Here you can download Top 50 questions for CMAT 2022 – important Quant Questions PDF by Cracku. Very Important Quant Questions for CMAT 2022 based on asked questions in previous exam papers. These questions will help your CMAT preparation. So kindly download the PDF for reference and do more practice.<\/p>\n
Download Quant Questions for CMAT <\/a><\/p>\n Enroll to CMAT crash course<\/a><\/p>\n Take CMAT mock tests here<\/a><\/p>\n Download CMAT previous papers PDF<\/a><\/p>\n Question 1:\u00a0<\/b>A three-digit number is such that the hundreds digit is double the units digit which is 66.67% less than the tens digit. If the tens and hundreds digits of the original number are interchanged, then the number obtained is 180 more than the original number. Find the number obtained by interchanging the units and hundreds digit of the original number.<\/p>\n a)\u00a0246<\/p>\n b)\u00a0642<\/p>\n c)\u00a0264<\/p>\n d)\u00a0286<\/p>\n e)\u00a0None of these<\/p>\n Question 2:\u00a0<\/b>What is the median of the given data? a)\u00a06.5<\/p>\n b)\u00a05<\/p>\n c)\u00a06<\/p>\n d)\u00a05.5<\/p>\n Question 3:\u00a0<\/b>What is the value of $\\frac{3\u00a0 of\u00a0 24 \\div 8 \\times 3 + 4 \\div 2 – 4 \\times 5}{36 \\div 12 \\times 4 \\div 2 + 5 \\times (6 – 4)}$?<\/p>\n a)\u00a0$\\frac{3}{10}$<\/p>\n b)\u00a0$\\frac{9}{16}$<\/p>\n c)\u00a0$\\frac{3}{4}$<\/p>\n d)\u00a0$\\frac{8}{15}$<\/p>\n Question 4:\u00a0<\/b>What is the value of $\\frac{\\frac{3}{4} \\div \\frac{9}{32} + \\frac{4}{3} \\times \\frac{2}{3}\u00a0 of\u00a0 \\frac{27}{16}}{\\frac{1}{2} \\times \\left(\\frac{8}{3} – 2\\right) \\div \\frac{4}{9} + \\left(\\frac{1}{3} + \\frac{1}{6}\\right)}$?<\/p>\n a)\u00a0$\\frac{13}{2}$<\/p>\n b)\u00a0$\\frac{10}{3}$<\/p>\n c)\u00a0$\\frac{31}{2}$<\/p>\n d)\u00a0$\\frac{25}{2}$<\/p>\n Question 5:\u00a0<\/b>What is the value of a)\u00a0-7<\/p>\n b)\u00a0-8<\/p>\n c)\u00a0-9<\/p>\n d)\u00a0-1<\/p>\n Get 5 CMAT mocks at just Rs.299<\/a><\/p>\n Instructions<\/b><\/p>\n In each of the following questions a question is asked followed by three statements.You have to study the questions and all the given three statements and all the statements and decide whether any information provided in the statement are redundant and can be dispensed with while answering the questions ?<\/p>\n Question 6:\u00a0<\/b>What will be the cost of the painting the four walls of a room with length ,width and height is 5,3,8 meters respectively. The room has one window and one door. a)\u00a0I only<\/p>\n b)\u00a0II only<\/p>\n c)\u00a0III only<\/p>\n d)\u00a0II or III only<\/p>\n e)\u00a0All are required to answer the question<\/p>\n Question 7:\u00a0<\/b>How much time will the train X take to cross another train Y running in the opposite direction ? a)\u00a0I only<\/p>\n b)\u00a0II only<\/p>\n c)\u00a0III only<\/p>\n d)\u00a0I or II only<\/p>\n e)\u00a0Data inadequate<\/p>\n Question 8:\u00a0<\/b>Three friends X,Y,Z started a partnership business investing money in the ratio of the 5 :4:2 respectively for a period of 3 years.What is the amount received by X as share in the total profit ? a)\u00a0I only<\/p>\n b)\u00a0II only<\/p>\n c)\u00a0III Only<\/p>\n d)\u00a0I and II only<\/p>\n e)\u00a0I or II or III<\/p>\n Question 9:\u00a0<\/b>what is the area of the given right angle triangle ? a)\u00a0I only<\/p>\n b)\u00a0II only<\/p>\n c)\u00a0II or III only<\/p>\n d)\u00a0II and III both<\/p>\n e)\u00a0Data inadequate<\/p>\n Question 10:\u00a0<\/b>A tank has two inlets : A and B. A alone takes 2 hours and B alone takes 3 hours to fill the empty tank completely when there is no leakage. A leakage was caused which would empty the full tank completely in \u2018x\u2019 hours when no inlet is open. Now, when only inlet A was opened, it took 3 hours to fill the empty tank completely. How much time will B alone take to fill the empty tank completely? (in hours)<\/p>\n a)\u00a04.5<\/p>\n b)\u00a07.5<\/p>\n c)\u00a03<\/p>\n d)\u00a09<\/p>\n e)\u00a06<\/p>\n Question 11:\u00a0<\/b>The time taken by a boat to travel a distance downstream is half the time taken by it to travel the same distance upstream. What is the speed of the boat downstream if it travels 7.5 km upstream in 1 hour 30 minutes ? (in km\/h)<\/p>\n a)\u00a07.5<\/p>\n b)\u00a05<\/p>\n c)\u00a09<\/p>\n d)\u00a010<\/p>\n e)\u00a0None of these<\/p>\n Question 12:\u00a0<\/b>To reach point B from point A, at 4pm, Sara will have to travel at an average speed of 18 kmph. She will reach point B at 3 pm if she travels at an average speed of 24 kmph. At what average speed should Sara travel to reach point B at 2 pm ?<\/p>\n a)\u00a036 kmph<\/p>\n b)\u00a028 kmph<\/p>\n c)\u00a025 kmph<\/p>\n d)\u00a030 kmph<\/p>\n e)\u00a032 kmph<\/p>\n Question 13:\u00a0<\/b>A boat running downstream covers a distance of 16 km in 2 hours while for covering the same distance upstream, it takes 4 hours. What is the speed of the boat in still water ?<\/p>\n a)\u00a04 kmph<\/p>\n b)\u00a06 kmph<\/p>\n c)\u00a08 kmph<\/p>\n d)\u00a03.5 kmph<\/p>\n e)\u00a0None of these<\/p>\n Question 14:\u00a0<\/b>A car starts at 11 am from point A towards point B at 36 kmph while another car starts at 1 pm from point B towards A at 44 kmph. They cover a distance of 592 km till meeting. At what time will they meet each other ?