CAT HCF and LCM Questions PDF [Most Expected with Solutions]

HCF and LCM for CAT are one of the key topics in the CAT Quant section. You can check out these HCF and LCM CAT Previous year questions.  In this article, we will look into some important HCF and LCM Questions for CAT. These are a good source for practice; If you want to practice these questions, you can download this CAT HCF and LCM Questions PDF, which is completely Free.

HCF and LCM CAT Questions form an important part of CAT Quant Number Systems. Applications of LCM and HCF questions in CAT were asked in the past years. These CAT questions will be very useful for quantitative aptitude for CAT.

Take a free CAT mock test and also try to solve CAT previous questions to get good understanding of the LCM and HCF CAT questions. This topic comes under the number system for CAT. Though the Least Common Multiple and Highest Common Factor is a very basic topic, their applications are very important.

You can download the HCF and LCM CAT Problems or you can go through the questions below.

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Question 1: A number N is randomly selected between 1 and 249 (both inclusive). What is the probability that the HCF of N and 249 is 1?

a) $\dfrac{164}{249}$

b) $\dfrac{163}{249}$

c) $\dfrac{2}{3}$

d) None of these

1) Answer (A)

Solution:

HCF of N will be 1 when the N and 249 are co-primes
Alternative 1:-
249 = 3*83
The number of co-primes of 249 = $249*(1-\frac{1}{3})*(1-\frac{1}{83})$ = $\dfrac{249*2*82}{3*83}$ = $164$
Thus, there are 164 numbers which are co-prime to 249.
Hence, the required probability = $\dfrac{164}{249}$
Alternative 2:-
249 = 3*83
All the multiples of 3 and 83 will have 3 and 83 as the HCF with 249.
The number of multiple of 3 between 1 and 249 = 83
The number of multiple of 83 between 1 and 249 = 3
Thus, the total number of numbers which are not co-prime to 249 and lying between 1 and 249 is 83+3-1(as we have counted 249 twice) = 85
Thus, the number of co-primes = 249-85 = 164
Thus, the required probability = $\dfrac{164}{249}$
Hence, option A is the correct answer.

Question 2: Sum of two non co-prime numbers a, b and their HCF gives 77. What is the number of possible values of (a, b)?

2) Answer: 6

Solution:

Let the HCF be h, then the numbers can be expressed as,  a=hx, b=hy , where x,y are co-primes.

h+hx+hy = 77
h(1+x+y) = 77
h can be 1 or 7 or 11
h $\neq$ 1    (as a,b are non co-primes)
If h = 7, (1+x+y) = 11
x+y = 10
Now we have to select the values of x,y such that they are co-prime to each other.
x = 1, y = 9
x = 3, y = 7
x = 7, y = 3,
x = 9, y = 1
Hence when HCF is 7, there are 4 possible pairs of (x, y)
If HCF = 11, (1+x+y) = 7
x+y = 6
x = 1, y = 5
x = 5, y = 1
There are two possible values of (x,y).

Total values of a and b = 4+2=6
Hence 6 is the correct answer.

Question 3: If the roots of a quadratic equation f(x) = 0 have an HCF of 5 and an LCM of 40, find the approximate distance between the roots of the equation whose roots are the arithmetic mean and the harmonic mean of the roots of the equation f(x) = 0.

a) 14.5

b) 13.6

c) 12.0

d) 11.4

3) Answer (B)

Solution:

HCF = 5, LCM = 40

Let the numbers be 5x and 5y.

25xy = 200

xy = 8

x = 1, y = 8

x = 2, y = 4 (Not possible since they have to be co-prime)

Hence, the roots are 5 and 40.

AM = 45/2 = 22.5

HM = $\frac{2\times\ 5\times\ 40}{40+5}=\frac{80}{9}$ = 8.88 OR 8.9

Hence, distance between the roots = difference = 22.5 – 8.9 = 13.6

Question 4: Find the number of unordered pairs of 2-digit numbers, such that their LCM is twice their HCF.

4) Answer: 40

Solution:

Let the HCF be h. Hence, LCM = 2h.

Hence, product of the numbers = LCM X HCF = $2h^2$

Let the numbers be hx and hy, such that x and y are co-prime.

Hence, product = $h^2xy$

Equating, we get xy = 2.

It is only possible when one of the numbers is twice the other.

Hence, the lowest possible group of such numbers is (10, 20), and the highest is (49, 98).

Hence, total possible unordered pairs = 49 – 10 + 1 = 40.

Question 5: What is the HCF of $3^{80}-1$ and $3^{60}-1$?

a) $ 3^{5}-1$

b) $ 3^{10}-1$

c) $ 3^{20}-1$

d) $ 3^{30}-1$

5) Answer (C)

Solution:

HCF { $3^{80}-1,\ 3^{60}-1$}

HCF { $3^{80}-1,\ 3^{80}-1-3^{60}+1$}

HCF { $3^{80}-1,\ 3^{80}-3^{60}$}

HCF { $\left(3^{20}-1\right)\left(3^{60}+3^{40}+3^{20}+1\right),\ 3^{60}\left(3^{20}-1\right)$}

$\left(3^{60}+3^{40}+3^{20}+1\right)\ \&\ \ 3^{60}\ $ are coprimes

Thus the HCF of both the terms= $ 3^{20}-1$

Option C

HCF & LCM Tricks For CAT

Question 6: The value of a natural number N lies between 105 and 315. If the HCF of N and 315 is 1, what is the number of possible values of N lying between 105 and 315?

a) 128

b) 116

c) 96

d) 112

e) 82

6) Answer (C)

Solution:

Given, HCF of N and 315 is 1 i.e. N and 315 are co-primes.
315 = $5 \times 7 \times 3^2$
Hence, N should not be a multiple of 3 or 5 or 7.
Number or natural number of between 105 and 315 = 315-105-1 = 209
Number of multiples of 3 between 105 and 315 = (315-105)/3 -1 = 69
Number of multiples of 5 between 105 and 315 = (315-105)/5 -1 = 41
Number of multiples of 7 between 105 and 315 = (315-105)/7 -1 = 29
Number of multiples of 3*5 between 105 and 315 = (315-105)/15 -1 = 13
Number of multiples of 3*7 between 105 and 315 = (315-105)/21 -1 = 9
Number of multiples of 5*7 between 105 and 315 = (315-105)/35 -1 = 5
Number of multiples of 3*5*7 between 105 and 315 = (315-105)/105 -1 = 1

Hence number of Natural numbers between 105 and 315 which are not divisible by 3,5 or 7  = 209 – 69 – 41 – 29 + 13 + 9 + 5 – 1 = 96

Alternative solution:
105 = $ 5 \times 7 \times 3$
Since both 315 and 105 have the same prime factors N would also be co-prime to 105. Thus, the value of N = All co-primes of 315 – All co-primes of 105
= 315( 1 – 1/7)(1 – 1/5)(1 – 1/3) – 105(1 – 1/7)(1 – 1/5)(1 – 1/3)
= 48 x 3 – 48 = 96

Question 7: Find the number of positive integral solutions of integers x, y (x > y) such that HCF (x, y) + LCM (x, y) = 51

a) 3

b) 4

c) 5

d) 6

7) Answer (B)

Solution:

Let HCF be h x=ha,y=hb where a and b are co-prime. LCM of(x,y)=h*a*b h+h*a*b=51 h(1+ab)=1*3*17 h=1,ab=50.Only 2 pairs (1,50),(2,25) h=3,ab=16.Only 1 pair (1,16) as rest are not co-prime h=17,ab-2.Only 1 pair (2,1) Hence 4 pairs of positive integral solutions are possible.

