TISSNET Percentage questions with a PDF by Cracku. Practice TISSNET solved Percentage Questions paper tests, and these are the practice question to have a firm grasp on the Percentage topic in the TISSNET exam. Very Important Percentage Questions for TISSNET based on the questions asked in the previous TISSNET exam papers. Click on the link below to download the TISSNET Percentage Questions with answers PDF.<\/p>\n
Download Percentage Questions for TISSNET<\/a><\/p>\n Enroll to TISSNET 2023 Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>Sachin\u2019s income is 25% more than Dileep\u2019s income. By how much percentage is Dileep\u2019s income less than Sachin\u2019s income ?<\/p>\n a)\u00a015%<\/p>\n b)\u00a020%<\/p>\n c)\u00a018%<\/p>\n d)\u00a022%<\/p>\n 1)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the income of Sachin = S<\/p>\n Income of Dileep = D<\/p>\n Given,\u00a0 Sachin’s income is 25% more than Dileep’s income<\/p>\n $=$> \u00a0$\\text{S}=\\frac{125}{100}\\text{D}$<\/p>\n $=$> \u00a0$\\text{S}=\\frac{5}{4}\\text{D}$<\/p>\n $\\therefore\\ $Required Percentage = $\\frac{S-D}{S}\\times100$<\/p>\n =\u00a0$\\frac{\\frac{5}{4}D-D}{\\frac{5}{4}D}\\times100$<\/p>\n = $\\frac{\\frac{D}{4}}{\\frac{5D}{4}}\\times100$<\/p>\n =\u00a0$\\frac{1}{5}\\times100$<\/p>\n = 20%<\/p>\n $\\therefore\\ $Dileep’s income is 20% less than Sachin’s income<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 2:\u00a0<\/b>The volume of the water in two tanks, A and B,is in the ratio of 6 : 5. The volume of water in tank A is increased by 30%. By what percentage should the volume of water in tank B be increased so that both the tanks have the same volume of water?<\/p>\n a)\u00a056%<\/p>\n b)\u00a030%<\/p>\n c)\u00a018%<\/p>\n d)\u00a015%<\/p>\n 2)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the volume of water in tanks A and B are $6p$ and $5p$ respectively<\/p>\n Volume of water in tank A after increasing 30% = $\\frac{130}{100}\\times6p=7.8p$<\/p>\n Volume of water to be increased in tank B to have same volume as tank A = $7.8p-5p=2.8p$<\/p>\n $\\therefore\\ $Percentage increase in volume of water in tank B = $\\frac{2.8p}{5p}\\times100=56\\%$<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 3:\u00a0<\/b>Sachin scored 120 runs, which included 6 boundaries and 4 sixes. What percentage of his total score did he make by running between the wickets?<\/p>\n a)\u00a0$45 %$<\/p>\n b)\u00a0$46 \\frac{4}{9} %$<\/p>\n c)\u00a0$60 %$<\/p>\n d)\u00a0$33 \\frac{1}{3} %$<\/p>\n 3)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Total runs scored by Sachin =120<\/p>\n Runs scored in boundaries =\u00a0$\\left(6\\times4\\right)+\\left(4\\times6\\right)=24+24=48$<\/p>\n Runs scored by running between wickets =\u00a0$120-48=72$<\/p>\n $\\therefore\\ $Required Percentage = $\\frac{72}{120}\\times100=60\\%$<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 4:\u00a0<\/b>Anu spends 68% of her monthly income. If her monthly income increases by 20% and her monthly savings increase by $9\\frac{3}{8}\\%$, then the percentage increase in her monthly expenditure is:<\/p>\n a)\u00a020%<\/p>\n b)\u00a025%<\/p>\n c)\u00a022%<\/p>\n d)\u00a032%<\/p>\n 4)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the initial salary of Anu be 100%.