**Answer –** Option

**Explanation –**

Let the number of selected and unselected candidates be 6x and x respectively.

Then, number of candidates applied = 6x + x = 7x

According to the question,

If 90 less applied and 30 less selected, the ratio of selected to unselected candidates = \(\frac {8}{1}\), i.e.

New number of selected candidates = 6x – 30

And new number of applied candidates = 7x – 90

∴new number of candidate unselected = 7x – 90 – (6x – 30)

= x – 60

Thus, \(\frac {6x – 30}{x – 60} = \frac {8}{1}\)

\(\Rightarrow\) 6x – 30 = 8x – 480

\(\Rightarrow\) x = 225

Thus, the number of candidates applied for the process = 7x

= 7 × 225

= 1575

**2. The price of gold is directly proportional to square of its weight. A person broke down the gold 3:2:1 and face a loss of Rs 4,620. Find the initial price of gold.**

** Answer –** Option A

**Explanation –**

3:2:1 \(\rightarrow\) 6 (According to Question) \(\rightarrow\) 36 (On squaring)

In this question if we brake it in the ratio and then squaring it then we get 14 (9 + 4 + 1)

So, we loss 22 (36 – 14) if we broke it

Loss = Rs. 4620

Initial price = \(\frac {4620}{22} \times 36\)

= Rs 7560

**3. The no. of mangoes on three baskets are in the ratio of 3 : 5 : 7. In which ratio the no. of mangoes in first two baskets must be increase so that the new ratio becomes 5 : 6 : 8.**

**Answer –** Option C

**Explanation –**

A.T.Q.

\({B}_{1} : {B}_{2} : {B}_{3} = 3X : 5X : 7X\)

and again,

\({B}_{1} : {B}_{2} : {B}_{3} = 5X : 6X : 8X\)

According to question there is increase in first two baskets only. It means the no. of mangoes remain constant in the third basket.

So, 7x = 8y or x = \(\frac {8}{7} y\) Hence, 3x : 5x : 7x = \(3 \times \frac {8}{7} y : 5 \times \frac {8}{7} y : 7 \times \frac {8}{7} y \)

= 24y : 40y : 56y and 5y : 6y : 8y \(\Rightarrow\) 35y : 42y: 56y

i.e, increase in first basket = 11 increases in second basket = 2 Required ratio = 11 : 2

**4. In a mixture of 60 liters, the ratio of milk and water 2:1. If this ratio is to be 1:2 , then the quantity of water to be further added is:**

**Answer –** Option D

**Explanation –**

Quantity of milk =\(60 \times \frac {2}{3}\) = 40 liters

Quantity of water = 60 – 40 = 20 liters

Let x liters of water to be added so as to get the ratio 1:2. Then,

\(\frac {40}{20 + x} = \frac {1}{2}\)

x = 80 – 20 = 60 liters

**5. Two numbers are in the ratio 3: 5. If 9 is subtracted from each number, then they are in the ratio of 12: 23. What is the second number?**

**Answer –** Option B

**Explanation –**

Let the numbers be 3x & 5x (since they are in the ratio of 3:5)

So, subtraction 9, we get

3x – 9 and 5x – 9

Now, the ratio is 12 :12: (given)

\(\frac {3x – 9}{5x – 9} = \frac {12}{23}\)

(By cross multiplication)

23(3x – 9) = 12(5x -9)

69x – 207 = 60x – 108

9x = 99

x = 11

So, the original numbers are 3x = 3 \(\times\) 11 = 33 and 5 \(\times\) 11 = 55 , with the second number being 5.

**Answer –** Option C

**Explanation –**

Ratio of income = 6x : 7x

Ratio of expenditure = 5y : 4y

A.T.Q

6x – 5y = 700

7x – 4y = 2100

On Solving we get x = 700, y = 700

Required

Monthly Expenditure of anurag = 4y = 4x 700 = Rs.2800

**2. In Shyam’s wallet there are total Rs 36, consisting of 10 paise, 20 paise and Re 1 coins such that there is at least one coin of each denomination. The ratio of coins of 10 paise to 20 paise is 8 : 5. Find the minimum no. of Rs 1 coins is**

** Answer –** Option A

**Explanation –**

Let the coins of 10 paise, 20 paise and Rs 1 be x, y & z respectively.

