Qu. 01: Raju earns twice as much in March as in each of the other months of the year. What part of his annual earnings he earns in that month? (a) 1/5 (b) 5/7 (c) 2/13 (d) 1/10. CLAT 2008

Sol: Let Raju earns Rs. 2P in the month of March, and earns Rs. P in every other month. Hence, he earns Rs. 13P in an year. Thus he earns 2P/13P = 2/12 in the month of March. Option (c) is the answer.

Qu. 02: Sanjay sold his watch for Rs. 1140 and thereby loses 5%. In order to gain 5% he has to sell the watch for: (a) Rs. 1254 (b) Rs. 1260 (c) Rs. 1197 (d) Rs. 1311. CLAT 2008

Sol: This question can be solved by observation only within a couple of second. As 5% of CP of 1200 is 60, which when reduced from 1200 (as there is loss), the SP is 1140, the SP when there is profit of 5%, i.e. of Rs. 60, the new SP shall be Rs. 1260.

In solving this question traditionally, let us say the CP is 100P, then, as given, 95P = 1140. Hence, P = 12. Thus, CP = 1200. Thus new SP at the profit of 5% is 1260. Option (b) is the answer.

Qu. 03: A mixture of 40 liters of milk and water contains 10% of water. How much water is to be added to the mixture so that the water may be 20% in the new mixture? (a) 5 Liters

(b) 4 Liters (c) 6.5 Liters (d) 7.5 Liters. CLAT 2008

Sol: Given that the mixture contains 4 liter of water. Let us add P liter of water to make it a 20% solution. Thus, (4+P) = 20% of (40+P). This gives, 5(4+P) = 40+P, this is, 4P = 20, P = 5. Option (a) is the answer.

Qu. 04: A train 100 meters long running at 54 km/hr takes 20 seconds to pass a bridge. The length of the bridge is: (a) 50 mt

(b) 150 mt (c) 200 mt

(d) 620 mt. CLAT 2008

Sol: 54 km/hr = (54x1000)/36oo m/s = 15 m/s.

Let the length of the bridge be P meter. Then, as we know, distance = speed x time, we have, (100+P) = 15 x 20. Thus, 100+P = 300; P = 200. Option (c) is the answer.

Qu. 05 Sameer is as much younger to Mohan as he is older to Arun. If the sum of the ages of Mohan and Arun is 48, the age of Sameer is: (a) 20 years (b) 24 years (c) 30 years (d) cannot be determined. CLAT 2008

Sol: Let the age of Sameer is P years. Then, as give, X – P = P – Y, where X and Y are the respective ages of Mohan and Arun. Thus, 2P = X – Y, and, as given, X + Y = 48. Thus, 2P = 48, P = 24.

Option (b) is the answer.

## CLAT 2008

There were 10 question of Mathematics in CLAT 2008

Qu. 06: A tank can be filled up by two pipes A and B in 2 hours and 3 hours respectively. A third pipe C can empty the full tank in 6 hours. If all the taps can be turned on at the same time, the tank will be full in:

(a) 1 hour (b) 40 minutes (c) 1.5 hours (d) 3 hours. CLAT 2008

Sol: The part of tank to be filled in one hour is:

Hence, the tank will be filled in 3/2 hours = 1.5 hours. Option (c) is the answer.

Qu. 07: Of the three numbers, the first is one third of the second and twice the third. The average of these numbers is 27. The largest of these numbers is: (a) 18 (b) 36 (c) 54 (d) 108. CLAT 2008

Sol: Let the first number be 2P (Note, 2P is divisible by 2). Given, then, the second number is 6P and third one is P. Further given, (2P+6P+P)/3 = 27, or, 3P = 27, or, P = 9. Thus the numbers are: 18, 54, 9. Option (c) is the answer.

Qu. 08: The length of a square is increased by 15% and breadth decreased by 15%. The area of the rectangle so formed: (a) neither increases nor decreases

(b) decreases by 2.25% (c) increases by 2.25% (d) decreases by 22.5%. CLAT 2008

Sol: Let the original length and breadth of the square be 100 and 100. Then its area is: 100 x 100 = 10000.

Modified length and breadth of the square are given to be 115 and 85.

Then its area is: 115 x 85 = 9775. Thus, there is a reduction of 10000 - 9775 = 225 square meter in area of square whose original area was 10000.

The percentage reduction is (225/10000)x100 %= 2.25%.

Option (b) is the answer.

Qu. 09: The ratio of milk and water in 60 Liters of adulterated milk is 2: 1. If the ratio of milk and water is to be 1 : 2, then the amount of water to be added further is: (a) 20 Liters

(b) 30 Liters (c) 40 Liters (d) 60 Liters. CLAT 2008

Sol: Given that the quantity of milk and water in 60 Liters of adulterated milk is 2: 1, hence the quantity of milk is 40 liter and that of water is 20 liter.

Let us add P liter of water to make the ratio of milk and water equal to 1: 2.

Then, we have, 40: (20+P) = 1: 2 This gives, 40/(20+P) = 1/2.

This gives 80 = 20+P, or, P = 60. Option (d) is the answer.

Qu. 10: A piece of cloth costs Rs. 70. If the piece is 4 meter longer and each meter costs Rs. 2 less, the cost remains unchanged. The length of the piece is: (a) 8 mt (b) 9 mt (c) 10 mt (d) 12 mt. CLAT 2008

Sol: This question is most easily solved by observation, as, 10 x 7 = 70 = (10 + 4) x (7 – 2). Hence correct answer is 10.

