IPMAT Indore 2019 MCQ PYQs

Q1:ipmat indore 2019QA › Simple & Compound InterestEasyMCQ · MCQ

If the compound interest earned on a certain sum for 2 years is twice the amount of simple interest for 2 years, then the rate of interest per annum is _______ percent

A200%
B2%
C4%
D400%

Pick an option to attempt

The Setup: Let's speedrun this finance glitch. Compound interest beats simple interest only because of interest earned *on* the first year's interest - so demanding that CI be exactly twice SI is a very aggressive ask, and the rate it forces is correspondingly extreme. Step 1: Define the base formulas. Let

P

be the principal and

r

the annual rate as a decimal.

SI=P\cdot r\cdot 2=2Pr, \qquad CI=P(1+r)^2-P

Step 2: Equate and expand. The condition is

CI=2\times SI

:

P(1+r)^2-P=2(2Pr)

Divide through by

P

(a principal of zero is meaningless) and expand:

1+2r+r^2-1=4r \implies r^2+2r=4r

Step 3: Solve for the rate.

r^2-2r=0 \implies r(r-2)=0

The root

r=0

would mean no interest at all, making

CI=SI=0

- technically satisfying the equation but describing no loan. Discard it, leaving

r=2

as a decimal, i.e.

200\%

. Step 4: Sanity-check, because 200% looks absurd. Take

P=100

at

r=200\%

: * Simple interest:

100\times 2\times 2=400

* Compound:

100(1+2)^2-100=900-100=800

And

800=2\times 400

exactly. The answer really is 200% - the reason it feels wrong is that doubling CI relative to SI is a far harsher demand than it sounds, and only a triple-per-year growth factor achieves it. Option 2 is the trap: solve correctly, get

r=2

, and then read that 2 as a percentage instead of as the decimal it is. Final Answer: 200%

Q2:ipmat indore 2019QA › FactorisationEasyMCQ · MCQ

The maximum value of the natural number

n

for which

21^n

divides

50!

is

A6
B7
C8
D9

Pick an option to attempt

The Setup: Factorials scale up like a boss-level mob. To count how many times

21

divides

50!

, break 21 into primes and count each prime's supply separately - whichever runs out first is the bottleneck. Step 1: Prime factorise the base.

21=3\times 7

Each copy of 21 consumes one 3 and one 7, so

21^n

needs

n

of each. Step 2: Count the supply of each prime with Legendre's formula. The exponent of a prime

p

in

n!

is

\sum_{i\geq 1}\left\lfloor\frac{n}{p^i}\right\rfloor

.

E_7(50!)=\left\lfloor\frac{50}{7}\right\rfloor+\left\lfloor\frac{50}{49}\right\rfloor=7+1=8

E_3(50!)=\left\lfloor\frac{50}{3}\right\rfloor+\left\lfloor\frac{50}{9}\right\rfloor+\left\lfloor\frac{50}{27}\right\rfloor=16+5+1=22

Step 3: Take the minimum. The larger prime really is the scarcer one, and that is a theorem rather than a guess: for primes

p<q

we have

p^i<q^i

, so

\left\lfloor\frac{n}{p^i}\right\rfloor\geq\left\lfloor\frac{n}{q^i}\right\rfloor

term by term, and summing gives

E_p(n!)\geq E_q(n!)

. Still, computing both counts is worth the ten seconds - it turns a general principle into the concrete numbers the answer needs:

n=\min\left(E_3,\ E_7\right)=\min(22,\ 8)=8

With 22 threes available and only 8 sevens, the sevens are exhausted first - we can assemble 8 copies of 21, and the 14 spare threes are useless without partners. Final Answer: 8

Q3:ipmat indore 2019QA › RemainderMediumMCQ · MCQ

The remainder when

(29^{29})^{29}

is divided by

9

is

A1
B2
C3
D4

Pick an option to attempt

The Setup: Modular arithmetic is overpowered for shrinking absurd numbers. Rather than computing a skyscraper of digits, reduce the base mod 9 and exploit the fact that its powers cycle. Step 1: Read the exponent tower correctly. This is the first trap:

\left(29^{29}\right)^{29}

has brackets, so the exponents multiply:

\left(29^{29}\right)^{29}=29^{29\times 29}=29^{841}

This is emphatically not

29^{\left(29^{29}\right)}

, which would be an astronomically larger number. When the tower is bracketed from the bottom, multiply. Step 2: Simplify the base modulo 9. Since

29=27+2

and

27

is a multiple of 9:

29\equiv 2\pmod 9

So the task reduces to finding

2^{841}\bmod 9

. Step 3: Find the cycle. List powers of 2 modulo 9 until they repeat:

2^1\equiv 2,\quad 2^2\equiv 4,\quad 2^3\equiv 8,\quad 2^4\equiv 7,\quad 2^5\equiv 5,\quad 2^6\equiv 1

Hitting 1 at the sixth power means the cycle length is exactly 6, consistent with

\phi(9)=6

. Step 4: Reduce the exponent modulo the cycle length.

841=6\times 140+1 \implies 841\equiv 1\pmod 6

So

2^{841}

sits at the same position in the cycle as

2^1

:

2^{841}\equiv 2^1\equiv 2\pmod 9

Final Answer: 2

Q4:ipmat indore 2019QA › Divisibility RulesEasyMCQ · MCQ

Placing which of the following two digits at the right end of

4530

makes the resultant six digit number divisible by

6, 7

and

9

:

A96
B78
C42
D54

Pick an option to attempt

The Setup: Rather than running three separate divisibility side quests, fuse them into one by taking the LCM - a number divisible by 6, 7 and 9 is exactly a number divisible by their least common multiple. Step 1: Find the master divisor. Factorising:

6=2\times 3

,

7=7

,

9=3^2

. The LCM takes the highest power of each prime:

\text{LCM}(6,7,9)=2\times 3^2\times 7=126

Note it is not

6\times 7\times 9=378

- 6 and 9 share a factor of 3, so the product over-counts. Step 2: Build the number. Appending two digits to 4530 gives

453000+x

, where

x

runs from 00 to 99. Step 3: Find the gap. Divide the base by 126:

453000=126\times 3595+30

The base overshoots a multiple of 126 by 30, so to reach the next one we must add:

126-30=96

That gives

453096=126\times 3596

. And since the following candidate would be

96+126=222

- far beyond two digits - 96 is the only possible answer in range, which is worth noting rather than stopping at the first hit. Step 4: Confirm the distractors fail. All three wrong options clear the divisibility by 6, so that test alone separates nothing - 78 and 42 both fall at the 7, and 54 falls at the 9:

Appended	Number	by 6	by 7	by 9
96	453096	yes	yes	yes
78	453078	yes	no	yes
42	453042	yes	no	yes
54	453054	yes	yes	no

Final Answer: 96

Q5:ipmat indore 2019QA › Set TheoryEasyMCQ · MCQ

In a school

70\%

of the boys like cricket and

50\%

like football. If

x\%

like both Cricket and Football, then

A $20 \leq x \leq 50$
B $x \leq 20$
C $x \geq 50$
D $10 \leq x \leq 70$

Pick an option to attempt

The Setup: Two overlapping sets, and we need the floor and ceiling on the overlap

x

. Both bounds come from hard constraints, not from assumptions about the school. Step 1: The ceiling. An intersection can never exceed either of the sets forming it. Only 50% like football, and everyone who likes both is among them:

x\leq\min(70,\ 50)=50

Step 2: The floor - derived, not assumed. The union of the two groups cannot exceed the whole school, so

n(C\cup F)\leq 100

. By inclusion-exclusion:

n(C\cup F)=70+50-x=120-x

Combining the two:

120-x\leq 100 \implies x\geq 20

This is worth stating as an inequality rather than as *suppose everyone likes at least one sport*. That supposition is only the extreme case; the real reason

x

cannot drop below 20 is that 70 and 50 together already claim 120 percentage points of a 100-point school, so at least 20 points must be double-counted. The overlap is forced by arithmetic, whether or not anyone likes neither sport. Step 3: Combine. Both bounds are attainable -

x=50

when every football fan also likes cricket, and

x=20

when the two groups spread across the whole school with nobody left out:

20\leq x\leq 50

Final Answer:

20 \leq x \leq 50

Q6:ipmat indore 2019QA › Set TheoryMediumMCQ · MCQ

In a class of 65 students 40 like cricket, 25 like football and 20 like hockey. 10 students like both cricket and football, 8 students like football and hockey and 5 students like all three sports. If all the students like at least one sport, then the number of students who like both cricket and hockey is

A7
B8
C10
D12

Pick an option to attempt

The Setup: A triple-set problem where the union is handed to us - every one of the 65 students likes at least one sport - so inclusion-exclusion has exactly one unknown and can simply be solved. Step 1: The master formula. With

C

,

F

,

H

for cricket, football and hockey:

n(C\cup F\cup H)=n(C)+n(F)+n(H)-n(C\cap F)-n(F\cap H)-n(C\cap H)+n(C\cap F\cap H)

Step 2: Substitute. Since every student likes at least one sport, the union is the full class of 65. Let

y=n(C\cap H)

:

65=40+25+20-10-8-y+5

Step 3: Solve.

