## CAT Online Test Series 1, CAT Free Mock Test | Free CAT Quiz

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**CAT Online Test in English Series 1, CAT Free Mock Test Series 1.** Common Admission Test (CAT) is one of the most challenging and competitive MBA entrance test in our country. Check your level of preparation for *CAT* with free All India *Mock test* 2019. RRB Exam Online Test 2019, CAT Free Mock Test Exam 2019. CAT Exam Free Online Quiz 2019, CAT Full Online Mock Test **Series 1st** in English. RRB Online Test for All Subjects, CAT Free Mock Test Series in English. CAT Free Mock Test **Series 1.** CAT English Language Online Test in English **Series 1st**. Here we are providing** CAT Full Mock Test Paper in English. CAT **Mock Test **Series 1st** 2019. Now Test your self for CAT Exam by using below quiz…

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- Question 1 of 50
##### 1. Question

A question paper consists of five problems, each problem having three internal choices. In how many ways can a candidate attempt one or more problems?

CorrectGiven that, the question paper consists of five problems. For each problem, one or two or three or none of the choices can be attempted.

Hence, the required number of ways = 4^{5}– 1.

= 2^{10}– 1 = 1024 – 1 = 1023IncorrectGiven that, the question paper consists of five problems. For each problem, one or two or three or none of the choices can be attempted.

Hence, the required number of ways = 4^{5}– 1.

= 2^{10}– 1 = 1024 – 1 = 1023 - Question 2 of 50
##### 2. Question

The denominator of a fraction is 1 less than twice the numerator. If the numerator and denominator are both increased by 1, the fraction becomes 3/5. Find the fraction?

CorrectLet the numerator and denominator of the fraction be ‘n’ and ‘d’ respectively.

d = 2n – 1

(n + 1)/(d + 1) = 3/5

5n + 5 = 3d + 3

5n + 5 = 3(2n – 1) + 3 => n = 5

d = 2n – 1 => d = 9

Hence the fraction is : 5/9.IncorrectLet the numerator and denominator of the fraction be ‘n’ and ‘d’ respectively.

d = 2n – 1

(n + 1)/(d + 1) = 3/5

5n + 5 = 3d + 3

5n + 5 = 3(2n – 1) + 3 => n = 5

d = 2n – 1 => d = 9

Hence the fraction is : 5/9. - Question 3 of 50
##### 3. Question

The number of arrangements that can be made with the letters of the word MEADOWS so that the vowels occupy the even places?

CorrectThe word MEADOWS has 7 letters of which 3 are vowels.

-V-V-V-

As the vowels have to occupy even places, they can be arranged in the 3 even places in 3! i.e., 6 ways. While the consonants can be arranged among themselves in the remaining 4 places in 4! i.e., 24 ways.

Hence the total ways are 24 * 6 = 144.IncorrectThe word MEADOWS has 7 letters of which 3 are vowels.

-V-V-V-

As the vowels have to occupy even places, they can be arranged in the 3 even places in 3! i.e., 6 ways. While the consonants can be arranged among themselves in the remaining 4 places in 4! i.e., 24 ways.

Hence the total ways are 24 * 6 = 144. - Question 4 of 50
##### 4. Question

64 boys and 40 girls form a group for social work. During their membership drive, same number of boys and girls joined the group. How many members does the group have now, if the ratio of boys to girls is 4 : 3?

CorrectLet us say x boys and x girls joined the group.

(64 + x) / (40 + x) = 4/3

192 + 3x = 160 + 4x => x = 32

Number of members in the group = 64 + x + 40 + x

= 104 + 2x = 168.IncorrectLet us say x boys and x girls joined the group.

(64 + x) / (40 + x) = 4/3

192 + 3x = 160 + 4x => x = 32

Number of members in the group = 64 + x + 40 + x

= 104 + 2x = 168. - Question 5 of 50
##### 5. Question

The length of a rectangular plot is thrice its breadth. If the area of the rectangular plot is 867 sq m, then what is the breadth of the rectangular plot?

CorrectLet the breadth of the plot be b m.

