Directions (1 – 5): What approximate value should come in place of the question mark (?) in the following questions?
27.03 × 523.93 ÷ 4 + 39.11 – 23.11 = ?
Correct
Incorrect
Unattempted
Question 3 of 35
3. Question
1 points
Directions (1 – 5): What approximate value should come in place of the question mark (?) in the following questions?
2256.123 – 1864.986 + 567.09 – 258.89 = ?
Correct
Incorrect
Unattempted
Question 4 of 35
4. Question
1 points
Directions (1 – 5): What approximate value should come in place of the question mark (?) in the following questions?
16.06% of 250.05 + 24.99 * 15.09 = 45.05% of 1000 – √?
Correct
Incorrect
Unattempted
Question 5 of 35
5. Question
1 points
Directions (1 – 5): What approximate value should come in place of the question mark (?) in the following questions?
25.09 × 7.01 + 23.92% of 249.93 – √492 = ?
Correct
Incorrect
Unattempted
Question 6 of 35
6. Question
1 points
There are 4 red, 3 black and 5 green balls in a box. 3 balls are chosen at random. What is the probability that at least 2 balls are green in color?
Correct
Case 1 : 2 green and 1 non-green
So probability = ^{5}C_{2 }* ^{7}C_{1}/^{12}C_{3}
Case 1 : all 3 green
So probability = ^{5}C_{3}/^{12}C_{3}
Add the two cases
Incorrect
Case 1 : 2 green and 1 non-green
So probability = ^{5}C_{2 }* ^{7}C_{1}/^{12}C_{3}
Case 1 : all 3 green
So probability = ^{5}C_{3}/^{12}C_{3}
Add the two cases
Unattempted
Case 1 : 2 green and 1 non-green
So probability = ^{5}C_{2 }* ^{7}C_{1}/^{12}C_{3}
Case 1 : all 3 green
So probability = ^{5}C_{3}/^{12}C_{3}
Add the two cases
Question 7 of 35
7. Question
1 points
Ratio of age of A 3 years hence and B 7 years ago is 4 : 3. The average of ages of B and C is 27 years. Twice the age of A 6 years ago is equal to the age of C 1 year hence. Find the sum of ages of all three A, B and C.
Correct
(A+3)/(B-7) = 4/3
B+C = 54
2(A-6) = C+1
Solve all equations
A = 21, B = 25, C = 29
Incorrect
(A+3)/(B-7) = 4/3
B+C = 54
2(A-6) = C+1
Solve all equations
A = 21, B = 25, C = 29
Unattempted
(A+3)/(B-7) = 4/3
B+C = 54
2(A-6) = C+1
Solve all equations
A = 21, B = 25, C = 29
Question 8 of 35
8. Question
1 points
There are two filling pipes A and B. If A fills bottom 3/4^{th} tank and B fills the rest tank, then they can fill the tank in 18 minutes. If B fills bottom 3/4^{th} tank and A fills the rest tank, then they can fill the tank in 22 minutes. What is the time taken by both the pipes to fill the tank together?
Correct
Let A can fill complete tank in x minutes, then 3/4 in 3/4 * x minutes
Let B can fill tank in y minutes, so 1/4 in 1/4 * x minutes
So from given information:
3x/4 + y/4 = 18
and x/4 + 3y/4 = 22
Solve both the equations, x = 16, y = 24
So together they complete 1/16 + 1/24 = 5/48 tank in 1 minute or whole tank in 48/5 minutes
Incorrect
Let A can fill complete tank in x minutes, then 3/4 in 3/4 * x minutes
Let B can fill tank in y minutes, so 1/4 in 1/4 * x minutes
So from given information:
3x/4 + y/4 = 18
and x/4 + 3y/4 = 22
Solve both the equations, x = 16, y = 24
So together they complete 1/16 + 1/24 = 5/48 tank in 1 minute or whole tank in 48/5 minutes
Unattempted
Let A can fill complete tank in x minutes, then 3/4 in 3/4 * x minutes
Let B can fill tank in y minutes, so 1/4 in 1/4 * x minutes
So from given information:
3x/4 + y/4 = 18
and x/4 + 3y/4 = 22
Solve both the equations, x = 16, y = 24
So together they complete 1/16 + 1/24 = 5/48 tank in 1 minute or whole tank in 48/5 minutes
Question 9 of 35
9. Question
1 points
A boat can cover 24 km against the current of the stream in 4 hours and can cover 32 km along the current in 2 hours. How long will it take to go 5 km in still water?
