There are some cows, hens and goats in Baburam's fam His 8 year-old son, Vistinu, is fond of counting One day, he counts me legs of all the animals and tells us tather that the sum is 210 Next day. Baburam adds more hens and goats to his farm The number of animals added by Babura half The number of animals which were earlier there in the farm. With this addition, the number of cows is same as the number of goats in the farm Now Vishnu tells that the sum of all the animals legs o 300.
Next day, Baburam neighbor Bhola, who has same number of cows as Baburam sends this cows to Baburam farm Now, Vishnu counts and tells tis father that the total number of animals in their tarm is 135 How many goats were newly added by Baburant in the tarm?
A. 15
B. 7
C. 10
D. 5
Answers
Solution :-
Let us assume that, initially their were C cows, H hens and G goats in Baburam's farm.
we know that, a cow and a goat has total 4 legs and a hen as 2 legs.
So,
→ First day total legs = 4C + 4G + 2H = 210 ----------- Eqn.(1)
Now, baburam adds more hens and goats equals to half the number of animals earlier .
So,
→ G1 + H1 = (C + H + G)/2
→ 2G1 + 2H1 = C + H + G ----------- Eqn.(2)
Also,
→ Number of cows = Number of goats now,
→ C = G + G1 . ------------ Eqn.(3)
Than,
→ sum of all legs now = 300
→ 4C + 4(G + G1) + 2(H + H1) = 300
→ 4C + 4G + 2H + 4G1 + 2H1 = 300
→ (4C + 4G + 2H) + 2G1 + (2G1 + 2H1) = 300
Putting value of Eqn.(1) and Eqn.(2) ,
→ 210 + C + H + G + 2G1 = 300
→ C + H + G + 2G1 = 300 - 210
→ C + H + G + 2G1 = 90
Putting value of G1 from Eqn.(3) now,
→ C + H + G + 2(C - G) = 90
→ C + H + G + 2C - 2G = 90
→ 3C + H - G = 90 --------- Eqn.(4)
Now, his neighbour send equal number of cows and now their are total 135 animals.
So,
→ 2C + (G + G1) + (H + H1) = 135
→ 2C + G + H + (G1 + H1) = 135
→ 2C + G + H + (C + G + H/2) = 135
→ 4C + 2G + 2H + C + G + H = 135*2
→ 6C + 3G + 3H = 270
→ 3C + (3C + 3G + 3H) = 270
→ 3C + 3(C + G + H) = 270
→ 3(C + G + H) = 270 - 3C
→ 3(C + G + H) = 3(90 - C)
→ (C + G + H) = (90 - C)---------- Eqn.(5)
Adding Eqn.(1) and Eqn.(4) ,
→ 4C + 3C + 4G - G + 2H + H = 210 + 90
→ 7C + 3G + 3H = 300
→ 4C + (3C + 3G + 3H) = 300
→ 4C + 3(C + G + H) = 300
Putting value of Eqn.(5) here, we get,
→ 4C + 3(90 - C) = 300
→ 4C - 3C + 270 = 300
→ C = 300 - 270
→ C = 30 .
Putting value of C in Eqn.(5) now,
→ C + G + H = 90 - C
→ 30 + G + H = 90 - 30
→ 30 + G + H = 60
→ G + H = 60 - 30
→ G + H = 30.
Also, when we put value of C in Eqn.(4),
→ 3C + H - G = 90
→ 3*30 + H - G = 90
→ 90 + H - G = 90
→ H - G = 90 - 90
→ H = G .
Therefore,
→ G + H = 30
→ H = G
we can conclude that,
→ H = G = 15.
Hence, Putting value of C and G in Eqn.(3) now, we get,
→ C = G + G1
→ 30 = 15 + G1
→ G1 = 30 - 15
→ G1 = 15.(Ans.) (Option A).
∴ 15 goats were newly added by Baburant in the farm.
(Excellent Question.)