Maths

et the present ages be X, Y, and Z years (X – 5): (Y – 5): (Z – 5) = 2 : 4 : 5 and taking pairwise values. Substitute these values in the equation

2a + 4a + 5a = x + y + z – 15

11a = 165

a = 15

Five years ago ages were 30, 60, 75

Present ages are 35, 65, 80.

Let the number of men be 100.

Then, Men * Time = Work

100 * 1 = 100 unit

Amount of work increased by 50%.

So, New work = 150 unit

as the planned time remains same i.e. 1

Then men required will be 150 i.e., 50 more workers but since new workers are 25% more efficient i.e. $\frac{5}{4}$ times efficient as existing workers.

$\therefore$ Actual number of workers = $\frac{50}{5/4}$ = 40 men

$\therefore$ Required percent = $\frac{40}{100}*100$ = 40%

81! when divided by 83 leaves a remainder +1 (as 83 is prime) Hence the total remainder is +1 * 81 = 81.

Average salary of each temporary employee is 600.

1000 temporary employee,

Total salary = 6,00,000

let teaching departement = Rs. x / staff and cleaning department salary = x + 100

Now, 600 (x + 100) + 400 (x)  = 6,00,000

x=540

Hence answer, x+100= 540+100=640

We have (A – 20) = 0.4 (B + 20), i.e. A – 0.4B = 28

And (B – 40) = 0.4 (A + 40), i.e. B – 0.4A = 56

Solving we get, A = Rs. 60

Let speeds of P, Q & R be P, Q & R km/hr respectively.

Thus 4P = 2R

= $\frac{P}{R}=\frac{1}{2}$ . (1)

= 5Q = 4R = $\frac{Q}{R}=\frac{4}{5}$ . (2)

From (1) & (2),

= $\frac{P/R}{Q/R}=\frac{1/2}{4/5}$

= $\frac{P}{Q}=\frac{5}{8}$

Since M is the midpoint of side PQ, the length of MQ is 2.

Hence, the area of? MQR = $\frac{1}{2}$ * 2 * 4 = 4.

Also area of? NSR = 4. Thus, the unshaded area of the figure = 4 + 4 = 8.

Hence, the area of quadrilateral PMRN

= Area of the square PQRS – The unshaded area of the figure

= 16 – 8 = 8

Let n = the number of terms.

Then,  $25\frac{7}{16}=\frac{n}{2}\left[17+\left(-12\frac{3}{8}\right)\right]=\frac{n}{2}*4\frac{5}{8}$

$\Rightarrow \frac{407}{16}=37\frac{n}{16}$ $\Rightarrow n=\frac{407}{37}=11$

Let d be the common difference.

Then $-12\frac{3}{8}$  (= the eleventh term) = 17 + 10d

$\Rightarrow 10d=-12\frac{3}{8}-17=-\frac{235}{8}$

$\Rightarrow d=\frac{-235}{8*10}=-\frac{47}{16}$

Let oil in containers be A & B.

After 1st operation

Container A = 0.4 A

Container B = 0.6 A + B

After 2nd operation

Container A = 0.4 A + 0.3 A + 0.5 B

Container B = 0.3 A + 0.5 B

= $\frac{(0.7A+B)}{11}=\frac{(0.3A+B)}{7}$

1.6A = 2B $\Rightarrow$ $\frac{A}{5}=\frac{B}{4}$

Volume of A : B = 5 : 4.