Otherwise, let the length of the piece of cloth (with fixed width) be P meter. The cost of this piece is Rs. 70. Then the cost per meter = Rs. 70/P. Given,

This is: (P – 10) (P + 14) = 0, or P = 10, -14. Option (c) is the answer.

## CLAT 2016

Qu. 01: The number of 'three digit numbers' which are multiples of 9 are: (A) 98 (B) 101 (C) 100 (D) 99. CLAT 2016

Sol: The first three digit number, and a multiple of 9 shall be 12x9 = 108; and last three digit number and a multiple of 9 shall be 111x9 = 999. Thus, there are 100 (corresponding to numbers 12 to 111) three digit numbers and simultaneously a multiple of 9 .

Option (C) is the answer.

Qu. 02: A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days, she has to pay Rs. 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs. 1180 as hostel charges. Find the fixed charges and the cost of food per day.

(A) 300, 30 (B) 400, 40 (C) 200, 20 (D) 400, 30. CLAT 2016

Sol: Let fixed charges be Rs. P and variable charges be Rs. Q per day.

Given, P+20Q = 1000 and P+26Q = 1180. Subtracting first equation from the second one, we have 6Q = 180, Q = 30 per day. Hence P = 1000 – (20x30) = 1000-600 = 400. Option (D) is the answer.

Qu. 03: What is the sum of all the natural numbers from 1 to 100?

(A) 5000 (B) 5050 (C) 6000 (D) 5052. CLAT 2016

Sol: Let S = 1+ 2+ 3 + ……… + 100. Again S = 100 + 99 + 09 + ….. + 1

Adding these two, 2S = 100x101; S = 50x101 = 5050.

Option (B) is the answer.

Or, since it is an AP with first term equal to 1 and common difference equal to 1, the sum of first 100 terms = (100/2){2+(100-1)1} = 50x101 = 5050

Note: The sum of first n terms of an AP with initial term ‘a’ and common difference ‘d’, is: (n/2){2a+(n-1)d}

Qu. 04: A library has an average of 510 visitors on Sundays and 240 on other days. What is the average number of visitors per day in the month of June beginning with a Sunday? (A) 250

(B) 280 (C) 285 (D) 276. CLAT 2016

Sol: Since first day of June is Sunday, there shall be 5 Sundays, and 25 other weekdays in June.

The total number of visitors in June = {(5x510) + (25x240)} = 2550 + 6000 = 8550. The average is: 8550/30 = 285. Option (C) is the answer.

Qu. 05: 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.

(A) 7/12 (B) 11/12 (C) 10/12 (D) 9/12. CLAT 2016

Sol: Obviously the required probability = Total number of good pens/total number of pens = 132/(132+12) = 132/144 = 11/12. Option (B) is the answer.

## CLAT 2017

Qu. 01: Keerthi’s father gave him some money to buy books. He spent half of the money equally to buy books and entertaining his friends. Whatever amount left with him, he deposited half in his savings account and gave Rs. 5 to a poor person as charity. Finally, Keerthi was left with Rs. 20 which he returned to his father. What amount did his father give him initially? 1. Rs. 120 2. Rs. 100 3. Rs. 200 4. Rs. 160. CLAT 2017

Sol: Let Keerthi have Rs. 4P. He bought books for half the amount, i.e. for Rs. 2P. Half of the amount left with him, i.e. half of Rs. 2P, i.e. Rs P were deposited in saving account. Given, P = 20+5 = 25, hence 4P = 100. Option (2) is the answer.

Note: I took amount with Keerthi Rs. 4P as I had to halve it two times.

Qu. 02: Praveen has Rs. 4,662 in the form of 2, 5 and 10 rupee notes. If these notes are in the ratio of 3:5:8, the number of five rupees notes with him is: 1. 250 2. 336 3. 210 4. 84 CLAT 2017

Sol: Let the number of notes of 2, 5 and 10 be 3P, 5P and 8P respectively. Then, as given, (2x3P) + (5x5P) + (10x8P) = 6P+25P+80P = 4662. 111P = 4662. That is: P = 42. Hence 5P = 210. Option (3) is the answer.

Qu. 03: Gold and copper are as heavy as water by 19 and 9 times respectively. The ratio in which these two metals be mixed so that the mixture is 17 times as heavy as water is: 1. 2:3 2. 3:2 3. 4:1. 4. 3:4 CLAT 2017

Sol: Let us mix Gold and Copper in the quantity of G and C respectively. Then as given, 19G + 9C = 17(G+P). 2G = 8C.

G/C = 4/1. The required ratio is 4:1. Option (3) is the answer.

Qu. 04: A piece of cloth costs rupees 75. If the piece is four meters longer and each meter costs rupees 5 less, the cost remains unchanged. What is the length of the piece?

1. 8 meters 2. 10 meters 3. 12 meters 4. 6 meters. CLAT 2017

Sol: Let the length of the cloth with fixed width (say W) be P meter. The cost of this cloth is Rs. 75.

Thus the cloth is Rs. 75/P per meter. Extended length is: P+4. With this length, the cost of cloth = (P+4)x{(75/P)-5} = 75 (given). Solving this, we get P = 6. Option (4) is the answer.

Note: Keeping in view the constricted timeliness of exam, this is a lengthy process, hence, my suggestion to you is to go through the options, and find the answer.

