Solutions Grade VI Maths

Math Problems Solutions

Q8. We need to find the greatest common divisor (GCD) of 24, 36, and 60. \\ The GCD of 24, 36, and 60 is 12. So, the minimum number of rows needed is $\frac{24}{12} + \frac{36}{12} + \frac{60}{12} = 10$.

Q9. The dimensions of the room in centimeters are 250 cm x 150 cm. The GCD of 250 and 150 is 50 cm. So, the side of the largest possible square slab is 50 cm. The number of slabs is $\frac{37500 cm^{2}}{2500 cm^{2}} = 15$ slabs.

Q10. The GCD of 680, 850, and 1020 is 170. So, the greatest number that will divide 690, 875, and 1050 leaving remainders 10, 25, and 30 respectively is 170.

Q11. The least common multiple (LCM) of 15, 20, 36, and 48 is 720. So, the least number which when divided separately by 15, 20, 36 and 48 leaves 3 as remainder is 720 + 3 = 723.

Q12. The LCM of 12, 15, 18, and 27 is 540. So, the largest 4-digit number exactly divisible by 12, 15, 18, and 27 is 9720.

Q13. The LCM of 8, 18, and 24 is 72. So, the smallest 4-digit number divisible by 8, 18, and 24 is 1008.

Q14. The least common multiple (LCM) of the lengths into which they cut their ribbons is 3360 cm, or 33.6 meters. So, the shortest possible length of ribbon they could have been given is 33.6 meters.

Q15. The least common multiple (LCM) of their firing intervals is 840 s, or 14 minutes. So, they will all hit their targets simultaneously again at 9:14 a.m.