{"id":67142,"date":"2023-11-11T10:25:00","date_gmt":"2023-11-11T04:55:00","guid":{"rendered":"https:\/\/www.practicemock.com\/blog\/?p=67142"},"modified":"2023-11-07T12:26:28","modified_gmt":"2023-11-07T06:56:28","slug":"quant-mensuration-tips-practice-questions","status":"publish","type":"post","link":"https:\/\/www.practicemock.com\/blog\/quant-mensuration-tips-practice-questions\/","title":{"rendered":"Quant Mensuration Tips & Practice Questions"},"content":{"rendered":"\n

Mensuration is yet another important concept for banking exam preparation and is the study of different dimensions including length, breadth, height, area, and volume of various geometric shapes. These kinds of questions contain a significant weightage in bank exams and thus, candidates must learn good preparation strategy in order to ace the exams especially the one which is coming up that is RBI Assistant prelims exam<\/a>. We are here with a quick revision of a few Quant Mensuration Tips & Practice Questions.\u00a0<\/p>\n\n\n\n

Mensuration:<\/strong><\/p>\n\n\n\n

Mensuration deals with measurement of areas and volumes of different figures. These figures include both \u20182D\u2019 (two dimensional) and \u20183D\u2019 (three dimensional) figures (also known as solids).<\/p>\n\n\n\n

Note: By definition, \u20182D\u2019 figures exist in a two dimensional world and hence have no volume.<\/p>\n\n\n\n

Let\u2019s take a brief look at the formulae for areas of various \u20182D\u2019 figures, and areas and volumes of \u20183D\u2019 figures.<\/p>\n\n\n\n

2D Figures:<\/p>\n\n\n\n

Polygons:<\/strong> A Polygon is a closed figure made up of at least 3 line segments in a two-dimensional plane.<\/p>\n\n\n\n

Triangle:<\/strong> A triangle is a polygon containing 3 sides. The area of a triangle is represented by the figure \u0394.<\/p>\n\n\n\n

Formulas associated with triangles:<\/p>\n\n\n\n

Area:<\/p>\n\n\n\n

I. For any triangle in general;<\/p>\n\n\n\n

For a triangle having sides of length \u2018a\u2019 cm, \u2018b\u2019 cm and \u2018c\u2019 cm;<\/p>\n\n\n\n

Perimeter of triangle = Sum of length of three sides = (a + b + c) cm.<\/p>\n\n\n\n

Area = \u221a{s \u00d7 (s \u2013 a) \u00d7 (s \u2013 b) \u00d7 (s \u2013 c)}<\/p>\n\n\n\n

Where \u2018s\u2019 is the semi-perimeter of the triangle. \u2018s\u2019 = {(a + b + c)\/2}<\/p>\n\n\n\n

This formula is universally known as heron\u2019s formula.<\/p>\n\n\n\n

II. For a triangle whose length of base and length of altitude (Height) to that base is given;<\/p>\n\n\n\n

Area = (1\/2) \u00d7 base \u00d7 height<\/p>\n\n\n\n

III. Equilateral triangle: If all three sides of a triangle are equal, then it is an equilateral triangle.<\/p>\n\n\n\n

For an equilateral triangle with side \u2018a\u2019 cm, Area = (\u221a3\/4) \u00d7 a2<\/sup><\/p>\n\n\n\n

Using (II), we have; (1\/2) \u00d7 a \u00d7 height = (\u221a3\/4) \u00d7 a2<\/sup><\/p>\n\n\n\n

Or, height = (\u221a3\/2) \u00d7 a<\/p>\n\n\n\n

IV. Isosceles triangle: If any two sides of a triangle are equal, then it is an isosceles triangle.<\/p>\n\n\n\n

Area = (a\/4)\u221a(4c2<\/sup> \u2013 a2<\/sup>) {where \u2018c\u2019 is the length of each of the two equal sides and \u2018a\u2019 is the length of the third (un-equal) side}<\/p>\n\n\n\n

RBI Assistant prelims free mock test<\/a><\/p>\n\n\n\n

Quadrilaterals:  <\/strong>Any polygon having for sides is called a quadrilateral.<\/p>\n\n\n\n

I. Area of a normal quadrilateral with vertices \u2018A\u2019, \u2018B\u2019, \u2018C\u2019 and \u2018D\u2019 (As shown in the figure)<\/p>\n\n\n\n

Area of quadrilateral ABCD = (1\/2) \u00d7 Length of BD \u00d7 (AF + DE), where AF and DE are perpendicular to BC.<\/p>\n\n\n\n

II. Trapezium:<\/strong> A trapezium is a quadrilateral having exactly two parallel sides.<\/p>\n\n\n\n

Area of trapezium = (1\/2) \u00d7 sum of parallel sides \u00d7 Height of the trapezium.<\/p>\n\n\n\n

III. Parallelogram:<\/strong>  A parallelogram is a quadrilateral whose opposite sides are parallel as well as equal to each other.<\/p>\n\n\n\n

Area = length of base \u00d7 height (perpendicular distance)<\/p>\n\n\n\n

IV. Square:<\/strong> A square is a parallelogram whose all four sides are of equal length and angles are of 90o<\/sup> each.<\/p>\n\n\n\n