Qu. 05 There are two urns. One contains two white balls and four red balls; the other contains three white and nine red balls. All balls are of the same shape and size. From each urn, one ball is drawn. What is the probability of getting both the balls of the same color?

1. 1/24 2. 7/12 3. 1/12 4. 1 / 2. CLAT 2017

Sol: The event consists of two mutually exclusive events; either both the balls are white or both the balls are red. Required probability = {(2/6)x(3/12)} + {(4/6)x(9/12)} = (6/72) + (36/72) = 42/72 = 7/12. Option (2) is the answer.

## CLAT 2018

1. When 13511, 13903 and 14589 are divided by the

greatest number ‘n’, the remainder in each case is ‘m’. The value of (n + m) is: (A) 183 (B) 182 (C) 181 (D) 179. (CLAT 2018)

Given, 13511 == m mod n (1);

And, 13903 == m mod n (2)

And, 14589 == m mod n (3);

Hence, 392 == 0 mod n (4),

{By reducing (1) from (2)}

Hence, 686 == 0 mod n (5), reducing (2) from (3)

Hence, 1078 == 0 mod n (6), reducing (1) from (3)

Obviously, n is the HCF of 392, 686, 1078.

Now, 392 = 4 x 98 = 4 x 7 x 14

And, 686 = 2 x 343 = 2 x 7 x 7 x 7

And, 1078 = 2 x 539 = 2 x 7 x 7 x 11

The HCF of three numbers = 7 x 14 = 98

Now, 13511 = (137 x 98) + 85 {137 x 98 = 13426}

Hence, m = 85. Hence, n + m = 98 + 85 = 183.

Option (a) is the answer.

2. If a person travels at a speed of 40 km per hour, he will reach his destination on time. He covered half the journey in 2/3 of the time. At what speed in kilometer per hour should he travel to cover the remaining journey to reach the destination on time? (a) 48 (b) 50 (c) 60 (d) 72 (CLAT 2018)

Let the distance be 2d, and time taken to cover complete journey be 3t. Given, that first half distance is covered in 2t time, hence the second half distance is covered in t time.

Then, given, (2d/3t) = 40, or, (d/t) = 60.

Now, the first half distance, i.e. d, is covered in time 2t hour, hence, speed = (d/2t) = 30 km/h.

And, the second half of the distance, i.e. d, is covered in time t hour, hence, speed = (d/t) = 60 km/h.

Option (c) is the answer.

3. The parallel sides of a trapezium shaped field are 25 m and 10 m and non-parallel sides are 14 m and 13 m. What is the area (in sq m) of the field? (a) 204 (b) 196 (c) 156 (d) 144 (CLAT 2018)

Since CD = EF, we have, EF = 10. Let AE = P, then, FB = 15 – P.

We have, CE = DF. Hence, in two right angle triangles, CEA and DFB, we have, using Pythagoras theorem,

This is: (15 – P – P) (15 – P + P) = 27, or, (15 – 2P) x 15 = 27

ALTERNATIVELY:

We may consider the height CE = A, then,

It is as obtained at (1). Option (d) is the answer.

## AILET 2014

AILET 2014: 01: Mohan credits 15% of his salary into his bank and spend 30% of the remaining amount on the household articles. If cash on hand is Rs. 2380, what is his salary in rupees?

Sol: Let the salary of Mohan be Rs. 100P.

Given, 15P + {(30%) of 85P} + 2380 = 100P

85P - {(30%) of 85P} = 2380.

This is: (70%) x 85P = 2380

85P = 3400; 5P = 200; 100P = 40000.

Option (c) is the answer.

AILET 2014: 02: The average marks of a student in ten papers are 80. If the highest and lowest score are not considered, the average is 81. If the highest score is 92, what is the lowest score? (a) 55 (b) 60 (c) 62 (d) 61

Sol: Let the lowest marks be P.

Average marks of 10 papers = 80.

Hence, total marks in 10 papers = 80x10 = 800

Total marks in 8 papers (excluding highest and lowest marks) = 800 – 92 – P = 708 – P

Average marks of 8 papers = 81. Hence, total marks in 8 papers = 81x8 = 648

Hence, 708 – P = 648; P = 708 – 648 = 60. Option (b) is the answer.

AILET 2014: 03: Ten years ago, Sunil was half of Sudip’s age. If the age of both at present is in the ratio of 3:4, what will be the total of the present age? (a) 20 (b) 30 (c) 35 (d) 45

Sol: Let the present age of Sunil and Sudip be 3P and 4P respectively.

Then, as given,

(3P - 10) = (1/2)(4P - 10),

or 3P – 10 = 2P – 5; P = 5.

The sum of their present age is 7P = 35. Option (c) is the answer.

AILET 2014: 04: A and B started a business with the total capital of Rs. 30,000. At the end of the year, they shared the profit in the ratio of their investments. If their capitals were interchanged, then A would have received 175% more than what he actually received. Find out the capital of B. (a) Rs. 20,000 (b) Rs. 22,000 (C) Rs. 21,000 (d) Rs. 23,000

Sol: Let capital of B be Rs. 100P.

Then capital of A is 30000 – 100P.

Profit Sharing of A and B is; A:B = (30000 – 100P):100P

After interchange of capitals, this ratio is; A:B = 100P: (30000 – 100P)

And, as given, 175% more of (30000 – 100P) becomes 100P.

