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Your Final Score is : 0 Directions (1-5): Study the following carefully to answer the questions that follow: Some students participated in three different activities – painting, swimming and dancing. Find the ratio between number of students who took part in painting and dancing only and those who took part in swimming and dancing only Who took part in only painting = a So here we need d : f Who took part in only painting = a So here we need d : f Who took part in only painting = a So here we need d : f Directions (1-5): Study the following carefully to answer the questions that follow: Some students participated in three different activities – painting, swimming and dancing. What is the number of people who took part in more than one activity? a = 2500, b = 175, c = 3200, d = 575, e = 250, f = 375, g = 3300 So who took part in more than one activity = b + d + e + f = 175 + 575 + 250 + 375 = 1375 a = 2500, b = 175, c = 3200, d = 575, e = 250, f = 375, g = 3300 So who took part in more than one activity = b + d + e + f = 175 + 575 + 250 + 375 = 1375 a = 2500, b = 175, c = 3200, d = 575, e = 250, f = 375, g = 3300 So who took part in more than one activity = b + d + e + f = 175 + 575 + 250 + 375 = 1375 Directions (1-5): Study the following carefully to answer the questions that follow: Some students participated in three different activities – painting, swimming and dancing. What is the number of people who took part in only one activity? a = 2500, b = 175, c = 3200, d = 575, e = 250, f = 375, g = 3300 People who took part in only one activity = a + c + g = 2500 + 3200 + 3300 = 9000 a = 2500, b = 175, c = 3200, d = 575, e = 250, f = 375, g = 3300 People who took part in only one activity = a + c + g = 2500 + 3200 + 3300 = 9000 a = 2500, b = 175, c = 3200, d = 575, e = 250, f = 375, g = 3300 People who took part in only one activity = a + c + g = 2500 + 3200 + 3300 = 9000 Directions (1-5): Study the following carefully to answer the questions that follow: Some students participated in three different activities – painting, swimming and dancing. What is the different between the number of people who took part in all three activities and who took part in painting and swimming only? a = 2500, b = 175, c = 3200, d = 575, e = 250, f = 375, g = 3300 Required difference = e – b = 250 – 175 = 75 a = 2500, b = 175, c = 3200, d = 575, e = 250, f = 375, g = 3300 Required difference = e – b = 250 – 175 = 75 a = 2500, b = 175, c = 3200, d = 575, e = 250, f = 375, g = 3300 Required difference = e – b = 250 – 175 = 75 Directions (1-5): Study the following carefully to answer the questions that follow: Some students participated in three different activities – painting, swimming and dancing. Number of people who took part in painting and dancing only is what percent greater than number of people who took part in swimming and dancing only? a = 2500, b = 175, c = 3200, d = 575, e = 250, f = 375, g = 3300 Required % = (575- 375)/375 * 100 = 53 1/3% a = 2500, b = 175, c = 3200, d = 575, e = 250, f = 375, g = 3300 Required % = (575- 375)/375 * 100 = 53 1/3% a = 2500, b = 175, c = 3200, d = 575, e = 250, f = 375, g = 3300 Required % = (575- 375)/375 * 100 = 53 1/3% Directions (6-8): Each question below contains a statement followed by Quantity I and Quantity II. Find both to find the relationship among them. Mark your answer accordingly. Quantity I: Cost Price of an article if it is sold making a loss of 20% given that if the cost price was 15% less, a loss of Rs 6 was made. I: Let CP = Rs 100, then at 20% loss, SP = Rs 80 I: Let CP = Rs 100, then at 20% loss, SP = Rs 80 I: Let CP = Rs 100, then at 20% loss, SP = Rs 80 Directions (6-8): Each question below contains a statement followed by Quantity I and Quantity II. Find both to find the relationship among them. Mark your answer accordingly. Quantity I: Speed of boat in still water if in a total of 12 hours, it can cover a distance of 36 km going downstream and back. Given the speed of current is 4 km/hr I: Let speed of boat = x km/hr I: Let speed of boat = x km/hr I: Let speed of boat = x km/hr Directions (6-8): Each question below contains a statement followed by Quantity I and Quantity II. Find both to find the relationship among them. Mark your answer accordingly. Quantity I: Time taken by A to complete 1/5th of work if B takes 6 days to complete 3/5thof work and together they take 5 days to complete 3/4th of work I: A and B completes 3/4th work in 5 days, so complete 1 work in 4/3 * 5 = 20/3 days I: A and B completes 3/4th work in 5 days, so complete 1 work in 4/3 * 5 = 20/3 days I: A and B completes 3/4th work in 5 days, so complete 1 work in 4/3 * 5 = 20/3 days Directions (9-11): Study the following carefully to answer the questions that follow: A, B, and C started a business by investing Rs X, Rs (X+800), and Rs (X+1200) respectively. After 8 months of the start of business, A added Rs 600 and B withdrew Rs 600 respectively in their respective investments. After a year, out of the total profit of Rs 23,000, B got his shares equal to Rs 7500. Find X. Ratio of shares of A : B : C is Ratio of shares of A : B : C is Ratio of shares of A : B : C is Directions (9-11): Study the following carefully to answer the questions that follow: A, B, and C started a business by investing Rs X, Rs (X+800), and Rs (X+1200) respectively. After 8 months of the start of business, A added Rs 600 and B withdrew Rs 600 respectively in their respective investments. After a year, out of the total profit of Rs 23,000, B got his shares equal to Rs 7500. What is the difference between the shares of A and C from the total profit? X = Rs 2400 X = Rs 2400 X = Rs 2400 Directions (9-11): Study the following carefully to answer the questions that follow: A, B, and C started a business by investing Rs X, Rs (X+800), and Rs (X+1200) respectively. After 8 months of the start of business, A added Rs 600 and B withdrew Rs 600 respectively in their respective investments. After a year, out of the total profit of Rs 23,000, B got his shares equal to Rs 7500. If all of A, B and C would have invested their initial investments for 12 months, and also another partner partner D joined them at the same time by investing Rs 2800, then find the share of B and D together from the total annual profit of Rs 22,800? Ratio of shares of A : B : C : D is Ratio of shares of A : B : C : D is Ratio of shares of A : B : C : D is Directions (12-16): Study the following bar graph carefully and answer the questions that follow: If the cost price of P and R is equal to Rs 960 each. Find the difference in their Market Prices. For P => For P => For P => Directions (12-16): Study the following bar graph carefully and answer the questions that follow: If the market price of Q is 250% of market price of R, then find the selling price of R, if market price of Q is Rs 3500. For Q: For Q: For Q: Directions (12-16): Study the following bar graph carefully and answer the questions that follow: Find the ratio of selling price of Q and T if the cost price of both is same. For Q: For Q: For Q: Directions (12-16): Study the following bar graph carefully and answer the questions that follow: Find the market price of P if the ratio of cost price of P and T is 10 : 3 and the cost price of T is Rs 3960. For P: For P: For P: Directions (12-16): Study the following bar graph carefully and answer the questions that follow: If the cost price of R and T is same, then find how much % the selling price of R is more or less than that of T. For R: For R: For R: Directions (17-21): Study the following table to answer the questions that follow: If number of males and females who left the company from department A is equal, find the total number of males from departments C and A who did not left the job? females who left from A = 12% of 350 = 42 females who left from A = 12% of 350 = 42 females who left from A = 12% of 350 = 42 Directions (17-21): Study the following table to answer the questions that follow: The number of males who left the company from departments C and D is in the ratio 8 : 9 respectively. Total number of employees in department D and who left from D is 870 and 48 respectively. What % of males did not left from department D? Males who left from C = 5% of 480 = 24 Males who left from C = 5% of 480 = 24 Males who left from C = 5% of 480 = 24 Directions (17-21): Study the following table to answer the questions that follow: Number of graduate males in department C is 127 more than number of graduate females in department B. If 33 1/3% of males who left from department C are post-graduates, then find the number of post-graduates who did not left from department C? Number of graduate females in B = 55% of 380 = 209 Number of graduate females in B = 55% of 380 = 209 Number of graduate females in B = 55% of 380 = 209 Directions (17-21): Study the following table to answer the questions that follow: 29% of number of females who left from department E are graduates. If % of number of graduate females who left from department E with respect to total number of graduate females from department E is 4%, then find the number of females from department E who left the company. (Number of females in E = 400). Let x% females left the company from dept E Let x% females left the company from dept E Let x% females left the company from dept E Directions (17-21): Study the following table to answer the questions that follow: If there are total 1000 employees in department C, find the total number of female employees in departments B and C or company. Total employees in C = 1000 Total employees in C = 1000 Total employees in C = 1000 Directions (22-26): In each of the following questions, a question is followed by three statements numbered I, II and III. Read all the statements and answer accordingly. What is the speed of boat in still water? Let speed of boat = u, and stream = v Let speed of boat = u, and stream = v Let speed of boat = u, and stream = v Directions (22-26): In each of the following questions, a question is followed by three statements numbered I, II and III. Read all the statements and answer accordingly. A person borrowed some money at compound interest for 2 years. What will be the amount required to return after 2 years? From I: SI = 600, R = 5 years From I: SI = 600, R = 5 years From I: SI = 600, R = 5 years Directions (22-26): In each of the following questions, a question is followed by three statements numbered I, II and III. Read all the statements and answer accordingly. What was the profit earned by selling an article? From I and III: From II: CP = 540, discount = 10% From I and III: From II: CP = 540, discount = 10% From I and III: From II: CP = 540, discount = 10% Directions (22-26): In each of the following questions, a question is followed by three statements numbered I, II and III. Read all the statements and answer accordingly. What is the cost of flooring a rectangular hall? Using any two statement, the length and breadth can be found. And Rate of flooring is also available. Using any two statement, the length and breadth can be found. And Rate of flooring is also available. Using any two statement, the length and breadth can be found. And Rate of flooring is also available. Directions (22-26): In each of the following questions, a question is followed by three statements numbered I, II and III. Read all the statements and answer accordingly. What is the principal amount? From I : Pr2/1002 = 450. So P can be found. From I : Pr2/1002 = 450. So P can be found. From I : Pr2/1002 = 450. So P can be found. Directions (27-31): The following bar graph depicts the percentage wise distribution of what people eat in their breakfast in 5 states. Study the bar graph and answer the questions that follow: If the ratio of number of people who eat oats in states A and D is 5 : 9 respectively, then what is the ratio of number of people who eat wheat in these two states respectively? Let number of people in state A = x, and that in D = y Let number of people in state A = x, and that in D = y Let number of people in state A = x, and that in D = y Directions (27-31): The following bar graph depicts the percentage wise distribution of what people eat in their breakfast in 5 states. Study the bar graph and answer the questions that follow: The number of people who eat cornflakes in state C is four-fifth of number of people who eat cornflakes in state E. If the number of people who eat wheat in state C is 96,000, then find the total number of people in state E? Let total number of people in state C = x Let total number of people in state C = x Let total number of people in state C = x Directions (27-31): The following bar graph depicts the percentage wise distribution of what people eat in their breakfast in 5 states. Study the bar graph and answer the questions that follow: The number of people who eat millet in state B is 4200 more than number of people who eat millet in state D. If the total number of people in state B is 10,000 more than total number of people in state D, find the total number