Area = (a)2<\/sup> {where \u2018a\u2019 is the length of each side}<\/p>\n\n\n\n

V. Rhombus:<\/strong> A Rhombus is a parallelogram whose all four sides are equal.<\/p>\n\n\n\n

Area = (1\/2) \u00d7 product of length of diagonals<\/p>\n\n\n\n

VI. Rectangle: <\/strong>A rectangle is a parallelogram that has all four angles of 90o<\/sup><\/p>\n\n\n\n

For a rectangle with length \u2018L\u2019 cm and Breadth \u2018B\u2019 cm,<\/p>\n\n\n\n

Area = L \u00d7 B<\/p>\n\n\n\n

And, Perimeter = 2(L + B)<\/p>\n\n\n\n

Circle:<\/strong> <\/p>\n\n\n\n

For a circle with radius \u2018r\u2019 units<\/p>\n\n\n\n

Area = \u03c0r2<\/sup> <\/p>\n\n\n\n

Circumference = 2\u03c0r<\/p>\n\n\n\n

Area of sector = (\u03b8o<\/sup>\/360o<\/sup>) \u00d7 \u03c0r2<\/sup> {Where \u03b8 is the angle subtended inside the sector}<\/p>\n\n\n\n

Length of arc = (\u03b8o<\/sup>\/360o<\/sup>) \u00d7 2\u03c0r<\/p>\n\n\n\n

RBI Assistant prelims free mock test<\/a><\/p>\n\n\n\n

Area and Volumes of solids:<\/strong><\/p>\n\n\n\n

Cube: <\/strong>A cube is a solid shape with six square faces.<\/p>\n\n\n\n

Total surface area of cube = 6a2<\/sup> {Where \u2018a\u2019 is the length of edge of the cube}<\/p>\n\n\n\n

Volume of cube = a3<\/sup><\/p>\n\n\n\n

Cuboid: <\/strong>A cuboid is bound by 6 rectangular faces. The opposite faces of a cuboid are equal rectangles lying in parallel planes.<\/p>\n\n\n\n

Volume of a cuboid with length \u2018x\u2019 cm, breadth \u2018y\u2019 cm and height \u2018h\u2019 cm = (x \u00d7 y \u00d7 h)  <\/p>\n\n\n\n

Total surface area of cuboid = 2(xy + yh + xh)<\/p>\n\n\n\n

Lateral surface area of a cuboid = 2xh + 2yh = 2h(x + y)<\/p>\n\n\n\n

Cone:<\/strong> A cone is a solid which has a circle at its base and a slanting lateral surface that converges at the apex.<\/p>\n\n\n\n

Curved surface area of the cone = \u03c0 \u00d7 r \u00d7 s {Where \u2018r\u2019 is radius and \u2018s\u2019 is slant height}<\/p>\n\n\n\n

Slant height of a cone = \u221a(r2<\/sup> + h2<\/sup>), where \u2018h\u2019 is height of the cone.<\/p>\n\n\n\n

Total surface area of cone = Curved surface area + area of base = \u03c0rs + \u03c0r2<\/sup> = \u03c0r(r + s)<\/p>\n\n\n\n

Frustum of a cone: <\/strong>If a cone is cut into two parts by a plane parallel to the base, the portion with the upper tip of the cone still remains a cone; however the portion that contains the base is called the frustum of the cone.<\/p>\n\n\n\n

Curved surface area of frustum = \u03c0 \u00d7 s(R + r) {Where \u2018s\u2019 is slant height ,\u2018R\u2019 is radius of the base and \u2018r\u2019 is the radius of the top.}<\/p>\n\n\n\n

Total surface area of the frustum = \u03c0(R2 <\/sup>+ r2<\/sup> + Rs + rs)<\/p>\n\n\n\n

Volume of the frustum = (1\/2) \u00d7 \u03c0 \u00d7 h \u00d7 {R2<\/sup> + r2<\/sup> + Rr} {Where \u2018h\u2019 is the height of the cone}<\/p>\n\n\n\n

s2<\/sup> = (R2<\/sup> \u2013 r2<\/sup>) + h2<\/sup><\/p>\n\n\n\n

Cylinder: <\/strong>Cylinder is a three-dimensional solid that holds two parallel bases joined by a curved surface, at a fixed distance.<\/p>\n\n\n\n

Curved surface area of a cylinder = 2\u03c0rh {Where \u2018r\u2019 is the radius of the base and \u2018h\u2019 is the height of the cylinder}<\/p>\n\n\n\n

Total surface area of the cylinder = 2\u03c0rh + 2\u03c0r2<\/sup> = 2\u03c0r(r + h)<\/p>\n\n\n\n

Volume of cylinder = \u03c0r2<\/sup>h<\/p>\n\n\n\n

RBI Assistant prelims free mock test<\/a><\/p>\n\n\n\n

Sphere: <\/strong>Sphere is a three dimensional solid, that has all its surface points at equal distances from the centre. The distance between the centre and a surface point of the sphere is called its radius (r).<\/p>\n\n\n\n

Volume of sphere = (4\/3) \u00d7 \u03c0r3<\/sup><\/p>\n\n\n\n

Surface area of a sphere = 4\u03c0r2<\/sup><\/p>\n\n\n\n

Hemisphere: <\/strong>Where a sphere is divided into two equal and identical parts, the new formed parts are called hemispheres.<\/p>\n\n\n\n

Volume of hemisphere = (2\/3)\u03c0r3<\/sup><\/p>\n\n\n\n

Total surface area of hemisphere = 3\u03c0r2<\/sup><\/p>\n\n\n\n

Curved surface area of hemisphere = 2\u03c0r2<\/sup><\/p>\n\n\n\n

RBI Assistant prelims free mock test<\/a><\/p>\n\n\n\n

Quant Mensuration Tips & Practice Questions<\/h2>\n\n\n\n

Question 1:<\/strong> If circumference of the base of a right circular cone is 220 cm and volume of the cone is 15400 cm3<\/sup>, then slant height of the cone is:<\/p>\n\n\n\n

