CMAT 2018 Quantitative Techniques and Data Interpretation Question Paper with Answer Key PDFs (January 20 - Afternoon Session)

Step 1: Find the total number of outcomes.

When two dice are thrown, each die has 6 outcomes.

So total possible outcomes = \(6 \times 6 = 36\).


Step 2: Find outcomes that give sum 9.

Sum 9 occurs for the pairs:
\((3,6), (4,5), (5,4), (6,3)\)

So, number of outcomes for sum 9 = 4.


Step 3: Find outcomes that give sum 10.

Sum 10 occurs for the pairs:
\((4,6), (5,5), (6,4)\)

So, number of outcomes for sum 10 = 3.


Step 4: Compute probability.

Total favourable outcomes = \(4 + 3 = 7\).
\[ P = \frac{7}{36} \]
Quick Tip: For two dice, total possible outcomes are always \(36\). Count favourable cases and divide by \(36\).

Step 1: Assume breadth and length.

Let breadth \(= x\) meters.

Then length \(= x + 2\) meters.


Step 2: Write the original area.
\[ A = x(x+2) \]


Step 3: Apply new dimensions.

New length \(= (x+2)+4 = x+6\).

New breadth \(= x-2\).

New area:
\[ A' = (x+6)(x-2) \]


Step 4: Since area remains same, equate.
\[ x(x+2) = (x+6)(x-2) \]
\[ x^2 + 2x = x^2 + 4x - 12 \]
\[ 2x = 4x - 12 \]
\[ 2x = 12 \Rightarrow x = 6 \]


So breadth \(= 6\)m and length \(= 8\)m.


Step 5: Find surface area of walls.

Height \(h = 3\)m.

Surface area of four walls:
\[ 2h(l+b) = 2 \times 3 \times (8+6) \]
\[ = 6 \times 14 = 84m^2 \]
Quick Tip: Surface area of four walls of a room = \(2h(l+b)\).

Step 1: Total distance covered.
\[ 50 + 50 + 30 = 130 km \]


Step 2: Find time for each part.

First part: 50 km in 40 minutes.

Second part: speed = 2 km/min
\[ Time = \frac{50}{2} = 25 minutes \]

Third part: speed = 1 km/min
\[ Time = \frac{30}{1} = 30 minutes \]


Step 3: Total time.
\[ 40 + 25 + 30 = 95 minutes \]


Step 4: Convert time into hours and find average speed.
\[ 95 minutes = \frac{95}{60} hours \]
\[ Average speed = \frac{130}{95/60} = \frac{130 \times 60}{95} = 82.1 kmph \]
Quick Tip: Average speed always depends on total distance and total time, not individual speeds.

Step 1: Convert revolutions per minute into time per revolution.

First wheel: 60 rev/min means 1 rev/sec.

So it completes 1 revolution in 1 second.


Second wheel: 36 rev/min
\[ 36 rev in 60 sec \Rightarrow 1 rev in \frac{60}{36} = \frac{5}{3} sec \]


Third wheel: 24 rev/min
\[ 24 rev in 60 sec \Rightarrow 1 rev in \frac{60}{24} = \frac{5}{2} sec \]


Step 2: Find LCM of these time intervals.

We need the smallest time when all complete an integer number of revolutions together.

So required time = LCM of \(1, \frac{5}{3}, \frac{5}{2}\).

LCM \(= 5\) seconds.


Step 3: Conclusion.

Thus, after 5 seconds, all wheels will again be in the same starting position.
Quick Tip: In wheel/revolution problems, convert to time per revolution and take LCM.

Step 1: Use compound interest formula.
\[ A = P\left(1+\frac{R}{100}\right)^n \]

Given \(A = 2000\), \(R = 10\), \(n = 2\).


Step 2: Substitute values.
\[ 2000 = P\left(1+\frac{10}{100}\right)^2 \]
\[ 2000 = P(1.1)^2 = P(1.21) \]


Step 3: Solve for P.
\[ P = \frac{2000}{1.21} \approx 1653 \]
Quick Tip: For CI, principal = \(\dfrac{Amount}{(1+R/100)^n}\).

