Quantitative Aptitude (QA) Test - 35

Cost price of the book = Rs. 720

Selling price of the book = Rs. 720 X (100 + P)% = Rs. (720 + 7.2P)

Profit on the book = 720 + 7.2P - 720 = Rs. 7.2P

Profit percentage (P%) = (7.2P/720) X 100

After interchanging the cost price and Selling price,

Cost price of the book = Rs. (720 + 7.2P)

Selling price of the book = Rs. 720

Loss on the book = Rs. (720 + 7.2P) - 720 = Rs. 7.2P

Loss percentage (L%) = [7.2P/(720 + 7.2P)] X 100

Ratio of L to P = 80:100 = 4:5

{[7.2P/(720 + 7.2P)] X 100}:[(7.2P/720) X 100] = 4:5

720 X 5 = 4 X (720 + 7.2P)

3600 = 2880 + 28.8P

720 = 28.8P

P = 25

L = [7.2 X 25/(720 + 7.2 X 25)] X 100 = (180/900) X 100 = 20

For I:

The selling price of the book if the shopkeeper wants to earn the profit of (L + P)%, is Rs. 1044.

(L + P)% = (20 + 25)% = 45%

Selling price of the book = Rs. 720 X 145% = Rs. 1044

Therefore, this statement is true.

For II:

If the marked price of the book is 2P% more than that of cost price and L% of discount is given, the selling price of the book will be Rs. 864.

Marked price of the book = 720 X (100 + 50)% = Rs. 1080

Selling price of the book = 1080 X 80% = Rs. 864.

Therefore, this statement is true.

For III:

If the ratio of the cost price to selling price of the book is 10:7, the loss on the book will be (P + 10)%.

Cost price of the book = Rs. 720

Selling price of the book = (720/10) X 7 = Rs. 504

Loss on the book = 720 - 504 = Rs. 216

Loss percentage = (216/720) X 100 = 30%

So, (P + 10)% = 25 + 10 = 35%

Therefore, this statement is not true.

Hence, option 4 is correct.

Let speed of Shikha = Speed of Rubi = 'x' m/s

In a race of 2700 m, Rupal beats Shikha by 200 metres

Hence distance covered by Shikha = 2700 - 200 = 2500 metres

Since, time taken by Rupal and Shikha will be the same.

So, time taken by Rupal = time taken by Shikha

(2700/speed of Rupal) = (2500/speed of Shikha)

(2700/speed of Rupal) = (2500/x)

Speed of Rupal = (27x/25) m/s

Similarly, in the race of 3300 m, Sharmila beats Rubi by 300 metres

Hence distance covered by Rubi = 3300 - 300 = 3000 metres

Since, time taken by Sharmila and Rubi will be the same.

So, time taken by Sharmila = time taken by Rubi

(3300/speed of Sharmila) = (3000/speed of Rubi)

(3300/ speed of Sharmila) = 3000/x

Speed of Sharmila = (11x/10) m/s

Time taken by Sharmila to complete the race = 4400/(11x/10) = 4000/x sec

Distance covered by Rupal in the same time = (27x/25) X (4000/x) = 4320 metres

So, Sharmila beats Rupal by (4400 - 4320) = 80 metres

Hence, option 2 is correct.

For I:

Ratio of the profits received by 'A', 'B' and 'C'

= (2000 X 12):{(2400 X 6) + (2000 X 6)}:(1600 X 12) = 10:11:8

Therefore, profit share of 'C' = 109620 X (8/29) = Rs. 30,240

Therefore, I is true.

For II:

Ratio of the profits received by 'A', 'B' and 'C'

(2000 X 12):{(2400 X 6) + (1600 X 6)}:(1600 X 12) = 5:5:4

Therefore, profit share of 'C' = 109620 X (4/14) = Rs. 31,320

Therefore, II is true.

For III:

Ratio of the profits received by 'A', 'B' and 'C'

(2000 X 12):{(2400 X 6) + (1200 X 6)}:(1600 X 12) = 10:9:8

Therefore, profit share of 'C' = 109620 X (8/27) = Rs. 32,480

Therefore, III is false.

Hence, option 2 is correct.

Let the marks required to get a grade of C and B be 50p, and 60p ( As the ratio is given 5:6)

The passing marks required to pass the test = 60% of the marks required to get a B grade

The passing marks required to pass the test = 60% of 60p = 36p

It is given that if he had scored 35 marks more, he would have managed to just get a C grade.

Therefore, 36p+35 = 50p

=> p = 2.5

Therefore, marks required to obtain a B grade = 2.5 X 60 = 150 marks

It is given that to obtain an A grade; marks should be 40% more marks than B grade marks = 150 X (1.4) = 210 marks

Also, an A grade is obtained at 84% marks, which means 84% of the total marks will get an A grade

= 84% of total marks of unit test= 210

Therefore, total marks of unit test = (210/84) X 100 = 250 marks

For I:

Let the present age of 'P' = x years

Present age of 'R' = 2x years

Sum of the present ages of 'Q' and 'R' = 36 X 2 = 72 years

Present age of 'Q' = (72 - 2x) years

Present age of 'S' = [(72 - 2x)/6] X 13 years

After eight years, the age of 'P' = (x + 8) years

After eight years, the age of 'T' = [(x + 8)/7] X 13 years

Present age of 'T' = [(x + 8)/7] X 13 - 8 years

Sum of the ages of 'P', 'S' and 'T = (40 X 5) - 72 = 128 years

x + {[(72 - 2x)/6] X 13} + {[(x + 8)/7] X 13} - 8= 128

x + (936 - 26x)/6 + (13x + 104)/7 - 8= 128

42x + 6552 - 182x + 78x + 624 - 336 = 128 X 42

62x = 1464

x = 1464/62

Therefore, I cannot be true.