Length of the plot = 3 b m

(3b)(b) = 867

3b^{2}= 867

b^{2}= 289 = 17^{2}(b > 0)

b = 17 m.IncorrectLet the breadth of the plot be b m.

Length of the plot = 3 b m

(3b)(b) = 867

3b^{2}= 867

b^{2}= 289 = 17^{2}(b > 0)

b = 17 m. - Question 6 of 50
##### 6. Question

Find the cost of fencing around a circular field of diameter 28 m at the rate of Rs.1.50 a meter?

Correct2 * 22/7 * 14 = 88

88 * 1 1/2 = Rs.132Incorrect2 * 22/7 * 14 = 88

88 * 1 1/2 = Rs.132 - Question 7 of 50
##### 7. Question

A and B can do a piece of work in 6 2/3 days and 5 days respectively. They work together for 2 days and then A leaves. In how many days after that B will complete the work alone.

Correct3/20 * 2 + (2 + x)/5 = 1

x = 1 ½ daysIncorrect3/20 * 2 + (2 + x)/5 = 1

x = 1 ½ days - Question 8 of 50
##### 8. Question

Find the one which does not belong to that group ?

CorrectExcept in 862, in all other numbers sum of first two digits is same as the last digit.

IncorrectExcept in 862, in all other numbers sum of first two digits is same as the last digit.

- Question 9 of 50
##### 9. Question

Simplify the following:

(169/121)^{-3/2}* 27/2 * (13/22)^{-1}Correct(169/121)

^{-3/2}* 27/2 * (13/22)^{-1}

(13^{2}/11^{2})^{-3/2}* 27/2 * 22/13 = 13^{-3}/11^{-3}* 27/2 * 22/13

= 13^{-4}11^{4}3^{3}Incorrect(169/121)

^{-3/2}* 27/2 * (13/22)^{-1}

(13^{2}/11^{2})^{-3/2}* 27/2 * 22/13 = 13^{-3}/11^{-3}* 27/2 * 22/13

= 13^{-4}11^{4}3^{3} - Question 10 of 50
##### 10. Question

How much time will a train of length 200 m moving at a speed of 72 kmph take to cross another train of length 300 m, moving at 36 kmph in the same direction?

CorrectThe distance to be covered = Sum of their lengths = 200 + 300 = 500 m.

Relative speed = 72 -36 = 36 kmph = 36 * 5/18 = 10 mps.

Time required = d/s = 500/10 = 50 sec.IncorrectThe distance to be covered = Sum of their lengths = 200 + 300 = 500 m.

Relative speed = 72 -36 = 36 kmph = 36 * 5/18 = 10 mps.

Time required = d/s = 500/10 = 50 sec. - Question 11 of 50
##### 11. Question

(56

^{2}– 24^{2}) * 1/32 + ?% of 1200 = 146Correct(56

^{2}– 24^{2}) * 1/32 + ?% of 1200 = 146

=> (3136 – 576) * 1/32 + ?/32 of 1200 = 146

=> 2560/32 + ? * 12 = 146 => ? * 12 = 146 – 80

=> ? = 66/12 => 5 1/2Incorrect(56

^{2}– 24^{2}) * 1/32 + ?% of 1200 = 146

=> (3136 – 576) * 1/32 + ?/32 of 1200 = 146

=> 2560/32 + ? * 12 = 146 => ? * 12 = 146 – 80

=> ? = 66/12 => 5 1/2 - Question 12 of 50
##### 12. Question

The profit earned by selling an article for Rs. 832 is equal to the loss incurred when the same article is sold for Rs. 448. What should be the sale price for making 50% profit?

CorrectLet C.P. = Rs. x.

Then, 832 – x = x – 448

2x = 1280 => x = 640

Required S.P. = 150% of Rs. 640 = 150/100 * 640 = Rs. 960.IncorrectLet C.P. = Rs. x.

Then, 832 – x = x – 448

2x = 1280 => x = 640

Required S.P. = 150% of Rs. 640 = 150/100 * 640 = Rs. 960. - Question 13 of 50
##### 13. Question

A train running at a speed of 36 kmph crosses an electric pole in 12 seconds. In how much time will it cross a 350 m long platform?

