Enroll to 15 SSC MTS 2022 Mocks At Just Rs. 149<\/a><\/p>\n Question 1:\u00a0<\/b>If the height of a triangle is decreased by 40% and it\u2019s base is increased by 40% , what will be the effect on its area?<\/p>\n a)\u00a0No change<\/p>\n b)\u00a016 % increase<\/p>\n c)\u00a08% decrease<\/p>\n d)\u00a016% decrease<\/p>\n e)\u00a0None of these<\/p>\n 1)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n area of triangle = 1\/2(base*height)<\/p>\n after decreasing height by 40%, new height = 0.6H<\/p>\n after increasing base by 40%, new base = 1.4B<\/p>\n now, area of triangle = 1\/2(1.4B*0.6H) = 0.84(1\/2*B*H)<\/p>\n Area decreased by 100-84%= 16%<\/p>\n so the answer is option D.<\/p>\n Question 2:\u00a0<\/b>A circular Ground whose diameter is 35 meters has a 1.4 meters broad garden around it.What is the area of the garden in square meters ?<\/p>\n a)\u00a0160.16<\/p>\n b)\u00a06.16<\/p>\n c)\u00a01122.66<\/p>\n d)\u00a0Data Inadequate<\/p>\n e)\u00a0None of these<\/p>\n 2)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Radius of circular ground = $r = \\frac{35}{2}=17.5$ m and width of broad garden = 1.4 m<\/p>\n => Radius of outer circle\u00a0(ground+garden), R = 17.5 + 1.4 = 18.9 m<\/p>\n Area of garden = Area of outer circle – Area of inner circle<\/p>\n = $\\pi R^2-\\pi r^2 = \\pi(R-r)(R+r)$<\/p>\n = $\\frac{22}{7}(18.9-17.5)(18.9+17.5)$<\/p>\n = $\\frac{22}{7} \\times 1.4 \\times 36.4$<\/p>\n = $22 \\times 0.2 \\times 36.4 = 160.16$ $m^2$<\/p>\n => Ans – (A)<\/p>\n Question 3:\u00a0<\/b>The top of 15 meters high tower makes an angle of depression of 60 degrees with the bottom of a electric pole and an angle of 30 degrees with the top of the pole. What is the height of the electric pole?<\/p>\n a)\u00a05 meters<\/p>\n b)\u00a08 meters<\/p>\n c)\u00a010 meters<\/p>\n d)\u00a012 meters<\/p>\n e)\u00a0None of these<\/p>\n 3)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n AD is the tower = 15 m and CE is the electric pole<\/p>\n Let AB = $x$ m and DE = BC = $y$ m<\/p>\n Also, $\\angle$ AED = 60\u00b0 and\u00a0$\\angle$\u00a0ACB = 30\u00b0<\/p>\n In\u00a0$\\triangle$\u00a0ADE, => $tan(\\angle AED)=\\frac{AD}{DE}$<\/p>\n => $tan(60)=\\sqrt{3}=\\frac{15}{y}$<\/p>\n => $y=\\frac{15}{\\sqrt{3}}=5\\sqrt{3}$ m<\/p>\n In\u00a0$\\triangle$\u00a0ABC, => $tan(\\angle ACB)=\\frac{AB}{BC}$<\/p>\n => $tan(30)=\\frac{1}{\\sqrt{3}}=\\frac{x}{5\\sqrt{3}}$<\/p>\n => $x=5$ m<\/p>\n $\\therefore$ CE = AD – AB = 15 – 5 = 10 meters<\/p>\n => Ans – (C)<\/p>\n Question 4:\u00a0<\/b>If the height of a triangle is decreased by 40% and it\u2019s base is increased by 40% , what will be the effect on its area?<\/p>\n a)\u00a0No change<\/p>\n b)\u00a016 % increase<\/p>\n c)\u00a08% decrease<\/p>\n d)\u00a016% decrease<\/p>\n e)\u00a0None of these<\/p>\n 4)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let base and height of the triangle be 10 cm each<\/p>\n Area of original triangle, $\\triangle = \\frac{1}{2} \\times 10 \\times 10 = 50$ $cm^2$<\/p>\n If the height of a triangle is decreased by 40%, => New height = $10-\\frac{40}{100} \\times 10=6$ cm<\/p>\n and it\u2019s base is increased by 40%, => New base =\u00a0$10+\\frac{40}{100} \\times 10=14$ cm<\/p>\n Area of new triangle, $\\triangle’ = \\frac{1}{2} \\times 6 \\times 14 = 42$ $cm^2$<\/p>\n $\\therefore$ Area decreases<\/strong> by = $\\frac{\\triangle – \\triangle’}{\\triangle} \\times 100$<\/p>\n = $\\frac{50-42}{50} \\times 100 = 8 \\times 2=16\\%$<\/p>\n => Ans – (D)<\/p>\n Question 5:\u00a0<\/b>When the length of the rectangular plot is increased by four times its perimeter becomes 480 meters and area becomes 12800 sq.m. What is its original length(in meters)?