Here you can download the important MAH – CET Quantitative Reasoning Questions PDF by Cracku. Very Important MAH – CET 2022 and These questions will help your MAH – CET preparation. So kindly download the PDF for reference and do more practice.<\/p>\n
Download Quantitative Reasoning Questions for MAH-CET PDF<\/a><\/p>\n Enroll to MAH-CET Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>The rate of onion increased from \u20b9 70 per kilogram to \u20b9 120 per kilogram within a month. By what percentage (approximate) did the rate of onion increase?<\/p>\n a)\u00a075.2%<\/p>\n b)\u00a065%<\/p>\n c)\u00a080%<\/p>\n d)\u00a071.5%<\/p>\n 1)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Percentage increase in the rate of onion =\u00a0$\\frac{120-70}{70}\\times100$<\/p>\n =\u00a0$\\frac{50}{70}\\times100$<\/p>\n =\u00a0$\\frac{500}{7}$<\/p>\n = 71.43%<\/p>\n = 71.5% (approximately)<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 2:\u00a0<\/b>Which two signs and two numbers should be interchanged to make the given equation correct? a)\u00a0$\\times$ and $+$; 15 and 25<\/p>\n b)\u00a0$\\div$ and $-$; 15 and 5<\/p>\n c)\u00a0$\\times$ and $-$; 8 and 25<\/p>\n d)\u00a0$\\times$ and $+$; 8 and 25<\/p>\n 2)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n By Trial and Error method,<\/p>\n Option A<\/u><\/p>\n $16\\times8+15-25\\div5=138$<\/p>\n $16\\times8+15-5=138$<\/p>\n $128+15-5=138$<\/p>\n $138=138$<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 3:\u00a0<\/b>Select the correct equation when the signs \u2018+\u2019 and $\u2018\\times\u2019$ and the numbers \u20184\u2019 and \u20188\u2019 are interchanged.<\/p>\n a)\u00a0$12 \\times 4 + 8 = 34$<\/p>\n b)\u00a0$6 + 8 \\times 4 = 38$<\/p>\n c)\u00a0$2 \\times 4 + 8 = 34$<\/p>\n d)\u00a0$4 + 8 \\times 2 = 32$<\/p>\n 3)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n By Trial and Error method,<\/p>\n Option A<\/u><\/p>\n $12+8\\times4=34$<\/p>\n $12+32=34$<\/p>\n $44=34$<\/p>\n Hence Option A is incorrect<\/p>\n Option B<\/u><\/p>\n $6\\times4+8=38$<\/p>\n $24+8=38$<\/p>\n $32=38$<\/p>\n Hence Option B is incorrect<\/p>\n Option C<\/u><\/p>\n $2+8\\times4=34$<\/p>\n $2+32=34$<\/p>\n $34=34$<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 4:\u00a0<\/b>Train \u2018A\u2019 runs at a speed of 80 kmph and leaves station \u2018X\u2019 at 11.00 a.m. Train \u2018B\u2019 leaves station \u2018X\u2019 at 11.15 a.m.in the same direction, on the same day . At what speed train \u2018B\u2019 should run in order to catch train \u2018A\u2019 at station \u2018Y\u2019 located at a distance of 60 km?<\/p>\n a)\u00a0120 km\/h<\/p>\n b)\u00a0125 km\/h<\/p>\n c)\u00a0115 km\/h<\/p>\n d)\u00a0110 km\/h<\/p>\n 4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Speed of train A = 80 km\/h<\/p>\n Distance between station X and station Y = 60 km<\/p>\n Time taken by train A to reach station Y = $\\frac{60}{80}$ hours = $\\frac{3}{4}$ hours\u00a0$\\frac{3}{4}\\times$60 minutes = 45 minutes<\/p>\n Since train B starts 15 min late, it should reach station Y in 30 minutes to catch train A at station Y.