65=85-18+5-y=72-y \implies y=72-65=7

Step 4: Verify the Venn diagram is actually consistent. A value that satisfies the equation can still be nonsense if it forces a negative region, so break the class into its seven disjoint parts. Subtracting the triple overlap of 5 from each pair:

Region	Count
All three	5
Cricket & football only	$10-5=5$
Football & hockey only	$8-5=3$
Cricket & hockey only	$7-5=2$
Cricket only	$40-5-5-2=28$
Football only	$25-5-5-3=12$
Hockey only	$20-5-3-2=10$

Every region is non-negative, and they total

5+5+3+2+28+12+10=65

- exactly the class size. The answer is not just algebraically valid but genuinely realisable. Final Answer: 7

Q7:ipmat indore 2019QA › InequalitiesHardMCQ · MCQ

If

x \in (a, b)

satisfies the inequality

\dfrac{x - 3}{x^2 + 3x + 2} \geq 1

, then the largest possible value of

b - a

is

A3
B1
C2
DNo real values of x satisfies the inequality

Pick an option to attempt

The Setup: A rational inequality. We cannot cross-multiply, because the denominator's sign is unknown and multiplying by a negative would flip the inequality. Everything goes to one side instead. Step 1: Reframe. Move the 1 across and combine over a common denominator:

\frac{x-3}{x^2+3x+2}-1\geq 0 \implies \frac{x-3-\left(x^2+3x+2\right)}{x^2+3x+2}\geq 0 \implies \frac{-x^2-2x-5}{x^2+3x+2}\geq 0

Step 2: Kill the numerator's sign ambiguity. Multiply by

-1

and reverse the inequality:

\frac{x^2+2x+5}{x^2+3x+2}\leq 0

Check the numerator's discriminant:

\Delta=2^2-4(1)(5)=4-20=-16<0

With

\Delta<0

and a positive leading coefficient,

x^2+2x+5

is **strictly positive for every real

x

- it never touches zero. So the numerator can never make the fraction zero or negative on its own. Step 3: Force the denominator negative.** A positive over something, being

\leq 0

, requires the denominator to be strictly negative (zero is not allowed - the expression would be undefined):

(x+1)(x+2)<0

An upward parabola sits below zero strictly between its roots:

x\in(-2,-1)

Note the interval is open because the endpoints

-2

and

-1

make the denominator zero, where the original expression is undefined. Separately worth seeing: the numerator

x^2+2x+5

never vanishes, so the fraction is never exactly zero - meaning equality in the original

\geq 1

is never attained, and the condition behaves throughout as a strict

>1

. Step 4: Compute the width. With

a=-2

and

b=-1

:

b-a=-1-(-2)=1

Option 4 - claiming no real

x

works - is the trap for anyone who mishandles the sign flip in Step 2 and concludes the fraction can never be negative. Final Answer: 1

Q8:ipmat indore 2019QA › IdentitiesHardMCQ · MCQ

If

a, b, c

are real numbers and

a^2 + b^2 + c^2 = 1

, then the set of values

ab + bc + ca

can take is:

A[-1,2]
B[- $\frac{1}{2}$ , 2]
C[-1,1]
D[- $\frac{1}{2}$ , 1]

Pick an option to attempt

The Setup: We need the exact range of

ab+bc+ca

on the unit sphere

a^2+b^2+c^2=1

. Both bounds fall out of the fact that a real square is never negative - but a bound is only the answer if it is actually reached, so we must exhibit a witness for each. Step 1: The lower bound. Expand a perfect square, which cannot be negative:

(a+b+c)^2\geq 0 \implies a^2+b^2+c^2+2(ab+bc+ca)\geq 0

Substituting the constraint:

1+2(ab+bc+ca)\geq 0 \implies ab+bc+ca\geq -\frac{1}{2}

Step 2: The upper bound. Use the sum of pairwise squared differences, also non-negative:

(a-b)^2+(b-c)^2+(c-a)^2\geq 0 \implies 2\left(a^2+b^2+c^2\right)-2(ab+bc+ca)\geq 0

2(1)-2(ab+bc+ca)\geq 0 \implies ab+bc+ca\leq 1

Step 3: Show both ends are attained. Inequalities alone would allow a smaller range, so produce explicit points on the sphere: * Upper: take

a=b=c=\frac{1}{\sqrt{3}}

. Then

a^2+b^2+c^2=3\cdot\frac{1}{3}=1

✓, and

ab+bc+ca=3\cdot\frac{1}{3}=1

. Equality holds because

(a-b)^2+(b-c)^2+(c-a)^2=0

exactly when all three are equal. * Lower: take

a=\frac{1}{\sqrt{2}}

,

b=-\frac{1}{\sqrt{2}}

,

c=0

. Then

a^2+b^2+c^2=\frac{1}{2}+\frac{1}{2}+0=1

✓, and

ab+bc+ca=-\frac{1}{2}+0+0=-\frac{1}{2}

. Equality holds because

a+b+c=0

. Step 4: Conclude. The expression is continuous on a connected sphere, so it sweeps every value between the two attained extremes - the set is the closed interval, endpoints included:

-\frac{1}{2}\leq ab+bc+ca\leq 1

Final Answer: [-

\frac{1}{2}

, 1]

Q9:ipmat indore 2019QA › LogarithmsMediumMCQ · MCQ

The inequality

\log_{2} \frac{3x - 1}{2 - x} < 1

holds true for

A $x \in \left( \frac{1}{3}, 1 \right)$
B $x \in \left( \frac{1}{3}, 2 \right)$
C $x \in \left( 0, \frac{1}{3} \right) \cup \left( 1, 2 \right)$
D $x \in \left( -\infty, 1 \right)$

Pick an option to attempt

The Setup: Before touching the inequality we must respect the logarithm's domain. A log of a non-positive number does not exist, so any

x

that breaks the argument is disqualified before the inequality is even considered - and skipping this step is exactly how the wrong option gets picked here. Step 1: Determine the domain. The argument must be strictly positive:

\frac{3x-1}{2-x}>0

The critical points are

x=\frac{1}{3}

(numerator zero) and

x=2

(denominator zero). Testing the three regions: for

x<\frac{1}{3}

the numerator is negative and denominator positive, giving a negative; for

\frac{1}{3}<x<2

both are positive; for

x>2

the numerator is positive and denominator negative. So:

\text{Domain}=\left(\frac{1}{3},\ 2\right)

Step 2: Unpack the inequality. The base is

2>1

, so the logarithm is increasing and the inequality direction is preserved when exponentiating:

\frac{3x-1}{2-x}<2^1=2

Move everything to one side rather than cross-multiplying, since

2-x

changes sign:

\frac{3x-1}{2-x}-2<0 \implies \frac{3x-1-2(2-x)}{2-x}<0 \implies \frac{5x-5}{2-x}<0

Step 3: Solve. Factor and tidy the sign:

\frac{5(x-1)}{-(x-2)}<0 \implies \frac{x-1}{x-2}>0

A quotient is positive when both parts share a sign, giving

x<1

or

x>2

. Step 4: Intersect with the domain - this is where the question is won or lost.

\left[(-\infty,1)\cup(2,\infty)\right]\cap\left(\frac{1}{3},2\right)=\left(\frac{1}{3},\ 1\right)

The branch

x>2

solves the algebra perfectly but lies entirely outside the domain, where the logarithm does not exist at all. Discarding it is not optional bookkeeping: carry it through and the answer set is simply wrong, no matter how clean the algebra behind it looked. Final Answer:

x \in \left( \frac{1}{3}, 1 \right)

Q10:ipmat indore 2019QA › InequalitiesMediumMCQ · MCQ

The set of values of

x

which satisfy the inequality

0.7^{(2x^2 - 3x + 4)} < 0.343

is

A $\left( \frac{1}{2}, 1 \right)$
B $\left( \frac{1}{2}, \infty \right)$
C $\left( -\infty, \frac{1}{2} \right)$
D $\left( -\infty, \frac{1}{2} \right) \cup \left( 1, \infty \right)$

Pick an option to attempt

The Setup: An exponential inequality with a trap built into the base. Because

0.7<1

, the function

0.7^t

is decreasing, so comparing exponents reverses the inequality - an UNO reverse card most people forget to play. Step 1: Match the bases. Since

7^3=343

, scaling the decimal places gives:

0.343=(0.7)^3 \implies 0.7^{\left(2x^2-3x+4\right)}<0.7^3

Step 2: Drop the bases and flip. For a base in

(0,1)

, a smaller value corresponds to a larger exponent:

2x^2-3x+4>3

Step 3: Solve the quadratic inequality.

2x^2-3x+1>0

Factor by splitting the middle term:

2x^2-2x-x+1>0 \implies 2x(x-1)-1(x-1)>0 \implies (2x-1)(x-1)>0

The roots are

x=\frac{1}{2}

and

x=1

. The leading coefficient is positive, so the parabola opens upward and is positive outside its roots and negative between them. Step 4: Write the interval.

x<\frac{1}{2} \quad\text{or}\quad x>1

Option 1, the interval

\left(\frac{1}{2},1\right)

, is precisely what you get by not flipping the sign in Step 2: that error yields

2x^2-3x+1<0

, whose solution is the region *between* the roots rather than outside them. The whole question turns on that single step. Final Answer:

\left( -\infty, \frac{1}{2} \right) \cup \left( 1, \infty \right)

Q11:ipmat indore 2019QA › CirclesHardMCQ · MCQ

A chord is drawn inside a circle, such that the length of the chord is equal to the radius of the circle. Now, two circles are drawn, one on each side of the chord, each touching the chord at its midpoint and the original circle. Let k be the ratio of the areas of the bigger inscribed circle and the smaller inscribed circle, then k equals

A $2 + \sqrt{3}$
B $1 + \sqrt{2}$
C $7 + 4\sqrt{3}$
D $97 + 56\sqrt{3}$

Pick an option to attempt

The Setup: Pure equilateral-triangle energy. With the chord equal to the radius, joining the centre to the chord's endpoints makes an equilateral triangle, which hands us the centre-to-chord distance. From there the two inscribed circles are pinned down exactly. Step 1: Find the distance to the chord. Take

R=1

to keep the arithmetic light. A chord of length 1 subtends an equilateral triangle with the centre, and the perpendicular distance from centre to chord is that triangle's height:

d=\sqrt{R^2-\left(\frac{R}{2}\right)^2}=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}

Step 2: Pin down each inscribed circle - this is the load-bearing step. Put the centre

O

at the origin with the chord horizontal at height

d

, so its midpoint is

M=\left(0,\tfrac{\sqrt{3}}{2}\right)

and the perpendicular through

M

is the

y

-axis. A circle touching the chord **at

M

** must have its centre on that perpendicular, at distance

r

from

M

. * Above the chord (away from

O

): centre

\left(0,d+r\right)

. Internal tangency to the big circle means the centres are

R-r

apart, so

d+r=R-r

, giving

2r=R-d

- its **diameter is

R-d

**. * Below the chord (past

O

): centre

\left(0,d-r\right)

with

r>d

, so the distance to

O

is

r-d

, and internal tangency gives

r-d=R-r

, i.e.