A) 39 cm<\/p>\n\n\n\n

B) 38 cm<\/p>\n\n\n\n

C) 41 cm<\/p>\n\n\n\n

D) 37 cm<\/p>\n\n\n\n

E) None of these<\/p>\n\n\n\n

Question 2:<\/strong> The difference between the length and breadth of a rectangular field is 27 m. If the cost of fencing the field at the rate of Rs. 2\/metre is Rs. 492, then find the ratio of the length to breadth of the field.<\/p>\n\n\n\n

A) 26:15<\/p>\n\n\n\n

B) 24:17<\/p>\n\n\n\n

C) 2:1<\/p>\n\n\n\n

D) 12:5<\/p>\n\n\n\n

E) 25:16<\/p>\n\n\n\n

Question 3:<\/strong> A cube having each edge equals to 20 cm is cut into several smaller cubes having each edge equals to 4 cm. Find the number of smaller cubes which are the part of the faces of the larger cube but not the part of edges of the larger cube.<\/p>\n\n\n\n

A) 56<\/p>\n\n\n\n

B) 64<\/p>\n\n\n\n

C) 32<\/p>\n\n\n\n

D) 54<\/p>\n\n\n\n

E) 72<\/p>\n\n\n\n

RBI Assistant prelims free mock test<\/a><\/p>\n\n\n\n

Question 4:<\/strong> Each side of square \u2018A\u2019 is 6 cm less than that of square \u2018B\u2019. The perimeter of a rectangle is 3 times the difference of the perimeters of the two squares. If the length of the rectangle is 20 cm, then find its breadth.<\/p>\n\n\n\n

A) 19 cm<\/p>\n\n\n\n

B) 14 cm<\/p>\n\n\n\n

C) 12 cm<\/p>\n\n\n\n

D) 18 cm<\/p>\n\n\n\n

E) None of these<\/p>\n\n\n\n

Question 5:<\/strong> Ratio of radius to height of right circular cylinder is 4:9 respectively. Find the area of a rectangle whose breadth is 37.5% more than radius of the cylinder while ratio of length of rectangle to height of cylinder is 3:4 respectively. Given that curved surface area of the cylinder is 3456 cm2<\/sup>. (Take \u03c0 = 3)<\/p>\n\n\n\n

A) 884 cm<\/p>\n\n\n\n

B) 584 cm2<\/sup><\/p>\n\n\n\n

C) 594 cm2<\/sup><\/p>\n\n\n\n

D) 624 cm2<\/sup><\/p>\n\n\n\n

E) None of these<\/p>\n\n\n\n

Question 6:<\/strong> The diagonal of a square is 24\u221a2 cm. The perimeter of the square is 24 cm less than a rectangle whose length to breadth ratio is 7:5. Find the area of the rectangle.<\/p>\n\n\n\n

A) 725 cm2<\/sup><\/p>\n\n\n\n

B) 875 cm2<\/sup><\/p>\n\n\n\n

C) 915 cm2<\/sup><\/p>\n\n\n\n

D) 625 cm2<\/sup><\/p>\n\n\n\n

E) 1015 cm2<\/sup><\/p>\n\n\n\n

Question 7:<\/strong> The ratio of length to breadth of a rectangle is 8:5. When the length and breadth of the rectangle is increased by 25% and decreased by 40% respectively, then its perimeter becomes 390 cm. Find the original area of the rectangle.<\/p>\n\n\n\n

A) 8400 cm2<\/sup><\/p>\n\n\n\n

B) 9000 cm2<\/sup><\/p>\n\n\n\n

C) 7200 cm2<\/sup><\/p>\n\n\n\n

D) 6400 cm2<\/sub><\/p>\n\n\n\n

E) 10500 cm2<\/sup><\/p>\n\n\n\n

RBI Assistant prelims free mock test<\/a><\/p>\n\n\n\n

Question 8:<\/strong> The perimeter of a square is 36 metres more than that of a rectangle. The ratio of the length to breadth of the rectangle is 9:4. If the area of the square is 3721 m2<\/sup>, then find the cost of painting one surface of the rectangle at the rate of Rs. 2.5\/m2<\/sup>.<\/p>\n\n\n\n

A) Rs. 5760<\/p>\n\n\n\n

B) Rs. 4320<\/p>\n\n\n\n

C) Rs. 5220<\/p>\n\n\n\n

D) Rs. 6300<\/p>\n\n\n\n

E) Rs. 5040<\/p>\n\n\n\n

Question 9:<\/strong> Volume of a cylinder is 4928 m3<\/sup> while cost of painting its outer surface (except base and top) at a rate of Rs. 8 per m2 <\/sup>is Rs. 5632. Find the ratio of radius of base to height of the cylinder.<\/p>\n\n\n\n

A) 7:6<\/p>\n\n\n\n

B) 7:8<\/p>\n\n\n\n

C) 7:4<\/p>\n\n\n\n

D) 7:2<\/p>\n\n\n\n

E) None of these<\/p>\n\n\n\n

Question 10:<\/strong> Area of the base of a cone is 616 cm2<\/sup> and curved surface area of a cylinder is 3168 cm2<\/sup>. Find the height of the cylinder if radius of base of cylinder is 4 cm more than that of the cone.<\/p>\n\n\n\n

