Mock Test: Numerical Ability - 6 - AFCAT MCQ

►Efficiency of A = 20% = 100 / 5

►Efficiency of B = 10% = 100 / 10

►Efficiency of A,B and C = 50%

►∴ Efficiency of C = (Efficiency of A,B and C) – (Efficiency of A and B) = (50) - (20 + 10) = 20%

►∴ Number of days required by C to work alone = 100 / 20 = 5 days

►Alternatively: Go through options and satisfy the values.

4x – 2y = 5000 and 5x – 3y = 5000.
X = 2500, so income = 10000 and 12500.

If the distance between house and office is d km
Then time taken to go from home to office = d/3 hr (Time = Distance/Speed)
and Time taken to come back from office to home= d/2 hr
So total time taken = (d/3 +d/2) hour
But according to question , time taken = 5 hour
so we have
d/3+d/2=5
or (2d+3d)/6= 5
or 5d/6=5
or, d=5 x6/5= 6km
So distance between home and office = 6km

C.P. = (80000 + 5000 + 1000)
⇒ 86000
As we know that Profit = 25%
S.P. = 125% of 86000
⇒ (125/100 * 86000)
⇒ 107500 Rs.

► According to question, A:B:C:D = 6:5:10:14
► Now in D there are 70 chapters
Therefore, A:B:C:D = 30:25:50:70      [ Multiplying Each by 5 ]
L.C.M of 30, 25, 50, 70 is 1050
So each part should have 1050 pages.
So total pages = 4 x 1050 = 4200 ans.

Each term at an even place in the series is the product of its two adjacent terms.

Thus, if the missing term be x, then we have : 7 x x = 56 or x = 56 / 7 = 8.

Let the amount invested by Q = q.

Answer – C.36.45 Explanation : 50 * 45/50 * 45/50 *45/50 = 36.45L

Simple interest = P * r * t /100, where, P is the Principal, r is the rate of interest and t is the time period such that the rate of interest and the time period have mutually compatible units like r % per annum and t years.

Here, in the first case, let the Principal be Rs.P and rate of interest be r % per annum. Time period is given as 8 years . Interest is given as 80 P / 100
So, 80 P / 100 = P * r * 8 / 100
Or, 80 = 8 r
So, r = 10 % per annum.

Now, in the second case, Principal is given as Rs. 14,000, rate of compound interest is 10 % per annum as determined above; and t is 3 years.
Compound interest = P * ( 1 + r/100)t - P 
= 14,000 * ( 1+ 10 /100)3 - 14,000
= 14,000 * ( 1 + 0.1)3 - 14,000
= 14,000 *(1.1)3 - 14,000
(14,000 * 1.331) - 14,000
= 18,634 - 14,000 = 4,634
So, the compound interest on Rs.14,000 at the same rate of interest as in case 1 in 3 years will be Rs. 4,634.

At 2 O’ clock Aflatoon has already covered 16 km @ 8 km/h, Bablajoon starts running in the same direction @ 10 km/h. The relative speed is 2 km/h. 

Therefore to be 5km apart bablajoon has to cover 11 km. relative speed=2km/hr, therefore for 11km it will take 5hr 30 min more.  Therefore they will meet at 7:30 the same day. And similarly 12.30 a.m. on the next day.

LetL= 100, B = 100 A= 10000
New length = 120, new breadth =150 New Area = 18000

Let the present age of father be x and mother be y

∴ Ram age = (x – 30)

And Ram’s sister = (y – 25)

We know that

Average = Sum of all/total number of terms

⇒ 26.25 = Sum of all/4

Sum of their ages will be = 105

∴ x + y + (x - 30) + (y – 25) = 105

⇒ x + y = 80     ----(i)

After four years their total ages will be (excluding daughter age)

∴ (x + 4) + (y + 4) + (x – 30) + 4 = 107

⇒ 2x + y = 125     ----(ii)

By solving equation (i) and (ii) we get

⇒ x = 45 and y = 35

∴ The age of Ram’s father will be 45 years 

This is a simple alternating addition and subtraction series.

In the first pattern, 3 is added. In the second, 2 is subtracted.

► 7 + 3 = 10,
► 10 - 2 = 8,
► 8 + 3 = 11,
► 11 - 2 = 9,
► 9 + 3 = 12,
► 12 - 2 = 10.