Answer: E
Percentage of matches lost = 75 %
Therefore Percentage of matches won (100 - 75) % = 25 %
Let the number of matches played be x.
Then 25 % of x = 45
25/100 × x =45
x = (45 × 100)/25
x = (4500)/25
x = 180
Therefore, the total number of matches played is 180.
Description : A candidate scoring 50% in an examination fails by 60 marks , while another candidate scores 75 % mark, gets 40 marks more than the minimum pass marks . Find the minimum pass mark. a) 125 b) 220 c) 140 d) 260 e) 298
Last Answer : Answer: B Let x be the maximum marks, Then (50% of x)+60 = (75% of x)-40 x/2 +60 = 3x/4 -20 60+20 = 3x/4 – x/ 2 X=320 Hence maximum marks = 320 Minimum pass marks = 320/2 + 60 = 220
Description : Nesha scored 120 out of 150 in English, 120 out of 180 in mathematics and 160 out of 200 in Science. Find Nesha’s score as percentage: (i) in Mathematics (ii) in all the three subjects (on the whole). a) 66 2/3, 78 22/51 b) 71 23/5, 45 7/31 c) 54 13/33, 82 3/4 d) 60 3/4, 76 32/41
Last Answer : Answer: A (i) Percentage scored in Mathematics = 120/180 100 = 12000/180 =1200/18 = 66 2 /3 % (ii) Total maximum of all the three subjects = 150 + 160 + 200 = 510 and Total score in the ... percentage on the whole = (400/510 100) = (40000/510) = 4000/51 = 78 22/51 %
Description : If the production of a factory grows at a 16% p.a., what will be its production for the year 2016 if its production in 2014 was 140 lakh tonnes? a) 222.734 b) 149.597 c) 188.384 d) 28 3.812 e) 180.534
Last Answer : Answer: C Required Production=140(1 + 16/100)2 lakh tones =140(1 + 4/25)2 =140((25 + 4) /25)² =140(29/25)² =140(1.16)² =140×1.3456 =188.384 lakh tonnes
Description : In final exam of class IX there are 130 students 20 % students failed. How many students passed to class X? a) 105 b) 112 c) 104 d) 117 e) 104.5
Last Answer : Answer: C) Percentage of students passed to class X = (100 % - 20 %) of 130 = 80 % of 130 =>80 % of 130 => 80/100 × 130 =>10400/100 => 104 Therefore, 104 students passed to class X
Description : A number is increased by 70 % and then decreased by 70 %. Find the net increase or decrease per cent. a) 25 b) 56 c) 45 c) 49 e) 98
Last Answer : Answer: D Let the number be 100. Increase in the number = 70 % = 40 % of 100 = (40/100 100) = 70 Therefore, increased number = 100 + 70 = 170 This number is decreased by 70 % Therefore, decrease ... = 100 - 51 = 49 Hence, net percentage decrease = (49/100 100) = (4900/100) = 49 %
Description : Two train travel in opposite directions at 36 kmph and 45 kmph and a man sitting in slower train passes the faster train in 8 seconds. Then length of the faster train is: A.120 m B.140 m C.160 m D.180 m E.None of these
Last Answer : Answer – D (180 m) Explanation – When OPP. direction-PLUS Relative speed = (36 + 45) km/hr = [81 x 5/18] m/sec = [45/2] m/sec. Length of train = [45/2 x 8] m = 180 m.
Description : Find the number which when decreased by 24 % becomes 594. a) 625.8 b) 360.9 c) 465.7 d) 781.5 e) 398.5
Last Answer : Answer: D Let the number be m. Decrease = 24 % of m = 24/100 × m = 6m /25 Therefore, decrease number = m – 6m/25 = (25m – 6m)/25 = 19m/25 According to the question 19m/25 = 594 19m = 594 × 25 19m = 14850 m = 14850/19 m =781.5
Description : By what number must the given number be multiplied to increase the number by 25 %. a) 25 b) 60 c) 50 d) None e) 98
Last Answer : Answer: A Let the number be m Increase in its value = 25 % of m = 25/100 × m = m/4 Therefore, increase value = m + m/4 = (4m + m)/4 = 5m/4 Therefore, the given number must be multiplied by 5/4 to increase the number by 25 %.
