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

Step 1: Use the volume formula for a sphere.

The volume of a sphere of radius \(R\) is:
\[ V=\frac{4}{3}\pi R^{3} \]


Step 2: Find the volume of the big sphere.

Here, \(R=8\) cm, so:
\[ V=\frac{4}{3}\pi (8)^{3}=\frac{4}{3}\pi (512) \]


Step 3: Find the volume of one small iron ball.

Each small ball has radius \(r=1\) cm, so its volume is:
\[ v=\frac{4}{3}\pi (1)^{3}=\frac{4}{3}\pi \]


Step 4: Find the number of iron balls formed.

Let \(n\) balls be formed. Since volume is conserved:
\[ n \times v = V \]
\[ n \times \frac{4}{3}\pi = \frac{4}{3}\pi (512) \]
\[ n = 512 \]
Quick Tip: Whenever an object is melted and reshaped, total volume remains the same. Divide total volume by volume of one piece.

Step 1: Use the given condition.

We are given:
\[ 0 < a < 1 \]


Step 2: Check statement (i) \(a^{2}-1>0\).

Squaring \(0 < a < 1\), we get:
\[ 0 < a^{2} < 1 \]

Subtract \(1\) from all sides:
\[ -1 < a^{2}-1 < 0 \]

So \(a^{2}-1\) is negative, hence statement (i) is false.


Step 3: Check statement (ii) \(a^{2}+1>0\).

Since \(a^{2}>0\), adding \(1\) gives:
\[ a^{2}+1>0 \]

So statement (ii) is true.


Step 4: Check statement (iii) \(a^{2}-a>0\).

Factor:
\[ a^{2}-a=a(a-1) \]

Here \(a>0\) but \(a-1 < 0\), so product is negative:
\[ a(a-1) < 0 \]

Hence statement (iii) is false.
Quick Tip: If \(0 < a < 1\), then squaring makes it smaller: \(a^{2} < a\).

Step 1: Assume total revenue in 2010 as 100.

Let total revenue in 2010 be \(100\).


Step 2: Find revenue of company \(T\) in 2010.

From pie chart, \(T=14%\).

So revenue of \(T\) in 2010 is:
\[ 14 \]


Step 3: Increase in revenue of \(T\) in 2011.

Revenue increases by \(20%\):
\[ 20% of 14 = \frac{20}{100}\times 14 = 2.8 \]


Step 4: Total revenue in 2011.

Only \(T\) increases, others remain same.

So total revenue becomes:
\[ 100 + 2.8 = 102.8 \]


Step 5: Percentage increase in total revenue.
\[ \frac{102.8-100}{100}\times 100 = 2.8% \]
Quick Tip: If only one component increases and the rest remain constant, the increase in total equals that component's increase.

Step 1: Condition for doubling in simple interest.

If money doubles, final amount \(A=2P\).

So interest earned:
\[ I=A-P=2P-P=P \]


Step 2: Use simple interest formula.
\[ I=\frac{PTR}{100} \]


Step 3: Substitute values.

Here \(P=200000\), \(R=11.5\), \(I=P=200000\):
\[ 200000=\frac{200000 \times T \times 11.5}{100} \]


Step 4: Solve for \(T\).

Cancel \(200000\):
\[ 1=\frac{T \times 11.5}{100} \]
\[ T=\frac{100}{11.5}=8.695 \]


So time is approximately \(8.7\) years, which lies between \(8\) and \(9\).
Quick Tip: For doubling in simple interest: \(T=\frac{100}{R}\). It is independent of principal amount.

Step 1: Write the given relation.
\[ AB + C = D \]


Step 2: Substitute \(B=6\) and \(D=30\).
\[ 6A + C = 30 \qquad ...(1) \]


Step 3: Substitute \(B=8\) and \(D=36\).
\[ 8A + C = 36 \qquad ...(2) \]


Step 4: Subtract equation (1) from equation (2).
\[ (8A+C)-(6A+C)=36-30 \]
\[ 2A = 6 \Rightarrow A = 3 \]


Step 5: Substitute \(A=3\) in equation (1).
\[ 6(3)+C=30 \]
\[ 18+C=30 \Rightarrow C=12 \]
Quick Tip: To find unknowns quickly, form two equations and subtract them to eliminate one variable.