For II:

Let the present age of 'P' = x years

Present age of 'R' = 2x years

Sum of the present ages of 'Q' and 'R' = 32 X 2 = 64 years

Present age of 'Q' = (64 - 2x) years

Present age of 'S' = [(64 - 2x)/6] X 13 years

After eight years, the age of 'P' = (x + 8) years

After eight years, the age of 'T' = [(x + 8)/7] X 13 years

Present age of 'T' = [(x + 8)/7] X 13 - 8 years

Sum of the ages of 'P', 'S' and 'T = (36 X 5) - 64 = 116 years

x + {[(64 - 2x)/6] X 13} + {[(x + 8)/7] X 13} - 8= 116

x + (832 - 26x)/6 + (13x + 104)/7 - 8= 116

42x + 5824 - 182x + 78x + 624 - 336 = 116 X 42

62x = 1240

x = 20

Present age of 'T' = [(x + 8)/7] X 13 - 8 = 44 years

Present age of 'S' = [(64 - 2x)/6] X 13 = 52 years

Difference = 52 - 44 = 8 years

Therefore, 'II' can be true.

For III:

Let the present age of 'P' = x years

Present age of 'R' = 2x years

Sum of the present ages of 'Q' and 'R' = 20 X 2 = 40 years

Present age of 'Q' = (40 - 2x) years

Present age of 'S' = [(40 - 2x)/6] X 13 years

After eight years, the age of 'P' = (x + 8) years

After eight years, the age of 'T' = [(x + 8)/7] X 13 years

Present age of 'T' = [(x + 8)/7] X 13 - 8 years

Sum of the ages of 'P', 'S' and 'T = (30 X 5) - 40 = 110 years

x + {[(40 - 2x)/6] X 13} + {[(x + 8)/7] X 13} - 8= 110

x + (520 - 26x)/6 + (13x + 104)/7 - 8 = 110

42x + 3640 - 182x + 78x + 624 - 336 = 110 X 42

-62x = 692

x = -692/62

Age cannot be negative.

Therefore, III is false

Hence, option 1 is correct.

Let the time ratio to cover cities AB, BC and CD be 4t, 2t and yt

It is given that CD is 880 km, i.e. 4a = 880 km (As Distance ratio is 2:x:4)

a = 220 km

Hence, the distance between BC is 220x

Also, the time taken to cover AC is 24 hrs, i.e. 6t = 24 hrs

t = 4 hrs

Therefore, Average speed of car = (Total distance travelled)/(Total time taken)

=> Total distance = AB+BC +CD = 2*220+x*220+880 = 1320+220x km

=> Total time taken = 24+y*4 = 24+4y hrs

Therefore, average speed = (1320+220x)/(24+4y) = 55km/hr

=> (330+55x)/(6+y) = 55

=> 55x = 55y

=> x = y

Hence, option 1 is correct.

Dice D1 has outcomes = {1,2,3,4,5,6}

Dice D2 has outcomes = {2,4,6}

Total number of outcomes(m) when both of them are rolled = 6 x 3 = 18

Outcomes are (1,2), (1,4), (1,6), (2,2), (2,4), (2,6), (3,2), (3,4), (3,6), (4,2), (4,4), (4,6), (5,2), (5,4), (5,6), (6,2), (6,4), and (6,6).

The outcomes whose sum is less than 6 are (1,2), (1,4), (2,2), and (3,2).

The number of outcomes whose sum is less than 6 is 4.

Therefore, the number of outcomes (n) who sum is greater than or equal to 6 is 18-4, i.e. 14.

Required probability = n/m = 14/18 = 7/9

Let the number of students in sections A, B and C be a, b, and c

The average weight of students in section A = 56 kg

The average weight of students in section B = 48 kg

The average weight of students in section A = 58 kg

Also, the average weight of students from sections A and B = 54 kg

Total weight of section A and B students = 56a + 48b = 54(a+b)

=> 56a + 48b = 54a + 54b

=> 2a = 6b

=> a:b =3:1

Similarly, the average weight of students from sections B and C = 56 kg

Total weight of section B and C students = 48b + 58c = 56(b+c)

=> 48b + 58c = 56b + 56c

=> 2c = 8b

=> b:c = 1:4

=> a:b:c = 3:1:4

Therefore, the average weight of all students in three sections combined

We have given, (x + y) = √15, and (x - y) = √11

(x + y) = √15

By squaring both sides, we get,

x² + y² + 2xy = 15 ....(i)

Also, (x - y) = √11

By squaring both sides, we get,

x² + y² - 2xy = 11 ....(ii)

By adding equation (i) and (ii) we get,

2(x² + y²) = 26

Or, (x² + y²) = (26/2) = 13

Putting the value of (x² + y²) in equation (i) we get,

13 + 2xy = 15

Or, 2xy = 15 - 13

Or, xy = (2/2) = 1

xy(x² + y²) = 1 × 13 = 13