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1.

Quantity B: There are 28 students in a hostel. If the number of students is increased by 12, the expenses of the mess increased by Rs. 92 per day. While the average expenditure per head decreased by Rs. 1. Find per week original expenses of the mess.1). Quantity A > Quantity B2). Quantity A < Quantity B3). Quantity A ≥ Quantity B4). Quantity A ≤ Quantity B

Answer»

Quantity A:

Suppose the numbers are x, y and Z;

⇒ (x + y)/2 + z = 36---- 1

⇒ x + (y + z)/2 = 30---- 2

⇒ (x + z)/2 + y = 26---- 3

These equations can be written as

x + y + 2z = 72

2x + y + z = 60

x + z + 2y = 52

Adding the equations

⇒ 4 (x + y + z) = 184

⇒ (x + y + z) = 46---- 4

From equation 1, 2, 3 and 4

⇒ x = 14, y = 6, z = 26

∴ Product of the numbers = 14 × 6 × 26 = 2184

Quantity B:

Suppose the original per head EXPENSE of the mess was Rs. a

According to the given condition

⇒ 28a + 92 = 40(a – 1)

⇒ 12a = 132

⇒ a = 11

∴ Per week original expenses of the mess = 7 × 28 × 11 = 2156

∴ Quantity A > Quantity B
Discussion
2.

W(n) = 100n + 1 Kg1). 301 kg2). 401 kg3). 501 kg4). 601 kg

Answer»

Weight of FIRST vessel = W(1) = 100 × 1 + 1

∴ Sum of all weight’s = 100 × 1 + 1 + 100 × 2 + 1 + 100 × 3 + 1 + 100 × 4 + 1 + 100 × 5 + 1 = 100(1 + 2 + 3 + 4 + 5) + 1 × 5 = 100(5 × 6/2) + 5

∴ Average weight = (100 × (5 × 6/2) + 5)/5 = 100 × 6/2 + 1 = 301 Kg
Discussion
3.

If 5 students are transferred from A to B, the maximum possible average achieved for group A will be1). 20.662). 21.53). 20.754). 20.5

Answer»

For maximum achievable average of Group A STUDENTS with lowest MARKS should be transferred.

Lowest marks of Students in Group A = 18

Present total marks = 20 × 20 = 400

Average of 5 students transferred = 18 × 5 = 90

New total marks = 400 - 90 = 310

∴ Average maximum marks of Group A = $(\frac{{310}}{{15}} = 20.66)$

∴ 20.66 is the new maximum average of Class A.
Discussion
4.

If 5 students are transferred from B to A, what will be the minimum possible average of A?1). 20.692). 213). 20.754). 20.6

Answer»

Lowest score of B = 23

Present total MARKS of A = 20 × 20 = 400

Marks of 5 students of B = 23 × 5 = 115

New total marks of B = 400 + 115 = 515

New AVERAGE of B = $(\frac{{515}}{{25}} = 20.6)$

HENCE, 20.6 is the new minimum average of A.
Discussion
5.

The average weight of 4 friends Utkarsh, Ashu, Kamal & Himanshu is 100. If the average weight of Ashu & Kamal is 60, and that of Ashu & Himanshu is 72.5, then find the weight of Utkarsh if the weight of Kamal is 55?1). 1502). 1253). 1754). 200

Answer»

⇒ TOTAL weight of 4 FRIENDS = 100 × 4 = 400

⇒ Combined weight of Ashu & Kamal = 60 × 2 = 120

⇒ Weight of Ashu = 120 – 55 = 65

⇒ Combined weight of Himanshu & Ashu = 72.5 × 2 = 145

⇒ Weight of Himanshu = 145 – 65 = 80

∴ Weight of Utkarsh = 400 – (65 + 55 + 80) = 400 – 200 = 200 kgs
Discussion
6.

1). Rs. 10.122). Rs. 153). Rs. 16.254). Rs. 9.53

Answer»

⇒ Average = sum of elements / NUMBER of elements

Given,

Let the price of an orange and an apple be Rs. a and Rs. b.