of people who eat wheat in both states together? Let number of people in state B = x, and that in D = y Let number of people in state B = x, and that in D = y Let number of people in state B = x, and that in D = y Directions (27-31): The following bar graph depicts the percentage wise distribution of what people eat in their breakfast in 5 states. Study the bar graph and answer the questions that follow: The difference between the number of people who eat rice and cornflakes in state E is 1,62,000. What is the number of people in state 3 except the number of people who eat cornflakes in the same state? Let number of people in state E So required total = (4+20+48+ 16)% of 4,50,000 = 3,96,000 Let number of people in state E So required total = (4+20+48+ 16)% of 4,50,000 = 3,96,000 Let number of people in state E So required total = (4+20+48+ 16)% of 4,50,000 = 3,96,000 Directions (27-31): The following bar graph depicts the percentage wise distribution of what people eat in their breakfast in 5 states. Study the bar graph and answer the questions that follow: The ratio between the number of people who eat rice in states A and B is 10 : 9 and ratio between the number of people who eat oats in states B and D is 5 : 3. Find the ratio of number of people in state A : D : B respectively. Let number of people in state A = x, in B = y, and in D = z x/y = 14/15 and y/z = 4/3 Let number of people in state A = x, in B = y, and in D = z x/y = 14/15 and y/z = 4/3 Let number of people in state A = x, in B = y, and in D = z x/y = 14/15 and y/z = 4/3 Directions (32-33): Study the following information to answer the questions that follow: What is the probability of choosing one blue ball? Probability of choosing one blue ball = 1 – (1/2 + 1/4) = 1/4 Probability of choosing one blue ball = 1 – (1/2 + 1/4) = 1/4 Probability of choosing one blue ball = 1 – (1/2 + 1/4) = 1/4 Directions (32-33): Study the following information to answer the questions that follow: If the balls are numbered 1, 2, …. up to number of balls in the urn, what is the probability of choosing a ball containing a multiple of 3 and 5? Probability of choosing one blue ball is 1/4 There are 30 balls in the urn. Probability of choosing one blue ball is 1/4 There are 30 balls in the urn. Probability of choosing one blue ball is 1/4 There are 30 balls in the urn. Directions (34-35): Study the following pie-chart and answer the questions that follow: If there are 7440 employees in country D, then what is the number of employees in countries A and E together? Let x is the number of employees in all countries together Let x is the number of employees in all countries together Let x is the number of employees in all countries together Directions (34-35): Study the following pie-chart and answer the questions that follow: There are 43,200 employees in all 5 countries together. If the company opens it branch in one more country F, then the quadrant corresponding to country C becomes 84o then find the degree value (approximately) of quadrant formed by country F in given pie-chart. Employees in country C = 98/360 * 43200 = 11760
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1. Question
2 points
The total students who took part in painting is 3500 of which 5% students took part in both painting and swimming only. The number of students who took part in both dancing and swimming only is 50% greater than the number of students who took part in all 3 activities. The number of students who took part in only swimming is 100 less than the number of students who took part in only dancing. The total number of students who took part in dancing but not painting is 3675. The number of students who took part in only swimming is 2950 more than the number of students who took part in all three activities. The number of students who took part in painting only is 1925 more than the number of students who took part in painting and dancing only.
Who took part in painting + swimming = b
Who took part in painting = a + b + d + e
Who took part in all 3 = e
a + b + e + d = 3500 [Given]
Noe 5% of 3500 = b [Given]
So b = 175
So a + e + d = 3500 – 175 = 3325 …….. (1)
Next: f = 150/100 * e [Given]
c = g – 100 [Given]
f + g = 3675 [Given]
c – e = 2950 [Given]
Here we have 4 variables (c, e, f, g) and 4 equations
Solve all equations to find the values
c = 3200, e = 250, f = 375, g = 3300
So from (1), a + d = 3075
a = 1625 + d [Given]
Now from these 2 equations
a = 2500, d = 575
So
a = 2500, b = 175, c = 3200, d = 575, e = 250, f = 375, g = 3300
575 : 375 = 23 : 15
Who took part in painting + swimming = b
Who took part in painting = a + b + d + e
Who took part in all 3 = e
a + b + e + d = 3500 [Given]
Noe 5% of 3500 = b [Given]
So b = 175
So a + e + d = 3500 – 175 = 3325 …….. (1)
Next: f = 150/100 * e [Given]
c = g – 100 [Given]
f + g = 3675 [Given]
c – e = 2950 [Given]
Here we have 4 variables (c, e, f, g) and 4 equations
Solve all equations to find the values
c = 3200, e = 250, f = 375, g = 3300
So from (1), a + d = 3075
a = 1625 + d [Given]
Now from these 2 equations
a = 2500, d = 575
So
a = 2500, b = 175, c = 3200, d = 575, e = 250, f = 375, g = 3300
575 : 375 = 23 : 15
Who took part in painting + swimming = b
Who took part in painting = a + b + d + e
Who took part in all 3 = e
a + b + e + d = 3500 [Given]
Noe 5% of 3500 = b [Given]
So b = 175
So a + e + d = 3500 – 175 = 3325 …….. (1)
Next: f = 150/100 * e [Given]
c = g – 100 [Given]
f + g = 3675 [Given]
c – e = 2950 [Given]
Here we have 4 variables (c, e, f, g) and 4 equations
Solve all equations to find the values
c = 3200, e = 250, f = 375, g = 3300
So from (1), a + d = 3075
a = 1625 + d [Given]
Now from these 2 equations
a = 2500, d = 575
So
a = 2500, b = 175, c = 3200, d = 575, e = 250, f = 375, g = 3300
575 : 375 = 23 : 15
2. Question
2 points
The total students who took part in painting is 3500 of which 5% students took part in both painting and swimming only. The number of students who took part in both dancing and swimming only is 50% greater than the number of students who took part in all 3 activities. The number of students who took part in only swimming is 100 less than the number of students who took part in only dancing. The total number of students who took part in dancing but not painting is 3675. The number of students who took part in only swimming is 2950 more than the number of students who took part in all three activities. The number of students who took part in painting only is 1925 more than the number of students who took part in painting and dancing only.