Question 8: Let (a,b) be 2 numbers such that their lcm is  300. How many such ordered pairs are possible?

a) 38

b) 39

c) 75

d) 76

8) Answer (C)

Solution:

300 can be written as $2^2\times3\times5^2$

Thus $a$ and $b$ should also have 2,3 and 5 as their prime factors only.

For two such numbers a,b the number of ordered pairs whose LCM = $N\ =\ 2^p3^q5^r$ will be $\left(2p+1\right)\left(2q+1\right)\left(2r+1\right)$

Thus number of ordered pairs such that LCM 300 = $2^2\times3\times5^2$ is their lcm = $\left(2\left(2\right)+1\right)\left(2\left(1\right)+1\right)\left(2\left(2\right)+1\right)$

Thus number of ordered pairs = $5\times3\times5=75$

Question 9: How many pairs of numbers are there whose LCM is 20?

9) Answer: 8

Solution:

Let the two numbers be $a,b$ where $a \leq b$.
$20 = 2^{2}*5^{1}$
Let $a=2^{x_1}*5^{y_1}$ and $b=2^{x_2}*5^{y_2}$
The LCM of $a,b$ will be 20 if max($x_1,x_2$)=2 and max ($y_1,y_2$)=1
max($x_1,x_2$) = 2 is possible in 5 ways and max($y_1,y_2$) = 1 is possible in 3 ways.
Hence, the total number of ways is 5*3 = 15. But this includes duplicates. For example, (1,20) and (20,1) are considered separate cases. As the question asked for pairs alone, we should remove the duplicates.

Out of these 15 cases, we have 1 case where $a=b=20$, in all other cases, we have one number less than the other. Hence, the total number of pairs whose LCM is 20 is (15-1)/2 + 1 = 8

Question 10: The LCM of $7^{1403}-1$ and $7^{1403}+1$ when divided by 10 leaves a remainder of

10) Answer: 4

Solution:

$7^{1403}-1$ and $7^{1403}+1$ are two consecutive even numbers. Thus, their HCF has to be 2.
Now, HCF X LCM = Product of numbers
=> 2 X LCM= $7^{2806}-1$ =>LCM=$\frac{7^{2806}-1}{2}$
The remainder when this number is divided by 10 is the unit digit of the number.

As 7 has special cyclicity of 07, 49, 43 and 01.

2806/4 gives remainder 2 so the last 2 digit of $7^{2806}$ are 49.

which makes LCM/10 =( …48/2)/10

remainder is 4.

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Question 11: The LCM of three natural numbers P, Q and R is $2^6*3^7*5^5*7^7$. If the LCM of 2P, Q and 6R is also $2^6*3^7*5^5*7^7$, then find out the total number of values that Q can take?

11) Answer: 384

Solution:

Given that,  LCM of (P, Q, R) = $2^6*3^7*5^5*7^7$

LCM of (2P, Q, 6R) = $2^6*3^7*5^5*7^7$

We can see that we have multiplied by 2 in each of P and R even after that the power of 2 remains same in the LCM. This explains that the number of 2 in LCM depends upon the power of 2 available Q. Hence, we can say that Q is a multiple of $2^6$.

We have multiplied by 3 in R even after that power of 3 remains same in the LCM. This explains that the number of 3 in LCM depends upon the the power of 3 available in either P or Q. So, number of 3 can be anything in Q ranging from 0 to 7.

There is no restriction on number of 5 and 7 repeating in Q.

Therefore, Q = $2^6*3^a*5^b*7^c$   {Where 0 $\leq$ a $\leq$ 7,  0 $\leq$ b $\leq$ 5, 0 $\leq$ c $\leq$ 7)

Hence, we can say that total number of values that Q can take = 1*8*6*8 = 384.

Question 12: The LCM of $(17)_{n}$ and $(14)_{n}$ is $(330)_{4}$. Their GCD is $(11)_2$. Find n.

12) Answer: 8

Solution:

Given,
LCM = $(330)_{4}$ = $(60)_{10}$

GCD = $(11)_{2}$ = $(3)_{10}$

Product of the numbers = LCM * HCF

$(17)_{n}$ = n + 7

$(14)_{n}$ = n + 4

(n + 4) (n + 7) = 180

$n^2 + 11n – 152 = 0$

(n – 8) (n + 19) = 0

n = 8

Question 13: Rita has a tool which will give the LCM of two numbers. What is the minimum number of times she should run the tool to calculate the LCM of any 20 numbers?

13) Answer: 19

Solution:

Rita has to run the tool once to get the LCM of two numbers.

To find the LCM of two numbers say 2, 4, he has to run the tool once.

To find the LCM of three numbers say 2, 4, 6

Rita will run the tool once to find the LCM of 2, 4 and then run the tool once again with 6 and the LCM of 2, 4

On similar lines

To calculate the LCM of 20 numbers, she has to run the tool 19 times.

Question 14: The HCF of 2 numbers is 4 and the LCM is 1008. Further, it is known that one of the 2 numbers has an odd number of factors. The larger one among the 2 numbers is

a) 144

b) 112

c) 1008

d) Cannot be determined

14) Answer (D)

Solution:

We know that the product of 2 numbers = HCF * LCM
Let the 2 numbers be X and Y.
XY = $4*1008$
= $2^4*6^2*7$
We know that one of the numbers has an odd number of factors. Therefore, one of the numbers must be a perfect square.

However, the 2 numbers can be split as $2^2*6^2$ and $2^2*7$ or $2^2$ and $2^2*6^2*7.$
Therefore, we cannot determine the 2 numbers. Option D is the right answer.

Question 15: LCM of $\ \frac{\ 4}{3},\ \ \frac{\ 18}{5},\ \ \frac{\ 25}{3},\ \ \frac{\ 21}{41}$ is

a) 6300

b) 420

c) 2100

d) 10.24

15) Answer (A)

Solution:

LCM of fractions = $\frac{\text{ LCM of Numerator}}{\text{HCF of Denominator}}$

= $\ \frac{\ LCM\ of\ 4,\ 18,\ 25,\ 21}{HCF\ of\ 3,\ 5,\ 3,\ 41}$

= 6300 (LCM of 4, 18, 25, 21 is 6300 and HCF of 3, 5, 3, 41 is 1)

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Question 16: If LCM of two numbers is 231, HCF of two numbers is 11 and one number is 77, the other number is:

a) 33

b) 47

c) 37

d) 57

16) Answer (A)

Solution:

As we know that $one\ number\ \times\ other\ number\ =\ LCM\times\ HCF$

If LCM of two numbers is 231, HCF of two numbers is 11 and one number is 77.

$77\ \times\ other\ number\ =\ 231\times\ 11$

$7\ \times\ other\ number\ = 231$

other number = 33

Question 17: The product of 2 numbers, their HCF, and their LCM is 3600. If it is known that the LCM of the 2 numbers is 30, then the sum of the 2 numbers can be

a) 17

b) 16

c) 13

d) 19

17) Answer (B)

Solution:

Let the HCF of the 2 numbers be $h$.
Let us represent the numbers as $ha$ and $hb$.
=> $a$ and $b$ are co prime to each other and the LCM of the 2 numbers = $hab$.
Product of the 2 numbers = $h^2ab$
Product of the HCF and LCM of the 2 numbers = $h^2ab$
Product of the 2 numbers, their HCF and LCM = $h^4a^2b^2$
It has been given that $h^4a^2b^2$ = $3600$
Taking square root on both sides, we get,
$h^2ab=60$

Product of 2 numbers = 60.
The numbers can be  (1,60), (2,30), (3,20), (4,15), (5,12), and (6,10).
We know that the LCM of the 2 numbers is 30.
Therefore, we can eliminate (1,60), (3,20),(4,15), and (5,12).
The 2 numbers must be one among (2,30), and (10,6).
The sum of the 2 numbers can be 32, or 16.
Therefore, option B is the right answer.