Description : In an examination, 900 students appeared. Out of these students; 56 % got first division, 27 % got second division and the remaining just passed. Assuming that no student failed; find the number of students who just passed. a) 225 b) 153 c) 245 d) 148 e) 298
Last Answer : Answer: B The number of students with first division = 56 % of 900 = 56/100 × 900 = 50400/100 = 504 And, the number of students with second division = 27 % of 900 = 27/100 × 900 =24300/100 = 243 Therefore, the number of students who just passed = 900 – (504 + 243) = 900- 747 = 153
Description : In a college of 1335 students, 60 % are boys. Find the number of girls and number of boys in the college? a) 924,221 b) 691,404 c) 802,333 d) 441,564 e) 534,801
Last Answer : Answer: E Number of boys in the college = 60 % of 1335 = 60/100 × 1335 = 80100/100 = 801 Number of girls in the college = Total number of students in the college - Number of boys = 1335 – 801 =534
Description : Find the increase value if 450 is increased by 90 %. a) 525 b) 715 c) 645 d) 795 e) 855
Last Answer : Answer: E Increase = 90 % of 450 = 90/100 × 450 = 40500/100 = 405 Therefore, increase value = 450 + 405 = 855
Description : In an exam Aashika secured 1328 marks. If she secured 32 % marks, find the maximum marks. a) 2250 b) 3600 c) 4400 d) 4150 e) 1298
Last Answer : Answer: D Let the maximum marks be x. Aashika’s marks = 32% of x Aashika secured 1328 marks Therefore, 32% of x = 1328 ⇒ 32/100 × x = 1328 x = (1328 × 100)/32 x =132800/32 x = 4150 Therefore, Aashika got 1328 marks out of 4150 marks.
Description : mouli had $ 3600 left after spending 40 % of the money he took for shopping. How much money did he take along with him? a) 2500 b) 6000 c) 4500 d) 3000 e) 9800
Last Answer : Answer: B Let the money he took for shopping be x. Money he spent = 40 % of x = 40/100 x = 4/10 x Money left with him = x - 4/10 x = (10x - 4x)/10 = 6x/10 But money left with ... 6x/10 = 3600 x = 3600 10/6 x = 36000/6 x = 6000 Therefore, the money he took for shopping is 6000.
Description : Brother's weight is 25 % more than that of sister. What percent is brother’s weight less than sister's weight? a) 25% b) 60% c) 45% d) 20% e) 98%
Last Answer : Answer: D Let sister's weight be 100 kg. Then brother's weight = (100 + 25) kg = 125 kg If brother's weight is 125 kg, then sister's weight is 100 kg. If brother's weight is 1 kg, then ... then sister's weight = (100/125 100) kg Therefore, sister's weight is 20 % less than that of brother.
Description : An alloy contains 36 % of bronze. What quantity of alloy is required to get 340 g of bronze? a) 1500 b) 944.4 c) 950 d) 1000 e) 980.5
Last Answer : Answer: B Let the quantity of alloy required = x g Then 36 % of x =340 g ⇒ 36/100 × x = 340 g x = (340 × 100)/36 g x = 34000/36 g x = 944.4 g
Description : Sanjay gets 72 % marks in examinations. If these are 864 marks, find the maximum marks. a) 1050 b) 860 c) 1225 d) 1200 e) 1500
Last Answer : Answer: D Let the maximum marks be s Then 72 % of s = 864 72/100 × s = 864 s = (846 × 100)/72 s = 86400/92 s = 1200 Therefore, maximum marks in the examinations are 1200.
Description : One sixth of half of three fourth of a number is 25.What will be 30% of that number? a) 800 b) 120 c) 340 d) 300
Last Answer : Answer: B (1/6) ×(1/2)×(3/4)× x = 25 x/16 =25 x = 25×16 x = 400 30 % of 400 = 30/100 ×400 = 12000/100 = 120
Description : In a basket of eggs, 20% of them are rotten and 68 are in good condition. Find the total number of eggs in the basket. a) 85 b) 60 c) 75 d) 40 e) 98
Last Answer : Answer: A Let the total number of eggs in the basket be x 20 % of the eggs are rotten, and eggs in good condition are 68 Therefore, according to the question, 80% of x = 68 80/100 × x = 68 x = (68 × 100)/80 x = 6800 / 80 x = 85 Therefore, total number of eggs in the basket is 85.