<\/p>\n a)\u00a08 pm<\/p>\n b)\u00a06 : 30 pm<\/p>\n c)\u00a07 : 30 pm<\/p>\n d)\u00a05 : 30 pm<\/p>\n e)\u00a0None of these<\/p>\n Question 15:\u00a0<\/b>If the fraction $\\frac{4}{9},\\frac{2}{7},\\frac{3}{8},\\frac{6}{13} and \\frac{5}{11}$ are arranged in descending order which one will be second ?<\/p>\n a)\u00a0$\\frac{4}{9}$<\/p>\n b)\u00a0$\\frac{2}{7}$<\/p>\n c)\u00a0$\\frac{3}{8}$<\/p>\n d)\u00a0$\\frac{6}{13}$<\/p>\n e)\u00a0$\\frac{5}{11}$<\/p>\n Question 16:\u00a0<\/b>The digits of a two-digit number are in the ratio of 2 : 3 and the number obtained by interchanging the digit is bigger than the original number by 27. What was the original number ?<\/p>\n a)\u00a064<\/p>\n b)\u00a046<\/p>\n c)\u00a096<\/p>\n d)\u00a069<\/p>\n e)\u00a0None of these<\/p>\n Question 17:\u00a0<\/b>75% of a number is equal to three-seventh of another number. What is the ratio between the first number and the second number respectively ?<\/p>\n a)\u00a04:7<\/p>\n b)\u00a07:4<\/p>\n c)\u00a012:7<\/p>\n d)\u00a07:12<\/p>\n e)\u00a0None of these<\/p>\n Question 18:\u00a0<\/b>If the fraction $\\frac{4}{5},\\frac{9}{11},\\frac{7}{9},\\frac{5}{6},\\frac{11}{13}$ are arranged in ascending order which one will be the fourth ?<\/p>\n a)\u00a0$\\frac{9}{11}$<\/p>\n b)\u00a0$\\frac{7}{9}$<\/p>\n c)\u00a0$\\frac{5}{6}$<\/p>\n d)\u00a0$\\frac{4}{5}$<\/p>\n e)\u00a0$\\frac{11}{13}$<\/p>\n Question 19:\u00a0<\/b>Three-seventh of a number is equal to 45 per cent of another number What is the respective ratio between the first and the second numbers ?<\/p>\n a)\u00a021:20<\/p>\n b)\u00a020:21<\/p>\n c)\u00a045:100<\/p>\n d)\u00a045:300<\/p>\n e)\u00a0None of these<\/p>\n Question 20:\u00a0<\/b>Four of the following five parts numbered A,B,C,D,E are in the following equation are exactly equal.Which of the part is not equal to the other four.The number of that part is the answer<\/p>\n a)\u00a0$xy^{2}-x^{2}y+2x^{2}y^{2}$<\/p>\n b)\u00a0$xy^{2}(1-2x)+x^{2}y(2y-1)$<\/p>\n c)\u00a0$xy^{2}(1+2x)-x^{2}y(2y+1)$<\/p>\n d)\u00a0$xy[y(1+x)-x(1+y)]$<\/p>\n e)\u00a0$xy[(x+y)^{2}+y(1-y)-x(1+x)-2xy]$<\/p>\n Question 21:\u00a0<\/b>If 2x+y= 15, \u00a02y+z= 25 \u00a0 and \u00a0 2z+x =26 ,what is value of z?<\/p>\n a)\u00a04<\/p>\n b)\u00a07<\/p>\n c)\u00a09<\/p>\n d)\u00a012<\/p>\n e)\u00a0None of these<\/p>\n Question 22:\u00a0<\/b>I.\u00a0$x^{2}+3x-28=0$ a)\u00a0x > y<\/p>\n b)\u00a0x \u2265 y<\/p>\n c)\u00a0x < y<\/p>\n d)\u00a0x \u2264 y<\/p>\n e)\u00a0x = y or the relation cannot be established.<\/p>\n Question 23:\u00a0<\/b>I.\u00a0$6x^{2}+29x+35=0$ a)\u00a0x > y<\/p>\n b)\u00a0x \u2265 y<\/p>\n c)\u00a0x < y<\/p>\n d)\u00a0x \u2264 y<\/p>\n e)\u00a0x = y or the relation cannot be established.<\/p>\n Question 24:\u00a0<\/b>\u00a0I.\u00a0$2x^{2}+18x+40=0$ a)\u00a0x > y<\/p>\n b)\u00a0x \u2265 y<\/p>\n c)\u00a0x < y<\/p>\n d)\u00a0x \u2264 y<\/p>\n e)\u00a0x = y or the relation cannot be established.<\/p>\n Question 25:\u00a0<\/b>Profit earned by organisation is distributed among officers and clerks in the ratio of 5:3 If the number of officers is 45 and the number of clerks is 80 and the amount received by each officer is rs 25,000 what was the total amount of profit earned ?<\/p>\n a)\u00a0Rs. 22 lakh<\/p>\n b)\u00a0Rs. 18.25 lakh<\/p>\n c)\u00a0Rs. 18 lakh<\/p>\n d)\u00a0Rs. 23.25 lakh<\/p>\n e)\u00a0None of these<\/p>\n Download CMAT Previous Papers PDF<\/a><\/p>\n Question 26:\u00a0<\/b>An article was purchased for Rs.78,350. Its price was marked up by 30%. It was sold at a discount of 20% on the marked up price. What was the profit percent on the cost price?<\/p>\n a)\u00a04<\/p>\n b)\u00a07<\/p>\n c)\u00a05<\/p>\n d)\u00a03<\/p>\n e)\u00a06<\/p>\n Question 27:\u00a0<\/b>A shopkeeper sells two watches for Rs308 each. On one he gets 12% profit and on the other 12% loss. His profit or loss in the entire transaction was<\/p>\n a)\u00a01$\\frac{11}{25}$% loss<\/p>\n b)\u00a01$\\frac{11}{25}$% gain<\/p>\n c)\u00a03$\\frac{2}{25}$%loss<\/p>\n d)\u00a03$\\frac{2}{25}$%gain<\/p>\n e)\u00a0None of these<\/p>\n Question 28:\u00a0<\/b>A shopkeeper uses a weight of 960 g instead of 1000 g. What is his gain%?<\/p>\n a)\u00a04%<\/p>\n b)\u00a06%<\/p>\n c)\u00a04$\\frac{1}{6}$%<\/p>\n d)\u00a061\/4%<\/p>\n e)\u00a0None of these<\/p>\n Question 29:\u00a0<\/b>A person bought an article on 40% discount and sold it at 50% more than the marked price. What profit did he get?<\/p>\n a)\u00a0250 %<\/p>\n b)\u00a0150 %<\/p>\n c)\u00a0350 %<\/p>\n d)\u00a0200 %<\/p>\n e)\u00a0None of these<\/p>\n Question 30:\u00a0<\/b>In how many different ways can the letters of the word ‘STRESS’ be arranged<\/p>\n a)\u00a0360<\/p>\n b)\u00a0240<\/p>\n c)\u00a0720<\/p>\n d)\u00a0120<\/p>\n e)\u00a0None of these<\/p>\n Question 31:\u00a0<\/b>A bag contains 7 blue balls and 5 yellow balls. If two balls are selected at random, what is the probability that none is yellow?