Download CMAT 2022 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 <\/p>\n Take CMAT mock tests here<\/a><\/p>\n Download CMAT previous papers PDF<\/a><\/p>\n Question 1:\u00a0<\/b>The product of ages of Harris and Sharma is 240 .If twice the age of Sharma is more than Harris age by 4 years ,what is Sharma age in years?<\/p>\n a)\u00a012<\/p>\n b)\u00a020<\/p>\n c)\u00a010<\/p>\n d)\u00a014<\/p>\n e)\u00a0Data adequate<\/p>\n Question 2:\u00a0<\/b>If the ages of P and R added to twice the age of Q ,the total becomes 59 .If the age of Q and R are added to thrice age of P,the total becomes 68 and if the age of the P is added to thrice age of Q and thrice the age age of R the total becomes 108 .What is the age of P?<\/p>\n a)\u00a015 years<\/p>\n b)\u00a019 years<\/p>\n c)\u00a017 years<\/p>\n d)\u00a012 years<\/p>\n e)\u00a0None of these<\/p>\n Question 3:\u00a0<\/b>Jayesh is twice as old as Vijay and half as old as Suresh .If sum of Suresh\u2019s and Vijay\u2019s age is 85 years what is the Jayesh\u2019s age in years?<\/p>\n a)\u00a034<\/p>\n b)\u00a036<\/p>\n c)\u00a068<\/p>\n d)\u00a0Cannot determined<\/p>\n e)\u00a0None of these<\/p>\n Question 4:\u00a0<\/b>4 years ago, the respective ratio between ${1 \\over 2}$ of A\u2019s age at that time and four times of B\u2019s age at that time was 5: 12. Eight years hence ${1 \\over 2}$ of A\u2019s age at that time will be less than B\u2019s age at that time by 2 years. What is B\u2019s present age ?<\/p>\n a)\u00a010 years<\/p>\n b)\u00a014 years<\/p>\n c)\u00a012 years<\/p>\n d)\u00a05 years<\/p>\n e)\u00a08 years<\/p>\n Question 5:\u00a0<\/b>The sum of the present ages of P and Q is 25 years more than the age of R. The present age of Q is 5 years more than that of R. Find the present age\u00a0 of P<\/p>\n a)\u00a020 years<\/p>\n b)\u00a025 years<\/p>\n c)\u00a021 years<\/p>\n d)\u00a022 years<\/p>\n e)\u00a0None of these<\/p>\n Get 5 CMAT mocks at just Rs.299<\/a><\/p>\n Download CMAT Previous Papers PDF<\/a><\/p>\n Question 6:\u00a0<\/b>7.69% of 130 + 6.66% of 150 = ?<\/p>\n a)\u00a020<\/p>\n b)\u00a021<\/p>\n c)\u00a022<\/p>\n d)\u00a023<\/p>\n e)\u00a024<\/p>\n Question 7:\u00a0<\/b>$ \\frac{1}{7}$ of $2\\frac{5}{8} \\div \\frac{3}{4} + \\frac{1}{3} + \\frac{1}{6}$<\/p>\n a)\u00a01<\/p>\n b)\u00a00<\/p>\n c)\u00a0$\\frac{1}{7}$<\/p>\n d)\u00a02<\/p>\n e)\u00a0None of the above<\/p>\n Question 8:\u00a0<\/b>$ 125^{4} \\div 25^{4} * 5 ^{5} = 5 ^{?} $<\/p>\n a)\u00a07<\/p>\n b)\u00a09<\/p>\n c)\u00a011<\/p>\n d)\u00a013<\/p>\n e)\u00a0none of the above<\/p>\n Question 9:\u00a0<\/b>What is the LCM of 1\/3, \u00bc and 1\/5?<\/p>\n a)\u00a01<\/p>\n b)\u00a02<\/p>\n c)\u00a0\u00bd<\/p>\n d)\u00a01\/6<\/p>\n e)\u00a0None of the above<\/p>\n Question 10:\u00a0<\/b>134% of 3894 + 38.94 of 134 = ?<\/p>\n a)\u00a011452<\/p>\n b)\u00a010000<\/p>\n c)\u00a010452<\/p>\n d)\u00a01100<\/p>\n e)\u00a0None of these<\/p>\n Question 11:\u00a0<\/b>In order to pass in an exam a student is required to get 780 marks out of the aggregate marks. Sonu got 728 marks and was declared failed by 5 per cent. What are the maximum aggregate marks a student can get in the examination.<\/p>\n a)\u00a01040<\/p>\n b)\u00a01100<\/p>\n c)\u00a01000<\/p>\n d)\u00a0Cannot be determined<\/p>\n e)\u00a0None of these<\/p>\n Question 12:\u00a0<\/b>The average of four consecutive numbers A, B, C and D respectively is 49.5. What is the product of B and D?