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A Learning Portal from Recruitment India

If the product of two numbers is 84942 and their H.C.F. is 33, find their L.C.M.

A.

2500

B.

2574

C.

1574

D.

1365

Answer with explanation

Answer: Option BExplanation

Explanation:

HCF * LCM = 84942, because we know

Product of two numbers = Product of HCF and LCM

LCM = 84942/33 = 2574

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What will be the least number which when doubled will be exactly divisible by 12, 18, 21 and 30 ?

A.

196

B.

1260

C.

630

D.

2520

Answer with explanation

Answer: Option CExplanation

12 = 2 × 2 × 3

18 = 2 × 3 × 3

21 = 3 × 7

30 = 2 × 3 × 5

L.C.M. of 12, 18, 21 30 = 2 × 3 × 2 × 3 × 7 × 5 = 1260

Required number = 1260/2

= 630

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The H.C.F and L.C.M of two numbers are 50 and 250 respectively. If the first number is divided by 2, the qotient is 50. The second number is

A.

50

B.

100

C.

125

D.

75

Answer with explanation

Answer: Option CExplanation

as we know if two numbers say a and b are multiplied the product is equal to the product of their HCF and LCM

so A.T.Q

a/2=50

a=100

therefore

hcf(100,b)*lcm(100,b)=100b

50*250=100b

12500/100=b

125=b

the second no. is 125

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The h.c.f. of two numbers is 23 and the other two factors of their l.c.m. are 13 and 14. the larger of the two numbers is

A.

276

B.

299

C.

322

D.

345

Answer with explanation

Answer: Option CExplanation

HCF of two numbers is 23.So, Let the numbers be 23x & 23y.

The LCM => 23xy.

It is given that 13 & 14 are it other factors.

So, The value of x & y will be 13 & 14.

The largest number => 23 (14)

=> 322.

Workspace

Find the greatest number of 4 digits which is divisible by 15, 25, 40 and 75

A.

9600

B.

9400

C.

9800

D.

9000

Answer with explanation

Answer: Option AExplanation

Greatest number of 4-digits is 9999.

L.C.M. of 15, 25, 40 and 75 is 600.

On dividing 9999 by 600, the remainder is 399.

∴ Required number (9999 – 399) = 9600.

Workspace

Find the least number which when divided by 12, 15, 20 and 54 leaves 8 as remainder in each case.

A.

582

B.

548

C.

584

D.

854

Answer with explanation

Answer: Option BExplanation

LCM of 12, 15, 20, 54

12 = 2 x 2 x 3

15 = 3 x 5

20 = 2 x 2 x 5

54 = 2 x 3 x 3 x 3

LCM = 27 x 4 x 5 = 27 x 20

= 540

ADD 8

= 540 + 8 = 548

Workspace

The greatest number of four digits which is divisible by 15, 25, 40 and 75 is

A.

9600

B.

9400

C.

9800

D.

9000

Answer with explanation

Answer: Option AExplanation

Solution:

Greatest number of 4-digits is 9999.

L.C.M. of 15, 25, 40 and 75 is 600.

On dividing 9999 by 600, the remainder is 399.

∴ Required number (9999 – 399) = 9600.

Workspace

There are four prime numbers written in ascending order of magnitude. The product of the first three is 7429 and of the last three is 12673. Find the fourth number?

A.

17

B.

23

C.

31

D.

29

Answer with explanation

Answer: Option DExplanation

(7429/12673) = 17/29 Fourth Number is 29

Let the four prime numbers, in the ascending order of magnitude are a, b, c and d.

Given that the product of the first three is 7429 and of the last three is 12673

So a×b×c = 7429

b×c×d = 12673

Divide them:

Thus the first number is 17

The fourth number is 29

The middle two numbers are 19 and 23.

So your answer is D. 29

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John, Smith and Kate start at same time, same point and in same direction to run around a circular ground. John completes a round in 250 seconds, Smith in 300 seconds and Kate in 150 seconds. Find after what time will they meet again at the starting point?

A.

30 min

B.

25 min

C.

20 min

D.

15 min

Answer with explanation

Answer: Option BExplanation

L.C.M. of 250, 300 and 150 = 1500 sec Dividing 1500 by 60 we get 25, which mean 25 minutes. John, Smith and Kate meet after

25 minutes.

Workspace

The traffic lights at three different road crossings change after every 40 sec, 72 sec and 108 sec respectively. If they all change simultaneously at 5 : 20 : 00 hours, then find the time at which they will change simultaneously.

A.

5 : 28 : 00 hrs

B.

5 : 30 : 00 hrs

C.

5 : 38 : 00 hrs

D.

5 : 40 : 00 hrs

Answer with explanation

Answer: Option CExplanation

Traffic lights at three different road crossings change after every 40 sec, 72 sec and 108 sec respectively.

Therefore, find the L.C.M. of 40, 72 and 108.

L.C.M. of 40, 72 and 108 = 1080

The traffic lights will change again after 1080 seconds = 18 min

The next simultaneous change takes place at 5 : 38 : 00 hrs.

Workspace

A man was employed on the promise that he will be paid the highest wages per day. The contract money to be paid was Rs. 1189. Finally he was paid only Rs. 1073. For how many days did he actually work?

A.

39

B.

40

C.

37

D.

35

Answer with explanation

Answer: Option CExplanation

HCF of 1189, 1073 = 29

1073/29 = 37

Let man s every day s wages was X

He had to work for Y days

Total wages = XY =1189

But he worked for Z days

So he got XZ =1073

X is common in both so it HCF

HCF of 1189 & 1073 is X= 29

Total Days HE worked = 1073/29

=37

Answer – 37

Workspace

On dividing a number by 4, the remainder is 2. The quotient so obtained when divided by 5, leaves the remainder 3. Now the quotient so obtained when divided by 6, leaves the remainder 5. The last quotient is 7. The number was

A.

