**1. A person crosses a 600 metre long street in 5 minutes. What is his speed in km/hr?
**

Answer

Speed = 600/(5*60) m/sec = 2m/sec.

Converting m/sec to km/hr = 2*18/5 km/hr.

= 7.2 km/hr.

So, **Option B.**

**2. An airplane covers a certain distance at a speed of 240 kmph in 5 hours. To cover the same distance in 1 ⅔ hours, it must travel at a speed of:
**

Answer

Distance = 240*5 = 1200 km.

Speed = Distance/Time.

= 1200/(1 ⅔) = 1200/(5/3) = 720 kmph

So, **Option D.**

**3. If a person walks at 14 kmph instead of 10 kmph, he would have walked 20 km more. The actual distance traveled by him is:
**

Answer

Let the actual distance by x km.

Then, x/10 = (x + 20)/14.

Or, 14x = 10x + 200.

Or, 4x = 200.

Or, x = 50 km.

So, **Option A.**

**4. A train can travel 50% faster than a car. Both start from point A at the same time and reach point B, 75 kms away from A at the same time. On the way, however, the train lost about 12.5 minutes while stopping at the stations. The speed of the car is:
**

Answer

Let the speed of the car be x kmph.

Then, speed of the train = 150x/100 = 3x/2 kmph.

So, 75/x – 75/(3x/2) = 125/(10*60).

Or, x = 120 kmph.

So, **Option C.**

**5. Excluding stoppages, the speed of a bus is 54 kmph and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour?
**

Answer

Due to stoppages, it covers 9 kms less.

Time taken to cover 9 kms = 9*60/54 min = 10 min.

So, **Option B.**

**6. A bank offers 5% C.I. calculated on a half-yearly basis. A customer deposits Rs. 1600 each on 1st January and 1st July of a year. At the end of the year, the amount he would have gained by way of interest is:
**

Answer

Amount = Rs. (1600 * (1 + 5/200)^2 + 1600 * (1 + 5/200))

= Rs. (1600*41*81)/(40*40)

= Rs. 3321

So, C.I. = Rs. (3321 – 3200) = Rs. 121

So, **Option B.**

**7. The difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 4% p.a. is Re. 1. The sum (in Rs.) is:
**

Answer

Let the sum be Rs. X.

Then, C.I. = (x * (1 + 4/100)^2 – x) = 676x/625 – x = 51x/625.

S.I. = (4*2*x)/100 = 2x/25

So, 51x/625 – 2x/25 = 1.

Or, x = 625.

So, **Option A.**

**8. There is 60% increase in an amount in 6 years at simple interest. What will be the compound interest of Rs. 12000 after 3 years at the same time?
**

Answer

Let P = Rs. 100. Then, S.I. = Rs. 60 and T = 6 years.

So, R = (100*60)/(100*6) = 10% p.a.

Now, P = Rs. 12000. T = 3 years and R = 10% p.a.

So, C.I. = Rs. 12000 * ((1 + 10/100)^3 – 1)

= Rs. 3972

So, **Option B.**

**9. What is the difference between compound interests on Rs. 5000 for 1.5 years at 4% p.a. Compounded yearly and half-yearly?
**

Answer

C.I. when interest is compounded yearly = Rs. 5000 * (1 + 4/100) * (1 + 0.5*4/100) = Rs. 5304.

C.I. when interest is compounded half-yearly = Rs. 5000 * (1 + 2/100)^3 = Rs. 5306.04.

So, Difference = Rs. (5306.04 – 5304) = Rs. 2.04.

So, **Option A.**

**10. The C.I. on Rs. 30000 at 7% p.a. is Rs. 4347. The period (in years) is:
**

Answer

Amount = Rs. (30000 + 4347) = Rs. 34347.

Let time be n years.

Then, 30000 * (1 + 7/100)^n = 34347.

Or, n = 2 years.

So, **Option A.**

**11. An accurate clock shows 8 o’ clock in the morning. Through how many degrees will the hour hand rotate when the clock shows 2 o’ clock in the afternoon?
**

Answer

Angle traced by the hour hand in 6 hours = 360/12 * 6 = 180 degrees.

So, **Option D.**

**12. ****The reflex angle between the hands of a clock at 10.25 is:
**

Answer

Angle traced by hour hand in 125/12 hours = 360/12 * 125/12 = 312.5 degrees.

