MAT Mock Test Series 5 | MAT Online Test Series 5 | MAT Sample Papers
MAT Mock Test Series 5 | MAT Online Test Series 5 | MAT Sample Papers
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MAT Online Test Series 5, MAT Free Online Test Series 5. MAT Free Mock Test Exam 2019. MAT Exam Free Online Quiz 2019, MAT Full Online Mock Test Series 5th in English. MAT Free Mock Test Series in English. MAT Free Mock Test Series 5. MAT English Language Online Test in English Series 5th. Take MAT Online Quiz. The MAT Full online mock test paper is free for all students. MAT Question and Answers in English and Hindi Series 5. Here we are providing MAT Full Mock Test Paper in English. MAT Mock Test Series 5th 2019. Now Test your self for MAT Exam by using below quiz…
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Question 1 of 50
1. Question
8/15 / 1 1/3 * 3 ?/4 = 1.5
Correct
8/15 * 3/4 * 3 ?/4 = 1.5
=> 3 ?/4 = 3/2 * 5/2 = 15/4 = 3 3/4
=> ? = 3.Incorrect
8/15 * 3/4 * 3 ?/4 = 1.5
=> 3 ?/4 = 3/2 * 5/2 = 15/4 = 3 3/4
=> ? = 3. -
Question 2 of 50
2. Question
In finding the HCF of two numbers, the last divisor was 41 and the successive quotients, starting from the first, where 2, 4 and 2. The numbers are?
Correct
Incorrect
-
Question 3 of 50
3. Question
A group of 10 representatives is to be selected out of 12 seniors and 10 juniors. In how many different ways can the group be selected, if it should have 5 seniors and 5 juniors?
Correct
Here, five seniors out of 12 seniors can be selected in ¹²C₅ ways. Also, five juniors out of ten juniors can be selected ¹⁰C₅ ways. Hence the total number of different ways of selection = ¹²C₅ * ¹⁰C₅ = ¹²C₇ * ¹⁰C₅
= ¹²C₅ = ¹²C₇Incorrect
Here, five seniors out of 12 seniors can be selected in ¹²C₅ ways. Also, five juniors out of ten juniors can be selected ¹⁰C₅ ways. Hence the total number of different ways of selection = ¹²C₅ * ¹⁰C₅ = ¹²C₇ * ¹⁰C₅
= ¹²C₅ = ¹²C₇ -
Question 4 of 50
4. Question
The radius of the base of cone is 3 cm and height is 4 cm. Find the volume of the cone?
Correct
1/3 * π * 3 * 3 * 4 = 12 π
Incorrect
1/3 * π * 3 * 3 * 4 = 12 π
-
Question 5 of 50
5. Question
The average salary of a person for the months of January, February, March and April is Rs.8000 and that for the months February, March, April and May is Rs.8500. If his salary for the month of May is Rs.6500, find his salary for the month of January?
Correct
Sum of the salaries of the person for the months of January, February, March and April = 4 * 8000 = 32000 —-(1)
Sum of the salaries of the person for the months of February, March, April and May = 4 * 8500 = 34000 —-(2)
(2)-(1) i.e. May – Jan = 2000
Salary of May is Rs.6500
Salary of January = Rs.4500Incorrect
Sum of the salaries of the person for the months of January, February, March and April = 4 * 8000 = 32000 —-(1)
Sum of the salaries of the person for the months of February, March, April and May = 4 * 8500 = 34000 —-(2)
(2)-(1) i.e. May – Jan = 2000
Salary of May is Rs.6500
Salary of January = Rs.4500 -
Question 6 of 50
6. Question
A sum of money place at compound interest doubles itself in 4 years. In how many years will it amount to eight times itself?
Correct
100 —- 200 —- 4
400 —- 4
800 —- 4
——
12 yearsIncorrect
100 —- 200 —- 4
400 —- 4
800 —- 4
——
12 years -
Question 7 of 50
7. Question
The height of two right circular cones are in the ratio 1:2 and their perimeters of their bases are in the ratio 3:4, the ratio of their volume is?
