In an examination, x and y secured 732 and 864 marks respectively. If y secured `72%` marks then find the percentage of marks secured by x.

In an examination, x and y secured 732 and 864 marks respectively. If y secured `72%` marks then find the percentage of marks secured by x.
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In an examination, x and y secured 732 and 864 marks respectively. If y secured `72%` marks then find the percentage of marks secured by x.
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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

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

In an examination, Salim secured `52%` marks and got 114 marks more than the pass marks and A zam secured `39%` marks and got 36 marks more than the p

Description : In an examination, Salim secured `52%` marks and got 114 marks more than the pass marks and A zam secured `39%` marks and got 36 marks more than the p

Last Answer : In an examination, Salim secured `52%` marks and got 114 marks more than the pass marks and A zam ... (a) The pass mark. (b) The pass percentage.

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

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.

In an examination of a certain class, at least `70%` of the students failed in Physics, at least `72%` failed in Chemistry, at least `80%` failed in M

Description : In an examination of a certain class, at least `70%` of the students failed in Physics, at least `72%` failed in Chemistry, at least `80%` failed in M

Last Answer : In an examination of a certain class, at least `70%` of the students failed in Physics, at ... 15%` D. Cannot be determined due to insufficient data

Rupesh secures 495 marks out of 750 in his annual examination. Find the percentage of marks obtained by him. -Maths

Description : Rupesh secures 495 marks out of 750 in his annual examination. Find the percentage of marks obtained by him. -Maths

Last Answer : answer:

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

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

The cow’s milk contains how much amount of water in terms of percentage? A. 60% B. 65% C. 72% D. 80% (Answer)

Description : The cow’s milk contains how much amount of water in terms of percentage? A. 60% B. 65% C. 72% D. 80% (Answer)

Last Answer : D. 80% (Answer)

Consider efficiencies of hydro-turbines and generators. If the efficiency of a hydro-turbine is 80% and the efficiency of a generator is 90%, what is the overall efficiency of the unit? w) 85% x) 72% y) 90% z) 70%

Description : Consider efficiencies of hydro-turbines and generators. If the efficiency of a hydro-turbine is 80% and the efficiency of a generator is 90%, what is the overall efficiency of the unit? w) 85% x) 72% y) 90% z) 70% 

Last Answer : ANSWER: X -- 72%

A group of students of arithmetic mean of the marks in a test was 63. The brightest 30% of them secured a mean score of 70 and the dullest 15% a mean score of 41. The mean score of remaining 55 % is A) 63.675 B) 61.785 C) 65.181 D) 66.67

Description : A group of students of arithmetic mean of the marks in a test was 63. The brightest 30% of them secured a mean score of 70 and the dullest 15% a mean score of 41. The mean score of remaining 55 % is A) 63.675 B) 61.785 C) 65.181 D) 66.67

Last Answer : C) Let the requird mean score be ‘x’ Then (30*70)+(15*41)+55x = 63*100 2100+615+55x=6300 2715+55x=6300 55x=3585 X=65.181

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

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.

In an exam the average marks obtained by a candidate is 82 per paper. If he had obtained 32 marks more in science paper & 28more marks in social paper, then his average per paper is increased by 15 marks. Then how many papers when there in examination? A) 10 B) 6 C) 4 D) 8

Description : In an exam the average marks obtained by a candidate is 82 per paper. If he had obtained 32 marks more in science paper & 28more marks in social paper, then his average per paper is increased by 15 marks. Then how many papers when there in examination? A) 10 B) 6 C) 4 D) 8 

Last Answer : C)  let the number of paper be x total mark earned him = 82x then,, 82x+32+28= 97x 15x=60 x= 4 =number of subjects

Saravanan has scored an total of 90% in an examination with five subjects in the ratio 18:14:15:21:22. If 85% is the marks required to get an ‘A’ grade in each subject, then find in how many subjects did he get an ‘A’ grade. Given that the maximum marks in each subject is 120. A) 2 B) 3 C) 4 D) 5

Description : Saravanan has scored an total of 90% in an examination with five subjects in the ratio 18:14:15:21:22. If 85% is the marks required to get an ‘A’ grade in each subject, then find in how many subjects did he get an ‘A’ grade. Given that the maximum marks in each subject is 120. A) 2 B) 3 C) 4 D) 5 

Last Answer : Answer: B) Maximum total marks = 5 120 = 600 Marks scored by Saravanan = 90/100*600=540 Let the marks scored by him in the 5 subjects be 18x,14x,15x,21x and 22x respectively. 18x + 14x + 15x + 21x + ... A' grade i.e. 0.85 120 = 102 Therefore, he has got an A' grade in 3 subjects.

