**Directions(1-5):** What will come in place of question mark “?” in the following questions.

- (1575/63*24+48)*3 = ?^2*6^3
45683Option E

(1575/63*24+48)*3 = ?^2*6^3

=> (25*24+48)*3 = ?^2*216

=>?^2 = (648*3)/216 = 9

=> ? = 3 - 1610*(17.64)^1/2/49 = ?^2 – (961)^1/2
1813101614Option B

1610*(17.64)^1/2/49 = ?^2 – (961)^1/2

=> 1610*4.2/49 = ?^2 – 31

=> ?^2 = 138 – 31

=> ?^2 = 169

=> ? = 13 - (4095.8)^1/3 + 106.97% of 1599.8 = ?^3
815101213Option D

(4095.8)^1/3 + 106.97% of 1599.8 = ?^3

=> (4095)^1/3 + 107% of 1600 = ?^3

=> 16+1.07*1600 = ?^3

=> ?^3 = 1728

=> ? = 12 - 81^3 /27^5 * 9^7 = 3^?
913111417Option C

81^3 /27^5 * 9^7 = 3^?

=> 3^12 / 3^15*3^14 = 3^?

=> 3^11 = 3^?

=> ? = 11 - 32*45+16*35 + ? = 48*55+14*65
14401550130012501050Option B

32*45+16*35 + ? = 48*55+14*65

=> 1440+560+? = 2640+910

=> ? = 1550 - A boat goes 90 km upstream and comes back downstream to its original place in overall 16 hours. The speed of boat in still water is x km/hr. Find the value of x if the speed of stream is 3 km/hr.
1410151216Option D

Let speed of boat in still water be x km/hr. 90/[x+3]+ 90/[x-3] = 16 => 4x^2 -45x – 36 = 0

=> 4x(x-12)+3(x-12) = 0

=> x = 12,-3/4

Hence, the value of x = 12 km/hr. - A cricket player has an average of x runs after 10 matches. He scored 90 runs in his 11th match due to which the average of his runs becomes (x+7) in the 11 matches. Find his average runs in the 11 matches.
1820152224Option B

His average runs after 10 matches be x.

10x+90 = 11(x+7)

=> x= 13

Average of his runs after 11th match = 13+7 = 20 - Akash bought a laptop at Rs. x and marked it 40% above the cost price and sold it to Rajan after giving 15% discount thus made a profit of Rs. 9880. What would have been his amount of profit or loss if he had sold it at 30% discount.
10401250144012001000Option A

Let CP be x. MP = 1.40*x = Rs. 1.40x

SP = 0.85*1.40x= Rs. 1.19x

Profit = 1.19x – 1.40x = 9880

=> x = Rs. 52000

New SP = 0.70*1.40x = Rs. 0.98x

Loss = x – 0.98x = 0.02x = 0.02*52000 = Rs. 1040 - The present ages of Gaurav and Ashwani are in the ratio 11:13 resp. And the ratio of their age 12 years ago was 4:5 resp. If after Y years the ratio of their age would be 13:15 resp. The find the value of Y.
910685Option D

Now , [11x-12]/[13x-12] = 4/5

=> x = 4

Gaurav = 44 years and Ahswani = 52 years

After Y years, 44+Y/52+Y = 13/15

=>Y = 8 years - A bag contains 30 oranges out of which 8 are rotten. 3 oranges are drawn randomly and the probability that at most one orange is rotten is “p” . Find the value of “1-p”.
24/14123/14522/14320/14124/145Option E

At most one rotten orange means either no rotten orange or one rotten orange.

Probability = [22C3+22C2*8C1]/30C3 = 121/145

1-P = 1 – 121/145 = 24/145