CorrectLet the length of the train be x m.

When a train crosses an electric pole, the distance covered is its own length.

So, x = 12 * 36 * 5 /18 m = 120 m.

Time taken to cross the platform = (120 +350)/ 36 * 5/18 = 47 min.IncorrectLet the length of the train be x m.

When a train crosses an electric pole, the distance covered is its own length.

So, x = 12 * 36 * 5 /18 m = 120 m.

Time taken to cross the platform = (120 +350)/ 36 * 5/18 = 47 min. - Question 14 of 50
##### 14. Question

(3

^{-3}* 9^{5/2}) / (27^{2/3}* 3^{-4}) = ?Correct(3

^{-3}* 9^{5/2}) / (27^{2/3}* 3^{-4}) = [3^{-3}* (3^{2})^{5/2}] / [(3^{3})^{2/3}* 3^{-4}]

= (3^{-3}* 3^{5}) / (3^{2}* 3^{-4}) = (3^{-3+5})/(3^{2-4}) = 3^{2}/3^{-2}= 3^{2}* 3^{2}= 81.Incorrect(3

^{-3}* 9^{5/2}) / (27^{2/3}* 3^{-4}) = [3^{-3}* (3^{2})^{5/2}] / [(3^{3})^{2/3}* 3^{-4}]

= (3^{-3}* 3^{5}) / (3^{2}* 3^{-4}) = (3^{-3+5})/(3^{2-4}) = 3^{2}/3^{-2}= 3^{2}* 3^{2}= 81. - Question 15 of 50
##### 15. Question

The average runs scored by a batsman in 20 matches is 40. In the next 10 matches the batsman scored an average of 13 runs. Find his average in all the 30 matches?

CorrectTotal score of the batsman in 20 matches = 800.

Total score of the batsman in the next 10 matches = 130.

Total score of the batsman in the 30 matches = 930.

Average score of the batsman = 930/30 = 31.IncorrectTotal score of the batsman in 20 matches = 800.

Total score of the batsman in the next 10 matches = 130.

Total score of the batsman in the 30 matches = 930.

Average score of the batsman = 930/30 = 31. - Question 16 of 50
##### 16. Question

If 3 workers collect 48 kg of cotton in 4 days, how many kg of cotton will 9 workers collect in 2 days?

Correct(3 * 4)/48 = (9 * 2)/ x

x = 72 kgIncorrect(3 * 4)/48 = (9 * 2)/ x

x = 72 kg - Question 17 of 50
##### 17. Question

In a 1000 m race, A beats B by 50 m and B beats C by 100 m. In the same race, by how many meters does A beat C?

CorrectBy the time A covers 1000 m, B covers (1000 – 50) = 950 m.

By the time B covers 1000 m, C covers (1000 – 100) = 900 m.

So, the ratio of speeds of A and C =

1000/950 * 1000/900 = 1000/855 So, by the time A covers 1000 m, C covers 855 m.

So in 1000 m race A beats C by 1000 – 855 = 145 m.IncorrectBy the time A covers 1000 m, B covers (1000 – 50) = 950 m.

By the time B covers 1000 m, C covers (1000 – 100) = 900 m.

So, the ratio of speeds of A and C =

1000/950 * 1000/900 = 1000/855 So, by the time A covers 1000 m, C covers 855 m.

So in 1000 m race A beats C by 1000 – 855 = 145 m. - Question 18 of 50
##### 18. 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 at least one senior?

CorrectThe total number of ways of forming the group of ten representatives is ²²C₁₀.

The total number of ways of forming the group that consists of no seniors is ¹⁰C₁₀ = 1 way

The required number of ways = ²²C₁₀ – 1IncorrectThe total number of ways of forming the group of ten representatives is ²²C₁₀.

The total number of ways of forming the group that consists of no seniors is ¹⁰C₁₀ = 1 way

The required number of ways = ²²C₁₀ – 1 - Question 19 of 50
##### 19. Question

I. a

^{3}– 988 = 343,

II. b^{2}– 72 = 49 to solve both the equations to find the values of a and b?Correcta

^{3}= 1331 => a = 11

b^{2}= 121 => b = ± 11

a ≥ bIncorrecta

^{3}= 1331 => a = 11

b^{2}= 121 => b = ± 11

a ≥ b - Question 20 of 50
##### 20. Question

The slant height of a cone is 12 cm and radius of the base is 4cm, find the curved surface of the cone.

