Quantitative Aptitude Test - 4

Let us assume that Arun started running at 10 AM and Barun started at 12 noon. So, in these two hours distance traveled by arun is 60 km and the relative speed of barun w.r.t arun is 10 km/hr. so barun will overtake Arun after = 60/10 = 6 hours
So, Barun reaches there at 6 P.M
So, Kiranmala also overtakes t time to overtake Arun and the relative speed of Kiranmala w.r.t Arun is 30 km/hr and Arun ran for 8 hrs. So distance traveled by arun is while kiranmala distance traveled is t = 4 hours
So After 4 hrs kiranmala will start running

Initial rate developing question per week
= 2500/10 = 250 questions.
After six weeks total developed questions
= 250 x 6 = 1500 questions
Remaining questions = 2500 - 1500 = 1000
So, instead of developing 250 questions, the team has to develop 1000 question in the 7th weak.
Required increase in the rate = (1000 - 250) x 100/250 = 300%

Two digit multiples of 8 are 16,24,32,40,48,56,64,72,80,88 and 96. Out of these only two numbers - 40 and 48 satisfy the given condition. so the answer is 2.

a⁶ + b⁶ = (a² + b²)(a⁴ + b⁴ - a²b²)
Since a⁶ + b² is divisible by a² + b², it can be prime only in two cases:
(i) a² + b² is 1: This is not possible since both a and b are integers.
(ii) a⁶ + b⁶ = a² + b²: if a and b are distinct integers and a⁶ + b⁶ is prime, then this is possible only if
--> a = 1, b = -1
--> a = -1, b = 1
The sum of a and b is 0 in both the cases.

The product of two numbers is the product of the LCM and HCF of the two numbers. If the numbers are x and y, then x + y = 140, xy = 28 × 168.
Solving for x and y, the numbers are 84 and 56.
Alternative method:
The two numbers are 28X and 28Y, where X and Y are coprimes.
Also 28(X + Y) = 140
So X + Y = 5
Since LCM 28 × 6 = 168, XY = 6
So X = 3, Y = 2

Given:

\({AD}, {BE}=\) Medians

\({AD}=12 {~cm}\)

\({BE}=9 {~cm}\)

\({AD}\) perpendicular on \(BE.\)

We know that,

The Medians intersect each at a point called Centroid. The centroid divides the median in the ratio \(2 ∶ 1.\) The line joining the centroid and the three vertices of the triangle divide the original triangle into \(3\) smaller triangles of equal area.

Area of triangle \(=(\frac12) \times\) Base \(\times\) Height

\(\because\) Centroid divides the median in the ratio \(2: 1\);

\(\therefore {AO}=(\frac23) \times {AD}\)

\(\Rightarrow {AO}=(\frac23) \times 12\)

\(\Rightarrow {AO}=8 {~cm}\)

Similarly;

\({BO}=(\frac23) \times {BE}\)

\(\Rightarrow {BO}=6 {~cm}\)

\(\therefore\) Area of triangle \({AOB}=(\frac12) \times {BO} \times {AO}\)

\(=(\frac12) \times 8 {~cm} \times 6 {~cm}\)

\(=24 {~cm}^{2}\)

\(\because\) Area of triangle \({ABC}=3 \times(\) Area of triangle \({AOB})\)

\(=3 \times 24 {~cm}^{2}\)

\(=72 {~cm}^{2}\)

\(\therefore\) Area of triangle \({ABC}=72 {~cm}^{2}\)

Machine I:
Number of nuts produced in one minute = 100
To produce 1000 nuts time required = 10 min
Clearing time for nuts = 5 min
Over all time to produce 1000 nuts =15min
Over all time to produce 9000 = 138 min - 5min
= 133 min ....(1)
Machine II:
To produce 75 bolts time required = 1 min
To produce 1500 bolts time required = 20 min
Cleaning time for bolts = 10 min.
Effective time to produce 1500 bolts = 30 min
Effective time to produce 9000 bolts = 30 x 6 - 10 = 170 min .........(2)
From (1) and (2)
Minimum time = 170 minutes

Given:

\(5 a 7150 b\) is divisible by \(88.\)

We know that,

Divisibility rule of \(8\): A number is divisible by \(8\) if the last three digits of the number are divisible by \(8\).

Divisibility rule of \(11\): A number is divisible by \(11\) if the difference of the sum of digits at even places and the sum of digits at odd places is either \(0\) or a multiple of \(11.\)

If a number is divisible by \(88\) then it will be divisible by \(8\) and \(11.\)

Now,

Checking the divisibility of \(8\),

Last three digits of the number \(=50 {~b}\)

\(50 {~b}\) is divisible by \(8\) if \({b}=4\)

Sum of the digits at the odd places \(={b}+5+7+5\)

\(=4+5+7+5\)

\(=21\)

Checking the divisibility of \(11\),

Sum of the digits at the even places \(=0+1+a\)

\(=a+1\)

Difference of even and odd places digits sum \(=21-(a+1)\)

\(=20-a\)

\((20-a)\) will give a multiple of \(11\) if \((20-a)=11\)

\(a=20-11\)

\(\Rightarrow a=9\)

Putting the values of \(a\) and \(b\),

\(2 a-3 b=2 \times 9-3 \times 4\)

\(=18-12\)

\(= 6\)

\(\therefore\) The value of \(2 {a}-3 {b}\) is \(6.\)