Description : In an election three candidates received 2272 , 15272, and 23256 votes respectively. What percentage of the total votes did the winning candidate got? a) 25 b) 41 c) 75 d) 57 e) 93
Last Answer : Answer: D Total number of votes polled = (2272+15272+23256) =40800 Required percentage = 23256/40800 ×100 = 2325600/40800 = 57
Description : Mithun went to a shop and bought things worth Rs. 75, out of which 90 Paise went on sales tax on taxable purchases. If the tax rate was 18%, then what was the cost of the tax free items? a) 12.5 b) 69.80 c) 19.7 d) 34.5 e) 69.1
Last Answer : Answer: E Total cost of the items he purchased = Rs.75 Given that out of this Rs.75, 90 Paise is given as tax => Total tax incurred = 90 Paise = Rs.90/100 Let the cost of the tax free items = x Given that tax rate = ... −0.9 −x) = 90 ⇒ (75 − 0.9 − x) = 5 x = 75 − 0.9 - 5 X=69.1
Description : In India, which one of the following should be considered the right combination of the age in days and live body weight in kg for a lamb for its weaning ? (A) 20 days to 30 days and 6 kg to 7 kg (B) 45 days to 60 ... (D) 75 days to 90 days and 12 kg to 15 kg (E) 55 days to 70 days and 9 kg to 11 kg
Last Answer : (C) 105 days to 120 days and 18 kg to 21 kg
Description : The cost of a stencil is decreased by 30%. If the original cost is $140, find the decrease cost. a) 25 b) 68 c) 45 d) None e) 98
Last Answer : Answer: E Original cost = $140 Decrease in it = 30% of $140 = 30/100 × 140 = 4200/100 = $42 Therefore, decrease cost = $140 - $42 = $98
Description : The price of dhal is increased from $ 30 to $ 37.5 per kg. Find the percentage increase in price. a) 25 b) 60 c) 45 d) None e) 98
Last Answer : Answer: A Price of dhal before = $30 Price of dhal now = $ 37.5 Increase in dhal = current price - original price = $37.5 - $ 30 = $ 7.5 Therefore, percentage increase in price = Increase in price/ ... = 7.5 / 30 100 = 750/100 = 25 % Thus, increase in price= 25 %
Description : On a test consisting of 500 questions, Dhivya answered 80% of the first 250 questions correctly. What percent of the other 250 questions does she need to answer correctly for her grade on the entire exam to be 60% ? a) 20 b) 60 c) 45 d) 80 e) 40
Last Answer : Answer: E 60% of 500 = 300 80% of 250 = 200 No. of correct answers in remaining 250 questions = 300 – 200 = 100 Percentage = 100 x 100/ 250 = 40%
Description : Two students appeared at an examination. One of them secured 18marks more than the other and his marks was 72% of the sum of their marks. What are the marks obtained by them? a) 12.5,23.3 b) 26.7,16.0 c) 13.3,14.2 d) 11.45, 29.45 e) 29.8,15.4
Last Answer : Answer: D Let the marks secured by them be x and (x + 18) Then sum of their marks = x + (x + 18) = 2x + 18 Given that (x + 18) was 72% of the sum of their marks =>(x+18) = 72/100(2x+18) => ... 11x = 126 x = 11.45 Then (x + 18) = 11.45 + 18 = 29.45 Hence their marks are 11.45 and 29.45
Description : A mixture of 90 kg of rava and sugar contains 90% of sugar. The new mixture is formed by adding 30 kg of sugar. What is the percentage of sugar in the new mixture? a) 45 1/2 b) 60 1/5 c) 45 1/2 d) 92 1/2 e) 98 1/2
Last Answer : Answer: D 90 kg = 81 kg sugar 9 kg rava Adding = 30 kg sugar Total = 111kg sugar 9kg rava (new) 111/120 × 100 = 92.5 =92 1 /2
Description : A mixture of 60 kg of rice and dhal contains 60% of dhal. The new mixture is formed by adding 15 kg of dhal. What is the percentage of rice in the new mixture? a) 25 b) 64 c) 45 d) 32 e) 98
Last Answer : Answer: D 60 kg = 36 kg dhal 24 kg rice -------- 15 kg dhal Total = 15 kg dhal 24 kg rice (new) 24/75×100=32%
Description : What percent of a day in 12 hours? a) 25 b) 60 c) 45 d) 20 e) 50
Last Answer : Answer: E Total hours in a day = 24 Required percent = 12/24 ×100 = 50%
Description : Gugan obtained a total of 1313 marks out of 1400 in an examination. What is his approximate percentage in the examination? a) 25 b) 60 c) 45 d) None e) 98
Last Answer : Answer: D Required percentage = 1313/ 1400 × 100 =131300/1400 =93.7
Description : In a plot of 18000 sq. m., only 13500 sq. m. is allowed for construction. What percent of the plot is to be left without construction? a) 25 b) 60 c) 45 d) 40 e) 98
Last Answer : Answer: A Percentage of plot allowed for construction = (13500/18000 × 100) = 75 %. Thus, the percentage of plot to be left without construction = 100 % - 75 % = 25 %.