A) 21 cm<\/p>\n\n\n\n

B) 35 cm<\/p>\n\n\n\n

C) 28 cm<\/p>\n\n\n\n

D) 24 cm<\/p>\n\n\n\n

E) None of these<\/p>\n\n\n\n

RBI Assistant prelims free mock test<\/a><\/p>\n\n\n\n

\u092a\u094d\u0930\u0936\u094d\u0928 1:<\/strong> \u092f\u0926\u093f \u090f\u0915 \u0932\u0902\u092c \u0935\u0943\u0924\u094d\u0924\u0940\u092f \u0936\u0902\u0915\u0941 \u0915\u0947 \u0906\u0927\u093e\u0930 \u0915\u0940 \u092a\u0930\u093f\u0927\u093f 220 cm \u0939\u0948 \u0914\u0930 \u0936\u0902\u0915\u0941 \u0915\u093e \u0906\u092f\u0924\u0928 15400 cm3 <\/sup>\u0939\u0948, \u0924\u094b \u0936\u0902\u0915\u0941 \u0915\u0940 \u0924\u093f\u0930\u094d\u092f\u0915 \u090a\u0902\u091a\u093e\u0908 \u0915\u093f\u0924\u0928\u0940 \u0939\u0948:<\/p>\n\n\n\n

A) 39 cm<\/p>\n\n\n\n

B) 38 cm<\/p>\n\n\n\n

C) 41 cm<\/p>\n\n\n\n

D) 37 cm<\/p>\n\n\n\n

E) None of these<\/p>\n\n\n\n

\u092a\u094d\u0930\u0936\u094d\u0928 2:<\/strong> \u090f\u0915 \u0906\u092f\u0924\u093e\u0915\u093e\u0930 \u092e\u0948\u0926\u093e\u0928 \u0915\u0940 \u0932\u0902\u092c\u093e\u0908 \u0914\u0930 \u091a\u094c\u0921\u093c\u093e\u0908 \u0915\u0947 \u092c\u0940\u091a \u0915\u093e \u0905\u0902\u0924\u0930 27 m \u0939\u0948\u0964 \u092f\u0926\u093f Rs. 2\/metre \u0915\u0940 \u0926\u0930 \u0938\u0947 \u092e\u0948\u0926\u093e\u0928 \u092e\u0947\u0902 \u092c\u093e\u0921\u093c \u0932\u0917\u093e\u0928\u0947 \u0915\u0940 \u0932\u093e\u0917\u0924 Rs. 492 \u0939\u0948, \u0924\u094b \u092e\u0948\u0926\u093e\u0928 \u0915\u0940 \u0932\u0902\u092c\u093e\u0908 \u0914\u0930 \u091a\u094c\u0921\u093c\u093e\u0908 \u0915\u093e \u0905\u0928\u0941\u092a\u093e\u0924 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f\u0964<\/p>\n\n\n\n

A) 26:15<\/p>\n\n\n\n

B) 24:17<\/p>\n\n\n\n

C) 2:1<\/p>\n\n\n\n

D) 12:5<\/p>\n\n\n\n

E) 25:16<\/p>\n\n\n\n

\u092a\u094d\u0930\u0936\u094d\u0928 3:<\/strong> \u090f\u0915 \u0918\u0928, \u091c\u093f\u0938\u0915\u0940 \u092a\u094d\u0930\u0924\u094d\u092f\u0947\u0915 \u092d\u0941\u091c\u093e 20 cm \u0915\u0947 \u092c\u0930\u093e\u092c\u0930 \u0939\u0948, \u0915\u094b \u0915\u0908 \u091b\u094b\u091f\u0947 \u0918\u0928\u094b\u0902 \u092e\u0947\u0902 \u0915\u093e\u091f\u093e \u091c\u093e\u0924\u093e \u0939\u0948, \u091c\u093f\u0928\u092e\u0947\u0902 \u0938\u0947 \u092a\u094d\u0930\u0924\u094d\u092f\u0947\u0915 \u0915\u0940 \u092d\u0941\u091c\u093e 4 cm \u0915\u0947 \u092c\u0930\u093e\u092c\u0930 \u0939\u094b\u0924\u093e \u0939\u0948\u0964\u0909\u0928 \u091b\u094b\u091f\u0947 \u0918\u0928\u094b\u0902 \u0915\u0940 \u0938\u0902\u0916\u094d\u092f\u093e \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f \u091c\u094b \u092c\u0921\u093c\u0947 \u0918\u0928 \u0915\u0947 \u092b\u0932\u0915 \u0915\u093e \u092d\u093e\u0917 \u0939\u0948\u0902 \u0932\u0947\u0915\u093f\u0928 \u092c\u0921\u093c\u0947 \u0918\u0928 \u0915\u0947 \u092d\u0941\u091c\u093e\u0913\u0902 \u0915\u093e \u092d\u093e\u0917 \u0928\u0939\u0940\u0902 \u0939\u0948\u0902\u0964<\/p>\n\n\n\n