<\/p>\n Time taken by train B to reach station Y = 30 minutes =\u00a0$\\frac{1}{2}$ hour<\/p>\n $\\therefore\\ $Speed of train B =\u00a0$\\frac{\\text{Distance from station X to station Y}}{\\text{Time}}$ = $\\frac{60}{\\frac{1}{2}}$ =\u00a0120 km\/h<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 5:\u00a0<\/b>Which two numbers should be interchanged in the following equation to make it correct? a)\u00a036 and 48<\/p>\n b)\u00a06 and 14<\/p>\n c)\u00a048 and 15<\/p>\n d)\u00a02 and 6<\/p>\n 5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n By Trial and Error method,<\/p>\n Option A<\/u><\/p>\n $48\\div6-15\\times2+36=14$<\/p>\n $8-15\\times2+36=14$<\/p>\n $8-30+36=14$<\/p>\n $14=14$<\/p>\n Hence, the correct answer is Option A<\/p>\n Take Free MAH-CET mock tests here<\/a><\/p>\n Question 6:\u00a0<\/b>In a class, every student participated in one of the three activities i.e. folk song, kathak and quiz. 22% of the total students participated in folk song. 25% of the remaining students participated in kathak. The remaining students participated in quiz. What percentage of students participated in quiz?<\/p>\n a)\u00a053%<\/p>\n b)\u00a030.6%<\/p>\n c)\u00a041.5%<\/p>\n d)\u00a058.5%<\/p>\n 6)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the total number of students = T<\/p>\n 22% of the total students participated in folk song<\/p>\n Number of students participated in folk song =\u00a0$\\frac{22}{100}$T<\/p>\n Remaining students = T –\u00a0$\\frac{22}{100}$T =\u00a0$\\frac{78}{100}$T<\/p>\n 25% of the remaining students participated in kathak<\/p>\n Number of students participated in kathak = $\\frac{25}{100}\\times\\frac{78}{100}$T =\u00a0$\\frac{39}{200}$T<\/p>\n Remaining students participated in quiz.<\/p>\n Number of students participated in quiz = T – $\\frac{22}{100}$T – $\\frac{39}{200}$T = $\\frac{117}{200}$T<\/p>\n Percentage of students participated in quiz = $\\frac{\\frac{117}{200}T}{T}\\times100$ = 58.5%<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 7:\u00a0<\/b>A person is employed to a company on salary of \u20b9 6,000 per month. Through out the year, he remained absent from work for 15 days. If \u20b9 200 gets deducted for every single holiday, then how much money will he earn at the end of the year ?<\/p>\n a)\u00a0\u20b9 72,000<\/p>\n b)\u00a0\u20b9 70,000<\/p>\n c)\u00a0\u20b9 69,000<\/p>\n d)\u00a0\u20b9 75,000<\/p>\n 7)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Salary of the person per month =\u00a0\u20b9 6,000<\/p>\n Salary of the person per year =\u00a0$6000\\times12$ =\u00a0\u20b9 72,000<\/p>\n Deduction for 1 holiday =\u00a0\u20b9 200<\/p>\n Deduction for 15 holidays = $15\\times200$ = \u20b9 3,000<\/p>\n $\\therefore\\ $Money earned by the person at the end of the year = 72000 – 3000 = \u20b9 69,000<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 8:\u00a0<\/b>What will be the value of \u2018a + b\u2019 in the following