<\/p>\n a)\u00a02499<\/p>\n b)\u00a02352<\/p>\n c)\u00a02450<\/p>\n d)\u00a02550<\/p>\n e)\u00a0None of these<\/p>\n Question 13:\u00a0<\/b>Nandita scored 80 per cent marks in five subjects together viz, Hindi, Science, Maths, English and Sanskrit together wherein the maximum marks of each subject were 105. How many marks did Nandita score in Science if she scored 89 marks in Hindi, 92 marks in Sanskrit, 98 marks in Maths and 81 marks in English?<\/p>\n a)\u00a060<\/p>\n b)\u00a075<\/p>\n c)\u00a065<\/p>\n d)\u00a070<\/p>\n e)\u00a0None of these<\/p>\n Question 14:\u00a0<\/b>Find the average of the following sets of sources. a)\u00a0427<\/p>\n b)\u00a0413<\/p>\n c)\u00a0441<\/p>\n d)\u00a0490<\/p>\n e)\u00a0None of these<\/p>\n Question 15:\u00a0<\/b>Ramola\u2019s monthly income is three times Ravina\u2019s monthly income, Ravina\u2019s monthly income is fifteen percent more that Ruchika\u2019s monthly income. Ruchika\u2019s monthly income is Rs. 32,000. What is Ramola\u2019s annual income?<\/p>\n a)\u00a0Rs. 1, 10, 400<\/p>\n b)\u00a0Rs. 13, 24, 800<\/p>\n c)\u00a0Rs. 36, 800<\/p>\n d)\u00a0Rs. 52, 200<\/p>\n e)\u00a0None of these<\/p>\n Question 16:\u00a0<\/b>Seven men, five women and eight children were given an assignment of distributing 2000 books to students in a school over a period of three days. All of them distributed books on the first day. On the second day two women and three children remained absent and on the third day three men and five children remained absent. If the ratio of the number of books distributed in a day by a man, a woman and a child was 5: 4: 2 respectively, a total of \u2018\u2019approximately\u2019\u2019 how many books were distributed on the second day ?<\/p>\n a)\u00a01000<\/p>\n b)\u00a0800<\/p>\n c)\u00a0650<\/p>\n d)\u00a0900<\/p>\n e)\u00a0Cannot be determined<\/p>\n Question 17:\u00a0<\/b>Are all the four friends viz. A, B, C and D who are sitting around a circular table, facing the centre? a)\u00a0If the data in Statement I and II are sufficient to answer the question, while the data in Statement III are not required to answer the question.<\/p>\n b)\u00a0If the data in Statement I and III are sufficient to answer the question, while the data in Statement II are not required to answer the question.<\/p>\n c)\u00a0If the data in Statement II and III are sufficient to answer the question, while the data in Statement I are not required to answer the question.<\/p>\n d)\u00a0If the data in either Statement I alone or Statement II alone or Statement III alone are sufficient to answer the question.<\/p>\n e)\u00a0If the data in all the Statements I, II and III together are necessary to answer the question.<\/p>\n Question 18:\u00a0<\/b>How is ‘one’ coded in the code language? a)\u00a0If the data in Statement I and II are sufficient to answer the question, while the data in Statement III are not required to answer the question.<\/p>\n b)\u00a0If the data in Statement I and III are sufficient to answer the question, while the data in Statement II are not required to answer the question.<\/p>\n c)\u00a0If the data in Statement II and III are sufficient to answer the question, while the data in Statement I are not required to answer the question.<\/p>\n d)\u00a0If the data in either Statement I alone or Statement II alone or Statement III alone are sufficient to answer the question.<\/p>\n e)\u00a0If the data in all the Statements I, II and III together are necessary to answer the question.