Ask & Answer: India's Largest Education Community

Yes, the GATE GOAPS 2026 portal has been created. The GOAPS portal is goaps.iitg.ac.in. The portal was activated on August 28 for online GATE registrations 2026. 

$\frac{T1speed}{T2speed}=\sqrt[]{\frac{TimetakenbyT2toreachChennai}{TimetakenbyT1toreachYeshvantpur}}$

$or\frac{45}{T2speed}=\sqrt[]{\frac{200}{242}}=\frac{10}{11}$

T2 speed = $\frac{99}{2}$ $\frac{}{}$ =49.5 km/hr.

As per the problem, we have CP of 18? SP of 18 = CP of 3

CP of 15 = SP of 18

But CP of 15 toffees = Rs. 90

So SP of 18 toffees = Rs. 90

SP per toffee $=\frac{90}{18}=Rs.\text{?}5$

Let daily wages of A, B and C be 4d, 3d and 2d respectively.

Total wages = (8 * 4d) + (10 * 3d) + (15 * 2d) = 10764 or d = 117.

Now daily wages of each = Rs. 936, 1170, 1755

Let 100 units be B’s income and X units be B’s expenditure A’s income = 60 units

A’s expenditure =   $\frac{50X}{100}unitsBut60=\left(\frac{80}{100}\right)\times X\Rightarrow X=75$ $$

B’s saving = (100 – 75) units = 25 units

Hence, A’s saving = 

60 − $\left(\frac{50}{100}\right)\times 75units=22.5units$ $$

A’s saving : B’s saving = 22.5 : 25 = 9 : 10

f CP = 100, MP = (100 + 44) * $\frac{100}{90}$ $\frac{}{}$ = 160

If MP = 23200, CP = 23200 * $\frac{100}{160}$ $\frac{}{}$ = 14500

Speed become $\frac{2}{3}$ $\frac{}{}$

So, time also become $\frac{2}{3}*60$ $\frac{}{}$

= 40 minutes

Hence, bus will stop for 20 minutes per hour.

Y alone take 25 × 4 = 100 days

Total work LCM of (25&100) = 100

Y alone = $\frac{}{}\frac{}{}$ $\frac{100}{\left(4-1\right)}=\frac{100}{3}$

At 2:40 a.m., the hour hand is between 2 and 3 while the minute hand is at 8. Therefore, angle between the two hands at 2:40 a.m. = $\frac{}{}$ $\left(180-40\times \frac{1}{2}\right)$  = 160°.

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 = $\frac{40}{100}\times 100$ = 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

Given quadratic equation x2 + ax + b = 0

Then product of roots αβ = b

Sum of roots α + β = − a

Next quadratic equation bx2 + ax + 1 = 0

Then product of roots = $\frac{1}{b}=\frac{1}{\alpha \beta }$

Hence, clearly by visualising options

New roots we will be $\frac{1}{\alpha }$ and $\frac{1}{\beta }$

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),

= $\frac{P/R}{Q/R}=\frac{1/2}{4/5}$

= $\frac{P}{Q}=\frac{5}{8}$