3. Question
2 points
The total students who took part in painting is 3500 of which 5% students took part in both painting and swimming only. The number of students who took part in both dancing and swimming only is 50% greater than the number of students who took part in all 3 activities. The number of students who took part in only swimming is 100 less than the number of students who took part in only dancing. The total number of students who took part in dancing but not painting is 3675. The number of students who took part in only swimming is 2950 more than the number of students who took part in all three activities. The number of students who took part in painting only is 1925 more than the number of students who took part in painting and dancing only.
4. Question
2 points
The total students who took part in painting is 3500 of which 5% students took part in both painting and swimming only. The number of students who took part in both dancing and swimming only is 50% greater than the number of students who took part in all 3 activities. The number of students who took part in only swimming is 100 less than the number of students who took part in only dancing. The total number of students who took part in dancing but not painting is 3675. The number of students who took part in only swimming is 2950 more than the number of students who took part in all three activities. The number of students who took part in painting only is 1925 more than the number of students who took part in painting and dancing only.
5. Question
2 points
The total students who took part in painting is 3500 of which 5% students took part in both painting and swimming only. The number of students who took part in both dancing and swimming only is 50% greater than the number of students who took part in all 3 activities. The number of students who took part in only swimming is 100 less than the number of students who took part in only dancing. The total number of students who took part in dancing but not painting is 3675. The number of students who took part in only swimming is 2950 more than the number of students who took part in all three activities. The number of students who took part in painting only is 1925 more than the number of students who took part in painting and dancing only.
6. Question
1 points
Quantity II: Labeled Price of an article of cost price Rs 240 on which after giving a discount of 10% a profit of 20% is made.
Now if CP is 15% less, means CP = Rs 85, then loss is = 85 – 80 = Rs 5
but given loss is Rs 12
So if loss is Rs 5, then CP = Rs 100
If loss is Rs 12, then CP is 100/5 * 6 = Rs 120
II: Use formula MP = (100+profit%)/(100-dicount%) * CP
So MP = (100+20)/(100-10) * 240 = Rs 320
Hence II > I
Now if CP is 15% less, means CP = Rs 85, then loss is = 85 – 80 = Rs 5
but given loss is Rs 12
So if loss is Rs 5, then CP = Rs 100
If loss is Rs 12, then CP is 100/5 * 6 = Rs 120
II: Use formula MP = (100+profit%)/(100-dicount%) * CP
So MP = (100+20)/(100-10) * 240 = Rs 320
Hence II > I
Now if CP is 15% less, means CP = Rs 85, then loss is = 85 – 80 = Rs 5
but given loss is Rs 12
So if loss is Rs 5, then CP = Rs 100
If loss is Rs 12, then CP is 100/5 * 6 = Rs 120
II: Use formula MP = (100+profit%)/(100-dicount%) * CP
So MP = (100+20)/(100-10) * 240 = Rs 320
Hence II > I
7. Question
1 points
Quantity II: Average speed of a boy from his home to school if his average speed is 10 km/hr from school to home and he covers a total of 30 km in 3 hours.
So 36/(x+4) + 36/(x-4) = 12
Solve, x = 8 km/hr
II : Distance from home to school is 30/2 = 15 km
Let speed from home to school is x km/hr
So using distance = total time * average speed
15 = 3 *[10*x/(10+x)]
Solve, x = 10 km/hr
Hence II > I
So 36/(x+4) + 36/(x-4) = 12
Solve, x = 8 km/hr
II : Distance from home to school is 30/2 = 15 km
Let speed from home to school is x km/hr
So using distance = total time * average speed
15 = 3 *[10*x/(10+x)]
Solve, x = 10 km/hr
Hence II > I
So 36/(x+4) + 36/(x-4) = 12
Solve, x = 8 km/hr
II : Distance from home to school is 30/2 = 15 km
Let speed from home to school is x km/hr
So using distance = total time * average speed
15 = 3 *[10*x/(10+x)]
Solve, x = 10 km/hr
Hence II > I
8. Question
1 points
Quantity II: Time taken by a train to cross a platform of length 60 km given that it crosses a pole in two and a half hours running at 60 km/hr
