# IPPB Mains Mock Test Series 2nd, IPPB Mains Online Test Series 2

## IPPB Mains Mock Test Series 2nd, IPPB Mains Online Test Series 2

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IPPB Mains Mock Test Series 2nd, IPPB Mains Online Test Series 2. IPPB Mains Free Mock Test Exam 2019. IPPB Mains Exam Free Online Quiz 2019. IPPB Mains Full Online Mock Test **Series 2nd** in English. candidates who want to make sure whether their preparation level is up to date then they take this online test series without paying anything. CAknowledge provide **IPPB Online Test Series 2019 **in order to let you excel in the examination. The IPPB Mains Full online mock test paper is free for all students. IPPB Mains Question and Answers in English and Hindi **Series 2**. Here we are providing** IPPB Mains Full Mock Test Paper in English. IPPB Mains **Mock Test **Series 2nd** 2019. Now Test your self for IPPB Mains Exam by using below quiz…

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

I. x

^{2}+ 5x + 6 = 0,

II. y^{2}+ 9y +14 = 0 to solve both the equations to find the values of x and y?CorrectI. x

^{2}+ 3x + 2x + 6 = 0

=> (x + 3)(x + 2) = 0 => x = -3 or -2

II. y^{2}+ 7y + 2y + 14 = 0

=> (y + 7)(y + 2) = 0 => y = -7 or -2

No relationship can be established between x and y.IncorrectI. x

^{2}+ 3x + 2x + 6 = 0

=> (x + 3)(x + 2) = 0 => x = -3 or -2

II. y^{2}+ 7y + 2y + 14 = 0

=> (y + 7)(y + 2) = 0 => y = -7 or -2

No relationship can be established between x and y. - Question 2 of 50
##### 2. Question

At what rate percent per annum will the simple interest on a sum of money be 2/5 of the amount in 10 years?

CorrectLet sum = x. Then, S.I. = 2x/5, Time = 10 years.

Rate = (100 * 2x) / (x * 5 * 10) = 4%IncorrectLet sum = x. Then, S.I. = 2x/5, Time = 10 years.

Rate = (100 * 2x) / (x * 5 * 10) = 4% - Question 3 of 50
##### 3. Question

A 270 m long train running at the speed of 120 km/hr crosses another train running in opposite direction at the speed of 80 km/hr in 9 sec. What is the length of the other train?

CorrectRelative speed = 120 + 80 = 200 km/hr.

= 200 * 5/18 = 500/9 m/sec.

Let the length of the other train be x m.

Then, (x + 270)/9 = 500/9 => x = 230.IncorrectRelative speed = 120 + 80 = 200 km/hr.

= 200 * 5/18 = 500/9 m/sec.

Let the length of the other train be x m.

Then, (x + 270)/9 = 500/9 => x = 230. - Question 4 of 50
##### 4. Question

Two cylinders are of the same height. Their radii are in the ratio 1: 3. If the volume of the first cylinder is 40 cc. Find the volume of the second cylinder?

Correctr

_{1}= x r_{2}= 3x

Π * x^{2}* h = 40

Π9 x^{2}h = 40 * 9 = 360Incorrectr

_{1}= x r_{2}= 3x

Π * x^{2}* h = 40

Π9 x^{2}h = 40 * 9 = 360 - Question 5 of 50
##### 5. Question

A can do half the work in one day where as B can do it full. B can also do half the work of C in one day. Ratio in their efficiency will be?

CorrectWC of A: B = 1:2

B: C = 1:2

———————

A: B: C = 1:2:4IncorrectWC of A: B = 1:2

B: C = 1:2

———————

A: B: C = 1:2:4 - Question 6 of 50
##### 6. Question

In a fort, there are 1200 soldiers. If each soldier consumes 3 kg per day, the provisions available in the fort will last for 30 days. If some more soldiers join, the provisions available will last for 25 days given each soldier consumes 2.5 kg per day. Find the number of soldiers joining the fort in that case?