Correct
Upstream speed = 24/4 = 6 km/hr
Upstream speed = 32/2 = 16 km/hr
So speed of boat in still water = 1/2 (6+16) = 11 km/hr
So time taken to travel 5km in still water = 5/11 hours
Incorrect
Upstream speed = 24/4 = 6 km/hr
Upstream speed = 32/2 = 16 km/hr
So speed of boat in still water = 1/2 (6+16) = 11 km/hr
So time taken to travel 5km in still water = 5/11 hours
Unattempted
Upstream speed = 24/4 = 6 km/hr
Upstream speed = 32/2 = 16 km/hr
So speed of boat in still water = 1/2 (6+16) = 11 km/hr
So time taken to travel 5km in still water = 5/11 hours
Question 10 of 35
10. Question
1 points
At simple interest, an amount increases by 60% in 5 years. What will be the compound interest obtained on Rs 25,000 after 2 years at same rate percent per annum?
Correct
Is x is principal amount, SI = 60% of x
SO 60% of x = x*5*r/100
Solve, r = 12%
So
CI = 25000 [1 + 12/100]^{2} – 25000 = Rs 6360
Incorrect
Is x is principal amount, SI = 60% of x
SO 60% of x = x*5*r/100
Solve, r = 12%
So
CI = 25000 [1 + 12/100]^{2} – 25000 = Rs 6360
Unattempted
Is x is principal amount, SI = 60% of x
SO 60% of x = x*5*r/100
Solve, r = 12%
So
CI = 25000 [1 + 12/100]^{2} – 25000 = Rs 6360
Question 11 of 35
11. Question
1 points
Directions (11-15): Study the following pie-chart and table to answer the questions that follow:
Pie-chart shows the distribution of students from 6 different states who took part in a quiz.
Total number of boys from states B and C together is approximately what per cent of the total number of boys participated in quiz?
Correct
Total boys from B = 5/8 * 16% of 4500 = 450
Total boys from C = 360
Similarly find the number of boys from all states and add so total number of boys = 2353
So required % = 810/2353 * 100 = 34%
Incorrect
Total boys from B = 5/8 * 16% of 4500 = 450
Total boys from C = 360
Similarly find the number of boys from all states and add so total number of boys = 2353
So required % = 810/2353 * 100 = 34%
Unattempted
Total boys from B = 5/8 * 16% of 4500 = 450
Total boys from C = 360
Similarly find the number of boys from all states and add so total number of boys = 2353
So required % = 810/2353 * 100 = 34%
Question 12 of 35
12. Question
1 points
Directions (11-15): Study the following pie-chart and table to answer the questions that follow:
Pie-chart shows the distribution of students from 6 different states who took part in a quiz.
What is the difference between total number of boys and total number of girls from all states together?
Correct
Total boys = 2353
So girls = 4500 – 2353 = 2147
So required difference = 2353 – 2147 = 206
Incorrect
Total boys = 2353
So girls = 4500 – 2353 = 2147
So required difference = 2353 – 2147 = 206
Unattempted
Total boys = 2353
So girls = 4500 – 2353 = 2147
So required difference = 2353 – 2147 = 206
Question 13 of 35
13. Question
1 points
Directions (11-15): Study the following pie-chart and table to answer the questions that follow:
Pie-chart shows the distribution of students from 6 different states who took part in a quiz.
What is the ratio between the total number of boys from states A and D together and the total number of girls from same states together?
Correct
4/9 * 18 + 3/5 * 15 : 5/9 * 18 + 2/5 * 15
17 : 16
Incorrect
4/9 * 18 + 3/5 * 15 : 5/9 * 18 + 2/5 * 15
17 : 16
Unattempted
4/9 * 18 + 3/5 * 15 : 5/9 * 18 + 2/5 * 15
17 : 16
Question 14 of 35
14. Question
1 points
Directions (11-15): Study the following pie-chart and table to answer the questions that follow:Pie-chart shows the distribution of students from 6 different states who took part in a quiz.
Total number of boys from states A and B together is approximately what percent of the total number of girls from states E and F together?
Correct
boys from states A and B = 360+450 = 810
girls from states E and F = 320+432 = 752
So required % = 810*100/752= 108%
Incorrect
boys from states A and B = 360+450 = 810
girls from states E and F = 320+432 = 752
So required % = 810*100/752= 108%
Unattempted
boys from states A and B = 360+450 = 810
girls from states E and F = 320+432 = 752
So required % = 810*100/752= 108%
Question 15 of 35
15. Question
1 points
Directions (11-15): Study the following pie-chart and table to answer the questions that follow:
Pie-chart shows the distribution of students from 6 different states who took part in a quiz.