B complete 3/5 work in 6 days, so complete work in 5/3 * 6 = 10 days
So in 1 day A completes = 3/20 – 1/10 = 1/20
So to complete 1/5th work = 1/5 * 20 = 4 days
II: length of train = 2.5 * 60 = 150 km
So time taken to cross platform of length 60 km with speed 60 km/hr = (150+60)/60 = 3.5 hrs
Hence I > II
B complete 3/5 work in 6 days, so complete work in 5/3 * 6 = 10 days
So in 1 day A completes = 3/20 – 1/10 = 1/20
So to complete 1/5th work = 1/5 * 20 = 4 days
II: length of train = 2.5 * 60 = 150 km
So time taken to cross platform of length 60 km with speed 60 km/hr = (150+60)/60 = 3.5 hrs
Hence I > II
B complete 3/5 work in 6 days, so complete work in 5/3 * 6 = 10 days
So in 1 day A completes = 3/20 – 1/10 = 1/20
So to complete 1/5th work = 1/5 * 20 = 4 days
II: length of train = 2.5 * 60 = 150 km
So time taken to cross platform of length 60 km with speed 60 km/hr = (150+60)/60 = 3.5 hrs
Hence I > II
9. Question
1 points
X*8 + (X+600)*4 : (X+800)*8 + (X+200)*4 : (X+1200)*12
12X + 2400 : 12X + 7200 : 12X + 14400
(X+200) : (X+600) : (X+1200)
So according to question:
(X+600)/(X+200 + X+600 + X+1200) * 23000 = 7500
Solve, X = Rs 2400
X*8 + (X+600)*4 : (X+800)*8 + (X+200)*4 : (X+1200)*12
12X + 2400 : 12X + 7200 : 12X + 14400
(X+200) : (X+600) : (X+1200)
So according to question:
(X+600)/(X+200 + X+600 + X+1200) * 23000 = 7500
Solve, X = Rs 2400
X*8 + (X+600)*4 : (X+800)*8 + (X+200)*4 : (X+1200)*12
12X + 2400 : 12X + 7200 : 12X + 14400
(X+200) : (X+600) : (X+1200)
So according to question:
(X+600)/(X+200 + X+600 + X+1200) * 23000 = 7500
Solve, X = Rs 2400
10. Question
1 points
So Ratio of shares of A : B : C is
2400*8 + (2400+600)*4 : (2400+800)*8 + (2400+200)*4 : (2400+1200)*12
13 : 15 : 18
So difference between the shares of A and C = (18-15)/(13+15+18) * 23000 = Rs 2750
So Ratio of shares of A : B : C is
2400*8 + (2400+600)*4 : (2400+800)*8 + (2400+200)*4 : (2400+1200)*12
13 : 15 : 18
So difference between the shares of A and C = (18-15)/(13+15+18) * 23000 = Rs 2750
So Ratio of shares of A : B : C is
2400*8 + (2400+600)*4 : (2400+800)*8 + (2400+200)*4 : (2400+1200)*12
13 : 15 : 18
So difference between the shares of A and C = (18-15)/(13+15+18) * 23000 = Rs 2750
11. Question
1 points
2400 : 3200 : 3600 : 2800
6 : 8 : 9 : 7
So required answer = (8+7)/(6+8+9+7) * 22800 = Rs 11,400
2400 : 3200 : 3600 : 2800
6 : 8 : 9 : 7
So required answer = (8+7)/(6+8+9+7) * 22800 = Rs 11,400
2400 : 3200 : 3600 : 2800
6 : 8 : 9 : 7
So required answer = (8+7)/(6+8+9+7) * 22800 = Rs 11,400
12. Question
2 points
Profit% = 30% = 3/10 [CP = 10, SP = 10+3 = 13]
Discount = 20% = 1/5 [MP = 5, SP = 5-1 = 4]
Make SP same
CP : SP : MP is
40 : 52 : 65 ………………(1)
Now for R =>
Profit% = 20% = 1/5 [CP = 5, SP = 5+1 = 6]
Discount = 10% = 1/10 [MP = 10, SP = 10-1 = 9]
Make SP same
CP : SP : MP is
15 : 18 : 20……………….(2)
Make CP of P and R same by multiplying eq. (1) by 3 and (2) by 8, gives
For P: CP : SP : MP is
120 : 156 : 195
And For R: CP : SP : MP is
120 : 144 : 160
So 120 = 960 [as given in question]
=> 1 = 8
Difference in MP = 195-160 = 35, so => 35 = 35*8 = Rs 280
Profit% = 30% = 3/10 [CP = 10, SP = 10+3 = 13]
Discount = 20% = 1/5 [MP = 5, SP = 5-1 = 4]
Make SP same
CP : SP : MP is
40 : 52 : 65 ………………(1)
Now for R =>
Profit% = 20% = 1/5 [CP = 5, SP = 5+1 = 6]
Discount = 10% = 1/10 [MP = 10, SP = 10-1 = 9]
Make SP same
CP : SP : MP is
15 : 18 : 20……………….(2)
Make CP of P and R same by multiplying eq. (1) by 3 and (2) by 8, gives
For P: CP : SP : MP is
120 : 156 : 195
And For R: CP : SP : MP is
120 : 144 : 160
So 120 = 960 [as given in question]
=> 1 = 8
Difference in MP = 195-160 = 35, so => 35 = 35*8 = Rs 280
Profit% = 30% = 3/10 [CP = 10, SP = 10+3 = 13]
Discount = 20% = 1/5 [MP = 5, SP = 5-1 = 4]
Make SP same
CP : SP : MP is
40 : 52 : 65 ………………(1)
Now for R =>
Profit% = 20% = 1/5 [CP = 5, SP = 5+1 = 6]
Discount = 10% = 1/10 [MP = 10, SP = 10-1 = 9]
Make SP same
CP : SP : MP is
15 : 18 : 20……………….(2)
Make CP of P and R same by multiplying eq. (1) by 3 and (2) by 8, gives
For P: CP : SP : MP is
120 : 156 : 195
And For R: CP : SP : MP is
120 : 144 : 160
So 120 = 960 [as given in question]
=> 1 = 8
Difference in MP = 195-160 = 35, so => 35 = 35*8 = Rs 280
13. Question
2 points