<\/p>\n a)\u00a0160<\/p>\n b)\u00a040<\/p>\n c)\u00a020<\/p>\n d)\u00a0Cannot be determined<\/p>\n e)\u00a0None of these<\/p>\n 5)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the length of the plot be $l$ meters and breadth = $b$ meters<\/p>\n New length = $4l$ meters<\/p>\n Perimeter = $2(4l+b)=480$<\/p>\n => $4l+b=\\frac{480}{2}=240$<\/p>\n => $b=240-4l$ ———(i)<\/p>\n Area = $(4l \\times b)=12800$<\/p>\n => $lb=\\frac{12800}{4}=3200$<\/p>\n Substituting value of $b$ from equation (i)<\/p>\n => $l(240-4l)=3200$<\/p>\n => $240l-4l^2=3200$<\/p>\n => $l^2-60l+800$<\/p>\n => $l=\\frac{-(-60) \\pm \\sqrt{(-60)^2-(4 \\times 1 \\times 800)}}{2}$<\/p>\n => $l=\\frac{60 \\pm \\sqrt{3600-3200}}{2} = \\frac{60 \\pm \\sqrt{400}}{2}$<\/p>\n => $l=\\frac{60 \\pm 20}{2}$<\/p>\n => $l=\\frac{60+20}{2},\\frac{60-20}{2}$<\/p>\n => $l=\\frac{80}{2},\\frac{40}{2}$<\/p>\n => $l=40,20$ meters<\/p>\n => Ans – (D)<\/p>\n Take a free SSC MTS Tier-1 mock test<\/a><\/p>\n Download SSC CGL Tier-1 Previous Papers PDF<\/a><\/p>\n Question 6:\u00a0<\/b>Four circles having equal radii are drawn with center at the four corners of a square. Each circle touches the other two adjacent circle. If remaining area of the square is 168 cm, what is the size of the radius of the radius of the circle? (in centimeters)<\/p>\n a)\u00a014<\/p>\n b)\u00a01.4<\/p>\n c)\u00a038<\/p>\n d)\u00a021<\/p>\n e)\u00a03.5<\/p>\n 6)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Diameter of circle = side of square = $d$ cm<\/p>\n Area of square = $d^2$ sq. cm<\/p>\n Area of 1 quadrant = $\\frac{1}{4} \\times \\pi \\times (\\frac{d}{2})^2$<\/p>\n = $\\pi \\times \\frac{d^2}{16}$<\/p>\n => Area of 4 quadrants = $4 \\times \\pi \\times \\frac{d^2}{16} = \\frac{\\pi d^2}{4}$ sq. cm<\/p>\n Area of shaded region = 168<\/p>\n => $d^2 – \\frac{\\pi d^2}{4} = 168$<\/p>\n => $\\frac{1}{4} [d^2 (4 – \\pi)] = 168$<\/p>\n => $d^2 = \\frac{168 \\times 4}{4 – \\pi} = \\frac{168 \\times 4}{4 – \\frac{22}{7}}$<\/p>\n => $d^2 = \\frac{168 \\times 4 \\times 7}{28 – 22} = \\frac{168}{6} \\times 28$<\/p>\n => $d = \\sqrt{28 \\times 28} = 28$ cm<\/p>\n $\\therefore$ Radius = $\\frac{28}{2} = 14$ cm<\/p>\n Question 7:\u00a0<\/b>The area of a right angled triangle is 80 sq. cm. The ratio of the base and the height of the triangle is 4 : 5. Find the length of hypotenuse.<\/p>\n a)\u00a082cm<\/p>\n b)\u00a02 82cm<\/p>\n c)\u00a02 41cm<\/p>\n d)\u00a03 82cm<\/p>\n e)\u00a0None of these<\/p>\n 7)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let base and height be $4x$ and $5x$ respectively.