 2y^2 - 22y + 36 = 0 2 (y^2 - 11y + 18) = 0 2 (y - 2) (y - 9) = 0 y = 2 or 9 x = 2 or 9 Answer: 2 and 9. 4. Find two numbers whose sum is 16 and the sum of whose squares is 130. x + y = 16 or x = 16 - y x^2 + y^2 = 130 (16 - y)^2 + y^2 = 130 256 - 32y + y^2 + y^2 = 130 2y^2 - 32y + 126 = 0 2 (y^2 - 16y + 63) = 0 2 (y - 9) (y - 7) = 0 y = 9 or 7 x = 9 or 7 Answer: 9 and 7.

Answer 3

n+(n+1) =63 ==> 2n = 62 and n = 31

2) A = S+4 and A-8= 3(S-8) from the first S+4-8 = 3S-24
2S=20 S=10 A=14 years A(Anu) S(sunil)

Answer 4

The two consecutive positive integars whose sum is 63 are 31 and 32.
Lets Anu's present age = x
Sunil's present age = x-4
Eight years ago Anu's age= x-8
Eight years ago Sunils age=x-4-8
According to condition
x-8=3(x-4-8)
x-8=3(x-12)
-2x=-36+8
-2x=-28
x=14
Anu's age =14 years
Sunil's age=10 years

Answer 5

1)let the 2 integers be x and (x+1)
therefore x+(x+1)=63. Solving we get x as 31. hence 31 and 32 are the 2 required integers.
2)let sunils age be x. i.e anu's age is x+4
8 yrs ago, sunils age is (x-8). i.e anu's age is (x-8)+4=x - 4
also, anu's age was 3 times sunil's age
i.e (x - 4)= 3(x - 8)
Solving , we get x=10
therefore sunil's age is 10 and anu's , 14

Answer 6

Wait a minute. If Anu is FOUR years older, their ages are not going to be consecutive positive integers. The consecutive positive integers are 31 and 32. That's easy enough. But that's not Anu and Sunil's ages, as Anu is four years older.

Did you try to pack two questions into one?

Answer 7

Is this two questions?

for the first, let x = 1 number and x+1 = second

x + (x+1) = 63
2x+1 = 63
2x = 62
x = 31

for the second

let x = anu's age
x - 4 = sunil's age

8 years ago,
anu's age = x-8
and sunil's age = x-4 - 8 = x-12
and (x-8) = 3*(x-12)

x-8 = 3x-36

you do the rest...

Answer 8

ANS - 31 ,32
SOL - x +x +1 =63
2x + 1 =63
2x =62
x =62 / 2
x = 31
x+1 =32

Answer 9

Two consecutive positive integers whose sum is 63

Let

x = first positive integer

x + 1 = second positive integer

63 = the sum of the inteers

- -- - - - - - -

x + x + 1 = 63

2x + 1 = 63

2x + 1 - 1 = 63 - 1

2x = 63

2x / 2 = 62 / 2

x = 31

- - - - - - - -

The first positive integer = 31

The second positive integer x + 1 = 32

- - - - - - -s-

Answer 10

1)
let first no. be "x" and the second be "x+1"
(x) + (x+1) = 63
x+x+1=63
2x+1=63
2x=62
therefore x=62/2=31
so the first no. is (x)=31
and the second one is (x+1)=31+1=32.

2)
a=anu,s=sunil
a=4+s
a-8=3(s-8)subtracting 8 years from both of their ages.
as we know a=4+s so we will substitute:
4+s-8=3s-24
4+24-8=3s-s
so 2s=28-8
2s=20
s=10
so sunil's age is 10 years and anu's age is (s+4) so 10+4=14 years.

Answer 11

One way to solve this is just try different numbers, like this:
1+2 = 3. Thats too small
10+11 = 21. Still too small
20+21 = 41. Still too small.
30+31 = 61. Getting close
31+32 = 63. Just right!