2r=R+d

- its **diameter is

R+d

**. So the two circles have as diameters exactly the two pieces into which

M

cuts the vertical diameter of the original circle, running from

M

up to

(0,1)

and from

M

down to

(0,-1)

:

\text{smaller}=1-\frac{\sqrt{3}}{2}, \qquad \text{bigger}=1+\frac{\sqrt{3}}{2}

Step 3: Take the ratio of areas. Areas scale as the square of any linear measure, so the area ratio is the square of the diameter ratio:

k=\left(\frac{1+\frac{\sqrt{3}}{2}}{1-\frac{\sqrt{3}}{2}}\right)^2=\left(\frac{2+\sqrt{3}}{2-\sqrt{3}}\right)^2

Rationalise the inner fraction by multiplying top and bottom by

2+\sqrt{3}

, noting

(2-\sqrt{3})(2+\sqrt{3})=4-3=1

:

k=\left(\frac{(2+\sqrt{3})^2}{1}\right)^2=(2+\sqrt{3})^4

Step 4: Expand in two stages.

(2+\sqrt{3})^2=4+4\sqrt{3}+3=7+4\sqrt{3}

k=(7+4\sqrt{3})^2=49+56\sqrt{3}+48=97+56\sqrt{3}

Watch the option list:

7+4\sqrt{3}

is sitting there as option 3, and it is precisely the ratio of the diameters, not the areas. Stopping one squaring too early lands exactly on it. Final Answer:

97 + 56\sqrt{3}

Q12:ipmat indore 2019QA › QuadrilateralsMediumMCQ · MCQ

Points

P

,

Q

,

R

, and

S

are taken on sides

AB

,

BC

,

CD

, and

DA

of square

ABCD

respectively, so that

\frac{AP}{PB} = \frac{BQ}{QC} = \frac{CR}{RD} = \frac{DS}{SA} = \frac{1}{n}

. Then the ratio of the area of

PQRS

to the area of

ABCD

is

A $1 : (1 + n)$
B $1 : n$
C $1 + n^2 : (1 + n)^2$
D $(1 + n) : (1 + n^2)$

Pick an option to attempt

The Setup: A tilted quadrilateral inside a square. Rather than chase the inner side length directly, compute the four corner triangles and subtract them from the whole - the symmetry makes all four identical. Step 1: Define the segments. Let the big square have side

x

. The ratio

\frac{AP}{PB}=\frac{1}{n}

splits each side into

1+n

equal parts, one part then

n

parts:

AP=BQ=CR=DS=\frac{x}{n+1}, \qquad PB=QC=RD=SA=\frac{nx}{n+1}

Step 2: Compute one corner triangle, then argue the rest. Triangle

PBQ

sits in corner

B

, right-angled there, with legs

PB

and

BQ

:

[\,PBQ\,]=\frac{1}{2}\cdot\frac{nx}{n+1}\cdot\frac{x}{n+1}=\frac{nx^2}{2(n+1)^2}

The other three are congruent, not merely similar: at corner

C

the legs are

QC=\frac{nx}{n+1}

and

CR=\frac{x}{n+1}

- the same two lengths, and likewise at

D

and

A

. The whole configuration maps to itself under a

90^\circ

rotation, so:

\text{Total corner area}=4\times\frac{nx^2}{2(n+1)^2}=\frac{2nx^2}{(n+1)^2}

Step 3: Subtract.

[\,PQRS\,]=x^2-\frac{2nx^2}{(n+1)^2}=x^2\left(\frac{(n+1)^2-2n}{(n+1)^2}\right)=x^2\left(\frac{n^2+2n+1-2n}{(n+1)^2}\right)=x^2\cdot\frac{n^2+1}{(n+1)^2}

Step 4: Form the ratio. Dividing by

[\,ABCD\,]=x^2

, the side length cancels entirely - the answer depends only on

n

:

\frac{[\,PQRS\,]}{[\,ABCD\,]}=\frac{1+n^2}{(1+n)^2}

Sanity check at

n=1

, where every point is a midpoint: the ratio becomes

\frac{2}{4}=\frac{1}{2}

, the well-known result that joining the midpoints of a square halves its area. Final Answer:

1 + n^2 : (1 + n)^2

Q13:ipmat indore 2019QA › CirclesHardMCQ · MCQ

On a circular path of radius 6 m a boy starts from a point

A

on the circumference and walks along a chord

AB

of length 3 m. He then walks along another chord

BC

of length 2 m to reach point

C

. The point

B

lies on the minor arc

AC

. The distance between point

C

from point

A

is

A $\frac{\sqrt{15} + \sqrt{35}}{2}$ m
B $8$ m
C $\sqrt{13}$ m
D $6$ m

Pick an option to attempt

The Setup: Two chords laid end to end inside a circle. Each chord subtends a central angle we can extract with the cosine rule, and because

B

lies on the minor arc

AC

, those two angles simply add to give the angle subtending

AC

. Step 1: Find each central angle. Let

O

be the centre,

R=6

. Triangle

OAB

has sides

6,6,3

, so with

\angle AOB=\alpha

:

\cos\alpha=\frac{6^2+6^2-3^2}{2(6)(6)}=\frac{72-9}{72}=\frac{63}{72}=\frac{7}{8}

Triangle

OBC

has sides

6,6,2

, so with

\angle BOC=\beta

:

\cos\beta=\frac{6^2+6^2-2^2}{2(6)(6)}=\frac{72-4}{72}=\frac{68}{72}=\frac{17}{18}

Step 2: Add the angles. Since

B

lies on the minor arc

AC

, the two central angles sit side by side without overlapping, so

\angle AOC=\alpha+\beta

. Both are central angles in

(0,\pi)

, where sine is positive, so we may take the positive root:

\sin\alpha=\sqrt{1-\frac{49}{64}}=\frac{\sqrt{15}}{8}, \qquad \sin\beta=\sqrt{1-\frac{289}{324}}=\frac{\sqrt{35}}{18}

Now the cosine addition formula:

\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta=\frac{7}{8}\cdot\frac{17}{18}-\frac{\sqrt{15}}{8}\cdot\frac{\sqrt{35}}{18}=\frac{119-\sqrt{525}}{144}=\frac{119-5\sqrt{21}}{144}

using

\sqrt{525}=\sqrt{25\times 21}=5\sqrt{21}

. **Step 3: Cosine rule once more, on triangle

OAC

.**

AC^2=6^2+6^2-2(6)(6)\cos(\alpha+\beta)=72-72\cdot\frac{119-5\sqrt{21}}{144}=72-\frac{119-5\sqrt{21}}{2}

AC^2=\frac{144-119+5\sqrt{21}}{2}=\frac{25+5\sqrt{21}}{2}=\frac{50+10\sqrt{21}}{4}

Step 4: Recognise the perfect square. The surd untangles because

50+10\sqrt{21}

is a square:

\left(\sqrt{15}+\sqrt{35}\right)^2=15+35+2\sqrt{15\cdot 35}=50+2\sqrt{525}=50+10\sqrt{21}

AC=\frac{\sqrt{15}+\sqrt{35}}{2}\approx 4.89\text{ m}

A quick plausibility check:

AC

must be less than

AB+BC=5

(triangle inequality) and more than each chord, and

4.89

sits exactly there. The distractors fail this outright - 8 m and 6 m both exceed the straight-line distance

AB+BC=5

, which is impossible. Final Answer:

\frac{\sqrt{15} + \sqrt{35}}{2}

m

Q14:ipmat indore 2019QA › ModulusEasyMCQ · MCQ

The area enclosed by the curve

2|x| + 3|y| = 6

is

A12 square units
B3 square units
C4 square units
D24 square units

Pick an option to attempt

The Setup: Absolute values mirror the graph across both axes, so this curve is a quadrilateral centred at the origin. Find the intercepts, confirm the shape, and the area follows from one formula. Step 1: Find the intercepts. Setting

y=0

:

2|x|=6 \implies |x|=3 \implies x=\pm 3

Setting

x=0

:

3|y|=6 \implies |y|=2 \implies y=\pm 2

Step 2: Identify the shape. In each quadrant the equation is linear - for instance

2x+3y=6

when both are positive - so the curve is four straight segments joining

(3,0)

,

(0,2)

,

(-3,0)

and

(0,-2)

. Each side runs between an

x

-intercept and a

y

-intercept, so every side has length

\sqrt{3^2+2^2}=\sqrt{13}

: all four equal, making it a genuine rhombus with perpendicular diagonals along the axes. Step 3: Measure the diagonals.

d_1=3-(-3)=6, \qquad d_2=2-(-2)=4

Step 4: Compute the area. For a rhombus (indeed any quadrilateral with perpendicular diagonals), the area is half the product of the diagonals:

\text{Area}=\frac{1}{2}d_1d_2=\frac{1}{2}(6)(4)=12

Equivalently, the first-quadrant piece is a right triangle of legs 3 and 2, area

3

, and symmetry gives four of them:

4\times 3=12

. Two routes, same number. Final Answer: 12 square units

Q15:ipmat indore 2019QA › Straight LinesHardMCQ · MCQ

Two points on a ground are 1 m apart. If a cow moves in the field in such a way that its distance from the two points is always in ratio

