**Basic & Commercial Arithmetic-Solutions**

**1) Solution:**

^{(1/n)}– 1]Using the reverse gear approach, we can obtain the answer much faster Final/Initial=3000/780˜ 3.8 We need to look for something close to 3.8 in the answer options (Here n=5) Choose a middle answer option, assuming the answer is 30% (1.3)5 = (1.3)2 ×(1.3)3 = 17×17×1.3 = 3.75. Hence, the answer will be slightly higher than this. = 31%. Option (b)

**2) Solution:**

12% of profit = 1200 =>Remaining = 8800 After taking 1000 each, amount remaining = 6800.

Ram’s share = 3/8× 6800 = 2550 Ram’s profit = 2550+1000= 3550. Answer is option (d)

**3) Solution:**

If the number of shares remain constant, then p has to be greater than q. answer is option (c) You can verify this by taking some simple numbers

**4) Solution:**

option (d) Cost price= 40×65= 2600 and Cotton at 80 per kg= 32 x 80 = 2560 Cotton seed at 20 per kg = 4.8 x 20 =96.

Therefore, % change = 56/2600×100 = 21.32%

**5)** **Solution:**

If we assume that today’s population is 100, next year it would be 1.1 × 100 = 110, and the following year it would be 1.1 × 110 = 121.If this is double the population of one year ago, the population at that time must have been 0.5 × 121 = 60.5 . Because the problem seeks the “closest” answer choice, we can round 60.5 to 60. In this scenario, the population has increased from 60 to 100 over the last year, a net increase of 40 residents.To determine the percentage increase over the last year,divide the net increase by the initial population: 40/60 = 4/6=2/3, or roughly 67%.

**Option (d)**

**6) Solution**

Let x be the number of students, let n be the number of girls.

Then 40/100*x =n& 90 =x/100 *n. Thus, x= 150

**7) Solution**

Conventional method: Let x be the quantity of apples bought and p be their cost price per kg.Let the amount he sold on the first day be y, hence 2py=px, y=x/2= 0.5 x Let him sell z apples on the second day where 2py+1.5pz=1.3 px=>1.5 z=0.3 x=>z=0.2x So his net profit ={((1.3)px+(0.3 x)1.25 p)/px -1} X100 = (1.3 + 0.375 -1)X100 = 67.5% .

**Shortcut:-** Assumption: Let him buy 3 kg of apples at Rs 10 per kg. Hence his total cost price is Rs. 30.On the first day he should sell at 100% profit, i.e. at Rs 20 per kg, and he is supposed to break even, hence he can sell 1.5 kg of apples.Now on day 2 he sells at 50% profit, i.e. at Rs 15 per kg, and makes a profit of 30% overall. i.e. at the end of day 2 he should have Rs. 39.From day 1 he has Rs 30, so on day 2 he gets Rs. 9.Therefore he sells 6 kg at Rs. 15 to get 9 Rs.He is left with 0.9 kg which he has to sell at 12.5 Rs so he gets 11.25 Rs. In total he gets 30 + 9 + 11.25 = 50.25. Hence his profit is (20.25/30)×100 = 67.5 %.

**8) Solution**

Let the cost of an apple be a and that of an orange be b. Thenb/(2a+3b) = (profit PercentageofA)/100 = 20/100= 1/5 a = b.Let x and y be the cost price of an apple and an orange respectively for B.As a = b , x = y ( as CP of both fruits would have risen by the same ratio) Now profit % of B = x/(2x+2y) × 100 = 100/4= 25% 9) The amount after n years will be (1.1)(1.2)(1.3)….. (1+ 0.n) times the initial amount after n years.By observation we can see that (1.1)(1.2)(1.3)(1.4)>2 and(1.1)(1.2)(1.3) MP= 90. SP, when discount is 20% = 0.8x 90 =Rs 72

**12) Solution**

Here we can assume a value for x. Say x= 36. Dividing 36/6= 6 Multiplying 36×6 = 216 % change= (216-6)/6× 100 =210/6×100= 3500%

**13) Solution**

The winner will get 100-20= 80% of the votes 80% of the votes represent 128000 100% will represent = 128000×100/80= 160000 14) Original value = 392500 + 230000 = 622500. Final value = 3925000 Percentage depreciation = depreciation/(initial value)x 100 = 230000/622500 x 100= 36.95%

**15) Solution**

Use the formula r/(100+r)× 100 % = 20/120 x100 = 100/6 = 16.67%.

