Graphs, Scatter Plots and Trend Analysis Practice Questions for XAT: 210+ Solved Questions with Step-by-Step Solutions

Solve 210+ Graphs, Scatter Plots and Trend Analysis practice questions for XAT with answers and detailed solutions. Free sample questions below.

Introduction to Scatter Plots

Orientation

Introduction to Scatter Plots

Topic Hero: Visualize and analyze the relationship between two quantitative variables using a two-dimensional coordinate system.

What you will learn here:

Understand the structure of a scatter plot (X-axis and Y-axis representing two variables).
Identify the direction (positive/negative), form (linear/non-linear), and strength of a relationship.
Interpret trend lines and lines of best fit.
Compare multiple datasets or groups within the same plot.

Chapter: Graphs, Scatter Plots and Trend Analysis

The Three Pillars of Scatter Plot Analysis

Concept

The Three Pillars of Scatter Plot Analysis

The Core Principle

Every scatter plot relationship is defined by three characteristics:

Relationship = Direction + Form + Strength

1. Direction

Positive: X increases, Y increases (upward trend).
Negative: X increases, Y decreases (downward trend).
No Association: No clear pattern.

2. Form

Linear: Points follow a straight-line path.
Non-linear: Points follow a curved path (e.g., U-shaped).

3. Strength

Strong: Tightly clustered around the form.
Weak: Widely scattered with noise.

Graphs, Scatter Plots and Trend Analysis: Solved Questions with Step-by-Step Explanations (5 Problems)

Question 1 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

Refer to the hypothetical dataset describing 12 regions. Graph A plots "Per Capita Healthcare Spend ( $) " o n X - a x i s v s " L i f e E x p ec t an cy (Y e a r s) " o nY - a x i s . G r a p h B pl o t s " P er C a p i t a H e a l t h c a r e S p e n d ($ )" on X-axis vs "Patient Satisfaction Score (0-10)" on Y-axis.

Observations:

In Graph A, points form a steep positive curve for spends up to $3, 000, t h e n f l a tt e n co m pl e t e l y in t o ah or i z o n t a l l in e f or s p e n d s >$ 3,000.
In Graph B, points show a steady linear upward trend across the entire spend range ( $500 t o$ 8,000).

Based ONLY on these graphs, which inference is MOST valid?

A.
Increasing healthcare spend beyond $3,000 yields no additional benefit to patients.
B.
Life expectancy is biologically capped, whereas patient satisfaction is subjective and unbounded by physiological limits.
C.
For regions spending > $3,000, marginal gains in life expectancy are negligible, but marginal gains in satisfaction persist.
D.
Regions with highest satisfaction scores necessarily have the highest life expectancy.

Correct Answer:

Step-by-Step Solution

Key idea: This is a multi-graph synthesis question. It tests the ability to reconcile conflicting/divergent trends across two related visualizations and avoid over-generalization.

Step 1: Interpret Graph A (Life Expectancy).

Shape: Logarithmic/Saturation. Steep rise then plateau.

Meaning: Marginal utility of spend on LE drops to zero after $3k.

Step 2: Interpret Graph B (Satisfaction).

Shape: Linear. Constant positive slope.

Meaning: Marginal utility of spend on Sat remains constant/positive even at high spend.

Step 3: Synthesize.

At spend > $3 k :$ \Delta LE \approx 0 $, b u t$ \Delta Sat > 0$.

This directly supports Option C.

Step 4: Reject distractors.

A: False. Says "no additional benefit." Ignores Satisfaction benefit shown in Graph B.

B: Invalid. Introduces external biological/psychological theory not present in data. Data shows correlation, not cause/mechanism.

D: False. Correlation between Sat and LE is not established. High spend regions have high Sat but flat LE. They might have same LE as mid-spend regions.

Answer: C

Question 2 · Quantitative Aptitude and Data Interpretation (QA & DI) NAT

A scatter plot consists of 10 data points. The summary statistics for these 10 points are:

$\sum X = 50$ , $\sum Y = 110$ , $\sum X^{2} = 310$ , $\sum X Y = 730$ .

The line of best fit (OLS) for these 10 points is $Y = 2 X + 1$ .

One point is identified as an outlier and removed. The new line of best fit for the remaining 9 points is $Y = 3 X - 2$ .

What is the $Y$ -coordinate of the removed outlier point?

Correct Answer:

-7

Step-by-Step Solution

Key idea: This is a reverse engineering question using the algebraic properties of OLS summary statistics.

Step 1: Set up the equations for the means

Let the removed point be $(x_{0}, y_{0})$ .