Correct
Incorrect
-
Question 8 of 50
8. Question
The ratio of the number of boys and girls in a college is 7:8. If the percentage increase in the number of boys and girls be 20% and 10% respectively. What will be the new ratio?
Correct
Originally, let the number of boys and girls in the college be 7x and 8x respectively.
Their increased number is (120% of 7x) and (110% of 8x).
i.e., (120/100 * 7x) and (110/100 * 8x)
i.e., 42x/5 and 44x/5
Required ratio = 42x/5 : 44x/5 = 21:22Incorrect
Originally, let the number of boys and girls in the college be 7x and 8x respectively.
Their increased number is (120% of 7x) and (110% of 8x).
i.e., (120/100 * 7x) and (110/100 * 8x)
i.e., 42x/5 and 44x/5
Required ratio = 42x/5 : 44x/5 = 21:22 -
Question 9 of 50
9. Question
A rectangular lawn of dimensions 80 m * 60 m has two roads each 10 m wide running in the middle of the lawn, one parallel to the length and the other parallel to the breadth. What is the cost of traveling the two roads at Rs.3 per sq m?
Correct
Area = (l + b – d) d
(80 + 60 – 10)10 => 1300 m2
1300 * 3 = Rs.3900Incorrect
Area = (l + b – d) d
(80 + 60 – 10)10 => 1300 m2
1300 * 3 = Rs.3900 -
Question 10 of 50
10. Question
A bag contains five white and four red balls. Two balls are picked at random from the bag. What is the probability that they both are different color?
Correct
Two balls can be picked from nine balls in ⁹C₂ ways.
We select one white ball and one red ball from five white balls and four red balls. This can be done ⁵C₁ . ⁴C₁ ways.
The required probability = (5 * 4)/⁹C₂ = 20/36 = 5/9Incorrect
Two balls can be picked from nine balls in ⁹C₂ ways.
We select one white ball and one red ball from five white balls and four red balls. This can be done ⁵C₁ . ⁴C₁ ways.
The required probability = (5 * 4)/⁹C₂ = 20/36 = 5/9 -
Question 11 of 50
11. Question
8.008 + 0.8 + 80.8 + 800.8 + 0.08 = ?
Correct
8.008 + 0.8 + 80.8 + 800.8 + 0.08
= 8 + 0.008 + 0.8 + 80 + 0.8 + 800 + 0.8 + 0.08
= 8 + 80 + 800 + 0.008 + 0.08 + 0.8 + 0.8 + 0.8
= 888 + 2.488 = 890.488Incorrect
8.008 + 0.8 + 80.8 + 800.8 + 0.08
= 8 + 0.008 + 0.8 + 80 + 0.8 + 800 + 0.8 + 0.08
= 8 + 80 + 800 + 0.008 + 0.08 + 0.8 + 0.8 + 0.8
= 888 + 2.488 = 890.488 -
Question 12 of 50
12. Question
There is food for 760 men for 22 days. How many more men should join after two days so that the same food may last for 19 days more?
Correct
760 —- 22
760 —- 20
x —– 19
x*19 = 760*20
x = 800
760
——-
40Incorrect
760 —- 22
760 —- 20
x —– 19
x*19 = 760*20
x = 800
760
——-
40 -
Question 13 of 50
13. Question
A man on tour travels first 160 km at 64 km/he and the next 160 km at 80 km/hr. The average speed for the first 320 km of the tour is?
Correct
Total time taken = (160/64 + 160/8) = 9/2 hrs.
Average speed = 320 * 2/9 = 71.11 km/hr.Incorrect
Total time taken = (160/64 + 160/8) = 9/2 hrs.
Average speed = 320 * 2/9 = 71.11 km/hr. -
Question 14 of 50
14. Question
Rahim bought 65 books for Rs.1150 from one shop and 50 books for Rs.920 from another. What is the average price he paid per book ?
