1.When 15 is included in list of natural numbers, their mean is increased by 2. When 1 is included in this new list, the mean of the numbers in the new list is decreased by 1. How many numbers were there in the original list?
(1) 4
(2) 5
(3) 6
(4) 8
2.A shopkeeper marks his goods at such a price that after allowing a discount of 12.5% on the marked price, he still earns a profit of 10% The marked price of an article which costs him Rs. 4,900 is
(1) Rs. 5,390
(2) Rs. 6,160
(3) Rs. 5,490
(4) Rs. 5,390
3.Ravi can walk a certain distance in 40 days when he rests 9 hours a day. How long will he take to walk twice the distance, twice as fast and rest twice as long each days?
(1) 40 days
(2) 80 days
(3) 50 days
(4) 100 days
4.A solid is in the form of a right circular cylinder with hermisphereical ends. The total length of the solid is 35 cm. The diameter of the cylinder is 1/4 of its height. The surface area of the solid is (Take π=22/7)
(1) 462 cm^2
(2) 693cm^2
(3) 750 cm^2
(4) 770 cm^2
5.A sum of money lent out at compound interest increases in value by 50% in 5 years. A person wants to lend three different sums x, y and z for 10, 15 and 20 years respectively at the above rate in such a way that he gets back equal sums at the end of their respective periods. The ratio x : y : z is
(1) 6 : 9 : 4
(2) 9 : 4 : 6
(3) 9 : 6 : 4
(4) 6 : 4 : 9
6.The number (2^48 - 1) is exactly divisible by two numbers between 60 and 70. The numbers are :
(1) 63 and 65
(2) 63 and 67
(3) 61 and 65
(4) 65 and 67
7.The six -digit number 5ABB7A is a multiple of 33 for digits A and B. Which of the following could be possible value of A + B?
(1) 8
(2) 9
(3) 10
(4) 14
8.The product of two 2 -digit numbers is 2028 and their HCF is 13. The sum of the numbers is:
(1) 65
(2) 78
(3) 91
(4) 169
9.Cost of a diamond various directly as the square of its weight. A diamond broke into four places with their weight in the ratio 1 : 2 : 3 : 4. If the loss in the total value of the diamond was Rs. 70, 000, the price of the original diamond was
(1) Rs. 1, 00, 000
(2) Rs. 1, 40, 000
(3) Rs. 1, 50, 000
(4) Rs. 1, 75, 000
10.Two candles having the same lengths are such that one burns out completely in 3 hours at a uniform rate and the other in 4 hours. At what time (P.M.)., the lenghts of one is twice the length of the other?
(1) 1 : 24
(2) 1 : 30
(3) 1 : 36
(4) 1 : 32
Answers and Solution
1.(1) 15 = 2 * 5 + 5
SO Original mean = 5
New mean = 7
SO Sum of five numbers
= 7 * 5 = 35
After adding 1, new mean
= 36/6 = 6
2.(2) SP of the article
= 4900 * 110/100 = Rs. 5390
SO Marked Price = 100/87.5 * 5390
= Rs. 6160
3.(4) Working hours per day
= 24 - 9 = 15 hours
Total time
= 15 * 40 = 600 hours
Decrease in walking hours per day after increase in hours of rest = 24 - (2*9) = 6 hours
Required time = 600/6 = 100 days
4.(4)
5.(3) According to the question.
According to the question.
(3/2)^2 x = (3/2)^3 y = (3/2)^4 z =k let
=>x = (2/3)^2 k,y= (2/3)^3 k
and z = (2/3)^4 k
SO x:y:z = (2/3)^2 k∶ (2/3)^3 k∶ (2/3)^4 k
= 1 : 2/3 : (2/3)^2=9∶6∶4
6.(1) 2^48 - 1 = (2^24 + 1) (2^24 - 1)
= (2^24 + 1) (2^12 + 1) (2^12 - 1)
= (2^24 + 1) (2^12 + 1) (2^6 +1) (2^6 -1)
SO Required numbers
= 2^6 + 1 and 2^6 - 1
= 65 and 63
7.(2) A number is divisible by 33 if it is divisible by 3 and 11 both For divisiblity by 11,
so 5 + B + 7 = A + B + A
so 2A = 12
=>A = 6
so For divisibility of 5 6 B B 7 6 by 3, B = 3
so Number = 563376
so A + B = 6 + 3 = 9
8.(3) Let the number be 13x and 13y where x and y are prime to each other.
so 13x * 13y = 2028
=>xy = 2028/13 * 13 = 12 = 3 * 4
so Numbers = 13 * 13 = 39 and 13 * 4 = 52
so Sum of numbers
= 39 + 52 = 91
9.(1) Let the weights of the four prices of diamond be x, 2x, 3x and 4x units respectively.
