Circuitects Electronic Solutions Interview Questions, Process, and Tips

A’s speed=6 mph ,B’s speed=8 mph
Let, after x hrs, they will meet.
so, the distance traveled by A in x hrs should be the same as the distance traveled by B in (x-1/2)hrs [as B started the journey after 30 min of A]
Thus, 6x=8(x-1/2)[as distance=speed*time]
=>8x-6x=4
=>2x=4
=>x=2
after 2 hrs they will meet so time=(9+2)=11.00 a.m

(b) 16.66%

Earlier for ₹x we could purchase y gm of sugar.
Now we pay ₹1.2x for y gm of sugar

(As there was an increase in price so, x + 20%x = 1.2x)

At current rates for ₹x you can purchase y/1.2 gm of sugar

So the reduced consumption is y-(y/1.2)

Percentage change = (reduced consumption/ original consumption ) *100

That is (0.2/1.2) *100 = 16.66% (approx)

Answer is 45
First we need to subtract those reminders from the respective numbers, then we have to find the hcf of two numbers(numbers got from the subtraction) then you will get the answer.
So,
After subtraction you will get
3026-11 = 3015
5053-13 = 5040
HCF of these two numbers
5 | 3015 5040
3 | 603 1008
3 | 201 336
| 67 112
We can’t find a common diviser since 67 is a prime number
So the HCF = 5 * 3 * 3
= 45
5*3*3 = 45

2:3:4

24 days

if A is true B has to be true

Let the distance between each pole be x m. Then, distance up to 12th pole = 11x m

∴ Speed = 11x22m/s

∴ Time to cover total distance up to 20th pole

= 19x×2411x

= 41.45 s

31

B. The Secretary, Ministry of Finance is the right answer

In 3500 g rod contains 74% of silver = 3500 * 74/100 = 2590 g
Then 3500 g + 500 g of rod contains 84% of silver
Let x be the silver contained in 500 g of silver
(2590/3500 * 100) + (x/500 * 100) = 84
74 + x/5 = 84
(370 + x) /5 = 84
370 + x = 420
x = 50
Then the percentage of silver contained in 500 g of rod = 50/500 *100 =10%

1%=50
12%=600

value remains same