An article is sold making a loss of 10%. Had it been bought for 20% less and sold for Rs 18 more, there would have been a profit of 25%. What is the cost price of the article?
Correct
Let CP = Rs x
So SP = (90/100)*x = 9x/10
Now if CP = (80/100)*x = 4x/5
And SP is (9x/10 + 18), the profit is 25%
So (9x/10 + 18 â€“ 4x/5)/(4x/5) * 100 = 25
Solve, x = Rs 180
Incorrect
Let CP = Rs x
So SP = (90/100)*x = 9x/10
Now if CP = (80/100)*x = 4x/5
And SP is (9x/10 + 18), the profit is 25%
So (9x/10 + 18 â€“ 4x/5)/(4x/5) * 100 = 25
Solve, x = Rs 180
Unattempted
Let CP = Rs x
So SP = (90/100)*x = 9x/10
Now if CP = (80/100)*x = 4x/5
And SP is (9x/10 + 18), the profit is 25%
So (9x/10 + 18 â€“ 4x/5)/(4x/5) * 100 = 25
Solve, x = Rs 180
Question 12 of 35
12. Question
1 points
Ages of Avni and Bhumika is in the ratio 2 : 3 while that of Avni and Charru is in the ratio 4 : 9. If the difference in the ages of Avni and Charu is 25 years, what is the sum of ages of Avni and Bhumika?
Correct
B/A = 3/2 and A/C = 4/9
So B : A : C = 3*4 : 2*4 : 2*9 = 6 : 4 : 9
So (9-4)/(6+4+9) * x = 25
Solve, x = 95
Sum of ages of A and B = (6+4)/(6+4+9) * 95 = 50
Incorrect
B/A = 3/2 and A/C = 4/9
So B : A : C = 3*4 : 2*4 : 2*9 = 6 : 4 : 9
So (9-4)/(6+4+9) * x = 25
Solve, x = 95
Sum of ages of A and B = (6+4)/(6+4+9) * 95 = 50
Unattempted
B/A = 3/2 and A/C = 4/9
So B : A : C = 3*4 : 2*4 : 2*9 = 6 : 4 : 9
So (9-4)/(6+4+9) * x = 25
Solve, x = 95
Sum of ages of A and B = (6+4)/(6+4+9) * 95 = 50
Question 13 of 35
13. Question
1 points
Raghav borrowed a sum of Rs 5000 from his father at a certain rate of interest. After 2 years, he borrowed a sum of Rs 9000 again from his father at same interest rate. If after a further of 2 years he paid a total of Rs 18,560 to his father, what is the rate of percent at which he borrowed the money?
Correct
Total principal = 5000+9000 = 14000
So interest = 18560 â€“ 14000 = 4560
So (5000*r*4)/100 + (9000*r*2)/100= 4560
Solve, r= 12%
Incorrect
Total principal = 5000+9000 = 14000
So interest = 18560 â€“ 14000 = 4560
So (5000*r*4)/100 + (9000*r*2)/100= 4560
Solve, r= 12%
Unattempted
Total principal = 5000+9000 = 14000
So interest = 18560 â€“ 14000 = 4560
So (5000*r*4)/100 + (9000*r*2)/100= 4560
Solve, r= 12%
Question 14 of 35
14. Question
1 points
What is the probability of drawing 2 cards from a pack of 52 cards such that there is at least 1 king in the draw?
Correct
Case 1: 1 is king ^{4}C_{1}*^{48}C_{1} / ^{52}C_{2}
Case 2: both are king ^{4}C_{2} / ^{52}C_{2}
Add both cases.
Incorrect
Unattempted
Question 15 of 35
15. Question
1 points
From her monthly salary Sakshi spent 25% in shopping. Out of the remaining salary spent 20% on food and 4% on her cosmetics. If she is left with Rs 57,600 at the end of month, what is her monthly salary?