3: 2

then

Athe cow moves in a straight line
Bthe cow moves in a circle
Cthe cow moves in a parabola
Dthe cow moves in a hyperbola

Pick an option to attempt

The Setup: A locus question. When a point moves so that the ratio of its distances to two fixed points stays constant, the path is completely determined by that constant - this is the Circle of Apollonius. Step 1: State the condition. Let the fixed points be

A

and

B

with

AB=1

m, and let

P

be the cow. The condition is:

\frac{PA}{PB}=\frac{3}{2}

Step 2: Apply the Apollonius result. For

\frac{PA}{PB}=k

with

k>0

: * If

k=1

the locus is the perpendicular bisector of

AB

- a straight line, since equal distances means equidistant. * If

k\neq 1

the locus is a circle. Here

k=\frac{3}{2}\neq 1

, so the cow traces a circle. Step 3: See it concretely rather than just quoting the theorem. Put

A

at the origin and

B

at

(1,0)

, and let

P=(x,y)

. Then

PA^2=\tfrac{9}{4}PB^2

gives:

4\left(x^2+y^2\right)=9\left((x-1)^2+y^2\right) \implies 5x^2+5y^2-18x+9=0

x^2+y^2-\frac{18}{5}x+\frac{9}{5}=0 \implies \left(x-\frac{9}{5}\right)^2+y^2=\left(\frac{6}{5}\right)^2

That is unmistakably a circle - centre

\left(1.8,\,0\right)

, radius

1.2

m. The two points where it crosses the line

AB

are the internal and external division points of

AB

in the ratio

3:2

, namely

x=0.6

and

x=3

, which is exactly the diameter

[0.6,\,3]

of width

2.4

. Note the circle is not centred on

AB

's midpoint and does not pass through

A

or

B

- a useful check that this really is Apollonius and not the perpendicular bisector. Final Answer: the cow moves in a circle

Q16:ipmat indore 2019QA › TrigonometryHardMCQ · MCQ

Given that

\cos x + \cos y = 1

, the range of

\sin x - \sin y

is

A $[-1, 1]$
B $[-2, 2]$
C $[0, \sqrt{3}]$
D $[-\sqrt{3}, \sqrt{3}]$

Pick an option to attempt

The Setup: One equation constrains the pair

(x,y)

, and we want the range of a second expression. Squaring both and adding is the standard move, because the Pythagorean identity then collapses most of the terms. Step 1: Name the target. Let

S=\sin x-\sin y

, given

\cos x+\cos y=1

. Step 2: Square both and add.

(\cos x+\cos y)^2=1 \implies \cos^2x+2\cos x\cos y+\cos^2y=1

(\sin x-\sin y)^2=S^2 \implies \sin^2x-2\sin x\sin y+\sin^2y=S^2

Adding, and grouping

\sin^2+\cos^2=1

for each of

x

and

y

:

1+1+2\left(\cos x\cos y-\sin x\sin y\right)=1+S^2

The bracket is exactly the expansion of

\cos(x+y)

, so:

2+2\cos(x+y)=1+S^2 \implies S^2=1+2\cos(x+y)

**Step 3: Bound

\cos(x+y)

using the constraint.** Sum-to-product on the given equation, writing

u=\frac{x+y}{2}

and

v=\frac{x-y}{2}

:

2\cos u\cos v=1 \implies \cos u\cos v=\frac{1}{2}

Since

|\cos v|\leq 1

, we need

|\cos u|\geq\frac{1}{2}

, hence

\cos^2u\geq\frac{1}{4}

. The double-angle formula then gives:

\cos(x+y)=2\cos^2u-1 \geq 2\cdot\frac{1}{4}-1=-\frac{1}{2}

and of course

\cos(x+y)\leq 1

always. Substituting into

S^2=1+2\cos(x+y)

:

S^2\in[\,1-1,\ 1+2\,]=[0,\ 3]

Step 4: Confirm both extremes are actually reachable. Bounds are worthless unless attained, so produce explicit angles satisfying

\cos x+\cos y=1

: *

x=\frac{\pi}{3},\ y=-\frac{\pi}{3}

: then

\cos x+\cos y=\frac{1}{2}+\frac{1}{2}=1

✓ and

S=\frac{\sqrt{3}}{2}-\left(-\frac{\sqrt{3}}{2}\right)=\sqrt{3}

, the maximum. *

x=y=\frac{\pi}{3}

: then

\cos x+\cos y=1

✓ and

S=0

, the minimum of

|S|

. Swapping

x

and

y

negates

S

, so

-\sqrt{3}

is reached too, and since

S

varies continuously it sweeps everything between:

S\in\left[-\sqrt{3},\ \sqrt{3}\right]

Final Answer:

[-\sqrt{3}, \sqrt{3}]

Q17:ipmat indore 2019QA › TrigonometryHardMCQ · MCQ

If

sin \theta + cos \theta = m

, then

sin^6 \theta + cos^6 \theta

equals

A $\frac{3(m^2+1)}{4}$
B $\frac{3(m^2-1)}{4}$
C $1-\frac{3(m^2-1)}{4}$
D $1-\frac{3(m^2-1)^2}{4}$

Pick an option to attempt

The Setup: An algebraic identity wearing a trigonometry costume. We must climb from first powers to sixth powers, and the bridge is the single quantity

\sin\theta\cos\theta

, which squaring the given equation hands us immediately. Step 1: Extract the cross-term. Square both sides:

(\sin\theta+\cos\theta)^2=m^2 \implies \sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta=m^2

Since

\sin^2\theta+\cos^2\theta=1

:

1+2\sin\theta\cos\theta=m^2 \implies \sin\theta\cos\theta=\frac{m^2-1}{2}

Step 2: Reduce the sixth powers to that one quantity. Use the sum of cubes

a^3+b^3=(a+b)\left(a^2-ab+b^2\right)

with

a=\sin^2\theta

and

b=\cos^2\theta

:

\sin^6\theta+\cos^6\theta=\left(\sin^2\theta+\cos^2\theta\right)\left(\sin^4\theta-\sin^2\theta\cos^2\theta+\cos^4\theta\right)

The first bracket is 1. For the second, use the identity

a^2+b^2=(a+b)^2-2ab

on

\sin^4\theta+\cos^4\theta

, which turns

a^2-ab+b^2

into

(a+b)^2-3ab

:

\sin^4\theta-\sin^2\theta\cos^2\theta+\cos^4\theta=\left(\sin^2\theta+\cos^2\theta\right)^2-3\sin^2\theta\cos^2\theta=1-3\left(\sin\theta\cos\theta\right)^2

Step 3: Substitute. Note the cross-term gets squared, so the bracket from Step 1 is squared whole:

\sin^6\theta+\cos^6\theta=1-3\left(\frac{m^2-1}{2}\right)^2=1-\frac{3\left(m^2-1\right)^2}{4}

Option 3 drops that squaring and reads

1-\frac{3(m^2-1)}{4}

- the difference is one exponent, so check it before committing. Spot-check at

\theta=0

: then

m=\sin 0+\cos 0=1

, and the formula gives

1-\frac{3(1-1)^2}{4}=1

, matching

\sin^6 0+\cos^6 0=0+1=1

. Final Answer:

1-\frac{3(m^2-1)^2}{4}

Q18:ipmat indore 2019QA › Matrices & DeterminantsEasyMCQ · MCQ

If inverse of the matrix

\left[\begin{array}{cc}2 & -0.5 \\ -1 & x\end{array}\right]

is

\left[\begin{array}{ll}1 & 1 \\ 2 & 4 \end{array}\right]

, then the value of

x

is

A $0.5$
B $1$
C $2$
D $3$

Pick an option to attempt

The Setup: A matrix times its inverse is the identity, in either order. That single fact turns this into two-by-two multiplication with no determinant work at all. Step 1: Set up the product. With

A=\begin{bmatrix}2&-0.5\\-1&x\end{bmatrix}

and

A^{-1}=\begin{bmatrix}1&1\\2&4\end{bmatrix}

:

A^{-1}A=\begin{bmatrix}1&1\\2&4\end{bmatrix}\begin{bmatrix}2&-0.5\\-1&x\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}

**Step 2: Pick the entry that isolates

x

.** The

(1,2)

entry pairs row 1 of

A^{-1}

with column 2 of

A

, and must equal the identity's 0:

(1)(-0.5)+(1)(x)=x-0.5=0 \implies x=0.5

Step 3: Verify all four entries, not just one. A single equation can be satisfied by a value that breaks the others, so substitute

x=0.5

and multiply out completely:

\begin{bmatrix}1&1\\2&4\end{bmatrix}\begin{bmatrix}2&-0.5\\-1&0.5\end{bmatrix}=\begin{bmatrix}2-1&-0.5+0.5\\4-4&-1+2\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}

All four land correctly. Checking the other order too, since an inverse must work from both sides:

\begin{bmatrix}2&-0.5\\-1&0.5\end{bmatrix}\begin{bmatrix}1&1\\2&4\end{bmatrix}=\begin{bmatrix}2-1&2-2\\-1+1&-1+2\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}

Step 4: A determinant cross-check. Since

\det(A)\det\left(A^{-1}\right)

must be 1:

\det A=2(0.5)-(-0.5)(-1)=1-0.5=0.5

and

\det A^{-1}=(1)(4)-(1)(2)=2

, and

0.5\times 2=1

✓. Every other option fails -

x=1,2,3

each leave the

(1,2)

entry non-zero. Final Answer:

0.5

Q19:ipmat indore 2019QA › FunctionsHardMCQ · MCQ

The function

f(x) = \dfrac{x^3 - 5x^2 - 8x}{3}

is

Apositive and monotonically increasing for x $\in (-\infty, \frac{5-\sqrt{57}}{2})$ and x $\in (\frac{5+\sqrt{57}}{2}, +\infty)$
Bnegative and monotonically decreasing for x $\in (-\infty, \frac{5-\sqrt{57}}{2})$ and x $\in (\frac{5+\sqrt{57}}{2},+\infty)$
Cnegative and monotonically increasing for x $\in (-\infty, \frac{5-\sqrt{57}}{2})$ and positive and monotonically increasing for x $\in (\frac{5+\sqrt{57}}{2},+\infty)$
Dpositive and monotonically increasing for x $\in (-\infty, \frac{5-\sqrt{57}}{2})$ and negative and monotonically decreasing for x $\in (\frac{5+\sqrt{57}}{2},+\infty)$

Pick an option to attempt

The Setup: A two-part question: where the function is positive or negative, and where it is increasing or decreasing. The options pair those up, so both must be settled and then matched on the same intervals. Step 1: Find the roots - this fixes the sign. Factor out

x

:

f(x)=\frac{x\left(x^2-5x-8\right)}{3}=0

The roots are

x=0

together with the solutions of

x^2-5x-8=0

:

x=\frac{5\pm\sqrt{25+32}}{2}=\frac{5\pm\sqrt{57}}{2}

Write

\alpha=\frac{5-\sqrt{57}}{2}\approx -1.27

and

\beta=\frac{5+\sqrt{57}}{2}\approx 6.27

, so the three roots in order are

\alpha<0<\beta

. Being a cubic with positive leading coefficient,

f

alternates sign across them:

Interval	Sign of $f$
$x<\alpha$	negative
$\alpha<x<0$	positive
$0<x<\beta$	negative
$x>\beta$	positive

Step 2: Differentiate - this fixes the monotonicity.

f'(x)=\frac{3x^2-10x-8}{3}=\frac{(3x+2)(x-4)}{3}

The critical points are

x=-\frac{2}{3}

and

x=4

, and since the parabola opens upward,

f'>0

outside them:

f \text{ is increasing on } \left(-\infty,-\tfrac{2}{3}\right) \text{ and } (4,\infty), \text{ decreasing on } \left(-\tfrac{2}{3},4\right)

Step 3: Match the two pictures on the same intervals. The containments are what make the answer work, so check them numerically: * On

(-\infty,\alpha)

: here

\alpha\approx -1.27<-\frac{2}{3}\approx -0.67

, so this whole interval lies inside the increasing region

\left(-\infty,-\frac{2}{3}\right)

. From the table,

f

is negative there. So: negative and monotonically increasing. * On

(\beta,\infty)

: here

\beta\approx 6.27>4

, so this interval lies inside the increasing region

(4,\infty)

. From the table,

f

is positive there. So: positive and monotonically increasing. Both intervals are increasing, but they differ in sign - which is precisely what separates the correct option from the three that pair the signs the other way round or call one of them decreasing. Final Answer: negative and monotonically increasing for x

\in (-\infty, \frac{5-\sqrt{57}}{2})

and positive and monotonically increasing for x

\in (\frac{5+\sqrt{57}}{2},+\infty)

Q20:ipmat indore 2019QA › ModulusEasyMCQ · MCQ

For

a > b > c > 0

, the minimum value of the function

f(x) = |x - a| + |x - b| + |x - c|

is

A $2a - b - c$
B $a + b - 2c$
C $a + b + c$
D $a - c$

Pick an option to attempt

The Setup: The sum of distances from a moving point

x

to three fixed points

c<b<a

on a number line. The median minimises it - but rather than quote that rule, we can prove it in two lines and get the exact minimum for free. Step 1: Bound the outer pair. By the triangle inequality, for any real

x

:

|x-a|+|x-c|\geq |(x-c)-(x-a)|=a-c

with equality precisely when

x

lies between

c

and

a

, i.e.

x\in[c,a]

. Step 2: Bound the middle term. Trivially

|x-b|\geq 0

, with equality only at

x=b

. Step 3: Add the bounds. For every real

x

:

f(x)=|x-a|+|x-b|+|x-c|\geq (a-c)+0=a-c

Step 4: Show the floor is reached. The two equality conditions must hold simultaneously, and they can:

x=b

satisfies the second exactly, and since

c<b<a

it also lies inside

[c,a]

, satisfying the first. Substituting

x=b

, and using

a>b

so

|b-a|=a-b

, and

b>c

so

|b-c|=b-c

:

f(b)=(a-b)+0+(b-c)=a-c

The

b

terms cancel, which is why the answer depends only on the outermost two points - the middle value

b

affects where the minimum occurs but not what it is. The bound is attained, so it is the true minimum rather than merely a lower estimate. Final Answer:

a - c

Q21:ipmat indore 2019QA › Progression & SeriesHardMCQ · MCQ

Let

\alpha, \beta

be the roots of

x^2 - x + p = 0

and

\gamma, \delta

be the roots of

x^2 - 4x + q = 0

where p and q are integers. If

\alpha, \beta, \gamma, \delta

are in geometric progression then

p + q

is

A-34
B30
C26
D-38

Pick an option to attempt

The Setup: Two quadratics crossed with a geometric progression. Assign standard GP terms to the four roots, pull sum and product relations out with Vieta, and let the integrality of

p

and

q

pick the branch. Step 1: Assign the GP terms. The stem lists the roots in the order

\alpha,\beta,\gamma,\delta

, so writing the progression as

a,ar,ar^2,ar^3

:

\alpha=a,\quad \beta=ar \quad\text{(roots of the first quadratic)}, \qquad \gamma=ar^2,\quad \delta=ar^3 \quad\text{(roots of the second)}

Step 2: Apply Vieta to the sums. For

x^2-x+p=0

the roots sum to 1; for

x^2-4x+q=0

they sum to 4:

a+ar=a(1+r)=1, \qquad ar^2+ar^3=ar^2(1+r)=4

**Step 3: Divide to isolate

r

.** The first equation says

a(1+r)=1

, which is non-zero, so dividing by it is safe:

\frac{ar^2(1+r)}{a(1+r)}=\frac{4}{1} \implies r^2=4 \implies r=2 \text{ or } r=-2

Step 4: Let integrality choose the branch. Both roots of

r^2=4

satisfy the *sum* equations, so the deciding constraint is that

p

and

q

are integers: *

r=2

: then

a(1+2)=1

gives

a=\frac{1}{3}

, and the roots are

\frac{1}{3},\frac{2}{3},\frac{4}{3},\frac{8}{3}

. This yields

p=\frac{1}{3}\cdot\frac{2}{3}=\frac{2}{9}

and

q=\frac{4}{3}\cdot\frac{8}{3}=\frac{32}{9}

- neither an integer, so this branch is rejected. *

r=-2

: then

a(1-2)=1

gives

-a=1

, so

a=-1

, and the roots are

-1,\ 2,\ -4,\ 8

. **Step 5: Compute

p

and

q

from the products.**

p=\alpha\beta=(-1)(2)=-2, \qquad q=\gamma\delta=(-4)(8)=-32

p+q=-2+(-32)=-34

Step 6: Verify the whole construction. The four roots

-1,2,-4,8

do form a GP with common ratio

-2

throughout. And they genuinely solve their quadratics:

x^2-x-2=(x+1)(x-2)

has roots

-1

and

2

✓, while

x^2-4x-32=(x+4)(x-8)

has roots

-4

and

8

✓. A negative ratio is what makes the signs alternate, which is exactly why the integer branch is the one most people discard by reflex. Final Answer: -34

Q22:ipmat indore 2019QA › Progression & SeriesHardMCQ · MCQ

If

(1 + x - 2x^2)^6 = A_0 + \sum_{r=1}^{12} A_r x^r

, then the value of

A_2 + A_4 + A_6 + \cdots + A_{12}

is

A31
B32
C30
D29

Pick an option to attempt

The Setup: A coefficient-extraction trick. Expanding a sixth power of a trinomial by hand is a losing game; evaluating the polynomial at

x=\pm 1

collects the even-indexed coefficients in one move. **Step 1: Evaluate at

x=1

and

x=-1

.** Write

P(x)=(1+x-2x^2)^6=A_0+A_1x+A_2x^2+\cdots+A_{12}x^{12}

.

P(1)=\left(1+1-2\right)^6=0^6=0

P(-1)=\left(1-1-2\right)^6=(-2)^6=64

Note

P(1)=0

because

x=1

is a root of

1+x-2x^2

- the trinomial factors as

(1-x)(1+2x)

, so raising it to any power keeps that root. Step 2: Add them to kill the odd terms. In

P(-1)

every odd power picks up a minus sign while every even power is unchanged, so adding cancels the odd coefficients and doubles the even ones:

P(1)+P(-1)=2\left(A_0+A_2+A_4+\cdots+A_{12}\right)

0+64=2\left(A_0+A_2+\cdots+A_{12}\right) \implies A_0+A_2+\cdots+A_{12}=32

**Step 3: Strip out

A_0

.** The question starts the sum at

A_2

, so the constant term must be removed. Get it by evaluating at

x=0

, where every term with an

x

vanishes:

A_0=P(0)=(1+0-0)^6=1

A_2+A_4+\cdots+A_{12}=32-1=31

That last subtraction is the whole difficulty of the question - option 2 is 32, sitting there for anyone who stops one step early. Final Answer: 31

Q23:ipmat indore 2019QA › Progression & SeriesEasyMCQ · MCQ

The number of terms common to both the arithmetic progressions

2, 5, 8, 11, ..., 179

and

3, 5, 7, 9, ..., 101

is

A17
B16
C19
D15

Pick an option to attempt

The Setup: Where two arithmetic progressions overlap, the shared terms form a third AP whose common difference is the LCM of the two originals. Find its first term, its step, and where it has to stop. Step 1: Identify the two APs.