Step 1: Factorize the quadratic expression.
\[ y^{2}+3y-18 \geq 0 \]

We factorize:
\[ y^{2}+3y-18 = (y+6)(y-3) \]


Step 2: Use the sign rule for product inequality.
\[ (y+6)(y-3) \geq 0 \]

A product is non-negative when:
\[ y+6 \geq 0 and y-3 \geq 0 \quad or \quad y+6 \leq 0 and y-3 \leq 0 \]


Step 3: Solve both cases.

Case 1:
\[ y \geq -6 and y \geq 3 \Rightarrow y \geq 3 \]

Case 2:
\[ y \leq -6 and y \leq 3 \Rightarrow y \leq -6 \]


Step 4: Final solution interval.
\[ y \geq 3 \quad or \quad y \leq -6 \]
Quick Tip: For \((x-a)(x-b)\ge 0\), the solution is always \(x \le \min(a,b)\) or \(x \ge \max(a,b)\).

Step 1: Treat each color group as one unit.

Since balls of the same color must stay together, we consider:

- 3 red balls as 1 unit

- 2 blue balls as 1 unit

- 4 yellow balls as 1 unit

So, we have 3 units to arrange.


Step 2: Arrange the 3 units.

Number of ways to arrange these 3 units:
\[ 3! = 6 \]


Step 3: Arrange balls inside each unit.

Red balls can be arranged in:
\[ 3! = 6 \]

Blue balls can be arranged in:
\[ 2! = 2 \]

Yellow balls can be arranged in:
\[ 4! = 24 \]


Step 4: Multiply all arrangements.
\[ 3! \times 3! \times 2! \times 4! = 6 \times 6 \times 2 \times 24 = 1728 \]
Quick Tip: When items must stay together, treat them as a single block, then multiply by internal arrangements of each block.

Step 1: Read the values from the table for each given option.

We check courier charges for each route mentioned.


Step 2: Compare charges.

From the table:

- Ahmedabad to Jaipur = Rs. 10

- Mumbai to Bangalore = Rs. 25

- Jaipur to Bangalore = Rs. 10

- Kolkata to Mumbai = Rs. 7


Step 3: Select the minimum charge.

The least charge is Rs. 7 for Kolkata to Mumbai.
Quick Tip: For table-based questions, always locate exact row-column intersections carefully before comparing values.

Step 1: Represent the numbers in terms of a constant.

Given ratio \(X:Y:Z = 12:15:25\).

Let:
\[ X=12k,\quad Y=15k,\quad Z=25k \]


Step 2: Find required differences.

Difference between \(Y\) and \(X\):
\[ Y-X = 15k-12k = 3k \]

Difference between \(Z\) and \(Y\):
\[ Z-Y = 25k-15k = 10k \]


Step 3: Ratio of differences.
\[ (Y-X):(Z-Y) = 3k:10k = 3:10 \]
Quick Tip: In ratio questions, you can directly compare differences using the ratio values without finding actual numbers.

Step 1: Understand relative speed concept.

Since they move in opposite directions, relative speed is the sum of speeds.


Step 2: Calculate relative speed.

Speed of Ankush = 3 rounds/hour

Speed of Babulal = 5 rounds/hour
\[ Relative speed = 3 + 5 = 8 rounds/hour \]


Step 3: Find total time interval.

From 9:00 a.m. to 10:30 a.m. = 1.5 hours


Step 4: Find number of crossings.

They cross each other once every 1 round of relative motion.

So number of crossings:
\[ 8 \times 1.5 = 12 \]
Quick Tip: For opposite directions: \(Relative speed = v_1 + v_2\). Crossings \(=Relative speed \times Time\).

Step 1: Use savings ratio to find monthly savings.

Savings ratio of Amit and Bharat = \(37:18\).

Let savings be \(37k\) and \(18k\).

Total monthly savings:
\[ 37k+18k = 55k \]


Step 2: Convert annual savings to monthly savings.

Total annual savings = Rs. 1,32,000

So total monthly savings:
\[ \frac{132000}{12} = 11000 \]


Step 3: Find value of \(k\).
\[ 55k = 11000 \Rightarrow k = 200 \]


Step 4: Find individual monthly savings.

Amit's saving = \(37 \times 200 = 7400\)

Bharat's saving = \(18 \times 200 = 3600\)


Step 5: Use income and expense ratios.

Income ratio = \(5:4\). Let incomes be \(5x\) and \(4x\).

Expense ratio = \(19:21\). Let expenses be \(19y\) and \(21y\).


Savings = Income - Expense

So:
\[ 5x - 19y = 7400 \qquad ...(1) \]
\[ 4x - 21y = 3600 \qquad ...(2) \]


Step 6: Solve the equations.

