11) The ratio of income in two consecutive years is 2: 3 respectively. The ratio of their expenditure is 5: 9. Income of second-year is Rs 45000 and Expenditure of first-year is Rs 25000. Find the Savings in both years together.

Let the First-year income = 2x And, Second-year income = 3x But ATQ, the second-year income = 45000 So, x = 45000/3 = 15000 Then, first-year income = 2*15000=30000

Similarly, Let the first-year expenditure = 5y But ATQ, the first-year expenditure =25000 So, y = 25000/5 = 5000 Second-year expenditure = 9y= 9* 5000 = 45000

Now, total saving in two years = first-year savings + second-year savings Or, Total saving = 5000+0 = 5000.

12) Pervez, Sunny, and Ashu Bhati alone can complete a piece of work in 30, 50, and 40 days. The ratio of their salaries of each day is 4: 3: 2 respectively. The total income of Parvez is Rs 144. Find the total income of Sunny.

Let per day salary of Pervez = 4 Pervez can complete a piece of work in 30 days and his per day salary is 4 So, the total income of Pervez = 30* 4 = 120

Let per day salary of Sunny = 3

Similarly, Sunny can complete the same work in 50 days and his per day salary is 3 So, the total income of Sunny = 50 * 3 = 150

Let per day salary of Ashu Bhati = 2 Ashu Bhati can complete the same work in 40 days and his per day salary is 2 So, the total income of Ashu Bhati = 40* 2 = 80

Or, the ratio of total income of Pervez: Sunny: Ashu Bhati = 120: 150: 80 = 12: 15: 8 That means total income of Pervez = 12, but according to the question it is 144. On multiplying 12 by 12, we get the original value. So, multiply each by 12. Hence, the total income of Pervez = 144 Total income of Sunny = 15*12 = 180 Total income of Ashu Bhati = 8*12 = 96

13) A person covers the different distances by train, bus, and car in the ratio of 4: 3: 2. The ratio of the fair is 1: 2: 4 per km. The total expenditure as a fair is Rs 720. Find the total expenditure as fair on the train.

Distance covered in the ratio T: B: C = 4: 3: 2 Fair ratio per km. T: B: C = 1: 2: 4 So, the ratio of total fair T: B: C = 4: 6: 8

Sum of the ratio of total fair = 18 But ATQ, it is 720, so multiply 18 by 40. Now, multiply each and every ratio with 40. The total expenditure as fair on a train = 4*40=160

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14) The price of silver-biscuit is directly proportional to the square of its weight. A person broke down the silver-biscuit in the ratio of 3: 2: 1, and faces a loss of Rs 4620. Find the initial price of silver-biscuit.

Given ratio = 3: 2: 1 Or, 3x: 2x: x The initial price= (6x)^{2} = 36x^{2} After broke down the price = (3x)^{2}: (2x)^{2}: x^{2} = 9x^{2}: 4x^{2}: x^{2} = 14x^{2} After breakdown, a loss of Rs. 4620 occurs. i.e., loss = initial price – final price Loss = 36x^{2} – 14x^{2} = 22x^{2} 22x^{2} = 4620 Or, x2 =4620/22 = 210 The initial price of silver-biscuit = 36*210 = 7560

15) B is inversely proportional to the cube of A. If B=3, A=2. If B = 8/9. Find the value of A.

As per the question, B is inversely proportional to A^{3} i.e., B α 1/A^{3} Or, B =k/ A^{3} Given that, B=3, A=2 So, 3 = k/ 8, or k=24 Given that, B = 8/9 So, 8/9 = 24/A^{3} A^{3} = (9*24)/8 = 27 So, A = ^{3}√ 27 = 3

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16) Rs 7800 are distributed among A, B, and C. The share of “A” is the ¾ of the share of B, and the share of B is the 2/3 of the share of C. Find the difference between the share of B and C.