Correct
Average price per book = (1150 + 920) / (65 + 50) = 2070 / 115 = Rs.18
Incorrect
Average price per book = (1150 + 920) / (65 + 50) = 2070 / 115 = Rs.18
-
Question 15 of 50
15. Question
The radius of a cylindrical vessel is 7cm and height is 3cm. Find the whole surface of the cylinder?
Correct
r = 7 h = 3
2πr(h + r) = 2 * 22/7 * 7(10) = 440Incorrect
r = 7 h = 3
2πr(h + r) = 2 * 22/7 * 7(10) = 440 -
Question 16 of 50
16. Question
Two trains are moving in the same direction at 72 kmph and 36 kmph. The faster train crosses a man in the slower train in 27 seconds. Find the length of the faster train?
Correct
Relative speed = (72 – 36) * 5/18 = 2 * 5 = 10 mps.
Distance covered in 27 sec = 27 * 10 = 270 m.
The length of the faster train = 270 m.Incorrect
Relative speed = (72 – 36) * 5/18 = 2 * 5 = 10 mps.
Distance covered in 27 sec = 27 * 10 = 270 m.
The length of the faster train = 270 m. -
Question 17 of 50
17. Question
The ratio of numbers is 3:4 and their H.C.F is 4. Their L.C.M is:
Correct
Let the numbers be 3x and 4x.
Then their H.C.F = x. So, x = 4.
So, the numbers are 12 and 16.
L.C.M of 12 and 16 = 48.Incorrect
Let the numbers be 3x and 4x.
Then their H.C.F = x. So, x = 4.
So, the numbers are 12 and 16.
L.C.M of 12 and 16 = 48. -
Question 18 of 50
18. Question
Two persons A and B can complete a piece of work in 30 days and 45 days respectively. If they work together, what part of the work will be completed in 3 days?
Correct
A’s one day’s work = 1/30
B’s one day’s work = 1/45
(A + B)’s one day’s work = 1/30 + 1/45 = 1/18
The part of the work completed in 3 days = 3 (1/18) = 1/6.Incorrect
A’s one day’s work = 1/30
B’s one day’s work = 1/45
(A + B)’s one day’s work = 1/30 + 1/45 = 1/18
The part of the work completed in 3 days = 3 (1/18) = 1/6. -
Question 19 of 50
19. Question
There are some rabbits and peacocks in a zoo. The total number of their heads is 60 and total number of their legs is 192. Find the number of total rabbits?
Correct
Let the number of rabbits and peacocks be ‘r’ and ‘p’ respectively. As each animal has only one head, so r + p = 60 — (1)
Each rabbit has 4 legs and each peacock has 2 legs. Total number of legs of rabbits and peacocks, 4r + 2p = 192 — (2)
Multiplying equation (1) by 2 and subtracting it from equation (2), we get
=> 2r = 72 => r = 36.Incorrect
Let the number of rabbits and peacocks be ‘r’ and ‘p’ respectively. As each animal has only one head, so r + p = 60 — (1)
Each rabbit has 4 legs and each peacock has 2 legs. Total number of legs of rabbits and peacocks, 4r + 2p = 192 — (2)
Multiplying equation (1) by 2 and subtracting it from equation (2), we get
=> 2r = 72 => r = 36. -
Question 20 of 50
20. Question
A man can row with a speed of 15 kmph in still water. If the stream flows at 5 kmph, then the speed in downstream is?
Correct
M = 15
S = 5
DS = 15 + 5 = 20Incorrect
M = 15
S = 5
DS = 15 + 5 = 20 -
Question 21 of 50
21. Question
The dimensions of a room are 25 feet * 15 feet * 12 feet. What is the cost of white washing the four walls of the room at Rs. 5 per square feet if there is one door of dimensions 6 feet * 3 feet and three windows of dimensions 4 feet * 3 feet each?
Correct
Area of the four walls = 2h(l + b)
Since there are doors and windows, area of the walls = 2 * 12 (15 + 25) – (6 * 3) – 3(4 * 3) = 906 sq.ft.