<\/p>\n a)\u00a05\/33<\/p>\n b)\u00a05\/22<\/p>\n c)\u00a07\/22<\/p>\n d)\u00a07\/33<\/p>\n e)\u00a07\/66<\/p>\n Question 32:\u00a0<\/b>A die is thrown twice. What is the probability of getting a sum 7 from the two throws?<\/p>\n a)\u00a05\/18<\/p>\n b)\u00a01\/18<\/p>\n c)\u00a01\/9<\/p>\n d)\u00a01\/6<\/p>\n e)\u00a05\/36<\/p>\n Question 33:\u00a0<\/b>In how many different ways can the letters of the word ‘THERAPY’ be arranged so that the vowels never come together?<\/p>\n a)\u00a0720<\/p>\n b)\u00a01440<\/p>\n c)\u00a05040<\/p>\n d)\u00a03600<\/p>\n e)\u00a04800<\/p>\n Question 34:\u00a0<\/b>A bag contains 13 white and 7 black balls. Two balls are drawn at random. What is the probability that they are of the same colour?<\/p>\n a)\u00a0$\\frac{41}{190}$<\/p>\n b)\u00a0$\\frac{21}{190}$<\/p>\n c)\u00a0$\\frac{59}{290}$<\/p>\n d)\u00a0$\\frac{99}{190}$<\/p>\n e)\u00a0$\\frac{77}{190}$<\/p>\n Question 35:\u00a0<\/b>The simple interest on a certain principal is Rs. 1,280 in 2 years at the rate of 8 p.c.p.a. What would be the simple interest accrued on thrice of that principal at the rate of 10.75 p.c.p.a. at the end of two years ?<\/p>\n a)\u00a0Rs. 4960<\/p>\n b)\u00a0Rs. 4860<\/p>\n c)\u00a0Rs. 5060<\/p>\n d)\u00a0Rs. 5160<\/p>\n e)\u00a0None of these<\/p>\n Question 36:\u00a0<\/b>The perimeter of a rectangle is 60 cm and its breadth is 12 cm. What is the area of the rectangle ?<\/p>\n a)\u00a021.6 sq.cm<\/p>\n b)\u00a0225 sq.cm<\/p>\n c)\u00a0216 sq.cm<\/p>\n d)\u00a0224 sq.cm<\/p>\n e)\u00a0None of these<\/p>\n Question 37:\u00a0<\/b>12 men can complete a piece of work in 48 hours. In how many hours will 36 men complete the same piece of work ?<\/p>\n a)\u00a018<\/p>\n b)\u00a016<\/p>\n c)\u00a032<\/p>\n d)\u00a020<\/p>\n e)\u00a024<\/p>\n Question 38:\u00a0<\/b>If the numerator of a fraction is increased by 125% and the denominator of the fraction is increased by 120%, the resultant fraction is 9\/11 . What is the original fraction?<\/p>\n a)\u00a03\/5<\/p>\n b)\u00a04\/5<\/p>\n c)\u00a04\/7<\/p>\n d)\u00a07\/4<\/p>\n e)\u00a0None of these<\/p>\n Question 39:\u00a0<\/b>Krishna purchased an item for Rs. 16,200 and sold it at the gain of 16%. From that amount he purchased another item and sold it at the loss of 20%. What is his overall gain\/ loss ?<\/p>\n a)\u00a01166.40 loss<\/p>\n b)\u00a01166.40 profit<\/p>\n c)\u00a01266.40 profit<\/p>\n d)\u00a01266.40 loss<\/p>\n e)\u00a0None of these<\/p>\n Question 40:\u00a0<\/b>Q1: \u2019X\u2019 :The total distance travelled by kiran from point \u2018A\u2019 to point \u2018B\u2019 increasing his speed at 5 m\/s and decreasing his speed at 10 m\/s alternatively for every 5 sec till his speed becomes zero for the first time is X. Starting speed of kiran is 20 m\/s. a)\u00a0Q1 $\\geq$ Q2<\/p>\n b)\u00a0Q1 $\\leq$ Q2<\/p>\n c)\u00a0Q1 < Q2<\/p>\n d)\u00a0Q1 > Q2<\/p>\n e)\u00a0Q1= Q2<\/p>\n Instructions<\/b><\/p>\n Calculate the quantity I and the quantity II on the basis of the given information then compare them and answer the following questions accordingly.<\/p>\n Question 41:\u00a0<\/b>Quantity I: The average cost of 1 pen and 1 pencil is 3.5, 1 pen and 1 sharpener is 4.5 and 1 pencil and 1 sharpener is 3. What is the average cost of 1 pen, 1 pencil and 1 sharpener? a)\u00a0Quantity I > Quantity II<\/p>\n b)\u00a0Quantity I < Quantity II<\/p>\n c)\u00a0Quantity I $\\leq$ Quantity II<\/p>\n d)\u00a0Quantity I $\\geq$ Quantity II<\/p>\n e)\u00a0Quantity I = Quantity II or The relationship between Quantity I and Quantity II can\u2019t be determined<\/p>\n Instructions<\/b><\/p>\n Calculate the quantity I and the quantity II on the basis of the given information then compare them and answer the following questions accordingly.<\/p>\n Question 42:\u00a0<\/b>Quantity I: If three successive discounts of 20%,40% and 50% are given, then what is the effective discount percentage. a)\u00a0Quantity I > Quantity II<\/p>\n b)\u00a0Quantity I < Quantity II<\/p>\n c)\u00a0Quantity I $\\leq$ Quantity II<\/p>\n d)\u00a0Quantity I $\\geq$ Quantity II<\/p>\n e)\u00a0Quantity I = Quantity II or The relationship between Quantity I and Quantity II can\u2019t be determined<\/p>\n Question 43:\u00a0<\/b>Quantity I: The upstream speed of a boat is 12 km\/hr. The speed of stream and speed of boat downstream are in the ratio 1:3. Then, find the downstream speed of boat. a)\u00a0Quantity I > Quantity II<\/p>\n b)\u00a0Quantity I < Quantity II<\/p>\n c)\u00a0Quantity I $\\leq$ Quantity II<\/p>\n d)\u00a0Quantity I $\\geq$ Quantity II<\/p>\n e)\u00a0Quantity I = Quantity II or The relationship between Quantity I and Quantity II can\u2019t be determined<\/p>\n Question 44:\u00a0<\/b>Quantity I: Two persons A and B can do a piece of work in 8 days and 6 days respectively. With the help of C, they completed the work in 3 days. Then in how many days, C can complete the work alone a)\u00a0Quantity I > Quantity II<\/p>\n b)\u00a0Quantity I < Quantity II<\/p>\n c)\u00a0Quantity I $\\leq$ Quantity II<\/p>\n d)\u00a0Quantity I $\\geq$ Quantity II<\/p>\n e)\u00a0Quantity I = Quantity II or The relationship between Quantity I and Quantity II can\u2019t be determined<\/p>\n Question 45:\u00a0<\/b>Quantity I: If length