For the 10 points, $\overset{ˉ}{X} = 5, \overset{ˉ}{Y} = 11$ .

For the 9 points, the new means $\overset{ˉ}{X}^{'}, \overset{ˉ}{Y}^{'}$ must lie on the new line $Y = 3 X - 2$ .

$\overset{ˉ}{X}^{'} = \frac{50 - x _{0}}{9}$ , $\overset{ˉ}{Y}^{'} = \frac{110 - y _{0}}{9}$ .

Substituting into the new line equation:

$\frac{110 - y _{0}}{9} = 3 (\frac{50 - x _{0}}{9}) - 2$

$110 - y_{0} = 150 - 3 x_{0} - 18$

$110 - y_{0} = 132 - 3 x_{0} ⟹ y_{0} = 3 x_{0} - 22$ .

Step 2: Use the slope condition for the 9 points

The slope for the 9 points is $m^{'} = \frac{S _{X Y}^{'}}{S _{X X}^{'}} = 3$ .

$S_{X X}^{'} = \sum_{9} X^{2} - \frac{( \sum _{9} X ) ^{2}}{9} = (310 - x_{0}^{2}) - \frac{( 50 - x _{0} ) ^{2}}{9}$ .

$S_{X Y}^{'} = \sum_{9} X Y - \frac{( \sum _{9} X ) ( \sum _{9} Y )}{9} = (730 - x_{0} y_{0}) - \frac{( 50 - x _{0} ) ( 110 - y _{0} )}{9}$ .

Step 3: Simplify and solve for $x_{0}$

Substitute $110 - y_{0} = 132 - 3 x_{0}$ :

$S_{X X}^{'} = \frac{2790 - 9 x _{0}^{2} - ( 2500 - 100 x _{0} + x _{0}^{2} )}{9} = \frac{290 + 100 x _{0} - 10 x _{0}^{2}}{9}$ .

$S_{X Y}^{'} = 730 - x_{0} (3 x_{0} - 22) - \frac{( 50 - x _{0} ) ( 132 - 3 x _{0} )}{9}$

$= \frac{6570 - 27 x _{0}^{2} + 198 x _{0} - ( 6600 - 282 x _{0} + 3 x _{0}^{2} )}{9}$

$= \frac{- 30 + 480 x _{0} - 30 x _{0}^{2}}{9}$ .

We need $S_{X Y}^{'} = 3 S_{X X}^{'}$ :

$- 30 + 480 x_{0} - 30 x_{0}^{2} = 3 (290 + 100 x_{0} - 10 x_{0}^{2})$

$- 30 + 480 x_{0} - 30 x_{0}^{2} = 870 + 300 x_{0} - 30 x_{0}^{2}$ .

The quadratic terms cancel out perfectly!

$180 x_{0} = 900 ⟹ x_{0} = 5$ .

Step 4: Find $y_{0}$

$y_{0} = 3 (5) - 22 = 15 - 22 = - 7$ .

Answer: -7

Question 3 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

A scatter plot of 100 data points shows two distinct clusters, Group A and Group B. The line of best fit for Group A has a slope of 2, and for Group B has a slope of 4. If the overall line of best fit for all 100 points is calculated, which of the following is ALWAYS true about its slope?

A.
It can be any real number, including values outside the range [2, 4].
B.
It must be exactly 3.
C.
It must lie strictly between 2 and 4.
D.
It must be greater than 4.

Correct Answer:

Step-by-Step Solution

Key idea: This is a statement_truth question on cluster analysis and Simpson's Paradox in scatter plots, recognizable because it compares within-group trends to a pooled overall trend.

Step 1: Understand the setup. We have two groups with different internal slopes (2 and 4). The question asks about the slope of the pooled data (all 100 points combined).

Step 2: Recall the mechanics of the pooled slope. The overall slope in a simple linear regression is not just a weighted average of the within-group slopes. It also depends heavily on the distance between the means of the two groups (the "between-group" variation).

Step 3: Analyze the options.

Option B assumes a simple average, which is only true if the groups have identical variances and identical means, which is not guaranteed.
Option C assumes the overall slope must be a compromise between the two internal slopes. This is a common fallacy.
Option D assumes the overall slope must be steeper than both, which is also not guaranteed.
Option A states it can be any real number. This is true due to Simpson's Paradox. If Group B has a much higher mean $X$ and mean $Y$ than Group A, the line connecting the two cluster centers will be very steep, potentially making the overall slope much greater than 4. Conversely, if Group B has a higher mean $X$ but a lower mean $Y$ , the overall slope could even be negative.