\text{AP 1}: 2,5,8,11,\ldots,179 \quad (d_1=3), \qquad \text{AP 2}: 3,5,7,9,\ldots,101 \quad (d_2=2)

Step 2: Build the common AP. Scanning the openings, AP 1 gives

2,5,8,\ldots

and AP 2 gives

3,5,7,\ldots

- the first shared value is

5

. A number common to both must step by a multiple of 3 *and* of 2, so the common difference is:

\text{LCM}(3,2)=6

giving the sequence

5,11,17,23,\ldots

Step 3: Find where it must stop. A common term has to belong to both lists, so it cannot exceed either final term:

\text{upper limit}=\min(179,\ 101)=101

Step 4: Count the terms.

5+(n-1)6\leq 101 \implies (n-1)6\leq 96 \implies n-1\leq 16 \implies n\leq 17

Step 5: Check the endpoint is genuinely common. The count is only right if the 17th term,

5+16\times 6=101

, really appears in both lists - and it does:

101=2+33\times 3

sits in AP 1, and

101=3+49\times 2

is the last term of AP 2. So the bound is tight rather than an overshoot, and the count is exactly 17. Final Answer: 17

Q24:ipmat indore 2019QA › ProbabilityHardMCQ · MCQ

From a pack of 52 cards, we draw one by one, without replacement. If

f(n)

is the probability that an Ace will appear at the

n^{\text{th}}

turn, then

A $f(2) = \frac{1}{13} > f(3)$
B $\frac{1}{13} > f(2) > f(3)$
C $f(3) > f(2) = \frac{1}{13}$
D $f(2) = f(3) = \frac{1}{13}$

Pick an option to attempt

The Setup: This is built to bait a long conditional-probability chain - first draw, second draw given the first, and so on. Step back instead: with an unconditional question about a shuffled deck, every position is alike. Step 1: The symmetry argument. Deal all 52 cards face down in a row. Before anything is revealed, each of the 52 cards is equally likely to be sitting in any given position, so position

n

is no different from position 1. Formally, the number of orderings with an Ace at position

n

is the same for every

n

, so:

f(n)=\frac{\text{number of Aces}}{\text{number of cards}}=\frac{4}{52}=\frac{1}{13} \quad\text{for every } n=1,2,\ldots,52

Step 2: Why the chain reasoning does not contradict this. It is tempting to think the second draw *must* differ from the first, because if the first card was an Ace only 3 Aces remain. But we are not told what the first card was. Averaging over both possibilities restores the balance - take

n=2

:

f(2)=\underbrace{\frac{4}{52}\cdot\frac{3}{51}}_{\text{first was an Ace}}+\underbrace{\frac{48}{52}\cdot\frac{4}{51}}_{\text{first was not}}=\frac{12+192}{2652}=\frac{204}{2652}=\frac{1}{13}

The long route lands on precisely the same value the symmetry argument gives in one line. Conditioning on unknown information changes nothing. Step 3: Conclude. Since the probability is constant across all turns:

f(2)=f(3)=\frac{1}{13}

Options 1, 2 and 3 all assert that

f(2)

and

f(3)

differ, which is exactly the trap of imagining the deck depletes in a way you can observe. Final Answer:

f(2) = f(3) = \frac{1}{13}

Q25:ipmat indore 2019QA › ProbabilityMediumMCQ · MCQ

A die is thrown three times and the sum of the three numbers is found to be 15. The probability that the first throw was a four is

A $\frac{1}{6}$
B $\frac{1}{4}$
C $\frac{1}{5}$
D $\frac{1}{10}$

Pick an option to attempt

The Setup: Conditional probability with the answer already partly spoiled - we know the sum is 15. So the sample space shrinks to just those outcomes, and we count how many of them open with a 4. Step 1: Enumerate the outcomes summing to 15. Three dice, maximum 18, so 15 is near the top and the possibilities are few. Group by which values appear, then count orderings:

Multiset	Orderings	Count
$\{3,6,6\}$	one repeat, so $\frac{3!}{2!}$	3
$\{4,5,6\}$	all distinct, so $3!$	6
$\{5,5,5\}$	all identical	1

\text{Total}=3+6+1=10

Step 2: Count those starting with 4. A 4 can only appear in the

\{4,5,6\}

group - neither

\{3,6,6\}

nor

\{5,5,5\}

contains a 4 at all. Of its six orderings, those beginning with 4 are

(4,5,6)

and

(4,6,5)

:

\text{Favourable}=2

Step 3: Divide. Every one of the 10 outcomes is equally likely (each is a specific ordered triple with probability

\frac{1}{216}

), so the conditional probability is a plain ratio of counts:

P\left(\text{first is }4\ \middle|\ \text{sum is }15\right)=\frac{2}{10}=\frac{1}{5}

Final Answer:

\frac{1}{5}

Q26:ipmat indore 2019QA › ProbabilityHardMCQ · MCQ

In a given village there are only three sizes of families: families with 2 members, families with 4 members and families with 6 members. The proportion of families with 2, 4 and 6 members are roughly equal. A poll is conducted in this village wherein a person is chosen at random and asked about his/her family size. The average family size computed by sampling 1000 such persons from the village would be closest to

A4
B4.667
C4.333
D3.667

Pick an option to attempt

The Setup: A sampling-bias trap. The *families* are equally numerous, but the poll picks a person, and larger families put more people into the pool. Anyone who averages the three family sizes directly walks into it. Step 1: Build the population. Let each size have

N

families:

\text{people in 2-member families}=2N, \qquad 4N, \qquad 6N \implies \text{total}=12N

Note the families split

1:1:1

but the people split

2:4:6

, i.e.

1:2:3

. That mismatch is the entire question. Step 2: Probability a randomly chosen person comes from each group.

P(\text{size }2)=\frac{2N}{12N}=\frac{1}{6}, \qquad P(\text{size }4)=\frac{4N}{12N}=\frac{1}{3}, \qquad P(\text{size }6)=\frac{6N}{12N}=\frac{1}{2}

A person is three times as likely to be drawn from a 6-member family as from a 2-member one. Step 3: Take the expectation. Each respondent reports their own family's size:

E[X]=2\cdot\frac{1}{6}+4\cdot\frac{1}{3}+6\cdot\frac{1}{2}=\frac{1}{3}+\frac{4}{3}+3=\frac{5}{3}+3=\frac{14}{3}\approx 4.667

Step 4: Name the trap explicitly. Averaging the family sizes as though each family were one data point gives

\frac{2+4+6}{3}=4

- which is option 1. That is the average family size *per family*, a genuinely different quantity from the average size *reported by a random person*. The poll samples people, so the answer is the larger

4.667

. The sample of 1000 is a red herring: it only means the estimate is close to the true expectation, not that the size matters. Final Answer: 4.667

Q27:ipmat indore 2019QA › LogarithmsMediumMCQ · MCQ

The value of

(\log_{3} 30)^{-1} + (\log_{4} 900)^{-1} + (\log_{5} 30)^{-1}

is

A0.5
B30
C2
D1

Pick an option to attempt

The Setup: Inverted logarithms. The reciprocal rule

\frac{1}{\log_a b}=\log_b a

flips each term so they can be forced onto a shared base and merged. Step 1: Invert all three.

\left(\log_3 30\right)^{-1}=\log_{30}3, \qquad \left(\log_4 900\right)^{-1}=\log_{900}4, \qquad \left(\log_5 30\right)^{-1}=\log_{30}5

Step 2: Bring the middle term to base 30. The odd one out has base 900, but

900=30^2

. Using

\log_{b^n}x=\frac{1}{n}\log_b x

:

\log_{900}4=\log_{30^2}4=\frac{1}{2}\log_{30}4=\log_{30}\left(4^{1/2}\right)=\log_{30}2

Both the base and the argument cooperate here - halving the coefficient turns

4

into

\sqrt{4}=2

, exactly the factor needed later. Step 3: Merge and evaluate. With a common base, a sum of logs is the log of the product:

\log_{30}3+\log_{30}2+\log_{30}5=\log_{30}(3\times 2\times 5)=\log_{30}30=1

The three arguments

3

,

2

and

5

are precisely the prime factors of 30, which is why the expression collapses so cleanly - the question is built backwards from

30=2\times 3\times 5

. Final Answer: 1

Q28:ipmat indore 2019QA › LogarithmsMediumMCQ · MCQ

The inequality

\log_{a}{f(x)} < \log_{a}{g(x)}

implies that

A $f(x) > g(x) > 0$ for $0 < a < 1$ and $g(x) > f(x) > 0$ for $a > 1$
B $g(x) > f(x) > 0$ for $0 < a < 1$ and $f(x) > g(x) > 0$ for $a > 1$
C $f(x) > g(x) > 0$ for $a > 0$
D $g(x) > f(x) > 0$ for $a > 0$

Pick an option to attempt

The Setup: A logarithmic inequality is governed entirely by its base. Above 1 the function increases and the inequality is preserved; between 0 and 1 it decreases and the inequality reverses. On top of that, both arguments must be strictly positive for the logarithms to exist at all. Step 1: The domain condition, which applies in both cases. A logarithm is only defined on positive arguments, so regardless of the base:

f(x)>0 \quad\text{and}\quad g(x)>0

This is why every chain in the answer ends in

>0

rather than stopping at a comparison between

f

and

g

. **Step 2: Case

a>1

.** Here

\log_a

is strictly increasing, so the order of the arguments matches the order of their logarithms:

\log_a f(x)<\log_a g(x) \implies f(x)<g(x)

Combining with the domain condition, and noting

g

exceeds

f

which is already positive:

g(x)>f(x)>0

**Step 3: Case

0<a<1

.** Here

\log_a

is strictly decreasing - a larger argument gives a smaller logarithm - so the inequality reverses:

\log_a f(x)<\log_a g(x) \implies f(x)>g(x)