Total number of boys from states C and D is approximately what percent greater than the total number of girls from states B and D?
Correct
number of boys from states C and D = 360+405 = 765
number of girls from states B and D = 270+270 = 540
So required% = (765-540)/540 * 100 = 42%
Incorrect
number of boys from states C and D = 360+405 = 765
number of girls from states B and D = 270+270 = 540
So required% = (765-540)/540 * 100 = 42%
Unattempted
number of boys from states C and D = 360+405 = 765
number of girls from states B and D = 270+270 = 540
So required% = (765-540)/540 * 100 = 42%
Question 16 of 35
16. Question
1 points
Directions (15-20): In the following questions, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly-
I. 20x^{2} – 31x + 12 = 0, II. 6y^{2} – 7y + 2 = 0
Correct
20x^{2} – 31x + 12 = 0
20x^{2} – 16x – 15x + 12 = 0
So x = 3/4, 4/5
6y^{2} – 7y + 2 = 0
6y^{2} – 3y – 4y + 2 = 0
So y = 1/2, 2/3
Put on number line
1/2… 2/3… 3/4… 4/5
Incorrect
20x^{2} – 31x + 12 = 0
20x^{2} – 16x – 15x + 12 = 0
So x = 3/4, 4/5
6y^{2} – 7y + 2 = 0
6y^{2} – 3y – 4y + 2 = 0
So y = 1/2, 2/3
Put on number line
1/2… 2/3… 3/4… 4/5
Unattempted
20x^{2} – 31x + 12 = 0
20x^{2} – 16x – 15x + 12 = 0
So x = 3/4, 4/5
6y^{2} – 7y + 2 = 0
6y^{2} – 3y – 4y + 2 = 0
So y = 1/2, 2/3
Put on number line
1/2… 2/3… 3/4… 4/5
Question 17 of 35
17. Question
1 points
Directions (15-20): In the following questions, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly-
I. 6x^{2} – x – 2 = 0,
II. 5y^{2} – 18y + 9 = 0
Correct
6x^{2} – x – 2 = 0
6x^{2} + 3x – 4x – 2 = 0
So x = -1/2, 2/3
5y^{2} – 18y + 9 = 0
5y^{2} – 15y – 3y + 9 = 0
So y = 3/5, 3
Put on number line
-1/2 …. 3/5 ….2/3 …. 3
Incorrect
6x^{2} – x – 2 = 0
6x^{2} + 3x – 4x – 2 = 0
So x = -1/2, 2/3
5y^{2} – 18y + 9 = 0
5y^{2} – 15y – 3y + 9 = 0
So y = 3/5, 3
Put on number line
-1/2 …. 3/5 ….2/3 …. 3
Unattempted
6x^{2} – x – 2 = 0
6x^{2} + 3x – 4x – 2 = 0
So x = -1/2, 2/3
5y^{2} – 18y + 9 = 0
5y^{2} – 15y – 3y + 9 = 0
So y = 3/5, 3
Put on number line
-1/2 …. 3/5 ….2/3 …. 3
Question 18 of 35
18. Question
1 points
Directions (15-20): In the following questions, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly-
I. 3x^{2} – 10x + 8 = 0,
II. 3y^{2} + 8y – 16 = 0
Correct
3x^{2} – 10x + 8 = 0
3x^{2} – 6x – 4x + 8 = 0
So x = 2, 4/3
3y^{2} + 8y – 16 = 0
3y^{2} + 12y – 4y – 16 = 0
So y = -4, 4/3
Put on number line
-4 …. 4/3…. 2
Incorrect
3x^{2} – 10x + 8 = 0
3x^{2} – 6x – 4x + 8 = 0
So x = 2, 4/3
3y^{2} + 8y – 16 = 0
3y^{2} + 12y – 4y – 16 = 0
So y = -4, 4/3
Put on number line
-4 …. 4/3…. 2
Unattempted
3x^{2} – 10x + 8 = 0
3x^{2} – 6x – 4x + 8 = 0
So x = 2, 4/3
3y^{2} + 8y – 16 = 0
3y^{2} + 12y – 4y – 16 = 0
So y = -4, 4/3
Put on number line
-4 …. 4/3…. 2
Question 19 of 35
19. Question
1 points
Directions (15-20): In the following questions, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly-
I. 5x^{2} – 7x – 6 = 0,
II. 5y^{2} + 23y + 12 = 0
Correct
5x^{2} – 7x – 6 = 0
5x^{2} – 10x + 3x – 6 = 0
So x = -3/5, 2
5y^{2} + 23y + 12 = 0
5y^{2} + 20y + 3y + 12 = 0
So y = -4, -3/5
Put on number line
-4….. -3/5…. 2
Incorrect
5x^{2} – 7x – 6 = 0
5x^{2} – 10x + 3x – 6 = 0