Hence, (30000 – 100P) + (175/100) (30000 – 100P) = 100P

= (275/100) (30000 – 100P) = 100P

= 275 x ( 300 – P) = 100P;

This is: 375P = 275 x 300.

This is: 25P = 275 x 20; This is: 100P = 1100 x 20 = 22000.

Thus, the capital of B is Rs. 22,000.

Option (b) is the answer.

AILET 2014: 05: The ratio in which Aman and Bimal have contributed to the capital of a company is 3:4. Bimal has invested his capital for only 3 months and has received half as much profit as Aman at the end of the year. Find out how much time in months has Aman invested his capital in the company. (a) 8 (b) 14 (c) 15 (d) 12

Sol: Since ratio of investment is given, let the investments of Aman and Bimal be 3Q and 4Q respectively.

Profits of Aman and Bimal will be: 3QP and (4Q x 3), i.e. 3PQ and 12Q.

Given, 12Q = (1/2)(3PQ); P = 8. Aman invested his capital for 8 months. Option (a) is the answer.

AILET 2014: 06: Two whole numbers whose sum is 64 can be in the ratio of?

(a) 7:2 (b) 7:6 (c) 3:1 (d) 8:7

Sol: This type of questions is best done by observation method. Obviously options a, b and d are NOT the answers, as 64 is not divided by 9 (which is 7+2), not divided by 13 (which is 7+6) and not divided by 15 (which is 8+7). Option (c) is the answer as 64 is divided by 4 (which is 3+1).

AILET 2014: 07: In a call center 6 employees working for 10 hours complete a certain task. They started working at 11 am. This continued till 5 pm and after that, for each hour one more employee is added till the work gets completed. At what time will they complete the work? (a) 7:10 pm (b) 8:00 pm (c) 7:35 pm (d) 6:35 pm

Sol: The amount of work is 6 x 10 = 60 man-hours.

Till 5 pm from 11 am, i.e. for 6 hours there are 6 employees, hence 36 man-hours work was completed. Hence, the work of 24 man-hours remains to be completed.

From 5 pm onwards, for each hour one more employee is added till the work gets completed.

From 5 pm to 6 pm, i.e. for 1 hour, there are 7 employees; hence 7 man-hours work was completed. The work of 17 man-hours remains to be completed.

From 6 pm to 7 pm, i.e. for 1 hour, there are 8 employees; hence 8 man-hours work was completed. The work of 9 man-hours remains to be completed.

From 7 pm to 8 pm, i.e. for 1 hour, there are 9 employees; hence remaining 9 man-hours work was completed. Option (b) is the answer.

AILET 2014: 08: Amar is twice as fast as Rohit and Rohit is thrice as fast as Chanda is. The journey covered by Chanda in 42 minutes will be covered by Amar in: (a) 14 min 25 sec (b) 7 min (c) 28 min 37 sec (d) 54 min 35 sec

Sol: Given the conditions for the speeds of Amar, Rohit and Chanda, for some positive integer P, their speeds in meter per minute will be given by: 2P, P, P/3 respectively; or by 6P, 3P, P.

The distance covered by Chanda in 42 minutes = 42P meter (Distance = Speed x Time)

Time taken by Amar to cover this distance of 42P meter = 42P/6P = 7 minutes. Option (b) is the answer.

AILET 2014: 09: P, Q and R are three consecutive odd numbers in ascending orders. If the value of three times P is 3 less than the two times R, find the value of R.

(a) 5 (b) 7 (c) 9 (d) 11

Sol: Let P = 2n – 1, Q = 2n + 1 and

R = 2n + 3 be the three consecutive odd numbers in ascending orders.

Given, 3(2n – 1) + 3 = 2(2n + 3).

That is: 6n – 4n = 6, or 2n = 6; n = 3.

R = 2n + 3 = 9. Option (c) is the answer.

AILET 2014: 10: 72% of students in a class took Physics and 44% took Mathematics. If each student took Physics or Mathematics and 40 took both, the total number of students in the class would be: (a) 200 (b) 240

(c) 250 (d) 320

Sol: We have from the principle of Set Theory;

Let there be 100P students in the class. Then, using this principle,

we have; 100P = 72P + 44P – 40; That is: 116P – 100P = 40; 16P = 40: Or, 4P = 10; Or, 100P = 250.

Option (c) is the answer.

## AILET 2015

AILET 2015: 01: A piece of string is 40 centimeters long. It is cut into three pieces. The longest piece is 3 times as long as the middle-sized and the shortest piece is 23 centimeters shorter than the longest piece. Find length or the shortest piece (in cm). (a) 27 cm (b) 5 cm (c) 4 cm (d) 9 cm

Sol: Let shortest piece be of P meter. Then, as given: P + (P + 23) + (1/3)(P + 23) = 40

2P + (p/3) = 40 – 23 – (23/3): 7P/3 = 28/3; 7P = 28; P = 4. Option (C) is the answer.

AILET 2015: 02: Fresh grapes contain 90% water by weight while dried grapes contain 20% water by weight. What is the weight of dry grapes available from 20 kg of fresh grapes? (a) 2 kg (b) 2.4 kg (C) 2.5 kg (d) 2.6 kg

Sol: Obviously from 20 kg of fresh grapes, only the fiber content of 2kg (10% only as there is 90% water content in fresh grapes) converts into dried grapes which is 80% of dried grapes {or in other words, the fiber contents remains same in both fresh grapes and dried grapes}.