Question 18: If LCM of two numbers 28!*32! and 30!*31! is $\frac{a!*b!}{c!}$, where a, b (a > b) are two-digit numbers and c is one digit number. Find out the value of a -(b+c)?

18) Answer: 0

Solution:

Let us first calculate HCF of 28!*32! and 30!*31!

$\Rightarrow A = 28! *(32 *31!)$

$\Rightarrow A = 32*28! *31!$              … (1)

$\Rightarrow B = 30!*31!$

$\Rightarrow B = 30*29*28!*31!$           … (2)

From equation 1 and 2 it is clear that HCF = 2*28!*31!

We know that: Product of 2 number = LCM * HCF

$\Rightarrow LCM = \frac{Product of A and B}{HCF of A and B}$

$\Rightarrow LCM = \frac{(28!*32!)*(30!31!)}{2*28!*31!}$

$\Rightarrow LCM = \frac{32!*30!}{2} = \frac{32!*30!}{2!}$

Hence we can say that a = 30, b = 30 and c = 2

Hence a-(b+c) = 32 -(30 +2) = 0 (Answer :0)

Question 19: Find out the smallest number which when individually divided by 16, 17, and 20 leave remainders 13, 14 and 17 respectively.

a) 1357

b) 1460

c) 1268

d) 1456

19) Answer (A)

Solution:

16 – 13 = 3
17 – 14 = 3
20 – 17 = 3
Hence, we find out the LCM(16, 17, 20) = 16 X 17 X 5 = 1360.
Now, we subtract 3 from this LCM.
1360 – 3 = 1357.
Hence, 1357 is smallest possible number.

Question 20: Find the largest 3 digit number which when divided by 6, 7 and 8 leaves remainder of 4,5 and 6 respectively.

a) 672

b) 744

c) 838

d) 948

20) Answer (C)

Solution:

N = LCM(a,b,c)*k + LCM(a,b,c) – x
N = LCM(6,7,8)*k + LCM(6,7,8) – 2
= 168k + 168 – 2
= 168k + 166
Largest 3 digit number comes when k=4;
N = 838.

Question 21: 3 friends A, B and C run around a circular track of length 120 m continuously. All of them start at the same point. A and B run in the clockwise direction while C runs in the anti-clockwise direction. The speeds of A, B and C are in the ratio 13:5:7. At how many distinct points will all three of them meet?

a) 6

b) 4

c) 3

d) 5

21) Answer (B)

Solution:

Length of the track is 120 m.
Speeds of A and B are in the ratio 13:5. They are running in the same direction.
Therefore, they will meet at 13-5 = 8 distinct points.
Let us use the distance from the starting point to label the meeting points.
A and B will meet at points at a distance of 0 m, 15 m, 30 m, 45 m, 60 m, 75 m, 90 m, and 105 m from the starting point.

B and C run in the opposite directions. Their speeds are in the ratio 5:7.
Therefore, they will meet at 5+7 = 12 distinct points.
They will meet at points at a distance of 0 m, 10 m , 20 m, 30 m, 40 m, 50 m, 60 m, 70 m ,80 m, 90 m, 100 m and 110 m.

A and C run in the opposite directions. Their speeds are in the ratio 13:7.
Therefore, they will meet at 13+7 = 20 distinct points.
They will meet at points at a distance of 6 m, 12 m, 18 m, … 114 m.

All 3 of them will meet at points that are multiples of the LCM of 15, 10 and 6.
LCM (15,6,10) = 30 m.

Alternatively, Number of distinct points = HCF(8, 12, 20)

Therefore, in a sufficiently large number of rounds, all 3 of them will meet at points at a distance of 0 m, 30 m, 60 m and 90 m from the starting point. There are 4 points in total. Therefore, option D is the right answer.

Question 22: Raj buys an iron rod of unknown length. He cuts the rod into 4 equal parts. Then, he cuts one of the 4 parts into 6 equal pieces, another into 7 equal pieces, another into 8 equal pieces and the last one into 9 equal pieces. He notes that the lengths of all the smaller pieces of rods in centimetre are integers. The least possible length of the rod is

a) 504 cm

b) 496 cm

c) 2016 cm

d) 1512 cm

22) Answer (C)

Solution:

We know that each part of the rod can be divided into 6, 7, 8 and 9 pieces and each of the pieces will have an integral length. Therefore, the length of the least possible length of each part must be an integral multiple of 6,7,8 and 9 (or in other words LCM of 6,7,8 and 9).

The LCM of 6,7,8 and 9 is 504 cm.
Therefore, the length of each part must be at least 504 cm. There are 4 such parts in total. Therefore, the total length of the rod must be at least 4*504 = 2016 cm. Hence, option C is the right answer.

Question 23: A group of 20 friends went to a party. On their way they found a certain number of gold coins. The number of coins is such that if they divide them equally, then 18 coins will remain. Had there been 1 friend less, 17 coins would have been left after dividing all the coins equally. Similarly, if the number of friends had been 18, the number of coins left after equal division would have been 16 and so on. Also, it is known that the number of coins is the least possible value that satisfies all these conditions. What would be the number of coins left if there were 50 friends?

a) 18

b) 28

c) 8

d) 38

23) Answer (C)

Solution:

Let the total number of gold coins be ‘n’. We have been given that ‘n’ is of the form 20k + 18 = 19m + 17 + 18n + 16 = 15l + 13 . . . . . .
We can see that the terms are leaving a remainder of -2 on division with the different divisors. Hence, the required number of gold coins will be
LCM(1 to 20) – 2
LCM of first 20 numbers will be
Highest power of 2 among numbers from 1 to 20 = $2^4$
Highest power of 3 among numbers from 1 to 20 = $3^2$
Highest power of 5 among numbers from 1 to 20 = 5
Hence, the LCM will be
16*9*5*7*11*13*17*19
Hence, the required number will be 16*9*5*7*11*13*17*19 – 2
Now we need to find the remainder when this number is divided by 50.
{(16*9)(11*5)(7*13)(17*19) – 2} mod 50
=> {-6*5*-9*23 – 2} mod 50
=> (20*23 – 2) mod 50
=> 458 mod 50 = 8
Hence, the required answer is 8.

Question 24: Three arithmetic sequences $S_{1}$, $S_{2}$ and $S_{3}$ are given below.
$S_{1}$ = 1, 4, 7, 10, …., 991
$S_{2}$ = 2, 6, 10, 14, …., 1014
$S_{3}$ = 3, 8, 13, 18, …., 1008
If a sequence $S_{x}$ contains all terms which are common in all three sequences, then find the sum of terms of the sequence $S_{x}$.

24) Answer: 8128

Solution:

Let us assume $S_{y}$ is the series that contains all the common terms in $S_{1}$ and $S_{2}$.

