BANK & INSURANCE (RATIO AND PROPORTION) PART 3

Total Questions: 40

1. Amit has 5 more chocolates than Bharat but 9 less chocolates than Chamku and the ratio of the number of chocolates with Bharat to that with Dhiman is 5:4. If the sum of reciprocals of the number of chocolates with Amit and Bharat is equal to that with Chamku and Dhiman, then what is the ratio of the number of chocolates with Amit to that with Bharat?

Correct Answer: (b) 3:2
Solution:

Let the number of chocolates with Bharat be ‘x’

Number of chocolates with Amit = (x + 5)

Number of chocolates with Chamku
= x + 5 + 9 = x + 14

Number of chocolates with Dhiman = 4x/5

So, 1/(x + 5) + 1/x = 1/(x + 14) + 5/4x

⇒ (2x + 5)/x(x + 5) = (9x + 70)/4x(x + 14)

⇒ (x + 5)(9x + 70) = (2x + 5)(4x + 56)

⇒ x² − 17x + 70 = 0

⇒ (x − 10)(x − 7) = 0

⇒ x = 7 or 10

If x = 7, number of chocolates with Dhiman
= (4/5) × 7 [Not possible]

So, x = 10

Required ratio = 15 : 10 = 3 : 2

2. The ratio of marks obtained by Shalu and Priya is 5:9 and the ratio of marks obtained by Priya and Gautam is 3:4. If the sum of 4 times of marks obtained by Shalu and five times of marks obtained by Gautam is 288, then what is the average of marks obtained by Shalu, Priya and Gautam?

Correct Answer: (e) 31.2
Solution:

Let marks obtained by Shalu, Priya and Gautam are a, b and c, respectively.

a : b = 5 : 9 and b : c = 3 : 4

So, a : b : c = 5 : 9 : 12

Let a = 5x, b = 9x and c = 12x

4a + 5c = 288

⇒ 4 × 5x + 5 × 12x = 288

⇒ 80x = 288

⇒ x = 288/80

⇒ x = 3.6

Thus, a = 5 × 3.6 = 18, b = 9 × 3.6 = 32.4, and c = 12 × 3.6 = Rs. 43.2

So, required average = (18 + 32.4 + 43.2)/3 = 31.2

3. The average score in an examination taken by 52 students of a class is 85. If the scores of the best 5 performers are not considered, the average score of the remaining students falls by 2. If none of the first five highest scorers is not below 80 and if each of the 5 top scorers had distinct integral scores, find the ratio of maximum possible score of the topper to score of top 5 best students.

Correct Answer: (d) 193 : 519
Solution:

Let the score of the topper be T.

Total score of the 52 students = 52 × 85 = 4420

Total score of the remaining 47 students after scores of the best five performers are removed
= 47 × 83 = 3901

Total score of the top five students = 4420 − 3901 = 519

T + (total score of the next 4 top scores) = 519

T is the maximum when the total score of the next 4 top scorers is minimum.

Total score of the next 4 top scorers has a minimum value of
80 + 81 + 82 + 83 = 326 (since all the top 5 scores are distinct) and the least is 80.

T has a maximum value of 519 − 326 = 193

Reqd. ratio = 193 : 519

4. A bag contains a certain number of coins of different denominations. The ratio of the number of Rs. 1 coins to Rs. 2 coins is 5 : 7, respectively and the ratio of the number of Rs. 2 coins to Rs. 5 coins is 7 : 6 respectively. Find the total value of the Rs. 5 coins, if the total value of the Rs. 1 coins in the bag is Rs. 15.

Correct Answer: (b) Rs. 90
Solution:

Ratio of the number of coins of denominations Rs. 1 : Rs. 2 : Rs. 5 = 5 : 7 : 6.

Let, the number of coins of Rs. 1, Rs. 2 and Rs. 5 in the bag be 5x, 7x and 6x.

Since, the total value of Rs. 1 coin in the bag is Rs. 15

So, the number of coins of Rs. 1 in a bag = 15

5x = 15

⇒ x = 3

Therefore, number of Rs. 5 coins in the bag
= 6x = 18

So, required value of Rs. 5 coins = 6 × 3 × 5 = Rs. 90

5. Rohit distributed some chocolates among his four children and kept some with him. The eldest three children got chocolates in the ratio 3 : 11 : 7. The total number of chocolates with father and youngest child is three times the total chocolates with the three eldest children. The ratio of chocolates with father and that with all the children is 3 : 4. Find the sum of number of chocolates and the chocolate which father has, if the youngest child has 81 chocolates with him?

Correct Answer: (d) 360
Solution:

Let the children be P, Q, R and S and Father be F

Chocolates with P : Q : R = 3 : 7 : 11

Let the number of chocolates be 3k, 7k and 11k

Total chocolates with three eldest children = 21k

Chocolate with F and S = 3 × 21k = 63k

Total chocolates = (21k + 63k) = 84k

Chocolate with F : (P + Q + R + S) = 3 : 4

Total 7 units of chocolate = 84k

1 unit = 12k

Chocolate with F = 3 × 12k = 36k

Chocolate with S = (63k − 36k) = 27k

27k = 81

⇒ k = 3

Total number of chocolates = 84k = 84 × 3 = 252

Father = 36 × 3 = 108

Sum = 252 + 108 = 360

6. In class X of a school there are two sections A and B with strength ratio 2 : 1. The ratio of boys and girls in section A is 3 : 1 and that in section B is 3 : 5. Students of both the sections are made to stand in the ground in rows of boys and girls, with each row having equal number of students. If the maximum number of students possible in a row is 96 what is the difference between the number of boys in section A and B?