In a post graduate examination the marks obtained by a student is 75 per paper. If he had obtained 33 marks more in Evs paper & 27 more marks in science paper, then his average per paper is increased by 3 marks. Then how many papers were there in exam? A) 10 B) 12 C) 14 D) 20

Description : In a post graduate examination the marks obtained by a student is 75 per paper. If he had obtained 33 marks more in Evs paper & 27 more marks in science paper, then his average per paper is increased by 3 marks. Then how many papers were there in exam? A) 10 B) 12 C) 14 D) 20

Last Answer : Answer: D)  Let the number of paper be A. Then total marks earned him = 75A from questions, 75A + 33+ 27= 78A 3A = 60=> A=20 = number of subjects

In a test, minimum passing percentage for girls and boys is 60% and 25% respectively. A boy scored 560 marks and failed by 160 marks. How many more marks did a girl require to pass in the test if she scored 216 marks ? a) 2123 b) 1512 c) 2251 d) 1325 e) 3989

Description :  In a test, minimum passing percentage for girls and boys is 60% and 25% respectively. A boy scored 560 marks and failed by 160 marks. How many more marks did a girl require to pass in the test if she scored 216 marks ? a) 2123 b) 1512 c) 2251 d) 1325 e) 3989

Last Answer : Answer: B  Total marks in the test = (560 + 160) × 100 /25  = 720 ×100/25  = 72000/25  = 2880  Passing marks for girls =2880 × 60/100  = 1728  Required marks = 1728 – 216  = 1512

What is the strategy to get higher marks in the examination book ?

Description : What is the strategy to get higher marks in the examination book ?

Last Answer : The writing must be beautiful. Must be clearly written. You have to keep a certain distance from one line to another. What is asked in the question should be written in a short order, that ... point should be noted. Try to give something from yourself without writing the language of the whole book.

The marks obtained by 15 students in an examination are given below : 40, 20, 24, 19,20, 35, 12, 48, 29, 40, 45, 48, 42, 23, 35. Find the average mark

Description : The marks obtained by 15 students in an examination are given below : 40, 20, 24, 19,20, 35, 12, 48, 29, 40, 45, 48, 42, 23, 35. Find the average mark

Last Answer : The marks obtained by 15 students in an examination are given below : 40, 20, 24, 19,20, 35, 12, ... , 23, 35. Find the average mark of the students.

In a monthly test, 10 students were awarded marks in a Mathematics examination as follows: 23, 25, 15, 20, 17, 10, 24, 15, 19 If a student is selected

Description : In a monthly test, 10 students were awarded marks in a Mathematics examination as follows: 23, 25, 15, 20, 17, 10, 24, 15, 19 If a student is selected

Last Answer : In a monthly test, 10 students were awarded marks in a Mathematics examination as follows: 23, 25 ... the probability that he gets more than 18 marks?

The Average of marks obtained by 120 candidates in a certain examination is 35. If the average marks of passed candidates is 39 and that of failed candidates is 15, what is the number of candidates who passed the examination? A.90 B.100 C.108 D.115 E.None of these

Description : The Average of marks obtained by 120 candidates in a certain examination is 35. If the average marks of passed candidates is 39 and that of failed candidates is 15, what is the number of candidates who passed the examination? A.90 B.100 C.108 D.115 E.None of these

Last Answer : Answer – B (100) Explanation – Let the number of passed candidates be a Then total marks =>120 x 35 = 39 a + (120 – a) x 15 4200 = 39 a + 1800 – 15 a a = 100

In a Examination a candidate has to score minimum of 24 marks inorder to clear the exam. The maximum that he can score is 40 marks. Identify the Valid Equivalance values if the student clears the exam. A. 22,23,26 B. 21,39,40 C. 29,30,31 D. 0,15,22

Description : In a Examination a candidate has to score minimum of 24 marks inorder to clear the exam. The maximum that he can score is 40 marks. Identify the Valid Equivalance values if the student clears the exam. A. 22,23,26 B. 21,39,40 C. 29,30,31 D. 0,15,22

Last Answer : C. 29,30,31

Which of the following indicates evaluation? Options: A) Ram got 45 marks out of 200 B) Mohan got 38 percent marks in English C) Shyam got First Division in final examination D) All the above