of a rectangle is increased by 10% and breadth is increased by 20%, then change in area of the rectangle is? a)\u00a0Quantity I > Quantity II<\/p>\n b)\u00a0Quantity I < Quantity II<\/p>\n c)\u00a0Quantity I $\\leq$ Quantity II<\/p>\n d)\u00a0Quantity I $\\geq$ Quantity II<\/p>\n e)\u00a0Quantity I = Quantity II or The relationship between Quantity I and Quantity II can\u2019t be determined<\/p>\n Question 46:\u00a0<\/b>The average of 4 consecutive even numbers is 15. The smallest number is \u2018a\u2019. Quantity II: Find $\\large\\frac{a^{2}-1}{a-1}$<\/p>\n a)\u00a0Quantity I > Quantity II<\/p>\n b)\u00a0Quantity I < Quantity II<\/p>\n c)\u00a0Quantity I $\\leq$ Quantity II<\/p>\n d)\u00a0Quantity I $\\geq$ Quantity II<\/p>\n e)\u00a0Quantity I = Quantity II or The relationship between Quantity I and Quantity II can\u2019t be determined<\/p>\n Instructions<\/b><\/p>\n In the following question, Quantity 1 and Quantity 2 are given. Calculate the values and compare them and then choose the option accordingly<\/p>\n Question 47:\u00a0<\/b>Quantity 1:The length of a rope that connects the top of a lighthouse of height 40 m to a stake located 30 m away from the base of the lighthouse. a)\u00a0Quantity 1 $>$ Quantity 2<\/p>\n b)\u00a0Quantity 1 $<$ Quantity 2<\/p>\n c)\u00a0Quantity 1 $\\geq$ Quantity 2<\/p>\n d)\u00a0Quantity 1 $\\leq$ Quantity 2<\/p>\n e)\u00a0Quantity 1 $=$ Quantity 2 or no relation can be established<\/p>\n Question 48:\u00a0<\/b>Quantity 1:The speed of a train of length 300 m that crosses a pole in 40 seconds. a)\u00a0Quantity 1 $>$ Quantity 2<\/p>\n b)\u00a0Quantity 1 $<$ Quantity 2<\/p>\n c)\u00a0Quantity 1 $\\geq$ Quantity 2<\/p>\n d)\u00a0Quantity 1 $\\leq$ Quantity 2<\/p>\n e)\u00a0Quantity 1 $=$ Quantity 2 or no relation can be established<\/p>\n Question 49:\u00a0<\/b>Quantity 1: Interest earned on Rs.1000 invested at 10% p.a compound interest for 4 years (compounded annually) a)\u00a0Quantity 1 $>$ Quantity 2<\/p>\n b)\u00a0Quantity 1 $<$ Quantity 2<\/p>\n c)\u00a0Quantity 1 $\\geq$ Quantity 2<\/p>\n d)\u00a0Quantity 1 $\\leq$ Quantity 2<\/p>\n e)\u00a0Quantity 1 $=$ Quantity 2 or no relation can be established<\/p>\n Question 50:\u00a0<\/b>Quantity 1:Volume of a cone with base radius 3 cm and height 12 cm a)\u00a0Quantity 1 $>$ Quantity 2<\/p>\n b)\u00a0Quantity 1 $<$ Quantity 2<\/p>\n c)\u00a0Quantity 1 $\\geq$ Quantity 2<\/p>\n d)\u00a0Quantity 1 $\\leq$ Quantity 2<\/p>\n e)\u00a0Quantity 1 $=$ Quantity 2 or no relation can be established<\/p>\n Take CMAT Mock Tests<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Let the tens digit of the number be 3x. Then, the number = $100\\times2x+10\\times 3x+x = 200x+30x+x = 231x$<\/p>\n If the tens and hundreds digit are interchanged, then the number formed = $100\\times3x+10\\times2x+x = 300x+20x+x = 321x$<\/p>\n Given, 321x = 231x+180 Hence, The original number = 231x = 462. 2)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Firstly arrange the given no. Is asending order<\/p>\n 6, 2, 3, 5,9, 4, 8, 7<\/p>\n 2,3,4,5,6,7,8,9<\/p>\n Now find middle term<\/p>\n That is 5,6<\/p>\n Now take avg of it<\/p>\n $\\frac{5+6}{2}$<\/p>\n =5.5<\/p>\n 3)\u00a0Answer\u00a0(B)<\/strong><\/p>\n =$\\frac{72\\div 8 \\times 3 + 4 \\div 2 – 4 \\times 5}{36 \\div 12 \\times 4 \\div 2 + 5 \\times 2}$<\/p>\n = $\\frac{9 \\times 3 + 2 – 20}{3 \\times\u00a0 2 + 5 \\times 2}$<\/p>\n = $\\frac{27+ 2 – 20}{6+ 10}$<\/p>\n =$\\frac{9}{16}$<\/p>\n 4)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Firstly solve bracket and then apply bodmass rule<\/p>\n $\\frac{\\frac{3}{4} \\div \\frac{9}{32} + \\frac{4}{3} \\times \\frac{2}{3}\u00a0 of\u00a0 \\frac{27}{16}}{\\frac{1}{2} \\times \\left(\\frac{8}{3} – 2\\right) \\div \\frac{4}{9} + \\left(\\frac{1}{3} + \\frac{1}{6}\\right)}$<\/p>\n $\\frac{\\frac{3}{4} \\div \\frac{9}{32} + \\frac{4}{3} \\times \\frac{9}{8}}{\\frac{1}{2} \\times \\frac {2}{3} \\div \\frac{4}{9} + \\frac{1}{2}}$<\/p>\n $\\frac{\\frac{8}{3} + \\frac{3}{2}}{\\frac{3}{4} + \\frac{1}{2}}$<\/p>\n $\\frac{\\frac{25}{6}}{\\frac{5}{4}}$<\/p>\n $\\frac{10}{3}$<\/p>\n 5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n $7 \\div 2 – [3\u00a0 of\u00a0 7 \\div 4 \\div \\left\\{(2 \\div 5) \\times (25 \\div 8) \\div (5 \\div 2)\\right\\}]$<\/p>\n $7 \\div 2 – [3\u00a0 of\u00a0 7 \\div 4 \\div \\left\\{\\frac{2}{5} \\times \\frac{25}{8} \\times \\frac{2}{5}\\right\\}]$<\/p>\n $\\frac{7}{2} – [3\\times\\frac{7}{4}\\times\\frac{2}{1}]$<\/p>\n $\\frac{7}{2} – \\frac{21}{2}$<\/p>\n $\\frac{-14}{2}$ = $-7$<\/p>\n 6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Length = 5 m , Breadth = 3 m and Height = 8 m<\/p>\n To find the cost of painting the walls of the room, we should know the cost of painting and the area of door and window.<\/p>\n Area of 4 walls (Curved Surface Area of cuboid) = $2h(l+b)$<\/p>\n = $(2 \\times 8)(5+3) = 16 \\times 8 = 128$ $m^2$<\/p>\n From I & II : Cost = Rs 25 per square meter<\/p>\n Area of window = $2.25$ $m^2$<\/p>\n and Area of door = $2 \\times 2.25 = 4.5$ $m^2$<\/p>\n Thus, area to be painted = $128 – 2.25 – 4.5 = 121.25$ $m^2$<\/p>\n $\\therefore$ Cost of painting 4 walls = $25 \\times 121.25 = Rs$ $3031.25$<\/p>\n Thus, statement III is redundant.