Common difference of the series $S_{1}$ = 3

Common difference of the series $S_{2}$ = 4

We can see that the first term of $S_{y}$ = 10. Also common difference of the arithmetic sequence $S_{y}$ = LCM of (3, 4) = 12.

Hence, we can say that $S_{y}$ = 10, 22, 34, 46, 58, …

Now we can say that $S_{x}$ is the series that contains all the common terms in $S_{x}$ and $S_{3}$.

Common difference of the series $S_{y}$ = 12

Common difference of the series $S_{3}$ = 5

We can see that the unit digit of each number in $S_{3}$ is either 3 or 8. We also know that all terms in $S_{x}$ are even. Hence, the first common term of both the series will be the one which if the first term in $S_{x}$ that ends with 8 which is 58.

Hence, we can say that the first term of $S_{x}$ = 58. Also common difference of the arithmetic sequence $S_{y}$ = LCM of (12, 5) = 60.

Therefore, $S_{x}$ = 58, 118, 178, …

We know that the last term of $S_{1}$ is 991. Hence we can say that the last term of $S_{x}$ will be $\leq$ 991.

Let ‘p’ be the number of terms in the series $S_{x}$. Then we can say that,

$\Rightarrow$ 58 + (p – 1)*60 $\leq$ 991

$\Rightarrow$ p $\leq$ 16.55

Hence, the number of terms in the series $S_{x}$ = 16.

Therefore, the sum of all the terms of series $S_{x}$ = $\dfrac{16}{2}(2*58+(16-1)60)$ = 8128

Question 25: On a circular track of circumference 120 meters, three people, A, B, and C run at constant integral speeds (in m/sec). The speed of A is 5 m/sec and is greater than that of B, and when A and B start running on the track simultaneously in opposite directions, they can meet at a total of 8 distinct points on the track. When B and C start running on the track simultaneously in opposite directions, they meet at 9 distinct points. If A, B, and C start running simultaneously from the same point at a time after how many seconds do all three of them meet at the starting point?

a) 120 seconds

b) 96 seconds

c) 192 seconds

d) 108 seconds

25) Answer (A)

Solution:

Given the speed of A = 5 m/sec. Considering the speed to B = b m/sec and the speed of C to be c m/sec.

When two people with speeds a, b run on a track of length L in the opposite directions, the number of distinct points they can meet is: a+b

It has been given that the circumference is equal to  120 meters.

The number of distinct meeting points is 8. a+b = 8

Solving this we get multiple possibilities:

a = 5m/sec, b = 3m/sec or a = 5 m/sec, b = $\frac{25}{3}$ m/sec, a = 5m/sec, b = 35 m/sec.

But since it has been mentioned in the question that the speeds are integral and the speed of a is greater than that of b,

The possible value for b = 3m/sec.

Using the value of b and applying the same for B and C.

Since the number of distinct points is 9.

The possible ratios can be 1: 8, 2: 7, 4: 5, 5: 4, 7: 2, 8: 1.

There can be a single possible case to form the value of their reduced speeds to be equal to 9.

The case is b = 3m/sec and c = 24m/sec. This is when the ratio is 1: 8.

The time after which all three of them meet at the starting point is :

LCM of $\left(\frac{120}{5},\ \frac{120}{3},\ \frac{120}{24}\right)$

=(24, 40, 5)

We get the LCM as 120 seconds

Question 26: A number N leaves a remainder 4,5 and 13 when divided by 7,8 and 16 respectively.Find the number of values of N if it is known that N < 1000.

a) 10

b) 9

c) 8

d) 11

26) Answer (C)

Solution:

It is given that
N=7k+4
⇒N+3 =7k+7 =7K
N=8m+5
⇒ N+3 =8m+8 =8M
N=16n+13
⇒ N+3 =16n+16 =16N
So we can say
N+3 is of the form LCM(7K,8M,16N)
So N is of the form LCM(7K,8M,16N) -3
Therefore LCM (7,8,16) =112
So minimum value of N = 112-3=109
Now next value of N will be 109+LCM(7,8,16) =221
So values of N <1000 =109,221,333,445,557,669,781,893
So we have 8 values of N.

Question 27: In an equilateral triangle ABC, 3 line segments are drawn parallel to the three sides, such that the line parallel to BC, PQ divides AB in the ratio 4:3, the line parallel to AC, RS divides BC in the ratio 5:4 and the line parallel to AB, TU divides AC in the ratio 2:3. If the area of triangle ABC is 99225 sq cm, find the area of the triangle formed by the line-segments PQ, RS and TU.

a) $8324\ cm^2$

b) $7396\ cm^2$

c) $5642\ cm^2$

d) Data insufficient

27) Answer (B)

Solution:

AP = 4k and PB = 3k
BS = 5p and SC = 4p
AT = 2t and TC = 3t

LCM (7, 9, 5) = 315x

Let each side of the triangle be equal to 315x. So, AB = BC = CA = 315x => k = 315x/7 = 45x, p = 315x/9 = 35x and t = 315x/5 = 63x
So, PB = 135x, SC = 140x and AT = 126x
Triangles TUC, RBS and APQ are equilateral triangles.

So, TU = TC = 3t = 3*63x = 189x
PB = 3k = 3*45x = 135x
BPYU is a parallelogram. So, UY = BP = 135x
AB = BC and BR = BS => AB – BR = BC – BS => AR = SC = 4p = 4*35x = 140x
ARXT is a parallelogram => XT = AR = 140x

UT = 3/5 * 315x = 189x
So, XY = UY + XT – UT = 140x + 135x – 189x = 86x

Solving similarly, we get, XZ = ZY = 86x
Area of big triangle = $\sqrt3/4 * (315x)^2 = 99225$

Area of triangle XYZ = $\sqrt3/4 * (86x)^2 = 7396\ cm^2$

Question 28: Akash, Arun, Arpita and Akshara have followers on social media in the ratio $\frac{1}{3} : \frac{1}{5} : \frac{1}{7} : \frac{1}{11}$ respectively. If it is known that each of the friends has at least 200 followers, what is the minimum number of followers Arun could have?

28) Answer: 462

Solution:

The ratio of the number of followers with them is $\frac{1}{3} : \frac{1}{5} : \frac{1}{7} : \frac{1}{11}$
LCM of 3, 5, 7 and 11 is 1155
So, they must be having followers in the ratio
$\frac{1155}{3} : \frac{1155}{5} : \frac{1155}{7} : \frac{1155}{11}$
that is equivalent to
385 : 231 : 165 : 105
It is given they each one of them have at least 200 followers.
So, the minimum ratio can be
770 : 462 : 330 : 210
Therefore, the minimum number of followers Arun could have is 462.
Hence, 462 is the correct answer.

Question 29: Four bells ring simultaneously at a certain instant. Thereafter they ring at intervals of 6, 8, 10 and 12 seconds respectively. In how many minutes will they ring together again for the first time?

a) 2 minutes

b) $2\frac{1}{4}$ minutes

c) 1 minute

d) $1\frac{1}{2}$ minutes

29) Answer (A)

Solution:

Four bells ring simultaneously at a certain instant. Thereafter they ring at intervals of 6, 8, 10 and 12 seconds respectively
Now they will ring together again after LCM(6,8,10,12 ) seconds =120seconds = 2 minutes

Question 30: The dimensions of a floor are $18\times24$. What is the smallest number of identical square tiles that pave the entire floor without the need to break any tile?

a) 6

b) 24

c) 8

d) 12

30) Answer (D)

Solution:

We have dimensions : (24*18)
LCM (24,18) =72

HCF ( 24, 18) = 6

Now Area of floor: 432

In order to minimise the number of tiles considering a tile with a dimension equivalent to the HCF,  we have the dimension of the tile to 6*6 = 36 units.