Since 80% of dried grapes is given to be 2 kg, 100% of dried grapes (total weight of dried grapes) is 2.5 kg. Option (C) is the answer.

AILET 2015: 03: A group of men decides to do a job in 8 days but since 10 men drops out every day, the job got gets completed at the end of 12 day. How many men are there at the beginning? (a) 165 (b) 175 (C) 80 (d) None of these

Sol: Let there be P men at the beginning. Then, as given that the work was to be completed in 8 days, the amount of work is 8P mandays.

As given: P + (P-10) + (P-20) + (P-30) + …… + (P-110) = 8P {Given that work lasted for 12 days}

Using formula for sum of AP, i.e. S(n) = (n/2)(2A + (n-1)d), we have:

(12/2){2P + (12-1)(-10)} = 8P. Or, 12P – 660 = 8P, or, 4P = 660, or, P = 165. Option (A) is the answer. [The official answer to this question is option (D), don’t know, why?}

##### Take advantage of additional knack of Mathematics you possess. Do as much as you can do. Remember, though it is a matter of 10 marks only in AILET, it shall raise your rank in hundreds.

AILET 2015: 04: In a race of 200 m run, A beats S by 20 m and N by 40 m. If S and N are running a race of 100 m with exactly same speed as before, then by how many metres will S beat N? (a) 11.11 (b) 10 (C) 12 (d) 25

Sol: When S was at a distance of 180 m, N was at 160

Hence, when S is at 100m, N is at (160/180)x100 = 800/9

Thus, S beats N by 100 – (800/9) = 100/9 = 11.11m. Option (A) is the answer.

AILET 2015: 04: In a race of 200 m run, A beats S by 20 m and N by 40 m. If S and N are running a race of 100 m with exactly same speed as before, then by how many metres will S beat N? (a) 11.11 (b) 10 (C) 12 (d) 25

Sol: When S was at a distance of 180 m, N was at 160

Hence, when S is at 100m, N is at (160/180)x100 = 800/9

Thus, S beats N by 100 – (800/9) = 100/9 = 11.11m. Option (A) is the answer.

AILET 2015: 04: In a race of 200 m run, A beats S by 20 m and N by 40 m. If S and N are running a race of 100 m with exactly same speed as before, then by how many meters will S beat N? (a) 11.11 (b) 10 (C) 12 (d) 25

Sol: When S was at a distance of 180 m, N was at 160. Hence, when S is at 100m, N is at (160/180)x100 = 800/9. Thus, S beats N by 100 – (800/9) = 100/9 = 11.11 m. Option (A) is the answer.

AILET 2015: 05: Total expenses of a boarding house are partly fixed and partly varying linearly with the number of boarders. The average expense per boarder is Rs. 700 when there are 25 boarders and Rs. 600 when there are 50 boarders. What is the average expense per boarder when there are 100 boarders? (a) Rs. 550 (b) Rs. 580 (C) Rs. 540 (d) Rs. 570

Sol: Let varying expenses be Rs. P, proportional to number of boarders N. Thus P = kN, where k is the constant of proportionality.

Let fixed expenses be Rs. Q.

Given; Q + 25k = (25x700); and Q + 50k = (50x600).

These two equations give: Q + 25k = 17500; and Q + 50k = 30000.

These two equations give: 25k = 12500, k = 500.

This gives, Q = (25x700) – 25k = 17500 – 12500 = 5000.

Thus, P = 500N, and thus total expense for 100 boarders = Q + (500x100) = 5000 + 50000 = 55000.

Hence, average expense per boarder when there are 100 boarders = Rs. 550. Option (A) is the answer.

AILET 2015: 06: Three maths classes: X, Y and Z take an algebra test. The average score of class X is 83. The average score of class Y is 76 and the average score of class Z is 85. The average score of class X and Y is 79, and average score of class Y and Z is 81. What is the average score of classes X, Y and Z? (a) 81.5 (b) 80.5 (C) 83 (d) 78.

Sol: Let there be x, y and z number of students in three maths classes: X, Y and Z. Given:

Total score of class X = 83x, Total score of class Y = 76y, Total score of class Z = 85z,

Total score of class X and class Y= 83x + 76y = 79 (x+y), i.e. 4x = 3y

Total score of class Y and class Z= 76y + 85z = 81 (y+z), i.e. 5y = 4z

These two equations give: 20x = 15y = 15y = 12z; i.e. 20x = 15y = 12z; i.e. y = 4x/3, z = 5x/3

We have to find out: (83x + 76y + 85z)/(x + y + z)

= {83x + (304x/3) + (425x/3)}/{x + (4x/3) + (5x/3)}

= (249 + 304 + 425)/(12) = 978/12 = 81.5. Option (A) is the answer.

AILET 2015: 07: A man has 9 friends: 4 boys and 5 girls. In how many ways can he invite them, if there have to be exactly 3 girls in the invitee’s list? (a) 320 (b) 160 (C) 80 (d) 200

Sol: 3 girls out of 5 can be selected in C(5, 3) = 10 way; and he can invite in one of the following manners;

(3 girls + no boy) or (3 girls + 1 boy) or (3 girls + 2 boy) or (3 girls + 3 boys) or (3 girls + 4 boys) == 10 {C(4,0) + C(4,1) + C(4,2) + C(4,3) + C(4,4)} = 10 {1 + 4 + 6 + 4 + 1} = 160.

Thus Option (b) is the answer.

[Note, all these events separated with 'or' are mutually exclusive events, and hence their values have been added.]