Combining with positivity:

f(x)>g(x)>0

Step 4: Note that no single answer covers both. Options 3 and 4 each claim one chain holds 'for

a > 0

', i.e. for every admissible base at once. That cannot be right: the two cases give opposite orderings of

f

and

g

, so any statement quantified over all bases is false for half of them. Only a case-split answer survives. Final Answer:

f(x) > g(x) > 0

for

0 < a < 1

and

g(x) > f(x) > 0

for

a > 1

Q29:ipmat indore 2019QA › SolidsHardMCQ · MCQ

Three cubes with integer edge lengths are given. It is known that the sum of their surface areas is

564 \ \text{cm}^2

. Then the possible values of the sum of their volumes are

A $764 \ \text{cm}^3$ and $586 \ \text{cm}^3$
B $586 \ \text{cm}^3$ and $564 \ \text{cm}^3$
C $764 \ \text{cm}^3$ and $564 \ \text{cm}^3$
D $586 \ \text{cm}^3$ and $786 \ \text{cm}^3$

Pick an option to attempt

The Setup: A Diophantine equation dressed as geometry. Convert the surface-area condition into a sum of three squares, find every integer solution, then convert back to volumes. Step 1: Reduce the surface-area equation. A cube of edge

a

has surface area

6a^2

, so with edges

a,b,c

:

6a^2+6b^2+6c^2=564 \implies a^2+b^2+c^2=94

Step 2: Bound the search before starting it. Order the edges

a\leq b\leq c

, with each at least 1 (an edge of 0 is not a cube). Then

c

is the largest, so

3c^2\geq 94

and

c^2\leq 94

:

\frac{94}{3}\leq c^2\leq 94 \implies 31.3\leq c^2\leq 94 \implies 6\leq c\leq 9

That leaves exactly four cases to test, which is what makes the search genuinely exhaustive rather than a lucky hunt. Step 3: Test all four.

$c$	$c^2$	$a^2+b^2$ needed	Solutions with $a\leq b\leq c$
9	81	13	$4+9$ -> edges $\{2,3,9\}$
8	64	30	none ( $30=1+29,4+26,9+21,16+14,25+5$ - no square pairs)
7	49	45	$9+36$ -> edges $\{3,6,7\}$
6	36	58	none (needs both $\leq 36$ ; $58-36=22$ and $58-25=33$ are not squares)

So there are exactly two admissible triples, and no more. Step 4: Convert to volumes.

\{2,3,9\}: \quad 2^3+3^3+9^3=8+27+729=764

\{3,6,7\}: \quad 3^3+6^3+7^3=27+216+343=586

Both triples do give surface area

6(94)=564

✓, so both are genuine and the answer must list both values. Final Answer:

764 \ \text{cm}^3

and

586 \ \text{cm}^3

Q30:ipmat indore 2019QA › IndicesMediumMCQ · MCQ

Determine the greatest number among the following four numbers:

A $2^{300}$
B $3^{200}$
C $2^{100} + 3^{100}$
D $4^{100}$

Pick an option to attempt

The Setup: A power-scaling comparison. Computing these directly is hopeless, so rewrite every term with the same exponent - then only the bases need comparing. Step 1: Normalise to the exponent 100. Each exponent is a multiple of 100, so pull that out:

2^{300}=\left(2^3\right)^{100}=8^{100}, \qquad 3^{200}=\left(3^2\right)^{100}=9^{100}, \qquad 4^{100}=4^{100}

Step 2: Compare the three pure powers. With a common positive exponent,

t\mapsto t^{100}

is increasing on positive bases, so the ordering of the bases carries straight over:

9^{100}>8^{100}>4^{100} \implies 3^{200}>2^{300}>4^{100}

Step 3: Dispose of the sum. The remaining candidate

2^{100}+3^{100}

is not a pure power, so bound it. Since

2^{100}<3^{100}

:

2^{100}+3^{100}<3^{100}+3^{100}=2\cdot 3^{100}

Compare that generous overestimate against the leader:

\frac{3^{200}}{3^{100}}=3^{100} \quad\text{which is vastly larger than }2

so

2\cdot 3^{100}<3^{100}\cdot 3^{100}=3^{200}

. The sum loses even after being inflated, and the gap is not close -

3^{200}

exceeds it by a factor of roughly

\frac{3^{100}}{2}

. Step 4: Conclude. The full ordering is

3^{200}>2^{300}>4^{100}>2^{100}+3^{100}

, and the greatest is

3^{200}

. The lesson: adding two large powers is worth far less than multiplying the exponent.

2^{100}+3^{100}

is barely bigger than its larger half, while squaring

3^{100}

multiplies it by itself. Final Answer:

3^{200}

Q31:ipmat indore 2019QA › TrianglesMediumMCQ · MCQ

The number of points, having both coordinates as integers, that lie in the interior of the triangle with vertices

(0, 0), (0, 31),

and

(31, 0)

is

A435
B465
C450
D464

Pick an option to attempt

The Setup: We are hunting for lattice points strictly inside the triangle. The boundary lines are the axes (

x=0, y=0

) and the hypotenuse (

x+y=31

). To stay *interior*, our coordinates must be strictly positive integers that sum to less than 31. Boundary points fail the vibe check and get instantly disqualified. Step 1: Set the Constraints We need integers

x > 0

and

y > 0

such that

x + y < 31

. This means the absolute max limit for the sum is

x + y \le 30

. Step 2: Sum the Rows If

x = 1

,

y

can range from

1

to

29

(giving us 29 valid points). If

x = 2

,

y

can range from

1

to

28

(28 points). This sequence cascades linearly all the way down to

x = 29

, where

y

can only be

1

(1 point). Step 3: Calculate the Total We just need to sum the first 29 positive integers. We plug this into the classic arithmetic series sum formula

S = \frac{n(n+1)}{2}

.

S = \frac{29(30)}{2}

S = 29 \times 15 = 435

Final Answer: 435

Q32:ipmat indore 2019QA › Time, Speed & DistanceMediumMCQ · MCQ

Two small insects, which are

x

metres apart, take

u

minutes to pass each other when they are flying towards each other, and

v

minutes to meet each other when they are flying in the same direction. Then, the ratio of the speed of the slower insect to that of the faster insect is

A $\dfrac{u}{v}$
B $\dfrac{u}{v-u}$
C $\dfrac{v-u}{v+u}$
D $\dfrac{u}{v+u}$

Pick an option to attempt

The Setup: This is a relative speed speedrun. When they fly head-on, their speeds stack (addition). When it's a chase scene (same direction), their speeds counter each other (subtraction). We just set up the two relative speed equations and isolate the ratio like a standard system of equations. Step 1: Define the Speeds Let the faster insect's speed be

S_1

and the slower insect's speed be

S_2

. Flying towards each other (closing the gap

x

in time

u

):

S_1 + S_2 = \frac{x}{u}

Flying in the same direction (closing the gap

x

in time

v

):

S_1 - S_2 = \frac{x}{v}

Step 2: Isolate the Ratio Instead of solving for individual speeds and making it messy, we can just divide the two equations to instantly wipe the distance

x

from the board:

\frac{S_1 + S_2}{S_1 - S_2} = \frac{\frac{x}{u}}{\frac{x}{v}} = \frac{v}{u}

Step 3: Cross-Multiply and Solve

u(S_1 + S_2) = v(S_1 - S_2)

uS_1 + uS_2 = vS_1 - vS_2

Group the

S_1

terms on one side and

S_2

terms on the other:

uS_2 + vS_2 = vS_1 - uS_1

S_2(u + v) = S_1(v - u)

Now, isolate the ratio of the slower speed to the faster speed (

S_2/S_1

):

\frac{S_2}{S_1} = \frac{v - u}{v + u}

Final Answer:

\dfrac{v-u}{v+u}

Q33:ipmat indore 2019QA › Mixture & AlligationHardMCQ · MCQ

An alloy

P

has copper and zinc in the proportion of

5:2

(by weight), while another alloy

Q

has the same metals in the proportion of

3:4

(by weight). If these two alloys are mixed in the proportion of

a:b

(by weight), a new alloy

R

is formed, which has equal contents of copper and zinc. Then, the proportion of copper and zinc in the alloy

S

, formed by mixing the two alloys

P

and

Q

in the proportion of

b:a

(by weight) is

A $7:9$
B $9:7$
C $9:5$
D $5:9$

Pick an option to attempt

The Setup: We've got an alchemy minigame. First, we need to unlock the exact ratio (

a:b

) that balances the metals 1:1. Once we have those initial mix weights, we hit them with an Uno reverse card, swap the weights (

b:a

) for a new batch, and calculate the final drop rates for copper and zinc. **Step 1: Find the Initial Mix Ratio (

a:b

)** In alloy P, Copper is

5/7

and Zinc is

2/7

. In alloy Q, Copper is

3/7

and Zinc is

4/7

. Mix them in parts

a

and

b

to get equal Copper and Zinc:

\frac{5a}{7} + \frac{3b}{7} = \frac{2a}{7} + \frac{4b}{7}

Multiply by 7 to drop the denominators and balance the equation:

5a + 3b = 2a + 4b

3a = b \implies \frac{a}{b} = \frac{1}{3}

So the ratio

a:b

is strictly

1:3

. **Step 2: Form Alloy S (Ratio

b:a

)** Now we mix them using the reverse ratio: 3 parts of P and 1 part of Q. Calculate the total Copper:

\text{Cu} = 3\left(\frac{5}{7}\right) + 1\left(\frac{3}{7}\right) = \frac{15}{7} + \frac{3}{7} = \frac{18}{7}

Calculate the total Zinc:

\text{Zn} = 3\left(\frac{2}{7}\right) + 1\left(\frac{4}{7}\right) = \frac{6}{7} + \frac{4}{7} = \frac{10}{7}

Step 3: Final Proportion The ratio of Cu to Zn in alloy S is:

\frac{18}{7} : \frac{10}{7}

Multiply by 7 and simplify by dividing by 2:

18 : 10 = 9 : 5

Final Answer:

9:5

Q34:ipmat indore 2019QA › Permutation & CombinationMediumMCQ · MCQ

How many different numbers can be formed by using only the digits 1 and 3 which are smaller than 3000000?