A) 56<\/p>\n\n\n\n

B) 64<\/p>\n\n\n\n

C) 32<\/p>\n\n\n\n

D) 54<\/p>\n\n\n\n

E) 72<\/p>\n\n\n\n

\u092a\u094d\u0930\u0936\u094d\u0928 4:<\/strong> \u0935\u0930\u094d\u0917 ‘A’ \u0915\u0940 \u092a\u094d\u0930\u0924\u094d\u092f\u0947\u0915 \u092d\u0941\u091c\u093e \u0935\u0930\u094d\u0917 ‘B’ \u0915\u0940 \u092d\u0941\u091c\u093e \u0938\u0947 6 cm \u0915\u092e \u0939\u0948\u0964 \u090f\u0915 \u0906\u092f\u0924 \u0915\u093e \u092a\u0930\u093f\u0927\u093f \u0926\u094b \u0935\u0930\u094d\u0917\u094b\u0902 \u0915\u0940 \u092a\u0930\u093f\u0927\u093f \u0915\u0947 \u0905\u0902\u0924\u0930 \u0915\u093e \u0924\u0940\u0928 \u0917\u0941\u0928\u093e \u0939\u0948\u0964 \u092f\u0926\u093f \u0906\u092f\u0924 \u0915\u0940 \u0932\u0902\u092c\u093e\u0908 20 cm \u0939\u0948, \u0924\u094b \u0907\u0938\u0915\u0940 \u091a\u094c\u0921\u093c\u093e\u0908 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f\u0964<\/p>\n\n\n\n

A) 19 cm<\/p>\n\n\n\n

B) 14 cm<\/p>\n\n\n\n

C) 12 cm<\/p>\n\n\n\n

D) 18 cm<\/p>\n\n\n\n

E) \u0907\u0928\u092e\u0947\u0902 \u0938\u0947 \u0915\u094b\u0908 \u0928\u0939\u0940\u0902<\/p>\n\n\n\n

RBI Assistant prelims free mock test<\/a><\/p>\n\n\n\n

\u092a\u094d\u0930\u0936\u094d\u0928 5:<\/strong> \u090f\u0915 \u0932\u092e\u094d\u092c \u0935\u0943\u0924\u094d\u0924\u0940\u092f \u092c\u0947\u0932\u0928 \u0915\u0940 \u0924\u094d\u0930\u093f\u091c\u094d\u092f\u093e \u0914\u0930 \u090a\u0902\u091a\u093e\u0908 \u0915\u093e \u0905\u0928\u0941\u092a\u093e\u0924 \u0915\u094d\u0930\u092e\u0936\u0903 4:9 \u0939\u0948\u0964 \u090f\u0915 \u0906\u092f\u0924 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f \u091c\u093f\u0938\u0915\u0940 \u091a\u094c\u0921\u093c\u093e\u0908 \u092c\u0947\u0932\u0928 \u0915\u0940 \u0924\u094d\u0930\u093f\u091c\u094d\u092f\u093e \u0938\u0947 37.5% \u0905\u0927\u093f\u0915 \u0939\u0948 \u091c\u092c\u0915\u093f \u0906\u092f\u0924 \u0915\u0940 \u0932\u0902\u092c\u093e\u0908 \u0914\u0930 \u092c\u0947\u0932\u0928 \u0915\u0940 \u090a\u0902\u091a\u093e\u0908 \u0915\u093e \u0905\u0928\u0941\u092a\u093e\u0924 \u0915\u094d\u0930\u092e\u0936\u0903 3:4 \u0939\u0948\u0964 \u0926\u093f\u092f\u093e \u0917\u092f\u093e \u0939\u0948 \u0915\u093f \u092c\u0947\u0932\u0928 \u0915\u093e \u0935\u0915\u094d\u0930 \u092a\u0943\u0937\u094d\u0920\u0940\u092f \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 3456 cm2<\/sup> \u0939\u0948\u0964 (Take \u03c0 = 3)<\/p>\n\n\n\n

A) 884 cm<\/p>\n\n\n\n

B) 584 cm2<\/sup><\/p>\n\n\n\n

C) 594 cm2<\/sup><\/p>\n\n\n\n

D) 624 cm2<\/sup><\/p>\n\n\n\n

E) \u0907\u0928\u092e\u0947\u0902 \u0938\u0947 \u0915\u094b\u0908 \u0928\u0939\u0940\u0902<\/p>\n\n\n\n

\u092a\u094d\u0930\u0936\u094d\u0928 6:<\/strong> \u090f\u0915 \u0935\u0930\u094d\u0917 \u0915\u093e \u0935\u093f\u0915\u0930\u094d\u0923 24\u221a2 cm \u0939\u0948\u0964 \u0935\u0930\u094d\u0917 \u0915\u093e \u092a\u0930\u093f\u0927\u0940 \u0909\u0938 \u0906\u092f\u0924 \u0938\u0947 24 cm \u0915\u092e \u0939\u0948 \u091c\u093f\u0938\u0915\u0940 \u0932\u0902\u092c\u093e\u0908 \u0914\u0930 \u091a\u094c\u0921\u093c\u093e\u0908 \u0915\u093e \u0905\u0928\u0941\u092a\u093e\u0924 7:5 \u0939\u0948\u0964 \u0906\u092f\u0924 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f\u0964<\/p>\n\n\n\n