<\/p>\n Question 19:\u00a0<\/b>The height of a triangle is equal to the perimeter of a square whose diagonal is 14.14m and the base of the same triangle is equal to the side of the square whose area is 784 m2. What is the area of the triangle? (in m2)<\/p>\n a)\u00a0504<\/p>\n b)\u00a0560<\/p>\n c)\u00a0478<\/p>\n d)\u00a0522<\/p>\n e)\u00a0496<\/p>\n Question 20:\u00a0<\/b>The diameter of a wheel is 49 m. How many revolutions will it make to cover a distance of 3200 m?<\/p>\n a)\u00a017<\/p>\n b)\u00a027<\/p>\n c)\u00a024<\/p>\n d)\u00a021<\/p>\n e)\u00a018<\/p>\n Question 21:\u00a0<\/b>The area of a rectangle is equal to the area of a circle with circumference equal to 220 m What is the length of the rectangle of its breadth is 50 m ?<\/p>\n a)\u00a056 m<\/p>\n b)\u00a083 m<\/p>\n c)\u00a077 m<\/p>\n d)\u00a069 m<\/p>\n e)\u00a0None of these<\/p>\n Question 22:\u00a0<\/b>The perimeter of a square is one-fourth the perimeter of a rectangle If the perimeter of the square is 44 cm and the length of the rectangle is 51 cm what is the difference between the breadth of the rectangle and the side of the square ?<\/p>\n a)\u00a030 cm<\/p>\n b)\u00a018 cm<\/p>\n c)\u00a026 cm<\/p>\n d)\u00a032 cm<\/p>\n e)\u00a0None of these<\/p>\n Question 23:\u00a0<\/b>What would be the area of a square whose diagonal measures 28 cm?<\/p>\n a)\u00a0288 $cm^{2}$<\/p>\n b)\u00a0514 $cm^{2}$<\/p>\n c)\u00a0428 $cm^{2}$<\/p>\n d)\u00a0392 $cm^{2}$<\/p>\n e)\u00a0None of these<\/p>\n Question 24:\u00a0<\/b>0ne-eighth of a number is 17.25. What will 73% of the number be ?<\/p>\n a)\u00a0100.74<\/p>\n b)\u00a0138.00<\/p>\n c)\u00a096.42<\/p>\n d)\u00a082.66<\/p>\n e)\u00a0None of these<\/p>\n Question 25:\u00a0<\/b>\u2018 A\u2019 ,\u2019B\u2019 and \u2018C\u2019 are three consecutive even integers such that four times \u2018A\u2019 is equal to three times \u2018C\u2019. What is the value of B\u2019?<\/p>\n a)\u00a012<\/p>\n b)\u00a010<\/p>\n c)\u00a016<\/p>\n d)\u00a014<\/p>\n e)\u00a0None of these<\/p>\n Question 26:\u00a0<\/b>What is the LCM of the following fractions? 3\/11, 2\/5, 1\/9<\/p>\n a)\u00a00<\/p>\n b)\u00a01<\/p>\n c)\u00a02<\/p>\n d)\u00a03<\/p>\n e)\u00a06<\/p>\n Question 27:\u00a0<\/b>What will be the the compound interest acquired on sum of Rs 12,000\/- for 3 years at the rate of 10 % per annum ?<\/p>\n a)\u00a02,652<\/p>\n b)\u00a03,972<\/p>\n c)\u00a03,960<\/p>\n d)\u00a03852<\/p>\n e)\u00a0None of these<\/p>\n Question 28:\u00a0<\/b>Ms suchi deposits an amount of 24000 to obtain in a simple interest at the rate of S.I 14 p.c.p.a for 8 years .what total amont will ms suchi gets at the end of 8 years<\/p>\n a)\u00a052080<\/p>\n b)\u00a028000<\/p>\n c)\u00a050880<\/p>\n d)\u00a026880<\/p>\n e)\u00a0none of these<\/p>\n Question 29:\u00a0<\/b>What would be the simple interest obtained on a amount 5670 at the rate of 6 pcpa after 3 years?<\/p>\n a)\u00a01020.60<\/p>\n b)\u00a01666.80<\/p>\n c)\u00a01336 .80<\/p>\n d)\u00a01063.80<\/p>\n e)\u00a0none of these<\/p>\n Question 30:\u00a0<\/b>A, B and C started a business by investing Rs. 20,000 Rs. 28,000 and Rs. 36,000 respectively. After 6 months, A and B withdrew an amount of Rs. 8,000 each and C invested an additional amount of Rs. 8,000. All of them invested for equal period of time. If at the end of the year. C got Rs. 12,550 as his share of profit, what was the total profit earned ?<\/p>\n a)\u00a0Rs. 25,100<\/p>\n b)\u00a0Rs. 26,600<\/p>\n c)\u00a0Rs. 24,300<\/p>\n d)\u00a0Rs. 22,960<\/p>\n e)\u00a0Rs. 21,440<\/p>\n Question 31:\u00a0<\/b>Sarita started a boutique investing an amount of Rs. 50,000. Six months later Neeta joined her with an amount of Rs. 80,000. At the end of one year they earned a profit of Rs. 18,000. What is Sarita\u2019s share in the profit?