Description : Which of the following indicates evaluation? Options: A) Ram got 45 marks out of 200 B) Mohan got 38 percent marks in English C) Shyam got First Division in final examination D) All the above

Last Answer : D) All the above

What are the characteristics of Continuous and Comprehensive Evaluation ? (a) It increases the workload on students by taking multiple tests. (b) It replaces marks with grades. (c) It evaluates every aspect of the student. (d) It helps in reducing examination phobia. Select the correct answer from the codes given below: (A) (a), (b), (c) and (d) (B) (b) and (d) (C) (a), (b) and (c) (D) (b), (c) and (d)

Description : What are the characteristics of Continuous and Comprehensive Evaluation ? (a) It increases the workload on students by taking multiple tests. (b) It replaces marks with grades. (c) It evaluates every aspect of the student. (d) It helps in ... ) and (d) (C) (a), (b) and (c) (D) (b), (c) and (d)

Last Answer : Answer: D

Which of the following indicates evaluation ? (A) Ram got 45 marks out of 200 (B) Mohan got 38 percent marks in English (C) Shyam got First Division in final examination (D) All the above

Description : Which of the following indicates evaluation ? (A) Ram got 45 marks out of 200 (B) Mohan got 38 percent marks in English (C) Shyam got First Division in final examination (D) All the above 

Last Answer : (D) All the above 

The average of marks obtained by 60 students in a computer examination is 18. If the average marks of passed students is 20and that of the failed students is 8, what is the number of students who passed the examination? A) 100 B) 75 C) 50 D) 85

Description : The average of marks obtained by 60 students in a computer examination is 18. If the average marks of passed students is 20and that of the failed students is 8, what is the number of students who passed the examination? A) 100 B) 75 C) 50 D) 85 

Last Answer : C Let the number of passed students be x. Then total marks = 60 × 18= 20x + (60– x) × 8 1080= 20x + 480– 8x 12x = 600 ∴ x = 50 ∴ number of passed students = 50

In an examination mahi scores 64% of marks, nitesh scores 52% of marks and ritesh scores 48% of marks. The maximum mark of the exam is a three digit number, whose sum is 10 and the middle digit is equal to the sum of the other two digits. The number will decreased by 297, if its digit are reversed. approximately what is the average mark obtained by mahi, nitesh and ritesh? A) 247 B) 248 C) 264 D) 284

Description : In an examination mahi scores 64% of marks, nitesh scores 52% of marks and ritesh scores 48% of marks. The maximum mark of the exam is a three digit number, whose sum is 10 and the middle digit is equal to ... what is the average mark obtained by mahi, nitesh and ritesh? A) 247 B) 248 C) 264 D) 284

Last Answer : A) Maximum mark consist of a three digit number let's consider Unit digit place is Z,ten's place digit is Y and hundred's place digit is X. According to the question, Y=x+z---------1 X+Y ... Total number of Mahi, Nitesh and Ritesh =164/100 451=739.61 Required average =739.64/3=246.54~247.

In an competitive examination, the average was found to be 72 marks. After detecting the errors the marks of 94 candidates had to be charged from 90 to 66 each, and the average is reduced to 64 marks. Find the total number of candidates who took the exam. A) 282 B) 382 C) 828 D) 200

Description : In an competitive examination, the average was found to be 72 marks. After detecting the errors the marks of 94 candidates had to be charged from 90 to 66 each, and the average is reduced to 64 marks. Find the total number of candidates who took the exam. A) 282 B) 382 C) 828 D) 200 

Last Answer :  A)  let the number of candidates be A total marks of candidates = 72A after detecting error the change of marks for one candidate = 90 - 66 = 24 marks change of marks for 94 candidates = 94*24 = ... after detecting error of N candidates = 64A then, 72A- 2256= 64A , 8A=2256 A=282

In an examination it is required to get 672 aggregate marks to pass. A student gets 70% marks and is declared failed by 126 marks. What are the maximum aggregate marks a student can get? a) 780 b) 840 c) 741 d) 805 e) 983

Description : In an examination it is required to get 672 aggregate marks to pass. A student gets 70% marks and is declared failed by 126 marks. What are the maximum aggregate marks a student can get? a) 780 b) 840 c) 741 d) 805 e) 983

Last Answer : Answer: A Difference = 672-126 = 546 According to the question, 70% of total aggregate = 546 Total aggregate marks = 546 × 100 /70  = 54600/70  = 780

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

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

what is 864 rounded to the nearest hundred?