Correctπ * 12 * 4 = 48

Incorrectπ * 12 * 4 = 48

- Question 21 of 50
##### 21. Question

A wire in the form of a circle of radius 3.5 m is bent in the form of a rectangule, whose length and breadth are in the ratio of 6 : 5. What is the area of the rectangle?

CorrectThe circumference of the circle is equal to the permeter of the rectangle.

Let l = 6x and b = 5x 2(6x + 5x) = 2 * 22/7 * 3.5

=> x = 1

Therefore l = 6 cm and b = 5 cm Area of the rectangle = 6 * 5 = 30 cm^{2}IncorrectThe circumference of the circle is equal to the permeter of the rectangle.

Let l = 6x and b = 5x 2(6x + 5x) = 2 * 22/7 * 3.5

=> x = 1

Therefore l = 6 cm and b = 5 cm Area of the rectangle = 6 * 5 = 30 cm^{2} - Question 22 of 50
##### 22. Question

Find the one which does not belong to that group ?

CorrectSkin, Eye, Nose and Ear are sensory organs, while Leg is a limb.

IncorrectSkin, Eye, Nose and Ear are sensory organs, while Leg is a limb.

- Question 23 of 50
##### 23. Question

A plot ABCD is as shown in figure, where AF = 30 m, CE = 40 m, ED = 50 m, AE = 120 m. Find the area of the plot ABCD?CorrectArea of plot ABCD = Area of ADE + Area of AFB + Area of BCEF = 1/2 * 50 * 120 + 1/2 * 40 * 30 + 40 * 90 = 3000 + 600 + 3600 = 7200 sq.m

IncorrectArea of plot ABCD = Area of ADE + Area of AFB + Area of BCEF = 1/2 * 50 * 120 + 1/2 * 40 * 30 + 40 * 90 = 3000 + 600 + 3600 = 7200 sq.m

- Question 24 of 50
##### 24. Question

The average of first 10 prime numbers is?

CorrectSum of 10 prime no. = 129

Average = 129/10 = 12.9IncorrectSum of 10 prime no. = 129

Average = 129/10 = 12.9 - Question 25 of 50
##### 25. Question

? % of 400 + 40% of 160 = 17% of 400

Correct? % of 400 + 40% of 160 = 17% of 400

?/100 of 400 + 40/100 of 160 = 17/100 of 400

?(4) + 64 = 68 => ? = 1Incorrect? % of 400 + 40% of 160 = 17% of 400

?/100 of 400 + 40/100 of 160 = 17/100 of 400

?(4) + 64 = 68 => ? = 1 - Question 26 of 50
##### 26. Question

A laborer is engaged for 30 days on the condition that he receives Rs.25 for each day he works and is fined Rs.7.50 for each day is absent. He gets Rs.425 in all. For how many days was he absent?

Correct30 * 25 = 750

425

———–

325

25 + 7.50 = 32.5

325/32.5 = 10Incorrect30 * 25 = 750

425

———–

325

25 + 7.50 = 32.5

325/32.5 = 10 - Question 27 of 50
##### 27. Question

A and B starts a business with Rs.8000 each, and after 4 months, B withdraws half of his capital . How should they share the profits at the end of the 18 months?

CorrectA invests Rs.8000 for 18 months, but B invests Rs.8000 for the first 4 months and then withdraws Rs.4000. So, the investment of B for remaining 14 months is Rs.4000 only.

A : B

8000*18 : (8000*4) + (4000*14)

14400 : 88000

A:B = 18:11IncorrectA invests Rs.8000 for 18 months, but B invests Rs.8000 for the first 4 months and then withdraws Rs.4000. So, the investment of B for remaining 14 months is Rs.4000 only.

