- 4210
- 4200
- 4205
- 4195

**Correct Answers Is: 4**

**Concept **

1) Find the L.C.M of all the given divisors.

2) subtract the difference between the divisor and the remainder from the L.C.M of all the divisors.

**Calculation**

L.C.M of 15, 25, 35 and 40 is 4200 (i.e. 5 × 3 × 5 × 7 × 8)

Difference of divisor and remainder = 15 – 10

⇒ 5

25 – 20

⇒ 5

35 – 30

⇒ 5

40 – 35

⇒ 5

Difference is 5 in each case

so, subtract 5 from 4200

4200 – 5

⇒ 4195

**∴ The least number which when divided by 15, 25, 35, 40 leaves remainders 10, 20, 30, 35, respectively is 4195.**

__Shortcut Trick:__

In such type of question,

Where Numbers a, b, c and Remainders x, y, z

**Use format = LCM × k – D**,

Where D = a – x = b -y = c – z

So, LCM of (15, 25, 35, 40) = 4200

K = 1

D = 5

Least number = 4200 – 5 = 4195

- 221, 247
- 1700, 1900
- 209, 187
- 190, 170

**Correct Answers Is: 3**

**SOLUTION**

**Given:**

Ratio of two numbers = 19 : 17

HCF = 11

**Concept:**

HCF = Highest Common Multiple

**Calculation:**

Let the numbers be 19h and 17h respectively, where h is the HCF.

⇒ 19h = 19 × 11 = 209

⇒ 17h = 17 × 11 = 187

**∴ The numbers are 209 and 187.**

- 68, 104
- 24, 148
- 44, 128
- 64, 108

**Correct Answers Is: 4**

**SOLUTION**

**Given:**

HCF of two numbers = 4

Sum of the two numbers = 172

Difference of the two numbers = 44

**Calculation:**

Since HCF is 4, hence the numbers can be written as 4x and 4y; 4x ≥ 4y

According to the question:

4x + 4y = 172 —-(i)

4x – 4y = 44 —-(ii)

Adding equation (i) and equation (ii), we get

8x = 216

⇒ x = 27;

⇒ 4x = 108

⇒ 4y = 172 – 108 = 64

**∴ The numbers are 108 and 64.**

^{3}b – ab

^{3}, a

^{3}b

^{2}– a

^{2}b

^{3}, ab(a – b)?

- a
^{2}b^{2}(a^{2}+ b^{2}) - a
^{2}b^{2}(a^{2}– b^{2}) - a
^{2}b^{3}(a^{2}+ b^{2}) - a
^{3}b^{2}(a^{2}– b^{2})

**Correct Answers Is: 2**

**SOLUTION**

**Given:**

a3b – ab3, a3b2 – a2b3, ab(a – b)

**Concept Used:**

(a + b)(a – b) = (a2 – b2)

**Calculation:**

⇒ Let, x = a3b – ab^{3} = ab(a^{2 }– b^{2}) = ab(a – b)(a + b)

⇒ y = a3b2 – a2b^{3} = a^{2}b^{2}(a – b)

⇒ z = ab(a – b)

⇒ LCM = a2b2(a + b)(a – b) = a2b^{2}(a2 – b2)

**Therefore, the LCM of a3b – ab3, a3b2 – a2b3, ab(a – b) is a2b2(a2 – b2).**

*2/5*- 1/15
- 9/5
- 12/5

**Correct Answers Is: 4**

- 11
- 9
- 10
- 12

**Correct Answers Is: 3**

**SOLUTION**

**Given:**

There are 24 lambs, 36 goats and 60 sheep

Arranged: In a way, every row contains the same number of animals of only one type.

**Calculation:**

We know the minimum number of rows we arranged the maximum number of animals in each row.

So, we have to calculate the HCF of 24, 36, 60

So, the HCF of 24, 36, and 60 is 12.

∴ Lambs row = 24/12 = 2

Goats row = 36/12 = 3

Sheeps row = 60/12 = 5

∴ **Total number of rows = 5 + 3 + 2 = 10**

- After 3000 days
- After 2000 days
- After 1500 days
- After 1200 days

**Correct Answers Is: 1**

**SOLUTION**

**Given:**

Three planets revolve round the sun once in 200, 250, 300 days respectively.

**Concept used:**

**LCM: **The smallest multiple that two or more numbers have in common.

**Calculation:**

According to the question

Three planets revolve around the sun once in 200, 250, and 300

days respectively.

So, required time = LCM of (200, 250, 300)

⇒ (2 × 2 × 5 × 5 × 2 × 5 × 3) = 3000

Hence, after 3000 days they all come relatively to the same position at a certain point in time in their orbits.