Description : In a certain cricket tournament 45 matches were played. Each team played once against each of the other teams. The number of teams participated in the tournament is (A) 8 (B) 10 (C) 12 (D) 14
Last Answer : Answer: B
Description : If the numerator of a fraction is increased by 400% and the denominator of the fraction is increased by 300%, the resultant fraction is 18/35 . What is the original fraction? a) 72/ 175 b) 87/ 123 c) 56/232 d) 45/241
Last Answer : Answer: A Fraction is x/y (x + 400/100 x ) / (y + 300/100 y) = 18/35 (500/100) x(35) = (400/100)y (18) (5x) × 35 = 18 × 4 y 175 x = 72 y x/y = 72/175
Description : A school cricket team played 20 matches against another school. The first school won 25% of them. How many matches did the first school win?
Last Answer : over three to four days
Description : For batching 1:3:6 concrete mix by volume, the ingredients required per bag of 50 kg cement, are: (A) 70 litres of sand and 120 litres of aggregates (B) 70 kg of sand and 140 litres of ... C) 105 litres of sand and 140 litres of aggregates (D) 105 litres of sand and 210 litres of aggregates
Last Answer : Answer: Option D
Description : Q.61. CA Co manufactures a single product and has drawn up the following flexed budget for the year. 60% 70% 80% Rs Rs Rs Direct materials 120,000 140,000 160,000 Direct labour 90,000 105,000 120,000 Production ... activity? (a) Rs 330,300 (b) Rs 370,300 (c) Rs 373,300 (d) Rs 377,300
Last Answer : (b) Rs 370,300
Description : 24+(72 ÷ 48×8+14) × 36 ÷ (3×4) + 36 ÷ 3 ×4 a) 120 b) 140 c) 160 d) 180 e) None of these
Last Answer : 72÷48×8+14 = (72/48) × 8 + 14 = (3/2) × 8 + 14 = 12 +14 = 26 24 + (72 ÷ 48 × 8 + 14) ×36 ÷ (3 × 4) + 36 ÷ 3 × 4 = 24 + 26 × 36÷12 + 36 ÷ 3 × 4 = 24 + 26 × 3 + 12 × 4 = 24 + 78 + 48 = 150 Answer: e)
Description : The missing number in the series 40, 120, 60, 180, 90, ?, 135 is (A) 110 (B) 270 (C) 105 (D) 210
Description : The normal cholesterol level in human blood is - (1) 80 - 120 mg% (2) 120 - 140 mg% (3) 140 - 180 mg% (4) 180 - 200 mg%
Last Answer : (4) 180 - 200 mg% Explanation: Cholesterol is defined as a waxy alcohol, fat-like substance that occurs naturally in all areas of the human body. 100-200 mg/dL is considered as normal for ... as the ratio of LDL to HDL cholesterol is considered a more important factor in prediction of disease risk.
