XAT 2022 — QA & DI Question 22
Read the following scenario and answer the THREE questions that follow.
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The given candlestick chart depicts the prices of a particular stock over 10 consecutive days. A candlestick comprises of a rectangular box pieced by a line. The top and bottom ends of the line respectively indicate the maximum and minimum prices of the stock on that day, while the horizontal edges of the rectangle correspond to the stock's opening and closing prices. If the rectangle is white, the opening price is lower than the closing price, but if the rectangle is black, then it is the other way around.
Using the above information, answer the questions that follow:
Wilma, Xavier, Yaska and Zakir are four young friends, who have a passion for integers. One day, each of them selects one integer and writes it on a wall. The writing on the wall shows that Xavier and Zakir picked positive integers, Yaska picked a negative one, while Wilma’s integer is either negative, zero or positive. If their integers are denoted by the first letters of their respective names, the following is true:
W4 + X3 + Y2 + Z ≤ 4
X3 + Z ≥ 2
W4 + Y2 ≤ 2
Y2 + Z ≥ 3
Given the above, which of these can W2 + X2 + Y2 + Z2 possibly evaluate to?
Answer & solution
- A
9
- B
0
- C
4
6
- E
1
Given that X, Z are positive Y is negative and W can be either positive or zero or negative.
The given conditions are:
W4 + X3 + Y2 + Z ≤ 4
X3 + Z ≥ 2
W4 + Y2 ≤ 2
Y2 + Z ≥ 3
For W4 + Y2 ≤ 2. Since Y is negative but Y2 is always positive and must be less than 2 because W4 is a nonnegative value. Hence Y = -1 is the only possibility. For W this can take any value among -1, 0, 1.
Y2 + Z ≥ 3. Since Y = -1, Z must be at least equal to 2 so the value of Y2 + Z ≥ 3 is greater than 2.
X is a positive value and must at least be equal to 1.
The condition: W2 + X2 + Y2 + Z2 here has all the independent values:
X2, Y2, Z2, W2 are non-negative.
W4 + X3 + Y2 + Z ≤ 4:
Since the value of Z is at least equal to 2 the value of Y2 is equal to 1.
Since X is a positive number in order to have the condition of W4 + X3 + Y2 + Z ≤ 4 satisfied. The value of Z must be the minimum possible so that X3 + Y2 + Z to have a value equal to 4 when X takes the minimum possible positive value equal to 1.
Hence X must be 1. W must be equal to 0 so that:
W4 + X3 + Y2 + Z ≤ 4. The sum = (0 + 1 + 1 + 2) = 4. The only possible case.
The value of W2 + X2 + Y2 + Z2 = (0 + 1 + 1 + 4) = 6.