Angle traced by minutes hand in 25 minutes = 360/60 * 25 = 150 degrees.

So, Reflex angle = 360 degrees – (312.5 – 150) degrees = 197.5 degrees.

So, **Option D.**

**13. ****A clock starts at noon. By 10 minutes past 5, the hour hand has turned through:
**

Answer

Angle traced by hour hand in 12 hours = 360 degrees.

Angle traced by hour hand in 5 hours 10 mins or 31/6 hours = 360/12 * 31/6 = 155 degrees.

So, **Option C.**

**14. A watch which gains 5 seconds in 3 minutes was set right at 7 a.m. In the afternoon of the same day, when the watch indicated quarter past 4 o’ clock, the true time is:
**

b)

Answer

Time from 7 a.m. to 4.15 p.m. = 9 hours 15 minutes = 37/4 hours.

3 minutes 5 seconds of this clock = 3 minutes of the correct clock.

Or, 37/720 hours of this clock = 1/20 hours of the correct clock.

Or, 37/4 hours of this clock = 1/20 * 720/37 * 37/4 hours of the correct clock = 9 hours of the correct clock.

So, the correct time is 9 hours after 7 a.m. = 4 p.m.

So, **Option B.**

**15. How much does a watch lose per day, if its hands coincide every 64 minutes?
**

Answer

55 minutes spaces are covered in 60 minutes.

60 minutes spaces are covered in 60/55 * 60 minutes = 65 5/11 minutes.

Loss in 64 minutes = 65 5/11 – 64 = 16/11 minutes.

Loss in 24 hours = 16/11 * 1/64 * 24 * 60 = 32 8/11 minutes.

So, **Option A.**

**16. From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many days can it be done?
**

Answer

We may have (3 women and 2 women) or (4 men and 1 woman) or (5 men) only.

So, Required no. of ways = (7C3 * 6C2) + (7C4 * 6C1) + 7C5

= 525 + 210 + 21 = 756

So, **Option C.**

**17. In how many ways can letters of the word ‘CORPORATION’ be arranged so that the vowels always come together?
**

Answer

In the word ‘CORPORATION’, we treat the vowels OOAIO as one letter.

Thus, we have CRPRTN(OOAIO).

This has 6 + 1 letters of which R occurs 2 times and the rest are different.

No. of ways in which these letters can be arranged = 7!/2! = 2520.

Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged in 5!/3! = 20 ways.

So, required no. of ways = 2520 * 20 = 50400.

So, **Option D.**

**18. In how many ways can letters of the word ‘LEADING’ be arranged so that the vowels always come together?
**

Answer

The word ‘LEADING’ has 7 letters.

When the vowels EAI are always together, they can be supposed to form 1 letter.

Then, we have to arrange the letters LNDG(EAI).

Now, 4 + 1 letters can be arranged in 5! ways = 120 ways.

The vowels EAI can be arranged among themselves in 3! ways = 6 ways.

So, required no. of ways = 120*6 = 720 ways.

So, **Option B.**

**19. Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
**

Answer

No. of ways of selecting 3 out of 7 consonants and 2 out of 4 vowels = 7C3 * 4C2 = 210.

No. of groups, each having 3 consonants and 2 vowels = 210.

Each group contains 5 letters.

No. of ways of arranging 5 letters among themselves = 5! = 120

So, required no. of ways = 210*120 = 25200.

So, **Option C.**

**20. In how many ways can the letters of the word ‘LEADER’ be arranged?
**

Answer

The word ‘LEADER’ contains 6 letters.

So, required no. of ways = 6!/2! = 360.

So, **Option C.**

**21. Two trains are moving in opposite directions @ 60 km/hr and 90 km/hr. Their lengths are 1.10 km and 0.9 km respectively. The time taken by the slower train to cross the faster train in seconds is:
**

Answer

Relative speed = (60 + 90) km/hr.

= (150 x 5/18) m/sec.

= 125/3 m/sec

Distance covered = (1.10 + 0.9) km = 2 km = 2000 m.

Required Time = (2000 x 3/125) sec = 48 sec.

So, **Option C.**

**22. Two trains of equal length 140 meters are running in opposite directions on parallel tracks at the same speed. If the trains cross each other 14 seconds, find the speed of each train.
**

Answer

Let the speed of each train be x km/hr

Relative speed = x + x = 2x …………….. (1)

Also, Relative speed = (Total distance covered (sum of length of trains))/(Time taken by trains to cross each other).

