Basic Arithmetic Solutions

Basic & Commercial Arithmetic-Solutions

Basic Arithmetic
1) Solution:
We need to use Compounded Annual Growth Rate (CAGR) for this question Conventionally, CAGR is computed using the formula [(Final/Initial)(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

Basic Arithmetic
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)