**∴ The required answer is after 3000 days.**

- 2:05 pm
- 1:50 pm
- 2:10 pm
- 1:55 pm

**Correct Answers Is: 2**

**SOLUTION**

**Given:**

The four traffic lights of a crossroad change after the 30s, 45s, 1 minute, and 75s respectively.

These four lights changed simultaneously at 1:35 pm

**Concept used:**

Required time interval to change the four lights simultaneously = LCM of each light changing time

**Formula used:**

1 minute = 60s

**Calculation:**

The four traffic lights of a crossroad change after the 30s, 45s, 1 minute, and 75s respectively ⇒ 30s, 45s, 60s, and 75s respectively

Let’s calculate LCM of 30, 45, 60, and 75

Prime factors of 30 = 2 × 3 × 5

Prime factors of 45 = 3 × 3 × 5

Prime factors of 60 = 2 × 2 × 3 × 5

Prime factors of 75 = 3 × 5 × 5

∴ LCM of 30, 45, 60, and 75 = 2 × 2 × 3 × 3 × 5 × 5 = 900

900 seconds = 900/60 = 15 minutes

Then the four traffic lights of a crossroad change after 15 minutes.

The four traffic lights will be changed at 1:35 pm + 15 minutes = 1:50 pm

**∴ The four traffic lights of a crossroad will change simultaneously at 1:50 pm**

- 60
- 64
- 84
- 108

**Correct Answers Is: 3**

**SOLUTION**

**Given:**

LCM (m, n, s) = 1386

m : n : s = 3 : 7 : 11

Where, m, n, and s are three numbers

**Concept used:**

LCM = Lowest Common Factor

It is the product of the greatest power of each prime factor, involved in the numbers.

**Calculation**:

m = 3x

n = 7x

s = 11x

According to the question,

LCM (3x, 7x, 11x) = 1386

231x = 1386

x = = 6

Least number = 3x = 18

Greatest number = 11x = 66

Sum of least and greatest number = 18 + 66 = 84

**Hence, the correct answer is option 3).**

- 264
- 384
- 408
- 120

**Correct Answers Is: 2**

**SOLUTION**

**Given:**

Two numbers are in the ratio 5 ∶ 11.

HCF is 24.

**Formula used:**

HCF × LCM = Product of two numbers

**Calculations:**

Let the two numbers be 5a and 11a.

⇒ HCF = 24

⇒ LCM = 55a

Then, 24 × 55a = 55a^{2}

⇒ a = 24

Two numbers = 5 × 24 = 120

⇒ 24 × 11 = 264

∴ Required sum = 120 + 264 = 384

**∴ The answer is 384.**

- 167
- 168
- 176
- 186

**Correct Answers Is: 2**

**SOLUTION**

**Given:**

The six bells toll at intervals of 3, 4, 6, 7, 8, and 12 seconds

**Calculation:**

Prime factors of 3 = 3

Prime factors of 4 = 2 × 2

Prime factors of 6 = 2 × 3

Prime factors of 7 = 7

Prime factors of 8 = 2 × 2 × 2

Prime factors of 12 = 2 × 2 × 3

LCM of 3, 4, 6, 7, 8, and 12 = 2 × 2 × 2 × 3 × 7 = 168

**∴ The six bells will toll together again after 168 seconds**

^{2}× 2

^{3}× 7

^{3}and Y = 5 × 2

^{5}× 7, then find the HCF of X and Y.

- 5 × 2
^{3}× 7 - 23 × 7
- 5 × 2
^{3} - 5 × 7

**Correct Answers Is: 1**

**SOLUTION**

**Given:**

H.C.F of 5^{2} × 2^{3 }× 7^{3} and 5 × 2^{5} × 7

**Concept:**

H.C.F = product of common prime factors having least powers.

**Calculation:**

Let, the common factors of 5^{2} × 2^{3} × 7^{3} and 5 × 2^{5} × 7 be:

= 5 × 2^{3} × 7

**∴ The answer is 5 × 2 ^{3}× 7.**

- 8
- 10
- 12
- 9

**Correct Answers Is: 4**

**SOLUTION**

A, B, C, and D birds complete one round in 27 minutes, 30 minutes, 45 minutes, and 60 minutes, respectively.

Concept used:

LCM is the smallest common multiple of two or more numbers.

Calculation:

LCM (27, 30, 45, 60) = 540

Now, they will meet again at the starting position after 540 minutes.

⇒ 540 minutes = 540/60 = 9 hours

**∴ They will meet again at the starting position after 9 hours.**

- 3510
- 1350
- 1530
- 3150

**Correct Answers Is: 4**

**SOLUTION**

**Given:**

The number of trees in each row is the same,

There are 105 or 210 or 315 or 525 rows.