CP….SP…..MP is 36……..45………50 …………………..(1)
For R:
CP….SP…..MP is 15……..18………20 ………..…………(2)
MP of Q is 250% of R
250% of 20 = 50
Which is same as in equation (1)
CP………SP………MP is 36……..45………50
Now 50 = 3500
1 = 700
So 18 = 1260
CP….SP…..MP is 36……..45………50 …………………..(1)
For R:
CP….SP…..MP is 15……..18………20 ………..…………(2)
MP of Q is 250% of R
250% of 20 = 50
Which is same as in equation (1)
CP………SP………MP is 36……..45………50
Now 50 = 3500
1 = 700
So 18 = 1260
CP….SP…..MP is 36……..45………50 …………………..(1)
For R:
CP….SP…..MP is 15……..18………20 ………..…………(2)
MP of Q is 250% of R
250% of 20 = 50
Which is same as in equation (1)
CP………SP………MP is 36……..45………50
Now 50 = 3500
1 = 700
So 18 = 1260
14. Question
2 points
CP….SP…..MP is 36……..45………50 …………………..(1)
For T:
CP…..SP…..MP is 12 ……..15……….20 …………………(2)
Make CP same, Multiply eq. (2) by 3
(1) gives.,, 36……..45………50
(2) gives,, 36………..45…..……30
SP of S : R is 45 : 45 = 1 : 1
CP….SP…..MP is 36……..45………50 …………………..(1)
For T:
CP…..SP…..MP is 12 ……..15……….20 …………………(2)
Make CP same, Multiply eq. (2) by 3
(1) gives.,, 36……..45………50
(2) gives,, 36………..45…..……30
SP of S : R is 45 : 45 = 1 : 1
CP….SP…..MP is 36……..45………50 …………………..(1)
For T:
CP…..SP…..MP is 12 ……..15……….20 …………………(2)
Make CP same, Multiply eq. (2) by 3
(1) gives.,, 36……..45………50
(2) gives,, 36………..45…..……30
SP of S : R is 45 : 45 = 1 : 1
15. Question
2 points
CP…..SP…..MP is 40 ……..52……….65 …………………(1)
For T:
CP…..SP…..MP is 12 ……..15……….20 …………………(2)
Here already ratio of CP of P : T is 10 : 3
So,
12 == 3960
1 == 330
MP of P = 65 == 21450
CP…..SP…..MP is 40 ……..52……….65 …………………(1)
For T:
CP…..SP…..MP is 12 ……..15……….20 …………………(2)
Here already ratio of CP of P : T is 10 : 3
So,
12 == 3960
1 == 330
MP of P = 65 == 21450
CP…..SP…..MP is 40 ……..52……….65 …………………(1)
For T:
CP…..SP…..MP is 12 ……..15……….20 …………………(2)
Here already ratio of CP of P : T is 10 : 3
So,
12 == 3960
1 == 330
MP of P = 65 == 21450
16. Question
2 points
CP…..SP…..MP is 15………18……….20……………….(1)
For T:
CP…..SP…..MP is 12 ……..15……….20 ………………(2)
Make CP same, Multiply eq. (1) by 4 and (2) by 5
R => 60……..72………..80
T => 60……..75………..100
So required % = (75-72)/75 * 100 = 4%
CP…..SP…..MP is 15………18……….20……………….(1)
For T:
CP…..SP…..MP is 12 ……..15……….20 ………………(2)
Make CP same, Multiply eq. (1) by 4 and (2) by 5
R => 60……..72………..80
T => 60……..75………..100
So required % = (75-72)/75 * 100 = 4%
CP…..SP…..MP is 15………18……….20……………….(1)
For T:
CP…..SP…..MP is 12 ……..15……….20 ………………(2)
Make CP same, Multiply eq. (1) by 4 and (2) by 5
R => 60……..72………..80
T => 60……..75………..100
So required % = (75-72)/75 * 100 = 4%
17. Question
2 points
The table shows the employees information in 5 different departments of a company.
So males who left from A also equals 42
So males who did not left from A = 420 – 42 = 378
Males who did not left from C = 480 – 5% of 480 = 456
Required total = 378 + 456 = 834
So males who left from A also equals 42
So males who did not left from A = 420 – 42 = 378
Males who did not left from C = 480 – 5% of 480 = 456
Required total = 378 + 456 = 834
So males who left from A also equals 42
So males who did not left from A = 420 – 42 = 378
Males who did not left from C = 480 – 5% of 480 = 456
Required total = 378 + 456 = 834
18. Question
2 points
The table shows the employees information in 5 different departments of a company.
So in D = 9/8 * 24 = 27 [ratio is 8 : 9]
Let number of males in D = x, so females in D = (870-x)
Let y% of males from D left the company
So y% of x = 27
And
y% of x + 5% of (870-x) = 48
So 27 + 5% of (870-x) = 48
Solve, x = 450
Now y% of 450 = 27
Solve, y = 6
6% left, so who did not left = 94%
So in D = 9/8 * 24 = 27 [ratio is 8 : 9]
Let number of males in D = x, so females in D = (870-x)
Let y% of males from D left the company
So y% of x = 27
And
y% of x + 5% of (870-x) = 48
So 27 + 5% of (870-x) = 48
Solve, x = 450
Now y% of 450 = 27
Solve, y = 6
6% left, so who did not left = 94%
So in D = 9/8 * 24 = 27 [ratio is 8 : 9]
Let number of males in D = x, so females in D = (870-x)
Let y% of males from D left the company
So y% of x = 27
And
y% of x + 5% of (870-x) = 48
So 27 + 5% of (870-x) = 48
Solve, x = 450
Now y% of 450 = 27
Solve, y = 6
6% left, so who did not left = 94%
19. Question
2 points
The table shows the employees information in 5 different departments of a company.
