IPMAT Indore Solids PYQs

2 solved Solids previous year questions (PYQs) from IPMAT Indore past year papers — attempt each and check the answer.

Q1:ipmat indore 2019QA › SolidsMediumSA · TITA

The number of whole metallic tiles that can be produced by melting and recasting a circular metallic plate, if each of the tiles has a shape of a right-angled isosceles triangle and the circular plate has a radius equal in length to the longest side of the tile (Assume that the tiles and plate are of uniform thickness, and there is no loss of material in the melting and recasting process) is

Enter your answer to attempt

The Setup: Conservation of material dressed up as a 3D melting problem. Because the thickness is uniform and nothing is lost, the volume ratio equals the area ratio, and the whole thing reduces to plane geometry. Step 1: Total source material. With the plate's radius

R

\text{Plate area}=\pi R^2

Step 2: Area of one tile. The tile is a right-angled isosceles triangle, whose longest side is the hypotenuse, and the stem sets that hypotenuse equal to

R

. Letting each of the two equal legs be

x

, Pythagoras gives:

x^2+x^2=R^2 \implies 2x^2=R^2 \implies x^2=\frac{R^2}{2}

The legs are perpendicular, so they serve directly as base and height:

\text{Tile area}=\frac{1}{2}\cdot x\cdot x=\frac{x^2}{2}=\frac{1}{2}\cdot\frac{R^2}{2}=\frac{R^2}{4}

Step 3: Divide. Note

R

cancels completely - the answer does not depend on the plate's size, only on the stated relationship between the two:

n=\frac{\pi R^2}{\frac{R^2}{4}}=4\pi

Step 4: Filter for whole tiles. With

\pi\approx 3.14159

n\approx 4\times 3.14159=12.566

The stem asks for whole tiles, and you cannot cast a fraction of one, so the count floors to 12 - with material worth about

0.57

of a tile left unused. Rounding to 13 here is the trap: that would require metal that does not exist. Final Answer: 12

Q2: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

IPMAT Indore Solids — PYPs

Browse other topics · IPMAT Indore