CorrectAssume X soldiers join the fort. 1200 soldiers have provision for 1200(days for which provisions last them)(rate of consumption of each soldier)

= 1200(30)(3) kg

Also provisions available for (1200 + x) soldiers is (1200 + x)(25)(2.5) kg

As the same provisions are available

=> 1200(30)(3) = (1200 + x)(25)(2.5)

x = [1200(30)(3)]/[(25)(2.5)] – 1200

x = 528IncorrectAssume X soldiers join the fort. 1200 soldiers have provision for 1200(days for which provisions last them)(rate of consumption of each soldier)

= 1200(30)(3) kg

Also provisions available for (1200 + x) soldiers is (1200 + x)(25)(2.5) kg

As the same provisions are available

=> 1200(30)(3) = (1200 + x)(25)(2.5)

x = [1200(30)(3)]/[(25)(2.5)] – 1200

x = 528 - Question 7 of 50
##### 7. Question

Find the nearest to 25268 which is exactly divisible by 467?

CorrectIncorrect - Question 8 of 50
##### 8. Question

A shopkeeper sells 20% of his stock at 10% profit ans sells the remaining at a loss of 5%. He incurred an overall loss of Rs. 400. Find the total worth of the stock?

CorrectLet the total worth of the stock be Rs. x.

The SP of 20% of the stock = 1/5 * x * 1.1 = 11x/50

The SP of 80% of the stock = 4/5 * x * 0.95 = 19x/25 = 38x/50

Total SP = 11x/50 + 38x/50 = 49x/50

Overall loss = x – 49x/50 = x/50

x/50 = 400 => x = 20000IncorrectLet the total worth of the stock be Rs. x.

The SP of 20% of the stock = 1/5 * x * 1.1 = 11x/50

The SP of 80% of the stock = 4/5 * x * 0.95 = 19x/25 = 38x/50

Total SP = 11x/50 + 38x/50 = 49x/50

Overall loss = x – 49x/50 = x/50

x/50 = 400 => x = 20000 - Question 9 of 50
##### 9. Question

10 men and 15 women together can complete a work in 6 days. It takes 100 days for one man alone to complete the same work. How many days will be required for one woman alone to complete the same work?

Correct1 man’s 1 day work = 1/100

(10 men + 15 women)’s 1 day work = 1/6

15 women’s 1 day work = (1/6 – 10/100) = 1/15

1 woman’s 1 day work = 1/225

1 woman alone can complete the work in 225 days.Incorrect1 man’s 1 day work = 1/100

(10 men + 15 women)’s 1 day work = 1/6

15 women’s 1 day work = (1/6 – 10/100) = 1/15

1 woman’s 1 day work = 1/225

1 woman alone can complete the work in 225 days. - Question 10 of 50
##### 10. 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?

CorrectHere, 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 = ¹²C₇IncorrectHere, 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 11 of 50
##### 11. Question

A delegation of 5 members has to be formed from 3 ladies and 5 gentlemen. In how many ways the delegation can be formed, if 2 particular ladies are always included in the delegation?

CorrectThere are three ladies and five gentlemen and a committee of 5 members to be formed.

Number of ways such that two ladies are always included in the committee = ⁶C₃ = (6 * 5 * 4)/6 = 20.IncorrectThere are three ladies and five gentlemen and a committee of 5 members to be formed.

Number of ways such that two ladies are always included in the committee = ⁶C₃ = (6 * 5 * 4)/6 = 20. - Question 12 of 50
##### 12. Question

Two pipes can fill a tank in 18 minutes and 15 minutes. An outlet pipe can empty the tank in 45 minutes. If all the pipes are opened when the tank is empty, then how many minutes will it take to fill the tank?

CorrectPart of the filled by all the three pipes in one minute

= 1/18 + 1/15 – 1/45 = (5 + 6 – 2)/90 = 9/90 = 1/10

So, the tank becomes full in 10 minutes.IncorrectPart of the filled by all the three pipes in one minute

= 1/18 + 1/15 – 1/45 = (5 + 6 – 2)/90 = 9/90 = 1/10

So, the tank becomes full in 10 minutes. - Question 13 of 50
##### 13. Question

A man can do a job in 15 days. His father takes 20 days and his son finishes it in 25 days. How long will they take to complete the job if they all work together?