AILET 2015: 08: Number of students who have opted for the subjects A, B and C are 60, 84 and 108 respectively. The examination is to be conducted for these students such that only the students of the same subject are allowed in one room. Also the number of students in each room must be same. What is the minimum number of rooms that should be arranged to meet all these conditions? (a) 28 (b) 60 (C) 12 (d) 21.

Sol: Obviously, we have to fund out the HCF of the given three numbers.

60 = 12 x 5; 84 = 12 x 7; 108 = 12 x 9. Since 5, 7 and 9 are co-prime to one-another, the required HCF is 12. Option (c) is the answer.

AILET 2015: 09: You can collect rubies and emeralds as many as you can. Each ruby is worth Rs. 4 crore and each emerald is worth Rs. 5 crore. Each ruby weighs 0.3 kg and each emerald weighs 0.4 kg. Your bag can carry at the most 12 Kg. What should you collect to get the maximum wealth? (a) 20 rubies and 15 emirates (b) 40 rubies (c) 28 rubies and 9 emeralds (d) None of the above.

Sol: Let there be x number of rubies and y number of emeralds in the bag.

Then, Maximize Z = 4x + 5y

Subject to x >= 0, y >=0, 0.3x + 0.4y <= 12,

i.e. 3x + 4y <= 120

As per the Theorem on solving Linear Programming graphically, the optimal feasible solution shall be obtained at the corner point only; and corner points are; (0,0), (40,0), (0,30). Hence Z = 4x + 5y is maximum at (40,0). Thus Option (b) is the answer.

AILET 2015: 10: 10 points are marked on a straight line and 11 points are marked on another straight line. How many triangles can be constructed with vertices from among the above points? (a) 495 (b) 550 (c) 1045 (d) 2475

Sol: Obviously the two lines from one point on a line onto two points on other will form a triangle.

A point on first line can be selected in C(10,1) ways and two points on other line can be selected in C(11,2) ways.

Similarly, a point on the second line can be selected in C(11,1) ways and two points on other line can be selected in C(10,2) ways.

But these two events are mutually exclusive events.

Hence, the answer is:

{C(10,1) x C(11,2)} + { C(11,1) x C(10,2)}

= (10 x 55) + (11 x 45) = 550 + 495 = 1045.

Option (c) is the answer.

## AILET 2016

AILET 2016: 01: A dice is rolled twice. What is the probability that sum of the numbers on two faces is 5? (a) 5/12 (b) 1/9 (c) 1/6 (d) 5/36

Sol: The sample space has 6x6 = 36 items, and favorable items are 4, which are: (1, 4), (2, 3), (3, 2), (4, 1). Hence the required probability is: 4/36 = 1/9. Option (b) is the answer.

AILET 2016: 02: A Twenty five workers were employed to complete a compound wall in 12 days. Five workers left after working for 4 days. The remaining 20 workers completed the work. In how many days the total job was completed?

(a) 15 (b) 16 (c) 14 (d) 18

Sol: The amount of work is 25 x 12 = 300 man-days.

In 4 days, the work completed is: 25 x 4 = 100 man-days.

Hence, 200 man-days work remains to be completed, which is to be completed by remaining 20 workers.

This will be completed in 200/20 = 10 days.

Total number of days = 4 + 10 = 14. Option (c) is the answer.

AILET 2016: 03: In a garden there are 8 rows and 10 columns of papaya tree. The distance between the two trees is 2 m and a distance of 1 m is left from all sides of boundary of the garden. The length of the garden is: (a) 24 m (b) 14 m (c) 20 m (d) 18 m

Sol: The length will be determined by number of columns only. The first tree (planted at the left most corner, shown with star mark) and tenth tree (planted at the right most corner, shown with star mark) have been planted leaving a distance of 1 meter from sides of boundary, and the distance between all trees from number 1 to 10 is 2 meter each.

Hence, total length = (2x9) + 1 + 1 = 20. Option (c) is the answer.

{Note: I may be wrong in doing this question, don’t know where. The official answer to this question provided by NLU Delhi is option (d)}

AILET 2016: 04: A man can row 14 km per hour in still water. In the stream flowing with a speed of 10 km per hour he takes 4 hours to move with the stream and come back. Find the distance he rowed the boat.

(a) 11.71 km (b) 13.71 km (c) 14.71 km (d) 12.71 km

Sol: The speed of rowing with water = 14 + 10 = 24 Km/hour.

The speed of rowing against water = 14 – 10 = 4 Km/hour.

Let the man rowed a total distance of 2d

(d upwards, and d downwards)

Time = Distance / Speed.

Then total time = 4 = (d/24) + (d/4); 4 = (d/24) + (6d/24);

4 = 7d/24; d = 96/7 = 13.71

2d = 192/7 = 27.42 Km.

‘Row' means to cause a boat to move through water by pushing against the water with oars. Hence, the distance he rowed against the water shall only be considered, which is 27.42/2 = 13.71. Option (b) is the answer.

AILET 2016: 05: A man covers a certain distance between his house and office on a scooter. Having an average speed of 30 km per hour, he reaches office late by 10 minutes. However, with a speed of 40 km per hour, he reaches his office 5 minutes earlier. The distance between his house and office is: (a) 30 km (b) 10 km (c) 20 km (d) 40 km

Sol: Let the distance between the man’s house and office be d km. Time = Distance / Speed

d = 30 km. Option (a) is the answer.