Total cost = 906 * 5 = Rs. 4530Incorrect
Area of the four walls = 2h(l + b)
Since there are doors and windows, area of the walls = 2 * 12 (15 + 25) – (6 * 3) – 3(4 * 3) = 906 sq.ft.
Total cost = 906 * 5 = Rs. 4530 -
Question 22 of 50
22. Question
An automobile financier claims to be lending money at S.I., but he includes the interest every six months for calculating the principal. If he is charging an interest of 10%, the effective rate of interest becomes?
Correct
Let the sum be Rs. 100. Then,
S.I. for first 6 months = (100 * 10 *1) / (100 * 2) = Rs. 5
S.I. for last 6 months = (105 * 10 * 1) / (100 * 2) = Rs. 5.25
So, amount at the end of 1 year = (100 + 5 + 5.25) = Rs. 110.25
Effective rate = (110.25 – 100) = 10.25%.Incorrect
Let the sum be Rs. 100. Then,
S.I. for first 6 months = (100 * 10 *1) / (100 * 2) = Rs. 5
S.I. for last 6 months = (105 * 10 * 1) / (100 * 2) = Rs. 5.25
So, amount at the end of 1 year = (100 + 5 + 5.25) = Rs. 110.25
Effective rate = (110.25 – 100) = 10.25%. -
Question 23 of 50
23. Question
The salary of a typist was at first raised by 10% and then the same was reduced by 5%. If he presently draws Rs.1045.What was his original salary?
Correct
X * (110/100) * (95/100) = 1045
X * (11/10) * (1/100) = 11
X = 1000Incorrect
X * (110/100) * (95/100) = 1045
X * (11/10) * (1/100) = 11
X = 1000 -
Question 24 of 50
24. Question
A sum amount to Rs.1344 in two years at simple interest. What will be the compound interest on the same sum at the same rate of interest for the same period?
Correct
Incorrect
-
Question 25 of 50
25. Question
Two trains start from P and Q respectively and travel towards each other at a speed of 50 km/hr and 40 km/hr respectively. By the time they meet, the first train has traveled 100 km more than the second. The distance between P and Q is?
Correct
At the time of meeting, let the distance traveled by the second train be x km. Then, distance covered by the first train is (x + 100) km.
x/40 = (x + 100)/50
50x = 40x + 4000 => x = 400
So, distance between P and Q = (x + x + 100)km = 900 km.Incorrect
At the time of meeting, let the distance traveled by the second train be x km. Then, distance covered by the first train is (x + 100) km.
x/40 = (x + 100)/50
50x = 40x + 4000 => x = 400
So, distance between P and Q = (x + x + 100)km = 900 km. -
Question 26 of 50
26. Question
The diagonals of a rhombus are 15 cm and 20 cm. Find its area?
Correct
1/2 * 15 * 20 = 150
Incorrect
1/2 * 15 * 20 = 150
-
Question 27 of 50
27. Question
One train is traveling 45 kmph and other is at 10 meters a second. Ratio of the speed of the two trains is?
Correct
45 * 5/18 = 10
25:20 => 5:4Incorrect
45 * 5/18 = 10
25:20 => 5:4 -
Question 28 of 50
28. Question
The sum of the two numbers is 12 and their product is 35. What is the sum of the reciprocals of these numbers?
Correct
Let the numbers be a and b. Then,
a + b = 12 and ab = 35.
(a + b)/ab = 12/35
= (1/b + 1/a) = 12/35
Sum of reciprocals of given numbers = 12/35.Incorrect
Let the numbers be a and b. Then,
a + b = 12 and ab = 35.
(a + b)/ab = 12/35
= (1/b + 1/a) = 12/35
Sum of reciprocals of given numbers = 12/35. -
Question 29 of 50
29. Question
Sixty men can stitch 200 shirts in 30 days working 8 hours a day. In how many days can 45 men stitch 300 shirts working 6 hours a day?