Correct1 day work of the three persons = (1/15 + 1/20 + 1/25) = 47/300

So, all three together will complete the work in 300/47 = 6.4 days.-Incorrect1 day work of the three persons = (1/15 + 1/20 + 1/25) = 47/300

So, all three together will complete the work in 300/47 = 6.4 days.- - Question 14 of 50
##### 14. Question

Find the 37.5% of 976 =

Correct37.5 % of 976

= 37.5/100 * 976 = 375/1000 * 976 = 3/8 * 976

= 3 * 122 = 366.Incorrect37.5 % of 976

= 37.5/100 * 976 = 375/1000 * 976 = 3/8 * 976

= 3 * 122 = 366. - Question 15 of 50
##### 15. Question

If x is the interest on y and y is the interest on z, the rate and time is the same on both the cases. What is the relation between x, y and z?

CorrectX = (Y*NR)/100 Y = (Z*NR)/100

X/Y = NR/100 Y/Z = NR/100

X/Y = Y/Z

Y^{2}= XZIncorrectX = (Y*NR)/100 Y = (Z*NR)/100

X/Y = NR/100 Y/Z = NR/100

X/Y = Y/Z

Y^{2}= XZ - Question 16 of 50
##### 16. Question

A sum of money is put out at compound interest for 2 years at 20%. It would fetch Rs.482 more if the interest were payable half-yearly, then it were pay able yearly. Find the sum.

CorrectP(11/10)

^{4}– P(6/5)^{2}= 482

P = 2000IncorrectP(11/10)

^{4}– P(6/5)^{2}= 482

P = 2000 - Question 17 of 50
##### 17. Question

Two pipes A and B can separately fill a tank in 12 and 15 minutes respectively. A third pipe C can drain off 45 liters of water per minute. If all the pipes are opened, the tank can be filled in 15 minutes. What is the capacity of the tank?

Correct1/12 + 1/15 – 1/x = 1/15

x = 12

12 * 45 = 540Incorrect1/12 + 1/15 – 1/x = 1/15

x = 12

12 * 45 = 540 - Question 18 of 50
##### 18. 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?

CorrectTotal time taken = (160/64 + 160/8) = 9/2 hrs.

Average speed = 320 * 2/9 = 71.11 km/hr.IncorrectTotal time taken = (160/64 + 160/8) = 9/2 hrs.

Average speed = 320 * 2/9 = 71.11 km/hr. - Question 19 of 50
##### 19. Question

A and B together can complete a work in 12 days. A alone can complete it in 20 days. If B does the work only for half a day daily, then in how many days A and B together will complete the work?

CorrectB’s 1 day work = (1/12 – 1/20) = 1/30

Now, (A + B)’s 1 day work = (1/20 + 1/60) = 1/15

So, A and B together will complete the work in 15 days.IncorrectB’s 1 day work = (1/12 – 1/20) = 1/30

Now, (A + B)’s 1 day work = (1/20 + 1/60) = 1/15

So, A and B together will complete the work in 15 days. - Question 20 of 50
##### 20. Question

If the area of circle is 616 sq cm then its circumference?

Correct22/7 r

^{2}= 616 => r = 14

2 * 22/7 * 14 = 88Incorrect22/7 r

^{2}= 616 => r = 14

2 * 22/7 * 14 = 88 - Question 21 of 50
##### 21. Question

On dividing a number by 68, we get 269 as quotient and 0 as remainder. On dividing the same number by 67, what will be the remainder?

CorrectNumber = 269 * 68 + 0 = 18292

67) 18292 (273

18291

——–

1

Required number = 1.IncorrectNumber = 269 * 68 + 0 = 18292

67) 18292 (273

18291

——–

1

Required number = 1. - Question 22 of 50
##### 22. Question

A box contains nine bulbs out of which 4 are defective. If four bulbs are chosen at random, find the probability that exactly three bulbs are good.

CorrectRequired probability = (⁵C₃ . ⁴C₁)/⁹C₄ = (10 * 4)/126 = 20/63

IncorrectRequired probability = (⁵C₃ . ⁴C₁)/⁹C₄ = (10 * 4)/126 = 20/63

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

The average of first five prime numbers greater than 20 is?

Correct23 + 29 + 31 + 37 + 41 = 161/5 = 32.2

Incorrect23 + 29 + 31 + 37 + 41 = 161/5 = 32.2

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

What is the greater of the two numbers whose product is 2560, given that the sum of the two numbers exceeds their difference by 64?

