Quant Capsules by Shashank Prabhu - Set 11
CAT 100%iler, 5 times AIR 1, Director - Learningroots, Ex ITC, Pagalguy, TAS
The last digit of the LCM of 3^2003 - 1 & 3^2003 + 1 is
HCF = 2, so (3^2003-1)(3^2003+1)= 2 * LCM
2 * LCM = (3^2003-1)(3^2003+1)
3^2003 mod 10 is 7.
So, RHS is in the form of 6*8 and so, LCM will end in 4.
What is the sum of the roots of all the quadratic equations that can be formed such that both the roots of the quadratic equation are common with the roots of equation (x – a)(x – b)(x – c) = 0 ?
(1) 3(a + b + c)
(2) 2(a + b + c)
(3) (a + b + c)
(4) 4(a + b + c)
The roots of the equation are a, b, c. So, the quadratic equation formed can have these pairs of roots
- a, b
- a, c
- b, c
- a, a
- b, b
- c, c
Sum of all roots = 4(a+b+c)
Three runners A B C run a race ,with runner A finishing 24 metres ahead of runner B and 36 metres ahead of runner C, while runner B finishes 16 metres ahead of runner C. Each runner travels at a constant speed. What was the length of the race ?
(a) 72 metres
(b) 96 metres
(c) 120 metres
(d) 144 metres
Let the length be x. A covers x while B covers x-24 and C covers x-36. Similarly, when B covers x, C will cover x-16. So, we can get ratios of distances covered by B and C
(x-24)/(x-36) = x/(x-16)
When a two–digit number N is divided by the sum of its digits, the result is Q. Find the minimum possible value of Q.
For Q to be minimum, the denominator has to be maximum. So, b/a has to be maximum. The maximum value of b/a can be 9 and so, b=9 and a=1 and Q=1.9
Suresh and Ramesh are playing a game with a fair die marked 1, 2, 3, 4, 5 and 6 on its six faces. On his turn, a player rolls the die and notes the number s obtained. They roll the die turn by turn. The player, who gets 6 first wins the game. If Suresh gets the first turn to roll the die, then the probability of Ramesh winning the game is ?
Suresh rolls and he doesnt get a 6, probability of 5/6. Then Ramesh rolls and gets a 6, probability of 1/6. The probability of the game ending this way is 5/36
Suresh and Ramesh don't get a 6 in the first go, then Suresh again misses a 6 and Ramesh gets a 6. The probability of the game ending this way is (5/6)^3*1/6
With each addition to this game, the probability will keep on getting multiplied by (5/6)^2 and so, it is an infinite GP with a common ratio (5/6)^2 and the first term as (5/36). Applying the formula a/(1-r), we get 5/11 as the answer.
Yamini and Zora are standing 25 km apart. Zora starts moving towards Yamini. After 40 minutes Yamini also starts moving towards Zora. By the time Yamini covers 5 km, Zora has covered 15 km. They meet at a point 7 km from the starting point of Yamini. What is the speed of Yamini?
(a) 7.5 km/h
(b) 10.5 km/h
(c) 17.5 km/h
(d) 6 km/h
Yamini covered 5 km in say time t hours. So, Zora has covered 15 km in time (t+2/3) hours. So we get the ratios of their speeds. Also, the time in which Yamini covers 7 km will be 7t/5 and the time taken by Zora to cover 18 km will be 7t/5+2/3.
Equating the two speeds of Zora,
Speed of Yamini is 5/(2/3)=7.5 kmph
If the product of all the factors of a number is equal to the square of the number and the sum of all the factors of the number, other than the number itself is 13 ,then find the sum of all the values possible for the number.
If the number of factors of a number N is given by f, product of all the factors is given by N^(f/2). As is given, N^(f/2)=N^2. So, number of factors is 4. So, the number is either in the form of a^3 or a*b where a and b are the prime factors.
Case I: a^3... sum of all the factors except the number - 1+a+a^2=13. So, a=-4 or a=3. So a=3 and 27 is one of the numbers
Case II: a * b... sum of all the factors except the number - 1+a+b=13. So, a+b=12 where a and b are prime numbers. The only possible solution is (5, 7). So, 35 is another such number.
Sum of all values = 27+35=62
In how many ways you can climb up 8 steps if minimum and maximum numbers of steps you can take at a time are 1 and 6 respectively?
Let's say we represent n(x) as the number of ways of climbing x steps.
We have to remove the cases involving 7 and 8 from these. So, n(7) will have 1 case (7) and n(8) will have 2 cases (1, 7) and (7, 1). So, 3 removed. Remaining 125 ways.
“I am eight times as old as you were when I was as old as you are”, said a man to his son.
Find out their present ages if the sum of their ages is 75 years.
(a) 40 years and 35 years
(b) 56 years and 19 years
(c) 48 years and 27 years
(d) None of these
Let man's present age be m, son's age be s
When the man's age was equal to s, son's age would have been equal to s-(m-s) = 2s-m
Alternatively, use the options. When father was 27, son would have been 6. So, 8 times 6 is 48.