Multiply (1) by 4 and (2) by 5:
\[ 20x - 76y = 29600 \]
\[ 20x - 105y = 18000 \]

Subtract:
\[ 29y = 11600 \Rightarrow y = 400 \]

Substitute in (2):
\[ 4x - 21(400)=3600 \]
\[ 4x - 8400 = 3600 \Rightarrow 4x = 12000 \Rightarrow x = 3000 \]


Step 7: Find Amit's income.

Amit's income = \(5x = 5 \times 3000 = 15000\).
Quick Tip: When multiple ratios are given, convert each into equations using the relation: \(Savings = Income - Expense\).

Step 1: Use property of radii in a circle.

Since \(OA = OB\) (both are radii), triangle \(AOB\) is isosceles.


Step 2: Use angle sum property of a triangle.

Given central angle:
\[ \angle AOB = 114^\circ \]

So remaining two angles are equal:
\[ \angle ABO = \angle BAO \]
\[ \angle ABO = \frac{180^\circ - 114^\circ}{2} \]
\[ \angle ABO = \frac{66^\circ}{2} = 33^\circ \]
Quick Tip: In an isosceles triangle, angles opposite equal sides are equal. Always split the remaining angle equally.

Step 1: Represent the information using a Venn diagram.

We use three sets:

- TOI = 50

- IE = 60

- HT = 48









Step 2: Fill the intersections.

Given:
\[ TOI\cap IE = 20,\quad TOI\cap HT = 18,\quad IE\cap HT = 24 \]

And:
\[ TOI\cap IE\cap HT = 10 \]


So only two-set regions become:
\[ (TOI\cap IE) only = 20-10=10 \]
\[ (TOI\cap HT) only = 18-10=8 \]
\[ (IE\cap HT) only = 24-10=14 \]


Step 3: Find only one newspaper readers.

Only TOI:
\[ 50-(10+8+10)=22 \]

Only IE:
\[ 60-(10+14+10)=26 \]

Only HT:
\[ 48-(8+14+10)=16 \]


Step 4: Total who read only one newspaper.
\[ 22+26+16 = 64 \]
Quick Tip: In 3-set Venn problems, always subtract the common 3-set value first from all pairwise intersections.

Step 1: Understand the movement of the minute hand.

In 60 minutes, minute hand completes \(360^\circ\).

So in 30 minutes, it covers:
\[ \frac{30}{60}\times 360^\circ = 180^\circ \]


Step 2: Recognize it forms a semicircle.

A \(180^\circ\) sweep forms a semicircle.

Radius \(r = 42\) mm.


Step 3: Calculate area of semicircle.
\[ A = \frac{1}{2}\pi r^{2} \]
\[ A = \frac{1}{2}\times 3.14 \times 42^{2} \]
\[ A = \frac{1}{2}\times 3.14 \times 1764 \]
\[ A = \frac{5539.0}{2} = 2769.5 \]
Quick Tip: Minute hand makes \(360^\circ\) in 60 minutes, so in \(t\) minutes it makes \(\frac{t}{60}\times 360^\circ\).

Step 1: Use factor theorem.

If \((x+4)\) is a factor of \(f(x)\), then:
\[ f(-4)=0 \]


Step 2: Form the polynomial function.
\[ f(x)=x^{3}+2x^{2}+bx+68 \]


Step 3: Substitute \(x=-4\).
\[ f(-4)=(-4)^{3}+2(-4)^{2}+b(-4)+68=0 \]
\[ -64+2(16)-4b+68=0 \]
\[ -64+32-4b+68=0 \]
\[ 36-4b=0 \Rightarrow 4b=36 \Rightarrow b=9 \]
Quick Tip: Factor theorem: If \((x-a)\) is a factor of \(f(x)\), then \(f(a)=0\).

Step 1: Use the formula for average speed.

Average speed is given by:
\[ Average Speed=\frac{Total Distance}{Total Time} \]


Step 2: Find the total distance covered.
\[ 12 + 36 + 32 = 80 km \]


Step 3: Find the total time taken.

Time for 12 km at 6 km/hr:
\[ \frac{12}{6}=2 hr \]

Time for 36 km at 9 km/hr:
\[ \frac{36}{9}=4 hr \]

Time for 32 km at 4 km/hr:
\[ \frac{32}{4}=8 hr \]

Total time:
\[ 2+4+8=14 hr \]


Step 4: Calculate average speed.
\[ Average Speed=\frac{80}{14}=5.71 km/hr \]
Quick Tip: Average speed is always \(\frac{Total Distance}{Total Time}\), not the average of speeds.