So Number of graduate males in department C = 209 + 127 = 336
So post-graduate males in department C = 480 – 336 = 144
Total number of males who left from C = 5% of 480 = 24
Given 33 1/3% of 24 = post-graduates
So those males who left from C and are post-graduates = 33 1/3% of 24 = 8
So who did not left = 144 – 8 = 136
So Number of graduate males in department C = 209 + 127 = 336
So post-graduate males in department C = 480 – 336 = 144
Total number of males who left from C = 5% of 480 = 24
Given 33 1/3% of 24 = post-graduates
So those males who left from C and are post-graduates = 33 1/3% of 24 = 8
So who did not left = 144 – 8 = 136
So Number of graduate males in department C = 209 + 127 = 336
So post-graduate males in department C = 480 – 336 = 144
Total number of males who left from C = 5% of 480 = 24
Given 33 1/3% of 24 = post-graduates
So those males who left from C and are post-graduates = 33 1/3% of 24 = 8
So who did not left = 144 – 8 = 136
20. Question
2 points
The table shows the employees information in 5 different departments of a company.
Given:
Solve, x = 12%
So number of females from department F who left the company = 12/100 * 400 = 48
Given:
Solve, x = 12%
So number of females from department F who left the company = 12/100 * 400 = 48
Given:
Solve, x = 12%
So number of females from department F who left the company = 12/100 * 400 = 48
21. Question
2 points
The table shows the employees information in 5 different departments of a company.
So number of females = 1000 – 480 = 520
So required total = 55% of 380 + 80% of 520 = 625
So number of females = 1000 – 480 = 520
So required total = 55% of 380 + 80% of 520 = 625
So number of females = 1000 – 480 = 520
So required total = 55% of 380 + 80% of 520 = 625
22. Question
2 points
Statement I: The boat can cover 16 km downstream distance in 2 hours.
Statement II: Speed of the stream is one-third the speed of boat in still water.
Statement III: The boat can cover 16 km upstream distance in 4 hours.
From I: u + v = 16/2 = 8 km/hr
From II: v = u/3
From III: u – v = 16/4 = 4
So solving I and II ‘u’ can be found. Similarly by solving II and III and by solving I and III
From I: u + v = 16/2 = 8 km/hr
From II: v = u/3
From III: u – v = 16/4 = 4
So solving I and II ‘u’ can be found. Similarly by solving II and III and by solving I and III
From I: u + v = 16/2 = 8 km/hr
From II: v = u/3
From III: u – v = 16/4 = 4
So solving I and II ‘u’ can be found. Similarly by solving II and III and by solving I and III
23. Question
2 points
Statement I: If the amount was borrowed at simple interest, then after 5 years Rs 600 was required to pay as simple interest.
Statement II: The rate of interest is 6% per annum
Statement III: The sum of money borrowed is 10 times the amount required to be paid as simple interest after 2 years.
So 600 = P*R*5/100
So P*R = 12000
From II: R = 6%
So from I and II, get P, and then from P, R and T, get amount at CI
From III: P = 10 SI
SI = P*R*T/100
So SI = (10SI) * R * 2/100
From here , R = 5%
So from I and III, get P, and then from P, R and T, get amount at CI
So 600 = P*R*5/100
So P*R = 12000
From II: R = 6%
So from I and II, get P, and then from P, R and T, get amount at CI
From III: P = 10 SI
SI = P*R*T/100
So SI = (10SI) * R * 2/100
From here , R = 5%
So from I and III, get P, and then from P, R and T, get amount at CI
So 600 = P*R*5/100
So P*R = 12000
From II: R = 6%
So from I and II, get P, and then from P, R and T, get amount at CI
From III: P = 10 SI
SI = P*R*T/100
So SI = (10SI) * R * 2/100
From here , R = 5%
So from I and III, get P, and then from P, R and T, get amount at CI
24. Question
2 points
Statement I: The article was sold for Rs 950 after getting a discount of 5% on the labelled price.
Statement II: The article was bought for Rs 540 after getting a discount of 10%
Statement III: A profit of 2% is made on the labelled price in buying that article.
I: we can find the MP of article as (100/95) × 950 = Rs 1000
From III: we can find the CP of article by using the MP from statement I as (102/100) × 1000 = Rs 1020
Now we have both SP and CP to find the profit%
I: we can find the MP of article as (100/95) × 950 = Rs 1000
From III: we can find the CP of article by using the MP from statement I as (102/100) × 1000 = Rs 1020
Now we have both SP and CP to find the profit%
I: we can find the MP of article as (100/95) × 950 = Rs 1000
From III: we can find the CP of article by using the MP from statement I as (102/100) × 1000 = Rs 1020
Now we have both SP and CP to find the profit%
25. Question
2 points
Statement I: The ratio of length and breadth of the hall is 5:4.
Statement II: The length of the hall is 50 metres and the cost of flooring is Rs. 1000 per square metre.
Statement III: The perimeter of the hall is 180 metres and the cost of flooring is Rs. 1000 per square metre.
26. Question
2 points
Statement I: The Simple Interest obtained on the principal after 2 years at 8% rate of interest is Rs 450 less than the compound interest obtained on the same principal after 2 years at 8% per annum.
Statement II: The sum becomes double in 10 years at 6% per annum rate of simple interest.
Statement III: The compound interest obtained on the principal amount is Rs 4540 after 2 years at the rate of 8% compounded annually.