<\/p>\n => Ans – (C)<\/p>\n 7)\u00a0Answer\u00a0(E)<\/strong><\/p>\n To find the time taken by train X to cross train Y, we should know the sum of speed of the trains\u00a0(as they are moving in opposite directions) and the sum of lengths of the train<\/p>\n In statement II, ratio of speed of train\u00a0X : Y = 3\u00a0: 2<\/p>\n Thus, we cannot find the speeds of the two trains.<\/p>\n $\\therefore$ Data is inadequate<\/p>\n => Ans – (E)<\/p>\n 8)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Ratio of investment of\u00a0X : Y : Z = 5\u00a0: 4\u00a0: 2<\/p>\n To find the share of profit earned by X (or anyone), we need to find the total profit earned.<\/p>\n From statement III : Average profit for 3 years = Rs. 2750<\/p>\n => Total profit = $2750 \\times 3$ = Rs. 8250<\/p>\n Amount received by X = $\\frac{5}{(5+4+2)} \\times 8250$ = Rs. 3750<\/p>\n Thus, statements I and II are redundant.<\/p>\n => Ans – (D)<\/p>\n 9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n To find the area of right angled triangle, we need to know the base and height or the length of any side if its a\u00a030-60-90 triangle<\/p>\n From statement I and III, two angles are 90\u00b0 and 60\u00b0, => Third angle = 30\u00b0<\/p>\n In a 30-60-90 triangle, the hypotenuse is always twice as long as the side opposite the 30\u00b0 angle and the side opposite the 60\u00b0 angle is\u00a0\u221a3 times as long as the side opposite the 30\u00b0 angle.<\/p>\n The ratio of sides opposite 30\u00b0, 60\u00b0 and 90\u00b0 angles =1\u00a0: $\\sqrt{3}$ : 2<\/p>\n Length\u00a0of the side opposite the 90\u00b0 angle (hypotenuse) = 5 cm<\/p>\n => Length of side opposite the 30\u00b0 angle = $\\frac{5}{2}=2.5$ cm<\/p>\n and length of side opposite to 60\u00b0 angle = $2.5 \\sqrt{3}$ cm<\/p>\n $\\therefore$ Area of triangle = $\\frac{1}{2} \\times 2.5 \\times 2.5 \\sqrt{3} = 3.125\\sqrt{3}$ $cm^2$<\/p>\n Thus, statement II is redundant.<\/p>\n => Ans – (B)<\/p>\n 10)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Let the capacity of the tank be 6 litres. 11)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let the speed of boat in still water and speed of river be B km\/hr and R km\/hr respectively .<\/p>\n Speed of boat in downstream = B+ R km\/hr<\/p>\n Speed of boat in upstream = B-R km\/hr = $\\frac{7.5}{1.5}$= 5km\/hr<\/p>\n Let the same distance travelled in upstream and downstream be D km<\/p>\n So ,<\/p>\n $\\frac{D}{B-R}$ =2\u00d7 $\\frac{D}{B+R}$<\/p>\n B+R = 10 km\/hr<\/p>\n 12)\u00a0Answer\u00a0(A)<\/strong><\/p>\n average speed = (total distance ) \/(total time)<\/p>\n Let her time taken to finish race with 18km\/hr average speed be y hours<\/p>\n And distance covered in both cases is same<\/p>\n So<\/p>\n D = 18 \u00d7 y…..(1)<\/p>\n D = 24 x (y-1)….(2)<\/p>\n From equation 1 and 2<\/p>\n y = 4 hours<\/p>\n And D = 72 km<\/p>\n Now if she need to reach at 2pm i.e in 2 hours then the average speed = 72\/2 = 36 km\/hr<\/p>\n 13)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let the speed of boat in still water and speed of river be B km\/hr and R km\/hr respectively<\/p>\n speed of boat in downstream = (B+R) km\/hr<\/p>\n speed of boat in upstream = (B-R) km\/hr<\/p>\n Distance covered = 16 km<\/p>\n it is given that tie taken ti move 16 km downstream is 2 hours and for upstream it is 4 hours<\/p>\n 16 = (B+R) 2 ………….(1)<\/p>\n 16 = (B-R) 4……………(2)<\/p>\n from equation 1 and 2<\/p>\n B = 6 km\/hr<\/p>\n 14)\u00a0Answer\u00a0(C)<\/strong><\/p>\n distance travelled by car which started from point A from 11 am to 1 pm is = 36 \u00d72 = 72 km<\/p>\n Now it is given that total distance is 592 km<\/p>\n Distance left to be covered = 592 – 72 = 520 km<\/p>\n Speed of car started from B = 44 km\/hr<\/p>\n Soeed of car started from A = 36km\/hr<\/p>\n Relative speed = 80 km\/hr<\/p>\n Time taken = 520\/80 = 6.5 hours<\/p>\n So the two cars will meet at 7.30 pm<\/p>\n 15)\u00a0Answer\u00a0(E)<\/strong><\/p>\n In order to arrange the fractions in descending order, we need to have the same denominator for all the fractions. Let us represent all the numbers given with the same denominator, ie 72072. $\\frac{2}{7} = \\frac{20592}{72072}$<\/p>\n $\\frac{3}{8} = \\frac{27027}{72072}$<\/p>\n $\\frac{6}{13} = \\frac{33264}{72072}$<\/p>\n $\\frac{5}{11} = \\frac{32760}{72072}$<\/p>\n When arranged in descending order, the order will be as below 16)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let the two digit number be XY. So, the value of the number equals 10X+Y From the information given, we know that X:Y = 2:3 and (10Y+X)-(10X+Y) = 9Y-9X So, 3X=2Y and Y-X =3. 17)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Let the two numbers be X and Y respectively. 18)\u00a0Answer\u00a0(C)<\/strong><\/p>\n In order to arrange the fractions in ascending order, all of them should have the same denominator. Representing all the fractions with denominator 12870, the fractions look as below.