Q.2. What is the greatest number which divides 639, 1065 and 1491 exactly?
1. 193
2. 183
3. 223
4. 213
5. 233
Sol : Option 4
H.C.F. of 639 and 1065 is 213. H.C.F. of 213 and 1491 is 213.
Q.3. What is the H.C.F. of 4/9, 10/21 and 20/63?
1. 4/189
2. 6/63
3. 2/63
4. 20/21
5. None of these
Sol : Option 3
H.C.F of 4/9, 10/21 and 20/63 = H.C.F of 4,10 and 20 / L.C.M of 9,21 and 63
= H.C.F of 4, 10 and 20 = 2 & L.C.M. of 9, 21 and 63 = 63. Required H.C.F. = 2/63
Q.4. The H.C.F. of two numbers is 12 and their difference is 12. Which of the following can be the numbers?
1. 66, 77
2. 70, 84
3. 94, 108
4. 84, 96
5. 66, 106
Sol : Option 4
The difference of requisite numbers must be 12 and each should be divisible by 12. Checking the options given, only the fourth option satisfies.
Q.5. The HCF of two numbers is 29 & their sum is 174. The possible numbers are
1. 1,174
2. 74,100
3. 29, 154
4. 29, 145
5. None of these
Sol : Option 4
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.
Q.6. 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.
1. 56 cm
2. 42 cm
3. 38 cm
4. 34 cm
5. 48 cm
Sol : Option 4
The side of the square slab is the H.C.F. of 544 and 374 cm i.e. 34.
Q.7. The product of two numbers is 6760 and their H.C.F. is 13. How many such pairs can be formed?
1. 0
2. 3
3. 4
4. 1
5. 2
Sol : Option 5
Let the numbers be 13x and 13y. So, 13X ×13 Y = 6760 X×Y=40.
Possible values of (x, y) are (1, 40); (2, 20); (4, 10); (5, 8) Only two acceptable values are (1, 40) and (5, 8).
Q8. least perfect cube which is divisible by 2, 3, 4 and 6 is
Sol : Option 2
L.C.M. of 2, 3, 4, 6 = 12 ; 12 = 22 × 3. So, the required number =22 × 3 x 2 x 32 = 216.
Q9. The least perfect square number which is divisible by 3, 4, 5, 6 and 8 is
1. 900
2. 1200
3. 2500
4. 3600
5. 1600
Sol : Option 4
L.C.M. of 3, 4, 5, 6, 8 = 120 ; Required number =2 x 2 x 2 x 2 x 3 x 3 x 5 x 5 = 3600.
Q10. 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.
1. 28 cm
2. 14 cm
3. 42 cm
4. 63 cm
5. 11 cm
Sol : Option 5
Maximum possible length of such rod = (H.C.F. of 44, 22, 55)c m = 11cm.