962

B.

954

C.

946

D.

938

Answer with explanation

Answer: Option BExplanation

WKT,

Divisor x Quotient + Remainder = Number

6 x 7 + 5 = 47

5 x 47 + 3 = 238

4 x 238 + 2 = 954

Therefore, required number is 954.

Workspace

Find the side of the largest square slab which can be paved on the floor of a room 5 meters 44cm long and 3 meters 74 cm broad?

A.

56 cm

B.

42 cm

C.

38 cm

D.

34 cm

Answer with explanation

Answer: Option DExplanation

The side of the square slab is the H.C.F. of 544 and 374 cm i.e. 34.

Workspace

Three different containers contain 1365 litres, 1560 litres and 1755 litres of mixtures of milk and water respectively. What biggest measure that can measure all the different quantities exactly ?

A.

97 lit

B.

129 lit

C.

195 lit

D.

37 lit

Answer with explanation

Answer: Option CExplanation

containers contain mixtures of milk and water as 1365 lit, 1560 lit and 1755 lit.

Biggest measure that can measure all these different quantities exactly can be given by

HCF of (1365, 1560, 1755)

That can be found as

HCF of 1365, 1755 = 195

Now, HCF of 195 and 1560 = 195

Hence, biggest measure that can measure all the given different quantities exactly is 195 lit

Workspace

Given : Three numbers 17, 42 and 93

Find the largest number to divide all the three numbers leaving the remainders 4, 3, and 15 respectively at the end?

A.

13

B.

17

C.

21

D.

78

Answer with explanation

Answer: Option AExplanation

Here greatest number that can divide means the HCF

Remainders are different so simply subtract remainders from numbers

17 – 4 = 13; 42 – 3 = 39; 93 – 15 = 78

Now let’s find HCF of 13, 39 and 78

By direct observation we can see that all numbers are divisible by 13.

∴ HCF = 13 = required greatest number

Workspace

A milkman has 75 litres milk in one cane and 45 litres in another. The maximum capacity of container which can measure milk of either container exact number ?

A.

25 litres

B.

15 litres

C.

5 litres

D.

1 litre

Answer with explanation

Answer: Option BExplanation

75 litres, 45 litres

For maximum capacity take HCF (75, 45) = 15

Workspace

The product of two 2 digit numbers is 2028 and their HCF is 13. What are the numbers?

A.

26, 78

B.

39, 52

C.

13, 156

D.

36, 68

Answer with explanation

Answer: Option BExplanation

Let the numbers be 13x and 13y (∵ HCF of the numbers = 13)

13x×13y=2028⇒xy=12 co-primes with product 12 are (1, 12) and (3, 4) (∵ we need to take only co-primes with product 12. If we take two numbers with product 12,

but not co-prime, the HCF will not remain as 13)

Hence the numbers with HCF 13 and product 2028

= (13 × 1, 13 × 12) and (13 × 3, 13 × 4)

= (13, 156) and (39, 52)

Given that the numbers are 2 digit numbers. Hence numbers are 39 and 52.

Workspace

Find the least number which when divided by 2, 3, 4 and 5 leaves a remainder 3. But when divided by 9 leaves no remainder?

A.

B.

63

C.

183

D.

153

Answer with explanation

Answer: Option BExplanation

LCM of 2, 3, 4 and 5 is 30,

let number be 30k + 3 put k = 2,

we get 63 which is divisible by 9

Workspace

The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is:

A.

216

B.

389

C.

443

D.

548

Answer with explanation

Answer: Option DExplanation

Required number = (L.C.M. of 12, 15, 20, 54) + 8

= 540 + 8

= 548.

**(OR)**

Given numbers are 12, 15, 20, 54

To Find: least no. which is divided by given nos. and leaves remainder 8

Least no which is divisible by all given no is LCM of all nos.

LCM means the least common multiple.

First, we find LCM of 12, 15, 20, 54 by prime factorization method

12 = 2 × 2 × 3

15 = 3 × 5

20 = 2 × 2 × 5

54 = 2 × 3 × 3 × 3

LCM ( 12, 15 , 20 , 54 ) = 2 × 2 × 3 × 3 × 3 × 5 = 540

To find the required no. we add 8 to LCM

⇒ Required No. = 540 + 8 = 548

Thus, 548 is the least no which when divided by 12 15 20 and 54 leaves in each case a remainder of 8.

Workspace

The LCM of two numbers is 520 and their HCF is 4. If one of the number is 52, then the other number is

A.

42

B.

32

C.

50

D.

40

Answer with explanation

Answer: Option DExplanation

First number * second number = LCM * HCF

52*Second number = 520 * 4

Second number = (520*4)/52

= 40

Then the other number is 40.

Workspace

What is the number nearest to 10000 which is exactly divisible by 3, 4, 5, 6, 7, and 8?

A.

9956

B.

10080

C.

10096

D.

9240

Answer with explanation

Answer: Option BExplanation

Take LCM of 3, 4, 5, 6, 7, and 8. It will be = 840.

Now, 10000/840 = 760 (remainder).

840 – 760 = 80. If we add 80 in the given number, the number is exactly divisible by 840.

So, the required number is 10080.

Workspace

If LCM of two number is 693, HCF of two numbers is 11 and one number is 99, then find other

A.

34

B.

45

C.

77

D.

12

Answer with explanation

Answer: Option CExplanation

For any this type of question, remember

Product of two numbers = Product of their HCF and LCM

So Other number = 693×11/99 = 77

Workspace

The least number which when divided by 5, 6 , 7 and 8 leaves a remainder 3, 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 840*k + 3*

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

Required number = (840 x 2 + 3) = 1683.

Workspace

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