Correct
Let x is monthly salary
So 25% in shopping, remaining 75%
20% of 75% on furniture, so left with 80% of 75%
Now spent 4% of 80% of 75% on food
So left with 96% of 80% of 75% of x which is 57600
So (96/100)*(80/100)*(75/100)*x = 57,600
Solve, x = 100,000
Incorrect
Let x is monthly salary
So 25% in shopping, remaining 75%
20% of 75% on furniture, so left with 80% of 75%
Now spent 4% of 80% of 75% on food
So left with 96% of 80% of 75% of x which is 57600
So (96/100)*(80/100)*(75/100)*x = 57,600
Solve, x = 100,000
Unattempted
Let x is monthly salary
So 25% in shopping, remaining 75%
20% of 75% on furniture, so left with 80% of 75%
Now spent 4% of 80% of 75% on food
So left with 96% of 80% of 75% of x which is 57600
So (96/100)*(80/100)*(75/100)*x = 57,600
Solve, x = 100,000
Question 16 of 35
16. Question
1 points
Directions (16-20): Study the following table carefully to answer the following questions.
The table shows the number of students in Electronics and Computer department over the given years and the ratio of males to females in both departments over the given years
If the number of students in Computer Department in 2012 is 350 and the ratio of males to females in Computer Department in 2013 is 1 : 1, then what is the number of males in Computer department over these years?
Correct
(3/5)*350 + (1/2)*500
Incorrect
(3/5)*350 + (1/2)*500
Unattempted
(3/5)*350 + (1/2)*500
Question 17 of 35
17. Question
1 points
Directions (16-20): Study the following table carefully to answer the following questions.
The table shows the number of students in Electronics and Computer department over the given years and the ratio of males to females in both departments over the given years
If the number of students in Computer and Electronics department is 500 and 450 resp. in year 2010 and the ratio of males to females in Computer department in 2013 is same as given in question 21, then number of males in Electronics department in 2010 and 2013 together is approximately what percent more than the number of females in Computer department in the same years?
Correct
Females in C in 2010 and 2013 = (3/5)*500 + (1/2)*500 = 550
Males in E in 2010 and 2013 = (7/10)*450 + (2/5)*700 = 595
So required % = (595-550)/550 * 100
Incorrect
Females in C in 2010 and 2013 = (3/5)*500 + (1/2)*500 = 550
Males in E in 2010 and 2013 = (7/10)*450 + (2/5)*700 = 595
So required % = (595-550)/550 * 100
Unattempted
Females in C in 2010 and 2013 = (3/5)*500 + (1/2)*500 = 550
Males in E in 2010 and 2013 = (7/10)*450 + (2/5)*700 = 595
So required % = (595-550)/550 * 100
Question 18 of 35
18. Question
1 points
Directions (16-20): Study the following table carefully to answer the following questions.
The table shows the number of students in Electronics and Computer department over the given years and the ratio of males to females in both departments over the given years
In which of the following condition, the number of females is greater?
Directions (16-20): Study the following table carefully to answer the following questions.
The table shows the number of students in Electronics and Computer department over the given years and the ratio of males to females in both departments over the given years
If the number of students in Computer Department in 2012 is 350, then the number of females in Computer department in 2012 and 2014 together is approximately what percent of total students in Computer department for the same years?
Correct
2/5)*350 + (7/13)*650 = 490
Total = 350 + 650 = 1000
So required % = (490/1000)*100
Incorrect
2/5)*350 + (7/13)*650 = 490
Total = 350 + 650 = 1000
So required % = (490/1000)*100
Unattempted
2/5)*350 + (7/13)*650 = 490
Total = 350 + 650 = 1000
So required % = (490/1000)*100
Question 20 of 35
20. Question
1 points
Directions (16-20): Study the following table carefully to answer the following questions.
The table shows the number of students in Electronics and Computer department over the given years and the ratio of males to females in both departments over the given years
What is the total number of males in both departments in 2014 if the ratio of males to females in Electronics Department in 2014 is 8 : 5?
Correct
(6/13)*650 + (8/13)*650 = 700
Incorrect
(6/13)*650 + (8/13)*650 = 700
Unattempted
(6/13)*650 + (8/13)*650 = 700
Question 21 of 35
21. Question
1 points
Anil, Ajay and Aarti started a business by investing Rs 6000, Rs 5000 and Rs 6500 respectively. They all added the same amount as before after 4 months from the start of business. After a further of 4 months each of them took out Rs 1000. What is the ratio of their shares in total profit after a year?
Two persons P and Q start from city A at a speed of 30 km/hr and 20 km/hr respectively. P reached the city B and returns back and meets Q at point S. If the cities A and B are 90 km apart, find the distance between points A and S.