Correct Answer: (d) 288
Solution:

Let the strength of A = 16k and that of = 8k

Section A ratio of Boys : girls = 3 : 1

⇒ boys = 12k and girls = 4k

Section B ratio of boys : girls = 3 : 5

⇒ boys = 3k and girls = 5k

A + B boys = 15k and girls = 9k

Maximum number of students in a row = HCF (15k, 9k) = 3k

⇒ 3k = 96

⇒ k = 32

Difference between boys of section A and B = 12k − 3k = 9k

⇒ 9 × 32 = 288

7. Arun has Rs. 105 with him and Varun has Rs. P with him. Varun has coins in denominations of Rs. 2 and Rs. 5 in the ratio of 8:3, while Arun has denominations of 50 paise and Rs. 1 in the ratio of 1:3. The number of Rs. 5 coins is 10% of the number of Rs. 1 coins. Find the value of P.

Correct Answer: (e) None of these
Solution:

Let the 50 paise and Rs. 1 coins be a and 3a respectively

Now, a/2 + 3a = 105

7a/2 = 105

So, value of a = 30

Number of Rs. 5 coins = 10% × 3 × 30 = 9

Number of Rs. 2 coins = 8/3 × 9 = 24

Value of P = 24 × 2 + 5 × 9 = Rs. 93

8. The box A contains x number of red and blue balls and the number of red to blue balls from A in the ratio of 4:3 and The Box B contains 45 numbers of red and yellow balls and the number of red to yellow balls from B in the ratio of 7:8. If half of the red balls and one-fifth of the blue balls from A and one-fourth of red balls and 75% of the yellow balls in B are mixed, then the ratio of the number of red, blue and yellow balls becomes 14(3/8):3:15. Find the value of x?

Correct Answer: (d) 42
Solution:

Total number of balls in B = 45

Number of red balls in B = 45 × 7/15 = 21

Number of yellow balls in B = 45 × 8/15 = 24

(1/2 × 4x/7 + 21 × 1/4) : (1/5 × 3x/7) : (75/100 × 24)

= 14(3/8) : 3 : 15

(1/5 × 3x/7)/18 = 3/15

18 = 3x/7

x = 42

9. A group offers sweets to the people staying in flood relief camp. Every person receives Laddu, Jalebi and Barfi in the ratio 10 : 8 : 6 in terms of dozen but the weight of one Barfi is 16 grams and weight of one Laddu and one Jalebi is in the ratio of 4: 2. The weight of one Laddu is three times the weight of a Barfi. Find the ratio of the percentage contribution of all the sweets in terms of weight?

Correct Answer: (a) 62.5 : 25 : 12.5
Solution:

Ratio of sweets in terms of dozen = Laddu : Jalebi : Barfi = 10 : 8 : 6 = 5 : 4 : 3

Ratio of sweets in terms of weight = Laddu : Jalebi : Barfi = 48 : 24 : 16 = 6 : 3 : 2

Ratio of sweets in terms of weight = (5 × 6) : (4 × 3) : (3 × 2)

= 30 : 12 : 6 = 5 : 2 : 1

% contribution of Laddu = 5/8 = 62.5%

% contribution of Jalebi = 2/8 × 100 = 25%

% contribution of Barfi = 1/8 × 100 = 12.5%

Required ratio = 62.5 : 25 : 12.5

10. A man bought a few kgs of sugar of 3 varieties (X, Y, Z) in the ratio of their quantity 2 : 3 : 5 respectively and started to sell them at a profit of 20%, loss of 10% and profit of 32% respectively. He managed to sell only 80% of total sugar and the rest was returned at the same price as he bought. The price of this sugar is Rs. 30/kg, Rs. 40/kg and Rs. 50/kg respectively. Calculate his total profit amount, if the total sugar bought was 50 kg and he sold all the sugar of X and Y types.

Correct Answer: (c) Rs. 240
Solution:

He bought a total of 50 kg of sugar and the ratio of the quantity is given as 2:3:5.

Therefore, he has 10 kg of X, 15 kg of Y and 25 Kg of Z.

The cost price of X, Y & Z is Rs.30/kg, Rs.40/kg and Rs.50/kg

The profit earned on 1 kg of X is 20/100 × 30 = Rs.6

The loss earned on 1 kg of Y is 10/100 × 40 = Rs.4

The profit earned on 1 kg of Z is 32/100 × 50 = Rs.16

But he sold all of X & Y and 40 − (10 + 15) = 15 kg of Z.

∴ The total amount of profit earned is 6 × 10 − 4 × 15 + 16 × 15

= 60 − 60 + 240 = Rs.240