A : B

8000*18 : (8000*4) + (4000*14)

14400 : 88000

A:B = 18:11 - Question 28 of 50
##### 28. Question

If the sum of one-half of a number exceeds one-third of that number by 7 1/3, the number is:

CorrectLet the number be x. Then,

(1/2 x + 1/5 x) – 1/3 x = 22/3

11/30 x = 22/3

x = 20IncorrectLet the number be x. Then,

(1/2 x + 1/5 x) – 1/3 x = 22/3

11/30 x = 22/3

x = 20 - Question 29 of 50
##### 29. Question

What is the difference between the largest number and the least number written with the digits 7, 3, 1, 4?

Correct1347

7431

————

6084Incorrect1347

7431

————

6084 - Question 30 of 50
##### 30. Question

The area of the square formed on the diagonal of a rectangle as its side is 108 1/3 % more than the area of the rectangle. If the perimeter of the rectangle is 28 units, find the difference between the sides of the rectangle?

CorrectLet the sides of the rectangle be l and b respectively.

From the given data,

(√l^{2}+ b^{2}) = (1 + 108 1/3 %)lb

=> l^{2}+ b^{2}= (1 + 325/3 * 1/100)lb

= (1 + 13/12)lb

= 25/12 lb

=> (l^{2}+ b^{2})/lb = 25/12

12(l^{2}+ b^{2}) = 25lb

Adding 24lb on both sides

12l^{2}+ 12b^{2}+ 24lb = 49lb

12(l^{2}+ b^{2}+ 2lb) = 49lb

but 2(l + b) = 28 => l + b = 14

12(l + b)^{2}= 49lb

=> 12(14)^{2}= 49lb

=> lb = 48

Since l + b = 14, l = 8 and b = 6

l – b = 8 – 6 = 2m.IncorrectLet the sides of the rectangle be l and b respectively.

From the given data,

(√l^{2}+ b^{2}) = (1 + 108 1/3 %)lb

=> l^{2}+ b^{2}= (1 + 325/3 * 1/100)lb

= (1 + 13/12)lb

= 25/12 lb

=> (l^{2}+ b^{2})/lb = 25/12

12(l^{2}+ b^{2}) = 25lb

Adding 24lb on both sides

12l^{2}+ 12b^{2}+ 24lb = 49lb

12(l^{2}+ b^{2}+ 2lb) = 49lb

but 2(l + b) = 28 => l + b = 14

12(l + b)^{2}= 49lb

=> 12(14)^{2}= 49lb

=> lb = 48

Since l + b = 14, l = 8 and b = 6

l – b = 8 – 6 = 2m. - Question 31 of 50
##### 31. Question

The inverse ratio of 3: 2: 1 is?

Correct1/3: 1/2: 1/1 = 2:3:6

Incorrect1/3: 1/2: 1/1 = 2:3:6

- Question 32 of 50
##### 32. Question

The wheels revolve round a common horizontal axis. They make 15, 20 and 48 revolutions in a minute respectively. Starting with a certain point on the circumference down wards. After what interval of time will they come together in the same position?

CorrectTime for one revolution = 60/15 = 4

60/ 20 = 3

60/48 = 5/4

LCM of 4, 3, 5/4

LCM of Numerators/HCF of Denominators =

60/1 = 60IncorrectTime for one revolution = 60/15 = 4

60/ 20 = 3

60/48 = 5/4

LCM of 4, 3, 5/4

LCM of Numerators/HCF of Denominators =

60/1 = 60 - Question 33 of 50
##### 33. Question

If x % of 80 = 20% of y, then x = ? and y = ?

Correctx % of 80 = 20 % of y => x / 100 (80) = 20 / 100 (y)

=> 4x = y

i.e x / y = ¼Incorrectx % of 80 = 20 % of y => x / 100 (80) = 20 / 100 (y)

=> 4x = y

i.e x / y = ¼ - Question 34 of 50
##### 34. Question

The distance between Delhi and Mathura is 110 kms. A starts from Delhi with a speed of 20 kmph at 7 a.m. for Mathura and B starts from Mathura with a speed of 25 kmph at 8 p.m. from Delhi. When will they meet?