<\/p>\n $\\frac{4}{5} = \\frac{10296}{12870}$<\/p>\n $\\frac{9}{11} = \\frac{10530}{12870}$<\/p>\n $\\frac{7}{9} = \\frac{10010}{12870}$<\/p>\n $\\frac{5}{6} = \\frac{10725}{12870}$<\/p>\n $\\frac{11}{13} = \\frac{10890}{12870}$<\/p>\n Arranging them in ascending order, we get the following order So, the fourth number in order is $\\frac{10725}{12870}$\u00a0which is $\\frac{5}{6}$<\/p>\n 19)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Let the numbers be a&b.; We have, 3a\/7= 45b\/100.<\/p>\n so a\/7=15b\/100<\/p>\n a\/b=15*7\/100<\/p>\n Calculate for a\/b we get 21:20<\/p>\n 20)\u00a0Answer\u00a0(A)<\/strong><\/p>\n (A) :\u00a0$xy^{2}-x^{2}y+2x^{2}y^{2}$<\/p>\n = $xy(y-x+2xy)$<\/p>\n (B) :\u00a0$xy^{2}(1-2x)+x^{2}y(2y-1)$<\/p>\n = $(xy^2-2x^2y^2)+(2x^2y^2-x^2y)$<\/p>\n = $xy^2-x^y = xy(y-x)$<\/p>\n (C) :\u00a0$xy^{2}(1-2x)+x^{2}y(2y-1)$<\/p>\n = $(xy^2-2x^2y^2)+(2x^2y^2-x^2y)$<\/p>\n = $xy^2-x^y = xy(y-x)$<\/p>\n (D) :\u00a0$xy[y(1+x)-x(1+y)]$<\/p>\n = $xy(y+xy-x-xy) = xy(y-x)$<\/p>\n (E) :\u00a0$xy[(x+y)^{2}+y(1-y)-x(1+x)-2xy]$<\/p>\n = $xy[(x^2+y^2+2xy)+(y-y^2)+(-x-x^2)-2xy]$<\/p>\n = $xy(y-x)$<\/p>\n => Ans – (A)<\/p>\n 21)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Equation (i) : $2x+y=15$<\/p>\n Equation (ii) : $2y+z=25$<\/p>\n Equation (iii) : $2z+x=26$<\/p>\n Adding the three equations, => $3x+3y+3z=15+25+26$<\/p>\n => $3(x+y+z)=66$<\/p>\n => $x+y+z=\\frac{66}{3}=22$ ————(iv)<\/p>\n Subtracting equation (i) from (iv), => $(z)+(x-2x)+(y-y)=(22-15)$<\/p>\n => $z-x=7$ ————-(v)<\/p>\n Adding equations (v) and (iii), => $3z=26+7=33$<\/p>\n => $z=\\frac{33}{3}=11$<\/p>\n => Ans – (E)<\/p>\n 22)\u00a0Answer\u00a0(E)<\/strong><\/p>\n I.$x^{2} + 3x – 28 = 0$<\/p>\n => $x^2 + 7x – 4x – 28 = 0$<\/p>\n => $x (x + 7) – 4 (x + 7) = 0$<\/p>\n => $(x – 4) (x + 7) = 0$<\/p>\n => $x = 4 , -7$<\/p>\n II.$y^{2} – y – 20 = 0$<\/p>\n => $y^2 – 5y + 4y – 20 = 0$<\/p>\n => $y (y – 5) + 4 (y – 5) = 0$<\/p>\n => $(y + 4) (y – 5) = 0$<\/p>\n => $y = -4 , 5$<\/p>\n $\\therefore$ No relation can be established.<\/p>\n 23)\u00a0Answer\u00a0(C)<\/strong><\/p>\n I.$6x^{2} + 29x + 35 = 0$<\/p>\n => $6x^2 + 15x + 14x + 35 = 0$<\/p>\n => $3x (2x + 5) + 7 (2x + 5) = 0$<\/p>\n => $(2x + 5) (3x + 7) = 0$<\/p>\n => $x = \\frac{-7}{3} , \\frac{-5}{2}$<\/p>\n II.$3y^{2} + 11y + 10 = 0$<\/p>\n => $3y^2 + 6y + 5y + 10 = 0$<\/p>\n => $3y (y + 2) + 5 (y + 2) = 0$<\/p>\n => $(y + 2) (3y + 5) = 0$<\/p>\n => $y = -2 , \\frac{-5}{3}$<\/p>\n Hence $x < y$<\/p>\n 24)\u00a0Answer\u00a0(E)<\/strong><\/p>\n I.$2x^{2} + 18x + 40 = 0$<\/p>\n => $2x^2 + 8x + 10x + 40 = 0$<\/p>\n => $2x (x + 4) + 10 (x + 4) = 0$<\/p>\n => $(x + 4) (2x + 10) = 0$<\/p>\n => $x = -4 , -5$<\/p>\n II.$2y^{2} + 15y + 27 = 0$<\/p>\n => $2y^2 + 6y + 9y + 27 = 0$<\/p>\n => $2y (y + 3) + 9 (y + 3) = 0$<\/p>\n => $(y + 3) (2y + 9) = 0$<\/p>\n => $y = -3 , \\frac{-9}{2}$<\/p>\n $\\therefore$ No relation can be established.<\/p>\n 25)\u00a0Answer\u00a0(C)<\/strong><\/p>\n The total amount distributed among officers = 45*25000= 11,25,000<\/p>\n If the total amount is x, then this amount is (5\/8)x=11,25,000<\/p>\n Hence, x=18,00,000<\/p>\n 26)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Marked price = 78,350 + 30% of 78,350 = Rs 1,01,855 27)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Loss percent 28)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Gain% = (1000-960)\/960 x 100 = 25\/6=4 1\/6%<\/p>\n 29)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let the marked price be Rs. 100. 30)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Total number of letters = 6 of which 3 are the same.<\/p>\n Therefore, number of ways of arranging = $\\frac{6!}{3!}$ = $\\frac{720}{6}$ = 120<\/p>\n 31)\u00a0Answer\u00a0(C)<\/strong><\/p>\n None yellow => both blue => is possible in $^7C_2$ ways<\/p>\n => probability = $\\frac{^7C_2}{^{12}C_2}$ = $\\frac{7}{22}$<\/p>\n 32)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Number of ways in which the sum is \u00a07 = (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) => 6 ways<\/p>\n Total number of possibilities = 6 * 6 = 36<\/p>\n Probability that sum is 7 = $\\frac{6}{36}$ = $\\frac{1}{6}$<\/p>\n 33)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Number of ways of arranging seven letters = 7! 34)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Probability of getting two white balls = 13<\/sup>P2<\/sub> 35)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let the principal = Rs. $x$<\/p>\n S.I. = $\\frac{Principal \\times Rate \\times Time}{100}$<\/p>\n => $1280 = \\frac{x \\times 8 \\times 2}{100}$<\/p>\n => $x = 80 \\times 100 = 8,000$<\/p>\n Simple interest accrued on thrice of that principal<\/p>\n = $\\frac{24,000 \\times 10.75 \\times 2}{100}$<\/p>\n = Rs. $5,160$<\/p>\n 36)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Perimeter of rectangle = 60 cm<\/p>\n => $2 (l + b) = 60$<\/p>\n => $l + 12 = \\frac{60}{2} = 30$<\/p>\n => $l = 30 – 12 = 18$ cm<\/p>\n $\\therefore$ Area of rectangle = $(l \\times b)$<\/p>\n = $18 \\times 12 = 216 cm^2$<\/p>\n 37)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Using, $M_1 D_1 = M_2 D_2$<\/p>\n => $12 \\times 48 = 36 \\times D_2$<\/p>\n => $D_2 = \\frac{48}{3} = 16$ hours<\/p>\n 38)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let the original fraction = $\\frac{x}{y}$<\/p>\n $\\therefore \\frac{x \\times 225}{y \\times 220} = \\frac{9}{11}$<\/p>\n => $\\frac{x \\times 45}{y \\times 44} = \\frac{9}{11}$<\/p>\n => $\\frac{x}{y} = \\frac{9}{11} \\times \\frac{44}{45}$<\/p>\n => $\\frac{x}{y} = \\frac{4}{5}$<\/p>\n 39)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Case 1\u00a0: C.P. of 1st item = Rs. 16,200<\/p>\n S.P. of 1st item = $\\frac{116}{100} \\times 16,200$<\/p>\n = Rs. 18,792<\/p>\n Case 2\u00a0: C.P. of 2nd item = Rs. 18,792<\/p>\n S.P. of 2nd item = $\\frac{80}{100} \\times 18,792$<\/p>\n = Rs. 15,033.60<\/p>\n $\\therefore$ Overall loss = C.P. – S.P.