A total of $\frac{432}{36}$  = 12 tiles are required.

Minimum number of tiles that pave the entire floor :(LCM)^2/Area = 12

Question 31: What is the sum of all multiples of 3 less than 1000 which give an odd remainder when divided by 11 ?

31) Answer: 75837

Solution:

The LCM of 3 and 11 is 33. Thus, the number of numbers we find in the first 33 numbers will be the number of numbers that are there in every consecutive set 33 numbers.

Multiples of 3 less than 33 that have odd remainders when divided by 11 : 3,9,12,18,27

In the next set of 33 numbers, the numbers that satisfy the conditions are : 36,42,45,51,60 ie 33+3,33+9,33+12,33+18,33+27 respectively.

Below 1000, there will be $\dfrac{1000}{33} = 30 \dfrac{10}{33}$ which means there will be 30 such sets.

From 33×30 to 1000, ie from 990 to 1000, there are only 2 more numbers that satisfy the conditions ie 993 and 999.

To find sum of all the numbers, first we find the sum of the first series of numbers ie all numbers below 33 that satisfy the condition.

$S_1 = 3+9+12+18+27 = 69 $

The sum of the second series of numbers $S_2$ = (3+33)+(9+33)+(12+33)+(18+33)+(27+33) = 69 + (5 * 33) = 69+165

Similarly, the sum of the third series of numbers $S_3 = 69+(165 \times 2)$

Thus, sum of the $n^{th}$ series of numbers $S_n = 69+[165 \times (n-1)] $

Thus, total sum of the series till the $30^{th}$ set of numbers $S = (69 \times 30) + [165 \times (1+2+3…..29)] $

$ \Rightarrow S = 2070 + [165 \times \dfrac{29 \times 30}{2}] $

$ \Rightarrow S = 2070 + 71775 = 73845 $

We should remember that this series does not take into consideration the last 2 numbers.

Therefore the actual sum $S’ = 73845 + 993 + 999 = 75837 $

Question 32: A clock gains 12 minutes per hour and another clock loses 13 minutes in 1.5 hours.If both the clocks show the correct time on 13th May at 1 pm, find the day on which both the clocks show 1 o clock again.

a) August 10th

b) June 12th

c) May 28th

d) June 27th

32) Answer (D)

Solution:

The first clock gains 12 minutes per hour and the second clock loses 8 minutes 40 seconds every hour.

So, in 3 hours, the first clock will gain 36 minutes while the second clock will lose 26 minutes. Thus, the difference between the two clocks after 3 hours would be 36 + 26 = 62 minutes.

For both the clocks to show 1 o clock, again the difference should be a multiple of 720 minutes. (Since when the difference between the time shown by the two clocks is 12 hours, then both the clocks will essentially be showing the same time, except the AM, PM part).

Thus, the next time when both the clocks will show 1 clock will be LCM(720, 62) = 22320 minutes

Thus, the number of 3 hour cycles required = 22320/62 = 360

Hence, number of hours required = 360*3

Thus, the number of days = 360*3/24 = 45 days

45 days = 1 o clock, June 27th.

Question 33: Abhi is a very superstitious person. He went to an astrologer who suggested him that the next lottery winning ticket will be a 2-digit number which is less than 100, not a multiple of 2,3 or 5 and is not a perfect square or a perfect cube. Abhi wants to buy all the tickets which have a possibility of becoming the winning ticket. How many tickets should he buy ?

a) 22

b) 21

c) 24

d) 23

33) Answer (D)

Solution:

To find all the numbers that suits Abhi, we just subtract the ones that are not possible from the total number of possibilities.

Since it is a 2-digit number, so the number of possibilities = 99-9 = 90

Thus, within all the 2-digit numbers :

Number of multiples of 2 = 49 – 4 = 45

Number of multiples of 3 = 33 – 3=30

Number of multiples of 5 = 19 – 1 = 18

Number of perfect squares = 1 (only 49 matches the requirements)

Number of perfect cubes = 0 (all the cubes are already multiples of the prohibited numbers)

Here, we need to account for using a few numbers from the set more than once.

Thus, multiples of 6,10,15,30 (LCM of (2,3),(2,5),(3,5),(2,3,5) respectively) have been counted more than once.

Multiples of 6 = 16 – 1 = 15

Multiples of 10 = 9

Multiples of 15 = 6

Multiples of 30 = 3

Thus, we need to add the multiples of 6,10,15 to the sum and deduct the multiples of 30 to get the actual number of numbers.

Thus, number of numbers = 90 – (45+30+18+1) + (15+9+6) – 3 = 23 numbers

Alternatively,
There are 25 primes less than 100, 21 of which are 2 digit primes
These primes are relatively co-prime to 2, 3 and 5
We know the prime numbers are 2,3,5,7,11,13,17….
The lowest possible number which is not a prime,
not divisible by 2,3, and 5 and not a perfect square is 7*11=77
The only other possible number is 7*13=91
Hence, total such numbers = 21+2 = 23

Question 34: Abdul, Bimal, Charlie and Dilbar can finish a task in 10, 12, 15 and 18 days respectively. They can either choose to work or remain absent on a particular day. If 50 percent of the total work gets completed after 3 days, then, which of the following options is possible?

a) Each of them worked for exactly 2 days.

b) Bimal and Dilbar worked for 1 day each, Charlie worked for 2 days and Abdul worked for all 3 days.

c) Abdul and Charlie worked for 2 days each, Dilbar worked for 1 day and Bimal worked for all 3 days.

d) Abdul and Dilbar worked for 2 days each, Charlie worked for 1 day and Bimal worked for all 3 days.

e) Abdul and Charlie worked for 1 day each, Bimal worked for 2 days and Dilbar worked for all 3 days.

34) Answer (E)

Solution:

Let us assume the amount of work to be finished = LCM of {10, 12, 15, 18} = 180 units.

The amount of work which Abdul can complete in a day = $\dfrac{180}{10}$ = 18 units.

The amount of work which Bimal can complete in a day = $\dfrac{180}{12}$ = 15 units.

The amount of work which Charlie  can complete in a day = $\dfrac{180}{15}$ = 12 units.

The amount of work which Dilbar can complete in a day = $\dfrac{180}{18}$ = 10 units.

It is given that 50 percent of the total work gets completed after 3 days. Therefore, we can say that 90 units of work was completed in 3 days.

Let us check options.
Option A: Each of them worked for exactly 2 days.
In this case amount of work completed = 2*(10+15+12+18) = 110 units.

Option B: Bimal and Dilbar worked for 1 day each, Charlie worked for 2 days and Abdul worked for all 3 days.
In this case amount of work completed = 1*(10+15)+2*(12)+3*(18) = 103 units.

Option C: Abdul and Charlie worked for 2 days each, Dilbar worked for 1 day and Bimal worked for all 3 days.
In this case amount of work completed = 1*(10)+3*(15)+2*(18+12) = 115 units.

Option D: Abdul and Dilbar worked for 2 days each, Charlie worked for 1 day and Bimal worked for all 3 days.
In this case amount of work completed = 1*(12)+3*(15)+2*(18+10) = 113 units.

Option E: Abdul and Charlie worked for 1 day each, Bimal worked for 2 days and Dilbar worked for all 3 days.
In this case amount of work completed = 1*(18+12)+2*(15)+3*(10) = 90 units.