<\/p>\n Initially Expenditure of Anu = 68%<\/p>\n Saving = 100 -68 = 32%<\/p>\n Salary after 20% increment =\u00a0\u00a0100 $\\times 120\/100 = 120%$<\/p>\n Saving after increment = 32% $\\times \\frac{100 +\u00a09\\frac{3}{8}}{100}$ = 32% $\\times \\frac{109.375}{100}$ = 35%<\/p>\n Expenditure = 120 – 35 = 85%<\/p>\n Percentage increase in her monthly expenditure = $\\frac{85 – 68}{68} \\times 100 =\u00a0\\frac{17}{68} \\times 100$ = 25%<\/p>\n Question 5:\u00a0<\/b>The price of sugaris increased by 20%. By what percentage must one cut down on the consumption of sugar, so that no extra amount has to be incurred on sugar?<\/p>\n a)\u00a0$16\\frac{2}{3}\\%$<\/p>\n b)\u00a0$20\\%$<\/p>\n c)\u00a0$80\\%$<\/p>\n d)\u00a0$83\\frac{1}{3}\\%$<\/p>\n 5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Percentage decrements in the\u00a0consumption of sugar = $\\frac{rate}{100 + rate} \\times 100$<\/p>\n =\u00a0$\\frac{20}{100 + 20} \\times 100$ =\u00a0$\\frac{20}{120} \\times 100$ = $16 \\frac{2}{3}$%<\/p>\n Question 6:\u00a0<\/b>In an examination in which the full marks were 500, A scored 25% more marks than B, B scored 60% more marks than C and C scored 20% less marks than D. If A scored 80% marks, then the percentage of marks obtained by D is:<\/p>\n a)\u00a065%<\/p>\n b)\u00a060%<\/p>\n c)\u00a050%<\/p>\n d)\u00a054%<\/p>\n 6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given, Total marks = 500<\/p>\n A scored 80% marks in the examination<\/p>\n $\\Rightarrow$\u00a0 A =\u00a0$\\frac{80}{100}\\times500$ = 400<\/p>\n A scored 25% more marks than B<\/p>\n $\\Rightarrow$\u00a0 A =\u00a0$\\frac{125}{100}$B<\/p>\n $\\Rightarrow$\u00a0 400 =\u00a0$\\frac{125}{100}$B<\/p>\n $\\Rightarrow$\u00a0 400 = $\\frac{5}{4}$B<\/p>\n $\\Rightarrow$\u00a0 B = 320<\/p>\n B scored 60% more marks than C<\/p>\n $\\Rightarrow$\u00a0 B =\u00a0$\\frac{160}{100}$C<\/p>\n $\\Rightarrow$\u00a0 320 =\u00a0$\\frac{160}{100}$C<\/p>\n $\\Rightarrow$\u00a0 C = 200<\/p>\n C scored 20% less marks than D<\/p>\n $\\Rightarrow$\u00a0 C =\u00a0$\\frac{80}{100}$D<\/p>\n $\\Rightarrow$\u00a0 200 =\u00a0$\\frac{80}{100}$D<\/p>\n $\\Rightarrow$\u00a0 D = 250<\/p>\n $\\therefore\\ $Percentage of marks obtained by D =\u00a0$\\frac{250}{500}$ = 50%<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 7:\u00a0<\/b>A, B and C donate 8%, 7% and 9%,of their salaries, respectively to a charitable trust. The salaries of A and B are same and the difference between their donations is \u20b9259. Thetotal donation of A and B is \u20b91,185 more than that of C. The total donation ofA and C is what percentage ofthe total salaries of A, B and C? (Correct to one decimalplace)<\/p>\n a)\u00a06.2%<\/p>\n b)\u00a05.8%<\/p>\n c)\u00a06.4%<\/p>\n d)\u00a07.1%<\/p>\n 7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the salaries of A, B and C be x, x, y.<\/p>\n Donation of A, B and C = 0.08x, 0.07x and 0.09z<\/p>\n Difference between donations of A and B\u00a0= 259<\/p>\n 0.08x – 0.07x = 259<\/p>\n x = 259\/0.01 = 25900<\/p>\n Total donation of A and B = donation of A + 1185<\/p>\n 0.08x + 0.07x =\u00a00.09z + 1185<\/p>\n 0.15x =\u00a00.09z + 1185<\/p>\n 0.09z =\u00a00.15 $\\times $ 25900 – 1185 = 2700<\/p>\n z = 2700\/0.09 = 30000<\/p>\n Total donation of A and C =\u00a00.08x +\u00a00.09z<\/p>\n =0.08 $\\times$ 25900 + 2700 = 2072 + 2700 = 4772<\/p>\n Total salaries of A, B and C =\u00a025900 +\u00a025900 +30000\u00a0= 81800<\/p>\n Required percentage = $\\frac{4772}{81800} \\times 100 = 5.8%<\/p>\n Question 8:\u00a0<\/b>Sonu saves 15% of her income. If her income increases by 20% and she still saves the same amount as before, then what is the percentage increase in her expenditure? (correct to one decimal place)<\/p>\n a)\u00a023.5<\/p>\n b)\u00a022.8<\/p>\n c)\u00a023.8<\/p>\n d)\u00a0242<\/p>\n 8)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the initially income of Sonu be Rs.100.