Step 4: Conclusion. The overall slope depends on the internal slopes AND the separation of the cluster means. Therefore, it is not bounded by the internal slopes.

Common trap: Students often assume that the overall trend must be a "compromise" (between 2 and 4) of the two group trends. This ignores the impact of the vertical and horizontal distance between the clusters themselves.

Answer: A

Question 4 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

A scatter plot shows the relationship between the daily trading volume $V$ (in millions) and the candlestick body size $B$ (in dollars), given by $B = V + 10$ .

Over 3 consecutive days, the trading volumes form a geometric progression with a common ratio of 2. The stock is bullish on all 3 days, with the upper wick always 0 and the lower wick always equal to the body size.

If the opening price on Day 1 is 100 and the closing price on Day 3 is 172, what is the trading volume on Day 1 (in millions)?

A.
4
B.
5
C.
6
D.
7

Correct Answer:

Step-by-Step Solution

Key idea: This is a "Scatter Plot and Candlestick Synthesis" question. We must link the scatter plot model to the candlestick anatomy and trace the closing prices over the geometric progression of volumes.

Step 1: Express the body sizes in terms of Day 1 volume $v$ .

$V_{1} = v, V_{2} = 2 v, V_{3} = 4 v$ .

$B_{1} = v + 10$ .

$B_{2} = 2 v + 10$ .

$B_{3} = 4 v + 10$ .

Step 2: Trace the closing prices.

Since all candles are bullish and upper wick is 0, $C_{t} = H_{t}$ and $C_{t} = O_{t} + B_{t}$ .

Lower wick = $B_{t} ⟹ L_{t} = O_{t} - B_{t}$ .

Day 1: $O_{1} = 100$ .

$C_{1} = 100 + B_{1} = 110 + v$ .

Day 2: $O_{2} = C_{1} = 110 + v$ .

$C_{2} = O_{2} + B_{2} = 110 + v + 2 v + 10 = 120 + 3 v$ .

Day 3: $O_{3} = C_{2} = 120 + 3 v$ .

$C_{3} = O_{3} + B_{3} = 120 + 3 v + 4 v + 10 = 130 + 7 v$ .

Step 3: Solve for $v$ .

Given $C_{3} = 172$ .

$130 + 7 v = 172 ⟹ 7 v = 42 ⟹ v = 6$ .

Question 5 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

A stock's daily closing prices $C_{1}, C_{2}, C_{3}, C_{4}, C_{5}$ form an arithmetic progression. The 3-day moving averages of the closing prices for days 3, 4, and 5 are 100, 110, and 120 respectively.

The candlestick chart shows that for each day, the body size is exactly 10% of the closing price, the upper wick is equal to the body size, and the lower wick is equal to the upper wick. The stock is bullish on all 5 days.

If the opening price on Day 1 is 81, what is the highest price reached on Day 5?

A.
135
B.
140
C.
143
D.
150

Correct Answer:

Step-by-Step Solution

Key idea: This is a "Moving Average and Candlestick Anatomy" synthesis question. We must first use the moving averages to find the exact closing prices, and then apply the candlestick rules to find the high price.

Step 1: Find the closing prices using the moving averages.

For an arithmetic progression, the 3-day moving average at day $t$ is exactly the closing price at day $t - 1$ .

$M A_{3} = C_{2} = 100$ .

$M A_{4} = C_{3} = 110$ .

$M A_{5} = C_{4} = 120$ .

The common difference is $d = 10$ .

Therefore, $C_{1} = 90$ and $C_{5} = 130$ .

Step 2: Verify the Day 1 condition.

$C_{1} = 90$ . Body = $10% \times 90 = 9$ .

$O_{1} = C_{1} - 9 = 81$ . This matches the given condition perfectly.

Step 3: Calculate the High price for Day 5.

$C_{5} = 130$ .

Body = $10% \times 130 = 13$ .

Since the stock is bullish, $C_{5} > O_{5}$ .

Upper wick = Body = 13.

High = $C_{5} +$ Upper wick = $130 + 13 = 143$ .

Graphs, Scatter Plots and Trend Analysis Practice Questions for XAT: 210+ Solved Questions with Step-by-Step Solutions

Introduction to Scatter Plots

Introduction to Scatter Plots

What you will learn here:

The Three Pillars of Scatter Plot Analysis

The Three Pillars of Scatter Plot Analysis

The Core Principle

1. Direction

2. Form

3. Strength

Graphs, Scatter Plots and Trend Analysis: Solved Questions with Step-by-Step Explanations (5 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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