**Calculation:**

The number of trees in each row for 105 or 210 or 315 or 525 rows = L.C.M of (105, 210, 315, 525)

Prime factorisation of 105 = 3 × 5 × 7

Prime factorisation of 210 = 2 × 3 × 5 × 7

Prime factorisation of 315 = 3 × 3 × 5 × 7

Prime factorisation of 525 = 3 × 5 × 5 × 7

L.C.M of (105, 210, 315, 525) = 2 × 3 × 3 × 5 × 5 × 7 = 3150

**∴ The required number of trees is 3150.**

- 129
- 119
- 111
- 88

**Correct Answers Is: 1**

**SOLUTION**

**Given:**

The product of the two numbers is 5547 and the H.C.F is 43.

**Calculation:**

Let the two numbers are 43x and 43y

According to the question,

⇒ 43x × 43y = 5547

⇒ xy = 3

Now, co-prime with product 3 is (3,1)

⇒ The required number is (3 × 43, 1 × 43) = 129, 43

**∴ The greatest number is 129.**

- 299
- 676
- 426
- 364

**Correct Answers Is: 4**

**SOLUTION**

**Given:**

HCF of two numbers is 26, and the other two factors of their LCM are 13 and 14.

**Concept used:**

Multiplication of numbers = HCF of the numbers × LCM of the numbers

If HCF of numbers is **h** then numbers can be** hz** and **hy**.

**Calculation:**

Let, Larger number = a , smaller number = b

∵ HCF = 26, LCM = 26 × 13 × 14

∴ Larger number a = 26y , smaller number b = 26z .

Multiplication of numbers = HCF of the numbers × LCM of the numbers

⇒ 26y × 26z = 26 × 26 × 13 × 14

⇒ yz = 13 × 14

⇒ y = 14 or 182 , z = 13 or 1 (∵ y > z)

⇒ So, larger Number = a = 26y

⇒ larger Number = 26 × 14 = **364 Answer.**

- 49
- 44
- 54
- 39

**Correct Answers Is: 2**

**SOLUTION**

**Given:**

The three numbers are 988, 1637, and 2345.

**Concept Used:**

HCF of big numbers is equal to the HCF of their differences.

**Calculation:**

1637 – 988 = 649

2345 – 1637 = 708

2345 – 988 = 1357

Now the HCF of 649, 708, and 1357 is,

⇒ 649 = 59 × 11

⇒ 708 = 59 × 12

⇒ 1357 = 59 × 23

The HCF of above three numbers is 59.

If we divide 988, 1637, and 2345 by HCF we will get same reminder.

⇒ 988/59, then the reminder is 44.

⇒ 1637/59, then the reminder is 44.

⇒ 2345/59, then the reminder is 44.

**∴ The common reminder in each case is 44.**

18. Three cylindrical containers containing 871 litres, 737 litres and 1072 litres of ethanol respectively. Find the maximum capacity of a container that can measure the ethanol of the three containers exact number of time.

- 68
- 77
- 67
- 101

**Correct Answers Is: 3**

**SOLUTION**

**Given**

Capacities of containers are : 871, 737 and 1072

**Formula used**

HCF of the following numbers will give our required answer

**Calculation**

⇒ 871 = 67 × 13

⇒ 737 = 67 × 11

⇒ 1072 = 67 × 16

⇒HCF of 871, 737 and 1072 is 67

∴ 67 litre is the maximum required capacity.

- 104
- 124
- 114
- 106

**Correct Answers Is: 1**

**SOLUTION**

**Given:**

The ratio of four positive numbers is 5 : 8 : 12 : 18

LCM of the numbers is 2880

**Calculation:**

Let the numbers be 5x, 8x, 12x and 18x

LCM of 5x, 8x, 12x and 18x = 360x

According to the question,

360x = 2880

⇒ x = 2880/360 = 8

Difference between the greatest and the smallest number

⇒ (18 × 8) – (5 × 8)

⇒ 8 (18 – 5) = 104

**∴ The difference between the greater and the small number is 104.**

20. Find the greatest possible length of a scale that can be used to measure exactly, three pieces of cloth of the following lengths:

3 m, 5 m 10 cm, 12 m 90 cm

- 30 cm
- 60 cm
- 45 cm
- 35 cm

**option 1.**

**Given:** Length to be measured are 3 m, 5 m 10 cm, 12 m 90 cm

**Concept Used**: Taking HCF of these length

1 m = 100 cm

**Calculation:**

→3 m = 300 cm

→5 m 10 cm = 510 cm

→12 m 90 cm = 1290 cm

Now, HCF of (300 cm, 510 cm, 1290 cm)

Factors of these numbers

300 = 2^{2} x 3^{1} x 5^{2}

510 = 2 x 3 x 5 x 17

1290 = 2 x 3 x 5 x 43

So, HCF of these numbers = 2 x 3 x 5= 30 cm

**The greatest possible length of a scale can be used is 30 cm.**

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