<\/p>\n saving = 15%<\/p>\n Expenditure = 100 – 15 = 85%<\/p>\n Saving = 100 -85 = Rs.15<\/p>\n Expenditure = 85% of 100 = Rs.85<\/p>\n Income after increment = 100 $\\times \\frac{120}{100}$ = Rs.120<\/p>\n Saving = 15<\/p>\n Expenditure = 120 – 15 = 105<\/p>\n Increment in expenditure = 105 – 85 = Rs.20<\/p>\n Percentage\u00a0Increment in expenditure = $\\frac{20}{85} \\times$ 100 = 23.52%<\/p>\n Question 9:\u00a0<\/b>A is 75% less than B and C is 75% of the difference between A and B. C is what percentage more than A?<\/p>\n a)\u00a0125<\/p>\n b)\u00a0100<\/p>\n c)\u00a075<\/p>\n d)\u00a090<\/p>\n 9)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n As per the question,<\/p>\n Let B is = 100<\/p>\n So, as per the condition, A is 75% less than B, so A=25 Question 10:\u00a0<\/b>In an office, 70% of the total number of employees are females. 80% of the total number of employees, including 85 males, got promotion.If there are 105 female employees, then what percentage of female employees got promotion?<\/p>\n a)\u00a040%<\/p>\n b)\u00a0$33\\frac{1}{3}\\%$<\/p>\n c)\u00a035%<\/p>\n d)\u00a030%<\/p>\n 10)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n 105 female employee which is 70% of the total employee<\/p>\n so the total number of employee =$ 105 \\times \\dfrac{10}{7}$<\/p>\n $\\Rightarrow 150 $<\/p>\n 80% employee got a promotion so number of females who got a promotion = 120-85=35<\/p>\n so the %\u00a0 of female employee got is\u00a0 $\\dfrac{35}{105}\\times 100 $<\/p>\n $\\Rightarrow 33 \\dfrac{1}{3}$% Ans<\/p>\n Question 11:\u00a0<\/b>If A\u2019s income is 40% of B\u2019s income and B\u2019s income is 24% more than C\u2019s income, then by what percentage is C\u2019s\u00a0income more than A\u2019s income? (Your answer should be correct to one decimal place.)<\/p>\n a)\u00a075.6<\/p>\n b)\u00a0101.6<\/p>\n c)\u00a0104.2<\/p>\n d)\u00a050.4<\/p>\n 11)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let C income = 100<\/p>\n B’s income is 24% more than C<\/p>\n then B = 124<\/p>\n now A’s income is 40% of B income<\/p>\n so A income\u00a0 = $ 124 \\times \\dfrac{40}{100} $<\/p>\n $\\Rightarrow 49.6 $<\/p>\n so C’s income is 100- 49.6 = 50.4 (more than A)<\/p>\n then according to question $\\dfrac{50.4}{49.6}\\times 100 $<\/p>\n $\\Rightarrow 101.6 $Ans<\/p>\n Question 12:\u00a0<\/b>If A is 48% more than B and C is 60%less than the sum of A and B, then A is what percentage more than C? (Correct to one decimalplace.)