Description : The subcritical steam generators operate between a pressure ranges of? (rra) 120-160 bar (rrb) 100-110 bar ( c) 130-180 bar (rrd) 140- 200 bar
Last Answer : ( c) 130-180 bar
Description : The normal cholesterol level in human blood is (1) 80 – 120 mg% (2) 120 – 140 mg% (3) 140 – 180 mg% (4) 180 – 200 mg%
Last Answer : 180 – 200 mg%
Description : The maximum hourly consumption, is generally taken as (A) 110 % (B) 120 % (C) 140 % (D) 150 %
Last Answer : (D) 150 %
Description : In a college election between two candidates, one candidate got 25% of the total valid votes. 30% of the votes were invalid. If the total votes were 7600, what is the number of valid votes the other candidate got ? a) 2576 b) 2314 c) 5184 d) 3990 e) 9814
Last Answer : Answer: D Total valid votes = 70% of 7600 = 5320 Number of valid votes to other candidate = 75% of 5320 = 75/100 *5320 = 399000/100 = 3990
Description : In an election, candidate A got 40% of the total valid votes. If 55% of the total votes were declared invalid and the total numbers of votes is 280000, find the number of valid vote polled in favour of candidate. a) 32500 b) 56000 c) 35700 d) 50400 e) 29800
Last Answer : Answer: D Total number of invalid votes = 55 % of 280000 = 55/100 280000 = 15400000/100 = 154000 Total number of valid votes 280000 - 154000 = 126000 Percentage of votes polled in favour of candidate ... favour of candidate A = 40 % of 126000 = 40/100 126000 = 5040000/100 = 50400
Description : In an election between two candidates, one got 45% of the total valid votes, 40% of the votes were invalid. If the total number of votes was 15000, the number of valid votes that the other candidate got, was a) 2500 b) 3600 c) 4500 d) 4950 e) 9800
Last Answer : Answer: D Total number of votes = 15000 Given that 40% of Percentage votes were invalid => Valid votes = 60% Total valid votes = 15000* 60/100 1st candidate got 45% of the total valid votes. Hence ... votes that 2nd candidate got = total valid votes x 55/100 =15000*60/100*55/100 =4950
Description : A number is reduced by 10%. Its present value is 1080. What was its original value? a) 3250 b) 2600 c) 1200 d) 2300 e) 1980
Last Answer : Answer: C Original value is percentage = 100 %. Reduce amount in percentage = 10 % Therefore, Percent value in percentage = 100 % - 10 % = 90 %. According to the problem, 90 % of original value ... Therefore, 100 % of original value = 1080/100 90 = 1200. Thus, the original value was 1200.
Description : A number 168 was misread as 24. Find the reading error in per cent. a) 25.7 b) 42.8 c) 42.6 d) 85.7 e) 98.6
Last Answer : Answer: D Error = 168 – 24 = 144 Therefore, % error = 144/168 [Since, we know decrease% = decrease in value/original value × 100 %] = 144/168 × 100 = 85.7 %
Description : In a competitive examination in Pondicherry, 18% candidates got selected from the total appeared candidates. Tamilnadu and Pondicherry had an equal number of candidates appeared and in Tamilnadu 21% candidates got selected ... appeared from each State? a) 25000 b) 84000 c) 24000 d) 700000 e) 9800
Last Answer : Answer: C Pondichery and Tamilnadu had an equal number of candidates appeared In Pondicherry, 18% candidates got selected from the total appeared candidates In tamilnadu, 21% candidates ... > total appeared candidates in pondicherry = total appeared candidates in tamilnadu = 24000
Description : The population of a town is 378000. It decreases by 16% in the 1st year and increases by 10% in the 2nd year. What is the population in the town at the end of 2 years? a) 182574 b) 482576 c) 282674 d) 283574 e) 349272
Last Answer : Answer: E After 2 years required population is = 378000 (1-16/100) (1+10/100) = 378000 (84/100) (110/100) = 378000 (0.84) (1.1) = 3,49,272
Description : The population of a town was 200000 three years ago. If increased by 4%, 6% and 10% respectively in the last three years, then the present population of town is a) 1,10,313 b) 1,10,314 c) 2,42,528 d) 2,93,313 e) 2,10,313
Last Answer : Answer: C The present population = (1+4/100)×(1+6/100)×(1+10/100)×200000 = (104/100) × (106/100) × (110/100) × 200000 = 1.04×1.06×1.1×200000 = 2,42,528
Description : A wooden manufacturing company declares that a wooden is now available for $11200 as against $25200 one year before. Find the percentage reduction in the price of wooden offered by the company. a) 25 1/3 b) 691/3 c) 66 1/3 d) 33 1/3 e) 55 5/9
Last Answer : Answer: E Price of the wooden a year before = $25200 Price of the wooden after a year = $11200 Decrease in price = $(25200 - 11200) = $14000 Therefore, decrease % = 14000/25200 × 100 % = 55 5/9 %