A) 725 cm2<\/sup><\/p>\n\n\n\n

B) 875 cm2<\/sup><\/p>\n\n\n\n

C) 915 cm2<\/sup><\/p>\n\n\n\n

D) 625 cm2<\/sup><\/p>\n\n\n\n

E) 1015 cm2<\/sup><\/p>\n\n\n\n

\u092a\u094d\u0930\u0936\u094d\u0928 7:<\/strong> \u090f\u0915 \u0906\u092f\u0924 \u0915\u0940 \u0932\u0902\u092c\u093e\u0908 \u0914\u0930 \u091a\u094c\u0921\u093c\u093e\u0908 \u0915\u093e \u0905\u0928\u0941\u092a\u093e\u0924 8:5 \u0939\u0948\u0964 \u091c\u092c \u0906\u092f\u0924 \u0915\u0940 \u0932\u0902\u092c\u093e\u0908 \u0914\u0930 \u091a\u094c\u0921\u093c\u093e\u0908 \u092e\u0947\u0902 \u0915\u094d\u0930\u092e\u0936\u0903 25% \u0915\u0940 \u0935\u0943\u0926\u094d\u0927\u093f \u0914\u0930 40% \u0915\u0940 \u0915\u092e\u0940 \u0915\u0940 \u091c\u093e\u0924\u0940 \u0939\u0948, \u0924\u094b \u0907\u0938\u0915\u093e \u092a\u0930\u093f\u092e\u093e\u092a 390 cm \u0939\u094b \u091c\u093e\u0924\u093e \u0939\u0948\u0964 \u0906\u092f\u0924 \u0915\u093e \u0935\u093e\u0938\u094d\u0924\u0935\u093f\u0915 \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f\u0964<\/p>\n\n\n\n

A) 8400 cm2<\/sup><\/p>\n\n\n\n

B) 9000 cm2<\/sup><\/p>\n\n\n\n

C) 7200 cm2<\/sup><\/p>\n\n\n\n

D) 6400 cm2<\/sup><\/p>\n\n\n\n

E) 10500 cm2<\/sup><\/p>\n\n\n\n

RBI Assistant prelims free mock test<\/a><\/p>\n\n\n\n

\u092a\u094d\u0930\u0936\u094d\u0928 8:<\/strong> \u090f\u0915 \u0935\u0930\u094d\u0917 \u0915\u093e \u092a\u0930\u093f\u092e\u093e\u092a \u090f\u0915 \u0906\u092f\u0924 \u0915\u0947 \u092a\u0930\u093f\u092e\u093e\u092a \u0938\u0947 36 metres \u0905\u0927\u093f\u0915 \u0939\u0948\u0964 \u0906\u092f\u0924 \u0915\u0940 \u0932\u0902\u092c\u093e\u0908 \u0914\u0930 \u091a\u094c\u0921\u093c\u093e\u0908 \u0915\u093e \u0905\u0928\u0941\u092a\u093e\u0924 9:4 \u0939\u0948\u0964 \u092f\u0926\u093f \u0935\u0930\u094d\u0917 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 3721 m2<\/sup> \u0939\u0948, \u0924\u094b Rs. 2.5\/m2<\/sup> \u0915\u0940 \u0926\u0930 \u0938\u0947 \u0906\u092f\u0924 \u0915\u0940 \u090f\u0915 \u0938\u0924\u0939 \u0915\u094b \u0930\u0902\u0917\u0928\u0947 \u0915\u0940 \u0932\u093e\u0917\u0924 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f\u0964<\/p>\n\n\n\n

A) Rs. 5760<\/p>\n\n\n\n

B) Rs. 4320<\/p>\n\n\n\n

C) Rs. 5220<\/p>\n\n\n\n

D) Rs. 6300<\/p>\n\n\n\n

E) Rs. 5040<\/p>\n\n\n\n

\u092a\u094d\u0930\u0936\u094d\u0928 9:<\/strong> \u090f\u0915 \u092c\u0947\u0932\u0928 \u0915\u093e \u0906\u092f\u0924\u0928 4928 m3<\/sup> \u0939\u0948, \u091c\u092c\u0915\u093f \u0907\u0938\u0915\u0940 \u092c\u093e\u0939\u0930\u0940 \u0938\u0924\u0939 (\u0906\u0927\u093e\u0930 \u0914\u0930 \u0936\u0940\u0930\u094d\u0937 \u0915\u094b \u091b\u094b\u0921\u093c\u0915\u0930) \u0915\u094b Rs. 8 per m2 <\/sup>\u0915\u0940 \u0926\u0930 \u0938\u0947 \u0930\u0902\u0917\u0928\u0947 \u0915\u0940 \u0932\u093e\u0917\u0924 Rs. 5632 \u0939\u0948\u0964 \u092c\u0947\u0932\u0928 \u0915\u0947 \u0906\u0927\u093e\u0930 \u0915\u0940 \u0924\u094d\u0930\u093f\u091c\u094d\u092f\u093e \u0914\u0930 \u090a\u0902\u091a\u093e\u0908 \u0915\u093e \u0905\u0928\u0941\u092a\u093e\u0924 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f\u0964<\/p>\n\n\n\n

A) 7:6<\/p>\n\n\n\n

B) 7:8<\/p>\n\n\n\n

C) 7:4<\/p>\n\n\n\n

D) 7:2<\/p>\n\n\n\n

E) \u0907\u0928\u092e\u0947\u0902 \u0938\u0947 \u0915\u094b\u0908 \u0928\u0939\u0940\u0902<\/p>\n\n\n\n

