Question

Eena,Meena and Beena were born in three consecutive years. The product pf their ages 10 years ago was 1320. What is the sum of their ages today ?

Answer 1

Answer:

SUM OF THEIR AGES IS 63

10 YEARS AGO THE PRODUCT OF THEIR AGE IS 1320

SO THEIR AGES WERE 10,11,12 AS THEY WERE BORN IN CONSECUTIVE YEARS

AFTER 10 YEARS THEIR AGES IS AS IT IS ASKED OF TODAY

THERE AGES IS 20,21 ,22

SO SUM OF THEIR AGES IS 20+21+22 =63

Answer 2

Answer:

Sum of their ages today = 63

Step-by-step explanation:

Given data

Eena, Meena and Beena were born in three consecutive years

the product of their ages 10 years ago was 1320

Here we need to find sum of their present ages

given that Eena, Meena and Beena were born in three consecutive years

then their ages must be 3 consecutive numbers

Let x, x+1 and x+2 are be the 3 consecutive numbers and ages of Eena, Meena and Beena 10 years ago

then product of x, x + 1 and x + 2 = 1320                

⇒     x ( x + 1) ( x+2) = 1320  

⇒     x (x+1) (x+2)  = 132 × 10      [ 1320 = 132 × 10 ]  

⇒     x ( x+ 1) (x+2) = 10 × 11 × 12    [ 120 = 11 × 12 ]  

⇒ Ages of Eena, Meena and Beena (10 years ago) are 10, 11 and 12  

after ten years ages of Eena, Meena and Beena are  20, 21 and 22

⇒ sum of their present ages are = 20 + 21 + 22 = 63