June 23, 2019
Question 1
If xy = - 30 and x2 + y2 = 61, then find the value of (x + y).
2
3
1
4
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x2 + y2 = 61 - - - - Equation (1) xy = - 30 - - - - Equation (2) Multiplying Equation (2) by 2 and adding to Equation (1), we get, x2 + y2 + 2xy = 61 - 60 (x + y)2 = 1 ∴ x + y = ± 1
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Question 2
The product of two consecutive odd numbers is 4623. Which is the greater of the two numbers?
66
69
68
67
X (X + 2) = 4623 X² + 2X – 4623 = 0 X² + 69X – 67X – 4623 = 0 (X - 67)(X + 69) = 0 X = 67, Greater odd number = X + 2 => 67 + 2 = 69
Question 3
If a - b = 3 and a^2 + b^2 = 29, find the value of ab?
10
12
15
18
2ab = (a^2 + b^2) - (a - b)^2 = 29 - 9 = 20 ab = 10.
Question 4
The price of 10 chairs is equal to that of 4 tables. The price of 15 chairsand 2 tables together is Rs. 4000. The total price of 12 chairs and 3 tables is:
Rs. 3500
Rs. 3750
Rs. 3840
Rs. 3900
Let the cost of a chair and that of a table be Rs. x and Rs. y respectively. Then, 10x = 4y or y =5/2x. 15x + 2y = 4000 15x + 2*(5/2x) = 4000 20x = 4000 x = 200. So, y = (5/2) * 200 = 500. Hence, the cost of 12 chairs and 3 tables = 12x + 3y = Rs. (2400 + 1500) = Rs. 3900.
Question 5
Find the value of a/b + b/a, if a and b are the roots of the quadratic equation x2 + 8x + 4 = 0?
14
24
a/b + b/a = (a2 + b2)/ab = (a2 + b2 + a + b)/ab = [(a + b)2 - 2ab]/ab a + b = -8/1 = -8 ab = 4/1 = 4 Hence a/b + b/a = [(-8)2 - 2(4)]/4 = 56/4 = 14.
Question 6
The sum of the squares of two consecutive positive integers exceeds their product by 91. Find the integers?
9, 10
10,11
11, 12
12, 13
Let the two consecutive positive integers be x and x + 1 x2 + (x + 1)2 - x(x + 1) = 91 x2 + x - 90 = 0 (x + 10)(x - 9) = 0 => x = -10 or 9. As x is positive x = 9 Hence the two consecutive positive integers are 9 and 10.
Question 7
The sum of three consecutive odd natural numbers is 93. The smallest of three numbers is:
29
31
23
27
Let the three consecutive odd numbers be a, a + 2, a + 4. Given, sum of three consecutive odd natural numbers is 93. ∴ a + a + 2 + a + 4 = 93 ⇒ 3a = 87 ⇒ a = 29
Question 8
What is the sum of all the natural numbers from 1 to 35?
620
630
650
680
We know that Formula for Sum of natural no's beginning from 1 = n/2×(1+n) where n = last no. i.e. 35 ⇒ Sum = 35/2×(1+35) ⇒ Sum = 35 × 18 = 630
Question 9
Jayant gets 3 marks for each right sum and loses 2 marks for each wrong sum. He attempts 30 sums and obtains 40 marks. The number of sums attempted correctly is
25
20
26
suppose he attempted x sum correctly, then x*3 -2*(30-x) =40 5x =100 x=20
Question 10
There are 120 legs and 50 heads of some ducks and horses grazing in a field. Find the number of horses?
30
40
Let there be d ducks and h horses in the field. Then number of total heads in the field = (d + h), And number of total legs in the field = (2 × d + 4 × h). According to the question, d + h = 50 ----(1) 2d + 4h = 120 ----(2) Solving both equations, we get, 2h = 20, or, h = 10 And, d = 40. Hence, there are 10 horses in the field.
Question 11
If n^(n/2)=2 is true when n=2 in the same way what is the value of n if n^(n/2) =4?
6
n^(n/2)=4 Apply log N/2logn=log4 nlogn=2log4=log4^2=log16 Logn=log16 Now apply antilog n=16/n Now n=4...
Question 12
A man could buy a certain number of notebooks for Rs.300. If each notebook cost is Rs.5 more, he could have bought 10 notebooks less for the same amount. Find the price of each notebook?
8
7.5
Let the price of each notebook be Rs.x. Let the number of notebooks which can be brought for Rs.300 each at a price of Rs.x be y. Hence xy = 300 => y = 300/x (x + 5)(y - 10) = 300 => xy + 5y - 10x - 50 = xy =>5(300/x) - 10x - 50 = 0 => -150 + x2 + 5x = 0 multiplying both sides by -1/10x => x2 + 15x - 10x - 150 = 0 => x(x + 15) - 10(x + 15) = 0 => x = 10 or -15 As x>0, x = 10.