CorrectLet the greater and the smaller number be g and s respectively.

gs = 2560

g + s exceeds g – s by 64 i.e., g + s – (g – s) = 64

i.e., 2s = 64 => s = 32.

g = 2560/s = 80.IncorrectLet the greater and the smaller number be g and s respectively.

gs = 2560

g + s exceeds g – s by 64 i.e., g + s – (g – s) = 64

i.e., 2s = 64 => s = 32.

g = 2560/s = 80. - Question 25 of 50
##### 25. Question

Two same glasses are respectively 1/4th 1/5th full of milk. They are then filled with water and the contents mixed in a tumbler. The ratio of milk and water in the tumbler is?

Correct1/4 : 3/4 = (1:3)5 = 5:15

1/5 : 4/5 = (1:4)4 = 4:16

——

9:31Incorrect1/4 : 3/4 = (1:3)5 = 5:15

1/5 : 4/5 = (1:4)4 = 4:16

——

9:31 - Question 26 of 50
##### 26. 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?

CorrectWe have M

_{1}D_{1}H_{1}/ W_{1}= M_{2}D_{2}H_{2}/ W_{2}(Variation rule)

(60 * 30 * 8)/ 200 = (45 * D_{2}* 6) / 300

D_{2}= (60 * 30 * 8 * 300) / (200 * 45 * 6) => D_{2}= 80.IncorrectWe have M

_{1}D_{1}H_{1}/ W_{1}= M_{2}D_{2}H_{2}/ W_{2}(Variation rule)

(60 * 30 * 8)/ 200 = (45 * D_{2}* 6) / 300

D_{2}= (60 * 30 * 8 * 300) / (200 * 45 * 6) => D_{2}= 80. - Question 27 of 50
##### 27. Question

A train is 360 meter long is running at a speed of 45 km/hour. In what time will it pass a bridge of 140 meter length?

CorrectSpeed = 45 Km/hr = 45*(5/18) m/sec = 25/2 m/sec

Total distance = 360+140 = 500 meter

Time = Distance/speed

= 500 * (2/25) = 40 secondsIncorrectSpeed = 45 Km/hr = 45*(5/18) m/sec = 25/2 m/sec

Total distance = 360+140 = 500 meter

Time = Distance/speed

= 500 * (2/25) = 40 seconds - Question 28 of 50
##### 28. Question

4000 was divided into two parts such a way that when first part was invested at 3% and the second at 5%, the whole annual interest from both the investments is Rs.144, how much was put at 3%?

Correct(x*3*1)/100 + [(4000 – x)*5*1]/100 = 144

3x/100 + 200 – 5x/100 = 144

2x/100 = 56 → x = 2800Incorrect(x*3*1)/100 + [(4000 – x)*5*1]/100 = 144

3x/100 + 200 – 5x/100 = 144

2x/100 = 56 → x = 2800 - Question 29 of 50
##### 29. Question

A and B can do a work in 12 days, B and C in 15 days, C and A in 20 days. If A, B and C work together, they will complete the work in?

Correct(A + B)’s 1 day work = 1/12;

(B + C)’s 1 day work = 1/15;

(A + C)’s 1 day work = 1/20

Adding, we get: 2(A + B + C)’s 1 day work = (1/12 + 1/15 + 1/20) = 1/5

(A + B + C)’s 1 day work = 1/10

So, A, B and C together can complete the work in 10 days.Incorrect(A + B)’s 1 day work = 1/12;

(B + C)’s 1 day work = 1/15;

(A + C)’s 1 day work = 1/20

Adding, we get: 2(A + B + C)’s 1 day work = (1/12 + 1/15 + 1/20) = 1/5

(A + B + C)’s 1 day work = 1/10

So, A, B and C together can complete the work in 10 days. - Question 30 of 50
##### 30. Question

3 men or 6 women can do a piece of work in 20 days. In how many days will 12 men and 8 women do the same work?

Correct3M = 6W —- 20 days

12M + 8W —–?

24W + 8 W = 32W —?

6W —- 20 32 —–?

6 * 20 = 32 * x => x = 15/4 daysIncorrect3M = 6W —- 20 days

12M + 8W —–?

24W + 8 W = 32W —?

6W —- 20 32 —–?

6 * 20 = 32 * x => x = 15/4 days - Question 31 of 50
##### 31. Question

An engineering student has to secure 36% marks to pass. He gets 130 marks and fails by 14 marks. The maximum No. of marks obtained by him is?