Step 1: Volume of a cone.
\[ V = \frac{1}{3}\pi r^2 h \]


Step 2: Apply the changes.

Height increased by 200%:
\[ h' = h + 2h = 3h \]

Radius reduced by 50%:
\[ r' = \frac{r}{2} \]


Step 3: Compute new volume.
\[ V' = \frac{1}{3}\pi \left(\frac{r}{2}\right)^2 (3h) \]
\[ V' = \frac{1}{3}\pi \frac{r^2}{4} \cdot 3h \]
\[ V' = \frac{3}{4}\left(\frac{1}{3}\pi r^2 h\right) = \frac{3}{4}V \]


Step 4: Find percentage change.
\[ Decrease = 1 - \frac{3}{4} = \frac{1}{4} = 25% \]
Quick Tip: When radius changes, volume changes as \(r^2\), so radius has a stronger effect than height.

Step 1: Initial price.

Price = 600.


Step 2: Apply successive 10% reduction 3 times.

After first reduction:
\[ 600 \times 0.9 = 540 \]

After second reduction:
\[ 540 \times 0.9 = 486 \]

After third reduction:
\[ 486 \times 0.9 = 437.4 \]


Step 3: Final price.

So the current price is Rs. 437.40.
Quick Tip: Successive reductions are multiplied: \(600 \times 0.9^3\).

Step 1: Understand what the table represents.

The table provides age-wise distribution within each brand.

For example, for LG, 15% of LG mobiles are up to 1 year old, 45% are 1-2 years old, etc.


Step 2: Missing information.

We are given total mobiles sold last year = 1 crore.

But we are not told how many of these were LG mobiles.

Let total LG mobiles sold last year = \(A\).

Then 15% of \(A\) would represent LG mobiles up to 1 year old.

But \(A\) itself is unknown.


Step 3: Conclusion.

Since the number of LG mobiles sold last year is not provided, the exact number cannot be found.
Quick Tip: If only internal percentage distribution is given, totals cannot be calculated unless overall brand totals are provided.

Step 1: Rewrite the expression correctly.

From the given explanation in the image:
\[ \sqrt{188 + \sqrt{51} + \sqrt{169}} \]


Step 2: Evaluate the perfect square.
\[ \sqrt{169} = 13 \]


Step 3: Simplify further (as shown in the solution).
\[ \sqrt{188 + \sqrt{51} + 13} = \sqrt{188 + \sqrt{64}} = \sqrt{188 + 8} \]


Step 4: Final evaluation.
\[ \sqrt{188 + 8} = \sqrt{196} = 14 \]
Quick Tip: Always check if the expression under root can be converted into a perfect square for quick simplification.

Step 1: Identify rate per half-year and number of periods.

Compound interest is payable half-yearly.

So rate per half-year:
\[ \frac{4%}{2} = 2% \]

If time is \(n\) years, number of half-year periods = \(2n\).


Step 2: Apply compound interest formula.
\[ A = P\left(1+\frac{2}{100}\right)^{2n} \]

Given: \(A = 6632.55\), \(P = 6250\).


Step 3: Substitute values.
\[ 6632.55 = 6250(1.02)^{2n} \]
\[ \frac{6632.55}{6250} = (1.02)^{2n} \]
\[ 1.061 = (1.02)^{2n} \]


Step 4: Solve for \(n\).

From the given solution in the image:
\[ 1.061 = 1.02^3 \Rightarrow 2n = 3 \Rightarrow n = \frac{3}{2} \]


So time required = \(\dfrac{3}{2}\) years.
Quick Tip: For half-yearly CI:
Rate becomes \(R/2\) and time periods become \(2n\).

Step 1: Understand what is asked.

We need the average salary expenditure per annum for the given years.


Step 2: Add all salary expenditures.
\[ 576 + 682 + 648 + 672 + 740 = 3318 \]


Step 3: Divide by number of years.

There are 5 years.
\[ Average = \frac{3318}{5} = 663.6 \]
Quick Tip: Average = \(\dfrac{Sum of all values}{Number of values}\).

Step 1: Condition required.

We need:
\[ \frac{Refrigerators}{Cell Phones} \times 100 \approx 25% \]


Step 2: Check each option.


