BANK & INSURANCE (RATIO AND PROPORTION) PART 3

Total Questions: 40

21. The ratio between the three angles of a quadrilateral is 1:4:5 respectively. The value of the fourth angle of the quadrilateral is 60. What is the difference between the value of the largest and the smallest angles of the quadrilateral?

Correct Answer: (a) 120°
Solution:

x + 4x + 5x + 60 = 360°

10x = 300

⇒ x = 30

Required difference = 5x − x

= 4x = 4 × 30 = 120°

22. Mr. Pandit owned 950 gold coins all of which he distributed amongst his three daughters Lalita, Amita and Neeta. Lalita gave 25 gold coins to her husband, Amita donated 15 gold coins and Neeta made jewellery out of 30 gold coins. The new respective ratio of the coins left with them was 20 : 73 : 83. How many gold coins did Amita receive from Mr. Pandit?

Correct Answer: (a) 380
Solution:

Final ratio of gold coins among Lalita : Amita : Neeta

= 20 : 73 : 83

Let x be the common factor in the given ratio,

20x + 25 + 73x + 15 + 83x + 30 = 950

⇒ 176x = 880

⇒ x = 5

Gold coins received by Amita from Mr. Pandit

= 5 × 73 + 15 = 380

23. The angle of the quadrilateral are in the ratio of 3: 5: 9: 7. The second largest angle of the quadrilateral is equal to the largest angle of a triangle. One of the angles of the triangle is 25°. What is the value of the second largest angle of the triangle?

Correct Answer: (b) 50°
Solution:

3x + 5x + 9x + 7x = 360

⇒ 24x = 360

⇒ x = 15

Second largest angle of quadrilateral = 7 × 15 = 105

According to the question,

Largest angle of triangle = 105

Let the second largest angle be x

⇒ 105 + 25 + x = 180

⇒ 130 + x = 180

⇒ x = 50°

24. One rupee coins, 50 paise coins and 25 paise coins, whose numbers are proportional to 3, 3½ and 4 are together worth Rs.161. Find the number of 25 paise coins.

Correct Answer: (c) 112
Solution:

Coins are in the ratio 3 : 3½ : 4, i.e. 6 : 7 : 8

Their proportional value = 6 × 1 : 7 × 1/2 : 8 × 1/4

= 6 : 3½ : 2

= 12 : 7 : 4

The value of one rupee coins = (12 / 12 + 7 + 4) × 161 = Rs.84

The number of one rupee coins = 84 coins

The value of 50 paise coins = (7 / 12 + 7 + 4) × 161 = Rs.49

The number of 50 paise coins = 98 coins

The value of 25 paise coins = (4 / 12 + 7 + 4) × 161 = Rs.28

The number of 25 paise coins = 112 coins

25. In 3 cans, each having a capacity of 20 L, mixture of milk and water is filled. The ratios of milk and water in 3 cans respectively are 3:2, 5:2 and 1:3. If all the 3 cans are emptied into a single larger can, then find the proportion of milk and water in the larger can.

Correct Answer: (e) None of these
Solution:

Total milk in 3 cans : Total water in 3 cans

= 3/5 + 5/7 + 1/4 : 2/5 + 2/7 + 3/4

= (84 + 100 + 35)/140 : (56 + 40 + 105)/140

= 219 : 201 = 73 : 67

26. Rohit has some 50 paise coins, some Rs.2 coins, some Rs.1 coins and some Rs.5 coins. The value of all the coins is Rs.50. Number of Rs.2 coins is 5 more than the Rs.5 coins. 50 paise coins are double in number than Rs.1 coins. Value of 50 paise coins and Rs.1 coins is Rs.26. How many Rs.2 coins does he have?

Correct Answer: (c) 7
Solution:

Let he has x number of Rs.5 coins.

No. of Rs.2 coins = x + 5

Value of Rs.1 and 50 paisa coins = Rs. 26

The Value of all Coins = Rs. 50

The value of Rs. 2 and Rs. 5 coins = 50 − 26 = 24

⇒ 5x + (x + 5) × 2 = 24

⇒ 5x + 2x + 10 = 24

⇒ 7x = 14

⇒ x = 2

No. of Rs.2 coins = 2 + 5 = 7

27. A teacher distributed 600 chocolates among three students. 40 more than 2/5 times, the number of chocolates received by first student, 20 more than 2/7 times, the number of chocolates received by second student and 10 more than 9/17 times the number of chocolates received by third student are all equal. Then find the number of chocolates received by the first student.

Correct Answer: (c) 150
Solution:

Let the number of chocolates received by 1st student

= x

Let the number of chocolates received by 2nd student = y

Let the number of chocolates received by 3rd student = z

According to question:

2x/5 + 40 = 2y/7 + 20 = 9z/17 + 10 = K (Let)

X = 5(k − 40)/2

Y = 7(k − 20)/2

Z = 17(k − 10)/9

By solving the given equation, we get value of K = 100

Number of chocolates received by 1st student = x

= 5(k − 40)/2 = 150

28. A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced by water to make the ratio of syrup and water, 1:1?

Correct Answer: (c) 1/5
Solution:

Ratio of syrup and water = 5 : 3

Let syrup = 5x

Water = 3x and total mixture = 8x

Let 8y units of mixture is drawn off it means in 8y units

of mixture 5y units will be syrup and 3y units will be water (Because ratio of syrup and water is 5 : 3)

New quantity of syrup = 5x − 5y

New quantity of water = 3x − 3y + 8y = 3x + 5y (8y units of water is added)

They are equal to each other.

⇒ 5x − 5y = 3x + 5y

⇒ 2x = 10y

⇒ y/x = 2/10 = 1/5

⇒ 8y/8x = 1/5

So, 1/5th of the mixture must be taken out and replaced with water.

29. Chiku, Tipu and Pinku have some candies with each. Five times the number of candies with Pinku equals seven times the number of candies with Chiku while five times the number of candies with Chiku equals seven times the number of candies with Tipu. What is the minimum number of candies that can be there with all three of them put together?

Correct Answer: (c) 109
Solution:

Let the candies with Chiku, Pinku and Tipu be a, b and c respectively.

Given, 5b = 7a, and 5a = 7c

25b = 35a & 35a = 49c

25b = 35a = 49c ⇒ b/49 = a/35 = c/25

The least possible integral values for a,b & c will be

a = 35, b = 49 & c = 25

Total = 35 + 49 + 25 = 109

30. There are two vessels containing the mixture of milk and water. In the first vessel the water is 2/3 of the milk and in the second vessel water is just 40% of milk. in what ratio these are required to mix to make 24 litres mixture as 1 is to 2?

Correct Answer: (b) 5:7
Solution:

Concentration of water in first vessel = 2/5 = 40%

Concentration of water in second vessel = 2/7 = 28.57%

By allegation,

2/7    2/5
..........\       /
.............1/3
.........../       \
1/15   1/21

Hence the ratio will be, 21:15 = 7:5

Therefore, the ratio of first mixture to second mixture = 5:7