Correct
We have M1 D1 H1 / W1 = M2 D2 H2 / W2 (Variation rule)
(60 * 30 * 8)/ 200 = (45 * D2 * 6) / 300
D2 = (60 * 30 * 8 * 300) / (200 * 45 * 6) => D2 = 80.Incorrect
We have M1 D1 H1 / W1 = M2 D2 H2 / W2 (Variation rule)
(60 * 30 * 8)/ 200 = (45 * D2 * 6) / 300
D2 = (60 * 30 * 8 * 300) / (200 * 45 * 6) => D2 = 80. -
Question 30 of 50
30. Question
Mahesh marks an article 15% above the cost price of Rs. 540. What must be his discount percentage if he sells it at Rs. 496.80?
Correct
CP = Rs. 540, MP = 540 + 15% of 540 = Rs. 621
SP = Rs. 496.80, Discount = 621 – 496.80 = 124.20
Discount % = 124.2/621 * 100 = 20%Incorrect
CP = Rs. 540, MP = 540 + 15% of 540 = Rs. 621
SP = Rs. 496.80, Discount = 621 – 496.80 = 124.20
Discount % = 124.2/621 * 100 = 20% -
Question 31 of 50
31. Question
A candidate who gets 30% of the marks fails by 50 marks. But another candidate who gets 45% marks gets 25 marks more than necessary for passing. Find the number of marks for passing?
Correct
30% ———— 50
45% ———— 25
———————-
15% ————- 75
30% ————– ?
150 + 50 = 200 MarksIncorrect
30% ———— 50
45% ———— 25
———————-
15% ————- 75
30% ————– ?
150 + 50 = 200 Marks -
Question 32 of 50
32. Question
Excluding stoppages, the speed of a bus is 54 km/hr and including stoppages, it is 45 km/hr. For how many minutes does the bus stop per hour?
Correct
Due to stoppages, it covers 9 km less.
Time taken to cover 9 km = 9/54 * 60 = 10 min.Incorrect
Due to stoppages, it covers 9 km less.
Time taken to cover 9 km = 9/54 * 60 = 10 min. -
Question 33 of 50
33. Question
If a: b = 3:4, b:c = 7:9, c:d = 5:7, find a:d?
Correct
a/d = (3/4)*(7/9)*(5/7) => 5/12
Incorrect
a/d = (3/4)*(7/9)*(5/7) => 5/12
-
Question 34 of 50
34. Question
What is the least number, which is perfect square but contains 2700 as its factor?
Correct
900 * 3 * 3 = 8100
Incorrect
900 * 3 * 3 = 8100
-
Question 35 of 50
35. Question
If the L.C.M of two numbers is 750 and their product is 18750, find the H.C.F of the numbers.
Correct
H.C.F = (Product of the numbers) / (Their L.C.M) = 18750/750 = 25.
Incorrect
H.C.F = (Product of the numbers) / (Their L.C.M) = 18750/750 = 25.
-
Question 36 of 50
36. Question
In a hostel there were 100 students. To accommodate 20 more students the average is decreased by rupees 5. But total expenditure increased by Rs.400. Find the total expenditure of the hostel now?
Correct
100x + 400 = 12(x – 5)
x = 50
100 * 50 + 400 = 5400Incorrect
100x + 400 = 12(x – 5)
x = 50
100 * 50 + 400 = 5400 -
Question 37 of 50
37. Question
The weights of three boys are in the ratio 4 : 5 : 6. If the sum of the weights of the heaviest and the lightest boy is 45 kg more than the weight of the third boy, what is the weight of the lightest boy?
Correct
Let the weights of the three boys be 4k, 5k and 6k respectively.
4k + 6k = 5k + 45
=> 5k = 45 => k = 9
Therefore the weight of the lightest boy
= 4k = 4(9) = 36 kg.Incorrect
Let the weights of the three boys be 4k, 5k and 6k respectively.