For 2011:
\[ \frac{60}{336}\times 100 = 17.85% \]


For 2012:
\[ \frac{79}{404}\times 100 = 19.55% \]


For 2013:
\[ \frac{93}{411}\times 100 = 22.63% \]


For 2014:
\[ \frac{112}{442}\times 100 = 25.33% \]


Step 3: Conclusion.

2014 gives approximately 25%. Hence the correct option is 2014.
Quick Tip: To check “x is y% of z”, compute \(\dfrac{x}{z}\times 100\).

Step 1: Use property of diameter.

Since \(PQ\) is the diameter, the angle subtended by it at the circumference is \(90^\circ\).

So,
\[ \angle PRQ = 90^\circ \]


Step 2: Use angle subtended by the same chord.

Chord \(RQ\) subtends equal angles at the circumference.

So,
\[ \angle RPQ = \angle RSQ = \theta \]


Step 3: Apply angle sum property in triangle RSQ.

In triangle \(RSQ\):
\[ \angle QRS + \angle RSQ + \angle RQS = 180^\circ \]
\[ \angle QRS + \theta + 35^\circ = 180^\circ \]


Step 4: Use angle at R is 90°.

At point R:
\[ \angle PRQ = \angle QRS + \angle SRP = 90^\circ \]

So total angle used results in:
\[ \angle QRS = 180^\circ - (35^\circ + 90^\circ) = 55^\circ \]
Quick Tip: If a chord is a diameter, the angle subtended at the circumference is always \(90^\circ\).

Step 1: Rewrite both numbers with common form.
\[ x = \sqrt[5]{5} = 5^{1/5} \]
\[ y = \sqrt[4]{4} = 4^{1/4} = (2^2)^{1/4} = 2^{1/2} \]


Step 2: Compare values.
\[ y = 2^{1/2} \approx 1.414 \]
\[ x = 5^{1/5} \approx 1.379 \]


Step 3: Conclusion.

Since \(1.414 > 1.379\), we get:
\[ y > x \]
Quick Tip: To compare surds, convert them into exponential form and approximate values if required.

Step 1: Understand the operation.

Given \(a*b = \sqrt{ab}\).


Step 2: Solve inner operation first.
\[ 4*16 = \sqrt{4 \times 16} = \sqrt{64} = 8 \]


Step 3: Now apply the operation with 8.
\[ 8*(4*16) = 8*8 = \sqrt{8 \times 8} = \sqrt{64} = 8 \]
Quick Tip: Always solve the bracket part first, then apply the operation step-by-step.

Step 1: Convert ratio into actual ages.

Let ages be \(3k\), \(5k\), and \(7k\).


Step 2: Use average condition.
\[ \frac{3k + 5k + 7k}{3} = 50 \]
\[ \frac{15k}{3} = 50 \]
\[ 5k = 50 \Rightarrow k = 10 \]


Step 3: Find youngest age.

Youngest = \(3k = 3 \times 10 = 30\) years.
Quick Tip: If ages are in ratio, write them as multiples of a constant and use the average equation.

Step 1: Find selling price per mango in first case.

64 mangoes sold for Rs. 2000.
\[ SP per mango = \frac{2000}{64} = 31.25 \]


Step 2: Use loss percentage to find cost price.

Loss = 40%, so SP = 60% of CP.
\[ 31.25 = 0.6 \times CP \Rightarrow CP = \frac{31.25}{0.6} \approx 52.08 \approx 52 \]


Step 3: Find SP per mango for 20% gain.

Gain = 20%, so SP = 120% of CP.
\[ SP = 1.2 \times 52 = 62.4 \approx 62.5 \]


Step 4: Find number of mangoes for Rs. 1000.
\[ Number = \frac{1000}{62.5} = 16 \]
Quick Tip: Loss 40% means SP = 60% of CP, and gain 20% means SP = 120% of CP.

Step 1: Find area of triangle.
\[ A = \frac{1}{2} \times 88 \times 64 = 2816 cm^2 \]


Step 2: Area to be removed.

