Q.1. A room is 4 mts 37 cm long and 3 mts 23 cm broad. The floor needs to be paved with a minimum number of square slabs. Find the least number of slabs required for this purpose.

1. 485

2. 431

3. 391

4. 381

5. 415

Sol : Option 3
Length = 437 cm, Breadth = 323 cm. The side of the square slab is the H.C.F. of 437 and 323, i.e. 19 cm. Area of square slab = 19 cm x 19 cm = 361 cm2
The number of slabs
= Area of the room / Area of the slab
= 437 x 323cm2 / 361cm2 = 391

Q.2. Find the greatest number which will divide 3959, 4082 and 4164 leaving the same remainder in each case.

1. 30

2. 45

3. 41

4. 43

5. 47

Sol : Option 3
4082 - 3959 = 123, 4164 - 4082 = 82, 4164 - 3959 = 205. H.C.F. of 123, 82 and 205 is 41.

Q.3. Find the greatest number which will divide 1063 and 1403 leaving the remainder 9 in each case.

1. 48

2. 34

3. 40

4. 42

5. 36

Sol : Option 2
Subtract 4 from each of the numbers 1063 and 1403 and then take the H.C.F, i.e. H.C.F of 1054 and 1403 is 34.

Q.4. A merchant has three different types of milk- 324 litres, 351 litres and 459 litres. Find the minimum number of casks of equal size that can store the milk without mixing.

1. 17

2. 61

3. 42

4. 45

5. 51

Sol : Option 3
Since a minimum number of casks are required, the size of the cask is greatest. Also the cask in three cases is of equal size. The size of the cask is the H.C.F. of 324 litres, 351 litres and 459 litres which is 27. Now, the number of casks required for storing the milk = (324 + 351+ 459)÷ 27 = 42.

Q.5. Find the greatest number which will divide 21, 69 and 93 so as to leave the same remainder in each case.

1. 12

2. 18

3. 24

4. 32

5. 30

Sol : Option 3
69 - 21 = 48, 93 - 69 = 24, 93 - 21 = 72. H.C.F. of 48, 24 and 72 is 24.

Q.6. Find the greatest number that will divide 989, 1258 and 1359 leaving remainders 66, 51 and 10, respectively.

1. 71

2. 81

3. 68

4. 73

5. 77

Sol : Option 1
989 - 66 = 923, 1258 - 51 = 1207, 1359 - 10 = 1349
H.C.F. of 923 and 1207 is 71, H.C.F. of 71 and 1349 is 71.

Q.7. . The traffic lights at three different road crossings change after every 24 sec., 36 sec. and 72 sec. respectively. If they all change simultaneously at 8 : 20 : 00 hrs; then they will again change simultaneously at:

1. 8 : 21 : 12 hrs

2. 8 : 28 : 22 hrs

3. 8 : 27 : 36 hrs

4. 8 : 27 : 48 hrs

5. None of these

Sol : Option 1
Interval of change = (L.C.M. of 24, 36, 72) sec. = 72 sec. So, the lights will change after every 72 seconds, i.e. 1 min. 12 sec.
So, the next simultaneous change will take place at 8 : 21 : 12 hrs.

Q8. Find the least number which when divided by 6, 12, 16 and 125 leaves 7 as remainder in each case but when divided by 11 leaves no remainder.

1. 8007

2. 13007

3. 11007

4. 18017

5. 18007

Sol : Option 5
L.C.M of 6, 12, 16 and 125 is 6000. Required number = 6000 × K + 7 = (11 × 545 + 5)K + 7 = (7 × 545) K + 5K + 7. Now, (5K + 7) is divisible by 11 for K = 3. Required number = 6000 × 3 + 7 = 18007

Q9. A wholesale coffee dealer has 24 kilograms, 120 kilograms and 72 kilograms of three different qualities of coffee. He wants it all to be packed into various boxes of equal size without mixing. Find the capacity of the largest possible box.

1. 50

2. 36

3. 24

4. 12

5. 45

Sol : Option 3
The capacity of the box is H.C.F. of 24, 120 and 72, i.e. 24.

Q10. Find the least number of five digits which when divided by 4, 25, 12 and51 leaves remainder 8 in each case.

1. 10088

2. 10072

3. 10208

4. 10096

5. None of these

Sol : Option 3
The LCM of 4, 25, 12, 51 is 5100. So, the least number of 5 digits divisible by the L.C.M. will be 10200. But since the remainder is 8, so the number has to be 10200 + 8 = 10208.