<\/p>\n a)\u00a0Rs. 9000<\/p>\n b)\u00a0Rs. 8000<\/p>\n c)\u00a0Rs. 12000<\/p>\n d)\u00a0Rs. 10000<\/p>\n e)\u00a0None of these<\/p>\n Question 32:\u00a0<\/b>In 1 kg of a mixture of sand and iron, 20% is iron .How such sand should be added so that the proportion of iron becomes 10%<\/p>\n a)\u00a01 kg<\/p>\n b)\u00a0200gm<\/p>\n c)\u00a0800 gm<\/p>\n d)\u00a01.8 kg<\/p>\n e)\u00a0None of these<\/p>\n Question 33:\u00a0<\/b>P. Q and R have a certain amount of money with themselves. Q has 25% more than what P has, and R has ${1 \\over 5}$th of what Q has. If P. Q and R together have Rs. 150, then how much money does P alone have? (in Rs.)<\/p>\n a)\u00a040<\/p>\n b)\u00a070<\/p>\n c)\u00a080<\/p>\n d)\u00a060<\/p>\n e)\u00a050<\/p>\n Question 34:\u00a0<\/b>Among five people – A, B, C, D and E \u2014 each scoring different marks, only one person scored less marks than B. D scored more marks than B but less than A. A did not score the highest marks. Who scored the second highest marks?<\/p>\n a)\u00a0E<\/p>\n b)\u00a0Cannot be determined<\/p>\n c)\u00a0A<\/p>\n d)\u00a0C<\/p>\n e)\u00a0D<\/p>\n Question 35:\u00a0<\/b>A, B and C have a certain amount of money with themselves. C has ${3 \\over 4}$ of what A has and B has Rs. 50 less than C. If A, B and C together have Rs. 250, then how much does A alone have? (in Rs.)<\/p>\n a)\u00a075<\/p>\n b)\u00a0160<\/p>\n c)\u00a080<\/p>\n d)\u00a0120<\/p>\n e)\u00a0140<\/p>\n Question 36:\u00a0<\/b>If an amount of Rs. 97836 is distributed equally amongst 31 children, how much amount would each child get ?<\/p>\n a)\u00a0Rs. 3756<\/p>\n b)\u00a0Rs. 3556<\/p>\n c)\u00a0Rs. 3356<\/p>\n d)\u00a0Rs. 3156<\/p>\n e)\u00a0None of these<\/p>\n Question 37:\u00a0<\/b>12 men can finish a project in 20 days. 18 women can finish the same project in 16 days and 24 children can finish it in 18 days. 8 women and 16 children worked for 9 days and then left. In how many days will 10 men complete the remaining project ?<\/p>\n a)\u00a0$10\\frac{1}{2}$<\/p>\n b)\u00a010<\/p>\n c)\u00a09<\/p>\n d)\u00a0$11\\frac{1}{2}$<\/p>\n e)\u00a0$9\\frac{1}{2}$<\/p>\n Question 38:\u00a0<\/b>Sixteen men and twelve women can complete a work in 8 days, if 20 men can complete the same work in 16 days, in how many days 16 women can complete the same piece of work ?<\/p>\n a)\u00a012<\/p>\n b)\u00a08<\/p>\n c)\u00a010<\/p>\n d)\u00a015<\/p>\n e)\u00a020<\/p>\n Question 39:\u00a0<\/b>What will come in the place of question mark (?) in the following series ? a)\u00a096<\/p>\n b)\u00a0106<\/p>\n c)\u00a0126<\/p>\n d)\u00a0130<\/p>\n e)\u00a0None of these<\/p>\n Question 40:\u00a0<\/b>What will come in place of the question mark (?) in the following number series? a)\u00a01589<\/p>\n b)\u00a01598<\/p>\n c)\u00a01958<\/p>\n d)\u00a01985<\/p>\n e)\u00a01835<\/p>\n Take CMAT Mock Tests<\/a><\/p>\n Enroll to CMAT crash course<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(A)<\/strong><\/p>\n let age of sharma = x<\/p>\n let age of harris = y<\/p>\n xy = 240<\/p>\n 2x = y+4<\/p>\n on solving these two equations<\/p>\n y = 20 or -24<\/p>\n age must be positive, so y = 20<\/p>\n x = 12<\/p>\n answer is option A<\/p>\n 2)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let ages of P,Q and R be $p,q,r$ respectively<\/p>\n According to ques, => $(p+r)+2q=59$ ————(i)<\/p>\n and $(q+r)+3p=68$ ————-(ii)<\/p>\n