So x = -3/5, 2
5y^{2} + 23y + 12 = 0
5y^{2} + 20y + 3y + 12 = 0
So y = -4, -3/5
Put on number line
-4….. -3/5…. 2
Unattempted
5x^{2} – 7x – 6 = 0
5x^{2} – 10x + 3x – 6 = 0
So x = -3/5, 2
5y^{2} + 23y + 12 = 0
5y^{2} + 20y + 3y + 12 = 0
So y = -4, -3/5
Put on number line
-4….. -3/5…. 2
Question 20 of 35
20. Question
1 points
Directions (15-20): In the following questions, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly-
I. 3x^{2} + 16x + 20 = 0
II. 3y^{2} – 14y – 5 = 0
Correct
3x^{2} + 16x + 20 = 0
3x^{2} + 6x + 10x + 20 = 0
Gives x = -2, -10/3
3y^{2} – 14y – 5 = 0
3y^{2} – 15y + y – 5 = 0
Gives y = -1/3, 5
Put all values on number line and analyze the relationship
-10/3… -2…. -1/3…. 5
Incorrect
3x^{2} + 16x + 20 = 0
3x^{2} + 6x + 10x + 20 = 0
Gives x = -2, -10/3
3y^{2} – 14y – 5 = 0
3y^{2} – 15y + y – 5 = 0
Gives y = -1/3, 5
Put all values on number line and analyze the relationship
-10/3… -2…. -1/3…. 5
Unattempted
3x^{2} + 16x + 20 = 0
3x^{2} + 6x + 10x + 20 = 0
Gives x = -2, -10/3
3y^{2} – 14y – 5 = 0
3y^{2} – 15y + y – 5 = 0
Gives y = -1/3, 5
Put all values on number line and analyze the relationship
-10/3… -2…. -1/3…. 5
Question 21 of 35
21. Question
1 points
Price of 15 kg of wheat is Rs 450. Price of 20 kg of wheat is equal to price of 11 kg of rice. Price of 30 kg of rice is equal to price of 18 kg of pulses. Price of 16 kg of pulses is equal to price of 25 kg of barley. Find the price of 22 kg of barley.
Correct
Rs 450 = 15 kg of wheat
20 kg of wheat = 11 kg of rice
30 kg of rice = 18 kg of pulses
16 kg of pulses = 25 kg of barley
22 kg of barley = x
So
x = 450*20*30*16*22/15*11*18*25 = Rs 1280
Incorrect
Rs 450 = 15 kg of wheat
20 kg of wheat = 11 kg of rice
30 kg of rice = 18 kg of pulses
16 kg of pulses = 25 kg of barley
22 kg of barley = x
So
x = 450*20*30*16*22/15*11*18*25 = Rs 1280
Unattempted
Rs 450 = 15 kg of wheat
20 kg of wheat = 11 kg of rice
30 kg of rice = 18 kg of pulses
16 kg of pulses = 25 kg of barley
22 kg of barley = x
So
x = 450*20*30*16*22/15*11*18*25 = Rs 1280
Question 22 of 35
22. Question
1 points
A and B invested Rs 11200 and Rs 12600 respectively in a business. After 5 months A added Rs 1300 and added Rs 400 to their investments. If after a year share of B is Rs 11000, what is the total profit?
Correct
Ratio of shares A : B
11200*5 + 12500*7 : 12600*5 + 13000*7
112*5 + 125*7 : 126*5 + 130*7
112 + 25*7 : 126 + 26*7
16 + 25 : 18 + 26
41 : 44
So
44/85 * x = 11000
So x = Rs 21,250
Incorrect
Ratio of shares A : B
11200*5 + 12500*7 : 12600*5 + 13000*7
112*5 + 125*7 : 126*5 + 130*7
112 + 25*7 : 126 + 26*7
16 + 25 : 18 + 26
41 : 44
So
44/85 * x = 11000
So x = Rs 21,250
Unattempted
Ratio of shares A : B
11200*5 + 12500*7 : 12600*5 + 13000*7
112*5 + 125*7 : 126*5 + 130*7
112 + 25*7 : 126 + 26*7
16 + 25 : 18 + 26
41 : 44
So
44/85 * x = 11000
So x = Rs 21,250
Question 23 of 35
23. Question
1 points
A container contains a mixture of two liquids A and B in the ratio 7 : 5. When 9 litres of mixture are drawn off and the container is filled with B, the ratio of A and B becomes 7 : 9. How many litres of liquid A was contained in the container initially?
