1. In an army selection process, the ratio of selected to unselected was 6:1. If 90 less had applied and 30 less selected, the ratio of selected to unselected would have been 8:1. How many candidates had applied for the process?

A. 3150 B. 6300 C. 4725 D. 1575

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 = [latex]\frac {8}{1}[/latex], 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, [latex]\frac {6x - 30}{x - 60} = \frac {8}{1}[/latex]
[latex]\Rightarrow[/latex] 6x – 30 = 8x – 480
[latex]\Rightarrow[/latex] 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.

A. 7560 B. 6560 C. 7450 D. 7440

Answer - Option A
Explanation - 3:2:1 [latex]\rightarrow[/latex] 6 (According to Question) [latex]\rightarrow[/latex] 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 = [latex]\frac {4620}{22} \times 36[/latex]
= 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.

A. 3 : 13 B. 12 : 7 C. 11 : 2 D. 13 : 3

Answer - Option C
Explanation - A.T.Q.
[latex]{B}_{1} : {B}_{2} : {B}_{3} = 3X : 5X : 7X[/latex]
and again,
[latex]{B}_{1} : {B}_{2} : {B}_{3} = 5X : 6X : 8X[/latex]
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 = [latex]\frac {8}{7} y[/latex] Hence, 3x : 5x : 7x = [latex]3 \times \frac {8}{7} y : 5 \times \frac {8}{7} y : 7 \times \frac {8}{7} y [/latex]
= 24y : 40y : 56y and 5y : 6y : 8y [latex]\Rightarrow[/latex] 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:

A. 20 liters B. 30 liters C. 40 liters D. 60 liters

Answer - Option D
Explanation - Quantity of milk =[latex]60 \times \frac {2}{3}[/latex] = 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,
[latex]\frac {40}{20 + x} = \frac {1}{2}[/latex]
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?

A. 44 B. 55 C. 66 D. 77

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)
[latex]\frac {3x - 9}{5x - 9} = \frac {12}{23}[/latex]
(By cross multiplication)
23(3x - 9) = 12(5x -9)
69x - 207 = 60x - 108
9x = 99
x = 11
So, the original numbers are 3x = 3 [latex]\times[/latex] 11 = 33 and 5 [latex]\times[/latex] 11 = 55 , with the second number being 5.

1. Sachin and Anurag have monthly incomes in the ratio 6 : 7 and Their monthly expenditures in the ratio 5: 4. If they save Rs.700 and Rs.2100 respectively, find the monthly expenditure of Anurag.

A. 3200 B. 2480 C. 2800 D. 3140

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

A. 9 B. 7 C. 8 D. 11

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?

A. 7 B. 6 C. 5 D. 4

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

A. 75 B. 80 C. 60 D. 50

Answer - Option A
Explanation - A: B: C::20 [latex]\times 30 : 15\times 30 : 15 \times 20 = 4 : 3 :2 [/latex]
B = [latex]\frac {300}{4} \times 3 = 225[/latex]
c = [latex]\frac {300}{4} \times 2 = 150[/latex]
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)?

A. 1 : 2 B. 2 : 1 C. 1 : 3 D. 1 : 4

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

1. Before a battle there were the ratio of captains to soldiers was 2 : 7. During the war 25 captains and 100 soldiers were martyred. The new ratio of captains to soldiers became 3 : 10. What is the number of soldiers after the war?

A. 250 B. 200 C. 150 D. 100

Answer - Option A
Explanation - Let the No. of captain and soldiers = 2x & 7x
[latex]\frac {2x - 25}{7x - 100} = \frac {3}{10}[/latex]
20x - 250 = 21x - 300
x = 50
No. of soldiers after war = 7x - 100
= 7 [latex]\times[/latex]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.

A. 125 B. 250 C. 375 D. 500

Answer - Option C
Explanation -
Initially there were total 3x candidates and 2x were selected.
According to question
Ratio of selected to applied:
[latex]\frac {2x - 25}{3x - 50} = \frac {9}{13}[/latex]
26x – 25 [latex]\times[/latex] 13 = 27x – 9 [latex]\times[/latex] 50
27x – 26 x = 450 - 325
x = 125
Total candidates = 3x = 3 [latex]\times[/latex] 125 = 375
3. If 3A = 7B = 13C, then A : B : C is equal to

A. 21:39:91 B. 39:91:21 C. 13:7:3 D. 91:39:21

Answer - Option D
Explanation - Given: 3A = 7B = 13C
[latex]\frac {A}{B} = \frac {7}{3}[/latex]
And [latex]\frac {B}{C} = \frac {13}{7}[/latex]
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 = [latex]\frac {273}{3} : \frac {273}{7} : \frac {273}{13} = 91 : 39 : 21[/latex]
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?

A. 220 B. 260 C. 300 D. 340

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
[latex]\frac {13x - 40}{11x - 20} = \frac {11}{10}[/latex]
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)?

A. 5:8:7 B. 2:3:5 C. 5:8:10 D. 4:9:25

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