Correct
Time taken by P to reach city B is 3 hr. In 3 hr, distance covered by Q is 60 km. Now after this time taken by both to reach point S will be equal. Let from person B at x distance they meet, then A covered (90-60)-x = (30-x). So
x/20 = (30-x)/30.
Solve, x = 12
So distance A to S is 60 + 12
Incorrect
Time taken by P to reach city B is 3 hr. In 3 hr, distance covered by Q is 60 km. Now after this time taken by both to reach point S will be equal. Let from person B at x distance they meet, then A covered (90-60)-x = (30-x). So
x/20 = (30-x)/30.
Solve, x = 12
So distance A to S is 60 + 12
Unattempted
Time taken by P to reach city B is 3 hr. In 3 hr, distance covered by Q is 60 km. Now after this time taken by both to reach point S will be equal. Let from person B at x distance they meet, then A covered (90-60)-x = (30-x). So
x/20 = (30-x)/30.
Solve, x = 12
So distance A to S is 60 + 12
Question 23 of 35
23. Question
1 points
A work which can be completed by 20 men complete a work in 8 days can also be completed by 30 women in 7 days. At the start of work, there are 10 men. After they completed 3/4th of the work, 14 women also join them. In what time (approximate days) will the work get completed?
Correct
20 men in 8 days, so10 men in (20*8)/10 = 16 days
They completed 3/4 of the work so days required by them is (3/4) * 16 = 12 days
Now joined by 14 women, and they now have to complete remaining work i.e. 1 â€“ (3/4) = 1/4
Let x days are required to do this 1/4 work.
30 women in 7 days, so 14 women in (30*7)/14 = 15 days
Now 10 men and 14 women have to complete 1/4 work in x days. So
[1/16 + 1/15] * x = 1/4
Solve, x= 60/31 â‰ˆ 2
So total days 12 + 2 = 14 days
Incorrect
20 men in 8 days, so10 men in (20*8)/10 = 16 days
They completed 3/4 of the work so days required by them is (3/4) * 16 = 12 days
Now joined by 14 women, and they now have to complete remaining work i.e. 1 â€“ (3/4) = 1/4
Let x days are required to do this 1/4 work.
30 women in 7 days, so 14 women in (30*7)/14 = 15 days
Now 10 men and 14 women have to complete 1/4 work in x days. So
[1/16 + 1/15] * x = 1/4
Solve, x= 60/31 â‰ˆ 2
So total days 12 + 2 = 14 days
Unattempted
20 men in 8 days, so10 men in (20*8)/10 = 16 days
They completed 3/4 of the work so days required by them is (3/4) * 16 = 12 days
Now joined by 14 women, and they now have to complete remaining work i.e. 1 â€“ (3/4) = 1/4
Let x days are required to do this 1/4 work.
30 women in 7 days, so 14 women in (30*7)/14 = 15 days
Now 10 men and 14 women have to complete 1/4 work in x days. So
[1/16 + 1/15] * x = 1/4
Solve, x= 60/31 â‰ˆ 2
So total days 12 + 2 = 14 days
Question 24 of 35
24. Question
1 points
Circumference of a circle is 66m more than the perimeter of the rectangle. If the length and breadth of the rectangle are in the ratio 6 : 5 and the radius of circle and breadth of rectangle are in ratio 7 : 5, find the area of the rectangle.
Correct
l/b = 6/5 and b/r = 5/7
so l : b : r = 6 : 5 : 7
so 2* (22/7) * 7x = 66 + 2[6x+5x]
solve, x = 3
so area of rect = 6x * 5x = 30*9
Incorrect
l/b = 6/5 and b/r = 5/7
so l : b : r = 6 : 5 : 7
so 2* (22/7) * 7x = 66 + 2[6x+5x]
solve, x = 3
so area of rect = 6x * 5x = 30*9
Unattempted
l/b = 6/5 and b/r = 5/7
so l : b : r = 6 : 5 : 7
so 2* (22/7) * 7x = 66 + 2[6x+5x]
solve, x = 3
so area of rect = 6x * 5x = 30*9
Question 25 of 35
25. Question
1 points
A can work 2/3 times as fast as B and C together. A and C together can work twice as fast as B. If C alone takes 45 days to complete the work, how long will A take to complete the same work?