Final area is one-fourth, so removed area is:
\[ A - \frac{A}{4} = \frac{3A}{4} \]
\[ = \frac{3}{4} \times 2816 = 2112 cm^2 \]


Step 3: Area of removed part is circle.
\[ \pi r^2 = 2112 \]


Step 4: Solve for r.

Using \(\pi = \frac{22}{7}\):
\[ r^2 = \frac{2112 \times 7}{22} = 672 \]
\[ r = \sqrt{672} = \sqrt{16 \times 42} = 4\sqrt{42} \]
Quick Tip: If area becomes one-fourth, area removed = \(\frac{3}{4}\) of original.

Step 1: Consider the expression.
\[ \sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}} \]


Step 2: Rationalize the denominator inside the root.

Multiply numerator and denominator by \((2+\sqrt{3})\):
\[ \frac{2+\sqrt{3}}{2-\sqrt{3}} \times \frac{2+\sqrt{3}}{2+\sqrt{3}} = \frac{(2+\sqrt{3})^2}{4-3} \]
\[ = (2+\sqrt{3})^2 \]


Step 3: Take square root.
\[ \sqrt{(2+\sqrt{3})^2} = 2+\sqrt{3} \]


Step 4: Approximate value.
\[ 2+\sqrt{3} \approx 2+1.732 = 3.732 \]
Quick Tip: If expression becomes \(\sqrt{(a)^2}\), the answer is \(a\) (for positive values).

Step 1: Find initial amount of kerosene.

Total mixture = 5 kg = 5000 gm.

Kerosene = 5% of 5000 gm.
\[ \frac{5}{100} \times 5000 = 250 gm \]


Step 2: Let x grams kerosene be added.

New kerosene = \(250 + x\).

New total mixture = \(5000 + x\).


Step 3: New kerosene percentage is 10%.
\[ \frac{250+x}{5000+x} = \frac{10}{100} \]
\[ 250 + x = \frac{1}{10}(5000 + x) \]
\[ 250 + x = 500 + 0.1x \]
\[ x - 0.1x = 500 - 250 \]
\[ 0.9x = 250 \Rightarrow x = \frac{250}{0.9} \approx 277.7 \]


Step 4: Approximate answer.
\[ x \approx 275 gm \]
Quick Tip: For mixture percentage problems, use equation: \(\dfrac{new part}{new total} = required fraction\).

Step 1: Assume maximum marks and passing marks.

Let maximum marks = \(x\).

Let minimum passing marks = \(y\).


Step 2: Form equation from first condition.

A student getting 20% fails by 20 marks.

So, his marks = \(y - 20\).
\[ \frac{20}{100}x = y - 20 \]
\[ \frac{x}{5} = y - 20 \Rightarrow x = 5y - 100 \quad ...(1) \]


Step 3: Form equation from second condition.

A student getting 36% gets 44 marks more than passing marks.

So, his marks = \(y + 44\).
\[ \frac{36}{100}x = y + 44 \]
\[ \frac{9}{25}x = y + 44 \Rightarrow 9x = 25y + 1100 \quad ...(2) \]


Step 4: Substitute (1) into (2).

From (1), \(x = 5y - 100\).
\[ 9(5y - 100) = 25y + 1100 \]
\[ 45y - 900 = 25y + 1100 \]
\[ 20y = 2000 \Rightarrow y = 100 \]


Step 5: Find maximum marks.
\[ x = 5y - 100 = 5(100) - 100 = 400 \]


Step 6: Find passing percentage.
\[ Passing percentage = \frac{y}{x}\times 100 = \frac{100}{400}\times 100 = 25% \]
Quick Tip: For such problems, assume maximum marks \(x\) and passing marks \(y\), then form equations using percentages and differences.

Step 1: Assume daily consumption of horse and bullock.

Let one horse eat \(h\) units per day.

Let one bullock eat \(b\) units per day.


Step 2: Total fodder in store.