and $p+3q+3r=108$ ————(iii)<\/p>\n Multiplying equation (ii) by 3, => $3q+3r+9p = 204$ ———-(iv)<\/p>\n Subtracting equation (iii) from (iv), we get\u00a0:<\/p>\n => $(9p-p)=(204-108)$<\/p>\n => $8p=96$<\/p>\n => $p=\\frac{96}{8}=12$ years<\/p>\n => Ans – (D)<\/p>\n 3)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Let Vijay’s age = $x$ years<\/p>\n => Jayesh’s age = $2x$ years and Suresh’s age = $4x$ years<\/p>\n Sum of Suresh’s and Vijay’s ages = $(4x+x)=85$<\/p>\n => $x=\\frac{85}{5}=17$<\/p>\n $\\therefore$ Jayesh’s age = $2x=2 \\times 17=34$ years<\/p>\n => Ans – (A)<\/p>\n 4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Let B’s present age = $x$ years<\/p>\n and A’s present age = $y$ years<\/p>\n Acc. to ques, => $\\frac{\\frac{1}{2} (y – 4)}{4 (x – 4)} = \\frac{5}{12}$<\/p>\n => $6 (y – 4) = 20 (x – 4)$<\/p>\n => $6y – 24 = 20x – 80$<\/p>\n => $10x – 3y = 28$ ————(i)<\/p>\n Also, $(x + 8) – \\frac{1}{2} (y + 8) = 2$<\/p>\n => $2x + 16 – y – 8 = 4$<\/p>\n => $2x – y = -4$ ————-(ii)<\/p>\n Applying the operation, (i) – 3*(ii), we get\u00a0:<\/p>\n => $(10x – 6x) + (-3y + 3y) = 28 + 12$<\/p>\n => $x = \\frac{40}{4} = 10$ years<\/p>\n 5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Let present ages of P,Q and\u00a0R be $p,q,r$ years respectively.<\/p>\n => $q = r + 5$ ————(i)<\/p>\n and $p + q = r + 25$ ————(ii)<\/p>\n Subtracting eqn(i) from (ii), we get\u00a0:<\/p>\n => $p = 25 – 5 = 20$<\/p>\n $\\therefore$ Present age of P = 20 years<\/p>\n 6)\u00a0Answer\u00a0(A)<\/strong><\/p>\n 10+10 = 20 (7.69% = 1\/13 and 6.66%=1\/15)<\/p>\n 7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n $\\frac{1}{7}$ of $2\\frac{5}{8} \\div \\frac{3}{4}$ = 0.5<\/p>\n 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n $125^{4} is equal to 5^{12} and 25^{4} is equal to 5^{8}$<\/p>\n 9)\u00a0Answer\u00a0(A)<\/strong><\/p>\n LCM of fractions = LCM of numerators\/HCF of denominators 10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n [(3900 x 134)\u00a0\u00f7 100] + [134 x 39] Let the maximum aggregate marks be x 12)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Since the ages of A, B, C, D are consecutive<\/p>\n Let the ages of A, B, C, D be n, n+1,n+2,n+3<\/p>\n $\\frac{n+n+1+n+2+n+3}{4} = 49.5$<\/p>\n 4n+6 = 49.5*4 = 198<\/p>\n 4n = 192<\/p>\n n = 48<\/p>\n Ages of A, B, C, D = 48, 49, 50 ,51<\/p>\n Product of ages of B and D = 49*51 = (50-1)(50+1) =$50^2-1$ = 2500-1 = 2449.<\/p>\n 13)\u00a0Answer\u00a0(A)<\/strong><\/p>\n The maximum marks of each subjec is 105<\/p>\n 80% of 105 = 84<\/p>\n Since there are 5 subjects<\/p>\n Average marks =$\\frac{Sum\\ of\\ all\\ the\\ marks}{5} = 84$<\/p>\n Sum of all the marks = 84*5 = 420<\/p>\n 89+92+98+81+Science = 420<\/p>\n 360+Science = 420<\/p>\n Science = 420-360 = 60<\/p>\n 14)\u00a0Answer\u00a0(E)<\/strong><\/p>\n There are 9 numbers in the series:<\/p>\n Average =$\\frac{253+124+255+534+836+375+101+443+760}{9}$ = 3681\/9 = 409<\/p>\n Get 5 CMAT mocks at just Rs.299<\/a><\/p>\n 15)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Ruchika’s monthly income = Rs 32000 16)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Let the number of book distributed in a day by a man = 5x<\/p>\n woman = 4x & child = 2x<\/p>\n Day 1 : There were 7 men, 5 women, 8 children<\/p>\n => No. of books sold = (7 * 5x) + (5 * 4x) + (8 * 2x)<\/p>\n = 35x + 20x + 16x = 71x<\/p>\n Day 2 : There were 7 men, 3 women, 5 children [As, 2 women & 3 children were absent]<\/p>\n => No. of books sold = (7 * 5x) + (3 * 4x) + (5 * 2x)<\/p>\n = 35x + 12x + 10x = 57x<\/p>\n Day 3 : There were 4 men, 5 women, 3 children [As, 3 men & 5 children were absent]<\/p>\n => No. of books sold = (4 * 5x) + (5 * 4x) + (3 * 2x)<\/p>\n = 20x + 20x + 6x = 46x<\/p>\n Now, total books distributed on the course of three days = 71x + 57x + 46x = 2000<\/p>\n => x = 2000\/174<\/p>\n No. of books distributed on the second day = 57x = 57 * $\\frac{2000}{174}$ = 655.17 = ~650<\/p>\n 17)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Each of the 3 statements individually statements that all four of them aren’t facing the centre. Hence choice (d) is the correct option.<\/p>\n 18)\u00a0Answer\u00a0(E)<\/strong><\/p>\n To figure out how “one” is coded in that encryption, we need to identify the codes for every other word in “one of its kind”. That information is available in each of the 3 statements provided. Hence the correct option is (e)<\/p>\n 19)\u00a0Answer\u00a0(B)<\/strong><\/p>\n We know that, diagonal = 1.414* side Option B is correct option.<\/p>\n 20)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Circumference of th wheel = pi * Diameter = (22\/7) * 49 = 154 Ciircumference = 2*(22\/7)*r = 220 22)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Perimeter of rectangle = 4(44) = 176 cm 23)\u00a0Answer\u00a0(D)<\/strong><\/p>\n We know that, Let the number be $8x$<\/p>\n Acc to ques,<\/p>\n => $\\frac{1}{8} * 8x = 17.25$<\/p>\n => $x = 17.25$<\/p>\n $\\therefore$ 73 % of the number = $\\frac{73}{100} * 8x$<\/p>\n = 0.73 * 8 * 17.25 = 100.74<\/p>\n 25)\u00a0Answer\u00a0(D)<\/strong><\/p>\n let A be 2x ,B be 2x +2 and C be 2x +4<\/p>\n Given that 4A = 3C<\/p>\n 4(2x) = 3 (2x +4)<\/p>\n 8x = 6x +12<\/p>\n 2x = 12<\/p>\n x = 6<\/p>\n B = 2*6 + 2 = 14<\/p>\n 26)\u00a0Answer\u00a0(E)<\/strong><\/p>\n The LCM of the fractions = LCM of the numerators\/ HCF of the denominators 27)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Principal amount = Rs. 12,000<\/p>\n Time period = 3 years and rate of interest = 10% under compound interest.<\/p>\n =>\u00a0$C.I. = P [(1 + \\frac{R}{100})^T – 1]$<\/p>\n = $12,000 [(1 + \\frac{10}{100})^3 – 1]$<\/p>\n = $12,000 [(\\frac{11}{10})^3 – 1] = 12,000 (\\frac{1331 – 1000}{1000})$<\/p>\n = $12 \\times 331 = Rs. 3,972$<\/p>\n 28)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Amount deposited = Rs. 24,000<\/p>\n Rate = 14 % and time = 8 years under simple interest<\/p>\n => $S.I. = \\frac{P \\times R \\times T}{100}$<\/p>\n = $\\frac{24000 \\times 14 \\times 8}{100}$<\/p>\n = $240 \\times 112 = Rs. 26,880$<\/p>\n $\\therefore$Total amount will ms Suchi gets at the end of 8 years<\/p>\n = $24000 + 26880 = Rs. 50,880$<\/p>\n 29)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Principal amount = Rs. 5,670<\/p>\n Time period = 3 years and rate = 6% under simple interest.<\/p>\n => $S.I. = \\frac{P \\times R \\times T}{100}$<\/p>\n = $\\frac{5670 \\times 6 \\times 3}{100}$<\/p>\n = $56.7 \\times 18 = Rs. 1,020.6$<\/p>\n 30)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Amount invested by A,B and C =\u00a0Rs. 20,000 Rs. 28,000 and Rs. 36,000 respectively.