4k + 6k = 5k + 45
=> 5k = 45 => k = 9
Therefore the weight of the lightest boy
= 4k = 4(9) = 36 kg. -
Question 38 of 50
38. Question
A box contains nine bulbs out of which 4 are defective. If four bulbs are chosen at random, find the probability that atleast one bulb is good.
Correct
Required probability = 1 – 1/126 = 125/126
Incorrect
Required probability = 1 – 1/126 = 125/126
-
Question 39 of 50
39. Question
A man said to his son, “I was two-third of your present age when you were born”. If the present age of the man is 48 years, find the present age of the son?
Correct
Present age of the son be P, he was born P years ago.
The age of the man was: (48 – P).
His age when the son was born should be equal to 2/3 of P.
(48 – P) = 2/3 P
5P = 144 => P = 28.8Incorrect
Present age of the son be P, he was born P years ago.
The age of the man was: (48 – P).
His age when the son was born should be equal to 2/3 of P.
(48 – P) = 2/3 P
5P = 144 => P = 28.8 -
Question 40 of 50
40. Question
The speed of a car is 90 km in the first hour and 60 km in the second hour. What is the average speed of the car?
Correct
S = (90 + 60)/2 = 75 kmph
Incorrect
S = (90 + 60)/2 = 75 kmph
-
Question 41 of 50
41. Question
If Rs.450 amount to Rs.540 in 4 years, what will it amount to in 6 years at the same rate % per annum?
Correct
90 = (450*4*R)/100
R = 5%
I = (450*6*5)/100 = 135
450 + 135 = 585Incorrect
90 = (450*4*R)/100
R = 5%
I = (450*6*5)/100 = 135
450 + 135 = 585 -
Question 42 of 50
42. Question
3 men, 4 women and 6 children can complete a work in 7 days. A woman does double the work a man does and a child does half the work a man does. How many women alone can complete this work in 7 days?
Correct
Let 1 woman’s 1 day work = x.
Then, 1 man’s 1 day work = x/2 and 1 child’s 1 day work x/4.
So, (3x/2 + 4x + + 6x/4) = 1/7
28x/4 = 1/7 => x = 1/49
1 woman alone can complete the work in 49 days.
So, to complete the work in 7 days, number of women required = 49/7 = 7.Incorrect
Let 1 woman’s 1 day work = x.
Then, 1 man’s 1 day work = x/2 and 1 child’s 1 day work x/4.
So, (3x/2 + 4x + + 6x/4) = 1/7
28x/4 = 1/7 => x = 1/49
1 woman alone can complete the work in 49 days.
So, to complete the work in 7 days, number of women required = 49/7 = 7. -
Question 43 of 50
43. Question
Convert the 13/36 m/s into kilometers per hour?
Correct
13/36 m/s = 13/36 * 18/5 = 13/10 = 1.3 kmph.
Incorrect
13/36 m/s = 13/36 * 18/5 = 13/10 = 1.3 kmph.
-
Question 44 of 50
44. Question
A and B can do a work in 12 days and 36 days respectively. If they work on alternate days beginning with B, in how many days will the work be completed?
Correct
The work done in the first two days = 1/12 + 1/36 = 1/9
so, 9 such two days are required to finish the work.
i.e., 18 days are required to finish the work.Incorrect
The work done in the first two days = 1/12 + 1/36 = 1/9
so, 9 such two days are required to finish the work.
i.e., 18 days are required to finish the work. -
Question 45 of 50
45. Question
The volumes of two cubes are in the ratio 27: 125, what shall be the ratio of their surface areas?
Correct
a13 : a23 = 27 : 125
a1 : a2 = 3 : 5
6 a12 : 6 a22
a12 : a22 = 9 : 25Incorrect
a13 : a23 = 27 : 125
a1 : a2 = 3 : 5
6 a12 : 6 a22
a12 : a22 = 9 : 25 -
Question 46 of 50
46. Question
A can do a piece of work in 30 days. He works at it for 5 days and then B finishes it in 20 days. In what time can A and B together it?