Relative speed = (140 + 140)/14 = 280/14 = 20 m/s

From equation (1) and (2),

2x = 20

x = 10 m/s

We need the answer to be in km/hr.

So, x = 10 * 18/5 = 36km/hr.

So, **Option D.**

**23. A goods train runs at the speed of 72 kmph and crosses a 250 m long platform in 26 seconds. What is the length of the goods train?
**

Answer

Speed = 72* 5/18 m/sec = 20 m/sec.

Time = 26 seconds.

Let the length of the train be x metres.

Then, (x + 250)/26 = 20.

Or, x = 270 m.

So, **Option D.**

**24. A train 240 m long passes a pole in 24 seconds. How long will it take to pass a platform 650 m long?
**

Answer

Speed = 240/24 m/sec = 10 m/sec.

So, Required time = (240 + 650)/10 = 89 seconds.

So, **Option B.**

**25. A jogger running at 9 kmph alongside a railway track is 240 metres ahead of the engine of a 120 metres long train running at 45 kmph in the same direction. In how much time will the train pass the jogger?
**

Answer

Speed of train relative to jogger = (45 – 9) km/hr = 36 km/hr = 36 * 5/18 m/sec = 10 m/sec.

Distance to be covered = (240 + 120) m = 360 m.

So, Time taken = 360/10 seconds = 36 seconds.

So, **Option C.**

**26. Today is Monday. After 61 days it will be:
**

Answer

Each day of the week is repeated after 7 days.

So, after 63 days, it will be Monday.

After 61 days, it will be Saturday.

So, **Option B.**

**27. How many weekends are there in a year?
**

Answer

Weekend means Saturday & Sunday together. In total, we have 52 weeks in a year. So there are 52 weekends in a year.

Normally, we have 104 weekend days.

We know that each normal year has 365 days or 52 weeks plus one day, and each week has two weekend days, which means there are approximately 104 weekend days each year.

Whereas, in a leap year, we have 366 days; it adds one more day to the year. And what makes the change is the starting day of the year.

So, **Option B.**

**28. How many days are there in x weeks x days?
**

Answer

No. of days in x weeks x days = (7x + x) days = 8x days.

So, **Option B.**

**29. If 6th March, 2005 is Monday, what was the day on 6th March, 2004?
**

Answer

The year 2004 is a leap year. So, it has 2 odd days.

But, February 2004 is not included as we are calculating from March, 2004 to March, 2005 and so we have one odd day only.

So, the day on 6th March, 2005 will be 1 day beyond the day on 6th March, 2004.

So, 6th March, 2004 is one day before Monday, ie., Sunday.

So, **Option A.**

**30. ****Which of the following is not a leap year?
**

Answer

Centuries divisible by 400 are leap years.

So, 700 is not a leap year.

So, **Option A.**

**31. Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?
**

Answer

Here, S = {1,2,3,4,……,20}.

Let E = Event of getting a multiple of 3 or 5 = {3,6,9,12,15,18,5,10,20}.

So, P(E) = n(E)/n(S) = 9/20.

So, **Option D.**

**32. In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?
**

Answer

Total no. of balls = 8 + 7 + 6 = 21.

Let E = Event that the ball drawn is neither red nor green.

= Event that the ball drawn is blue.

So, n(E) = 7.

So, P(E) = n(E)/n(S) = 7/21 = 1/3.

So, **Option A.**

**33. What is the probability of getting a sum 9 from two throws of a dice?
**

Answer

In two throws of a dice, n(S) = 6*6 = 36.

Let E = Event of getting a sum of 9 = { (3,6) , (4,5) , (5,4) , (6,3) }.

So, P(E) = n(E)/n(S) = 4/36 = 1/9.

So, **Option C.**

**34. Three unbiased coins are tossed. What is the probability of getting at most 2 heads?
**

Answer

Here, S = {TTT, TTH, THT, HTT, HHT, HTH, THH, HHH}

Let E = Event of getting at most 2 heads.

Then E = {TTT, TTH, THT, HTT, HHT, HTH, THH}

So, P(E) = n(E)/n(S) = 7/8.

So, **Option D.**

**35. In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected is:
**

Answer

Let S be the sample space and E be the event of selecting 1 girl and 2 boys.