<\/p>\n = Rs. (16,200 – 15,033.60) = Rs. 1,166.40<\/p>\n 40)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Total distance is calculated by adding the distances travelled for every 5 seconds 41)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Quantity I: 42)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Quantity I: Hence, Quantity I = Quantity II<\/p>\n 43)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Quantity I: Given that the upstream speed of the boat = 12 km\/hr Then, speed of boat in still water = 3x-x = 2x km\/hr Hence, Downstream speed of boat = 3x = 3*12 = 36 km\/hr<\/p>\n Quantity II: 24 km\/hr 44)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Quantity I: Let the total work be 48 units (LCM of 8,6,3) Then, Efficiency of C = 16-(6+8) = 16-14 = 2 units\/day<\/p>\n Then, C can do the work alone in 48\/2 = 24 days 45)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Quantity I: 46)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Given $\\large\\frac{a+a+2+a+4+a+6}{4}$ $= 15$<\/p>\n => $\\large\\frac{4a+12}{4}$ $= 15$<\/p>\n => $4a+12 = 60$ Then, Quantity II: $\\large\\frac{a^{2}-1}{a-1} = \\frac{12^2-1}{12-1} = \\frac{144-1}{12-1} = \\frac{143}{11}$ $= 13$<\/p>\n Hence, Quantity I < Quantity II<\/p>\n 47)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Quantity 1:<\/p>\n The length of a rope that connects the top of a lighthouse of height 40 m to a stake located 30 m away from the base of the lighthouse.<\/p>\n The length of the rope will be minimum when it does not slack at any point. Applying Pythagoras theorem, we get, Therefore, the length of the rope should be greater than or equal to 50 m.<\/p>\n Quantity 2:<\/p>\n When the stick is rotated by its centre, a circle of area $625*\\pi$ is swept. As we can see, quantity 1 is greater than or equal to quantity 2. Therefore, option C is the right answer.<\/p>\n 48)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Quantity 1: Quantity 2: As we can see, quantity 1 is greater than quantity 2 and hence, option A is the right answer.<\/p>\n 49)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Since the principal is the same in both the cases, we will compare the amounts by the end of the given periods. Quantity 2: As we can see, quantity 1 is greater than quantity 2 and hence, option A is the right answer.<\/p>\n 50)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Volume of a cylinder = $\\pi*r^2*h$ Since we have to compare the values, we can ignore the $\\pi$ part and compare the constant value.<\/p>\n Volume of the cone = $\\pi*\\frac{1}{3}*3^2*12$ Volume of the cylinder = $\\pi*r^2*h$ As we can see, the volume of the cylinder is greater than the volume of the cone. Therefore, option E is the right answer.<\/p>\n
\n6, 2, 3, 5,9, 4, 8, 7<\/p>\n
\n$7 \\div 2 – [3\u00a0 of\u00a0 7 \\div 4 \\div \\left\\{(2 \\div 5) \\times (25 \\div 8) \\div (5 \\div 2)\\right\\}]$?<\/p>\n
\nI.Cost of painting per square meter is Rs 25
\nII.Area of window is 2.25 sq m which is half of the area of the door
\nIII.Area of the room is 15 sq meter<\/p>\n
\nI.Train X crosses a signal pole in 6 secs
\nII.Ratio of the speeds of trains X and Y is 3:2
\nIII.Length of the two trains together is 500 meters<\/p>\n
\nI.Total amount invested in the business is Rs.22,000
\nII.Profit was distributed after a period of 2 years
\nIII.The average amount of profit earned per year is Rs 2,750<\/p>\n
\nI. Length of the hypotenuse is 5 cm
\nII.perimeter of the triangle is four times its base
\nIII. one of the angles of the triangle is 60 degrees<\/p>\n
\nII.\u00a0$y^{2} -y-20=0$<\/p>\n
\nII.\u00a0$3y^{2} +11y+10=0$<\/p>\n
\nII.\u00a0$2y^{2} +15y+27=0$<\/p>\n
\nQ2: 575 meters<\/p>\n
\nQuantity II: 11\/4<\/p>\n
\nQuantity II: 76%<\/p>\n
\nQuantity II: 24 km\/hr<\/p>\n
\nQuantity II: 28 days<\/p>\n
\nQuantity II: 28%<\/p>\n
\nQuantity I: Find $\\large\\frac{a^{2}}{2a+1}$<\/p>\n
\nQuantity 2: The length of a stick which on rotating about its centre sweeps an area of $625\\pi m^2$.<\/p>\n
\nQuantity 2:The speed of a train of length 150 m that crosses a platform length 50 m in half a minute.<\/p>\n
\nQuantity 2:Interest earned on Rs.1000 invested at 20% p.a. compound interest for 2 years (compounded annually)<\/p>\n
\nQuantity 2:Volume of a cylinder with base radius 3 cm and height 4 cm<\/p>\n
\nUnits digit = (100-66.67)% of 3x = 33.33% of 3x = x.