Correct130

14

——-

361—— 144

100%——? => 400Incorrect130

14

——-

361—— 144

100%——? => 400 - Question 32 of 50
##### 32. Question

5/8 * 2

^{(2/5 – 5/4)}/ 1^{1/4 }= ?Correct(5/8 * 12/5 – 5/4)/(5/4) (3/2 -5/4)/(5/4) = 1/5

Incorrect(5/8 * 12/5 – 5/4)/(5/4) (3/2 -5/4)/(5/4) = 1/5

- Question 33 of 50
##### 33. Question

A vessel of capacity 90 litres is fully filled with pure milk. Nine litres of milk is removed from the vessel and replaced with water. Nine litres of the solution thus formed is removed and replaced with water. Find the quantity of pure milk in the final milk solution?

CorrectLet the initial quantity of milk in vessel be T litres.

Let us say y litres of the mixture is taken out and replaced by water for n times, alternatively.

Quantity of milk finally in the vessel is then given by [(T – y)/T]^{n}* T

For the given problem, T = 90, y = 9 and n = 2.

Hence, quantity of milk finally in the vessel

= [(90 – 9)/90]^{2}(90) = 72.9 litres.IncorrectLet the initial quantity of milk in vessel be T litres.

Let us say y litres of the mixture is taken out and replaced by water for n times, alternatively.

Quantity of milk finally in the vessel is then given by [(T – y)/T]^{n}* T

For the given problem, T = 90, y = 9 and n = 2.

Hence, quantity of milk finally in the vessel

= [(90 – 9)/90]^{2}(90) = 72.9 litres. - Question 34 of 50
##### 34. Question

There were two candidates in an election. Winner candidate received 62% of votes and won the election by 288 votes. Find the number of votes casted to the winning candidate?

CorrectW = 62% L = 38%

62% – 38% = 24%

24% ——– 288

62% ——– ? => 744IncorrectW = 62% L = 38%

62% – 38% = 24%

24% ——– 288

62% ——– ? => 744 - Question 35 of 50
##### 35. Question

A number is doubled and 9 is added. If the resultant is trebled, it becomes 75. What is that number?

CorrectLet the number be x. Then,

3(2x + 9) = 75

2x = 16 => x = 8IncorrectLet the number be x. Then,

3(2x + 9) = 75

2x = 16 => x = 8 - Question 36 of 50
##### 36. Question

A man saves 20% of his monthly salary. If an account of dearness of things he is to increase his monthly expenses by 20%, he is only able to save Rs. 200 per month. What is his monthly salary?

CorrectIncorrect - Question 37 of 50
##### 37. Question

The circumferences of two circles are 264 meters and 352 meters. Find the difference between the areas of the larger and the smaller circles.

CorrectLet the radii of the smaller and the larger circles be s m and l m respectively.

2∏s = 264 and 2∏l = 352

s = 264/2∏ and l = 352/2∏

Difference between the areas = ∏l^{2}– ∏s^{2}

= ∏{176^{2}/∏^{2}– 132^{2}/∏^{2}}

= 176^{2}/∏ – 132^{2}/∏

= (176 – 132)(176 + 132)/∏

= (44)(308)/(22/7) = (2)(308)(7) = 4312 sq mIncorrectLet the radii of the smaller and the larger circles be s m and l m respectively.

2∏s = 264 and 2∏l = 352

s = 264/2∏ and l = 352/2∏

Difference between the areas = ∏l^{2}– ∏s^{2}

= ∏{176^{2}/∏^{2}– 132^{2}/∏^{2}}

= 176^{2}/∏ – 132^{2}/∏

= (176 – 132)(176 + 132)/∏

= (44)(308)/(22/7) = (2)(308)(7) = 4312 sq m - Question 38 of 50
##### 38. Question

If Rs.3250 be divided among Ram, Shyam and Mohan in the ratio of 1/2:1/3:1/4 then the share of each are?

Correct1/2:1/3:1/4 = 6:4:3

Ram = 6/13 * 3250 = 1500

Shyam = 4/13 * 3250 = 1000

Mohan = 3/13 * 3250 = 750Incorrect1/2:1/3:1/4 = 6:4:3

Ram = 6/13 * 3250 = 1500

Shyam = 4/13 * 3250 = 1000

Mohan = 3/13 * 3250 = 750 - Question 39 of 50
##### 39. Question

How many numbers up to 100 and 300 are divisible by 11?

