Quant Boosters - Shashank Prabhu, CAT 100 Percentiler - Set 3

• The number has to be between 1000 and 8000. So, the first digit can take any value from 1-7 only

_ _ _ 1... as 1 has been fixed, first digit in 6 ways, 2nd in 8 ways and third in 7 ways -> 6 * 8 * 7
_ _ _ 3.. similar to the first case -> 6 * 8 * 7
_ _ _ 5.. similar to the first case -> 6 * 8 * 7
_ _ _ 7.. similar to the first case -> 6 * 8 * 7
_ _ _ 9.. first digit can be taken in 7 ways, 2nd in 8 ways and 3rd in 7 ways
6 * 8 * 7 * 4 + 7 * 8 * 7
56 * 31=1736

• Q11) How many unordered triplets of positive integers p, q, r exist such that p/q + q/r + r/p=2?
a. 0
b. 1
c. 2
d. More than 2

• (p/q+q/r+r/p)/3 >= (p/q * q/r * r/p)^1/3
(p/q+q/r+r/p) >= 3
So, it can never be equal to 2.

Can be done using simple substitution as well.

• Q12) If set S= {1,2,3,4,5,6,7,8,9,10} then how many sunsets of S are possible so that the sum of the elements is greater than 27?

• The simple approach is to figure out the total number of subsets possible i.e. 2^10. We know that the sum of all the elements will be 55. So, whatever set we form, if the sum is x, the sum of the elements in the complement set will be (55-x). So, half the cases will be greater than 55/2=27.5 and half will be lesser than 27.5. So, (2^10)/2 i.e 2^9=512 is our answer.

Take a smaller case... {1,2,3,4,5} and you want the sets that have sum greater than 7. You can form 32 sets or 16 pairs of complements.
{}...{1,2,3,4,5}
{1}...{2,3,4,5}
{2}...{1,3,4,5|
{3}...{1,2,4,5}
{4}...{1,2,3,5}
{5}...{1,2,3,4}
{1,2}...{3,4,5}
{1,3}...{2,4,5}
{1,4}...{2,3,5}
{1,5}...{2,3,4}
{2,3}...{1,4,5}
{2,4}...{1,3,5}
{2,5}...{1,3,4}
{3,4}...{1,2,5}
{3.5}...{1,2,4}
{4,5}...{1,2,3}
See that the second set is the complement of the first set. The sum total of the elements of the two sets will always be 15. If the first set totals to less than or equal to 7, the second will definitely by greater than 7. The elements are such that you will not have equal values in both the sets and so, you are left with half the cases. If the sum is an odd number and the total is +/-0.5 of the average then you can go with this method. Else, normal case making and permutations i guess

• Q13) A total of 8000 people were present for an exhibition. The ratio of men to women was 2:3 and ratio of boys to girls was 4:5. What can be the maximum ratio of males to females?
(1) 12:31
(2) 59:81
(3) 345:532
(4) 23:57
(5) None of these

• Q14) A watch showed 10 minutes past 6 o’ clock on Thursday morning when the correct time was 6 o’ clock. It loses uniformly and was observed to be 15 minutes slow at 8 o’ clock on Saturday morning. When did the watch show correct time ?

• As we need to increase the number of males, we need to increase boys instead of men. So, we need to find a multiple of (4+5) 9 which is close to 8000 so that the remaining number is a multiple of (2+3) 5.
As we know that 7920 is a multiple of 9, we have a point from where we can start. So
7929, 7938, 7947, 7956, 7965. We can't go above 7965. So the remaining 35 can be split in 2:3.
This gives us Boys = (7965)4/9 = 3540
And men = (35)2/5 = 14
3540 + 14 = 3554
8000 - 3554 = 4446
Answer: 3554 : 4446 = 1777 : 2223

This is the one wherein your worst nightmare comes true. Most of the time, we maintain that the fractions and ratios are there for a reason and by the end of it, things would cancel out. Well, this is the one where they don't. Beware of such questions. They might be rare but you will get at least a couple of these speed breakers.

• It loses 25 minutes over 50 hours in a uniform manner. It will match with the correct time when it has lost exactly 10 minutes (as it started 10 minutes ahead). So, 10/25=x/50 and so, x=20 hours past the starting time which is 2 am on Friday.

