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Mensuration questions for IPMAT, with solutions

10 free Mensuration practice questions for IPMAT Indore, Rohtak and JIPMAT, 1 from past papers, each with the answer and a full worked solution. No sign-in. Try each one first, then open the solution. The lessons for this chapter are in Mensuration on Learn.

1. Question 1Type the answerWritten by IPMATwayMensuration 2D

   The diagonal of a square is $10$ cm. Its area, in sq. cm, is

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   Answer: 50

   For a square of side $a$, the diagonal is $d=a\sqrt{2}$, so $a=\frac{10}{\sqrt{2}}$ and $a^2=\frac{100}{2}=50$. Equivalently, area $=\frac{d^2}{2}=\frac{100}{2}=50$ sq. cm.

2. Question 2Type the answerWritten by IPMATwayMensuration 2D

   The circumference of a circle is $88$ cm. Its area, in sq. cm, is (Take $\pi=\frac{22}{7}$.)

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   Answer: 616

   From $2\pi r=88$ with $\pi=\frac{22}{7}$: $\frac{44}{7}r=88$, so $r=14$ cm. Area $=\pi r^2=\frac{22}{7}\cdot 196=22\cdot 28=616$ sq. cm.

3. Question 3Type the answerWritten by IPMATwayMensuration 2D

   Instead of walking along two adjacent sides of a rectangular field $60$ m by $80$ m, a boy walks along the diagonal. The distance he saves, in m, is

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   Answer: 40

   Along two adjacent sides he would walk $60+80=140$ m. The diagonal is $\sqrt{60^2+80^2}=\sqrt{3600+6400}=\sqrt{10000}=100$ m. He saves $140-100=40$ m.

4. Question 4Type the answerWritten by IPMATwayMensuration 2D

   A wheel of diameter $70$ cm rolls along a road. How many revolutions does it make in covering $1.1$ km? (Take $\pi=\frac{22}{7}$.)

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   Answer: 500

   In one revolution the wheel covers its circumference: $\pi d=\frac{22}{7}\cdot 70=220$ cm. The distance is $1.1$ km $=1100$ m $=110000$ cm. Number of revolutions $=\frac{110000}{220}=500$.

5. Question 5Type the answerWritten by IPMATwayMensuration 3D

   The volume of a cube is $343$ cubic cm. Its total surface area, in sq. cm, is

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   Answer: 294

   The edge is $\sqrt[3]{343}=7$ cm, since $7^3=343$. The total surface area of a cube is $6a^2=6\cdot 49=294$ sq. cm.

6. Question 6Multiple choiceWritten by IPMATwayMensuration 3D

   The length of the longest rod that can be placed in a room $12$ m $\times$ $8$ m $\times$ $6$ m is

   1. A.$\sqrt{208}$ m
   2. B.$26$ m
   3. C.$\sqrt{244}+6$ m
   4. D.$2\sqrt{61}$ m
   Show the answer and solution

   Answer: D. $2\sqrt{61}$ m

   The longest rod lies along the space diagonal, of length $\sqrt{l^2+b^2+h^2}=\sqrt{12^2+8^2+6^2}=\sqrt{144+64+36}=\sqrt{244}$. Since $244=4\cdot 61$, this is $2\sqrt{61}\approx 15.6$ m.

7. Question 7Type the answerWritten by IPMATwayMensuration 3D

   A hall is $30$ m long, $12$ m wide and $6$ m high. If each person needs $8$ cubic m of space, the maximum number of persons the hall can accommodate is

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   Answer: 270

   The volume of the hall is $30\cdot 12\cdot 6=2160$ cubic m. Each person needs $8$ cubic m, so at most $\frac{2160}{8}=270$ persons can be accommodated.

8. Question 8Multiple choiceWritten by IPMATwayMensuration 3D

   A cylinder just encloses a sphere (the sphere touches its curved surface, top and bottom). The ratio of the volume of the cylinder to that of the sphere is

   1. A.$3:2$
   2. B.$2:3$
   3. C.$1:1$
   4. D.$4:3$
   Show the answer and solution

   Answer: A. $3:2$

   A cylinder that just encloses a sphere of radius $r$ has radius $r$ and height $2r$. Its volume is $\pi r^2\cdot 2r=2\pi r^3$. The sphere's volume is $\frac{4}{3}\pi r^3$. Ratio $=2:\frac{4}{3}=3:2$.

9. Question 9Multiple choiceWritten by IPMATwayMensuration 2D

   A circle and a rectangle have the same perimeter. The sides of the rectangle are $18$ cm and $26$ cm. The area of the circle, in sq. cm, is (Take $\pi=\frac{22}{7}$.)

   1. A.$154$
   2. B.$616$
   3. C.$2464$
   4. D.$468$
   Show the answer and solution

   Answer: B. $616$

   The rectangle's perimeter is $2(18+26)=88$ cm, which is also the circle's circumference. From $2\cdot\frac{22}{7}r=88$, $r=14$ cm. Area of the circle $=\frac{22}{7}\cdot 14^2=\frac{22}{7}\cdot 196=616$ sq. cm.

10. Question 10Multiple choiceAsked in JIPMAT 2023Mensuration 3D

    The radii of two cones are in the ratio $2:3$ and their volumes in the ratio $1:3$. Then the ratio of their heights is

    1. A.$3:2$
    2. B.$1:2$
    3. C.$3:4$
    4. D.$2:3$
    Show the answer and solution

    Answer: C. $3:4$

    Volume of a cone $=\frac{1}{3}\pi r^2h$, so $\frac{V_1}{V_2}=\frac{r_1^2}{r_2^2}\cdot\frac{h_1}{h_2}$. Substituting, $\frac{1}{3}=\frac{4}{9}\cdot\frac{h_1}{h_2}$. Hence $\frac{h_1}{h_2}=\frac{1}{3}\cdot\frac{9}{4}=\frac{3}{4}$, i.e. $3:4$.