Correct(300 – 100)/11 = 200/11 = 18 2/11

18 NumbersIncorrect(300 – 100)/11 = 200/11 = 18 2/11

18 Numbers - Question 40 of 50
##### 40. Question

Find the one which does not belong to that group ?

CorrectAll are blood relation except Daughter-in-law.

IncorrectAll are blood relation except Daughter-in-law.

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

Find the least number when successively divided by 2, 3 and 7 leaves remainders 1, 2 and 3 respectively?

Correct2) 65 (32 3) 32 (10 7) 10 (1

64 30 7

——– ——– – —–

1 2 3

=> 65Incorrect2) 65 (32 3) 32 (10 7) 10 (1

64 30 7

——– ——– – —–

1 2 3

=> 65 - Question 42 of 50
##### 42. 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?

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

(24 * 5 * 7 * 9) / ? = 216

Correct? = (24 * 5 * 7 * 9) / 216 = 35

Incorrect? = (24 * 5 * 7 * 9) / 216 = 35

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

One root of the quadratic equation x

^{2}– 12x + a = 0, is thrice the other. Find the value of a?CorrectLet the roots of the quadratic equation be x and 3x.

Sum of roots = -(-12) = 12

a + 3a = 4a = 12 => a = 3

Product of the roots = 3a^{2}= 3(3)^{2}= 27.IncorrectLet the roots of the quadratic equation be x and 3x.

Sum of roots = -(-12) = 12

a + 3a = 4a = 12 => a = 3

Product of the roots = 3a^{2}= 3(3)^{2}= 27. - Question 45 of 50
##### 45. Question

A car after covering ½ of a journey of 100 km develops engine trouble and later travels at ½ of its original speed. As a result, it arrives 2 hours late than its normal time. What is the normal speed of the car is?

Correct[50/x + 50/(x/2)] – 100/x = 2

x = 25Incorrect[50/x + 50/(x/2)] – 100/x = 2

x = 25 - Question 46 of 50
##### 46. Question

35 + 15 * 1.5 = ?

CorrectGiven Exp. = 35 + 15 * 3/2

= 35 + 45/2 = 35 + 22.5 = 57.5IncorrectGiven Exp. = 35 + 15 * 3/2

= 35 + 45/2 = 35 + 22.5 = 57.5 - Question 47 of 50
##### 47. Question

The sum of the squares of two consecutive positive integers exceeds their product by 91. Find the integers?

CorrectLet the two consecutive positive integers be x and x + 1

x^{2}+ (x + 1)^{2}– x(x + 1) = 91

x^{2}+ x – 90 = 0

(x + 10)(x – 9) = 0 => x = -10 or 9.

As x is positive x = 9

Hence the two consecutive positive integers are 9 and 10.IncorrectLet the two consecutive positive integers be x and x + 1

x^{2}+ (x + 1)^{2}– x(x + 1) = 91

x^{2}+ x – 90 = 0

(x + 10)(x – 9) = 0 => x = -10 or 9.

As x is positive x = 9

Hence the two consecutive positive integers are 9 and 10. - Question 48 of 50
##### 48. 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 49 of 50
##### 49. Question

A bag contains nine yellow balls, three white balls and four red balls. In how many ways can two balls be drawn from the bag?

CorrectTotal number of balls = 9 + 3 + 4

Two balls can be drawn from 16 balls in ¹⁶C₂ ways.IncorrectTotal number of balls = 9 + 3 + 4

Two balls can be drawn from 16 balls in ¹⁶C₂ ways. - Question 50 of 50
##### 50. Question

A person bought an article and sold it at a loss of 10%. If he had bought it for 20% less and sold it for Rs.75 less, he could have gained 25%. What is the cost price?

CorrectCP

_{1}= 100 SP_{1}= 90

CP_{2}= 80 SP_{2}= 80 * (140/100) = 112

22 —– 100

55 —– ? => Rs.250IncorrectCP

_{1}= 100 SP_{1}= 90

CP_{2}= 80 SP_{2}= 80 * (140/100) = 112

22 —– 100

55 —– ? => Rs.250