Question of the Week- 22 (Jack and Jill have certain number of ....)

Question

e-GMAT          Question of the Week #22

Jack and Jill have certain number of apples with them. The total number of apples with both of them is less than 80. If Jill gives a certain number of apples to Jack, then Jack will have four times the number of apples Jill has. If Jack gives the same number of apples to Jill, then Jack will have thrice the number of apples that Jill has. What can be the number of apples that Jack initially has?

A.  31
 B.  36
 C.  45
 D.  62
 E.  63

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EgmatQuantExpert · Accepted Answer

Solution 
  Given:  
 • Total number of apples with both of them = 80
• If Jill gives certain number of apples to Jack, then the number of apples with Jack = 4 * the number of apples with Jill
• If Jack gives the same number of apples to Jill, then the number of apples with Jack = 3 * the number of apples with Jill

To find:  
 • The number of apples that jack has initially

Approach and Working:   
 • Let us assume that the number of apples with Jack and Jill initially are P and Q
 o Implies, P + Q < 80

Now, when Jill gives x apples to Jack
 • The number of apples with Jack will be = P + x, and
• The number of apples with Jill = Q – x

As per given information, P + x = 4(Q - x)
 • Which implies that 4Q – P = 5x …………. (1)

And, when Jack gives x apples to Jill
 • The number of apples with Jack will be = P - x, and
• The number of apples with Jill = Q + x

And we are given that P - x = 3(Q + x)
 • Which implies that P - 3Q = 4x …………. (2)

Solving equations (1) and (2), we get,
 • Q = 9x and P = 31x

Thus, P must be a multiple of 31, since x is an integer
 • And, we know that P + Q < 80
• Implies, 31x + 9x < 80
• Thus, x < 2

Therefore, the only possible value of P = 31 * 1 = 31

Hence the correct answer is Option A.

Answer: A

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pandeyashwin · Answer

Let the number of apples with Jack and Jill be a & b respectively.
Let k be the number of apples transferred.

a+k = 4(b-k)
=> a + k = 4b - 4k
=> a - 4b = -5k ------ 1

and a-k = 3(b+k)
=> a - 3b = 4k ------ 2

Solving 1 & 2 ,
we get k = b/9 ------- 3

From 1 & 3 , we get

a = 31b/9

Now , a+b < 80
or , 31b/9 + b < 80
or b < 18

b can take only 1 value i.e 9 otherwise "k" will not be an integer.

k = 1 ; b = 9

Using this value in equation 1 , a = 31

Option A

AKY13 · Answer

Let K be the apples with Jack. Total apples 80-x. Hence apples with Jill (80-x-K)

Jill gives A apples to Jack. The new arrangement 
Jack = K+A 
Jill = (80-x-K)-A
Given that, (K+A) = 4(80-x-K)- A ------------Eq 1

Similarly, the other arrangement 
Jack = K-A 
Jill = (80-x-K)+A
Given that, (K-A) = 3(80-x-K)+ A--------------Eq2

Solving the 2 equations, we get 
A = k/31 
Plugging values, we get that K = 31(gives A =1) satisfies the equations Eq 1 & 2
Hence A

ScottTargetTestPrep · Answer

Solution:
 
Let K and L be the number of apples Jack and Jill have, respectively, and let a be the number of apples they give to each other. We can create the following equations and inequalities:

K + L < 80

4(L - a) = K + a

4L - 4a = K + a (Eq. 1)

3(L + a) = K - a

3L + 3a = K - a (Eq. 2)

If we subtract Eq. 2 from Eq. 1, we have:

L - 7a = 2a

L = 9a

If we add the two equations, we have:

7L - a = 2K

Substituting 9a for L, we have:

7(9a) - a = 2K

62a = 2K

31a = K

Substituting 31a for K and 9a for L in the inequality, we have:

31a + 9a < 80

40a < 80

a < 2

Since a must be a positive integer, then a = 1, and therefore, K = 31a = 31 x 1 = 31.

Answer: A

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Question of the Week- 22 (Jack and Jill have certain number of ....)

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