<\/p>\n a)\u00a050.8<\/p>\n b)\u00a049.2<\/p>\n c)\u00a050.2<\/p>\n d)\u00a049.8<\/p>\n 12)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let B = 100 $x$<\/p>\n Here A =$ 148 x$<\/p>\n C = 60% less than A+B<\/p>\n = $\\dfrac{40}{100} \\times 248x $<\/p>\n $\\Rightarrow \\dfrac{992}{100}\u00a0\u00a0x $<\/p>\n $\\Rightarrow 99.2x $<\/p>\n Now A is more than C = $148x – 99.2x $<\/p>\n $\\Rightarrow 48.8x$<\/p>\n % more than C = $\\dfrac{48.2x}{99.2x}\\times 100$<\/p>\n $\\Rightarrow 49.19 $ %<\/p>\n $\\Rightarrow 49.2$ % Ans<\/p>\n Question 13:\u00a0<\/b>The average weight of the boys in a class is 69.3 kg and that of the girls in the same class is 59.4 kg. If the average weight of all the boys and girls in the class is 63.8 kg, then the percentage of the number of boys in the class is:<\/p>\n a)\u00a0$55\\frac{5}{9}$<\/p>\n b)\u00a040<\/p>\n c)\u00a045<\/p>\n d)\u00a0$44\\frac{4}{9}$<\/p>\n 13)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let B be the number of Boys, G be number of girls<\/p>\n $ \\dfrac {total\u00a0 \u00a0weight\u00a0\u00a0 of\u00a0\u00a0 boys}\u00a0\u00a0{number\u00a0 of\u00a0 \u00a0boys} $= Any weight of boys<\/p>\n total weight of boys = (69.3)B<\/p>\n Similarly total weight of girls = (59.4) G<\/p>\n total weight of class = 69.3 B + 59.4 G<\/p>\n Any weight of class = $\\dfrac{total\u00a0 \u00a0weight\u00a0 \u00a0of\u00a0\u00a0 class}{total\u00a0 \u00a0Boys\u00a0 + Girls} $<\/p>\n $\\Rightarrow 63.8 = \\dfrac{ 63.3 B + 59.4 G} {B + G} $<\/p>\n $\\Rightarrow 63.8 B + 63.8 G = 69.3 B + 59.4 G $<\/p>\n $\\Rightarrow 4.4 G = 5.5 B $<\/p>\n $\\Rightarrow G = \\dfrac {5}{4} B $<\/p>\n % of Boys in class = $\\dfrac{B}{B+G} \\times 100 $<\/p>\n $\\Rightarrow \\dfrac{B}{B + \\dfrac{5}{4}B} \\times 100 $<\/p>\n $\\Rightarrow \\dfrac{B}{B+ \\dfrac{9}{4}B} \\times 100 $<\/p>\n $\\Rightarrow \\dfrac{4}{9} \\times 100 $<\/p>\n $\\Rightarrow \\dfrac {400}{9} $<\/p>\n $\\Rightarrow 44 \\dfrac{4}{9} $ Ans<\/p>\n Question 14:\u00a0<\/b>In a school, 60% of the numberof students are boys and the rest are girls. If 20% of the number of boys failed and 65% of the number of girls passed the examination, then the percentage of the total number of students who passed is:<\/p>\n a)\u00a068<\/p>\n b)\u00a072<\/p>\n c)\u00a074<\/p>\n d)\u00a078<\/p>\n 14)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let total number of students in school = 100<\/p>\n => Number of boys = 60 and number of girls = 40<\/p>\n Boys who passed = $\\frac{80}{100}\\times60=48$<\/p>\n Girls who passed = $\\frac{65}{100}\\times40=26$<\/p>\n $\\therefore$ Total percentage of students passed = $48+26=74\\%$<\/p>\n => Ans – (C)<\/p>\n Question 15:\u00a0<\/b>A man spends $\\frac{2}{3}rd$ of his income. If his income increases by 14% and the expenditure increases by 20%, then the\u00a0percentage increase in his savings will be<\/p>\n a)\u00a01%<\/p>\n b)\u00a02%<\/p>\n c)\u00a04%<\/p>\n d)\u00a06%<\/p>\n 15)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let initial income be Rs.\u00a0300<\/p>\n => Expenditure = Rs.\u00a0200 and savings = Rs. 100<\/p>\n Now, new income after 14% increase = $\\frac{114}{100}\\times300=Rs.$ $342$<\/p>\n Similarly, new expenditure = Rs. 240<\/p>\n => New savings = Rs. $(342-240)=Rs.$ $102$<\/p>\n $\\therefore$ Increase in savings = $\\frac{(102-100)}{100}\\times100=2\\%$<\/p>\n => Ans – (B)<\/p>\n Question 16:\u00a0<\/b>A is 40% less than B, and C is 40% of the sum of A and B. The difference between A and B is what percentage of C?