Given expression: $[∛(x^4 y)×\frac{1}{∜(x^3 y^8 )}]^{-6}$
We start by converting the roots to fractional exponents: $[(x^4 y)^{1/3} × (x^3 y^8 )^{-1/4}]^{-6}$
Then we distribute the exponents to each term inside the brackets: $[x^{4/3} y^{1/3} × x^{-3/4} y^{-2}]^{-6}$
We simplify by adding the exponents of like bases: $[x^{4/3 - 3/4} y^{1/3 - 2}]^{-6} = [x^{7/12} y^{-5/3}]^{-6}$
Finally, we distribute the -6 exponent to get the simplified form: $x^{-7/2} y^{10}$
Q8. There are 24 peaches, 36 apricots and 60 bananas and they have to be arranged in several rows in such a way that every row contains the same number of fruits of only one type. What is the minimum number of rows required for this to happen?
Q9. Find the side of the largest possible square slabs which can be paved on a floor of a room 2 m 50 cm long and 1 m 50 cm broad. Also, find the number of such slabs to pave the floor.
Q10. Find the greatest number that will divide 690,875 and 1050 leaving remainders 10,25 and 30 respectively.
Q11. Find the least number which when divided separately by 15,20,36 and 48 leaves 3 as remainder in each case.
Q12. Find the largest 4- digit number exactly divisible by each of the numbers 12,15,18 and 27.
Q13. Find the smallest 4-digit number which is divisible by 8,18 and 24.
Q14. Jiya, Reena and Armaan were each given a piece of Ribbon of equal length. Jiya cuts her ribbon into equal lengths 30 cm, Reena cuts her ribbon into equal lengths of 48 cm and Armaan cuts his ribbons into equal lengths of 56 cm. If there was no remainder in all these cases, find the shortest possible length of ribbon given to them?
Q15. In a firing range, 4 shooters are firing at their respective targets. The first, the second, the third and the fourth shooter hit the target once in every 5 s,6 s,7 s and 8 s respectively. If all of them hit their target at 9 a.m., when will they hit their target together again?
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.