<\/p>\n Area of triangle = $\\frac{1}{2} \\times$ base $\\times$ height = 80<\/p>\n => $\\frac{1}{2} \\times 4x \\times 5x = 80$<\/p>\n => $x^2 = \\frac{80}{10} = 8$<\/p>\n => $x = \\sqrt{8} = 2 \\sqrt{2}$<\/p>\n => Base = $8 \\sqrt{2}$ and Height = $10 \\sqrt{2}$<\/p>\n $\\therefore$ Hypotenuse = $\\sqrt{(8 \\sqrt{2})^2 + (10 \\sqrt{2})^2}$<\/p>\n = $\\sqrt{128 + 200} = \\sqrt{328}$<\/p>\n $\\approx 18$ cm<\/p>\n Question 8:\u00a0<\/b>Area of a rectangle is equal to the area of the circle whose radius is 21 cms. If the length and the breadth of the rectangle are in the ratio of 14 : 11 respectively, what is its perimeter?<\/p>\n a)\u00a0142 cms.<\/p>\n b)\u00a0140 cms.<\/p>\n c)\u00a0132 ems.<\/p>\n d)\u00a0150 cms.<\/p>\n e)\u00a0None of these<\/p>\n 8)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Radius of circle = 21 cm<\/p>\n => Area of circle = $\\pi r^2$<\/p>\n = $\\frac{22}{7} \\times 21 \\times 21 = 1386 cm^2$<\/p>\n Thus, area of rectangle = $1386 cm^2$<\/p>\n Let length and breadth of rectangle be $14x$ and $11x$ respectively.<\/p>\n => Area = $14x \\times 11x = 1386$<\/p>\n => $x^2 = \\frac{1386}{154} = 9$<\/p>\n => $x = \\sqrt{9} = 3$ cm<\/p>\n $\\therefore$ Perimeter = $2 (14x + 11x)$<\/p>\n = $2 \\times 25x = 50x$<\/p>\n = $50 \\times 3 = 150$ cm<\/p>\n Question 9:\u00a0<\/b>What would be the cost of building a fence around a square plot with area equal to 361 sq. ft. if the price per foot of building the fence is Rs. 62 ?<\/p>\n a)\u00a0Rs. 4026<\/p>\n b)\u00a0Rs. 4712<\/p>\n c)\u00a0Rs. 3948<\/p>\n d)\u00a0Cannot be determined<\/p>\n e)\u00a0None of these<\/p>\n 9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let side of square plot = $s$ ft<\/p>\n => Area = $s^2 = 361$<\/p>\n => $s = \\sqrt{361} = 19$ ft<\/p>\n Perimeter of square plot = $4 s = 4 \\times 19 = 76$ ft<\/p>\n Cost of building fence per foot = Rs. 62<\/p>\n $\\therefore$ Cost of building fence around the plot = $76 \\times 62 = Rs. 4712$<\/p>\n Question 10:\u00a0<\/b>The sum of the dimensions of a room (i.e. length, breadth and height) is 18 metres and its length, breadth and height are in the ratio of 3 : 2 : 1 respectively. If the room is to be painted at the rate of Rs. 15 per m2, what would be the total cost incurred on painting only the four walls of the room (in Rs.)?<\/p>\n a)\u00a03250<\/p>\n b)\u00a02445<\/p>\n c)\u00a01350<\/p>\n d)\u00a02210<\/p>\n e)\u00a02940<\/p>\n 10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the dimension of the room be $3x , 2x , x$ metres<\/p>\n Acc. to ques, => $3x + 2x + x = 18$<\/p>\n => $x = \\frac{18}{6} = 3$ metres<\/p>\n Curved surface area of the room = $2 h (l + b)$<\/p>\n = $2 \\times x \\times (3x + 2x) = 2x \\times 5x$<\/p>\n = $10 (x)^2 = 10 \\times (3)^2$<\/p>\n = $10 \\times 9 = 90 m^2$<\/p>\n $\\therefore$ Total cost incurred on painting only the four walls of the room\u00a0= $15 \\times 90$<\/p>\n = $Rs. 1,350$<\/p>\n Question 11:\u00a0<\/b>The sum of the dimensions of a room (i.e. length, breadth and height) is 24 metres and its length, breadth and height are in the ratio of 8: 7 : 5 respectively. If the room is to be painted at the rate of Rs. 12 per m2, what would be the total cost incurred on painting only the four walls of the room (in Rs.)?