AILET 2016: 06: In a certain class, 72% of the students prefer for cold coffees and 44% prefer for fruit juice. If each of them prefers cold coffee or fruit juice and 48 like both, the total number of students in the class is: (a) 240 (b) 200 (c) 300 (d) 250

Sol: We have from the principle of Set Theory;

Let there be 100P students in the class.

Then, using this principle,

we have; 100P = 72P + 44P – 48;

that is: 116P – 100P = 48; 16P = 48: Or, P = 3;

Or, 100P = 300. Option (c) is the answer.

AILET 2016: 07: What will be the difference in simple and compound interest on Rs. 2,000 after three years at the rate of 10% per annum? (a) Rs. 60 (b) Rs. 42 (c) Rs. 62 (d) Rs. 40

Sol: The The amount after three years allowing compound interest

Difference in simple and compound interest = 662 – 600 = 62.

Option (c) is the answer.

Easier Method:

Compound interest is interest levied on simple interest.

At the end of first year, the compound interest and simple interest shall be equal, and shall be Rs. 200.

At the end of second year, the interest on this amount of interest shall be Rs. 20, and total interest shall be Rs. 420.

At the end of third year, the interest on this amount of interest shall be Rs. 42, and thus total compound interest shall be Rs. (20 + 42) = Rs. 62.

###### Heading 6

AILET 2016: 08: A man covers a certain distance between his house and office on a scooter. Having an average speed of 30 km per hour, he reaches office late by 10 minutes. However, with a speed of 40 km per hour, he reaches his office 5 minutes earlier. The distance between his house and office is: (a) 30 km (b) 10 km (c) 20 km (d) 40 km

Sol: Let the number of 50 paisa coins in the box be 4P.

Then, the number of 25 paisa coins in the box be 2P.

Then, the number of 1 rupee coins in the box be P.

Given, P + (2P/4) + (4P/2) = 56; 7P/2 = 56;

P = 16; 4P = 64. Option (d) is the answer.

AILET 2016: 19: In The average price of 10 pens is Rs. 12 while the average price of 8 of these pens is Rs. 11.75. Of the remaining two pens, if the price of one pen is 60% more than the price of the other, what is the price of each of these two pens? (a) Rs. 12, Rs. 14 (b) Rs. 5, Rs. 7.50 (c) Rs. 8, Rs. 12 (d) Rs. 10, Rs. 16.

Sol: Total price of 10 pens = 10 x 12 = Rs. 120.

Total price of 8 pens = 8 x 11.75 = Rs. 94.

Cost of remaining 2 pens = Rs. 26.

If cost of one pen of these two pens is 10P, the cost of other pen is 60% more and is 16P

Thus, 10P + 16P = 26; 26P = 26; P = 1.

Costs of pens are: Rs. 10, Rs. 16.

Option (d) is the answer.

AILET 2016: 10: The price of 7 Banana is equal to the cost of 3 kiwis. The price of 2 kiwis is equal to the cost of 1 banana and 5 chikoos. If Rambo has just enough money to buy 30 chikoos, then how many bananas Rambo could buy with the same amount? (a) 22 (b) 20 (c) 25 (d) 11

Sol: Let the price of 7 banana be 21P, hence price of 1 banana is 3P. Since price of 3 kiwis is equal to the cost of 7, cost of one kiwi is 7P.

Let the price of 1 chikoo be Q.

Given that the price of 2 kiwis is equal to the cost of 1 banana and 5 chikoos.

Hence, 14P = 3P + 5Q; 11P = 5Q.

30 chikoos will cost 30 Q = 66P.

Since price of 1 banana is 3P, Rs. 66P will fetch 22 bananas. Option (a) is the answer.

## AILET 2017

## AILET 2018

AILET 2018: 01: Two cards are drawn together from a pack of 52 cards. The probability that one is a club and one is a diamond, is

(a) 13/102

(b) 46/104

(c) 12/21

(d) 13/200

Sol: The required probability is:

Option (a) is the answer.

AILET 2018: 02: Sanju has two watches with a 12-hour cycle. One of these watches, gains one minute a day and the other loses one and half minutes per day. If Sanju sets both the watches at the correct time, how long will it be before they again show the correct time together? (a) 482 days (b) 290 days (c) 1440 days (d) 730 days

Sol: It is a very easy question that demands a little brain-storming. Though, the answer is very clear from the options too, let us understand it this way.

24 hours means 1440 minutes.

Hence, after 1440 days, the faster watch would gain 1440 x 1 = 1440 minutes = 24 hrs.

And, the slower watch will lose 1440 x (-3/2) = -2160 minutes = -36 hrs.

-36 + 24 = -12

Remember, the watches are of 12-hour cycle. And this is the most important point for solving this problem.

Thus, if the watches were set at time 0:00 hrs. on some date, say on P, the faster watch will show the 0:00 on date (P+1440), the slower watch will show 12:00 noon. Hence Option (c) is the answer.

The question should have avoided the word ‘correct time”, as the two watches could never show the correct time again. The sentence of the type ‘display of same time’ was a better option.

AILET 2018: 03: To display 500 wax Statuettes in a Museum, it is required to construct a big rectangular hall, allowing 22.5 cubic meter space per statuette. The height of the hall is to be kept at 7.5 meter, while the total surface area of the walls must be 1,200 sq.m. Then the length and breadth of the hall, respectively, are: (a) 62 m and 22 m (b) 50 m and 30 m (c) 46 m and 35 m (d) 40 m and 32 m

Sol: Let the length and breadth of the hall be L and B respectively. Given, height of the hall = 7.5 m. Total volume required in cubic meter for 500 wax Statuettes = 500 x 22.5 = 11250 = 7.5xBXL. Hence, BL = 11250/7.5 = 1500. Only option (b) qualifies for the BL = 1500. Option (b) is the answer.