Therefore, we can say that option E is the correct answer.

Question 35: Car A starts from city P towards city Q. At the same time, another car B starts from the city Q towards city P. They will meet at point R if the ratio of the speed of car A and car B is 4:5. They meet at point S if the ratio of the speed of car A and car B is 3:4. The ratio of the distance between R and S and the distance between P and Q is m:n where m and n are co-primes. Find the value of m+n.

a) 62

b) 63

c) 64

d) 65

35) Answer (C)

Solution:

Assuming the distance between P and Q is 63 units (LCM of 4+5=9 and 3+4=7)
The fraction of the distance travelled by each car will be proportional to their speeds because the time taken is the same in each case.
Assume the speed of car A in the first case to be 4x and 5x
Hence, the distance of R from P = Distance travelled by A from P to R = Speed of A $\times$ Time taken to meet= $\dfrac{63}{4x+5x}\times\ 4x$ = 28 units
Similarly, let the speed of car A in the second case be 3x and 4x

Then, the distance of S from P = Distance travelled by A from P to S = Speed of A $\times$ Time taken to meet = $\dfrac{63}{3x+4x}\times\ 3x$ = 27 units

Hence, the distance between R and S = 1 unit
The ratio $\ \dfrac{\ m}{n}=\ \dfrac{\ 1}{63}$
Hence, the sum = 1+63 = 64

Question 36: For any two positive integer ‘a’ and ‘b’, what is the product of all possible values of ‘a’ for which 1/a + 1/b = 2/9 and a<b

a) 24

b) 30

c) 12

d) 48

36) Answer (B)

Solution:

1/a + 1/b = 2/9

Taking LCM, we get (a+b)/ab = 2/9

Cross multiplying, we get 9a+9b = 2ab

9a = 2ab-9b

9a = b(2a-9)

b = 9a/(2a-9). The only solutions for (a,b) are {(5,45) and (6,18)}

Question 37: Find the sum of digits of the largest five digit number that leaves remainders 4, 8, 10 and 14 when divided by 9, 13, 15 and 19 respectively.

a) 31

b) 27

c) 24

d) 26

37) Answer (A)

Solution:

The differences of the remainders with their respective divisors is 5 for all the four remainders.
Hence, the number is of the form LCM(9,13,15,19)-5
LCM(9,13,15,19) = 11115 => the number is of the form 11115k-5, where k is any positive integer.
The largest 5 digit number is 88915 => sum of digits = 31

Question 38: Series 1: 11, 2, -7, -16, -25,…… upto 1000 terms
Series 2: 25, 15, 5, -5, -15,…….. upto 1000 terms
What is the sum of all the terms common to both series?

a) -448000

b) -444000

c) -400400

d) -548000

38) Answer (A)

Solution:

The first common term of the 2 series is -25.

The first series has a common difference of -9, the second series has a common difference of -10.

Hence, common terms will be found at a common difference of -LCM(9,10) = -90.

Hence, the next common term is -115.

It continues in a similar way.

The last term of this new series must be present in both series.

Let us find out the last term of both series.

$t_{1000}$ = 11 – 999 x 9 = 11 – 8991 = -8980 for the first series.

$t_{1000}$ = 25 – 999 x 10 = 25 – 9990 = -9965 for the second series.

Hence, the last term of the new series must be greater than or equal to -8980.

-25 – 90 x (n-1) $\ge-8980$

-25 – 90n + 90 $\ge-8980$

65 – 90n $\ge-8980$

-90n $\ge-9045$

90n $\le9045$

n $\le100.5$

n = 100

100 terms in the new series.

Sum = $\frac{n}{2}\left[2a+\left(n-1\right)d\right]=\frac{100}{2}\left[2\times-25-99\times\ 90\right]$ = -448000

Question 39: A water tank has 3 taps attached to it. Tap 1 fills the entire tank alone in 8 hours and is at the bottom of the tank. Tap 2 alone empties the tank in 12 hours, but it is located at the middle of the tank and tap 3, which fills the entire tank alone in 4 hours is located at a distance of 25% of the height from the top. In how many hours will the tank be filled if all the three taps are open from the beginning?

a) 3 hours

b) $3\frac{\ 1}{21}\ $ hours

c) $3\frac{\ 1}{20}\ $ hours

d) $3\frac{\ 2}{21}\ $ hours

39) Answer (B)

Solution:

Tap 1 fills alone in 8 hours, tap 2 empties alone in 12 hours and tap 3 fills the entire tank alone in 4 hours.

Let there be 24 L of water(LCM of 8, 12 and 4) in the tank.

Tap 1 fills 3L in an hour, tap 2 empties 2L in an hour and tap 3 fills 6L in an hour.

Now irrespective of where taps 1 and 3 are located, they will begin filling the tank from the beginning but tap 2 will start working only when the tank is half filled.

So, let the time taken by tap 1 and 3 to fill the tank by 50% be ‘t’ hours.

(3+6)t= 12

=> t= $\frac{12}{9}=\frac{4}{3}\ $ hours.

Now, in 1 hour tap 1 and 3 will fill 9L of water and tap 2 will empty 2L in an hour. In total, all three operating together will fill 7L of water per hour.

.’. Time taken by all the three taps to fill the other half of the tank, ‘x’ can be found using 7x= 12L

.’.x = $\frac{12}{7}\ $ hours.

Total time to fill the tank= t+x= $\frac{12}{7}+\frac{4}{3}=\ \frac{\ 36+28}{21}=\frac{64}{21}=3\frac{\ 1}{21}\ $ hours

Question 40: Krishna and Shyam can finish work together in 18 days. After 6 days, Krishna left and Shyam finished the work alone. If the total time taken to finish the work in this way is 24 days, what is the ratio of work done by Krishna and Shyam?

a) 3:8

b) 4:9

c) 1:8

d) 2:9

40) Answer (C)

Solution:

It is given that Krishna and Shyam can finish work in 18 days. So after 6 days, one-third of work is finished.

Shyam finishes 2/3 of the work in 24-6=18 days.

Hence, Shyam can finish total work in 18*3/2 = 27 days

Assuming the total work to be 54 units.  (LCM(27,18)=54)

Work done by both Krishna and Shyam in 1 day = 54/18 = 3 units

Work done by Shyam in 1 day = 54/27 = 2 units

Hence, work done by Krishna = 3-2 = 1 units in 1 day

Then work done by Krishna in 6 days = 6 units

Work done by Shyam in 24 days = 24*2=48 units

The ratio = 6:48= 1:8

Question 41: A class which has x students had birthdays of three students – Seema, Raghav and Anand on a particular day. Seema bought 212 chocolates and distributed them equally among the other students and then had some chocolates left. Raghav bought 142 chocolates and distributed equally among the other students. He was also left with the same number of chocolates as Seema. Anand bought 352 chocolates did the same thing and had the same number of chocolates as the other two. What could be the maximum value of x ?