<\/p>\n Ratio of profits shared by A : B : C<\/p>\n = $[(20,000 \\times 6) + (12,000 \\times 6)]$ : $[(28,000 \\times 6) + (20,000 \\times 6)]$ : $[(36,000 \\times 6) + (44,000 \\times 6)]$<\/p>\n = $32 : 48 : 80 = 2 : 3 : 5$<\/p>\n Let total profit earned = $Rs. x$<\/p>\n Profit received by C = $\\frac{5}{(2 + 3 + 5)} \\times x = 12550$<\/p>\n => $\\frac{x}{2} = 12550$<\/p>\n => $x = 12550 \\times 2 = Rs. 25,100$<\/p>\n 31)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Amount invested by Sarita = Rs. 50,000 and amount invested by Neeta = Rs. 80,000<\/p>\n Ratio of amount invested by Sarita : Neeta = 5\u00a0: 8<\/p>\n Time period in which Sarita invested = 12 months and Neeta = 6 months<\/p>\n Ratio of time periods of Sarita :\u00a0Neeta = 2\u00a0: 1<\/p>\n => Ratio of profits earned by Sarita : Neeta = $(5 \\times 2)$:$(8 \\times 1)$<\/p>\n = 5\u00a0: 4<\/p>\n Total profit earned = Rs. 18,000<\/p>\n $\\therefore$\u00a0Sarita\u2019s share in the profit = $\\frac{5}{(5+4)} \\times 18,000$<\/p>\n = $5 \\times 2,000 = Rs. 10,000$<\/p>\n => Ans – (D)<\/p>\n 32)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Total mixture of sand and iron = 1 kg<\/p>\n Quantity of iron = $\\frac{20}{100} \\times 1 = 0.2$ kg<\/p>\n Let $x$ kg of sand should be added, thus total iron in the mixture<\/p>\n => $0.2=\\frac{10}{100} \\times (x+1)$<\/p>\n => $2=x+1$<\/p>\n => $x=2-1=1$ kg<\/p>\n => Ans – (A)<\/p>\n 33)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let P has = $Rs. 100x$<\/p>\n => Amount with Q = $100x + \\frac{25}{100} \\times 100x = Rs. 125x$<\/p>\n => Amount with R = $\\frac{1}{5} \\times 125x = Rs. 25x$<\/p>\n Total amount together = $100x + 125x + 25x = 150$<\/p>\n => $x = \\frac{150}{250} = \\frac{3}{5}$<\/p>\n => $x = 0.6$<\/p>\n $\\therefore$ Amount with P alone = $100 \\times 0.6 = Rs. 60$<\/p>\n 34)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Let us rank the person according to the marks scored by them, where 1 -> highest marks and 5 -> lowest marks.<\/p>\n Only one person scored less marks than B, => B = 4<\/p>\n Also, A\u00a0> D\u00a0>\u00a0B and $A \\neq 1$<\/p>\n => A = 2 and D = 3<\/p>\n Thus, ranking from 1-5 = C\/E , A , D , B , E\/C<\/p>\n $\\therefore$ A scored the second highest marks.<\/p>\n 35)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Amount with A = $Rs. 4x$<\/p>\n => Amount with C = $\\frac{3}{4} \\times 4x = Rs. 3x$<\/p>\n => Amount with\u00a0B = $Rs. (3x – 50)$<\/p>\n Total amount with A,B & C = $4x + 3x + (3x – 50) = 250$<\/p>\n => $10x = 250 + 50 = 300$<\/p>\n => $x = \\frac{300}{10} = 30$<\/p>\n $\\therefore$ Amount with A = $4 \\times 30 = Rs. 120$<\/p>\n 36)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Total amount = 97836<\/p>\n No. of children = 31<\/p>\n Since amount is distributed equally<\/p>\n => Amount each child will get = $\\frac{97836}{31}$ = Rs. 3,156<\/p>\n 37)\u00a0Answer\u00a0(B)<\/strong><\/p>\n 12 men can finish the project in 20 days.<\/p>\n => 1 day work of 1 man = $\\frac{1}{12 \\times 20} = \\frac{1}{240}$<\/p>\n Similarly,\u00a0=> 1 day work of 1 woman = $\\frac{1}{18 \\times 16} = \\frac{1}{288}$<\/p>\n => 1 day work of 1 children = $\\frac{1}{24 \\times 18} = \\frac{1}{432}$<\/p>\n 8 women and 16 children worked for 9 days<\/p>\n => Work done in 9 days = $9 \\times (8 \\times \\frac{1}{288}) + (16 \\times \\frac{1}{432})$<\/p>\n = $9 \\times (\\frac{1}{36} + \\frac{1}{27}) = 9 \\times \\frac{7}{108}$<\/p>\n = $\\frac{7}{12}$<\/p>\n => Work left = $1 – \\frac{7}{12} = \\frac{5}{12}$<\/p>\n $\\therefore$ Number of days taken by 10 men to complete the remaining work<\/p>\n = $\\frac{\\frac{10}{240}}{\\frac{5}{12}} = \\frac{1}{24} \\times \\frac{12}{5} = \\frac{1}{10}$<\/p>\n Thus, 10 men will complete the remaining the work in 10 days.<\/p>\n 38)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Let work done by 1 man be $x$ and 1 woman be $y$<\/p>\n Now, 16 men and 12 women complete work in 8 days.