Correct
7x and 5x
Let total mixture = 7x+5x+9 = 12x + 9
Now 9 l drawn and container filled again (with Q) so 9 l added
So
7x/(5x+9) = 7/9
Solve, x = 9/4
So total mixture = 12 * 9/4 + 9 = 36
So A = 7/12 * 36 = 21 l
Incorrect
7x and 5x
Let total mixture = 7x+5x+9 = 12x + 9
Now 9 l drawn and container filled again (with Q) so 9 l added
So
7x/(5x+9) = 7/9
Solve, x = 9/4
So total mixture = 12 * 9/4 + 9 = 36
So A = 7/12 * 36 = 21 l
Unattempted
7x and 5x
Let total mixture = 7x+5x+9 = 12x + 9
Now 9 l drawn and container filled again (with Q) so 9 l added
So
7x/(5x+9) = 7/9
Solve, x = 9/4
So total mixture = 12 * 9/4 + 9 = 36
So A = 7/12 * 36 = 21 l
Question 24 of 35
24. Question
1 points
A man bought an article and sold it at a gain of 10 %. If he had bought it at 5% less and sold it for Rs 16 more, he would have made a profit of 20%. The C.P. of the article was
Correct
Let original Cost price is x
Its Selling price = 110/100 * x = 11x/10
New Cost price = 95/100 * x = 19x/20
New Selling price = 120/100 * 19x/20 = 209x/200
[(114x/100) – (11x/10)] = 116
Solve, x = Rs 400
Incorrect
Let original Cost price is x
Its Selling price = 110/100 * x = 11x/10
New Cost price = 95/100 * x = 19x/20
New Selling price = 120/100 * 19x/20 = 209x/200
[(114x/100) – (11x/10)] = 116
Solve, x = Rs 400
Unattempted
Let original Cost price is x
Its Selling price = 110/100 * x = 11x/10
New Cost price = 95/100 * x = 19x/20
New Selling price = 120/100 * 19x/20 = 209x/200
[(114x/100) – (11x/10)] = 116
Solve, x = Rs 400
Question 25 of 35
25. Question
1 points
Find the area of trapezium whose parallel sides are 20 cm and 18 cm long, and the distance between them is 15 cm.
Correct
Area of a trapezium = 1/2 * (sum of parallel sides) * (perpendicular distance between them) = 1/2 * (20 + 18) * 15 = 285 cm^{2}
Incorrect
Area of a trapezium = 1/2 * (sum of parallel sides) * (perpendicular distance between them) = 1/2 * (20 + 18) * 15 = 285 cm^{2}
Unattempted
Area of a trapezium = 1/2 * (sum of parallel sides) * (perpendicular distance between them) = 1/2 * (20 + 18) * 15 = 285 cm^{2}
Question 26 of 35
26. Question
1 points
Directions (26-30): Study the following table to answer the questions that follow:
What is the average amount of interest per year which the company had to pay during this period?
Correct
Incorrect
Unattempted
Question 27 of 35
27. Question
1 points
Directions (26-30): Study the following table to answer the questions that follow:
The total amount of bonus paid by the company during the given period is approximately what percent of the total amount of salary paid during this period?
Correct
Incorrect
Unattempted
Question 28 of 35
28. Question
1 points
Directions (26-30): Study the following table to answer the questions that follow:
Total expenditure on all these items in 1998 was approximately what percent of the total expenditure in 2002?
Directions (26-30): Study the following table to answer the questions that follow:
The total expenditure of the company over these items during the year 2000 is?
Correct
Incorrect
Unattempted
Question 30 of 35
30. Question
1 points
Directions (26-30): Study the following table to answer the questions that follow:
The ratio between the total expenditure on Taxes for all the years and the total expenditure on Fuel and Transport for all the years respectively is approximately?
17 of 35 questions answered correctly
Your time: 00:24:00
appx k ques m kuch prob h
Which one ??
please check the 9 question’s solution it should b 5/11
Yes yes
mam 30th question main to ratio ayegi hi nhi 10:13 …uska total 451 and 586 ban rha hai
Ques me approx is asked
ok mam 🙂 thanku
sir/mam i appreciate and grateful to ur work plz provide more and more quiz on DI ..
Yes we ll provide