SHORTCUT

**You can also use the constant product rule. Increase of 20% (1/5)will result in a decrease of 1/6= 16.66%

so that the expenditure remains constant.

**16) option (d) **

She has scored 66+ 68 = 134 marks already. She needs 220 marks to gain admission into MSc Biological Science = 220-134 = 86 marks more. Given that she gains admission only into Biological Science,hence she can’t get more than 59 in Physics.To minimize Math, we need to maximize Physics = 59. Marks left = 86-59 = 27 marks for Math as a minimum

**17) option (c)**

Cost of living in 2000 = 25(12) +10(8) + 5(14) + 2(40) + 1.5 (60) = 620

Cost of living in 2005 = 25(15) + 10(12) + 5(16) + 2(36) + 1.5 (90) = 782

Taking 2000 as the base year, cost of living index for 2005= 782/620×100 = 126.3

**18) option (d).**

Let’s say the value of the index = 100. Then HBL = 7, Niposys = 13 & Brilliance= 1

Others = 100-21 = 79 Increased index value = 106 Increased individual values HBL = 1.09×7=7.63, Niposys = 14.3 & Brilliance = 1.04

Total = 22.97

Others = 106- 22.97=83.03

Increase = (83.1-79)/79 x 100 = 5.2%

**19) option (b)**

Go from answer options If x= 300, and interest = Rs 15 Using SI= PxRxT/100, we get R=5% Look at the answer options.

By plugging in values, only 300 & 5% will match. All other values will give a different rate of interest value.

**20) option (b)**

Let the CP of 1 litre = Rs 100, SP = Rs 75

SP of milk man includes = Milk man’s CP + 12.5% of milk man’s CP

75 = 1 + 1/8 CP ? 9/8 CP= 75 =>CP= 66.67.

For 100 rupees, he actually sells 666.7 ml. every litre has 666.7 ml of pure milk. For 2 litres = 666.7 x 2 = 1333.4 ml

**21) option (a)**

Go from answer options Let the Price of each chocolate= 50 Therefore, 15 chocolates= 750.

The order will be as follows 750(P) ? 825 (Q) ? 650 (R) ? 780 (S) ? 858 (T) ? 1287 (U)

**22) option (a)**

Out of 3000 books, 500 are given free and from the balance 2500, for every 24 books one book is given free. Hence, 2400 books are sold at 75% of 3.25. Thus, SP = 2400 x 3.25 x 0.75, CP = 2400. Gain percentage = 143.75.

**23) option (d)**

There are totally 600 voters, so after the number of people who initially rejected increases by 150%, it must still be below 600.

That is true only for option (d). Verifying the option:

Initially, Against = 200 & For = 400.

Hence, majority = 400-200 = 200 After discussions, against = 200 + 1.5×200 = 500, For = 600-500 = 100.

Hence, majority = 500-100 = 400, which is twice the initial majority. Thus, this is the correct answer.

**24) option (c)**

The best way to solve this question is by substituting answer options.

Let’s assume that he buys 5 kgs of 2nd variety of sugar. Total CP = 10×15 + 5×20 = 250

Profit of 20% implies a SP of 1.2 x 250 = 300 The mixture of 15 kg would be labeled as Rs 300 or Rs 20/kg He sells 10 kgs at Rs 15/kg. he gets Rs 150. His balance of 5 kgs, he sells at Rs 20, from which he gets Rs 100? the total cost (150 +100=250)

Hence, option (c)

**25) Option (b)**

To get a profit of 20%, he needs to get 300- 150 = Rs. 150 more from 5 kgs . 150/5 =Rs 30 / kg.

Answer is option (b)