41) Answer: 71

Solution:

Let n be the number of chocolates left with each of them after distributing equally and c1, c2 and c3 be the number of chocolates distributed by Seema, Raghav and Anand respectively. Let y = x – 1 (This is because each person distributing the chocolates distributes them only among the other children. The distributor himself does not take any chocolate)
According to the question, c1y + n = 212 -(1)
c2y + n = 142 -(2)
and c3y + n = 352 – (3)
subtracting (2) from (1) and (3) and (1) from (2) we get,
(c1 – c2)y = 70, (c3 – c2 )y = 210 and (c3 – c1)y = 140
The maximum value of y will occur at the HCF of (70, 140, 210)
Thus, the maximum value of y is 70. Therefore, the total no. of students is y + 1 = 71

Question 42: Ram has a certain number of chocolates with him. If he divided these chocolates among 12 children, he would be left with 9 chocolates. If he divides the chocolates among 17 children then he would be left with 14 chocolates. Similarly, if he decides to divide the chocolates among 7 people, he will be left with 4 chocolates. What is the minimum possible number of chocolates with Ram?

a) 116

b) 1425

c) 1723

d) 201

42) Answer (B)

Solution:

Let the total number of chocolates with Ram be ‘n’.
Then ‘n’ can be expressed as
n = 12k + 9 = 17m + 14 = 7p + 4
We can see that in each of the above cases, the remainder is -3.
Hence, the minimum number of chocolates with Ram will be LCM (12, 7, 17) – 3
Thus, the required number will be 12*7*17 – 3 = 1428 – 3 = 1425.

Instructions

A group of students surveyed a clothing store that sells only shoes, shirts, and ties for their upcoming MBA project. They surveyed for a week and noted down the numbers of customers who bought these products. Upon completing their survey the following observations were made:

1. The number of male customers who bought all 3 items and the number of female customers who bought all 3 are prime numbers, the difference of which is also an even prime number

2. Overall 40 of the customers bought only shoes which are also equal to the number of male customers who bought only shirts.

3. The total number of customers who bought only ties is equal to the number of female customers who bought only shirts.

4. 67 females bought shoes which is 1 less than the number of males who bought shoes.

5. Overall 29 females bought shoes and ties. Out of this, the number of females who bought only shoes and ties but no shirts is a non zero multiple of 5 which is also equal to the number of male employees who bought only ties.

6. Number of males who bought ties and shirts is 3 times the number of males who bought all 3 items

7. Number of males who bought ties is equal to the number of males who bought shoes.

8. 30 females only bought shirts which is double the number of females who bought only shoes.

9. Overall 56 females bought Ties.

Question 43: The number of customers who bought shoes but not shirts is $x$. Which of the following is true about $x$?

a) $x$ is a 2-digit prime number.

b) $x$ is the square of a prime number.

c) $x$ is the square of a composite number.

d) $x$ is the LCM of 2 prime numbers.

43) Answer (D)

Solution:

Let us form a Venn diagram to solve the question. We see that there are male and female are 2 clear separate entity with no overlaps. Thus the diagram can look like as follows:

Now we should start adding the values which we determine.

> 40 male customers bought only the shirt

>67 females bought shoes which is less than the number of males who bought shoes.

>30 females only bought the shirt which is double the number of females who bought shoes. Thus the number of females buying only shoes = 15

> Overall 40 of the customers bought the only shoe. 15 are female thus 25 are male

We are given the following info:

>The number of male customers who bought all 3 items and the number of female customers who bought all 3 are prime numbers, the difference of which is also an even prime number

>Overall 29 females bought shoes and tie. Out of which the number of females who bought only the shoe and tie but not the shirt is a multiple of 5.

Let a number of females who bought all 3 items be p (where p is prime) and a number of females who bought the shoe and ties but not the shirt is k(where k is a natural number and multiple of 5) The possible cases where p+k=29 are (p,k) = (24,5),(19,10),(14,15),(9,20) and(4,25) Out of which only one case p is prime. Thus p = 19 and k =10

Thus k = 10 is als0 equal to the number of male employees who bought only ties.

Only even prime number =2, thus the number of males who bought all 3 = 17 or 21. Out of which 17 is the prime number

> Number of males who bought tie and shirt is 3 times the number of males who bought all 3 items. Thus Number of males who bought tie and shirt = $3 \times 17 =51$ Out of which 17 bought all three. The remaining 51-17 = 34 bought Shirt and tie but not the shoe.

> The total number of employees who bought only ties is equal to the number of female customers who bought only shirts which is equal to 30. 10 are male thus the rest has to be female = 20. Thus the number of females who bought shirt and tie but no shoe = 56-(10+19+20)= 7

> Number of males who bought ties is equal to the number of males who bought shoes = 68. Thus Number of males who bought tie and shoes but not the shirt = 68-(10+17+34) = 7

Number of females who bought shirt and shoe but tie=67-(15+10+19) = 23

>Number of males who bought only ties is equal to the number of males who bought only shoes. Number of males who bought only shoe and tie = 68-(10+17+34) = 7

Number of males who like only shoes and Shirt = 68-(25+17+7) = 19

Customer who bought shoe but not Shirt = 25+7+15+10 = 57

57 is not a perfect square or a prime number. However LCM(3,19) = 57. Option D is correct.

Question 44: Find the number of terms common to the two series:
$S_1 :$ 1, 5, 9, 13, 17…….100 terms
$S_2 :$ 4, 9, 14, 19…… 100 terms

a) 18

b) 21

c) 19

d) 20

44) Answer (D)

Solution:

100th term of both the sequences
Last term of Sequence 1 -> 1 + 99*4 = 397
Last term of Sequence 2 -> 4 + 99*5 = 499
The common terms of both the sequences also form an AP with the first term being 9 and common difference, which is equal to LCM of common differences of both the sequences which is 4*5 = 20.
So nth term of common terms is 9 + (n-1) * 20 $\leq$ 397
==> n $\leq$ 20.3
==> n = 20.

Question 45: David buys a certain number of pencils. He decides to sell the pencils in packets consisting of ‘n’ pencils, earning a profit of 20%. Had David packed ‘n-1’ pencils per packet and sold the packets at the same price, the profit he would have earned is Rs P. If P is 20% more than the amount of profit(Rs) made by selling n pencils per packet, what is the least number of pencils that David could have bought?

a) 930

b) 900

c) 960

d) 620

45) Answer (A)

Solution:

Let us assume that $T$ is the total cost incurred by David in buying the pencils.
Let the total number of pencils be $X$.

Now, the number of packets will be $\frac{X}{n}$.
Let the cost of a packet be $c$.

Total revenue = Number of packets*cost of each packet.
=> Total revenue = $\frac{X}{n}*c$

We know that David earns a profit of 20% by selling ‘n’ pencils per packet.

=> $\dfrac{\frac{X}{n}*c}{T} = 1.2$
=> $T = \dfrac{\frac{X}{n}*c}{1.2}$ ———-(1)

If David packs $n-1$ pencils per packet, he will earn $20$% more profit. Therefore, the new profit will be 1.2P = 1.2*20 =  $24$%.

=> $\dfrac{\frac{X}{n-1}*c}{T} = 1.24$
=> $T = \dfrac{\frac{X}{n-1}*c}{1.24}$ ——–(2)

Equating (1) and (2), we get,

$\dfrac{\frac{X}{n}*c}{1.2} = \dfrac{\frac{X}{n-1}*c}{1.24}$
$1.2*n = 1.24*(n-1)$
$0.04n = 1.24$
=> $n = 31$

We know that the pencils can be grouped into packets of 31 or packets of 30. Therefore, the total number of pens should be a multiple of 30 and 31. Also, since the least possible value of the number of pens is asked, the answer will be the LCM of 30 and 31.

LCM of 30 and 31 is 930 and hence, 930 is the right answer.