equation? a)\u00a05<\/p>\n b)\u00a06<\/p>\n c)\u00a07<\/p>\n d)\u00a04<\/p>\n 8)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\left[4\\left(3+\\left\\{9\\times6\\div3+\\left\\{12\\div\\left(1\\div1\\times3\\right)\\right\\}\\right\\}\\right)\\right]=5^a\\times2^b$<\/p>\n $\\Rightarrow$ \u00a0$\\left[4\\left(3+\\left\\{9\\times6\\div3+\\left\\{12\\div\\left(1\\times3\\right)\\right\\}\\right\\}\\right)\\right]=5^a\\times2^b$<\/p>\n $\\Rightarrow$ \u00a0$\\left[4\\left(3+\\left\\{9\\times6\\div3+\\left\\{12\\div3\\right\\}\\right\\}\\right)\\right]=5^a\\times2^b$<\/p>\n $\\Rightarrow$ \u00a0$\\left[4\\left(3+\\left\\{9\\times6\\div3+4\\right\\}\\right)\\right]=5^a\\times2^b$<\/p>\n $\\Rightarrow$ \u00a0$\\left[4\\left(3+\\left\\{9\\times2+4\\right\\}\\right)\\right]=5^a\\times2^b$<\/p>\n $\\Rightarrow$ \u00a0$\\left[4\\left(3+\\left\\{18+4\\right\\}\\right)\\right]=5^a\\times2^b$<\/p>\n $\\Rightarrow$ \u00a0$\\left[4\\left(3+22\\right)\\right]=5^a\\times2^b$<\/p>\n $\\Rightarrow$ \u00a0$\\left[4\\left(25\\right)\\right]=5^a\\times2^b$<\/p>\n $\\Rightarrow$ \u00a0$100=5^a\\times2^b$<\/p>\n $\\Rightarrow$ \u00a0$25\\times4=5^a\\times2^b$<\/p>\n $\\Rightarrow$ \u00a0$5^2\\times2^2=5^a\\times2^b$<\/p>\n Comparing both sides,\u00a0 a = 2 and b = 2<\/p>\n $\\therefore\\ $a + b = 2 + 2 = 4<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 9:\u00a0<\/b>One pound is approximately equal to 0.453 kg. Jennifer bought 4.409 pounds of custard apple. How many kilograms of custard apple did she buy (approximately)?<\/p>\n a)\u00a02<\/p>\n b)\u00a01<\/p>\n c)\u00a01.5<\/p>\n d)\u00a02.5<\/p>\n 9)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given,\u00a0One pound is approximately equal to 0.453 kg<\/p>\n Custard apple bought by Jennifer = 4.409 pounds = 4.409 x 0.453 kg = 1.997 kg = 2 kg (approximately)<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 10:\u00a0<\/b>Select the correct combination of mathematical signs to replace ‘Y\u2019 sequentially and balance the following equation. a)\u00a0$\\times, =, -, \\div$<\/p>\n b)\u00a0$\\times, \\div, +, =$<\/p>\n c)\u00a0$\\div, \\times, =, -$<\/p>\n d)\u00a0$\\div, \\times, =, \\times$<\/p>\n 10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n By Trial and Error method,<\/p>\n Option A<\/u><\/p>\n $18\\times2=3-3\\div9$<\/p>\n $18\\times2=3-\\frac{1}{3}$<\/p>\n $36=3-\\frac{1}{3}$<\/p>\n $36=\\frac{8}{3}$<\/p>\n Hence Option A is incorrect<\/p>\n Option B<\/u><\/p>\n $18\\times2\\div3+3=9$<\/p>\n $18\\times\\frac{2}{3}+3=9$<\/p>\n $12+3=9$<\/p>\n $15=9$<\/p>\n Hence Option B is incorrect<\/p>\n Option C<\/u><\/p>\n $18\\div2\\times3=3-9$<\/p>\n $9\\times3=3-9$<\/p>\n $27=-6$<\/p>\n Hence Option C is incorrect<\/p>\n Option D<\/u><\/p>\n $18\\div2\\times3=3\\times9$<\/p>\n $9\\times3=3\\times9$<\/p>\n $27=27$<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 11:\u00a0<\/b>Which two signs and numbers should be interchanged to make the following equation correct? a)\u00a014 and 18, + and –<\/p>\n b)\u00a016 and 14, – and $\\times$<\/p>\n c)\u00a014 and 18, + and $\\times$<\/p>\n d)\u00a016 and 3, – and $\\div$<\/p>\n 11)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n By Trial and Error method,<\/p>\n Option A<\/u><\/p>\n $16\\times14-2+18\\div3=38$<\/p>\n $16\\times14-2+6=38$<\/p>\n $224-2+6=38$<\/p>\n $228=38$<\/p>\n Hence Option A is incorrect<\/p>\n Option B<\/u><\/p>\n $14-18+2\\times16\\div3=38$<\/p>\n $14-18+2\\times\\frac{16}{3}=38$<\/p>\n After solving the value will be decimal which is not possible<\/p>\n Option C<\/u><\/p>\n $16+14\\times2-18\\div3=38$<\/p>\n $16+14\\times2-6=38$<\/p>\n $16+28-6=38$<\/p>\n $38=38$<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 12:\u00a0<\/b>Which two signs should be interchanged to make the following equation correct? a)\u00a0+ and $\\times$<\/p>\n b)\u00a0+ and –<\/p>\n c)\u00a0$\\div$ and +<\/p>\n d)\u00a0$\\div$ and $\\times$<\/p>\n 12)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n By Trial and Error method,<\/p>\n Option A<\/u><\/p>\n $4\\div6\\times9-48+8=27$<\/p>\n $\\frac{2}{3}\\times9-48+8=27$<\/p>\n $6-48+8=27$<\/p>\n $-34=27$<\/p>\n Hence Option A is incorrect<\/p>\n Option B $4\\div6-9+48\\times8=27$<\/p>\n $\\frac{2}{3}-9+48\\times8=27$<\/p>\n After solving the value will be decimal which is not possible<\/p>\n Option C<\/u><\/p>\n $4+6\\div9-48\\times8=27$<\/p>\n $4+\\frac{2}{3}-48\\times8=27$<\/p>\n After solving the value will be decimal which is not possible<\/p>\n Option D<\/u><\/p>\n $4\\times6+9-48\\div8=27$<\/p>\n $4\\times6+9-6=27$<\/p>\n $24+9-6=27$<\/p>\n $27=27$<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 13:\u00a0<\/b>A and B can do a piece of work in 30 days and 18 days respectively. A started the work alone and then after 6 days B joined him till the completion of the work. In how many days has the whole work completed?<\/p>\n a)\u00a017<\/p>\n b)\u00a015<\/p>\n c)\u00a09<\/p>\n d)\u00a012<\/p>\n 13)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the total work be W<\/p>\n Number of days required for A to complete the work = 30 days<\/p>\n $\\Rightarrow$\u00a0 Work done by A in 1 day =\u00a0$\\frac{W}{30}$<\/p>\n Number of days required for B to complete the work = 18 days<\/p>\n $\\Rightarrow$ \u00a0Work done by B in 1 day = $\\frac{W}{18}$<\/p>\n Work done by A and B together in 1 day =\u00a0$\\frac{W}{30}+\\frac{W}{18}$ =\u00a0$\\frac{3W+5W}{90}$ =\u00a0$\\frac{4W}{45}$<\/p>\n Work done by A alone in 6 days =\u00a0$\\frac{W}{30}\\times6$ =\u00a0$\\frac{W}{5}$<\/p>\n Remaining work =\u00a0$W-\\frac{W}{5}$ =\u00a0$\\frac{4W}{5}$<\/p>\n Number of days required for both A and B to complete remaining work =\u00a0$\\frac{\\text{Remaining work}}{\\text{Work in 1 day}}$ = $\\frac{\\frac{4W}{5}}{\\frac{4W}{45}}$ = 9 days<\/p>\n $\\therefore\\ $Number of days required to complete the whole work = 6 + 9 = 15 days<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 14:\u00a0<\/b>Which two signs should be interchanged to make the given equation correct? a)\u00a0+ and –<\/p>\n b)\u00a0+ and $\\times$<\/p>\n c)\u00a0$\\times$ and $\\div$<\/p>\n d)\u00a0+ and $\\div$<\/p>\n 14)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n By Trial and Error method,<\/p>\n Option A<\/u><\/p>\n $(72\\div18)-30\\times8+4=20$<\/p>\n $4-30\\times8+4=20$<\/p>\n $4-240+4=20$<\/p>\n $-232=20$<\/p>\n Hence Option A is incorrect<\/p>\n Option B<\/u><\/p>\n $(72\\div18)\\times30+8-4=20$<\/p>\n $4\\times30+8-4=20$<\/p>\n $120+8-4=20$<\/p>\n $124=20$<\/p>\n Hence Option B is incorrect<\/p>\n Option C $(72\\times18)+30\\div8-4=20$<\/p>\n $1296+30\\div8-4=20$<\/p>\n $1296+3.75-4=20$<\/p>\n After solving the\u00a0 value will be decimal which is not possible<\/p>\n Option D<\/u><\/p>\n <\/u>$(72+18)\\div30\\times8-4=20$<\/p>\n $90\\div30\\times8-4=20$<\/p>\n $3\\times8-4=20$<\/p>\n $24-4=20$<\/p>\n $20=20$<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 15:\u00a0<\/b>Which two signs or numbers need to be interchanged to make the following equation correct? a)\u00a0$+, \\times$<\/p>\n b)\u00a0$+, \\div$<\/p>\n c)\u00a0$\\times, \\div$<\/p>\n d)\u00a0$18, 8$<\/p>\n 15)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n By Trial and Error method,<\/p>\n Option A<\/u><\/p>\n $(18\\div9)\\times9+8=24$<\/p>\n $2\\times9+8=24$<\/p>\n $18+8=24$<\/p>\n $26=24$<\/p>\n Hence Option A is incorrect<\/p>\n Option B<\/u><\/p>\n $(18+9)\\div9\\times8=24$<\/p>\n $27\\div9\\times8=24$<\/p>\n $3\\times8=24$<\/p>\n $24=24$<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 16:\u00a0<\/b>If \u2018#\u2019 means \u2018\u2014\u2019, \u2018&\u2019 means \u2018$\\div$\u2019, \u2018@\u2019 means \u2018$\\times$\u2019 and \u2018$\\div$\u2019 means \u2018+\u2019, then 15 @ 2 $\\div$ 900 & 30 # 10 = ?<\/p>\n a)\u00a021<\/p>\n b)\u00a0310<\/p>\n c)\u00a050<\/p>\n d)\u00a0600<\/p>\n 16)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n By applying the given conditions,<\/p>\n $15$ @ $2$ $\\div$ $900$ & $30$ # $10$ $=15\\times2\\ +900 \\div 30 -10$<\/p>\n $=15\\times2\\ +30\\ -10$<\/p>\n $=30+30\\ -10$<\/p>\n $=50$<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 17:\u00a0<\/b>Which two numbers should be interchanged in the following equation to make it correct? a)\u00a017 and 4<\/p>\n b)\u00a028 and 2<\/p>\n c)\u00a017 and 6<\/p>\n d)\u00a06 and 4<\/p>\n 17)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n By Trial and Error method,<\/p>\n Option A<\/u><\/p>\n $6+28\\div17-2\\times4=12$<\/p>\n $6+\\frac{28}{17}-2\\times4=12$<\/p>\n After solving the value will be decimal which is not possible. Hence Option A is incorrect.<\/p>\n Option B<\/u><\/p>\n $6+2\\div4-28\\times17=12$<\/p>\n $6+0.5-28\\times17=12$<\/p>\n After solving the value will be decimal which is not possible. Hence Option B is incorrect.<\/p>\n Option C<\/u><\/p>\n $17+28\\div4-2\\times6=12$<\/p>\n $17+7-2\\times6=12$<\/p>\n $17+7-12=12$<\/p>\n $12=12$<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 18:\u00a0<\/b>Aman has an equal number of one, five, ten and twenty rupee bank notes, to make a total of \u20b9 1,080. How many bank notes does Aman have in total?