\nHundreds digit = 2x.<\/p>\n
\n90x = 180
\nx = 2.<\/p>\n
\nNumber obtained by interchanging units and hundreds digit = 264.<\/p>\n
\nA’s efficiency = 3 litre\/hr
\nB’s efficiency = 2 litre\/hr
\nWith leakage A’s efficiency = 2 litre\/hr
\n$\\Rightarrow$ Leakage’s efficiency = 1 litre\/hr
\nWith leakage B will take = $\\frac{6}{2 – 1}$ = 6 hours<\/p>\n
\nSo, we need to find the LCM of all the denominators. The denominators are 9,7,8,13 and 11.
\nTheir LCM is 72072.<\/p>\n
\n$\\frac{4}{9} = \\frac{32032}{72072}$<\/p>\n
\n$\\frac{33264}{72072}$,\u00a0$\\frac{32760}{72072}$,$\\frac{32032}{72072}$,$\\frac{27027}{72072}$,$\\frac{20592}{72072}$
\nHence, the second number when the above fractions are arranged in descending order is $\\frac{32760}{72072}=\\frac{5}{11}$<\/p>\n
\nWhen the digits are interchanged, the number becomes YX. So, the value of the number equals 10Y+X<\/p>\n
\nHence, 9Y-9X = 27 or Y-X = 3<\/p>\n
\nSolving both the equations, we get X=6 and Y=9.
\nSo, the original number is 69<\/p>\n
\nSo, $75 \\% X = \\frac{3}{7}Y$
\nOr, $\\frac{3}{4}*X = \\frac{3}{7}*Y$
\nSo, the ratio of X:Y = 4:7<\/p>\n
\nFor this, we have to first calculate the LCM of the denominators.
\nThe LCM of 5,11,9,6 and 13 is 2574<\/p>\n
\n$\\frac{10010}{12870}$,\u00a0$\\frac{10296}{12870}$,\u00a0$\\frac{10530}{12870}$,\u00a0$\\frac{10725}{12870}$,\u00a0$\\frac{10890}{12870}$<\/p>\n
\nPrice after discount = 101855 -20% of 101855 = Rs 81,484
\nProfit = Rs 3,134<\/p>\n
\n= [$\\frac{loss}{10}$]2<\/sup> %
\n= [$\\frac{12}{10}$]2\u00a0<\/sup>%
\n= $\\frac{36}{25}$ %
\n= 1$\\frac{11}{25}$ %<\/p>\n
\nThen cost price = 100 – 40 = Rs. 60
\nSelling price = 100+ 50 = Rs. 150
\nProfit = 150 – 60 = 90
\nProfit%=90\/60\u00d7100=150%<\/p>\n
\nLet us consider the two vowels as a group
\nNow the remaining five letters and the group of two vowels = 6
\nThese six letters can be arranged in 6!2! ways( 2! is the number of ways the two vowels can be arranged among themselves)
\nThe number of ways of arranging seven letters such that no two vowels come together
\n=\u00a0Number of ways of arranging seven letters – Number of ways of arranging the letters with the two vowels being together
\n= 7! – (6!2!)
\n= 3600<\/p>\n
\nProbability of getting two black balls = 7<\/sup>P2<\/sub>
\nTotal probability = (13<\/sup>P2<\/sub>\u00a0+\u00a07<\/sup>P2<\/sub>)\/20<\/sup>P2<\/sub>\u00a0= 198\/380 = 99\/190<\/p>\n
\nSo total distance=(20*5)+(25*5)+(15*5)+(20*5)+(10*5)+(15*5)+(5*5)+(10*5)
\n=600<\/p>\n
\nLet the cost of pen, pencil and sharpener be p, c and s.
\nHence, p + c = 3.5*2 = 7
\np + s = 4.5*2 = 9
\nc + s = 3*2 = 6
\nAdding all three together
\n2p + 2c +2s = 7+9+6
\n2p+2c+2s = 22
\np+c+s = 11
\nAverage cost = (p+c+s)\/3 = 11\/3
\nQuantity II: 11\/4
\nHence, Quantity I > Quantity II<\/p>\n
\nLet the marked price be Rs.100
\nThen after three given successive discounts, Selling Price will be $100\\times\\large\\frac{80}{100}\\times\\frac{60}{100}\\times\\frac{50}{100}$ $= Rs.24$
\nHence, Effective discount percentage $= \\large\\frac{100-24}{100}\\times100$ $= 76$%
\nQuantity II: 76%<\/p>\n
\nRatio of speed of stream and downstream speed of boat = 1:3
\nLet speed of stream be \u2018x\u2019 km\/hr and speed of boat downstream be \u20183x\u2019 km\/hr<\/p>\n
\nUpstream speed of boat = 2x-x = x km\/hr = 12 km\/hr<\/p>\n
\nHence, Quantity I > Quantity II<\/p>\n
\nEfficiency of A = 48\/8 = 6 units\/day
\nEfficiency of B = 48\/6 = 8 units\/day
\nEfficiency of A,B and C together = 48\/3 = 16 units\/day<\/p>\n
\nQuantity II: 28 days
\nHence, Quantity I < Quantity II<\/p>\n
\nGiven that length of rectangle increased by 10% and breadth of rectangle increased by 20%.
\nPercentage change in area = $10+20+\\frac{10\\times20}{100} = 30+2 = 32$
\nQuantity II: 28%
\nHence, Quantity I > Quantity II<\/p>\n
\n=> $4a = 48$
\n=> $a = 12$<\/p>\n
\nQuantity I: $\\large\\frac{a^{2}}{2a+1} = \\frac{12^2}{2\\times12+1} = \\frac{144}{25}$ $= 5.76$<\/p>\n
\nAssuming the rope to be a straight line of length $l$, we get a right-angled triangle with sides 40 m (height of the lighthouse), 30 m (distance between the lighthouse and the stake) and the length of the rope.<\/p>\n
\n$l^2 = \\sqrt{30^2+40^2}$
\n=> $l=50$m<\/p>\n
\nLength of the stick = Diameter of the circle
\n$\\pi*r^2 = 625*\\pi$
\n=> $r=25$m
\nTherefore, the length of the stick is equal to $2*25=50$m<\/p>\n
\nThe speed of a train of length 300 m that crosses a pole in 40 seconds = 300\/40 = 7.5 m\/s<\/p>\n
\nThe speed of a train of length 150 m that crosses a platform length 50 m in half a minute = (150+50)\/30 = 200\/30 = 6.67 m\/s<\/p>\n
\nQuantity 1:
\nRs.1000 invested at 10% p.a compound interest for 4 years will amount to $1000*(1.1)^4$ = $1.4641*1000$<\/p>\n
\nRs.1000 invested at 20% p.a. compound interest for 2 years will amount to $1000*(1.2)^2$ = $1.44*1000$<\/p>\n
\nVolume of a cone = $\\frac{1}{3}*\\pi*r^2*h$<\/p>\n
\n= $36*\\pi$<\/p>\n
\n= $\\pi*3^2*4$
\n=$36*\\pi$<\/p>\n