Question 46: Three friends P, Q and R started running on a circular track of length 128 m at the same time and from the same spot. P and R were running in the clockwise direction with speeds 4 m/sec and 8 m/sec respectively. Q was running in counter clockwise direction with speed 6 m/sec. What is the time taken by all of them to meet together for the first time?

a) 32 seconds

b) 96 seconds

c) 64 seconds

d) 192 seconds

46) Answer (C)

Solution:

Ratio of speeds of P and R = 4 : 8 = 1 : 2

Both P and R are running in same direction. Hence, number of meeting points = |2-1| = 1

They meet at one point only, i.e they meet at starting point

Therefore, three of them will meet at starting point for the first time

Time taken by P to complete one round $\frac{128}{4}$ = 32 seconds

Time taken by R to complete one round = $\frac{128}{8}$ = 16 seconds

Time taken by Q to complete one round = $\frac{128}{6}$ = $\frac{64}{3}$ seconds

Time taken by all of them to meet together for the first time = LCM(32, 16, $\frac{64}{3}$) = $\ \frac{\ LCM\left(32,\ 16,\ 64\right)}{HCF\left(1,1,3\right)}$ = $\ \frac{64}{1}$ = 64 seconds

Alternate Explanation:

Length of the circular track = 128m

Speed of P, Q and R is 4 m/sec, 6 m/sec and 8 m/sec

Relative speed of P and R = 8 – 4 = 4 m/sec

Time taken by P and R to meet, t1 = $\frac{128}{4} =32$seconds

Relative speed of R and Q = 8 + 6 = 14 m/sec

Time taken by Q and R to meet, t2 = $\frac{128}{14} $ seconds

Time taken by all the three friends to meet together for the first time = $ LCM(t1, t2) $= $LCM(\frac{128}{4}, \frac{128}{14}) $= $\frac{LCM(128, 128)}{HCF(4, 14)} $=$ \frac{128}{2}$ = 64 seconds

Hence, option C is the right choice.

Question 47: Kriti’s, Karishma’s, Kajol’s and Kareena’s number of movie roles is in the ratio $\frac{1}{2}$:$\frac{1}{9}$:$\frac{1}{7}$:$\frac{1}{23}$ and for each movie they are paid in the ratio $\frac{1}{11}$:$\frac{1}{13}$:$\frac{1}{15}$:$\frac{1}{17}$ If the earnings of all 4 actors is a natural number, what is the minimum amount of money Kajol would have earned? (Enter 0 if the answer cannot be determined)

47) Answer: 1006434

Solution:

The ratio of movie roles of Kriti, Karishma, Kajol and Kareena are given in fractions. To convert the ratio into non-fractions we must take the LCM of the denominators. This is 2*9*7*23=2898
Their movie roles will be in the ratio of $\frac{2898}{2}$:$\frac{2898}{9}$:$\frac{2898}{7}$:$\frac{2898}{23}$= 1449:322:414:126

(The number of movie roles for any heroine can never be negative.)

Similarly, for incomes, the LCM is 11*13*15*17=36465
Their incomes from each movie will be in the ratio of $\frac{36465}{11}$:$\frac{36465}{13}$:$\frac{36465}{15}$:$\frac{36465}{17}$= 3315:2805:2431:2145.
It is given in the question that the earnings are all a natural number, all the ratios calculated above are in their lowest form.
So, the least amount Kajol could have earned is 414*2431=1006434

Question 48: Five pipes A, B, C, D, and E can fill a tank in 60 minutes, 20 minutes, 30 minutes, 8 minutes, 32 minutes respectively. Out of these five pipes, two have now been converted into emptying pipes such that their efficiency remains the same. When only one of filling pipes and one of the emptying pipes operate, the tank can be filled in 60 minutes. Another combination of one filling and one emptying pipes empties the tank in $\frac{8}{45}^{th}$ of the time taken by the previous combination to fill the tank. Which pipes are converted into emptying pipes?

a) A and D

b) C and D

c) B and D

d) Cannot be determined

48) Answer (D)

Solution:

Let the total work to be done be 480 units (LCM of 60,20,30,8,32)
So the respective efficiencies of different pipes are-
A- 8 units/min (480/60)
B- 24 units/min
C- 16 units/min
D- 60 units/min
E- 15 units/min
Now, one combination of emptying and filling pipe can fill the tank in 60 minutes, i.e. together they can do 480/60=8 units of work per minute.
Out of the combinations possible, only A-C and B-C have a difference of 8 units/min.
So, if A is an emptying the tank and C is a filling it, they will do together 8 units of work per min. When C is an emptying the tank and B is filling it, the work done per minute will be the same 8 units/min.

According to the question, the second combination takes 8/45th of time of first one, i.e 32/3 minutes. Only combination that can do so is when D is the emptying pipe and E is the filling pipe. So among A, B, and C, we cannot know for sure which pipe is the filling one and which one is the emptying one.

Thus, we can’t determine which pipes have been converted to emptying pipes.

Question 49: A car race has four laps. Each lap is covered in different speeds by a particular car, 50 km/hr,60 km/hr,70km/hr and 80km/hr respectively. Find the approximate average speed(in km/hr) of the car.

a) 66

b) 65

c) 64

d) 63

49) Answer (D)

Solution:

Average speed is total distance/total time. Let us consider lap length as l.
Total distance = 4l
Total time = l/50 + l/60+l/70+l/80.
On taking LCM and dividing, we get average speed as 63.03 km/hr.

Question 50: A tank has two taps attached to it. Tap A fills the empty tank in 10 hours and tap B empties the full tank in 12 hours. At time t = 0, the tank is empty and both the taps are open. Tap A is closed at t = 30 min, reopened at t = 60 min, closed again at t = 90 min and so on. Tap B is closed at t = 45 min, reopened at t = 90 min, closed at t = 135 min and so on. Both the taps are closed as soon as the tank is full. If the tank is full at t = x minutes. Find x.

a) 115.5 hours

b) 120 hours

c) 118 hours 22.5 minutes

d) 118.5 hours

50) Answer (A)

Solution:

Tap A takes 10 hours to fill an empty tank and tap B takes 12 hours to drain a full tank. So, the efficiencies of A and B are in the ratio 12:10. So, A does 12 units of work in 1 hour and B does 10 units of work in 1 hour.
The duration of one cycle (open-close-open) for tap A is 60 minutes and the duration of one cycle (open-close-open) for tap B is 90 minutes.
LCM (60, 90) = 180

Consider the time from t = 0 to t = 180 min:

Tap A is open for 90 minutes. So, it does a work of 1.5*12 = 18 units.
Tap B is open for 90 minutes. So, it does a work of 1.5*10 = 15 units.

So, effective work done by both the taps in 180 minutes = 3 units.

Total work needed to be done = 10*12 = 120 units.

So, time required = 120/3 * 180 minutes = 40 * 3 hours = 120 hours.

But, the tank will be full before 120 hours.

Consider the first cycle: Work done in the first half an hour = 1 unit. In every other cycle, work done in the next 15 minutes is -2.5 units by tap B. But, in the first cycle, there is only 1 unit to drain out in the next 15 minutes. So, in effect, there is a gain of 1.5 units of work in the first cycle compared to the other cycles. That is, 4.5 units of work is done in the first cycle.

At the end of the 38th cycle, 115.5 units of work is done. Now, in the 39th cycle, an effective work of 4.5 units is done in the first 90 minutes. This is when 120 units of work is done and the tank is full. Both the taps are now closed.

So, total time required to fill the empty tank = (38*3 + 1.5) hours = 115.5 hours

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