CorrectD = 110 – 20 = 90

RS = 20 + 25 = 45

T = 90/45 = 2 hours

8 a.m. + 2 = 10 a.m.IncorrectD = 110 – 20 = 90

RS = 20 + 25 = 45

T = 90/45 = 2 hours

8 a.m. + 2 = 10 a.m. - Question 35 of 50
##### 35. Question

A train 125 m long passes a man, running at 5 km/hr in the same direction in which the train is going, in 10 sec. The speed of the train is?

CorrectSpeed of the train relative to man = 125/10 = 25/2 m/sec.

= 25/2 * 18/5 = 45 km/hr

Let the speed of the train be x km/hr. Then, relative speed = (x – 5) km/hr.

x – 5 = 45 => x = 50 km/hr.IncorrectSpeed of the train relative to man = 125/10 = 25/2 m/sec.

= 25/2 * 18/5 = 45 km/hr

Let the speed of the train be x km/hr. Then, relative speed = (x – 5) km/hr.

x – 5 = 45 => x = 50 km/hr. - Question 36 of 50
##### 36. Question

The average weight of a group of persons increased from 48 kg to 51 kg, when two persons weighing 78 kg and 93 kg join the group. Find the initial number of members in the group?

CorrectLet the initial number of members in the group be n.

Initial total weight of all the members in the group = n(48)

From the data,

48n + 78 + 93 = 51(n + 2) => 51n – 48n = 69 => n = 23

Therefore there were 23 members in the group initially.IncorrectLet the initial number of members in the group be n.

Initial total weight of all the members in the group = n(48)

From the data,

48n + 78 + 93 = 51(n + 2) => 51n – 48n = 69 => n = 23

Therefore there were 23 members in the group initially. - Question 37 of 50
##### 37. Question

Two trains each 250 m in length are running on the same parallel lines in opposite directions with the speed of 80 kmph and 70 kmph respectively. In what time will they cross each other completely?

CorrectD = 250 m + 250 m = 500 m

RS = 80 + 70 = 150 * 5/18 = 125/3

T = 500 * 3/125 = 12 secIncorrectD = 250 m + 250 m = 500 m

RS = 80 + 70 = 150 * 5/18 = 125/3

T = 500 * 3/125 = 12 sec - Question 38 of 50
##### 38. Question

If a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. The actual distance traveled by him is?

CorrectLet the actual distance traveled be x km. Then,

x/10 = (x + 20)/14

4x – 200 =>x = 50 km.IncorrectLet the actual distance traveled be x km. Then,

x/10 = (x + 20)/14

4x – 200 =>x = 50 km. - Question 39 of 50
##### 39. 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?

CorrectLet 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.IncorrectLet 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 40 of 50
##### 40. Question

The speed of a train is 90 kmph. What is the distance covered by it in 10 minutes?

Correct90 * 10/60 = 15 kmph

Incorrect90 * 10/60 = 15 kmph

- Question 41 of 50
##### 41. Question

If p, q and r are positive integers and satisfy x = (p + q -r)/r = (p – q + r)/q = (q + r – p)/p, then the value of x is?

CorrectWhen two or more ratios are equal, each of the ratios are equal to sum of the numerators divided by the sum of the denominators, provided sum of the denominators is non-zero.

Hence, x = (p + q -r)/r = (p – q + r)/q = (q + r – p)/p

=> x = (p + q – r + p – q + r + q + r – p) / (r + q + p)

=> x = (r + q + p) / (r + q + p) = 1

p + q + r is non-zero.IncorrectWhen two or more ratios are equal, each of the ratios are equal to sum of the numerators divided by the sum of the denominators, provided sum of the denominators is non-zero.

Hence, x = (p + q -r)/r = (p – q + r)/q = (q + r – p)/p

=> x = (p + q – r + p – q + r + q + r – p) / (r + q + p)

=> x = (r + q + p) / (r + q + p) = 1

p + q + r is non-zero. - Question 42 of 50
##### 42. Question

Find the one which does not belong to that group ?

CorrectCrocodile, Turtle, Allegator and Frog are amphibians, while Chameleon is a terrestrial animal.