<\/p>\n a)\u00a060.5%<\/p>\n b)\u00a064%<\/p>\n c)\u00a062.5%<\/p>\n d)\u00a060%<\/p>\n 16)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let $B = 10$, => $A = 6$<\/p>\n => $C = \\frac{40}{100}\\times(10+6)=6.4$<\/p>\n $\\therefore$ Required % = $\\frac{(B-A)}{C}\\times100$<\/p>\n = $\\frac{4}{6.4}\\times100=62.5\\%$<\/p>\n => Ans – (C)<\/p>\n Question 17:\u00a0<\/b>The monthly salary of a person was \u20b950,000. He used to spend on three heads- personal and family expenses (E), taxes (T), philanthropy (P), and rest were his savings. E was 50% of the income, T was 20% of E and P was 15% of T. When his salary got raised by 40%, he maintained the percentage level of E, but T became 30% of E and P became 20% of T. By whatpercentage is the new savings more or less than the earlier savings? (correct up to one decimal place)<\/p>\n a)\u00a016.4% more<\/p>\n b)\u00a08.2% more<\/p>\n c)\u00a016.4% less<\/p>\n d)\u00a08.2% less<\/p>\n 17)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Initial salary of person = Rs. 50,000<\/p>\n Total % expenditure = $50+(\\frac{20}{100}\\times50)+(\\frac{15}{100}\\times\\frac{20}{100}\\times50)$<\/p>\n =\u00a0$50+10+1.5=61.5\\%$<\/p>\n => % savings initially = $100-61.5=38.5\\%$<\/p>\n Total savings = $\\frac{38.5}{100}\\times50,000=Rs.$ $19,250$<\/p>\n Now, new salary = Rs. 70,000<\/p>\n Similarly, total % expenditure =\u00a0$50+(\\frac{30}{100}\\times50)+(\\frac{20}{100}\\times\\frac{30}{100}\\times50)$<\/p>\n = $50+15+3=68\\%$<\/p>\n => % savings initially = $100-68=32\\%$<\/p>\n Total savings = $\\frac{32}{100}\\times70,000=Rs.$ $22,400$<\/p>\n $\\therefore$ New savings are\u00a0more<\/strong> by = $\\frac{22400-19250}{19250}\\times100\\approx16.4\\%$<\/p>\n => Ans – (A)<\/p>\n Question 18:\u00a0<\/b>If the word PHOTOGRAPH is spelt with \u2018F\u2019 in place of \u2018PH\u2019, then what would bethe percentage reduction in the number of letters?<\/p>\n a)\u00a025%<\/p>\n b)\u00a010%<\/p>\n c)\u00a020%<\/p>\n d)\u00a018%<\/p>\n 18)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Number of letters in PHOTOGRAPH = 10<\/p>\n Number of letters in FOTOGRAF = 8<\/p>\n => % reduction = $\\frac{10-8}{10}\\times100=20\\%$<\/p>\n => Ans – (C)<\/p>\n Question 19:\u00a0<\/b>If 25% of half of x is equal to 2.5 times the value of 30% of one-fourth of y. then x is what percent more or less than y?<\/p>\n a)\u00a0$33\\frac{1}{3}\\%$ more<\/p>\n b)\u00a050% more<\/p>\n c)\u00a0$33\\frac{1}{3}\\%$ less<\/p>\n d)\u00a050% less<\/p>\n 19)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n According to question,<\/p>\n x $\\times \\frac{1}{2} \\times \\frac{25}{100} =\u00a0y\u00a0\\times 2.5 \\times \\frac{1}{4} \\times \\frac{30}{100}$<\/p>\n $\\Rightarrow \\frac{x}{8} =\u00a0\\frac{3y}{40} \\times 2.5$<\/p>\n x = $\\frac{3y}{2}$<\/p>\n x =\u00a0$\\frac{3y}{2} \\times 100$ = 150% of y<\/p>\n x is 50% more than y.<\/p>\n Question 20:\u00a0<\/b>If the price of petrol increases by 19%, and Sunitha intends to spend only an additional 12% on petrol, by what percent should she reduce the quantity of petrol purchased (nearest to an integer)?<\/p>\n a)\u00a07<\/p>\n b)\u00a06<\/p>\n c)\u00a05<\/p>\n d)\u00a08<\/p>\n 20)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the price of petrol be Rs.100 and Sunitha spent Rs.100.
\nAs per the condition, c=$\\dfrac{(A-B)\\times 75}{100}=\\dfrac{(100-25)\\times 75}{100}=\\dfrac{225}{4}$
\nHence, the required percentage $=\\dfrac{\\dfrac{(225}{4}-25)\\times 100}{25}=\\dfrac{(225-100)\\times 100}{100}=125$<\/p>\n
\nAfter increment price of petrol = 119
\nExpenditure of petrol = 112
\nReduce quantity of petrol = $\\frac{119 – 112}{119} \\times 100 = \\frac{7}{119} \\times 100$ = 5.88 ~ 6%<\/p>\n