A64
B128
C190
D254

Pick an option to attempt

The Setup: This is a pure combinatorics flex. We are restricted to binary-style choices (only 1s and 3s) and capped by a ceiling of 3 million. We need to sum the possibilities for all digit lengths from 1 to 7 without busting the limit. Step 1: Count the Smaller Lengths (1 to 6 digits) For any

n

-digit number made of just two specific digits, there are exactly

2^n

combinations. Let's tally everything up to 6 digits (since all of them easily slide under

3{,}000{,}000

): * 1-digit:

2^1 = 2

* 2-digit:

2^2 = 4

* 3-digit:

2^3 = 8

* 4-digit:

2^4 = 16

* 5-digit:

2^5 = 32

* 6-digit:

2^6 = 64

Total combinations for lengths 1-6:

2+4+8+16+32+64 = 126

. Step 2: Handle the 7-Digit Boss Level To be a 7-digit number strictly smaller than 3,000,000, the first digit *must* be 1. (Starting with a 3 instantly busts the cap). First digit = locked at '1' (1 way). Remaining 6 digits = any combo of 1s and 3s (

2^6 = 64

ways). Valid 7-digit numbers =

1 \times 64 = 64

. Step 3: Final Sum Combine the two pools to get the total number of valid permutations:

\text{Total} = 126 + 64 = 190

Final Answer: 190

Q35:ipmat indore 2019QA › Progression & SeriesMediumMCQ · MCQ

There are numbers

a_1, a_2, a_3, \ldots, a_n

each of them being

+1

or

-1

. If it is known that

a_1 a_2 + a_2 a_3 + a_3 a_4 + \ldots a_{n-1} a_n + a_n a_1 = 0

then

A $n$ is a multiple of 2 but not a multiple of 4
B $n$ is a multiple of 3
C $n$ can be any multiple of 4
DThe only possible value of $n$ is 4

Pick an option to attempt

The Setup: We are dealing with a cyclic sum of products that perfectly zeroes out. Because the terms can only be +1 or -1, for the sum to be zero, there must be a perfectly balanced meta-game: an equal number of positive and negative terms. Step 1: Define the Terms Let

x_1 = a_1 a_2

,

x_2 = a_2 a_3, \dots, x_n = a_n a_1

. Each

x_i

is strictly either

+1

or

-1

. The given equation translates to

\sum x_i = 0

. This requires an equal number of

+1

s and

-1

s. Thus,

n

must be an even number. Let

n = 2k

, meaning there are exactly

k

positive terms and

k

negative terms. Step 2: Multiply Everything Together Let's look at the product of all

x_i

terms using two different perspectives. Method A (using the base variables):

\prod x_i = (a_1 a_2)(a_2 a_3)\dots(a_n a_1) = a_1^2 a_2^2 \dots a_n^2

Since every

a_i \in \{+1, -1\}

, their squares are all definitively

1

. So the overall product is exactly

1

. Method B (using the +1 and -1 counts): We know there are exactly

k

negative terms (

-1

) in the mix.

\prod x_i = (+1)^k (-1)^k = (-1)^k

Step 3: Match the Logic For Method A and Method B to agree without breaking math,

(-1)^k

must equal

1

. This means

k

MUST be an even integer. Let

k = 2m

. Since our total terms

n = 2k

, we substitute to get

n = 2(2m) = 4m

. This confirms

n

must strictly be a multiple of 4. Final Answer:

n

can be any multiple of 4

Q36:ipmat indore 2019LRDI › Bar GraphsEasyMCQ · MCQ

Analyze the given data for exports and imports of rubber in Rs. crores from 2006 to 2017 and answer the questions based on the analysis.

Average annual exports for the given period 2006-2017 was approximately question figure

ARs. 230 Cr
BRs. 220 Cr
CRs. 210 Cr
DRs. 190 Cr

Pick an option to attempt

\text{Average} = \dfrac{\text{Sum of annual exports across all years}}{\text{Number of years}}

=\dfrac{280+280+230+210+200+220+210+200+200+200+220+200}{12}

=\dfrac{2650}{12}\approx220

crores. Final Answer: Rs. 220 Cr

Q37:ipmat indore 2019LRDI › Bar GraphsEasyMCQ · MCQ

Analyze the given data for exports and imports of rubber in Rs. crores from 2006 to 2017 and answer the questions based on the analysis.

The percentage decline in exports during the period 2006-2011 is more than the percentage decline in exports during 2012-2017 by approximately question figure

A16.5
B20.5
C12.5
D21.5

Pick an option to attempt

Percentage decline during 2006-2011 period

=\frac{\text{Difference between 2006 and 2011 exports}}{\text{2006 exports}}\times100

=\frac{280-220}{280}\times100=\frac{60}{280}\times100=\text{approximately }21.42

% Now, percentage decline during 2012-2017 period will be calculated similarly.

=\frac{210-200}{210}\times100=\frac{10}{210}\times100=\text{approximately } 4.76

% As we can see, the first percentage is more than the second by approximately

16.5

, hence Option 1 is the answer. Final Answer: 16.5

Q38:ipmat indore 2019LRDI › Bar GraphsEasyMCQ · MCQ

Analyze the given data for exports and imports of rubber in Rs. crores from 2006 to 2017 and answer the questions based on the analysis.

The maximum difference between imports and exports is question figure

ARs. 60 Cr
BRs. 110 Cr
CRs. 120 Cr
DRs. 100 Cr

Pick an option to attempt

The phrase *difference between imports and exports* means the gap within a single year - the two bars standing side by side - so we compare year by year rather than picking the tallest bar from one year and the shortest from another. Reading the pairs off the chart and measuring the size of the gap between the two bars. Note that in 2006 and 2007 exports were the higher of the two, so the gap runs the other way in those years - what matters is how far apart the bars stand, not which is on top:

Year	Imports	Exports	Gap
2006	220	280	60
2007	240	280	40
2008	260	230	30
2009	280	210	70
2010	310	200	110
2011	290	220	70
2012	320	210	110
2013	300	200	100
2014	320	200	120
2015	310	200	110
2016	250	220	30
2017	240	200	40

The largest gap is in 2014, where imports peak at Rs. 320 Cr while exports sit at their floor of Rs. 200 Cr:

320 - 200 = \text{Rs. }120\text{ Cr}

A shortcut is tempting here - take the tallest import bar anywhere (320, in 2012 and 2014) and subtract the shortest export bar anywhere (200, in 2010, 2013, 2014, 2015 and 2017). It gives the right number only because 2014 happens to hold both extremes at once. On a chart where the highest import and the lowest export fell in different years, that shortcut would compare two bars that never coexist and overstate the answer. Scanning the pairs is the reliable route. Final Answer: Rs. 120 Cr

Q39:ipmat indore 2019LRDI › Bar GraphsEasyMCQ · MCQ

Analyze the given data for exports and imports of rubber in Rs. crores from 2006 to 2017 and answer the questions based on the analysis.

Balance of trade is defined as imports subtracted from exports ( = exports - imports). Which of the following blocks of three years has witnessed the largest average negative balance of trade? question figure

A2007-2009
B2015-2017
C2014-2016
D2010-2012

Pick an option to attempt

For 2007-2009: 2007:

280 - 240 = 40

2008:

230 - 260 = -30

2009:

210 - 280 = -70

Average =

\frac{40 - 30 - 70}{3} = -20

For 2015-2017: 2015:

200 - 310 = -110

2016:

220 - 250 = -30

2017:

200 - 240 = -40

Average =

\frac{-110 - 30 - 40}{3} = -60

For 2014-2016: 2014:

200 - 320 = -120

2015:

200 - 310 = -110

2016:

220 - 250 = -30

Average =

\frac{-120 - 110 - 30}{3} = -86.67

For 2010-2012: 2010:

200 - 310 = -110

2011:

220 - 290 = -70

2012:

210 - 320 = -110

Average =

\frac{-110 - 70 - 110}{3} = -96.67

After comparing, the period 2010-2012 has the largest negative average balance of trade at

-96.67

crores, making it the correct answer. Final Answer: 2010-2012

Q40:ipmat indore 2019LRDI › Bar GraphsEasyMCQ · MCQ

Analyze the given data for exports and imports of rubber in Rs. crores from 2006 to 2017 and answer the questions based on the analysis.

The percentage increase in imports over the previous year is maximum during question figure

A2009 to 2010
B2010 to 2011
C2013 to 2014
D2008 to 2009

Pick an option to attempt

The percentage increase in imports for each given option using the formula:

\frac{\text{New Value - Original Value}}{\text{Original Value}} \times 100

2009 to 2010: Imports 2009:

280

Imports 2010:

310

Percentage increase =

\frac{310 - 280}{280} \times 100 = \frac{30}{280} \times 100 = 10.71\%

2010 to 2011: Imports 2010:

310

Imports 2011:

290

This is a decrease:

\frac{290 - 310}{310} \times 100 = -6.45\%

2013 to 2014: Imports 2013:

300

Imports 2014:

320

Percentage increase =

\frac{320 - 300}{300} \times 100 = \frac{20}{300} \times 100 = 6.67\%

2008 to 2009: Imports 2008:

260

Imports 2009:

280

Percentage increase =

\frac{280 - 260}{260} \times 100 = \frac{20}{260} \times 100 = 7.69\%

Comparing all values, the maximum percentage increase was

10.71\%

during 2009 to 2010, making this the correct answer. Final Answer: 2009 to 2010

IPMAT Indore 2019 — MCQ

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