Question 13
The sum of the square of the three consecutive even natural numbers is 1460. Find the numbers?
18, 20, 22
20, 22, 24
22, 24, 26
24, 26, 28
Three consecutive even natural numbers be 2x - 2, 2x and 2x + 2. (2x - 2)2 + (2x)2 + (2x + 2)2 = 1460 4x2 - 8x + 4 + 4x2 + 8x + 4 = 1460 12x2 = 1452 => x2 = 121 => x = ± 11 As the numbers are positive, 2x > 0. Hence x > 0. Hence x = 11. Required numbers are 20, 22, 24.
Question 14
The sum of one-half, one-third and one-fourth of a number exceeds the number by 22. The number is
264
284
215
302
Let the number be 'x'. Then, from given data x/2 + x/3 + x/4 = x+22 13x/12 = x+22 x = 264
Question 15
A farmer divides his herd of k cows among his 4 sons so that one son gets one half of the herd, the second gets one-fourth, the third gets one-fifth and the fourth gets 9 cows. Then k is equal to?
120
140
160
180
Given total number of cows = k Now, 1st son share = k/2 2nd son share = k/4 3rd son share = k/5 4th son share = 9 (k) + (k/4) + (k/5) + 9 = k => k - (19k/20) = 9 => (20k-19k)/20 = 9 => k = 180.
Question 16
A man has only 20-paise and 25-paise coins in a bag. If he has 50 coins in all totaling to Rs.10.25, then the number of 20-paise coins is
42
45
38
Let number of 20 ps coins = x and number of 25 ps coins = y Given total coins in the bag = 50 x + y = 50.......(1) But the total money in the bag = Rs. 10.25 0.20x + 0.25y = 10.25 20x + 25y = 1025.........(2) Now multiplying (1) by 25 we get 25x+25y=1250.............(3) By solving (2) and (3) 20x + 25y = 1025; => x = 45; Then, the no. of 20 ps coins are 45.
Question 17
Hemavathi gets 3 marks for each right sum and loses 2 marks for each wrong sum. He attempts 35 sums and obtains 60 marks. The number of sums attempted correctly is ?
Let, Hema attempted 'k' sum correctly, then k x 3 -2 x(35-k) = 60 5k = 130 k = 26 so 26 correct sums.
Question 18
18 5 17 = x 64 8 11 = y What will be the value of x + y? Rules: (i) If an odd number is followed by another odd number, they are to be multiplied. (ii) If an even number is followed by another even number, the first number is to be divided by the second even number. (iii) If an even number is followed by the perfect square of an odd number, the first number is to be subtracted from the second number. (iv) If an odd number is followed by an even number the two are to be added. (v) If an even number is followed by an odd number which is not a perfect square, the square of the odd number is to be added to the even number.
189
129
169
(18+64)*5+8)(17+11)=129
Question 19
9 15 50 = x 12 25 24 = y What will be the value of x/y? Rules: (i) If an odd number is followed by another odd number, they are to be multiplied. (ii) If an even number is followed by another even number, the first number is to be divided by the second even number. (iii) If an even number is followed by the perfect square of an odd number, the first number is to be subtracted from the second number. (iv) If an odd number is followed by an even number the two are to be added. (v) If an even number is followed by an odd number which is not a perfect square, the square of the odd number is to be added to the even number.
5
9*15 50=x 135 50=x 135+50=x 185=x 25-12 24=y 13 24=y 35=y so, x/y=185/35=5.
Question 20
Two boys are playing on a ground. Both the boys are less than 10 years old. Age of the younger boy is equal to the cube root of the product of the age of the two boys. If we place the digit representing the age of the younger boy to the left of the digit representing the age of the elder boy, we get the age of the father of the younger boy. Similarly, we place the digit representing the age of the elder boy to the left of the digit representing the age of the younger boy and divide the figure by 2, we get the age of the mother of the younger boy. The mother of the younger boy is younger than his father by 3 years. Then, what are the ages of elder and younger boys?
E = 15 & Y = 3
E = 14 & Y = 12
E = 40 & Y = 22
E = 4 & Y = 2
Let the the age of the elder boy = E Let the the age of the younger boy = Y Given that Y = cube root of EY => Y^3= EY => E = Y^2 .....(1) By the condition of number replacement the age of the father is YE The Mother's age = EY/2 But she is 3 years less than father => EY/2 + 3 = YE 2YE = EY + 6 ......(2) Then now from the given options we can identify which satisfies the all the conditions. Here Y =2 and E = 4 satisfies all the conditions.
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