Quantitative Aptitude - Arithmetic Ability Questions

Let the bases of the two triangles be x and y and their heights be 3h and 4h respectively.
Then, 12×x×3h12×y×4h=43⇒xy=43×43=169 

Required ratio = 16 : 9.

M+A+S3=11111 

also S+M2=11111   (∵ M - 11111 = 11111 - S) 

=>A = 11111

Ratio of times taken by P & Q = 100 : 130 = 10:13

Let Q takes x days to do the work

Then, 10:13 :: 23:x

=> x = 23x13/10

=> x = 299/10

P's 1 day's work = 1/23

Q's 1 day's work = 10/299

(P+Q)'s 1 day's work = (1/23 + 10/299) = 23/299 = 1/13

Hence, P & Q together can complete the work in 13 days.

Let the number be 'x'

Then, according to the given data,

 52x-25x52xx100

= 2125x100

= 84%

Initial investment of Ram = Rs.2500.
After 2 months he withdraw Rs.1250 from his capital.
Therefore, we have, Ram invested Rs.2500 for 2 months and Rs.(2500-1250=) 1250 for 4 months.

Raj invested Rs. 2250 for 3 months and Rs.(2250-750=) 1500 for 3 months.

And, Rakesh invested Rs.3500 for 3 months;

Their investing ratio:
Ram:Raj:Rakesh = (2500x2 + 1250x4):(2250x3 + 1500x3):(3500x3)
= (10,000):(11,250):(10,500) = 1000:1125:1050 = 40:45:42

Total profit for 6 months = Rs.2540
Therefore, Ram's share = Rs.(2540 x 40/(40+45+42)) = Rs.(2540 x 40/127) = Rs.800
Raj's share = Rs.(2540 x 45/127) = Rs.900
Rakesh's share = Rs.(2540 x 42/127) = Rs.840

In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36.

Then, E= {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3,4),(3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1),6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

n(E) = 27.

P(E) = n(E)/n(S) = 27/36 = 3/4.

A runs 1000 meters while B runs 900 meters and C runs 800 meters.

Therefore, B runs 900 meters while C runs 800 meters.

So, the number of meters that C runs when B runs 1000 meters = (1000 x 800)/900 = 8000/9 = 888.88 meters

Thus, B can give C (1000 - 888.88) = 111.12 meters start

Quantity1-
Rakhi’s marks= 115
Meenakshi’s marks= 115 - 20 = 95
Puneeta’s marks= 95 + 65= 160
Ankita’s marks=160 - 35= 125
Simpy’s marks= 125+ 67= 192
Total maximum marks= 192 + 108= 300
Required percentage marks of Simpy
= 192300×100=64%
Quantity2-
Let length and breadth be 100.

After increase in length it become 160, then
reduction in breadth be ‘x’
Now, 160*x= 100*100
Hence, x = 10000160=62.5

100-62.5=37.5%