Description : what is 864 rounded to the nearest hundred?

Last Answer : It'll be 900

How many sets of 8 in 864?

Description : How many sets of 8 in 864?

Last Answer : 18

What is 3.864 rounded?

Description : What is 3.864 rounded?

Last Answer : The answer will depend on the degree to which the number isrounded.

How many times can 2 go into 864?

Description : How many times can 2 go into 864?

Last Answer : Exactly 432 times

How many times can 2 go into 864?

Description : How many times can 2 go into 864?

Last Answer : Exactly 432 times

A soil sample of mass specific gravity 1.92 has moisture content 30%. If the specific gravity of solids is 2.75, the void ratio, is (A) 0.858 (B) 0.860 (C) 0.862 (D) 0.864

Description : A soil sample of mass specific gravity 1.92 has moisture content 30%. If the specific gravity of solids is 2.75, the void ratio, is (A) 0.858 (B) 0.860 (C) 0.862 (D) 0.864

Last Answer : Option C

Students of two university appeared for a common test of maximum 60 marks. The average of their marks for university 1 & university 2 are 39 & 42 respectively. If the no of students of university 1 is twice the no of students of university 2. Then what is the average marks of all students of both the university? A) 40 B) 42 C) 26 D) 36

Description : Students of two university appeared for a common test of maximum 60 marks. The average of their marks for university 1 & university 2 are 39 & 42 respectively. If the no of students of university 1 is twice the no ... the average marks of all students of both the university? A) 40 B) 42 C) 26 D) 36

Last Answer : Answer: A) Let number of students of university2 be N and the no of students of university 1 be 2N the average of university1 and university 2 is 39 and 42 total marks of university 1 students and university 2 ... 78N and N*42 =42N average of both university,  =(78N+42N) /(N+2N) = 40 marks

Coordinate of â–¡ABCD is WCS are: lowermost corner A(2,2) & diagonal corner are C(8,6). W.r.t MCS. The coordinates of origin of WCS system are (5,4). If the axes of WCS are at 600 in CCW w.r.t. the axes of MCS. Find new vertices of point A in MCS. a.(4.268, 6.732) b.(5.268, 6.732) c.(4.268, 4.732) d.(6.268, 4.732)

Description : Coordinate of â- ABCD is WCS are: lowermost corner A(2,2) & diagonal corner are C(8,6). W.r.t MCS. The coordinates of origin of WCS system are (5,4). If the axes of WCS are at 600 in CCW w.r.t. the axes of MCS. Find new ... in MCS. a.(4.268, 6.732) b.(5.268, 6.732) c.(4.268, 4.732) d.(6.268, 4.732)

Last Answer : a.(4.268, 6.732)

Rationalise the denominator in each of the following and hence evaluate by taking √2 = 1.414, √3 = 1.732 and √5 = 2.236 up to three places of decimal. -Maths 9th

Description : Rationalise the denominator in each of the following and hence evaluate by taking √2 = 1.414, √3 = 1.732 and √5 = 2.236 up to three places of decimal. -Maths 9th

Last Answer : Rationalisation of denominator

Rationalise the denominator in each of the following and hence evaluate by taking √2 = 1.414, √3 = 1.732 and √5 = 2.236 up to three places of decimal. -Maths 9th

Description : Rationalise the denominator in each of the following and hence evaluate by taking √2 = 1.414, √3 = 1.732 and √5 = 2.236 up to three places of decimal. -Maths 9th

Last Answer : Rationalisation of denominator

If `sqrt(2)=1.414, sqrt(3)=1.732` and `sqrt(5)=2.236`, find the values of the following. (i) `sqrt(98)` (ii) `sqrt(300)` (iii) `sqrt(128)+sqrt(48)` (i

Description : If `sqrt(2)=1.414, sqrt(3)=1.732` and `sqrt(5)=2.236`, find the values of the following. (i) `sqrt(98)` (ii) `sqrt(300)` (iii) `sqrt(128)+sqrt(48)` (i

Last Answer : If `sqrt(2)=1.414, sqrt(3)=1.732` and `sqrt(5)=2.236`, find the values of the following. (i) ` ... `sqrt(128)+sqrt(48)` (iv) `2sqrt(75)-3sqrt(12)`

In conversion of 3-phase to 2-phase supply with T connection, the transformation ratio of the teaser will be__ times to that of main transformer: a) 1.732 b) 1.15 c) 0.866 d) Unity