Then, n(S) = no. of ways of selecting 3 students from 25 students = 25C3 = 2300.

And, n(E) = 10C1 * 15C2 = 10 * 105 = 1050.

So, P(E) = n(E)/n(S) = 1050/2300 = 21/46.

So, **Option A.**

**36. A hall is 15 m long and 12 m broad. If the sum of the areas of the floor and the ceiling is equal to the sum of the areas of four walls, the volume of the hall is:
**

Answer

2 (15 + 12) * h = 2 (15 * 12).

h = 180/27 = 20/3 m.

Volume = (15 * 12 * 20/3) m cubic = 1200m^3.

So, **Option C.**

**37. The dimensions of a hall are 40 m, 25 m and 20 m. If each person requires 200 cubic meters, find the number of persons who can be accommodated in the hall.
**

Answer

Length of the hall = 40 m.

Breadth of the hall = 25 m.

Height of the hall = 20 m.

Volume of the hall = l * b * h.

= 40 * 25 * 20 = 20,000 m^3.

Space occupied by each person = 200 m^3.

Number of persons that can be accommodated in the hall =

Volume of the hall/Space occupied by one person = 20000/200 = 100 persons.

So, **Option D.**

**38. A right triangle with sides 3 cm, 4 cm and 5 cm is rotated the side of 3 cm to form a cone. The volume of the cone so formed is:
**

Answer

We have r = 3 cm and h = 4 cm.

So, Volume = ⅓*pi*r^2*h = ⅓*pi*3^2*4 = 12 pi cm^3.

So, **Option A.**

**39. In a shower, 5 cm of rain falls. The volume of water that falls on 1.5 hectares of ground is:
**

Answer

1 hectare = 10000 m^2.

So, Area = 10000*1.5 = 15000 m^2.

Depth = 5/100 m = 1/20 m.

So, Volume = Area * Depth = 15000 * 1/20 = 750 m^3.

So, **Option B.**

**40. 50 men took a dip in a water tank 40 m long and 20 m broad on a religious day. If the average displacement of water by a man is 4 m^3, then the rise in the water level in the tank will be:
**

Answer

Total volume of water displaced = 4*50 m^3 = 200 m^3.

So, Rise in water level = 200/(40 * 20) = 0.25 m = 25 cm.

So, **Option B.**

**41. Two successive discounts of 10% and 5% are equivalent to a single discount of
**

Answer

Discount D_{1} = 10%

Discount D_{2} = 5%

equivalent discount = D_{1 } + D_{2 } – D_{1}.D_{2}/100

= 10 + 5 – 10*5/10 = 15 – 0.5 = 14.5%

So, **Option C**

**42. P makes a profit of 12% on a transaction. Had he bought the article at 10% less and sold it for 4% less, by what percent his profit percentage would have increased?
**

Answer

Let the cost price of the article = ₹ 100

Selling price = 100 + 100 * 12/100 = ₹112

When cost price = 100 – 100 * 10/100 = ₹90

and selling price = 112 – 112 * 4/100 = ₹107.52

Profit % = ((107.52-90)/90) * 100 = 19.47%

Increase in profit percent

((17.47-12)/12) * 100 = 62.25%

So, **Option B.**

**43. A sum of INR 15000 is put on compound interest. This sum becomes 2 times in 5 years at this rate. What would the original sum become after 20 years?**

**a) 60000 ****b) 225000 ****c) 140000 ****d) None of these**

Answer

Principal (P) = ₹ 15,000.

Time (T) = 5 years.

Amount (A) = 2 × 15000 = 2P.

So, 2P = P(1 + r/100)^5 => (1 + r/100)^5 = 2

Now, (1 + r/100)^20 = 2^{4} = 16 times

Hence, amount becomes = 16 × 15000 = ₹2,40,000

So, **Option D**

**44. A sum of INR 27800 is invested for 2 years in 2 different schemes, both on simple interest. The rates of interest are 14% and 11% in the 1st and the 2nd schemes respectively. The total interest after 2 years is INR 7016. What sum was invested in the 2nd scheme?**

**a) ₹ 12800 b) ****₹ 16200 c) ****₹ 14600 d) ****None of these**

Answer

Let amount invested in first and second schemes are x and (27800 – x) respectively.

