Quantitative Aptitude Test - 7

From statement 1, we can calculate the radius of M.
From this we can get the circumference of M, which will give us the circumference of N.
The circumference of N will give us the radius of N, which in turn will give us the area of N.
So, SUFFICIENT Statement 2 does not help because we haven't been given an actual figure for the diameter to work with.
So, NOT SUFFICIENT.

50! = 50 × 49 × 48………………× 1
So, we basically need to check how many powers of 3 are in this list.
The highest multiple of 3 in the list is 48 so we have $\frac{48}{3}$ = 16 powers of 3.
However, note that the following numbers in the list have more than 1 power of 3.
9 = 3 × 3
18 = 3 × 3 × 2
27 = 3 × 3 × 3
36 = 3 × 3 × 4
45 = 3 × 3 × 5
Thus the total powers of 3 become 16 + 6 = 22.

Commission on books sold at Rs 850 = 850 × 10 × 0.15 = 1275
Commission on books sold at Rs 750 = 750 × 5 × 0.15 = 562.5
Commission on books sold at Rs 750 = 700 × 10 × 0.15 = 1050
Therefore total commission earned by Satish = 1275 + 562.5 + 1050 = 2887.5
Total commission Satish would have earned had he sold all the books for Rs 850 = 25 × 850 × 0.15 =3187.5
Therefore he would have made (3187.5 - 2887.50) = Rs 300 more worth of commission.

Harsh can complete the entire project in 6 x $\frac{3}{2}$ = 9 days
Sumit can complete the entire project in 8 x $\frac{3}{1}$ = 24 days
Mini can complete the entire project in 12 x $\frac{4}{3}$ = 16 days
Total work done by all 3 in one day = $\frac{1}{9}$ + $\frac{1}{24}$ + $\frac{1}{16}$ = $\frac{31}{144}$
Therefore, total work done by all 3 in 4 days = $\frac{31}{144}$ x 4 = $\frac{31}{36}$
Therefore, work remaining = 1 – $\frac{31}{36}$ = $\frac{5}{36}$
Time taken by Sumit to complete $\frac{5}{36}$ of the work = $\frac{5}{36}$ x 24 = 3.33 days.

Hence, the correct option is (c).

For numbers 1–9:
9 keystrokes
For numbers 10–99:
90 × 2 = 180 keystrokes
For numbers 100–999:
900 × 3 = 2700 keystrokes
For numbers 1000–1001:
4 × 2 = 8 keystrokes
Thus total number of keystrokes = 2897

Let the quantity of Type 1 flour in mixture be x
So, total CP = (42x + (24 × 25) = 42x + 600
Total SP = 40 (x + 25) = 40x + 1000
As per the question,(42x + 600) × 1.25 = 40x + 1000
Or, 52.5x + 750 = 40x + 1000
Or, 12.5x = 250
Therefore x = $\frac{250}{12.5}$ = 20
Thus, the required difference in quantities = 25 – 20 = 5 kg.

Statement 1 tells us that Ramu sold his items for no profit no loss.
But, we do not know anything about Somu. So, NOT SUFFICIENT.
Statement 2 tells us that Somu sold his items for a slight loss.
To understand this, tick any number as the selling price for each of Somu's items.
Let's say, Somu sold each item for Rs. 100 and let q = 10% Then,
Somu's total SP = 200
CP for 1 item sold at 10% profit = 91 approx.
CP for 2nd item sold at 10% loss = 111 approx.
Thus total CP = 91 + 111 = 202
Thus loss = 202 -200 = 2
Note that statement 2 still does not tell us anything about Ramu.
Hence, NOT SUFFICIENT However, by combining the two statements, we can conclude that Ramu's deal was better.
Hence, SUFFICIENT

Let the ratio of mid-cap to small-cap for Paul be x:y.
Now all small cap stocks have same weightage,
So, the value of Paul's portfolio must be x(stock 4)+y(stock 1+stock2+stock3)
So, 36,000 = x(5,000) + y(6,500+7,500+7,000)
36,000 = x(5,000) +y(21,000)
Now, x>y (as weightage of mid cap is more than small cap stocks)
If we check values, then only option D and E have x>y.
So, we check only D and E Using option D,
x=3 and y=1 give the correct value of the portfolio.
=3 : 1.

Let the value of stock 3 which Sheela holds be x
So, the value of her portfolio is 1(4,800+x) + N(8,900)
The value of her portfolio is 39,500
So, we have 1(4,800+x)+N(8,900) = 39,500
For, x to be maximum, N must be minimum and we know the weightage of mid-cap stock must be more than that of small-cap.
So, N>1. Also, given N must be an integer.
The smallest integer > 1 is 2 & hence for x to be maximum, N=2
So, we have 1(4,800+x)+2(8,900) = 39,500
Solving it we get, x=Rs.16,900.

Let the weightage of small to mid cap stocks for Bindu be x:ywhere y>x
So, sum of value of stock 1 & 3 is x(7,300+6,700)
Value of mid cap stock 4 is y(6,300)
As, value of small cap and mid cap stocks is same
Hence, x(7,300+6,700) = y(6,300)
14,000 x= 6,300y
Therefore, x:y = 6,300:14,000
= 63:140
=9:20.