• Q15) An institute conducts 32 tests. The number of tests attempted by three students studying in the class is as follows:
Neil – 16
Nitin – 18
Mukesh – 20
The number of tests written by more than one students is at least:
a) 8
b) 11
c) 13
d) 16
e) 15

• Let a be the number of students who have taken exactly 1 test, b be the number of students who have taken exactly 2 tests, c be the number of students who have taken exactly 3 tests.
Total number tests will be equal to a+2b+3c=16+18+20=54
Also, 32 tests have been conducted in total. So a+b+c=32
From the two equations
b+2c=22
To minimize overlap, we need to maximize c.. so c=11.

• Q16) Find the number of positive integral solutions of x + y + z + w = 20 such that x, y, z, w have different values

• a+b+c+d=20
Give 1 to a, 2 to b, 3 to c and 4 to d.
a+b+c+d=10
To maintain the difference created, we need to find values such that a < = b < = c < = d
(0,0,0,10), (0,0,1,9).... (0,0,5,5)=6 cases
(0,1,1,8), (0,1,2,7).... (0,1,4,5)=4 cases
(0,2,2,6), (0,2,3,5), (0,2,4,4)=3 cases
(0,3,3,4)=1 case
(1,1,1,7), (1,1,2,6).... (1,1,4,4)=4 cases
(1,2,2,5), (1,2,3,4)=2 cases
(1,3,3,3)=1 case
(2,2,2,4), (2,2,3,3)=2 cases
Total 23 cases.
23*4!=552.

• Q17) Shalini multiplies 987 by a certain number and obtains 559981 as her answer. If in the answer, both 9s are wrong but the other digits are correct and at their original positions, then the correct answer will be

• Trial n error is of course the way to go. But a bit of smartness will reduce your work by a lot.
987 * abc ~ 559981
Now 987 has to be multiplied by a 3 digit number as if it is multiplied by a two digit number, you will get a value less than 98700 and if it is multiplied by a four digit number, it will be greater than or equal to 987000.
Now, the only problem lies with the middle two digits. As the last digit is 1, the value of c should be 3. So we get ab3 as the number.

The first digit of the answer is 5 and as 9 * 5 = 45 and 9 * 6 = 54, a should be either of 5 or 6.
987 * 5b3
002961
xxxxxx0
493500
559981

So, we get the middle line to end in 20. So, the only possible value for b is 6 as only 6 * 7 ends in 2.
987 * 6b3
002961
xxxxxx0
592200
559981

This is not possible as the last line exceeds the eventual sum.

So, 563 it is. With practice, you need not do it as elaborately as i have explained here. This is just for your understanding.

• Q18) A dishonest dealer marks up the price of his goods by 20% and gives a discount of 10% to the customer. Besides, he also cheats both his supplier and buyer by 100 grams while buying or selling 1 kilogram. Find the percentage of profit earned by the shopkeeper. (assuming the dealer mark up the price on the basis of the actual cost price)

• The dealer buys 1000 gm for Rs, 1000 but because of the cheat weight, gets 1100 gm for Rs. 1000. Now, he says that his cost price is Rs. 1000 and so, he will sell the entire thing for Rs. 1080 (post the 20 up and 10 down thing). Now, he has put up his sales price at Rs. 1080 for 1100 gm. But, he says that he gives 1000 gm when in reality, he has given 900 gm. So, when in reality he has given 1100 gm, he will say that he has given 11000/9 gm. So, corresponding sales price will be 11000/9*1080/1100=Rs. 1200. So, profit of Rs. 200 on Rs. 1000 and so, 20% profit.

• Q19) ABCD is a trapezium with AB parallel to CD. M is the midpoint of AD, Angle MCB = 150, BC = 4 and MC = 12. The area of ABCD is

• Extend CM to a point N such that N-A-B. CMD is congruent to NMA. So, NM=CM=12. So, CN=24.
Area of trapezium = area AMCB + area CDM = area AMCB + area NMA = area of triangle NCB.
So, 0.5 * 24 * 4 * sin150 = 48 * 1/2 = 24

• Q20) At Pizza-Hut pizzas are made only on an automatic pizza-making machine. The machine continually makes different sorts of pizzas by adding different sorts of toppings on a common base. The machine makes the pizzas at the rate of 1 pizza per minute. The various toppings are added to the pizza in the following manner. Starting from every pizza, every fifth pizza is topped with pepperoni, every seventh with olive and baby corn, every eighth with mushroom, and the rest with cheese and tomatoes. The machine works for 13 hours per day without any breaks in between. How many pizzas per day are made with cheese and tomatoes as topping?
a) 418
b) 458
c) 478
d) none of the foregoing

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