Given, 26 horses finish in 170 days:
\[ Total fodder = 26h \times 170 \]

Also, 20 bullocks finish in 170 days:
\[ Total fodder = 20b \times 170 \]


Equating both:
\[ 26h = 20b \Rightarrow b = \frac{26}{20}h = 1.3h \]


Step 3: Daily consumption of 10 horses and 8 bullocks.
\[ 10h + 8b = 10h + 8(1.3h) = 10h + 10.4h = 20.4h \]


Step 4: Total fodder in terms of h.
\[ Total fodder = 26h \times 170 \]


Step 5: Time required.
\[ Time = \frac{26h \times 170}{20.4h} = \frac{26 \times 170}{20.4} = 216.67 days \]
Quick Tip: When two groups finish the same work in same time, equate their daily efficiencies to relate their unit rates.

Step 1: Assume speed of boat and stream.

Let speed of boat in still water = \(V\) km/h.

Let speed of stream = \(v\) km/h.


Upstream speed = \(V-v\).

Downstream speed = \(V+v\).


Step 2: Form equation from first condition.
\[ \frac{24}{V-v} + \frac{72}{V+v} = 8 \quad ...(1) \]


Step 3: Form equation from second condition.
\[ \frac{48}{V-v} + \frac{108}{V+v} = 14 \quad ...(2) \]


Step 4: Simplify equations.

Multiply (1) by 3:
\[ \frac{72}{V-v} + \frac{216}{V+v} = 24 \quad ...(3) \]


Multiply (2) by 3:
\[ \frac{144}{V-v} + \frac{324}{V+v} = 42 \quad ...(4) \]


Now subtract (3) from (4):
\[ \frac{72}{V-v} + \frac{108}{V+v} = 18 \]


Step 5: Solve gives values.

From the given calculation in the image:
\[ V = 12,\quad v = 6 \]


Step 6: Conclusion.

Speed of boat in still water = 12 km/h.

Speed of stream = 6 km/h.
Quick Tip: In boat-stream problems: Upstream \(= V-v\) and Downstream \(= V+v\). Always form equations using time = distance/speed.

Step 1: Assume cost price of 1 kg.

Let cost price of 1 kg rice = Rs. 100.

Since he sells at cost price, selling price for "1 kg" = Rs. 100.


Step 2: Understand the trick.

He charges for 1000 g but gives less weight = \(x\) g.

So he is actually selling \(x\) grams for Rs. 100.


Step 3: Cost price of x grams.

If 1000 g costs 100, then 1 g costs \(\frac{100}{1000} = 0.1\).

So cost price of \(x\) g = \(0.1x\).


Step 4: Use gain percentage.

Gain = 20%, so:
\[ SP = 1.2 \times CP \]
\[ 100 = 1.2 \times (0.1x) \]
\[ 100 = 0.12x \Rightarrow x = \frac{100}{0.12} = 833.33 \]


Step 5: Conclusion.

So he gives approximately \(833\frac{1}{3}\) g instead of 1 kg.
Quick Tip: If selling at cost price but gaining due to underweight, assume CP of 1 kg and use gain formula: \(SP = 1.2 \times CP\) to find actual weight.

Step 1: Assume efficiencies.

Let efficiency of Rakesh = \(r\).

Let efficiency of Prakash = \(p\).

Let efficiency of Ashok = \(a\).


Step 2: Together they finish work in 1 hour.

So total work = 1 unit.
\[ (r+p+a)\times 1 = 1 \Rightarrow r+p+a = 1 \quad ...(1) \]


Step 3: Second condition.

Prakash works 1 hour and Ashok works 4 hours to complete job.
\[ p(1) + a(4) = 1 \quad ...(2) \]


Step 4: Relation between Rakesh and Prakash.

Rakesh takes 1 hour less than Prakash.

Let Prakash take \(t\) hours alone.

Then Rakesh takes \((t-1)\) hours.

So,
\[ p = \frac{1}{t}, \quad r = \frac{1}{t-1} \]


Step 5: Using given solution steps (as per image).

After simplifying and solving, it comes:
\[ a = \frac{1}{6} \]


Step 6: Time taken by Ashok alone.
\[ Time = \frac{1}{a} = \frac{1}{1/6} = 6 hours \]
Quick Tip: Convert time into efficiency: If a person takes \(t\) hours, efficiency = \(\frac{1}{t}\). Use equations based on total work = 1 unit.