The share of A: B is 3: 4 The share of B: C is 2: 3

Note: Whenever such form is given, multiply a to b, then b to b, and then b to c.

i.e., A: B: C = 3*2: 4*2: 4*3 Or, A: B: C = 6: 8: 12 Or, A: B: C = 3: 4: 6 Sun of ratios = 13 Now, the share of B = [4/13] * 7800 = 2400 Share of C = [6/13]* 7800 = 3600

The difference between the share of B and C = 3600- 2400 = 1200

17) A bag contains Rs 410 in the form of Rs 5, Rs 2, and Rs 1 coins. The number of coins is in the ratio 4: 6: 9. So, find the number of 2 Rupees coins.

ATQ, if the ratio of coins = 4: 6: 9 That means if Rs 5 coins are 4, Rs 2 coins are 6, and then Rs 1 coins are 9.

According to the given ratio, the ratio of amounts = 5*4: 6*2: 9*1 = 20: 12: 9 The sum of the ratios of the amounts = 20+12+9 = Rs 41 But ATQ, it is Rs 410, which means multiply each ratio by 10 i.e., new ratio = 40: 60: 90 Now, 40*5: 60*2: 90*1 = 200: 120: 90

The total amount in the form of two rupees coins = 120 So, the two rupees coins = 120/2= 60

18) The ratio of copper and zinc in a 63 kg alloy is 4: 3. Some amount of copper is extracted from the alloy, and the ratio becomes 10: 9. How much copper is extracted?

Amount of copper in alloy = [copper ratio/ sum of ratios]* total quantity of alloy

Copper = {4/7}* 63 = 36 kg Similarly, the amount of zinc in alloy = [3/7] * 63 = 27 kg Let the extracted copper from alloy = x kg Remaining copper in alloy = 36-x The new ratio = 10: 9 i.e., 36-x: 27 = 10: 9 36-x: 3 = 10: 1 36-x = 10* 3 36-x = 30 x = 36-30 = 6 kg. Hence, the extracted amount of copper is 6 kg.

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19) The ratio of land and water on earth is 1: 2. In the northern hemisphere, the ratio is 2: 3. What is the ratio in the southern hemisphere?

The ratio of land: water = 1: 2 In the northern hemisphere the ration of land: water = 2: 3

Note: Earth is divided equally into two hemispheres called northern hemisphere and southern hemisphere.

i.e., the northern hemisphere is 50% of the total earth.

We can say southern hemisphere = total area- northern hemisphere.

To make the northern hemisphere 50% of the total area Multiply the ratio of the earth by 10 and northern hemisphere by 3 Now, earth’s ratio Land: water = 1*10: 2*10 = 10: 20…………… (i) Northern hemisphere’s ratio = 2*3: 3*3 = 6: 9…………………. (ii)

Subtract equation i by ii

Hence, southern hemisphere = 10- 6: 20-9 = 4: 11

20) Vessels A and B contain mixtures of milk and water in the ratios 4: 5and 5: 1respectively. In what ratio should quantities of the mixture be taken from A and B to form a mixture in which milk to water is in the ratio 5: 4?

Let the quantity of vessels A and B in the ratio x: y to form a mixture of water and milk in the ratio 4: 5. Or, Milk: water = 5: 4. From vessel A, milk: water = 4x/9: 5x/9 From vessel B, milk: water = 5y/6: 1y/6

Now, ATQ quantities of the mixture should be taken from A and B. Ratio of milk: water = [4x/9 +5y/6]: [5x/9+ 1y/6] = 5/4 Take the LCM of 9 and 6 = 54 Or, [24x + 45y]/54: [30x+9y]/54 = 5/4 Or, [8x + 15 y]: [10x + 3y] = 5/4 Or, [8x + 15y] * 4 = [10x + 3y] * 5 Or, (60-15) y = (50-32) x Or, 45y = 18x Or, 5y = 2x Or, x: y = 5: 2 Hence, the quantity of mixture should be taken from A and B in the ratio 5: 2.