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Three men start together to travel the same way around a circular track of 11 kilometers in circumference. Their speeds are 4, 5 and 8 kilometers per hour respectively. When will they meet at a starting point?

A.

11 hours

B.

12 hours

C.

220 hours

D.

22 hours

Answer with explanation

Answer: Option AExplanation

Time for one round = 11/4, 11/5, 11/8

LCM of 11/4, 11/5, 11/8 = 11/1 = 11

Workspace

What is the smallest number, which, when divided by 12, 15, 18, and 27, leaves remainder of 8, 11, 14 and 23, respectively?

A.

781

B.

536

C.

581

D.

None

Answer with explanation

Answer: Option BExplanation

Prime factorizations first.

12 = 2 x 2 x 3

15 = 3 x 5

18 = 2 x 3 x 3

27 = 3 x 3 x 3

LCM = 2 x 2 x 3 x 3 x 3 x 5 = 540

Since all of the remainders are 4 less than the divisors, the number will be 540 – 4 = 536

Workspace

A merchant has three different types of milk: 435 liters, 493 liters and 551 liters. Find the least number of casks of equal size required to store all the milk without mixing.

A.

51

B.

61

C.

47

D.

45

Answer with explanation

Answer: Option AExplanation

HCF of 435, 493, 551 = 29

(453/29) + (493/29) + (551/29) = 51

Workspace

The sum of two non co–prime numbers added to their HCF gives us 91. How many such pairs are possible?

A.

2

B.

4

C.

3

D.

6

Answer with explanation

Answer: Option CExplanation

Let HCF of the numbers be h. The numbers can be taken as ha + hb, where a, b are coprime.

h + ha + hb = 91

h(1 + a + b) = 91

h ≠ 1

h = 7

=> 1 + a + b = 13 a + b = 12

h = 13

=> 1 + a + b = 7

=> a + b = 6

**Case 1:** h = 7, a + b = 12

(1, 11), (5, 7) => Only 2 pairs are possible as a, b have to be coprime.

**Case 2:** h = 13, a + b = 6

(1, 5) only one pair is possible as a, b have to be coprime.

Overall, 3 pairs of numbers are possible – (7, 77) (35, 49) and (13, 65)

The question is **“How many such pairs are possible?”**

Workspace

Sum of two numbers x, y = 1050. What is the maximum value of the HCF between x and y?

A.

350

B.

700

C.

1050

D.

525

Answer with explanation

Answer: Option DExplanation

If the question states x, y have to be distinct, then the best solution would be x = 350, y = 700, HCF = 350.

So the HCF is 525.

The question is **“What is the maximum value of the HCF between x and y?”**

Workspace

Find the smallest number that leaves a remainder of 4 on division by 5, 5 on division by 6, 6 on division by 7, 7 on division by 8 and 8 on division by 9?

A.

2519

B.

5039

C.

1079

D.

979

Answer with explanation

Answer: Option AExplanation

Note: When a number is divided by 8, a remainder of 7 can be thought of as a remainder of -1. This idea is very useful in a bunch of questions. So, N = 5a – 1 or N + 1 = 5a

N = 6b – 1 or N + 1 = 6b

N = 7c – 1 or N + 1 = 7c

N = 8d – 1 or N + 1 = 8d

N = 9e – 1 or N + 1 = 9e

N + 1 can be expressed as a multiple of (5, 6, 7, 8, 9)

N + 1 = 5a*6b*7c*8d*9e

Or N = (5a*6b*7c*8d*9e) – 1

Smallest value of N will be when we find the smallest common multiple of (5, 6, 7, 8, 9)

or LCM of (5, 6, 7, 8, 9)

N = LCM (5, 6, 7, 8, 9) – 1 = 2520 – 1 = 2519.

Workspace

The HCF of two numbers is 29 & their sum is 174. The possible numbers are 1. 1,174

A.

1,174

B.

74,100

C.

29, 154

D.

29, 145

Answer with explanation

Answer: Option DExplanation

Let the numbers be 29a and 29b. Then, 29a + 29b= 174 or 29(a + b) = 174 or, a + b = 174/29 = 6. Values of co-primes (with sum 6) is(1, 5).

So, the possible pairs of numbers is (29 x 1, 29 x 5) i.e. 29 and 145, which will become the answer.

Workspace

3 different pieces of iron are of varying length are given to a student which are 44cm, 22 cm,55 cm respectively.He has to form rods of maximum length such that no iron waste is left.Find the maximum length of such rod.

A.

14 cm

B.

42 cm

C.

63 cm

D.

11 cm

Answer with explanation

Answer: Option DExplanation

Maximum possible length of such rod = (H.C.F. of 44, 22, 55)c m = 11cm.

Workspace

The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then sum of the number is:

A.

28

B.

32

C.

40

D.

64

Answer with explanation

Answer: Option CExplanation

Let the numbers be 2*x* and 3*x*.

Then, their L.C.M. = 6*x*.

So, 6*x* = 48 or *x* = 8.

The numbers are 16 and 24.

Hence, required sum = (16 + 24) = 40.

Workspace

Three numbers which are co-prime to each other are such that the product of the first two is 551 and that of the last two is 1073. The sum of the three numbers is:

A.

75

B.

81

C.

85

D.

89

Answer with explanation

Answer: Option CExplanation

Since the numbers are co-prime, they contain only 1 as the common factor.

