Quiz 1A

SP of trader A = 1000 (1 + x).

Profit of trader A = 1000 (1 + x) – 1000.

MP of trader B = 2000 (1 + 2x).

SP of trader B = 2000 (1 + 2x) (1 – x).

 

Profit of trader B = 2000(1 + 2x) (1 – x) – 2000.

Both make the same profit => 1000(1 + x) – 1000 = 2000(1 + 2x) (1 – x) – 2000

1000x = 2000 – 4000x2 + 4000x – 2000x – 2000

4000xx2 -1000x = 0

1000x (4x – 1) = 0

=> x = 25%

Hence, the answer is 25%.

Choice A is the correct answer

Let's say the price of a maruti car is x rs.

 Sales = y revenue = xy 

Changed price = 1.3x 

changed value of sales = 0.8y 

new revenue = 1.04 xy

Percentage change in revenue = 4%

Let number of girls in class A = x

Let number of boys in class A = y

Total number of students = x + y

Proportion of girls = x/x+y

 

Number of boys in class B = x

Total number of students in class B = 1.5(x + y)

Proportion of girls = 1-x/1.5(x+y)

Percentage of boys in the overall student community = x+y/2.5*(x+y)*100 = 40

40% of the student group are boys

Hence, the answer is 40%

CP = 80 × 40

Profit from the n objects = n% × 40 × n.

Profit from the remaining objects = (100 – n)% × 40 × (80 – n).

We need to find the minimum possible value of n% × 40 × n + (100 – n)% × 40 × (80 – n).

Or, we need to find the minimum possible value of n2 + (100 – n) (80 – n).

Minimum of n2 + n2 – 180n + 8000

Minimum of n2 – 90n + 4000

Minimum of n2 – 90n + 2025 – 2025 + 4000

We add and subtract 2025 to this expression in order to create an expression that can be expressed as a perfect square.

Minimum of n2 – 90n + 2025 + 1975 = (n – 45)2 + 1975

This reaches minimum when n = 45.

When n = 45, the minimum profit made

45% × 40 × 45 + 55% × 40 × 35

18 × 45 + 22 × 35 = 810 + 770 = 1580

Let us assume he buys n goods.

Total CP = 20n

Total SP = 2 + 4 + 6 + 8 ….n terms

Total SP should be at least 40% more than total CP

2 + 4 + 6 + 8 ….n terms ≥ 1.4 * 20 n

2 (1 + 2 + 3 + ….n terms) ≥ 28n

n(n + 1) ≥ 28n

n2 + n ≥ 28n

n2 - 27n ≥ 0

n ≥ 27

Let the C.P. be x and S.P. be y.

 Given

3(y-x) = 2y – x

=> 3y – 3x = 2y – x

=> y = 2x

Original profit = Rs y – x

= Rs 2x – x (Since, y = 2x)

= Rs x

 Original profit % = x/x ∗ 100

= 100 %

Let track length be equal to T.

Time taken to meet for the first time =T/relative speed

Time taken for a lap for A  = T/6

Time taken for a lap for B = T/b

So, time taken to meet for the first time at the starting point =LCM(T/6,T/b)

Number of meeting points on the track = Time taken to meet at starting point/Time taken for first meeting = Relative speed / HCF (6,b).

So, in essence we have to find values for b such that 

6-b/ HCF(6,b)

b = 2, 10, 18 satisfy this equation. So, there are three different values that b can take.

Hence, the answer is 3.

Let track length be equal to T. When A completes a lap, let us assume B has run a distance of (t - d). At this time, C should have run a distance of (t - 2d).

After 3 laps C is in the same position as B was at the end of one lap. So, the position after 3t - 6d should be the same as t - d. Or, C should be at a distance of d from the end of the lap. C will have completed less than 3 laps (as he is slower than A), so he could have traveled a distance of either t - d or 2t - d.

=> 3t - 6d = t - d

=> 2t = 5d

=> d = 0.4t

The distances covered by A, B and C when A completes a lap will be t, 0.6t and 0.2t respectively. Or, the ratio of their speeds is 5 : 3 : 1.

In the second scenario, 3t - 6d = 2t - d => t = 5d => d = 0.2t.

The distances covered by A, B and C when A completes a lap will be t, 0.8t and 0.6t respectively. Or, the ratio of their speeds is 5 : 4 : 3.

Let us assume Car A travels at a speed of a and Car B travels at a speed of b.

Further, let us assume that they meet after t minutes.

Distance traveled by car A before meeting car B = a * t. Likewise distance traveled by car B before meeting car A = b * t.

Distance traveled by car A after meeting car B = a * 54. Distance traveled by car B after meeting car A = 24 * b.

Distance traveled by car A after crossing car B = distance traveled by car B before crossing car A (and vice versa).

=> at = 54b ---------- (1)

and bt = 24a -------- (2)

Multiplying equations 1 and 2

we have ab * t2 = 54 * 24 * ab

=> t2 = 54 * 24

=> t = 36

So, both cars would have traveled 36 minutes prior to crossing each other. Or, B would have taken 36 + 24 = 60 minutes to travel the whole distance.

If the overall average weight has to increase after the new people are added, the average weight of the new entrants has to be higher than 40.

So, n > 40

 

Consequently, m has to be < 10 (as n + m = 50)

students would be (n - 40) each, above the already existing average of 40. m(n - 40) is the total extra additional weight added, which is shared amongst 40 + m students.

So m*(n-40)/(m+40) has to be maximum for the overall average to be maximum.

At this point, use the trial and error approach (or else, go with the answer options) to arrive at the answer.

The maximum average occurs when m = 5, and n = 45

And the average is 40 + (45 – 40) *5/45 = 40+5/9 = 40.56

Let's say that the students are named a, b, c, d and e, in increasing order of weights. The average of a, b, c and d is 40 kg, whereas the average of b, c, d and e is 45 kg.

The sum of a, b, c and d is 160 kg, and the sum of b, c, d and e is 180 kg.

 

What is the total weight of all the students?

There are two ways of looking at this.

1. a) 160 + e


2. b) 180 + a

Let his marks be 10, 9, 8, 7 and 6 in the five subjects. Hence, totally he has scored 40 marks. This constitutes only 60% of the total marks. Hence, total marks 40/0.6 = 66.7 or 67 approx. , which is the maximum marks in all 5 subjects.

Since the total marks in each subject is the same, hence maximum marks in each subject will be 67/5 = 13 approx. Out of this 50% is the passing marks . In other words to pass in a subject, he needs to score 6.5 marks. We can see that only in 1 subject, he scored less than this viz. 6. Hence, he passed in 4 subjects.