Correct
5/30 + 20/x = 1
x = 24
1/30 + 1/24 = 3/40
40/3 = 13 1/3 daysIncorrect
5/30 + 20/x = 1
x = 24
1/30 + 1/24 = 3/40
40/3 = 13 1/3 days -
Question 47 of 50
47. Question
Two trains of length 100 m and 200 m are 100 m apart. They start moving towards each other on parallel tracks, at speeds 54 kmph and 72 kmph. After how much time will the trains meet?
Correct
They are moving in opposite directions, relative speed is equal to the sum of their speeds.
Relative speed = (54 + 72)*5/18 = 7*5 = 35 mps.
The time required = d/s = 100/35 = 20/7 sec.Incorrect
They are moving in opposite directions, relative speed is equal to the sum of their speeds.
Relative speed = (54 + 72)*5/18 = 7*5 = 35 mps.
The time required = d/s = 100/35 = 20/7 sec. -
Question 48 of 50
48. Question
The sum of four consecutive even numbers is 36. Find the sum of the squares of these numbers?
Correct
Let the four numbers be x, x + 2, x + 4 and x + 6.
=> x + x + 2 + x + 4 + x + 6 = 36
=> 4x + 12 = 36 => x = 6
The numbers are 6, 8, 10 and 12.
Sum of their squares = 62 + 82 + 102 + 122 = 36 + 64 + 100 + 144 = 344.Incorrect
Let the four numbers be x, x + 2, x + 4 and x + 6.
=> x + x + 2 + x + 4 + x + 6 = 36
=> 4x + 12 = 36 => x = 6
The numbers are 6, 8, 10 and 12.
Sum of their squares = 62 + 82 + 102 + 122 = 36 + 64 + 100 + 144 = 344. -
Question 49 of 50
49. Question
A box contains 3 blue marbles, 4 red, 6 green marbles and 2 yellow marbles. If two marbles are picked at random, what is the probability that they are either blue or yellow?
Correct
Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles. Probability that both marbles are blue = ³C₂/¹⁵C₂ = (3 * 2)/(15 * 14) = 1/35
Probability that both are yellow = ²C₂/¹⁵C₂ = (2 * 1)/(15 * 14) = 1/105
Probability that one blue and other is yellow = (³C₁ * ²C₁)/¹⁵C₂ = (2 * 3 * 2)/(15 * 14) = 2/35
Required probability = 1/35 + 1/105 + 2/35
= 3/35 + 1/105 = 1/35(3 + 1/3)
= 10/(3 * 35) = 2/21Incorrect
Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles. Probability that both marbles are blue = ³C₂/¹⁵C₂ = (3 * 2)/(15 * 14) = 1/35
Probability that both are yellow = ²C₂/¹⁵C₂ = (2 * 1)/(15 * 14) = 1/105
Probability that one blue and other is yellow = (³C₁ * ²C₁)/¹⁵C₂ = (2 * 3 * 2)/(15 * 14) = 2/35
Required probability = 1/35 + 1/105 + 2/35
= 3/35 + 1/105 = 1/35(3 + 1/3)
= 10/(3 * 35) = 2/21 -
Question 50 of 50
50. Question
Mohit sold an article for Rs. 18000. Had he offered a discount of 10% on the selling price, he would have earned a profit of 8%. What is the cost price of the article?
Correct
Let the CP be Rs. x.
Had he offered 10% discount, profit = 8%
Profit = 8/100 x and hence his SP = x + 8/100 x = Rs. 1.08x = 18000 – 10/100(18000) = 18000 – 1800 = Rs. 16200
=> 1.08x = 16200
=> x = 15000Incorrect
Let the CP be Rs. x.
Had he offered 10% discount, profit = 8%
Profit = 8/100 x and hence his SP = x + 8/100 x = Rs. 1.08x = 18000 – 10/100(18000) = 18000 – 1800 = Rs. 16200
=> 1.08x = 16200
=> x = 15000