From III: using the compound interest formula, P can be find out
From III: using the compound interest formula, P can be find out
From III: using the compound interest formula, P can be find out
27. Question
2 points
So
(10% of x)/(12% if y) = 5/9
Solve, x/y = 2/3
So required ratio is
48% of x : 20% of y
Put value of x or y from above
Ratio is 8 : 5
So
(10% of x)/(12% if y) = 5/9
Solve, x/y = 2/3
So required ratio is
48% of x : 20% of y
Put value of x or y from above
Ratio is 8 : 5
So
(10% of x)/(12% if y) = 5/9
Solve, x/y = 2/3
So required ratio is
48% of x : 20% of y
Put value of x or y from above
Ratio is 8 : 5
28. Question
2 points
So 40% of x = 96000
Solve, x = 2,40,000
Now
15% of 2,40,000 = 4/5 of 12% of y
Solve, y = 3,75,000 = total number of people in state E
So 40% of x = 96000
Solve, x = 2,40,000
Now
15% of 2,40,000 = 4/5 of 12% of y
Solve, y = 3,75,000 = total number of people in state E
So 40% of x = 96000
Solve, x = 2,40,000
Now
15% of 2,40,000 = 4/5 of 12% of y
Solve, y = 3,75,000 = total number of people in state E
29. Question
2 points
So
7% of x = 4200 + 6% of y
Also x= 10,000 + y
Solve both equations
x = 3,60,000 and y = 3,50,000
So required total = 47% of 360000 + 20% of 350000 = 169200 + 70000 = 2,39,200
So
7% of x = 4200 + 6% of y
Also x= 10,000 + y
Solve both equations
x = 3,60,000 and y = 3,50,000
So required total = 47% of 360000 + 20% of 350000 = 169200 + 70000 = 2,39,200
So
7% of x = 4200 + 6% of y
Also x= 10,000 + y
Solve both equations
x = 3,60,000 and y = 3,50,000
So required total = 47% of 360000 + 20% of 350000 = 169200 + 70000 = 2,39,200
30. Question
2 points
Given: (48-12)% of x = 162000
Solve, x = 4,50,000
Given: (48-12)% of x = 162000
Solve, x = 4,50,000
Given: (48-12)% of x = 162000
Solve, x = 4,50,000
31. Question
2 points
So
(25% of x)/(21% of y) = 10/9
Solve, x/y = 14/15
Also
(15% of y)/(12% of z) = 5/3
Solve, y/z = 4/3
So x : y : z = 14*4 : 15*4 : 15*3 = 56 : 60 : 45
So required ratio is 56 : 45 : 60
So
(25% of x)/(21% of y) = 10/9
Solve, x/y = 14/15
Also
(15% of y)/(12% of z) = 5/3
Solve, y/z = 4/3
So x : y : z = 14*4 : 15*4 : 15*3 = 56 : 60 : 45
So required ratio is 56 : 45 : 60
So
(25% of x)/(21% of y) = 10/9
Solve, x/y = 14/15
Also
(15% of y)/(12% of z) = 5/3
Solve, y/z = 4/3
So x : y : z = 14*4 : 15*4 : 15*3 = 56 : 60 : 45
So required ratio is 56 : 45 : 60
32. Question
1 points
An urn contains some balls colored white, blue and green. The probability of choosing a white ball is 1/2 and the probability of choosing a green ball is 1/4. There are 10 blue balls.
33. Question
1 points
An urn contains some balls colored white, blue and green. The probability of choosing a white ball is 1/2 and the probability of choosing a green ball is 1/4. There are 10 blue balls.
And total blue balls are 10. So with 10/40 we get probability as 1/4
So total balls must be 40
Multiples of 3 up to 40 = 40/3 = 13 (take only whole number before the decimal part)
Multiples of 5 up to 40 = 40/5 = 18
Multiples of 15 (3×5) up to 40 = 40/15 = 2
So total such numbers are = 13 + 18 – 2 = 29
So required probability = 29/40
And total blue balls are 10. So with 10/40 we get probability as 1/4
So total balls must be 40
Multiples of 3 up to 40 = 40/3 = 13 (take only whole number before the decimal part)
Multiples of 5 up to 40 = 40/5 = 18
Multiples of 15 (3×5) up to 40 = 40/15 = 2
So total such numbers are = 13 + 18 – 2 = 29
So required probability = 29/40
And total blue balls are 10. So with 10/40 we get probability as 1/4
So total balls must be 40
Multiples of 3 up to 40 = 40/3 = 13 (take only whole number before the decimal part)
Multiples of 5 up to 40 = 40/5 = 18
Multiples of 15 (3×5) up to 40 = 40/15 = 2
So total such numbers are = 13 + 18 – 2 = 29
So required probability = 29/40
34. Question
1 points
The pie-chart shows the degree wise distribution of number of employees of a company in 5 different countries – A, B, C, D and E.
So 62/360 * x = 7440
Solve, x = 43200
So no. of employees in A+E = (35+55)/360 * 43200 = 10800
So 62/360 * x = 7440
Solve, x = 43200
So no. of employees in A+E = (35+55)/360 * 43200 = 10800
So 62/360 * x = 7440
Solve, x = 43200
So no. of employees in A+E = (35+55)/360 * 43200 = 10800
35. Question
1 points
The pie-chart shows the degree wise distribution of number of employees of a company in 5 different countries – A, B, C, D and E.
So now after adding country F to pie-chart, Let total number of employees in 6 countries together is now x
So
84/360 * x = 11760
Solve, x = 50400
So number of employees in F = 50400 – 43200 = 7200
So y/360 * 50400 = 7200
Solve, y = 51o
can you please explain question number 20..??
which step u dint get ?
mam actually it was just the misprint in the question that made me confuse… there is F in place of E…kindly correct it
yes its E
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尼卡西奥公司 III 作弊者