Description : In conversion of 3-phase to 2-phase supply with T connection, the transformation ratio of the teaser will be__ times to that of main transformer: a) 1.732 b) 1.15 c) 0.866 d) Unity

Last Answer : Ans: C

The angle of minimum deviation of an equiangular prism, having a refractive index of 1.732 is (a) 60° (b) 30° (c) 45° (d) 0°

Description : The angle of minimum deviation of an equiangular prism, having a refractive index of 1.732 is (a) 60° (b) 30° (c) 45° (d) 0°

Last Answer : Ans:(a)

For a critical angle of 60 degree and the refractive index of the first medium is 1.732, the refractive index of the second medium is a) 1 b) 1.5 c) 2 d) 1.66

Description : For a critical angle of 60 degree and the refractive index of the first medium is 1.732, the refractive index of the second medium is a) 1 b) 1.5 c) 2 d) 1.66

Last Answer : b) 1.5

Find the ratio of refractive index of medium 2 to that of medium 1, when the Brewster angle is 60 degree. a) 0.707 b) 1.5 c) 0.866 d) 1.732

Description : Find the ratio of refractive index of medium 2 to that of medium 1, when the Brewster angle is 60 degree. a) 0.707 b) 1.5 c) 0.866 d) 1.732

Last Answer : d) 1.732

The magnitude of the conduction current density for a magnetic field intensity of a vector yi + zj + xk will be a) 1.414 b) 1.732 c) -1.414 d) -1.732

Description : The magnitude of the conduction current density for a magnetic field intensity of a vector yi + zj + xk will be a) 1.414 b) 1.732 c) -1.414 d) -1.732

Last Answer : b) 1.732

Saina spent Rs. 44,668/- on her air tickets, Rs.56,732 on buying gifts to the family members and the remaining 22% of the total amount she had as cash with her. What was the total amount ? a) Rs. 1,25,800/- b) Rs. 1,30,000/- c) Rs. 1,10,500/- d) Rs. 1,68,300/- e) Rs. 1,50,000/-

Description : Saina spent Rs. 44,668/- on her air tickets, Rs.56,732 on buying gifts to the family members and the remaining 22% of the total amount she had as cash with her. What was the total amount ? a) Rs. 1,25,800/- b) Rs. 1,30,000/- c) Rs. 1,10,500/- d) Rs. 1,68,300/- e) Rs. 1,50,000/- 

Last Answer : Saving = 22% Spent = a00 - 22) = 78% Spending money = 44668 + 56732 = 101400 78%- 101400 100% = 101400 100/78 = 130000 Answer: b)

72% of incarcerated juveniles come from a single-parent household - What is a good explanation for this?

Description : 72% of incarcerated juveniles come from a single-parent household - What is a good explanation for this?

Last Answer : Single parent households have less money for a good lawyer. There is a difference between being convicted and being guilty and between being exhonerated and being innocent.

Net protein utilisation of egg protein is (A) 75% (B) 80% (C) 91% (D) 72%

Description : Net protein utilisation of egg protein is (A) 75% (B) 80% (C) 91% (D) 72%

Last Answer : Answer : C

The iron ore which contains 72% of iron is – (1) Magnetite (2) Limonite (3) Haematite (4) Siderite

Description : The iron ore which contains 72% of iron is – (1) Magnetite (2) Limonite (3) Haematite (4) Siderite

Last Answer : (1) Magnetite Explanation: Iron (Fe) is a metallic element and composes about 5% of the Earth's crust. When pure it is a dark, silvery-gray metal. It is a very reactive element ... alloy. Iron-nickel meteorites are believed to represent the earliest material formed at the beginning of the universe.

The Iron ORE which contains 72% of Iron is –

Description : The Iron ORE which contains 72% of Iron is –

Last Answer : Magnetite

Average yearly load on the tractor should not exceed its rated load up to a) 70% b) 72% c) 75 % d) 78%

Description : Average yearly load on the tractor should not exceed its rated load up to a) 70% b) 72% c) 75 % d) 78%

Last Answer : c) 75 %

Upto a maximum of 72% of iron, is available in (A) Magnetite (B) Limonite (C) Siderite (D) Iron pyrites

Description : Upto a maximum of 72% of iron, is available in (A) Magnetite (B) Limonite (C) Siderite (D) Iron pyrites

Last Answer : Answer: Option A