Verbal Reasoning :: Arithmetical Reasoning :: Calculation Based Problems

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At a farm, there are hens, cows and bullocks, and the keepers to look after them. There are 69 heads less than legs; the number of cows and hens is the same and there is one keeper per ten birds and cattle. The total number of hens plus cows and bullocks, and their keepers does not exceed 50. How many cows are there?

Answer:

Explanation:

 Let H, C, B and K represent the number of hens, cows, bullocks and keepers respectively. Then, as given, we have:

H + C + B + K < 50 .......(i)

 C = 2B  .........(ii)

C = H .........(iii)

$K= \frac{H+C+B}{10}$ .........(iv)

From , (ii),(iii),(iv), we have:

10K = H + C+ B  <=> 10K = 2C + B  = 2 X 2B  <=> 10K = 5B

B = 2K

So, B = 2K ,C = 2B =4K , H = C 4K

Total number of heads =  H + C + B + K .

Total number of legs = 2H + 4C + 4B + 2K

(2H + 4C + 4B + 2K) - (H + C + B + K ) = 69

H + 3C + 3B + K = 69

4K + 12K + 6K + K =69

K = 3

Hence number of cows = C = 4K =(4 X 3) = 12

A man has a certain number of small boxes to pack into parcels. If he packs 3, 4, 5 or 6 in a parcel, he is left with one over; if he packs 7 in a parcel, none is left over. what is the number of boxes, he may have to pack?

Answer:

Explanation:

Clearly, the required number would be such that it leaves a remainder of 1 when divided by 3,4,5 or 6 and no remainder when divided by 7. Thus, the number must be of the form (L.C.M. of 3, 4,5,6) x + 1 i'e. (60x+ 1) and a multiple of 7. Clearly, for x = 5, the number is a multiple of 7. So, the number is 301.

In a certain office, 1/3 of the workers are women, 1/2 of the women are married and 1/3 of the married women have children.If 3/4 of men are married and  2/3 of the married men have children, What part of the workers are without children?

Answer:

Explanation:

let the total number of workers be x

Number of women = x/3 and number of men = (x-x/3) = 2x/3

Number of women having children = 1/3 of 1/2 of x/3 = x/18

Number of men having children = 2/3 of 3/4 of 2x/3 = x/3

Number of workers having children = (x-7x/18) = 11x/18 = 11/18 of all workers 

Five children take part in a tournament. Each one has to play every other one. How many g€unes must they play

Answer:

Explanation:

(i) matches of the first player with other 4 players

(ii) matches of the second player with 3 players other than the first player

(iii) matches of the third player with 2 players other than the first and second players

(iv) match the fourth player with one player other than the first three players. So, number of matches played during the to[rnament =4 +3 +2 + 1= 10.

• Arithmetical Reasoning - Calculation Based Problems
• Arithmetical Reasoning - Data Based
• Arithmetical Reasoning - Problems on Ages
• Arithmetical Reasoning - Venn Diagram