model 3 find hcf of numbers Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (c)

Using Rule 7,

(The largest number which when divide the numbers a, b and c give remainders as p, q, r respectively is given by H.C.F. of (a – p), (b – q) and (c – r).)

We have to find HCF of

(1657 – 6 = 1651) and (2037 – 5 = 2032)

1651 = 13 × 127

2032 = 16 × 127

∴ HCF = 127

So, required number will be 127.

Answer: (a)

Length of the floor

= 15 m 17 cm = 1517 cm

Breadth of the floor

= 9m 2 cm = 902 cm.

Area of the floor

= 1517 × 902 $cm^2$

The number of square tiles will be least, when the size of each tile is maximum.

∴ Size of each tile = HCF of 1517 and 902 = 41

∴ Required number of tiles

=${1517× 902}/{41× 41}$ = 814

Answer: (a)

Using Rule 7,

(The largest number which when divide the numbers a, b and c give remainders as p, q, r respectively is given by H.C.F. of (a – p), (b – q) and (c – r). )

Required number

= HCF of 2300 – 32

= 2268 and 3500 – 56 = 3444

∴ HCF = 84

Answer: (b)

Maximum quantity in each can

= HCF of 21, 42 and 63 litres

= 21 litres

Required least number of cans

= $21/21 + 42/ 21 + 63/21$

= 1 + 2 + 3 = 6

Answer: (d)

Minimum number of rows = Maximum number of fruits in each row

∴ HCF of 24, 36 and 60 = 12

∴ Minimum number of rows

= $24/12 + 36/12 + 60/12$

= 2 + 3 + 5 = 10

Answer: (c)

Using Rule 7,

The greatest number N = HCF of (1305 – x ), (4665 – x ) and (6905 – x),

where x is the remainder

= HCF of (4665 – 1305), (6905– 4665) and (6905 – 1305)

= HCF of 3360, 2240 and 5600

∴ N = 1120

Sum of digits = 1 + 1 + 2 + 0 = 4