IncorrectCrocodile, Turtle, Allegator and Frog are amphibians, while Chameleon is a terrestrial animal.

- Question 43 of 50
##### 43. Question

A train 150 m long running at 72 kmph crosses a platform in 25 sec. What is the length of the platform?

CorrectD = 72 * 5/18 = 25 = 500 – 150 = 350

IncorrectD = 72 * 5/18 = 25 = 500 – 150 = 350

- Question 44 of 50
##### 44. Question

64 is what percent of 80?

CorrectLet x percent of 80 be 64.

80 * x/100 = 64 => x = (64 * 100)/80 => x = 80.

80% of 80 is 64.IncorrectLet x percent of 80 be 64.

80 * x/100 = 64 => x = (64 * 100)/80 => x = 80.

80% of 80 is 64. - Question 45 of 50
##### 45. Question

A brick measures 20 cm * 10 cm * 7.5 cm how many bricks will be required for a wall 25 m * 2 m * 0.75 m?

Correct25 * 2 * 0.75 = 20/100 * 10/100 * 7.5/100 * x

25 = 1/100 * x => x = 25000Incorrect25 * 2 * 0.75 = 20/100 * 10/100 * 7.5/100 * x

25 = 1/100 * x => x = 25000 - Question 46 of 50
##### 46. Question

The sum of five consecutive odd numbers of set p is 435. What is the sum of five consecutive numbers of another set q. Whose largest number is 45 more than the largest number of set p?

CorrectLet the five consecutive odd numbers of set p be 2n – 3, 2n – 1, 2n + 1, 2n + 3, 2n + 5.

Sum of these five numbers

= 2n – 3 + 2n – 1 + 2n + 1 + 2n + 3 + 2n + 5

= 10n + 5 = 435 => n = 43

Largest number of set p = 2(43) + 5 = 91

The largest number of set q = 91 + 45 = 136

=> The five numbers of set q are 132, 133, 134, 135, 136.

Sum of above numbers = 132 + 133 + 134 + 135 + 136 = 670.IncorrectLet the five consecutive odd numbers of set p be 2n – 3, 2n – 1, 2n + 1, 2n + 3, 2n + 5.

Sum of these five numbers

= 2n – 3 + 2n – 1 + 2n + 1 + 2n + 3 + 2n + 5

= 10n + 5 = 435 => n = 43

Largest number of set p = 2(43) + 5 = 91

The largest number of set q = 91 + 45 = 136

=> The five numbers of set q are 132, 133, 134, 135, 136.

Sum of above numbers = 132 + 133 + 134 + 135 + 136 = 670. - Question 47 of 50
##### 47. Question

36 * 48 ÷ 64 + 36 ÷ 12 = ?

Correct36 * 48 / 64 + 36/12 = 27 + 3 = 30

Incorrect36 * 48 / 64 + 36/12 = 27 + 3 = 30

- Question 48 of 50
##### 48. Question

The average age of three boys is 15 years and their ages are in proportion 3:5:7. What is the age in years of the youngest boy?

Correct3x + 5x + 7x = 45

x =3

3x = 9Incorrect3x + 5x + 7x = 45

x =3

3x = 9 - Question 49 of 50
##### 49. Question

Sides of a rectangular park are in the ratio 3: 2 and its area is 3750 sq m, the cost of fencing it at 50 ps per meter is?

Correct3x * 2x = 3750 => x = 25

2(75 + 50) = 250 m

250 * 1/2 = Rs.125Incorrect3x * 2x = 3750 => x = 25

2(75 + 50) = 250 m

250 * 1/2 = Rs.125 - Question 50 of 50
##### 50. Question

If A got 80 marks and B got 60 marks, then what percent of A’s mark is B’s mark?

CorrectA’s marks = 80 ; B’s marks = 60.

Let x% of A = B => x/100 * 80 = 60

=> x = (60 * 100)/80 = 75

B’s marks is 75% of A’s marks.IncorrectA’s marks = 80 ; B’s marks = 60.

Let x% of A = B => x/100 * 80 = 60

=> x = (60 * 100)/80 = 75

B’s marks is 75% of A’s marks.