Given,

The average price of 40 ORANGES is 80% of average price of 20 apples in a basket.

The average price of 40 oranges = (40 × a)/40 = a

The average price of 20 apples = (20 × b)/20 = b

⇒ a = (80/100) × b

⇒ a = 4b/5

⇒ b = 5a/4

Given,

The price of a pineapple is 40% of price of an orange.

Let the price of a pineapple be Rs. m.

⇒ a = (40/100) × m

⇒ a = 2m/5

⇒ m = 5a/2

Given,

a = Rs. 8

Then,

⇒ b = 5a/4 = (5 × 8)/4 = Rs. 10

⇒ m = 5a/2 = (5 × 8)/2 = Rs. 20

If 5 more pineapples are added into the basket of price Rs. 20 each then,

⇒ TOTAL price of 5 pineapples = 5 × 20 = Rs. 100

⇒ Total price of 40 oranges = 40 × 8 = Rs. 320

⇒ Total price of 20 apples = 20 × 10 = Rs. 200

Total price of 40 oranges, 20 apples and 5 pineapples in a basket =

= 100 + 320 + 200

= 620

Average price of a basket contains 40 oranges, 20 apples and 5 pineapples =

= 620/(40 + 20 + 5)

= 620/65

= 9.53

Average price of a basket contains 40 oranges, 20 apples and 5 pineapples is Rs. 9.53.

Alternate Solution: Price of an orange = 8

So, price of an apple will be = 10

And Price of a pineapple will be = 20

Average price of a fruit = (8 × 40 + 10 × 20 + 20 × 5)/65 = 620/65 = 9.53
Discussion
7.

1). 8502). 10003). 9004). 800

Answer»

LET the AGGREGATE marks be X

Minimum PASSING marks = 240 + 32 = 272

⇒ 32% of X = 272

⇒ X = 272 × 100/32 = 850

∴ The aggregate marks = 850
Discussion
8.

What will be the minimum possible average of Group A if 5 students are transferred from A to B?1). 19.332). 21.53). 22.54). 20.5

Answer»

Present total MARKS of Group A = 20 × 20 = 400

Highest score of Group A = 22

Total marks of 5 students of Group A = 22 × 5 = 110

NEW total marks = 400 - 110 = 290

∴ New AVERAGE = $(\frac{{290}}{{15}} = 19.33)$

HENCE, 19.33 is the new average.
Discussion
9.

Quantity B: Five years ago, the average age of husband and wife was 27 years and today, the average of husband, wife and their child is 22 years. What is the age of the child?1). Quantity A > Quantity B2). Quantity A < Quantity B3). Quantity A ≥ Quantity B4). Quantity A ≤ Quantity B

Answer»

Quantity A:

Difference in temperatures of Monday and Friday = (39 × 4) – (37 × 4) = 8

The temperature of Friday is 4/5 TIMES that of Monday;

If X is the temperature of Monday;

∴ X – 4X/5 = 8

⇒ X/5 = 8

⇒ X = 40

∴ 5% of X = 40 × 0.05 = 2

Quantity B:

Five YEARS ago, the AVERAGE age of husband and wife was 27 years;

∴ Total age of husband and wife today = 27 × 2 + 10 = 64 years

Since the average of husband, wife and their child is 22 years today;

∴ Total age of husband, wife and child today = 22 × 3 = 66 years

∴ Age of the child = 66 – 64 = 2 years

∴ Quantity A = Quantity B
Discussion
10.

What is the maximum possible average achievable for group B?1). 252). 24.53). 25.54). 24

Answer»

Present AVERAGE of B = 25 × 25 = 625

Number of students transferred from A to B = 5

For maximum average B’s HIGHEST score students should get transferred = 22

Max score of A = 22

Max marks for 5 students in A = 22 × 5 = 110

New marks = 625 + 110 = 735

New number of students in Class B = 25 + 5 = 30

∴ Average number of marks of B = $(\FRAC{{735}}{{30}} = 24.5)$

∴ 24.5 is the maximum average achievable for GROUP B.
Discussion