Quant Boosters  Kamal Lohia  Set 6

Q21) In a class there are three divisions. The number of students and the average marks in mathematics in the three divisions are 30, 40, 30 and 40%, 30% and 50% respectively. What are the average marks in mathematics of the class?
a. 39%
b. 40%
c. 41%
d. 49%
e. 51%OA : a (39%)

Q22) Five years ago average age of P and Q was 15 years. Average age of P, Q and R today is 21 years. How old R will be after 10 years.
a. 31
b. 32
c. 33
d. 34
e. 35OA : c (33)

Q23) Shweta took a loan of Rs.24000 for purchasing a TV. After two years, she paid it off by Rs.27840. Find the rate of simple interest.
a. 12%
b. 8%
c. 10%
d. 15%
e. 6%OA : b (8%)

Q24) There are two equal capacity vessels full of milk solution with milk and water in the ratio 1:3 and 3:5 respectively. If both are mixed in the ratio 3:2, what would be the ratio of milk and water in the new mixture?
a. 3 : 10
b. 4 : 7
c. 3 : 7
d. 9 : 19
e. None of theseOA : c (3 : 7)

Q25) Ram prepares solutions of alcohol in water according to customers' needs. This morning Ram has prepared 27 liters of a 12% alcohol solution and kept it ready in a 27 liter delivery container to be shipped to the customer. Just before delivery, he finds that the customer had asked for 27 liters of 21% alcohol solution. To prepare what the customer wants, Ram replaces a portion of 12% solution by 39% solution. How many liters of 12% solution is replaced? [XAT2012]
a. 5
b. 9
c. 10
d. 12
e. 15OA : b (9)

Q26) A man borrows Rs.6000 at 5% interest, on reducing balance, at the start of the year. If he repays Rs.1200 at the end of each year, find the amount of loan outstanding at the beginning of third year. [XAT2012]
a. 3162.75
b. 4145.00
c. 4155.00
d. 5100.00
e. 5355.00OA : c (4155)

Q27) After allowing a discount of 11.11%, a trader still makes a gain of 14.28%. At how many percent above the cost price does he mark his goods? [CAT1997]
a. 28.56%
b. 35%
c. 22.22%
d. None of theseOA : a (28.56%)

Q28) A man invests Rs 3000 at a rate of 5% per annum. How much more should he invest at a rate of 8% per annum, so that he can earn a total of 6% per annum? [CAT1995]
a. 1200
b. 1300
c. 1500
d. 2000OA : c (1500)

Q29) Last year Mr. Basu bought two scooters. This year he sold both of them for Rs.30,000 each. On one, he earned 20% profit, and on other he made a loss of 20%. What was his net profit or loss? [XAT2005]
a. He gained less than 2000
b. He gained more than 2000
c. He lost less than 2000
d. He lost more than 2000OA : d (He lost more than 2000)

Q30) What sum of money lent out at compound interest will amount to Rs.968 in 2 years at 10% p.a. Interest being charged annually.
a. 850
b. 900
c. 800
d. 950
e. 750OA : c (800)

So we write 1, 2, 3, φ, 4, 5..
1  10 : +1
11  20 : +1
21  30 : +1
31  φ0 : +2 (3φ, φ0)
φ1  40 : +10 (φ1, φ2, φ3, φφ, φ4, φ5, φ6, φ7, φ8,φ9)
41  50 : +1
51  60 : +1
61  70 : +1
71  80 : +1
81  90 : +1
91  94 : +1
Total = +21
So Age of alien = 94 + 21 = 115

Here the expression has odd number of factors.
Which means it is a perfect square.Now if N is even expression is odd(e+e+o =o)
And if n is odd expression is still odd(o+o+o=o)
Thus, the square is of an odd positive number.Putting expression equal to 11^2 satisfies it
And n comes out to be 9
As n(n+3)=108But ,how to ensure that there is no other solution besides this for higher squares????


the gcd cant be greater than the difference of the two consecutive terms of the sequence
Diff= 2n+1
Equating to 41 gives n =20 , nd the numbers are 410 and 451...whose gcd is 41..59 on equating doesnt give two numbers which hv gcd as 59
Hence 41.

X=[10^20000/10^100+2]
X= 10^200002^200/ 10^100+2
X=2^200(A1)/2(B+1)
X=2^199(some expression)Hence highest power is 199
Is my solution right?

Remainder of (P3)!/p=(p1)/2
Remainder of (p4)!×(p3)/p =(p1)/2
Rem(p4)!/p = (1p)/6Hence rem( 6*(p4)!/p)= rem((1p)/p)= 1

12100 =2^2 ×5^2 ×11^2
110 =2×5×11Number of factors of 12100 less than 110 =
3×3×31/2= 13
Numbr of factors of 110 less than 110 =7Hence total factors of 12100 which are less than 110 but do not divide it are 137 =6

What is the logic to be used for tgis prob? Could anyone guide?

It will satisfy for 11th digit x=0,7

Could anyone suggest how to estimate that the series like
10^2 > 19^2 gives 1, 2, 3
20^2 > 29^2 gives 4, 5, 6, 7, 83 + 5 + 7 + 11 ....... which gives 100.
So It continues like this but fails for higher squares...which gives answer as 76 ...because the series of integers generated is not continuous.
So what is the right way to solve it?