ATQ,

x * 2 * 14/100 + (27800 – x) * 2 * 11/100 = 7016

6x/100 = 7016 – 6116

x = 900 * 100/6 = 15000

Hence, amount invested into second scheme = 27800 – 15000 = 12,800

So, **Option A.**

**45. A and B can finish a work in 10 days and 15 days respectively. They together work on it for 5 days and then the rest of the work is finished by C in 2 days.They get INR 450 for finishing this work. What should be the shares of A, B and C respectively?
**

Answer

Work done by A and B in 5 days

= 5(1/10 + 1/15) = 5((3+2)/30) = 5/6

Rest work = 1 – 5/6 = 1/6

Amount of work finished by C in 2 day = 1/6

C will finish whole work in 6 × 2 = 12 days.

Now, L.C.M. of 10, 15 and 12 = 60.

Unit of work done by them in one days are 60/10, 60/15 and 60/12 i.e. 6, 4 and 5 units respectively.

As A and B works for 5 days and C works for 2 days. Ratio of work done by them= 6 × 5 : 4 × 5 ; 5 × 2 = 3:2:1.

Now, 3x + 2x + x = 450

x = 75

Amount received by A = 3x = 3 × 75 = ₹ 225

Amount received by B = 2x = 2 × 75 = ₹150

Amount received by C = x = 75 = ₹ 75.

So, **Option C.**

**46. There are 10 people in a family: the average age of the 2 grandparents is 64 years; the average age of 2 uncles is 42 years; the average age of the 2 aunts is 38 years and the average age of 4 cousins is 16 years. What is the average age of the family members?
**

Answer

Sum of the age of two grandparents

= 2 × 64= 128 years.

Sum of the age of two uncles

= 2 × 42 = 84 years.

Sum of the age of two Aunts

= 2 × 38 = 76years

Sum of the age of four cousins

= 4 × 16 = 64 years.

Average age of the family

= (125 + 84 + 76 +64)/10 = 35.2

So, **Option C.**

**47. The median salaries of 3 B-Schools A, B and C are in the ratio 7:6:5 in the year 2019. The median salaries of these 3 B-schools are in the ratio 3:3:4 in the year 2020. If the median salary of A increased by 20% from 2019 to 2020, by what percent did the median salary of B increase?
**

Answer

Let median salaries of A, Band C in 2019 are 7x, 6x and 5x respectively.

A’s salary in 2020 is 7x * 120/100 = 8.4x

Ratio of A and B salary in 2020 = 3 : 3.

Therefore, 3/3 = 8.4x/B’s salary

B’s salary in 2020 = 8.4x

% increase in B’s salary

= ((8.4x – 6x)/6x) * 100 = 40%

So, **Option A.**

**48. The combined age of a mother and daughter duo is 42 years. The product of their ages, 5 years back, was 60 years. What is the present age of the mother?**

**a) 40 years b) ****38 years c) ****32 years d) ****None of these**

Answer

Let present age of Mother and daughter are M and D respectively.

ATQ,

M + D = 42 …(i)

(M – 5)(D – 5) = 60

MD –5(M + D) + 25 = 60.

MD – 5 (42) + 25 = 60.

MD – 210 + 25 = 60

MD = 245

Now, M – D = sqrt((M + D)^2 – 4MD)

= sqrt((42)^2 – 4 * 245)

therefore, M–D=28

From (i) and (ii), M = 35

Hence, Mother’s age = 35 years.

So, **Option D.**

**49. A is 2 times B and B is 2 times C. The average of the reciprocals of A, B and C is 7/12. What is the value of A?**

**a) 2 b) ****4 c) ****1 d) ****None of these**

Answer

ATQ, A = 2B and B = 2C.

Average = (1/A + 1/B + 1/C)/3 = 7/12

Therefore, 1/A + 1/B + 1/C = 7/4 1/4C + 1/2C + 1/C = 7/4

C = 1, B = 2 and A = 4.