Step 1: Convert filling times into rates.

Rate of pipe A:
\[ \frac{1}{120} tank/min \]

Rate of pipe B:
\[ \frac{1}{150} tank/min \]

Let rate of outlet C be \(\frac{1}{x}\) tank/min (it empties, so it will be subtracted).


Step 2: Use combined rate formula.

All three together fill the tank in 100 minutes:
\[ \frac{1}{120}+\frac{1}{150}-\frac{1}{x}=\frac{1}{100} \]


Step 3: Solve the equation.

LCM of \(120,150,100\) is \(600\).
\[ \frac{5}{600}+\frac{4}{600}-\frac{1}{x}=\frac{6}{600} \]
\[ \frac{9}{600}-\frac{1}{x}=\frac{6}{600} \]
\[ \frac{3}{600}=\frac{1}{x} \]
\[ x=200 minutes \]


Step 4: Convert minutes into hours and minutes.
\[ 200 minutes = 3 hours 20 minutes \]
Quick Tip: In pipe problems, always convert time into rate. Outlet rate is negative since it empties the tank.

Step 1: Identify work and rest hours in week 1.

From the table for 1st week:

Work = 6 hours/day, Rest = 3 hours/day.

Since work \(>\) rest, wage per working hour = Rs. 60.

Rest wage is half of working wage:
\[ Rest wage per hour = \frac{60}{2}=30 \]


Step 2: Find daily earning for week 1 and week 3.

Daily earning:
\[ (60 \times 6) + (30 \times 3) = 360 + 90 = 450 \]

Each week has 6 days, so week earning:
\[ 450 \times 6 = 2700 \]

Weeks 1 and 3 have same pattern, so total for week 1 and 3:
\[ 2700 \times 2 = 5400 \]


Step 3: Identify work and rest hours in week 2 and week 4.

Opposite pattern means:

Work = 3 hours/day, Rest = 6 hours/day.

Now rest \(>\) work, so wage per working hour becomes \(\frac{60}{3}=20\).

Rest wage = half of 20 = 10.


Step 4: Find daily earning for week 2 and week 4.

Daily earning:
\[ (20 \times 3) + (10 \times 6) = 60 + 60 = 120 \]

Week earning:
\[ 120 \times 6 = 720 \]

Weeks 2 and 4 total:
\[ 720 \times 2 = 1440 \]


Step 5: Total salary for the 1st month (4 weeks).
\[ 5400 + 1440 = 6840 \]
Quick Tip: In pattern wage problems, compare work and rest hours first, then apply wage-rate conditions week-wise.

Step 1: Find total number of ways to choose 3 balls from 12 balls.

Total balls = \(8+4=12\).
\[ Total ways=\binom{12}{3} \]
\[ \binom{12}{3}=\frac{12 \times 11 \times 10}{3 \times 2 \times 1}=220 \]


Step 2: Find favorable outcomes.

Choose 2 yellow balls out of 8:
\[ \binom{8}{2}=28 \]

Choose 1 black ball out of 4:
\[ \binom{4}{1}=4 \]

Favorable ways:
\[ 28 \times 4 = 112 \]


Step 3: Find probability.
\[ P=\frac{112}{220}=\frac{28}{55} \]
Quick Tip: In probability without replacement, use combinations: \(\frac{favorable selections}{total selections}\).

Step 1: Let \(\tan A = x\).

Then:
\[ \cot A = \frac{1}{x} \]

Given:
\[ x+\frac{1}{x}=\sqrt{5} \]


Step 2: Cube both sides.
\[ \left(x+\frac{1}{x}\right)^{3}=(\sqrt{5})^{3} \]
\[ x^{3}+\frac{1}{x^{3}}+3\left(x+\frac{1}{x}\right)=5\sqrt{5} \]


Step 3: Substitute \(x+\frac{1}{x}=\sqrt{5}\).
\[ x^{3}+\frac{1}{x^{3}}+3\sqrt{5}=5\sqrt{5} \]
\[ x^{3}+\frac{1}{x^{3}}=2\sqrt{5} \]


Step 4: Convert back to trigonometric form.
\[ \tan^{3}A+\cot^{3}A = 2\sqrt{5} \]
Quick Tip: Use \((a+b)^{3}=a^{3}+b^{3}+3ab(a+b)\). Here \(ab=\tan A \cdot \cot A = 1\).

Step 1: Let marked price of two balls be \(X\).

A discount of \(15%\) means selling price is \(85%\) of marked price.


