Quant Capsules by Shashank Prabhu  Set 16

Solve both equations and choose the correct option
i) x^2 + 0.2x = 4.83
ii) y^2 + 4.62 = 4.3y
a) x > y
b) x ≥ y
c) x < y
d) x < y
e) x = y483=161×3=7×23×3=21×23
(x+2.3)(x2.1)=0
462=2×231=2×3×77=2×3×7×11=22×21
(y2.2)(y2.1)=0
So x < = yThe cost price of an article increases by Rs. 100. The selling price increases by 10%. If the profit decreases from 15% to 10%, what is the original cost price?
original CP> x , final x+100
original SP>y , final 1.1y
yx / x = 0.15 ; 1.1yx100 / x+100 = .1
Solve for x = 2000/3In a race, A beats B by 100 m and C by 180 m. If B beats C by 90 m, what is the length (in m) of the race?
When A is at Finish Point, gap between B and C is 80m. When B covers the remaining 100m, the gap increases by 10m. Thus to make the gap of 10m, B has to cover 100m.
Hence, to make the gap of 90m, B must cover 900mA bag contains 9 red cards numbered 1, 2, 3 …, 9 and 9 black cards numbered 1, 2, 3 …, 9. In how many ways can we choose 9 out of the 18 cards so that there are exactly 3 duos, where a duos means a red card and a black card with the same number?
a. 80640
b. 13440
c. 18480
d. 1680The 3 numbered cards can be chosen in 9c3 = 84 ways. Post that, we need to select cards such that the numbers don't repeat. So, the first card can be chosen from the remaining 12 cards in 12 ways, the second card in 10 ways and the third card in 8 ways. However, as the order is not important, we will have to divide it by 3!. So, 84 * 12 * 10 * 8/3! = 84 * 16 * 10 = 13440
The angles of a convex pentagon are in an arithmetic progression. Which of the following can never be the value of any of its angles?
(1) 36°
(2) 35°
(3) 34°
(4) All of these4 is perfect. If the angles are a2d, ad, a, a+d and a+2d, a=108. The largest angle will be less than 180 as it is a convex pentagon. So, 108+2d greater than 36
The number of distinct points at which the curve y^3  5y^2 + x^2 + 6y  5x = 0 intersects either the xaxis or the yaxis is
If x=0, y^35y^2+6y=0, y=0 or 2 or 3
If y=0, x^25x=0, x=0 or 5
(0,0)(0,2)(0,3)(5,0) are the only points (0,0) was common to both the cases.A number x is such that it can be expressed as a+b+c=x where a,b, c are factors of x. How many numbers below 200 have this property?
Just 3 factors or sum of all factors except the number should equal the number. If it's 3 factors then 6 is the only such number I think. 28 also equals 1+2+4+7+14. The other perfect numbers are 496 and 8128.
Gautam decided to go to a temple only on first and the last day of a year. He continues to go to the temple in this fashion till the time he finds that he has visited the temple atleast once on each of the different days of a week. The minimum number of days required to achieve this is
Starting from the last day of a non leap year you get x and x+1, if the next year is a leap year, you get x+2 as the last day. Then x+3, x+3, x+4, x+4, x+5, x+5, as the first and the last days and finally x+6 as the first day of the new year. 4 years and 2 days... 1463 days.
What is the smallest possible positive integer such that the product of all its digits equals 9! ?
9! = 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9
2^7 * 3^4 * 5 * 7
5 and 7 cannot be increased further so they have to be present as it is. We can reduce the numbers by joining various powers of 2 and 3. The highest power of 2 that is a single digit is 8 and the highest power of 3 that is a single digit is 9. So, we are left with 2 * 8 * 8 * 9 * 9 * 5 * 7. We cannot tweak these numbers any further and so, it will be 2578899.Three elements a, b and c are selected from the set A = {2, 3, 5, 6, 7} to form a threedigit number ‘abc’, where a < b < c. Similarly, two elements p and q are selected from the set B = {0, 1, 8, 9} to form a twodigit number ‘pq’, where p > q. Let, M be the total number of all the possible values of ‘abc’ and N be the total number of all the possible values of ‘pq’. What is the value of (MN)?
From set A, the number of three digit numbers that have unequal digits is 5c3×3!=60. But, out of every 6 cases, only one would be in the form of a>b>c. So, number of cases satisfying the criteria is 60/6=10. For set B, the number if two digit numbers with distinct digits that can be formed is 4c1×3c1=12... half of these will have p>q and so, 6 cases in total. So, 106=4.
Understand that i have not considered the first digit to be non 0 in the second case because there won't be a number starting with a 0 that will satisfy p>q and the fact that it reduces our work.