Total weight of the original diamond
= x + 2x + 3x + 4x = 10x units
Price of the original diamond
k (100x^2) where k is a constant. Cost of four pieces of diamond
= k(x^2 + 4x^2 + 9x^2 + 16x^2)
= k.30x^2
Loss in the cost of diamond
=k(100x^2 = 30x^2) = k.70x^2
so k.70x^2 = 70000
=>kx^2 = Rs. 1000
so Cost of the original diamond
= k(100x^2)
= 1000 * 100 = Rs. 100000
10.(3) If the length of each candle be l then rate of burning of the first
= l/4 and that of the second = l/3
According to the question.
l - l*t/4 = 2(l - lt/3)
=>l - lt/4 = 2l - 2lt/3
=> 2/3t - t/4 = 2 - 1
=>((8 -3)/12)t = 1
=>5t/12 = 1
so t = 12/5 hours
= 2 hours 24 minutes
So Required time = 4 hours - 2 hours 24 minutes = 1 hour 36 minutes
Answers and Solution
1.(1) 15 = 2 * 5 + 5
SO Original mean = 5
New mean = 7
SO Sum of five numbers
= 7 * 5 = 35
After adding 1, new mean
= 36/6 = 6
2.(2) SP of the article
= 4900 * 110/100 = Rs. 5390
SO Marked Price = 100/87.5 * 5390
= Rs. 6160
3.(4) Working hours per day
= 24 - 9 = 15 hours
Total time
= 15 * 40 = 600 hours
Decrease in walking hours per day after increase in hours of rest = 24 - (2*9) = 6 hours
Required time = 600/6 = 100 days
4.(4)
5.(3) According to the question.
According to the question.
(3/2)^2 x = (3/2)^3 y = (3/2)^4 z =k let
=>x = (2/3)^2 k,y= (2/3)^3 k
and z = (2/3)^4 k
SO x:y:z = (2/3)^2 k∶ (2/3)^3 k∶ (2/3)^4 k
= 1 : 2/3 : (2/3)^2=9∶6∶4
6.(1) 2^48 - 1 = (2^24 + 1) (2^24 - 1)
= (2^24 + 1) (2^12 + 1) (2^12 - 1)
= (2^24 + 1) (2^12 + 1) (2^6 +1) (2^6 -1)
SO Required numbers
= 2^6 + 1 and 2^6 - 1
= 65 and 63
7.(2) A number is divisible by 33 if it is divisible by 3 and 11 both For divisiblity by 11,
so 5 + B + 7 = A + B + A
so 2A = 12
=>A = 6
so For divisibility of 5 6 B B 7 6 by 3, B = 3
so Number = 563376
so A + B = 6 + 3 = 9
8.(3) Let the number be 13x and 13y where x and y are prime to each other.
so 13x * 13y = 2028
=>xy = 2028/13 * 13 = 12 = 3 * 4
so Numbers = 13 * 13 = 39 and 13 * 4 = 52
so Sum of numbers
= 39 + 52 = 91
9.(1) Let the weights of the four prices of diamond be x, 2x, 3x and 4x units respectively.
Total weight of the original diamond
= x + 2x + 3x + 4x = 10x units
Price of the original diamond
k (100x^2) where k is a constant. Cost of four pieces of diamond
= k(x^2 + 4x^2 + 9x^2 + 16x^2)
= k.30x^2
Loss in the cost of diamond
=k(100x^2 = 30x^2) = k.70x^2
so k.70x^2 = 70000
=>kx^2 = Rs. 1000
so Cost of the original diamond
= k(100x^2)
= 1000 * 100 = Rs. 100000
10.(3) If the length of each candle be l then rate of burning of the first
= l/4 and that of the second = l/3
According to the question.
l - l*t/4 = 2(l - lt/3)
=>l - lt/4 = 2l - 2lt/3
=> 2/3t - t/4 = 2 - 1
=>((8 -3)/12)t = 1
=>5t/12 = 1
so t = 12/5 hours
= 2 hours 24 minutes
So Required time = 4 hours - 2 hours 24 minutes = 1 hour 36 minutes
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