So, **Option B.**

**50. What is the value of log 8/log 64?**

**a) 25 b) ****1/3 c) ****5 d) ****None of these**

Answer

log8/log64 = log8/log(8)^2 = 1/2 * log8/log8 = 0.5

So, **Option C.**

**51. Which of the following is a factor of?**

**a)** **X ^{2 }– 1**

Answer

f(x) = x^{4 }+ 4x^{2} + 3

= x^{4 }+ 3x^{2} + x^{2 } + 3

= (x^{2} + 3)(x^{2} + 1)

Hence, factors are (x^{2} + 3) and (x^{2} + 1)

So, **Option B.**

**52. If one of the sides of a square is increased by 20% and the other side of decreased by 20% to get a rectangle, what percent of the area of the square will be the area of this rectangle?
**

Answer

Let side of the square is l.

then, area A = 1 * 1 = 1^{2
}When one side increases by 20% and other decreases by 20%

New area A’ = (1 + 20*1/100) * (1 – 20*1/100)

= 1.2*1 * 0.8*1 = 0.96 * 1^{2
}Required percent =

A’/A * 100 = (0.96*1^{2}/1^{2}) * 100 = 96%

So, **Option A.**

**53. There is a cube of edge 8 cm. This cube is immersed in a water-filled vessel with the dimension of its rectangular base (12cm × 20 cm). By how many cms will the level of water raise in the vessel?
**

Answer

Volume of water displaced = Volume of the cube = 12 × 20 × height (h) = 8 × 8 × 8.

Height (h) = 8*8*8/12*20 = 2.13cm

So, **Option D.**

**54. What is the value of sin + tan ?
**

Answer

sin 45^{o} + tan 45^{o
}= 1/sqrt(2) + 1 = (sqrt(2) + 1)/sqrt(2)

So, **Option C.**

**55. What is the value of sin / cos?**

**a) 5 b) ****1 c) ****(sqrt(2) – 1)/sqrt(2) d) ****None these**

Answer

sin 42^{o}/cos 48^{o} = sin 42^{o}/cos(90^{o} – 42^{o}) = sin 42^{o}/sin 42^{o} = 1

So, **Option B.**

**56. What will be the missing term in the series:**

**11,88,……,6336, 57024, 456192?**

**a) 704 b) ****862 c) ****764 d) ****None of these**

Answer

So, **Option D.**

**57. A bag has a total of 108 toys. These toys are of 3 different colours: black, green and white. If a toy is taken out randomly, the probabilities of getting a black and a green are 1/3 and 4/9 respectively. What is the number of white toys?**

**a) 36 b) ****24 c) ****51 d) ****None of these**

Answer

Let number of black and green toys in the box are m and n respectively.

then, probability of getting a black toy

= ^{m}C_{1}/^{108}C_{1 }= m/108 = 1/3 (given) therefore, m = 108/3 = 36

Probability of getting a green toy

= ^{n}C_{1}/^{108}C_{1 }= n/108 = 4/9 therefore, n = 108*4/9 = 12*4 = 48

Hence, number of white toy = 108 – 36 – 48 = 24.

So, **Option B.**

**58. The Average weight of three men A, B and C is 84 kg. D joins them and the average weight of the four becomes 80 kg. If E whose weight is 3 kg more then that of D replaces A, the average weight of B, C, D and E becomes 79 kg. The weight of A is.**

**a) 65 kg b) ****70kg c) ****75 kg d) ****80 kg**

Answer

Total weight of A, B and C = 84 × 3 = 252 kg.

D’s weight = 80 × 4 – 84 × 3 = 68 kg.

E’s weight = 68 + 3 = 71 kg.

Total weight of B, C, D and E = 79 × 4 = 316 kg.

Then, total weight of B and C = 316 – 68 – 71 = 177kg.

Hence. A’s weight = 252 – 177 = 75 kg.

So, **Option C.**

**59. The sum of the digits of a number is 10. The difference between this number and a number formed with the same digits, but in the reverse order, is 18. What is the square of formed by reversing its digits.**

**a) 2116 b) ****4096 c) ****1764 d) ****None of these**

Answer

Let two digits are x and y.

ATQ, x + y = 10 and, (10x + y) – (10y + x) = 18.

(x – y) = 2

From (i) and (ii), we get

x = 6 and y = 4.

Hence, number = 64

Hence, square of (46) = (46)^{2 }= 2116.

So, **Option A.**

**60. We have 60, 76 and 84 loaves of bread of companies A, B and C respectively. We have to supply these bread to retail stores so that a store has the bread of just one of the companies and tail the stores have the same number of loaves. What can be the minimum number of stores?**

**a) 4 b) ****44 c) ****55 d) ****None of these**

Answer

H.C.E of 60,76 and 84 = 4.

Minimum number of stores

= 60/4 + 76/4 + 84/4 = 15 + 19 +21 = 55

So, **Option C.**