<\/p>\n a)\u00a0100<\/p>\n b)\u00a090<\/p>\n c)\u00a0120<\/p>\n d)\u00a0105<\/p>\n 18)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the number of one rupee notes with the Aman = a<\/p>\n Aman has an equal number of one, five, ten and twenty rupee bank notes<\/p>\n $\\Rightarrow$\u00a0 Total number of notes with Aman = 4a<\/p>\n Total amount with Aman = \u20b9 1,080<\/p>\n $\\Rightarrow$ \u00a0$\\left(1\\times a\\right)+\\left(5\\times a\\right)+\\left(10\\times a\\right)+\\left(20\\times a\\right)=1080$<\/p>\n $\\Rightarrow$ \u00a0$36a=1080$<\/p>\n $\\Rightarrow$ \u00a0$a=30$<\/p>\n $\\therefore\\ $Total number of notes with Aman = 4a = 4 x 30 = 120<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 19:\u00a0<\/b>Which two signs should be interchanged to make the given equation correct? a)\u00a0$+$ and $-$<\/p>\n b)\u00a0$+$ and $\\div$<\/p>\n c)\u00a0$\\div$ and $-$<\/p>\n d)\u00a0$\\times$ and $+$<\/p>\n 19)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n By Trial and Error method,<\/p>\n Option A<\/u><\/p>\n $(560\\div80)-90\\times8+38=600$<\/p>\n $7-90\\times8+38=600$<\/p>\n $7-720+38=600$<\/p>\n $-675=600$<\/p>\n Hence Option A is incorrect<\/p>\n Option B<\/u><\/p>\n $(560+80)\\div90\\times8-38=600$<\/p>\n $640\\div90\\times8-38=600$<\/p>\n $\\frac{64}{9}\\times8-38=600$<\/p>\n The value will be decimal after solving which is not possible<\/p>\n Option C<\/u><\/p>\n $(560-80)+90\\times8\\div38=600$<\/p>\n $480+90\\times8\\div38=600$<\/p>\n $480+720\\div38=600$<\/p>\n $480+\\frac{360}{19}=600$<\/p>\n The value will be decimal after solving which is not possible<\/p>\n Option D $(560\\div80)\\times90+8-38=600$<\/p>\n $7\\times90+8-38=600$<\/p>\n $630+8-38=600$<\/p>\n $600=600$<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 20:\u00a0<\/b>After giving a discount of 25% on the marked price, a laptop is sold at \u20b9 22,500. What is its marked price?<\/p>\n a)\u00a0\u20b9 32,000<\/p>\n b)\u00a0\u20b9 36,500<\/p>\n c)\u00a0\u20b9 30,000<\/p>\n d)\u00a0\u20b9 29,000<\/p>\n 20)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the marked price be M<\/p>\n Discount = 25%<\/p>\n Selling price = \u20b9 22,500<\/p>\n $\\Rightarrow$\u00a0 $\\frac{75}{100}\\text{M}=22500$<\/p>\n $\\Rightarrow$\u00a0 $\\text{M}=30000$<\/p>\n $\\therefore\\ $Marked price =\u00a0\u20b9 30,000<\/p>\n Hence, the correct answer is Option C<\/p>\n
\n$16 + 8 \\times 25 – 15 \\div 5 = 138$<\/p>\n
\n$36 \\div 6 – 15 \\times 2 + 48 = 14$<\/p>\n
\n$\\left[4 \\left(3 + \\left\\{9 \\times 6 \\div 3 + \\left\\{12 \\div \\left(1 \\div 1 \\times 3\\right)\\right\\}\\right\\}\\right)\\right] = 5^a \\times 2^b$<\/p>\n
\n18 Y 2 Y 3 Y 3 Y 9<\/p>\n
\n$16 \\times 18 + 2 – 14 \\div 3 = 38$<\/p>\n
\n$4 \\div 6 + 9 – 48 \\times 8 = 27$<\/p>\n
\n<\/u><\/p>\n
\n$(72 \\div 18) + 30 \\times 8 – 4 = 20$<\/p>\n
\n<\/u><\/p>\n
\n$(18 \\div 9) + 9 \\times 8 = 24$<\/p>\n
\n$6 + 28 \\div 4 – 2 \\times 17 = 12$<\/p>\n
\n$(560\\div80)+90\\times8-38=600$<\/p>\n
\n<\/u><\/p>\n