Step 2: Form the equation.
\[ \frac{85}{100}X = 37.40 \]


Step 3: Solve for \(X\).
\[ X = \frac{37.40 \times 100}{85} = 44 \]


Step 4: Find marked price of each ball.
\[ \frac{44}{2}=22 \]
Quick Tip: After discount: \(Selling Price = \frac{100-d}{100} \times Marked Price\).

Step 1: Use modulus definition.
\[ |x| = \begin{cases} x, & x \ge 0
-x, & x < 0 \end{cases} \]

Here \(x=2a-3\).


Step 2: Case 1: \(2a-3 \ge 0\).
\[ 2a-3 \ge 0 \Rightarrow a \ge \frac{3}{2} \]

Then:
\[ |2a-3| = 2a-3 \]

Equation becomes:
\[ 2a-3 = 3a+2 \Rightarrow -a = 5 \Rightarrow a = -5 \]

But \(a=-5\) does not satisfy \(a \ge \frac{3}{2}\).

So this case is rejected.


Step 3: Case 2: \(2a-3 < 0\).
\[ 2a-3 < 0 \Rightarrow a < \frac{3}{2} \]

Then:
\[ |2a-3| = -(2a-3)=3-2a \]

Equation becomes:
\[ 3-2a = 3a+2 \Rightarrow 1 = 5a \Rightarrow a = \frac{1}{5} \]

This satisfies \(a < \frac{3}{2}\), so it is valid.
Quick Tip: For modulus equations, always solve in two cases and then verify if the solution satisfies the assumed condition.

Step 1: Identify fraction removed each time.

Each time 4 litres is removed from 32 litres, so fraction removed is:
\[ \frac{4}{32}=\frac{1}{8} \]

So fraction of wine remaining after each operation is:
\[ 1-\frac{1}{8}=\frac{7}{8} \]


Step 2: Apply repeated replacement formula.

After 3 operations, wine left:
\[ 32 \left(\frac{7}{8}\right)^{3} \]
\[ = 32 \times \frac{343}{512} = \frac{343}{16} = 21.4375 \]


Step 3: Find water in the jar.

Total volume remains 32 litres.
\[ Water = 32 - 21.4375 = 10.5625 \]


Step 4: Find ratio wine : water.
\[ 21.4375 : 10.5625 \]

Multiply both by 16:
\[ 343 : 169 \]
Quick Tip: In repeated replacement, wine left after \(n\) operations \(= V\left(1-\frac{x}{V}\right)^n\).

Step 1: Use formula for geometric mean.

Geometric mean between \(a\) and \(b\) is:
\[ \sqrt{a \times b} \]


Step 2: Simplify the two given numbers.
\[ 30+\sqrt{200}=30+10\sqrt{2}=10(3+\sqrt{2}) \]
\[ 54-\sqrt{648}=54-18\sqrt{2}=18(3-\sqrt{2}) \]


Step 3: Multiply the two expressions.
\[ a \times b = 10(3+\sqrt{2}) \times 18(3-\sqrt{2}) \]
\[ =180 \left((3+\sqrt{2})(3-\sqrt{2})\right) \]
\[ =180(3^{2}-(\sqrt{2})^{2}) =180(9-2) =180 \times 7 =1260 \]


Step 4: Take the square root.
\[ \sqrt{1260}=\sqrt{36 \times 35}=6\sqrt{35} \]
Quick Tip: Use \((a+b)(a-b)=a^{2}-b^{2}\) to simplify products involving surds.

Step 1: Convert efficiency relation into time relation.

Anil is twice as efficient as Bharat.
\[ Efficiency of Anil : Efficiency of Bharat = 2:1 \]

Time is inversely proportional to efficiency, so:
\[ Time of Anil : Time of Bharat = 1:2 \]


Step 2: Assume Bharat takes \(t\) minutes.

Then Anil takes \(t-30\) minutes.

So:
\[ \frac{t-30}{t} = \frac{1}{2} \]


Step 3: Solve for \(t\).
\[ 2(t-30)=t \Rightarrow 2t-60=t \Rightarrow t=60 \]

So Bharat takes 60 min and Anil takes 30 min.


Step 4: Find combined time.

Combined work rate:
\[ \frac{1}{30}+\frac{1}{60}=\frac{2+1}{60}=\frac{3}{60}=\frac{1}{20} \]

So together time:
\[ 20 minutes \]
Quick Tip: If two people take \(x\) and \(y\) time, together time \(=\frac{xy}{x+y}\).