<\/p>\n => $16x + 12y = \\frac{1}{8}$ ———Eqn(i)<\/p>\n Also, $20x = \\frac{1}{16}$<\/p>\n => $16x = \\frac{1}{20}$<\/p>\n Putting it in eqn(i), we get\u00a0:<\/p>\n => $\\frac{1}{20} + 12y = \\frac{1}{8}$<\/p>\n => $12y = \\frac{1}{8} – \\frac{1}{20} = \\frac{3}{40}$<\/p>\n => $y = \\frac{3}{40 \\times 12} = \\frac{1}{160}$<\/p>\n Thus, 16 women can complete the work in = $16 \\times \\frac{1}{160} = \\frac{1}{10}$<\/p>\n $\\therefore$ 16 women can complete the work in 10 days.<\/p>\n 39)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Each number is of the form $(n^3+1)$ where $n$ is a natural number<\/p>\n $1^3+1$ = 2<\/p>\n $2^3+1$ =\u00a09<\/p>\n $3^3+1$ =\u00a028<\/p>\n $4^3+1$ =\u00a065<\/p>\n $5^3+1$ =\u00a0126<\/strong><\/p>\n => Ans – (C)<\/p>\n 40)\u00a0Answer\u00a0(A)<\/strong><\/p>\n The pattern followed is\u00a0:<\/p>\n 9 $\\times 1 + 1^2$ = 10<\/p>\n 10\u00a0$\\times 3 + 3^2$ =\u00a039<\/p>\n 39\u00a0$\\times 5 + 5^2$ =\u00a0220<\/p>\n 220\u00a0$\\times 7 + 7^2$ =\u00a01589<\/strong><\/p>\n 1589\u00a0$\\times 9 + 9^2$ =\u00a014382<\/p>\n
\n253, 124, 255, 534, 836, 375, 101, 443, 760<\/p>\n
\nI) B sits second to right of D. D faces the centre. C sits to immediate right of both B and D.
\nII) A sits to immediate left to B. C is not an immediate neighbour of A. C sits to immediate right of D.
\nIII) D is an immediate neighbour of both A and C. B sits to the immediate left of A. C sits to the immediate right of B.<\/p>\n
\nI) ‘one of its kind’ is coded as ‘zo pi ko fe’ and ‘in kind and cash’ is coded as ‘ga to ru ko’
\nII) ‘Its point for origin’ is coded as ‘ba le fe mi’ and ‘make a point clear’ is coded as ‘yu si mi de’
\nIII) ‘make money and cash’ is coded as ‘to mi ru hy’ and ‘money of various kind’ is coded as ‘qu ko zo hy’.<\/p>\n
\n2 \u00a09 \u00a028 \u00a065 ?<\/p>\n
\n9\u00a0 10\u00a0 39\u00a0 220\u00a0 ?\u00a0 14382<\/p>\n
\nHCF of (3, 4, 5) = 1
\nSo, LCM of the three fractions = 1\/1 = 1<\/p>\n
\n= 10452
\n11)\u00a0Answer\u00a0(A)<\/strong><\/p>\n
\n780\/x – 728\/x = 5\/100
\n52\/x = 1\/20
\nx = 52*20 = 1040<\/p>\n
\nRavina’s monthly income\u00a0= 32000 x (1 + $\\frac{15}{100}$) = 32000 x $\\frac{115}{100}$ = Rs. 36800
\nRamola’s monthly income = 3 x 36800\u00a0= 110400
\nRamola’s annual income = 12 x 110400 =\u00a01324800<\/p>\n
\nHere, diagoal = 14.14, side = 10 m
\nHeight = Perimeter of square = 10*4 = 40m
\nNow, base = side of square with 784 $m^2$ area.
\nbase = 28
\nArea of triangle = .5*28*40 = 560 $m^2$<\/p>\n
\nNumber of revolutions of the wheel = 3200\/154 = 20.77~21
\nThe wheel will make around 21 revolutions.
\n21)\u00a0Answer\u00a0(C)<\/strong><\/p>\n
\nHence, r = 35
\nNow, area of circle = (22\/7)*35*35 = 3850
\nArea of rectangle is same as that of circle
\nArea of rectangle = 3850
\nLength * breadth = 3850
\nlength = 3850\/50 = 77m
\nTherefore, option C is correct answer.<\/p>\n
\nNow, 2*(length+breadth) = 176
\n2*(51+breadth) = 176
\n51 + breadth = 88
\nbreadth = 88-51 = 37
\nDifference between side of square and breadth of rectangle = 37 – 11 = 26 cm
\nTherefore, the correct option is option C.<\/p>\n
\nDiagonal = side * 1.414
\nTherefore, side = 28\/1.414 = 19.8
\nArea of square = 19.8 *19.8 = 392.04 ~ 392
\nTherefore, the correct option is option D.
\n24)\u00a0Answer\u00a0(A)<\/strong><\/p>\n
\nLCM of the numerators = LCM of 3, 2 and 1 = 6
\nHCF of the denominators = HCF of 11, 5 and 9 = 1
\nSo, the required LCM = 6\/1 = 6<\/p>\n