Correct
Days ratio –
A : (B+C) = 3 : 2 => 3x and 2x
and B : (A+C) = 2 : 1 => 2y and y
So
1/2x – 1/45 = 1/2y
and
1/y – 1/45 = 1/3x
Solve both equations, x = 10
So A will take 3x = 30 days
Incorrect
Days ratio –
A : (B+C) = 3 : 2 => 3x and 2x
and B : (A+C) = 2 : 1 => 2y and y
So
1/2x – 1/45 = 1/2y
and
1/y – 1/45 = 1/3x
Solve both equations, x = 10
So A will take 3x = 30 days
Unattempted
Days ratio –
A : (B+C) = 3 : 2 => 3x and 2x
and B : (A+C) = 2 : 1 => 2y and y
So
1/2x – 1/45 = 1/2y
and
1/y – 1/45 = 1/3x
Solve both equations, x = 10
So A will take 3x = 30 days
Question 26 of 35
26. Question
1 points
Directions (26-30): In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answer –
I. 3x^{2} + 10x â€“ 8 = 0,
II. 2y^{2} â€“ 13y + 6 = 0
Directions (31-35): Study the following table carefully and answer the questions that follow:
What is the total number of students from State B who have taken part in Painting and from state D who have taken part in Indoor Play?
Correct
(15/100)*1800 + (45/100)*2000
Incorrect
(15/100)*1800 + (45/100)*2000
Unattempted
(15/100)*1800 + (45/100)*2000
Question 32 of 35
32. Question
1 points
Directions (31-35): Study the following table carefully and answer the questions that follow:
Total number of students in Dance and Painting from State A is approximately what percent more than number of students participating in Cycling from State C?
Correct
Dance and Painting from State A = (15/100)*2000 + (26/100)*1800 = 768
Cycling from State C = (28/100)*2400 = 672
Required% = (768-672)/672 *100
Incorrect
Dance and Painting from State A = (15/100)*2000 + (26/100)*1800 = 768
Cycling from State C = (28/100)*2400 = 672
Required% = (768-672)/672 *100
Unattempted
Dance and Painting from State A = (15/100)*2000 + (26/100)*1800 = 768
Cycling from State C = (28/100)*2400 = 672
Required% = (768-672)/672 *100
Question 33 of 35
33. Question
1 points
Directions (31-35): Study the following table carefully and answer the questions that follow:
What is the ratio of total number of students participating in dancing from States A and B to total number of students participating in singing from States B and C?
Correct
dancing from States A and B = (15+35)/100 * 2000 = 1000
singing from States B and C = (25+35)/100 * 2200 = 1320
1000 : 1320
Incorrect
dancing from States A and B = (15+35)/100 * 2000 = 1000
singing from States B and C = (25+35)/100 * 2200 = 1320
1000 : 1320
Unattempted
dancing from States A and B = (15+35)/100 * 2000 = 1000
singing from States B and C = (25+35)/100 * 2200 = 1320
1000 : 1320
Question 34 of 35
34. Question
1 points
Directions (31-35): Study the following table carefully and answer the questions that follow:
If a total of 1200 students win from State B, then students participating in Cycling from state B are what percent of winners from State B?
Correct
Cycling from state B = (20/100)*2400 = 480
Required % = (480/1200)*100
Incorrect
Cycling from state B = (20/100)*2400 = 480
Required % = (480/1200)*100
Unattempted
Cycling from state B = (20/100)*2400 = 480
Required % = (480/1200)*100
Question 35 of 35
35. Question
1 points
Directions (31-35): Study the following table carefully and answer the questions that follow:
What is the difference between number of students who took part in Singing in state D and Indoor Play from State A?
Correct
Singing in state D = (22/100)*2200 = 484
Indoor Play from State A = (15/100)*2000 = 300
Incorrect
Singing in state D = (22/100)*2200 = 484
Indoor Play from State A = (15/100)*2000 = 300
Unattempted
Singing in state D = (22/100)*2200 = 484
Indoor Play from State A = (15/100)*2000 = 300
TQ ðŸ™‚
palsssssssssssh :))
OO- /
mam 12th ques main kuch wrong hai plz check
statement main B and C ki place pe A and C ka difference hona chahiye
Oh yes. Typing mistake
ok mam
thanku
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thankuu mam
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