\u092a\u094d\u0930\u0936\u094d\u0928 10:<\/strong> \u090f\u0915 \u0936\u0902\u0915\u0941 \u0915\u0947 \u0906\u0927\u093e\u0930 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 616 cm2<\/sup> \u0939\u0948 \u0914\u0930 \u090f\u0915 \u092c\u0947\u0932\u0928 \u0915\u093e \u0935\u0915\u094d\u0930 \u092a\u0943\u0937\u094d\u0920\u0940\u092f \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 3168 cm2<\/sup> \u0939\u0948\u0964 \u092c\u0947\u0932\u0928 \u0915\u0940 \u090a\u0901\u091a\u093e\u0908 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f \u092f\u0926\u093f \u092c\u0947\u0932\u0928 \u0915\u0947 \u0906\u0927\u093e\u0930 \u0915\u0940 \u0924\u094d\u0930\u093f\u091c\u094d\u092f\u093e \u0936\u0902\u0915\u0941 \u0915\u0940 \u0924\u094d\u0930\u093f\u091c\u094d\u092f\u093e \u0938\u0947 4 cm \u0905\u0927\u093f\u0915 \u0939\u0948\u0964<\/p>\n\n\n\n

A) 21 cm<\/p>\n\n\n\n

B) 35 cm<\/p>\n\n\n\n

C) 28 cm<\/p>\n\n\n\n

D) 24 cm<\/p>\n\n\n\n

E) \u0907\u0928\u092e\u0947\u0902 \u0938\u0947 \u0915\u094b\u0908 \u0928\u0939\u0940\u0902<\/p>\n\n\n\n

RBI Assistant prelims free mock test<\/a><\/p>\n\n\n\n

ANSWER KEYS and SOLUTIONS:<\/strong><\/h4>\n\n\n\n
1) – D)<\/td>2) – E)<\/td>3) – D)<\/td>4) – E)<\/td>5) – C)<\/td>6) – B)<\/td><\/tr>
7) – B)<\/td>8) – A)<\/td>9) – C)<\/td>10) – C)<\/td> <\/td> <\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Solution 1: D)<\/strong><\/p>\n\n\n\n

Let radius of the cone is \u2018r\u2019 cm.<\/p>\n\n\n\n

So, 2\u03c0r = 220<\/p>\n\n\n\n

Or, x = 35 cm<\/p>\n\n\n\n

And, (1\/3) \u00d7 \u03c0r2<\/sup>h = 15400<\/p>\n\n\n\n

Or, (1\/3) \u00d7 (22\/7) \u00d7 35 \u00d7 35 \u00d7 h = 15400<\/p>\n\n\n\n

Or, h = 12 cm<\/p>\n\n\n\n

So, slant height = \u221a(122<\/sup> + 352<\/sup>) = \u221a1369 = 37 cm<\/p>\n\n\n\n

Hence, option d.<\/p>\n\n\n\n

Solution 2: E)<\/strong><\/p>\n\n\n\n

Let the length and breadth of the field be \u2018l\u2019 metres and \u2018b\u2019 metres, respectively<\/p>\n\n\n\n

According to the question,<\/p>\n\n\n\n

(l \u2013 b) = 27\u2026.. (1)<\/p>\n\n\n\n

Also, perimeter of the rectangular field = 492\/2 = 246<\/p>\n\n\n\n

Or, 2(l + b) = 246<\/p>\n\n\n\n

Or, (l + b) = 123\u2026\u2026 (2)<\/p>\n\n\n\n

On solving equation (1) and (2), we get<\/p>\n\n\n\n

Length of the rectangle = l = 75 metres<\/p>\n\n\n\n

Breadth of the field = b = 48 cm<\/p>\n\n\n\n

Required ratio = 75:48 = 25:16<\/p>\n\n\n\n

Hence, option e.<\/p>\n\n\n\n

RBI Assistant prelims free mock test<\/a><\/p>\n\n\n\n

Solution 3: D)<\/strong><\/p>\n\n\n\n

From each face of the larger cube, the number of smaller cubes obtained = (20 \u00d7 20)\/(4 \u00d7 4) = 25<\/p>\n\n\n\n

Total number of smaller cubes obtained from the edges of each face = 5 + 5 + 3 + 3 = 16 cubes<\/p>\n\n\n\n

Therefore, number of smaller cubes obtained from each face that doesn\u2019t lie on edge = 25 \u2013 16 = 9<\/p>\n\n\n\n

Required number of cubes = 9 \u00d7 6 = 54<\/p>\n\n\n\n

Hence, option d.<\/p>\n\n\n\n

Solution 4: E)<\/strong><\/p>\n\n\n\n

Let each side of the square \u2018A\u2019 be \u2018a\u2019 cm<\/p>\n\n\n\n

Therefore, each side of square \u2018B\u2019 = (a + 6) cm<\/p>\n\n\n\n

Difference between their perimeters = 4(a + 6) \u2013 4a = 24 cm<\/p>\n\n\n\n

Therefore, perimeter of the rectangle = 3 \u00d7 24 = 72 cm<\/p>\n\n\n\n

Let the breadth of the rectangle be \u2018b\u2019 cm<\/p>\n\n\n\n

According to the question,<\/p>\n\n\n\n

2 \u00d7 (20 + b) = 72<\/p>\n\n\n\n

Or, 20 + b = 36<\/p>\n\n\n\n

Or, b = 16 cm<\/p>\n\n\n\n

Hence, option e.<\/p>\n\n\n\n

Solution 5: C)<\/strong><\/p>\n\n\n\n

Let radius and height of the cylinder is \u20184x\u2019 cm and \u20189x\u2019 cm respectively.<\/p>\n\n\n\n

