**Directions(1-5):** Find the missing term “?” of the following series.

- 27,40,66,105,?,222
142150147157140Option D

+13*1

+13*2

+13*3

+13*4

+13*5

? = 157 - 230,240,260,?,380,540
300220310280200Option A

+10

+20

+40

+80

+160

? = 300 - 57,?,63,69,77,87
5942404450Option A

+2,+4,+6,+8,+10

? = 59 - 35,36,74,225,?,4525
904950888910890Option A

+1*1

+2*2

+3*3

+4*4

+5*5

? = 904 - 350,349,341,314,?,125
200250210245230Option B

-1^3, -2^3, -3^3, -4^3, -5^3

? = 250 - Karan invested a sum of Rs.4800 partially in two schemes A and B in the ratio of 2:x resp. Scheme A and scheme B is offering simple interest of 8% and 6% per annum resp. If the simple interest earned from scheme A after 3 years is equal to the simple interest earned from scheme B after 8 years , then find the value of x?
32451Option E

Amount invested in scheme A = 2/(2+x)*4800 = Rs.9600/(2+x)

Amount invested in scheme B = x/(2+x)*4800 = Rs.4800x/(2+x)

Now, 9600/[2+x]*8%*3 = 4800x/[2+x]*6%*8

=> x = 1 - 720 litres of milk and water in the ratio of 5:3 resp. If (x+5)% of this mixture is taken out and is mixed with 6 litres of water in a bucket such that the quantity of water in the bucket is 60 litres, then find the value of x ?
1015121316Option B

Quantity of milk in the solution = (5/8)*720 = 450 litres

Quantity of water in the solution = 720 â€“ 450 = 270 litres

Now, (x+5)% of 270 + 6 = 60

=> x = 15 - The total time taken by a boatman to travel a distance of (x+60) km upstream and (x-30) km downstream is 26 hours. If the boat speed and speed of stream is 12 km/hr. and 3 kmhr. resp. find the time taken by the boatman to travel (x+42) km upstream.
18 hours16 hours14 hours13 hours10 hoursOption A

Downstream speed of boat = 12+3 = 15 km/hr.

Upstream speed of boat = 12- 3 = 9 km/hr.

Therefore, (x+60)/9 + (x-30)/15 = 26

=> (10x + 600 + 6x – 180)/90 = 26

=> x = 1920/16 = 120

Required time = (120+42)/9 = 18 hours - The length of the cuboidal room is equal to the length of side of square of area is 576 m^2. If the breadth and height of the cuboidal room is 12 m and 8 m resp. find the maximum length of a stick that can be placed inside the room.
25 m20 m28 m23 m22 mOption C

Length of the cuboidal room = (576)^1/2 = 24 m

Breadth = 12 m

Height = 8 m

Required length of the stick = (24^2+12^2+8^2)^1/2

= 28 m - A student bought a pen and notebook for Rs. 200. He sold the notebook at a profit of 20% and the pen at a loss of 30%. If in the whole transaction, he made a profit of Rs.15 find the cost price of notebook.
160140100150130Option D

Let the CP of notebook be x.

CP of pen = (200 – x)

Now, 0.2x â€“ 0.3(200 â€“ x) = 15

=> 0.2x â€“ 60 + 0.3x = 15

=> x = 150