A.T.Q.

10x + 20y + 100z = 3600

x : y = 8 : 5

80y + 100y + 100z = 3600

180y + 100z = 3600

9y + 5y = 180

Putting z = 1, 2, 3, 4, 5…., we get z = 9, y = 15 (an integer)

Hence, minimum 9 coins of `1 will be there.

**3. In an army camp ration is available for 100 soldiers for 10 days. After 2 days, 60 soldiers joined. Then for how many more days will the remaining ration last?**

**Answer –** Option C

**4. If 15A = 20B = 30C, and A is 300, then find the value of B + C – A:**

**Answer –** Option A

**Explanation –**

A: B: C::20 \(\times 30 : 15\times 30 : 15 \times 20 = 4 : 3 :2 \)

B = \(\frac {300}{4} \times 3 = 225\)

c = \(\frac {300}{4} \times 2 = 150\)

B + C – A = 225 + 150 – 300 = 75

**5. If x – y : y – z : z – k = 2 : 3 : 5, then what is the value of z -k : (x – k)?**

**Answer –** Option A

**Explanation –**

Given: x – y : y – z : z – k = 2 : 3 : 5

Let x – y = 2x ———(i)

y – z = 3x ———(ii)

And z – k = 5x ———(iii)

Adding eq. (i), (ii) and (iii), we get

x – k = 10x

z -k : (x – k) = 5x : 10x = 1 : 2

**Answer –** Option A

**Explanation –**

Let the No. of captain and soldiers = 2x & 7x

\(\frac {2x – 25}{7x – 100} = \frac {3}{10}\)

20x – 250 = 21x – 300

x = 50

No. of soldiers after war = 7x – 100

= 7 \(\times\)50 – 100 = 250

**2. If 50 less had applied and 25 less selected, the ratio of selected to unselected would have been 9 : 4. So how many candidates had applied if the ratio of selected to unselected was 2 : 1.**

** Answer –** Option C

**Explanation –**

Selected | Unselected | Total | Results |
---|---|---|---|

2 | 1 | 3 | Initial |

9 | 4 | 13 | Final |

Initially there were total 3x candidates and 2x were selected.

According to question

Ratio of selected to applied:

\(\frac {2x – 25}{3x – 50} = \frac {9}{13}\)

26x – 25 \(\times\) 13 = 27x – 9 \(\times\) 50

27x – 26 x = 450 – 325

x = 125

Total candidates = 3x = 3 \(\times\) 125 = 375

**3. If 3A = 7B = 13C, then A : B : C is equal to**

**Answer –** Option D

**Explanation –**

Given: 3A = 7B = 13C

\(\frac {A}{B} = \frac {7}{3}\)

And \(\frac {B}{C} = \frac {13}{7}\)

To get the relation between A, B and C,

Taking LCM of 3, 7 and 13, i.e. 273

Now, divide the given as

A : B : C = \(\frac {273}{3} : \frac {273}{7} : \frac {273}{13} = 91 : 39 : 21\)

**4. In an MBA selection process, the ratio of selected to unselected was 11:2. If 40 less had applied and 20 less selected, the ratio of selected to unselected would have been 10:1. How many candidates had applied for the process?**

**Answer –** Option B

**Explanation –**

Let the selected candidates = 11x

Unselected candidates = 2x

Total candidates = 11x + 2x = 13x

New applied candidates = 13x – 40

New selected candidates = 11x – 20

Ratio between the selected and unselected candidates = 10:1

ratio of total candidates to selected candidates = 11:10

\(\frac {13x – 40}{11x – 20} = \frac {11}{10}\)

130x – 400 = 121x – 220

9x = 180

X = 20

Total number of applied candidates = 13x = 13 × 20 = 260

**5. If P : Q : R = 2 : 3 : 5, then what is the value of (P + Q) : (Q + R) : (R + P)?**

**Answer –** Option A

**Explanation –**

P:Q:R = 2:3:5

Then (P + Q):(Q + R):(R + P) = (2 + 3): (3 + 5): (5 + 2) = 5:8:7