According to question;<\/p>\n\n\n\n

2\u03c0rh = 3456<\/p>\n\n\n\n

2 \u00d7 3 \u00d7 4x \u00d7 9x = 3456<\/p>\n\n\n\n

Or, x2<\/sup> = 16<\/p>\n\n\n\n

Or, x = 4<\/p>\n\n\n\n

Radius of cylinder 4 \u00d7 4 = 16 cm<\/p>\n\n\n\n

Breadth of rectangle = 1.375 \u00d7 16 = 22 cm<\/p>\n\n\n\n

Height of cylinder = 4 \u00d7 9 = 36 cm<\/p>\n\n\n\n

Length of rectangle = (3\/4) \u00d7 36 = 27 cm<\/p>\n\n\n\n

Area of the cylinder = 22 \u00d7 27 = 594 cm2<\/sup><\/p>\n\n\n\n

Hence, option c.<\/p>\n\n\n\n

Solution 6: B)<\/strong><\/p>\n\n\n\n

Let each side of the square be \u2018a\u2019 cm<\/p>\n\n\n\n

Therefore, \u221a2a = 24\u221a2<\/p>\n\n\n\n

Or, a = 24 cm<\/p>\n\n\n\n

Perimeter of the square = 4 \u00d7 24 = 96<\/p>\n\n\n\n

Let the length and breadth of the rectangle be 7x cm and 5x cm respectively<\/p>\n\n\n\n

Therefore, 2 \u00d7 (7x + 5x) = 96 + 24 = 120<\/p>\n\n\n\n

Or, 12x = 60<\/p>\n\n\n\n

Or, x = 5<\/p>\n\n\n\n

Therefore, area of the rectangle = 7x \u00d7 5x = 35x2<\/sup> = 875 cm2<\/sup><\/p>\n\n\n\n

Hence, option b.<\/p>\n\n\n\n

Solution 7: B)<\/strong><\/p>\n\n\n\n

Let the original length and breadth of the rectangle be 8x cm and 5x cm, respectively<\/p>\n\n\n\n

According to the question,<\/p>\n\n\n\n

2 \u00d7 {(1.25 \u00d7 8x) + (0.6 \u00d7 5x)} = 390<\/p>\n\n\n\n

Or, 10x + 3x = 195<\/p>\n\n\n\n

Or, x = 195\/13 = 15<\/p>\n\n\n\n

Therefore, original area of the rectangle = 8x \u00d7 5x = 40x2<\/sup> = 9000 cm2<\/sup><\/p>\n\n\n\n

Hence, option b.<\/p>\n\n\n\n

Solution 8: A)<\/strong><\/p>\n\n\n\n

Each side of the square = \u221a3721 = 61 m<\/p>\n\n\n\n

Therefore, perimeter of the square = 61 \u00d7 4 = 244 m<\/p>\n\n\n\n

Let the length and breadth of the rectangle be 9x m and 4x m, respectively.<\/p>\n\n\n\n

According to the question,<\/p>\n\n\n\n

2 \u00d7 (9x + 4x) = 244 \u2013 36<\/p>\n\n\n\n

Or, 13x = 208\/2 = 104<\/p>\n\n\n\n

Or, x = 104\/13 = 8<\/p>\n\n\n\n

Therefore, area of rectangle = 9x \u00d7 4x = 36x2<\/sup> = 2304 m2<\/sup><\/p>\n\n\n\n

Cost of painting = 2.5 \u00d7 2304 = Rs. 5760<\/p>\n\n\n\n

Hence, option a.<\/p>\n\n\n\n

RBI Assistant prelims free mock test<\/a><\/p>\n\n\n\n

Solution 9: C)<\/strong><\/p>\n\n\n\n

Let, radius and height of the cylinder is \u2018r\u2019 metres and \u2018h\u2019 metres, respectively.<\/p>\n\n\n\n

So, \u03c0r2<\/sup>h = 4928\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.(1)<\/p>\n\n\n\n

CSA of cylinder = 5632\/8 = 704 m2<\/sup><\/p>\n\n\n\n

So, 2\u03c0rh = 704\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..(2)<\/p>\n\n\n\n

Dividing equation (1) by equation (2), we get<\/p>\n\n\n\n

r = 7 \u00d7 2 = 14<\/p>\n\n\n\n

So, 2 \u00d7 22\/7 \u00d7 14 \u00d7 h = 704<\/p>\n\n\n\n

Or, h = 8<\/p>\n\n\n\n

Desired ratio = 14:8 = 7:4<\/p>\n\n\n\n

Hence, option c.<\/p>\n\n\n\n

Solution 10: C)<\/strong><\/p>\n\n\n\n

Let radius of the cone be \u2018r\u2019 cm<\/p>\n\n\n\n

So, 22\/7 \u00d7 r2<\/sup> = 616<\/p>\n\n\n\n

Or, r2<\/sup> = 196<\/p>\n\n\n\n

Or, r = 14 cm<\/p>\n\n\n\n

Radius of the cylinder = 14 + 4 = 18 cm<\/p>\n\n\n\n

Let height of the cylinder is \u2018h\u2019 cm.<\/p>\n\n\n\n

So, 2 \u00d7 (22\/7) \u00d7 18 \u00d7 h = 3168<\/p>\n\n\n\n

Or, h = 3168 \u00d7 (7\/792)<\/p>\n\n\n\n

Or, h = 28 cm<\/p>\n\n\n\n

Hence, option c.<\/p>\n\n\n\n

RBI Assistant prelims free mock test<\/a><\/p>\n\n\n\n

Preparing for upcoming RBI Assistant prelims exam? Read & practice these Quant Mensuration Tips & Practice Questions. <\/p>\n","protected":false},"excerpt":{"rendered":"

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