<\/p>\n a)\u00a02592<\/p>\n b)\u00a02648<\/p>\n c)\u00a02848<\/p>\n d)\u00a02120<\/p>\n e)\u00a01956<\/p>\n 11)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the dimension of the room be $8x , 7x , 5x$ metres<\/p>\n Acc. to ques, => $8x + 7x + 5x = 24$<\/p>\n => $x = \\frac{24}{20} = 1.2$ metres<\/p>\n Curved surface area of the room = $2 h (l + b)$<\/p>\n = $2 \\times 5x \\times (8x + 7x) = 10x \\times 15x$<\/p>\n = $150 (x)^2 = 150 \\times (1.2)^2$<\/p>\n = $150 \\times 1.44 = 216 m^2$<\/p>\n $\\therefore$ Total cost incurred on painting only the four walls of the room\u00a0= $12 \\times 216$<\/p>\n = $Rs. 2,592$<\/p>\n Question 12:\u00a0<\/b>It takes Rs. 3159 to plant synthetic grass in a square lawn, ${1 \\over 4}$ of which is paved (and thus does not require grass). If each side of this lawn measures 18m, what is the rate that the gardener charges for planting synthetic grass? (in Rs.\/m\u00b2)<\/p>\n a)\u00a018<\/p>\n b)\u00a011<\/p>\n c)\u00a016<\/p>\n d)\u00a015<\/p>\n e)\u00a013<\/p>\n 12)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n Side of square lawn = 18 m<\/p>\n Part of the lawn that requires grass = $1 – \\frac{1}{4} = \\frac{3}{4}$<\/p>\n => Area of the lawn that requires grass = $\\frac{3}{4} \\times (18)^2$<\/p>\n = $3 \\times 9 \\times 9 = 243 m^2$<\/p>\n Total amount to plant grass = Rs. 3159<\/p>\n $\\therefore$ Rate that the gardener charges for planting synthetic grass = $\\frac{3159}{243} = 13$<\/p>\n Question 13:\u00a0<\/b>If (11) is subtracted from the square of a number, the answer so obtained is 430. What is the number?<\/p>\n a)\u00a036<\/p>\n b)\u00a028<\/p>\n c)\u00a021<\/p>\n d)\u00a048<\/p>\n e)\u00a0None of these<\/p>\n 13)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the number be $x$<\/p>\n According to ques,<\/p>\n => $x^2-11 = 430$<\/p>\n => $x^2=430+11=441$<\/p>\n => $x=\\sqrt{441}=21$<\/p>\n => Ans – (C)<\/p>\n Question 14:\u00a0<\/b>Area of circle is equal to the area of a rectangle having perimeter of 50 cms. and length more than the breadth by 3 cms. What is the diameter of the circle?<\/p>\n a)\u00a07 cms.<\/p>\n b)\u00a021 cms.<\/p>\n c)\u00a028 cms.<\/p>\n d)\u00a014 cms.<\/p>\n e)\u00a0None of these<\/p>\n 14)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the breadth of rectangle be $x$ cm<\/p>\n => Length of rectangle = $(x + 3)$ cm<\/p>\n Now, perimeter = $2 (x + x + 3) = 50$<\/p>\n => $2x + 3 = 25$<\/p>\n => $x = \\frac{22}{2} = 11$<\/p>\n Thus, breadth = 11 cm and length = 14 cm<\/p>\n $\\because$ Area of circle = Area of rectangle<\/p>\n => $\\pi r^2 = 14 \\times 11$<\/p>\n => $r^2 = \\frac{14 \\times 11 \\times 7}{22}$<\/p>\n => $r = \\sqrt{7 \\times 7} = 7$ cm<\/p>\n $\\therefore$ Diameter = 2 * 7 = 14 cm<\/p>\n Question 15:\u00a0<\/b>When 3888 is divided by the square of a number and the answer so obtained is multiplied by 21, the final answer obtained is 252. What is the number ?<\/p>\n a)\u00a0324<\/p>\n b)\u00a016<\/p>\n c)\u00a0256<\/p>\n d)\u00a0144<\/p>\n e)\u00a0None of these<\/p>\n 15)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the number be $x$<\/p>\n Acc to ques,<\/p>\n => $\\frac{3888}{x^2} \\times 21 = 252$<\/p>\n => $\\frac{3888}{x^2} = 12$<\/p>\n => $x^2 = \\frac{3888}{12} = 324$<\/p>\n => $x = \\sqrt{324} = 18$<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/9966.PNG\")
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