Also, the given two products have the middle number in common.

So, middle number = H.C.F. of 551 and 1073 = 29;

First number = | 551 | = 19; Third number = | 1073 | = 37. | ||||

29 | 29 |

Required sum = (19 + 29 + 37) = 85.

Workspace

A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and c in 198 seconds, all starting at the same point. After what time will they again at the starting point ?

A.

26 minutes and 18 seconds

B.

42 minutes and 36 seconds

C.

45 minutes

D.

46 minutes and 12 seconds

Answer with explanation

Answer: Option DExplanation

L.C.M. of 252, 308 and 198 = 2772.

So, A, B and C will again meet at the starting point in 2772 sec. *i.e.,* 46 min. 12 sec.

Workspace

The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is

A.

74

B.

94

C.

184

D.

364

Answer with explanation

Answer: Option DExplanation

L.C.M. of 6, 9, 15 and 18 is 90.

Let required number be 90*k* + 4, which is multiple of 7.

Least value of *k* for which (90*k* + 4) is divisible by 7 is *k* = 4.

Required number = (90 x 4) + 4 = 364.

Workspace

Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together ?

A.

4

B.

10

C.

15

D.

16

Answer with explanation

Answer: Option DExplanation

L.C.M. of 2, 4, 6, 8, 10, 12 is 120.

So, the bells will toll together after every 120 seconds(2 minutes).

In 30 minutes, they will toll together | 30 | + 1 = 16 times. |

2 |

Workspace

A rectangular courtyard 4.55 meters long and 5.25 meters wide is paved exactly with square tiles of same size. Find the largest size of the tile used for this purpose?

A.

25 cm

B.

45 cm

C.

21 cm

D.

35 cm

Answer with explanation

Answer: Option DExplanation

Here, we are asked to find the largest size of tile. Therefore, calculate H.C.F.

**Step 1: **Covert numbers without decimal places i.e 455 cm and 525 cm

**Step 2: **Find the H.C.F. of 455 and 525

H.C.F. of 455 and 525 = 35 cm

Hence, the largest size of the tile is 35 cm

Workspace

H.C.F. of two numbers is 13. If these two numbers are in the ratio of 15: 11, then find the numbers.

A.

230, 140

B.

215, 130

C.

195, 143

D.

155, 115

Answer with explanation

Answer: Option CExplanation

H.C.F. of two numbers = 13

2) The numbers are in the ratio of 15 : 11

Let the two numbers be 15y and 11y

H.C.F. is the product of common factors

Therefore, H.C.F. is y. So y = 13

The two numbers are:

15y = 15 × 13 = 195

11y = 11 × 13 = 143

**We can cross-check the answer using the trick. **(Product of two numbers = Product of their H.C.F. and L.C.M.)

Product of H.C.F. and L.C.M. = 13 × 2145 = 27885

Product of two numbers = 195 × 143 = 27885

Workspace

Find the largest number of 4-digits divisible by 12, 15 and 18.

A.

9900

B.

9750

C.

9450

D.

9000

Answer with explanation

Answer: Option AExplanation

Largest 4-digit number is 9999.

Remember: The question may be asked in a tricky way. Here, largest number does not mean H.C.F.. We have to find a number which is divisible by 12, 15 and 18

Required largest number must be divisible by the L.C.M. of 12, 15 and 18

L.C.M. of 12, 15 and 18

12 = 2 × 2 × 3

15 =5 × 3

18 = 2 × 3 × 3

L.C.M. = 180

Now divide 9999 by 180, we get remainder as 99

The required largest number = (9999 – 99) =9900

**Number 9900 is exactly divisible by 180.**

Workspace

The least number, which when divided by 5, 6, 7 and 8 leaves a remainder 3 in each case, but when divided by 9 leaves no remainder, is:

A.

1677

B.

1683

C.

2523

D.

3363

Answer with explanation

Answer: Option BExplanation

L.C.M. of 5,6,7,8 = 840.

Required number is of the form 840k + 3

Least value of k for which (840k + 3) is divisible by 9 is k = 2.

Required number = (840 X 2 + 3) = 1683

Workspace

If the sum of two numbers is 55 and the H.C.F. and L.C.M. of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to?

A.

16/41

B.

14/57

C.

13/125

D.

11/120

Answer with explanation

Answer: Option DExplanation

Let the numbers be a and b.

We know that product of two numbers = Product of their HCF and LCM

Then, a + b = 55 and ab = 5 x 120 = 600.

=> The required sum = (1/a) + (1/b) = (a+b)/ab

=55/600 = 11/120

Workspace

The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:

A.

127

B.

236

C.

123

D.

305

Answer with explanation

Answer: Option AExplanation

Required number = H.C.F. of (1657 – 6) and (2037 – 5)

= H.C.F. of 1651 and 2032 = 127.

Workspace

There are three numbers, these are co-prime to each other are such that the product of the first two is 551 and that of the last two is 1073. What will be the sum of three numbers :

A.

87

B.

85

C.

83

D.

81

Answer with explanation

Answer: Option BExplanation

As given the questions these numbers are co-primes, so there is only 1 as their common factor.

It is also given that two products have the middle number in common.

So, middle number = H.C.F. of 551 and 1073 = 29;

So first number is : 551/29 = 19

Third number = 1073/29 = 37

So sum of these numbers is = (19 + 29 + 37) = 85

Workspace

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