Math MCQs

From 3 to 4 pm, A fills 1/3. From 4 to 5 pm, A+B add 1/3+1/4=7/12, so the tank contains 11/12 at 5 pm. With all three open, net rate=1/3+1/4−1=−5/12 per hour. Emptying 11/12 takes (11/12)/(5/12)=11/5=2.2 hours, so the cistern empties at 7:12 pm.

Let the original price and consumption both be represented by convenient units. After a 20% price rise, the price factor is 1.20. To keep expenditure constant, consumption must become 1/1.20 = 5/6 of the original, a reduction of 1/6 = 16 2/3%.

Each numerator term is the square of ten times the corresponding denominator-base term, so the numerator sum is 100 times the denominator sum. The ratio under the radical is therefore 100, its square root is 10, and multiplying by 1/2 gives 5.

If d divides both a+b and a-b, then d divides their sum 2a and difference 2b. Since a and b are coprime, d can have no odd prime factor common to both; therefore d divides 2. Hence the HCF is either 1 or 2, depending on the parity of a and b.

Dividing 19999 by 21111 gives approximately 0.9473. Rounded to three decimal places, the value is 0.947. Because the numerator is slightly smaller than the denominator, the result must be below 1, which also helps reject implausible choices.

Align the decimal points and add: 0.007+0.777=0.784; 17.838+310.022=327.860; total = 328.644. This cleaned version preserves the decimal-addition result intended by the source while avoiding ambiguous recurring-decimal notation in the scan.

Let the fixed charge be F and the additional rate be r per kilometre beyond 5 km. Then F + 5r = 350 and F + 20r = 800. Subtracting gives 15r = 450, so r = ₹30. Then F = 350 - 150 = ₹200. For 30 km, the charge is F + 25r = 200 + 750 = ₹950.

Factor each expression. The first contains 36x^2(x+2)(3x-1); the second is 18x(x+1)(3x-1); and the third is 54x(3x-1)(9x^2+3x+1). The numerical HCF is 18, while the common algebraic factors are x and (3x-1). Therefore the HCF is 18x(3x-1).

After n hours, the first car covers 10n km. The second covers an arithmetic-series distance: n/2[2×8+(n−1)×0.5]. Equating the two distances gives n=9 as the positive non-zero solution, so the second car overtakes the first after 9 hours.

The remaining one-third is half as much work as the completed two-thirds. The completed part used 120×64 man-days, so the remaining part needs 3840 man-days. With 60 days left, 64 men are sufficient. Thus 120−64=56 men can be discharged.