Note: There is no need for considering given surface area of the walls. It was only necessary if for two options, LB would have been 1500.

AILET 2018: 04: The ratio between the length and the breadth of a rectangular jogging park is 3:2. If Ashish cycling along the boundary of the park at the speed of 12 km per hour completes one round in 8 minutes, then the area of the park in square meter is: (a) 3,07,200

(b) 30,720 (c) 1,53,600 (d) 15,360

Sol: The distance covered in 8 minutes = time x speed = (8/60) x 12 = 8/5 km = 1600 meter.

Given, for some positive integer P, length of park = 3P and breadth = 2P. Hence, 2(3P + 2P) = 1600, or, 5P = 800, or, P = 160. Area of the park = 6PxP = 6x160x160 = 153600. Option (c) is the answer.

AILET 2018: 05: A certain sum of money amounts to Rs. 1008 in 2 years and to Rs. 1164 in 3.5 years. Find the sum and the rate of interest. (a) Rs. 900 and 12% (b) Rs. 800 and 12% (c) Rs. 700 and 13% (d) Rs. 800 and 13%

Note: Keeping in view the strict timelines of the exam, I shall do it using the options.

Sol: Let the sum and the rate of interest be S and R respectively.

The difference of interest of 1.5 years = 1164 – 1008 = 156.

Since Sx(R%)xT = Interest, where T is time, we have, Sx(R%)x(3/2) = 156, hence, SxR%x(1) = 156x(2/3) = 104, which is obtained from option (d) only.

AILET 2018: 06: Azhar can row 9.333 kmph in still waters and find that it takes him thrice as much time to row up to than as to row down in the same distance in the river. The speed of the current of the river in km/h is: (a) 3.333 (b) 3.111 (c) 4.666 (d) 4.5

Sol: Time = Distance/ Speed. If the speed of river is P kmph, then, as given, D/(9.333 – P) = 3D/(9.333 + P), or, (9.333 + P) = 3x(9.333 – P), or,

4P = 2x9.333. Thus, P = 9.333/2 = 4.666 kmph. Option (c) is the answer.

AILET 2018: 07: A wheel that has 6 cogs is meshed with a larger wheel of 14 cogs. When the smaller wheel has made 21 revolutions, then the number of revolutions made by the larger wheel is: (a) 7 (b) 9 (c) 14 (d) 42

Sol: The smaller wheel with less number of teeth has more revolutions per minute than the one with more number of teeth. Hence number of teeth N and number of revolutions per minute, say R, are inversely proportional to each other. Hence NR = Constant. Let number of revolutions made by the larger wheel per minute be P. Then, 6 x 21 = 14 x P; or, P = 126/14 = 9.

Option (b) is the answer.

AILET 2018: 09: A wooden almirah is sold at a certain price. By selling it at two third of that price one loses 10%. Find the profit percent at the original price. (a) 15% (b) 25% (c) 35% (d) 45%

Sol: Let the selling price be 135P (NOTE). Then by selling it at two third of that price, i.e. at Rs. 90P, loss is 10%. Hence, CP is 100P. Hence profit percent at Rs. 100P when it is sold for Rs. 135P, is 35%. Option (a) is the answer.

The calculations have rather been easy by taking selling price equal to 135P.

AILET 2018: 08: In a University for each Rs. 200 spent by the Cultural Committee, Debating Committee spends Rs. 20, and for every Rs. 400 spent by Debating Committee, the Student Welfare Committee spends Rs. 150. The triple ratio of money spent by the Cultural Committee to the money spent by the Debating Committee to the money spent by the Student Welfare Committee can be expressed as: (a) 80 : 8 : 3

(b) 60 : 8 : 3 (c) 40 : 4 : 3 (d) 20 : 8 : 3

Sol: Given Cultural Committee to Debating Committee 200 : 20; and Debating Committee to the Student Welfare Committee spends 400 : 150.

Hence, by multiplying by 20 in the first ratio, Cultural Committee to Debating Committee 4000 : 400; and Debating Committee to the Student Welfare Committee spends 400 : 150.

Hence, Cultural Committee : Debating Committee : Student Welfare Committee = 4000 : 400: 150 = 80: 8: 3. Option (a) is the answer.

AILET 2018: 10: The ratio between the present age of Anu and Balbir is 5 : 3 respectively. The ratio between Anu’s age 4 years ago and Balbir’s age 4 years hence is 1 : 1. What is the ratio between Anu’s age 4 years hence and the Balbir’s age 4 years ago? (a) 4 : 1 (b) 2 : 4 (c) 3 : 2 (d) 3 : 1

Sol: Let the present age of Anu and Balbir be 5P and 3P respectively. Then, given, (5P – 4)/(3P + 4) = 1/1.

Hence, 5P – 4 = 3P + 4; or, 2P = 8, P = 4. Hence, present age of Anu and Balbir are 20 and 12 years respectively.

The ratio between Anu’s age 4 years hence and the Balbir’s age 4 years ago = (20 + 4)/(12 – 4) = 24/8 = 3 : 1.

Option (d) is the answer.